Just found on the net:
http://www.theverylastpageoftheinternet.com/forsale/plans/elsa/ELSA.htm (http://www.theverylastpageoftheinternet.com/forsale/plans/elsa/ELSA.htm)
Looks interesting - do you think it could work?
-- Andi.
call him,get sustainable answers and then try it !
I tried allready via eMail - he has entered 2 or 3 eMail addresses - all mails came back...
The concept will not work.
All masses, that have gone down in the gravity field, that have not stored their
potential energy inside a spring,so they could be pulled back automatically,
will have lost their energy.
Only if you store the energy inside a spring or other storage device you can re-use
it again later and in the meantime use it to imbalance a wheel or something else...
Regards, Stefan.
You miss something. The energy is here stored inside the compressed piston. The only question is, would the water that is transported during the way down and up enough to compress the piston for a next ride.
-- Andi.
hi all,
i found this on the internet by accident. how did this get buried? did hartiberlin's post scare everyone off? surely , if you are a diver, you've used a BC to get you to the surface.
all of us crying for free energy should really be looking at this one very close. for me (i live on a boat) it would be a little heavy to get the needed power to move my home around. however, a smaller version aimed at only suppling my living needs, i think quite possible. most of you live in a house with at least a little yard (or garden as you might call it) where you could construct your power station. an ocean, lake or pond, etc, is really not necessary. a self contained unit constructed above ground would work the same. it may not be 30+ feet, but what you lose in depth, you can make up in diameter.
this unit may have a lot of drawbacks, maintance, large volume for power ratio, etc., but it has one large plus....FREE ENERGY! isn't that what this forum is looking for?
if you are truely not a "na-sayer", you should get out your drawing board and post some workable designs people can build at home (or with min. outside work).
if anyone can see why this really won't work, please enlighten me.
sure, we maybe looking at the model-t of the free energy efforts, but we have to START someplace.
just from looking at the stuff posted by Don Adsitt of John Herring's work, i would say the recompression of the shuttle piston would be the hardest to do. but, when you think about it, you are only trying to create a small amount of pressure (depending on depth shuttle piston travels to) in the shuttle piston. rather than use the leverage as John Herring suggest, i would use the leverage to raise a wieght that would be heavy enough to compress the shuttle piston from the top.
i think attention should be given to how your waterfall (that's what you are really making here) should best be used. hydroturbines seem like an expensive item and water wheels seem like they waste a lot of availabe engery. LET'S BE CLEVER!!
don't you think it deserves as much attention as has been given to some of the other NON-WORKING topics? the replies to this post will show if (and who) this group is really interested in free engery. just because you may not use it yourself, doesn't mean you can't help those who will.
if i have overlooked some basic reason this device can't work, i apologize. if we can't find a reson, we all need to apologize to John Herring.
tbird
http://www.overunity.com/index.php?PHPSESSID=598512917ff281a4da3cf3b2fb9f0414&topic=210.msg1379#msg1379
hi FreeEnergy,
thanks for the link. it just goes to show how trusting and easily put off some people are. if anybody believed Jim_Mich, they should be ashamed of themselves. he is so far off on the displacement value, i wouldn't believe anything he said. 1 cubic foot of water weights 64.7 pounds (salt water more). terry5732 makes a point of the water wieght above. did he forget the water pressure filling in behind? all the others were talking with emotions or off topic. the whole link seemed pretty useless to me.
i think a simple test to see if the water could be lifted by, in our case what we call the shuttle piston, all we have to do is take a ping pong ball under water with a length of pipe (doesn't have to be long, one end out of water by 1 inch) with id the same as ball diameter. if the ball went up and pushed out the water, great. if not, the proof it doesn't work is there (as long as you didn't try to push the water too high). anybody up for the job?
tbird
hi all,
couldn't stand the wait. went out this a.m. and did the test myself. i took a piece of 1.25" id pvc pipe, 12" long and a plug of styrofoam wrapped with duct tape (about 2" long), went to the edge of the harbor, submerged the pipe vertically except for the last 1.5" to 2". i inserted the plug from the bottom and without hesitation it not only went to the top, pushing water out all the way, it came out of the tube.
anybody else want to tell me why this won't work?
QuoteYou miss something. The energy is here stored inside the compressed piston. The only question is, would the water that is transported during the way down and up enough to compress the piston for a next ride.
ooandioo, you wrote the last post before i stepped in. the amount of water can be almost any amout simply because you can deliver it almost any distance away. lets say you need 100#s to compress the shuttle. if you have 10#s of water (just over a gallon) 10 feet away with a lever and fulcrum, you now have enough force to raise the 100#s, right? how many gallons in 1 cubic foot? just under 8. if you took it 20 feet away, you only need 5#s. sounds like plenty of water to me. the one thing you have to be aware of is the height the smaller weight needs to move the 100#s the distance needed for operation.
if all are clear on this point, what's next?
tbird
Quote from: tbird on August 22, 2006, 10:56:24 AM
hi all,
couldn't stand the wait. went out this a.m. and did the test myself. i took a piece of 1.25" id pvc pipe, 12" long and a plug of styrofoam wrapped with duct tape (about 2" long), went to the edge of the harbor, submerged the pipe vertically except for the last 1.5" to 2". i inserted the plug from the bottom and without hesitation it not only went to the top, pushing water out all the way, it came out of the tube.
anybody else want to tell me why this won't work?
QuoteYou miss something. The energy is here stored inside the compressed piston. The only question is, would the water that is transported during the way down and up enough to compress the piston for a next ride.
ooandioo, you wrote the last post before i stepped in. the amount of water can be almost any amout simply because you can deliver it almost any distance away. lets say you need 100#s to compress the shuttle. if you have 10#s of water (just over a gallon) 10 feet away with a lever and fulcrum, you now have enough force to raise the 100#s, right? how many gallons in 1 cubic foot? just under 8. if you took it 20 feet away, you only need 5#s. sounds like plenty of water to me. the one thing you have to be aware of is the height the smaller weight needs to move the 100#s the distance needed for operation.
if all are clear on this point, what's next?
tbird
now how much water in weight did you get out? would it be heavy enough to push the piston all the way back down?
I also spent some time thinking about the gravity mill - still I am not able to say its working or not. You are right, if the pipe could really be as large as possible there would be an imbalance that can be used for overunity. I think some more practical tests will be needed in order to find this out (tbird - thanks for the first live-test). We should clear out how high the water can be pumped out of the pipe.
Andi.
Quotenow how much water in weight did you get out? would it be heavy enough to push the piston all the way back down?
hi FreeEnergy,
i moved all the water in the pipe above the styrofoam except what went by the loose fit. as long as you have a positive buoyancy you will push out all the water above except what gets by the piston. the closer the fit, the less loss. this brings us to "head pressure". most pumps give you a head value. this is the height the pump can pump the fuild in a given size pipe before it can't pump. this will be dependent on the size of pipe used from the outlet (in our case, above water level). the smaller the pipe, the less liquid, the less weight, the higher the lift. in our case, it does not change the amount, just how long to get it. less flow = longer time. still will be all the water above except losses by piston (and any leaks in the plumbing). AND, this is only half the cycle. as far as using all the water to compress the piston, i wouldn't plan on using much more than 10% of the water available. you will have plenty of time to fill a reservoir (small) that will work the lever that will lift and park the weight that does the recompression. this leaves 80+% of the water to do work. are you with me?
ooandioo ,
i wouldn't call it an "imbalance", i would call it a positive pressure supplying a reservoir (or 2) that supplies constant water flow to do work. your own personal water fall.
what other test would you like to have done?
before you can tell how high, you have to know the size of the mill (this doesn't seem like the right name for it). anyway, we also need to know how we will use the energy. this will tell us how much electricty (if that's what you want to produce) we need, which would tell us (we could work out the math) how much flow, etc... so, maybe we need to focus on a common size. we can always up or down size it later. do we want to start with a proof of concept or go right to a useful unit?
i fear my post will be fairly long winded, but if you will read, i will write.
tbird
Hi TBird,
I had another look at all the graphics , which were
a bit confusing the first time I looked at them..
Hmm, maybe this concept is really having some merit ?
Many thanks for making this experiment.
How much water can you press out of the pipe ?
At what height is the pipe over the water surface ?
The main question is now, how much water-weight x distance to move ( lever action) do you need
to recompress the swimmer-piston ?
I guess there is pretty much energy needed, right ?
Can somebody calculate this and show, if the potential energy of the water:
water-mass x g x heightdifference is then bigger than the energy needed to recompress this
swimmer-piston ?
Many thanks.
Regards, Stefan.
Hi
I?m sure that this won?t work as long this SHUTTLE is piston like!
Well I begin with the downward phase.
A body will only sink if it is heavier then the weight of the amount of displaced water, equivalent to the body?s volume. In reverse case is this the reason why big tankers are able to swim.
As long the shuttle is piston like with seals it won?t displace water in any kind, because the body is not surrounded by water.
So nothing would happen. If the colum of water has a height of 20 m (10m in water and 10m over surface), by diameter of 1 m, and the shuttle"piston" is 1 m high at surface position, so its is trying to push 9 m colum of water against the pressure in the deep. More the IMPOSSIBLE!
Therefor any drop of water would reach the surface through this small pipe aside.
Now the upward phase!
Assuming that shuttle sinks against all rules and we have enough pressure to doubble its volume, the shuttle has now a height of 2 m.
Nothing will happen, too! Because 2m water colum, equivalent to the body?s volume is now trying for vaine to lift a 18m water colum :o (now through this "magical valve"!!) even against the friccion of the pistons seal.
I think that in overunity.com there are more inventions / setups / machines that are more worth to think about it!
By
2Tiger
Since I found the gravity mill, the idea flashes in my head. Some times ago, I allready had some interesting contact with physics and mathmatics experts in german about one part of the gravity mill (http://www.wasser.de/aktuell/forum/index.pl?job=thema&tnr=100000000002584&seite=1&begriff=auftrieb&tin=&kategorie=).
I'm happy that the idea is now back in the "charts". Lets see, what we can make out of it.
Andi.
hi Stefan,
it's good to see a na-sayer onboard ;D. i'll try to kept the proof coming so you stay onboard. i certainly understand your confusion. the inventor tried a little too hard to get his message across. once you get the hang of what he is trying to say, then the drawings become clearer.
QuoteHow much water can you press out of the pipe ?
At what height is the pipe over the water surface ?
let's think about a syringe. the body holding the fluid and the needle the pipe above water. the plunger would be our shuttle piston. if you have ever used a syringe, you know the harder you push, the further the fluid goes. no pressure, nothing comes out, negative pressure, you suck stuff in. now if you put a larger diameter needle on, the fluid comes out easier, more of it, but not as far for same pressure applied on smaller diameter needle. if you increase the pressure proportionate to the increase in needle diameter (volume), you will get the same distance with more volume. thus the body empties faster. appling this to our device, if you have 10 psi positive pressure on a surface area of 12 square inches, you could raise (10x12=120lbs) 1 inch above water line. if you decreased the exit diameter by half, you double the heigth to 2 inches and so on. if you got down to 1 sq. in. exit tube size, the the lift would be 12 inches. that's 120 lbs thru a 1inch squre pipe to a height of 12 inches. every time you reduce the volume, you reduce the weight, which increases the height needed to weight 120 lbs. at 120lbs, the piston will stop moving. at 119lbs., the piston will go up, 121lbs the piston will go down.
QuoteThe main question is now, how much water-weight x distance to move ( lever action) do you need
to recompress the swimmer-piston ?
I guess there is pretty much energy needed, right ?
here again comes back to my previous post. leverage is the key. the water being moved by the shuttle piston sideways has, basicly, no effect on the amount of pressure being applied. the load comes from being higher than water level. the pressure needed to change the volume of the shuttle piston depends on how deep and how much force you want the shuttle piston to go and supply, respectively. in the 12" square example above, the figures i used were to keep it simple, but not accurate. fresh water does have a specific weight. 1 cubic foot weighs 64.7 pounds (real close). so 1/12th (1inch x 12inches x 12inches) of that is 5.39 pounds. in the netural state (doesn't rise or sink) the shuttle piston would displace the same amount of water as it weights. if we give our shuttle piston a 12" depth we would now have a 1 cubic ft piston weighing 64.7 pounds. if we compress it 1 inch, it will have a negative buoyacy of 5.39 pounds which = (5.39lbs/12cubic.in.) .449psi. not much pressure, but it will move water above water level. now if we allow our piston to fall to a depth of 11.33 ft (1/3 of an atmosphere (14.7lbs psi)), we will have an increase of (14.7/3) 4.9psi. if we allow our piston to return to its orginal size, it won't do it. the pressure surrounding it will be greater than the .449psi we compressed it to in the beginning. ok...what do we do? we increase the pressure in the piston to 5psi to start with. now we need to compress it to 5.449psi so it will sink with the afore mentioned pressure of .449psi. now when we release the piston at 11.33 ft, it will expand to not only the orignal 1 cubic ft, but to a plus .1psi. this will give it a positive buoyancy, so it will float back up. the higher it travels, the more it will expand (if not contained), creating more pressure above. this would be ok if we didn't expect continuous work to be done on this half of the cycle. BUT, we are greedy. so, let's go back to the start and increase the pressure inside the piston by the amount to produce the work we want. anybody figure it out already? 5.349 would be right. if we want the work to be exactly = in both directions, we wll have to limit our piston volume to 13 inches high. once back to the top of the run, it will be time to recompress the shuttle piston. now we know how much pressure is inside (at compressed state), we need to set up the leverage to recompress the shuttle piston. now, how much weigh do we need to exert 5.349psi to 12 sq.in.? 12 x 5.349=64.188lbs. that amounts to just under 1 cubic ft of water. since we allowed our shuttle piston to go down 11.33ft and return that same 11.33ft, we should have 22.66 cubic ft of water available. now the question becomes, how much do we take away from doing our main work? if we allow for a little loss, we could use 1 cubic ft at 1 to 1 leverage and still leave 21.66 cubic ft to do work. we must remember we didn't set this up to produce much pressure (.449psi). as a result, the exit pipe would have to be fairly small to get to any favorable height. the smaller the pipe, the slower the flow for a given pressure.
armed with this example, you can now think more clearly about the real project to build.
if i made any math mistakes, please rap me on the knuckles and let me know.
ooandioo & 2tiger, more to come.
hi 2tiger,
i'm sure your heart is in the right place, but your phyics isn't or maybe you don't understand completely how the unit works.
QuoteAs long the shuttle is piston like with seals it won?t displace water in any kind, because the body is not surrounded by water.
hopefully you are right about not being totally surrounded. that would mean our seals weren't doing a good job. however, the lift comes from the bottom, not the sides. since in the up movement the water is allowed to fill in behind the piston, the water above the piston is neturalized. remember the ouside water goes all the way to the top. so with the piston now displacing more weight than it weighs, it goes up. didn't you read my post about this test? do you not believe me? if you are not too lazy and really want to be fair, do it yourself. i'm sure the water where you are will give the same result.
your only problem maybe you didn't realize the job of the 2 (of 4)checkvalves was to allow outside water to fill in behind the shuttle piston, in both directions (1 for up, 1 for down).
either way, please don't come back with such negative definitive statements without logical and defenitive proof.
hi Andi,
i'm sorry i wasn't interested in this before. i'll try to make it up to you and John Herring.
the link you sent might be of help, if i could read it. i'm afraid my first and only language is english. i'll leave the connection to them up to you, ok? if their specialty is math, i'm sure we will need them.
you said before you thought more tests would be needed. did the post to stepan help any?
tbird
Hi tbird.
Tomorrow, if I find time, I'm going to do some tests myself. I'm thinking about only using the second part of the mill - the move upward. Also I think about if its possible to use a baloon thats going to be filled with compressed air at the bottom of the pipe in order to take the shuttle upward. Maybe inserting the compressed air could also be done from the outside of the whole system...
I'll get back to you later.
Andi.
hi all,
at this time would like to say a few things in general.
we don't know when someone post a comment where their loyalities really lie until they use definitive statements without the definitive proof. i, myself have probably done this in the heat of the moment. it does no good for anybody. please, all of us, refrain from doing this here. if we are barking up the wrong tree, it will come out soon enough.
also, it would be nice to have a little humor along the way. i think it would be a good thing if at least once a page someone would write something to bring a smile to our face. if you agree, you'll get a kick from my next.
along time ago my dad had a friend named joe. seems joe was feeling a bit run down, so he went to see the old doc. after telling the doc about not having any energy, the doc said, "sounds like you need to cut back on your love life." to which fred responded, "which part doc, the thinking about it or the talking about?"
have a good day and good luck in your searches.
tbird
hi Andi,
i have a few questions that i hope you have asked yourself and answered.
is only using half the cycle a benefit? will this air supply down under be equal to or less energy intent than anything else? will the system now depend on some other system to work? if so, is it as or more reliable than the orignal design?
i'm sure there will be more, but before they can be asked, let's get past these.
tbird
Hi tbird.
I think, using only one part of the cycle is a benefit as the shuttle doesn't have to be that heavy in order to went down and push the water up. It only has to be a little heavier than the water itself displaces. On the way up theres also a benefit because of the weight of the shuttle.
The idea with the air supply is because of the practical use. Perhaps its easier to implement a baloon that can be filled with air under water. The needed energy will be the same i think.
Andi
hi again Andi,
let me see if i follow.
if you have a piston that consist of a disk (shape of the pipe, be it round or square) with a balloon attached that would deflate at the surface and allow the disk to turn vertical so they both can reach the bottom again in the fastest time (no water delivery), then once there, inflate the balloon to the size we talked about before (1 cubic ft.) and turn the disk horizontal again to form the seal. since there is hardly any weight to be lifted, the balloon assy. would effectively increase the pressure 11-12 times the other system. this would allow you to push the water higher or at least faster to the surface. to get the same amout of water out, you would have to do this twice in the same time as the other system would take for 1 cycle. this might happen because of the speed coming down and if the water wasn't lifted any higher, faster going up. in order to have the air pressure at the bottom to inflate the ballon, you would have a compressor running topside constantly to provide enough volume at the pressure needed. this work would be done with the same (or less) 1 cubic ft of water per 2 cycles we set aside for 1 cycle of the other system.
does that sound about right?
tbird
i dont think air sealed shuttle is a good idea but i can be wrong. i think a sponge/foam shuttle is easier to push back down under water.
Quote from: tbird on August 23, 2006, 11:05:02 AM
here again comes back to my previous post. leverage is the key. the water being moved by the shuttle piston sideways has, basicly, no effect on the amount of pressure being applied. the load comes from being higher than water level. the pressure needed to change the volume of the shuttle piston depends on how deep and how much force you want the shuttle piston to go and supply, respectively. in the 12" square example above, the figures i used were to keep it simple, but not accurate. fresh water does have a specific weight. 1 cubic foot weighs 64.7 pounds (real close). so 1/12th (1inch x 12inches x 12inches) of that is 5.39 pounds. in the netural state (doesn't rise or sink) the shuttle piston would displace the same amount of water as it weights. if we give our shuttle piston a 12" depth we would now have a 1 cubic ft piston weighing 64.7 pounds. if we compress it 1 inch, it will have a negative buoyacy of 5.39 pounds which = (5.39lbs/12cubic.in.) .449psi. not much pressure, but it will move water above water level. now if we allow our piston to fall to a depth of 11.33 ft (1/3 of an atmosphere (14.7lbs psi)), we will have an increase of (14.7/3) 4.9psi. if we allow our piston to return to its orginal size, it won't do it. the pressure surrounding it will be greater than the .449psi we compressed it to in the beginning. ok...what do we do? we increase the pressure in the piston to 5psi to start with. now we need to compress it to 5.449psi so it will sink with the afore mentioned pressure of .449psi. now when we release the piston at 11.33 ft, it will expand to not only the orignal 1 cubic ft, but to a plus .1psi. this will give it a positive buoyancy, so it will float back up. the higher it travels, the more it will expand (if not contained), creating more pressure above. this would be ok if we didn't expect continuous work to be done on this half of the cycle. BUT, we are greedy. so, let's go back to the start and increase the pressure inside the piston by the amount to produce the work we want. anybody figure it out already? 5.349 would be right. if we want the work to be exactly = in both directions, we wll have to limit our piston volume to 13 inches high. once back to the top of the run, it will be time to recompress the shuttle piston. now we know how much pressure is inside (at compressed state), we need to set up the leverage to recompress the shuttle piston. now, how much weigh do we need to exert 5.349psi to 12 sq.in.? 12 x 5.349=64.188lbs. that amounts to just under 1 cubic ft of water. since we allowed our shuttle piston to go down 11.33ft and return that same 11.33ft, we should have 22.66 cubic ft of water available. now the question becomes, how much do we take away from doing our main work? if we allow for a little loss, we could use 1 cubic ft at 1 to 1 leverage and still leave 21.66 cubic ft to do work. we must remember we didn't set this up to produce much pressure (.449psi). as a result, the exit pipe would have to be fairly small to get to any favorable height. the smaller the pipe, the slower the flow for a given pressure.
armed with this example, you can now think more clearly about the real project to build.
if i made any math mistakes, please rap me on the knuckles and let me know.
ooandioo & 2tiger, more to come.
Hi TBird,
many thanks.
What kind of formulars did you use to calculate this ?
I am pretty busy with other things right now and don?t have time
to look it up myself.
Maybe you can post as a summary all the basic formulas with
also the units in MSI system ( Meter, Kg , Bar etc..)
so we can try to calculate specific systems ?
Maybe we can just use as the swimmer lifter unit an
inverted "U" shaped case, where we press via a hose at the bottom
of the water container air into it and if it has risen to the top,
we open a valve at the top of this inverted "U" case and all the air goes
out at the top and this device can dive again to the ground.
Maybe someone can calculate how much energy you need to press e.g. 1 Liter of air
via an about 1 cm (0.39 inch) diameter hose e.g. 10 Meters deep unter water ?
Thanks.
hi free energy,
air filled shuttles will have their problems, but a well designed unit will perform well (think about all the hydraulics at work today). i think we will be forced to use them simply because of why you want to use a sponge/foam shuttle. if it's easy to "push back down under water", it won't deliver much work. if all you want is a small toy or proof of concept, that would be fine (as long as it can be compressed).
i have some thoughts of parts and design, but am not opposed to exploring other suggestions, like ooandioo's. that's why we have the forum. many heads are better than one.
hi stepan,
not sure what formulas you are referring to, but most of the figures i used just came from simple math. this site
http://en.wikipedia.org/wiki/Buoyancy
does have good info about buoyancy we might use at a later date.
as far as changing all the numbers to metric, i think it would be a waste of time, because everytime something is modified, i would have to change the metric numbers too. what i would suggest is download this Calculator...
http://www.calculator.org/
it's free and does the conversions you would like, as well as a number of other useful things.
after responding to andi's post, i thought of your "U" shaped case. not sure what type valve you would use, but i think if the top of the "u" were 1 or 2 doors (opening out) that would seal when closed, this could work well. when it hit bottom, the door(s) would close and latch (from gravity). at the same time a spring loaded latch would be holding the unit down until enough force (air in container) trips the hold down latch and away it goes. at the top, the door latch could be triped by a fixed arm.
food for thought.
tbird
What exact formula did you use for the compression of air ?
Thanks.
hi stefan,
"compression of air " is measured in pounds per square inch (psi). all you need to know is area. if you want to compress it, just add weight. divide weight by area=psi. just basic math. 100 pounds placed on 10 square inches = 10psi. what ever mass you placed on it (100lbs.), now has 10psi more applied. the weight of air at sea level is considered to be 14.7 pounds per square inch (this column goes to outer space). 34 ft in fresh water (33ft in salt water) is considered 1 atmosphere (1 square inch column of water 34 ft deep=14.7psi). most pressure gauges you see are reading gauge pressure and not Absolute pressure. the later would be showing the atmosphereic pressure all the time, then add whatever you are checking the pressure of.
wouldn't really call these formulas, mostly just givens. hope this helps.
tbird
ps ifound some formulas for you. look here http://en.wikipedia.org/wiki/Pressure
TBird, Okay, thanks for the info.
But how many centimeters or inches must then the swimmer
unit be compressed at what size dimension ?
And how much energy does this use up ?
Maybe someone can post a step by step calculation
of an example with all mathematical steps to calculate that,
so it it getting more clear ?
Many thanks.
Quote from: tbird on August 23, 2006, 07:17:39 PM
hi free energy,
air filled shuttles will have their problems, but a well designed unit will perform well (think about all the hydraulics at work today). i think we will be forced to use them simply because of why you want to use a sponge/foam shuttle. if it's easy to "push back down under water", it won't deliver much work. if all you want is a small toy or proof of concept, that would be fine (as long as it can be compressed).
i have some thoughts of parts and design, but am not opposed to exploring other suggestions, like ooandioo's. that's why we have the forum. many heads are better than one.
hi stepan,
not sure what formulas you are referring to, but most of the figures i used just came from simple math. this site
http://en.wikipedia.org/wiki/Buoyancy
does have good info about buoyancy we might use at a later date.
as far as changing all the numbers to metric, i think it would be a waste of time, because everytime something is modified, i would have to change the metric numbers too. what i would suggest is download this Calculator...
http://www.calculator.org/
it's free and does the conversions you would like, as well as a number of other useful things.
after responding to andi's post, i thought of your "U" shaped case. not sure what type valve you would use, but i think if the top of the "u" were 1 or 2 doors (opening out) that would seal when closed, this could work well. when it hit bottom, the door(s) would close and latch (from gravity). at the same time a spring loaded latch would be holding the unit down until enough force (air in container) trips the hold down latch and away it goes. at the top, the door latch could be triped by a fixed arm.
food for thought.
tbird
you are right.
i also had an idea that might help in this project. the shuttle would be air sealed and in its center would have a one way valve so when it's being pushed down it can recharge the tube/pipe with water. the pipe/tube would be half way in the water. i.e 3 feet in water and 3 feet outside of water in an open tank...this is still half baked idea.
tbird, you are right, that is the way I'm thinking about.
hartiberlin, the inverted "U" is also a good idea.
I have calculated something: Never mind if we fill a baloon under water or a inverted "U" the pressure we need to do that depends on the water pressure+airpressure. Air pressure is 1 bar and waterpressure is 1 bar for every 10 m. Filling a baloon at 10 m under water we will need pressure more than 2 bar (29psi).
p=F/A (pressure=force/area)
F=m x a (force=mass x acceleration)
We now should see, if the amount of water thats beeing pressed upward is enough to fith these formulas, thats the crux.
Andi.
hi stefan,
i'm not sure why you keep asking
QuoteBut how many centimeters or inches must then the swimmer
unit be compressed at what size dimension ?
And how much energy does this use up ?
without knowing what size unit you are talking about, no answer can be given. it's like asking how much gas does it take to drive across country without saying what transport will be used and how far across you want to go. those answers can be figured out if we know what the size of the unit is. this in itself needs to be determined by the user/builder. how much work do you want to do?
QuoteMaybe someone can post a step by step calculation
of an example with all mathematical steps to calculate that,
so it it getting more clear ?
i have already done this in reply 15. except for figuring out how much work the water we moved aboved could do. i also said in that reply
Quoteif i made any math mistakes, please rap me on the knuckles and let me know.
why haven't i had a knuckle rapping? are you not reading the post? are you not following my explanations? if not, why not ask me to re-explain where you got lost? i can't lay the blame on you. if you don't understand, it's only because i didn't do a good job. for that i apologize.
the mistake i made was "now when we release the piston at 11.33 ft, it will expand to not only the orignal 1 cubic ft, but to a plus .1psi." if you read on, you will see the figure to gain equal work was less than the pressure we increased to after figuring out the shuttle piston wouldn't expand at depth. when i suggested the pressure to compensate for the depth (11.33ft.) should be 5psi, this was wrong. all that was needed was 4.9psi. then adding the amount of pressure to do the work (move water) .449psi we would end up with a 5.39psi (that was right).
the figures i used in this example were based on a size i created in the example. this unit, for it's size would hardly produce any work. it was a starting point. work is not cheap. the more work we want, the more we have to pay. to get more work from the example; we could compress the shuttle piston farther (more work in, higher psi). with 21.66 cubic ft of water available per cycle and/or extending the leverage of the compressing unit, we have the resource available to do extra work.
hope this clears up things a bit. forgive me if i was too tuff on you. btw, did you check out the link in my last (ps at the end)?
tbird
Here some more calulations in this:
Lets assume you have pipe, 10m under water and 1m over water. Its diameter is 1m.
Our shuttle has to only lift the amount of water thats in 1m above the water, thats:
V=A*h, A=r?*Pi=2500cm*Pi, V=785.398cm? this weights 785,4kg, its force is F=m*g=7.704 N
That means, the lifting power has to be greater then 7.704 N.
If this is the fact, the shuttle will push the water thats in the pipe under water 1m above the water.
This will be V=7854cm?*1000cm=785.398.000cm? this weights 785.398kg, its force is F=7.704.754 N.
Looks very good, as we need quite less lifting power.
One more, the pumped water has its energy:
E=m*g*Dh/3.600.000J/kWh
E=785.398kg*9,81*0,5m/3.600.000J/kWh=1,07kWh
The maximum energy we could take out of the water 1m above sealevel is 1,07kWh (if we alow 0,5m fall down).
Andi
hi stefan,
a light just came on. are you trying to get me to WRITE a formula that you could stick in whatever variables you want and come up with answers for the unit you have in mind? if so, you give me way too much credit. i make too many mistakes. they maybe simple mistakes, but the formula you seek can't have any. it has to work for all samples. maybe there is somebody out there watching that can. please step up if you are willing to help.
tbird
hi FreeEnergy,
not sure what you are saying. are you suggesting the shuttle piston be 6ft long or the tube the shuttle piston rides in? also don't understand the function (purpose) of the one way valve. can you try again?
tbird
hi andi,
you're moving right along. well done.
i have a question or 2.
"This will be V=7854cm?*1000cm=785.398.000cm? this weights 785.398kg"
"V=785.398cm? "
is the first V value just another way to write the second (never saw this before)?
"its force is F=m*g=7.704 N" what is g?
also not sure what you mean "Looks very good, as we need quite less lifting power." less than what? what are you comparing it to?
"E=m*g*Dh/3.600.000J/kWh" is Dh distance of height?
one last. if the fall distace were .75m, does that mean power would be 1/2 again more (1.6kwh)?
tbird
Quote from: tbird on August 24, 2006, 09:34:41 AM
hi andi,
you're moving right along. well done.
i have a question or 2.
"This will be V=7854cm?*1000cm=785.398.000cm? this weights 785.398kg"
"V=785.398cm? "
is the first V value just another way to write the second (never saw this before)?
QuoteI mean, 785.398.000cm? equals 785.398 kg water mass.
"its force is F=m*g=7.704 N" what is g?
Quoteg=gravitational acc.=9.81m/s?
also not sure what you mean "Looks very good, as we need quite less lifting power." less than what? what are you comparing it to?
QuoteI'm comparing to the mass thats above sealevel, thats the mass that has to be pumped with the shuttle.
"E=m*g*Dh/3.600.000J/kWh" is Dh distance of height?
Quoteright
one last. if the fall distace were .75m, does that mean power would be 1/2 again more (1.6kwh)?
Quotealso right.
tbird
hi andi,
got most of your answers.
still not sure what mass above sealevel you are comparing. we are talking about your design that only works on the up stroke, right? are you comparing the weight to produce the pressure needed to compress the shuttle piston in the first design to.... something? i'm stuck.
help!
tbird
hi tbird,
the shuttle has to be lighter than the water it displaces, then it wents up. In a closed pipe, heading 1m above sealevel, it also has to overcome the weight of the water in the pipe above sealevel, that is 785,4kg. It has to be lighter, or its volume has to be bigger than its weight in compare of the weight of the water it displaces+the water it shall displace in the pipe above the sealevel (i think i don't understand the sentence myself...).
Andi
andi,
i think i see why we confused each other. the water being pushed out doesn't stay above the pipe (creating back pressure). it could be delivered to a holding device supported above sealevel (you used that word) but lower than the max height the shuttle can push the water to. once the water starts to fall, work can then be extracted. after we get water delivery (a waterfall) accomplished, then we can design the best way to make work using this falling water. does that change your figures any?
tbird
Hi TBird, okay, I have to reread your message 15 again, but am online via PDA now posting this..Maybe you can post again a summary with all steps of Mathematics of a model setup you are sure will work and also tell all dimension of pipes and swimmer, so we can see, that it works.. many thanks.
andi,
just converted your numbers to ones i understand. i think you have an error in the kwh figure. before i start barking up the wrong tree, let me make sure of a couple of things. kwh does stand for kilowatts per hour, right? if a hour is the time frame for power, what is the time frame for delivering 785.398kg of water. do we know how long it will take to collect this much? oh heck! i can't keep it to myself. unless you were going to limit the shuttle to rise only 1 meter, you will get almost 10 meters worth of water (7,853.98kg) per cycle. it won't be that much because the shuttle can't travel the full meter of tube that is out of the water. so you might only get 2/3rd of that last meter.
maybe we can't figure kwh until we know how much water will be available per hour.
tbird
Maybe the easiest mechanical setup would be to have piston like swimmer unit which uses internally compressed springs to have its volume size small and use a double latch magnet relay to switch/toggle electrically via short voltage pulses between bigger and smaller volume. What do you think? You could power this control circuit via an accumulator and use a waterwheel and generator to recharge the accumulator. This then should run forever and also deliver additional output energy.
Quote from: ooandioo on August 24, 2006, 08:38:08 AM
Here some more calulations in this:
Lets assume you have pipe, 10m under water and 1m over water. Its diameter is 1m.
Our shuttle has to only lift the amount of water thats in 1m above the water, thats:
V=A*h, A=r?*Pi=2500cm*Pi, V=785.398cm? this weights 785,4kg, its force is F=m*g=7.704 N
That means, the lifting power has to be greater then 7.704 N.
Okay, so far okay with me.
Quote
If this is the fact, the shuttle will push the water thats in the pipe under water 1m above the water.
This will be V=7854cm?*1000cm=785.398.000cm? this weights 785.398kg, its force is F=7.704.754 N.
Looks very good, as we need quite less lifting power.
One more, the pumped water has its energy:
E=m*g*Dh/3.600.000J/kWh
E=785.398kg*9,81*0,5m/3.600.000J/kWh=1,07kWh
The maximum energy we could take out of the water 1m above sealevel is 1,07kWh (if we alow 0,5m fall down).
Andi
Andi I don?t understand this part ! Did you have a typo in it ?
How do you get now more than 785 tons of water moved ??
I think there is an error somewhere and should this water go up or down ???
Hi all.
First to tbird. If the pipe reaches 1m outside the water, there is ALWAYS 785,4kg of water in the upper pipe part, needing to be moved.
The calculated energy is kWh - if you leave the /36.000.000 you have Joule. I thought about its easier to understand with kWh.
hartiberlin - the idea with spring compression is worth thinking about it.
785 tons of water are in the pipe part under water.
Andi
ok i have drawn a picture hope this help. crappy picture but hope you can get the diea.
see the shuttle has a one way water valve in the center so when its going down the pipe the water can easily go above the shuttle.
picture this shuttle starting to move up the pipe from the buttom to the middle of the pipe. half the pipe in water and half out.
Thanks for your time drawing the picture, FreeEnergy.
Do you also think its a benefit using only the "going upward" part? I think, the system can be much easier arranged without respecting the "going downward" - the shuttle has not to be that heavy.
Andi.
when the water is saved up in the upper half of the pipe, that water saved up will provide some kind of mechanical kick to bring down the one-way-valve-shuttle back to its initial position. when the shuttle goes up, the valve closes, and when it comes down, it opens.
sorry had a typo in my previous post. i fixed it :)
i for to say...
in the middle of the pipe where the water in pushed up there will be a one way water valve so water can only go up.
hi stefan,
i think if i do as you suggested in your reply #40, i might be getting the cart before the horse. if you re-read my post 25 & 27 and go to the links on those post, and take a few minutes to think about it, most of your lack of understanding what the design is, will go away.
QuoteAndi I don?t understand this part ! Did you have a typo in it ?
How do you get now more than 785 tons of water moved ??
I think there is an error somewhere and should this water go up or down
i noticed the same thing, but if you do the math, his weight per meter of water in the tube is right. (maybe i made a mistake so the answer came out right) i think people from diffferent parts of the world use the . & , with numbers to mean different things. i use the . for decimals and the , to seperate the larger whole numbers (every 3 places). is that ok with everyone here?
QuoteMaybe the easiest mechanical setup would be to have piston like swimmer unit which uses internally compressed springs to have its volume size small and use a double latch magnet relay to switch/toggle electrically via short voltage pulses between bigger and smaller volume. What do you think? You could power this control circuit via an accumulator and use a waterwheel and generator to recharge the accumulator. This then should run forever and also deliver additional output energy.
this sounds like a clever arrangement, but i'm not too excited about putting electronic underwater. maybe if i were as familar with the device as you are,i would feel different.
hi andi,
QuoteFirst to tbird. If the pipe reaches 1m outside the water, there is ALWAYS 785,4kg of water in the upper pipe part, needing to be moved.
i think we need to get this cleared up. water in a pipe that is 1 meter in diameter and 1 meter long will weight 785.4 kg or 1728 pounds. do we agree? so if you have 10 meters of water to be moved, you will be moving 7,854kg or 17,280 pounds. right?
QuoteThe calculated energy is kWh - if you leave the /36.000.000 you have Joule. I thought about its easier to understand with kWh
i don't have any problem with this. i just think that if it takes 2 hours to move whatever the correct amount of water is, the kwh would be different, as compared to only taking 30 minutes, wouldn't it?
hi FreeEnergy,
i'm getting the picture now. of course the valve would lock in place at the bottom so the air can't get out. now, this means you only have water move out on the up stroke, right? in your 6ft long pipe, how long would the shuttle be? remember whatever size the shuttle is will reduce the amount of water between it and the surface. so if the shuttle is 1 ft long, it only has 2 feet of water to push. but being only 1 foot long you can only push up 1 foot if the pipe remains the same size above water. if the shuttle is 1.5 feet long, you will have 1.5 feet of water to push up and it will rise the majority of the 3 feet out of the water. it will be shy by the amount the shuttle displaces at the surface.
do you figure to fill it with a compressor like andi plans?
maybe in the future we can refer to the type unit we will be talking about when we start. i suggest using type 1 for the double action unit and type 2 for the single action type. what do you think?
Hi tbird.
Right, I'm using "." as seperator and "," for decimals - sorry for that.
Quotei think we need to get this cleared up. water in a pipe that is 1 meter in diameter and 1 meter long will weight 785.4 kg or 1728 pounds. do we agree? so if you have 10 meters of water to be moved, you will be moving 7,854kg or 17,280 pounds. right?
Right, we are moving 7,854kg, but we do only lift the weight of water thats in the upper 1m (785kg), whatever the way from the bottom to the top is, if not - there would be no overunity.
Quotei don't have any problem with this. i just think that if it takes 2 hours to move whatever the correct amount of water is, the kwh would be different, as compared to only taking 30 minutes, wouldn't it?
The unit kWh only shows what is possible - surely we have to divide it by the time we will need to lift the water up. E only shows what amount of energy lies in the lifted water at all.
Andi
Okay Andi and TBird lets try again step by step to calculate it,
if we can get more potential energy from moving the water into an upper
reservoir: potential-Energy= watermass x earthaccel. g x heightdifference,
than needing for the compression of the swimmer unit !
Okay, let?s stay with Andi?s example:
A pipe 10 Meter deep inside the seawater and 1 Meter above the seawater level.
diameter = 1 Meter
Now let us just use the lifting not the sinking to produce the energy.
Okay, now the START condition.
All water in the pipe is in the seawaterlevel position and the
piston like swimmer body must be 2 Meters under seawater level
to be able to go up to seawaterlevel and thus
pump 1 Meter of volume over the 1 Meter high pipe into an upper reservoir.
Okay, how big will be the potential energy we win, if we succeed to pump this
volume of 1 Meter height of 1 Meter diameter water up:
Volume=Area x height, Area=r?*Pi=2500cm*Pi,
V=785398cm? this means the water weights 785.4kg
Thus we have a potential energy stored of:
Energy= 785.4 Kg x 9.81 x 1 Meter heightdifference= 7705 Joule or Wattseconds.
That means put into Watthours we must divide it through 3600, so
we get about 2.14 Watthours of potential energy stored 1 Meter above sealevel with
about 785.4 kg= 785.4 Liter of water !
Now let us see, if this would be enough energy to compress the piston
swimmer body again, so he can go down again 1 Meter now under seawaterlevel.
As we needed 2 Meter volume of water= 2 x 785.4 Kg = 1570.8 Kg watermass to be pushed up 1 Meter
ontop of the pistion swimmer unit,
we needed a force of F= watermass x g to push it up , that is
about 15410 Newton of force.
That means, the lifting power or positive boyuancy has to be greater then 15410 Newton.
Now how big and what volume must the piston like swimmer boddy have ?
diameter= 1 Meter
What height ?
As Archimedes's principle says:
The buoyant force is equal to the weight of the displaced fluid,
we must use very light piston swimmer body and if we do not count in its weight
we must have then also 2 Meters of height of the piston swimmer body, because it must have the
same force as the Mass force of the 2 Meter water volume above it.
So to get on the positive site, the swimmer body must be a little over 2 Meters high himself
also depending on its own weight.
Okay, now at the starting point, the swimmer unit is 2 Meters under seawaterlevel
with its upper surface and 4 Meters under seawaterlevel
with its lower surface.
When it has risen up the 2 Meters , 1 Meter of volume of water is pushed up into the
upper reservoir and the swimmer is with its upper surface at seawaterlevel
and ontop of it we still have a volume of 1 Meter water, which now
weights still 785,4kg, which is a force of about 7705 Newton onto the upper surface
of the swimmer unit.
The swimmer unit gets from the buttom a buyoancy force of 2 x 7705= 15410 Newton,
so we have to reduce the positive buyoancy to lower than 7705 Newton to get the swimmer
body again to sink.
This can only be done by compressing it to a little bit less than 1 Meter height, so the
1 Meter height water above it has enough weight to sink it down again.
Now let us calculate the pressure inside the swimmer body, which we assume will be a balloon
for the moment.
At the starting point 2 Meters under seawater level the 2 Meter high balloon must have an airpressure of
P= F / A of P= 15410 Newton / 7853.98 squarecm =1.96 bar
( that is about the pressure in my car wheel ! for comparison only)
So this is the pressure to keep the volume of 2 Meters water above the ballon inside the 1 Meter diameter pipe,
so that the balloon will not be compressed and still has its shape of 2 Meter height itsself.
Now in the end position, when the upper balloon surface area is at seawaterlevel and only 1 Meter
of Watervolume is still ontop of it, the ballon will have a pressure of:
P= F / A of P= 7705 Newton / 7853.98 squarecm =0.98 bar
So now we must compress the volume of the balloon from 2 meter height to 1 Meter height,
this means we have to press the upper surface of the balloon 1 Meter down, so the pressure
inside the ballon rises again from 0.98 bar to 1.96 bar.
So we must apply 7705 Newton at the start to 15410 Newton at the end of the compression phase of 1 Meter
distance to do that .
But how much energy is this ?
If we do it graphically we see, that the area under Force over distance is a triangle and the
energy area under it is half the value of the full area, that the endposition would give.
So the energy required is:
Compression-Energy=15410 Newton x 1 Meter / 2 = 7705 Newtonmeter = 7705 Wattseconds = 2.14 Watthours !
So we need the same energy to compress the balloon cylinder as we have won in the first place !
So as we also have frictional and seal losses, this does not work !
Also due the hydrostatic paradoxon we will also not be able to pump
air under water into any balloon or other container, as we have to overcome
the full height of the water inside the hose, so this also needs too much energy
to pump air under water !
So this gravity mill just does not work... otherwise it would have been used probably much
earlier...
Too bad.
Regards, Stefan.
Hmm, I am just asking myself, what happens, when the balloon has 2 Meter height
and 1.96 bar at the starting point 2 Meters under water and then rises 2 Meter and
has 0.96 bar at the end.
Then its volume must have also doubled , right ?
So must it now be 4 Meter in height ?
I am a bit confused now... Hmm...
I guess one can not just work with a balloon, that can expand
and compress itsself due to pressure !
One must use a swimmer body, that can be put into 2 volume positions
via electric relays. If you use a vaccuum inside the swimmer body,
it will be easy to recompress the swimmer from 2 Meters volume height
to 1 Meter volume height.
But then it is the question, how much energy you need to
restretch the volume of the swimmer body again from
1 Meter to 2 Meters when it has sunken again...?
Probably also more as you have won to move the 1 Meter volume water 1 Meter up !???
hi stefan,
nice try, but no cigar. must be frustrating to go thru all those calcs and still come up wrong.
most of the setup you used is for a unit that travels to 1 atomsphere (10 meters) and not just 2 (or 4 as you threw in) meters. maybe you have come up with type 3.
if you only go down 1/5th of an atmosphere, you only need to compress to .2 bar (not use to these measurements). so everything you based on .98 and 1.96 is wrong.
there are lots of holes in your summary. it would take awhile, but if you want me to, i will try to point them all out.
i will have to say, i probably could not have done a summary nearly as well as you did yours.
let me know about the corrections.
tbird
Quote from: FreeEnergy on August 24, 2006, 05:09:12 PM
when the shuttle goes up, the valve closes, and when it comes down, it opens.
it's not opening for air passege but for water. when the one-way-valve-shuttle goes down the pipe it opens for water to come in, then it closes when it goes up to push the water to the upper half of the pipe.
Hi stefan.
Thanks for the good calculation - very good and exact job.
There is one thing that makes the whole thing running: You calculated the energy gain and loss while the swimmer is 2 meters under the water, so it has to go 2 meters upward and pump 1m water in a reservoir. Then your calculations are right then energy gained=energy lost when compressing. I think, never mind how deep the swimmer is and how much water he has to push up, the needed energy is always the same, its always the energy we need for pumping 1m water in the reservoir, the 1m above sealevel. As long as sealevel is not reached by the shuttle, it has the same buoyancy which needs to be greater than the force of 1m water above the sealevel. Correct me if I'm wrong or simply tell your criticism!
I think, the problem is - the deeper the shuttle goes, the more pressure we need to reinflate it...
Andi
no air being pumped in my idea
Well I am now convinced that it does not work.
It is the same analogy, as if you would take a U shaped tube,
fill it with water , so that the water is 10 cm high in each leg of the tube.
Then use a syringe in the right leg to press air into the upper right leg above
the water. You need Force x distance energy to press on the syringe
piston. Then the water in the left leg will go up, but it will only
have the same height difference as much energy you added via Force x distance
via the syringe.
The height difference can be calculated via:
Left water leg Potential-Energy = Syringe-Energy
watermass x g x heightdifference = Force x distance.
So there is no overunity to gain in here.
If you see the watermolecules as small balls it is clear, that
you don?t gain any energy out of lifting a few balls up, but you have
to pay the same energy prize for doing this by pumping the same energy
air amout unter water !
So if you can find an error in my ramblings, please let me know.
Thanks !
Hi TBird,
let me know, where my errors should be.
Thanks.
As the boyuancy force must be greater than the volume
of the water above the balloon swimmer unit, also if you go deeper,
say 10 Meters below seawater level you also have to lift 10 Meters of
water with the balloon, so you need at least a balloon filled with air that is also
10 Meters high in the pipe ! Otherwise you can not lift 10 Meters of water volume
above it !
If the balloon would only be 9 Meters high inside the pipe, the
water above it would be too heavy to get lifted and all stays in place !
Is this correct ?
Quote from: ooandioo on August 24, 2006, 06:31:43 PM
As long as sealevel is not reached by the shuttle, it has the same buoyancy which needs to be greater than the force of 1m water above the sealevel. Correct me if I'm wrong or simply tell your criticism!
Andi
`
Andi this is wrong, you have to lift the whole mass above the balloon,
so when it is 10 Meters deep you also have to lift 10 Meters of volume
of water above it , not just 1 Meter of Water volume !
This is, where all people over here seem to make the mistake I guess...
Am I right ?
Or did I forget the water pressure from the lower balloon side ?
hi FreeEnergy,
Quoteit's not opening for air passege but for water. when the one-way-valve-shuttle goes down the pipe it opens for water to come in, then it closes when it goes up to push the water to the upper half of the pipe.
Quoteno air being pumped in my idea
in order to sink, the air either has to go out or be compressed. if you are not pumping air in.... tell me about the air. how do you handle it?
andi,
QuoteThanks for the good calculation - very good and exact job.
There is one thing that makes the whole thing running: You calculated the energy gain and loss while the swimmer is 2 meters under the water, so it has to go 2 meters upward and pump 1m water in a reservoir. Then your calculations are right then energy gained=energy lost when compressing. I think, never mind how deep the swimmer is and how much water he has to push up, the needed energy is always the same, its always the energy we need for pumping 1m water in the reservoir, the 1m above sealevel. As long as sealevel is not reached by the shuttle, it has the same buoyancy which needs to be greater than the force of 1m water above the sealevel. Correct me if I'm wrong or simply tell your criticism!
I think, the problem is - the deeper the shuttle goes, the more pressure we need to reinflate it...
did you miss my last post to stefan? let me address some things here.
to begin with, you can not get water out at the total lift height. if lift = 1 meter, the exit would have to be < 1 meter and the storage area (if used) lower than that. space available for the storage device is the only limitation for the size. the shallower you can make it, the higher it can be mounted (water exit height from system max). if the shuttle is at 10meters and has expanded to displace a positive 1 meter, the 10 meters above will be displaced to the system just under 1 meter above sealevel. this does make a difference. you now have almost 10 time the weight to do work. that's a lot! also the designs you are talking about don't reduce the exit pipe size, you don't take advantage of leverage for recompression, and stefan wouldn't let it go all the way to the top (actually could go above surface).
this is why i didn't want to get the cart before the horse. there are a lot of ways to make this work, there are a lot of ways to make it fail, but before we draw up plans we should see how all fits together. clearly type 1 and type 2 use different parts, so combining the 2 would just produce a type 3 that probably doesn't work. maybe we should stick to the orignal design until everyone understands how it works. then if we want to modify, for the better, everybody will know why it will or won't work.
tbird
it gets mechanically compressed (some type of lever) some how so that the one-way-water-shuttle returns to the buttom of the pipe.
There is a difference if a cylindrical balloon is under water inside a pipe
or inside the water free floating at the same water deepth !
If you have it in a pipe it can not lift up, if the water volume above it
will be more than its own balloon volume.
In the free floating case you can have it several hundred Meters under water
and it will go up to the top, so there is the difference, that there the
water can move to the side ontop if it, so the boyuancy force F is always there.
For instance, if you have a balloon 100 Meters under water and hold it via a thread
and let it go to the top you just can get the Energy: boyuancy force F x distance 100 Meters.
In the case of the gravity mill the violation would be, that it would be able to shift
10 times the weight force of all the water volume above it, which is
not the case, otherwise the energy would be e.g. 10 x boyuancy force F x distance,
which is not the case.
Hmm, I just came across this nice Cartesian diver applet:
http://lectureonline.cl.msu.edu/~mmp/applist/f/f.htm
Hold the red top with your mouse and drag it.
Can this be used to generate overunity energy ?
We only need to apply a short compression or release
of the pressure and then we could earn the whole
distance x boyuancy force F as the energy output !
If we make the water volume very high, e.g. 100 Meters,
the distance it travels is pretty high and we can have
alot of energy output !
The input energy needed to compress the water is only a pulse and
can be latched when compressed and released ?
What do you think ?
The good thing is, you could put the electrical generator into the
water there too sealed and the diver would pull via a string the axis of the generator
via a clutch ratched mechanism and out of the container could
come already electrical energy via wires, so the only outside "control circuit"
would be a weight, that would be placed upon the water surface and
lifted a bit to release the water pressure.
stefan,
you have to stop. you are telling people things that are not true.
let's start over with the basics.
if you have a fixed object that displaces 1 pound more of water than it weights, what will it do at 10 meters? if you said rise, you were right. how far will it rise? if you said, all the way to the surface and expose 1 pound worth of its mass, you were right. how deep can you take this object before it won't rise? if you said, there is no such depth, you would be right. it has a fixed shape so it would still displace 1 pound more than it weights. any questions?
now, if you place this object in a tube, how high can the tube be above water before the object would not rise? if you said, the distance that would contain 1 pound of water, then you would be right. if we doubled the mass, but not weight, how far could the tube be above water level before the object would stop rising and pushing water out? to answer this question, we have to have more info. lets say our object is 1 cubic ft. (12x12x12 inches) and the tube is 10 meters long and fits the object snug. for our work here, we'll say no water gets by in either direction. if it is 1 pound less than it is displacing (fresh water), that makes it weigh 63.7 pounds. now back to the question. if you said, at least 12 inches plus enough for 1 pound, you would be right. i doubt if many here really got this right. if they did, it was probably for the wrong reason. if the top of the tube were lowered so only 11 inches was above water, the object would rise. not because it can push over 9 meters, 12 inch square of water weighting over 600 pounds, because it can not. it can only push 65.7 pounds (1 cubic ft (64.7lbs) we increased the mass plus our orignal positive 1 pound) up. so how can it rise? the water coming in behind the object has the balance of the force due to the pressure exserted by an external colume of water, 12 inches square (the tube prevents any more than that to be applied), from the surface. this is why if the object were removed, the water level in the pipe would be the same as outside the pipe.
it's getting late here now. study the above and if you have any questions about what we just discussed, feel free to post here. if all are clear, i will continue tomorrow with what i believe to be problem areas.
tbird
Quote from: tbird on August 24, 2006, 09:05:39 PM
stefan,
you have to stop. you are telling people things that are not true.
let's start over with the basics.
if you have a fixed object that displaces 1 pound more of water than it weights, what will it do at 10 meters? if you said rise, you were right. how far will it rise? if you said, all the way to the surface and expose 1 pound worth of its mass, you were right. how deep can you take this object before it won't rise? if you said, there is no such depth, you would be right. it has a fixed shape so it would still displace 1 pound more than it weights. any questions?
You are right here, but only, if this thing is not inside a pipe, so the water can also enter and flow
in and out from the side !
Quote
now, if you place this object in a tube, how high can the tube be above water before the object would not rise? if you said, the distance that would contain 1 pound of water, then you would be right. if we doubled the mass, but not weight, how far could the tube be above water level before the object would stop rising and pushing water out? to answer this question, we have to have more info. lets say our object is 1 cubic ft. (12x12x12 inches) and the tube is 10 meters long and fits the object snug. for our work here, we'll say no water gets by in either direction. if it is 1 pound less than it is displacing (fresh water), that makes it weigh 63.7 pounds. now back to the question. if you said, at least 12 inches plus enough for 1 pound, you would be right. i doubt if many here really got this right. if they did, it was probably for the wrong reason. if the top of the tube were lowered so only 11 inches was above water, the object would rise. not because it can push over 9 meters, 12 inch square of water weighting over 600 pounds, because it can not. it can only push 65.7 pounds (1 cubic ft (64.7lbs) we increased the mass plus our orignal positive 1 pound) up. so how can it rise? the water coming in behind the object has the balance of the force due to the pressure exserted by an external colume of water, 12 inches square (the tube prevents any more than that to be applied), from the surface. this is why if the object were removed, the water level in the pipe would be the same as outside the pipe.
Hmm, I don?t understand this.
Could you please try to explain it with the example dimensions I used in my previous example ?
Many thanks.
Holy shit, this thing really works ! :P :-X
I just tried it in my bathtube !
The plastic-air-cylinder I had in a bit larger diameter tube (pipe) could lift at least 3 times
its own volume of water above it.. !
My tube(pipe) was too short, so I could not lift more water !
But it really works !
I have to go out and get some longer tubes ! This is really amazing !
If you go very deep you can really "pump" so much water with it upwards !
Now I also saw my fault.
As the hydrostatic preessure force is always bigger at the lower surface of the air-cylinder,
there is always a positive pressure and thus force to lift up the whole stack-volumeof water
above the aircylinder ! Thus if you have it 100 Meters deep, it will lift the whole 100 Meter column
of water above the aircylinder !
This has really a very big overunity factor !
Stay tuned.
Regards, Stefan.
;)
Okay, here is now a quick video I just recorded
of my experiment via my PDA with just one hand , so it is a bit shaky !
The attached AVI movie has Microsoft MPEG4-V2 video codec and GSM6.10
audio codec at 320x240 res.
You can only download it, if you are logged into the forum.
Let me know, if you can replicate it simularly !
Many thanks to TBIRD to getting me onto the right track !
Regards, Stefan.
All,
one easy design criterium is, that you can lift the water up over the
level of the main bassin as high as the swimmer body has as height !
So if your cylindrical swimmer piston body is 1 Meter high, you can also
lift the whole water above it 1 Meter high.
So if you have the 1 Meter high and 1 Meter diameter swimmer piston body 100 Meters deep inside the water,
you can lift 100 x 785.4 kg water= 78540 Kg of water 1 Meter high, which will be an energy of about 214 Watthours.
If you make the swimmer body itsself 10 Meter high, you can also lift the 78540 Kg of water 10 Meters high above
the main level, so you already have 2140 Watthours of energy.
So it is wise to maximize the height of the swimmer body to lift all of the water the maximum height
over the main water level.
P.S: To overcome 100 Meters of water pressure you need to pump
the air with about 10 bar in the hose to get down to 100 Meters deepth.
Now the question is, how much energy do we need to push a volume of 10 Meter x 1 Meter diameter
volume 10 bar air down there 100 Meters ?
Ok, now you all are at the point i was long time ago. The swimmer is able to pump up as much water as you like, as long its able to pump up the 1m above sealevel.
hartiberlin - thats the question I'm asking all the time. Will the amount of stored water be enough to alow recompression or reinflating the shuttle?
Andi
hi stefan,
it sure feels good having you back onboard. extra well done for really trying to get the definative and not just say "it doesn't work"!! with this attiude, we can do anything!
QuoteThus if you have it 100 Meters deep, it will lift the whole 100 Meter column
of water above the aircylinder !
we should restudy John Herrings designs. using his method, you can produce the same weight (lift the same amount of water) in only 50 meters. after all this topic is called "Gravity Mill" (his work).
Quoteone easy design criterium is, that you can lift the water up over the
level of the main bassin as high as the swimmer body has as height !
this may be true, but it is not limited to this. the size of the exit pipe determines how high you can push it. remember the shuttle piston will displace above water level the positive buoyancy. if your exit pipe were twice as large, it would only raise it half as high. you still get the same weight. if the exit pipe is half the volume, you can push it twice as high. as a result though, you change the time it takes to get all the water out. bigger pipe, more volume. smaller pipe, less volume. being able to lift the water to the height you want is a nice advantage.
QuoteNow the question is, how much energy do we need to push a volume of 10 Meter x 1 Meter diameter
volume 10 bar air down there 100 Meters ?
this can probably be done. that's why i didn't boohoo andi's design before, but why go to 100meters? if you stay with John Herrings designs, you can have a continues (pretty much) flow from a more reasonable depth. remember, the deeper you go, the more pressure you have to create. plus if you drop something, who's going to pick it up at 100 meters?
andi
Quotehartiberlin - thats the question I'm asking all the time. Will the amount of stored water be enough to alow recompression or reinflating the shuttle?
if you use the John Herring design, this is quite easy. he uses leveage (could be big time) to do this work. from the examples before if you took 1meter (not all 10) of water 10 feet away and attached it to a lever, the other end would now feel 10 times the weight. the only question is how much of your water supply do you want (or have to because of space resriction) to use to do this job. having said that, you should be able to use this same leverage to work your compressor. it may not end up looking like a conventional compressor, but there should be a way.
i sure there are still unanswered questions and i probably just created more, so let's get to it.
tbird
Hi all, Here is another site with som more drawings. http://www.icestuff.com/energy/elsa/
Keytronic
Hi I am comming into this one a little late but has one of these devices ever been built and if so are there any photo's
Kind Regards
mark
http://www.icestuff.com/energy/energy21/
this was the end ...... (in past)
hi keytronic88,
great addition to our cause. the Don Adsitt site has to get a gold star too. if John Herring had sent as much info as in on your link, there might not have been as much confusion.
good to have you onboard too.
hi mark australia,
better late than never.
to my knowledge, the only units that have been built would have been by the inventor, John Herring. at the link above, in keytronic88 post, Mr. Herring mentions showing his unit to a school or two. the pages there don't show a complete unit, but there are a lot of drawings of verious parts with some explanation. if you can read thru those pages, you should come away with a fairly good picture of the type and size right for you.
good to have you onboard too.
tbird
ps has everyone had a smile today?
It is directly here at icestuff:
http://www.icestuff.com/energy/elsa/
Best start with looking at picture #770
as the others are very confusing sometimes,
if you have not yet understand the basic principle in
#770
The main advantageof the ELSA design is,
that you can pump all water over the moving swimmer piston
body into an upper tank, where you then have stored potential energy
from the water there and it needs only a fraction of this energy
to recompress the swimmer piston body for another cycle.
http://www.icestuff.com/energy/elsa/page_29_-_770.jpg
direct link to #770
Anybody made some new calculationst? Will it be possible to compress the shuttle with the gained waterpower?
I think, this will be the sticking point. If we are not able to give enough power to the shuttle that it is able to reinflate in e.g. 10m depth, the whole overunity will be gone.
Has anybody an idea?
Maybe we can adopt this cartesian diver to extract energy from
the diving and lifting of the swimmer body ?
You only need a short relay latch push to put a weight onto the bottle
and then remove it again to restart the cycle, but you can earn all
the boyuancy force x distance upwards and downwars as output energy,
if you couple the diver unit to some kind of energy output production unit,
for instance a magnet-coil generator it will slide through.
Also if you make the bottle more deep, the more output you will get at the
same input.
If you mix that with the ELSA principle, you would only need
to compress the outside water surface, so the center tube, where all
the water is pushed up does not need to be modified.
Have a look at this cartesian diver video:
http://video.google.com/videoplay?docid=-3177135341125519060
Here is another video:
http://video.google.com/videoplay?docid=3460425418602417891
Okay, mix this with the ELSA principle and
we will have a winner !
Look at the attached animation picture. ( you must be logged into the forum)
If you put the egg into a pipe and the pipe goes a bit over the watersurface,
this will be the ideal combination.
hi andi,
QuoteAnybody made some new calculationst? Will it be possible to compress the shuttle with the gained waterpower?
I think, this will be the sticking point. If we are not able to give enough power to the shuttle that it is able to reinflate in e.g. 10m depth, the whole overunity will be gone.
Has anybody an idea?
sorry my answer in reply 74 didn't satisfy you. give me some numbers and method (way you want to recompress) you want to use and i'll try to do the math for you.
tbird
hi
stefan,boy, your mind is really racing!
not sure if i follow your unit, but i get the impression you want to put those electronics under water again, is that right?
in this case, do you want to make electricy or move water? don't forget at depth your "egg" still has to be able to expand enough to be buoyant. if you are making electricy, won't take as much as pushing water. do you think you can make a gen that will put out as much at that speed as you would get from the water? maybe you want to do both at the same time?
I am just trying to correct my old posting to calculate now
with a 10 Meters deep inside the water 1 Meter high swimmer body.
Does anyone know, how to calculate the work-energy
required to compress normal air to
785,4 Liter of air with 1 bar pressure ?
How much normal air with normal seawaterlvel pressure
do you need for this ?
Habe a look at :
http://www.peter-junglas.de/fh/vorlesungen/thermodynamik1/html/kap2-3.html
Do I just have to calculate the area under the PV diagramm ?
But with how much air do I have to start in the begining to get
785,4 Liter of air with 1 bar pressure ?
Many thanks.
Regards, Stefan.
Well it seems the energy required to compress air in a airpump
to higher pressure with smaller volume is this:
W= - (Integral px dV ) + Pu ( V2 - V1)
With Pu being the pressure of the normail air outdoors at sealevel
and V1 being my 785,4 Liter at 1 bar pressure.
So how can I calculate the Work Energy W, if I don?t don?t know the starting
Volume V2 ?
hi stefan,
i could not understand anything on your last link (not my language).
the metric system is pretty cool, but.....iget lost pretty quick with this stuff.
in psi and pounds, we might be able to figure something. like how much water. if we know that, you can figure how much work you can do with the leftovers. would that be good?
hi
Quote
stefan,boy, your mind is really racing!
not sure if i follow your unit, but i get the impression you want to put those electronics under water again, is that right?
Right, as this is easy and it only needs to be encapsuled, so no water comes
into the generator, that would be not too complicated.
That has the advantage, that you need no seal, to transport any
mechanical push-pull-string-thread action outside the water case.
Quote
in this case, do you want to make electricy or move water? don't forget at depth your "egg" still has to be able to expand enough to be buoyant. if you are making electricy, won't take as much as pushing water. do you think you can make a gen that will put out as much at that speed as you would get from the water? maybe you want to do both at the same time?
You can also build a linear magnet coil-generator to extract energy from the
movement ofthe Cartesian diver directly along its path up and down.
If you use strong neodymium magnets you can also extract quite nice
power, also if the diver does not move too fast...
Quotealso if the diver does not move too fast...
did you say what you meant?
tbird
Quote from: tbird on August 25, 2006, 07:19:36 PM
hi stefan,
i could not understand anything on your last link (not my language).
the metric system is pretty cool, but.....iget lost pretty quick with this stuff.
in psi and pounds, we might be able to figure something. like how much water. if we know that, you can figure how much work you can do with the leftovers. would that be good?
Okay, work something out in the PSI and pounds US system and I will
convert it to Metric SI system units.
I really need to now, how much work it takes to compress normal
outside air to 1 Bar pressure = 14,5 psi at 785,4 Liters= 207,5 gallons.
Quote from: tbird on August 25, 2006, 07:24:23 PM
Quotealso if the diver does not move too fast...
did you say what you meant?
tbird
Sorry,
It must say: also if it does not move so fast...
so if it moves slowly, you have to at least have big magnet forces
and low coil resistances, so you get big amperage at low speed and convert
this electronically up to higher voltages.
Okay guys, lets try again to calculate it step by step and remove my old bugs.
Pay attention, that I use now the Komma (,) as a seperator to the
decimal point for the numbers below 0 and use NO point between thousands points.
This time I go for a balloon sized swimmer pistion like body,
that gets inflated at 10 Meters deep inside the water via an aircompressor
to pump air into 10 Meters balloon, so it will get a cylindrical balloon
with 1 Meter height and 1 Meter diameter of air and as this is at 10 Meters
deep we need 1 bar of air pressure to do this.
We must try get more potential energy from moving the water into an upper
reservoir: potential-Energy= watermass x earthaccel. g x heightdifference,
than needing for the compression of the swimmer unit !
Okay, let?s stay with Andi?s example:
A pipe 12 Meter deep inside the seawater and 1 Meter above the seawater level.
diameter = 1 Meter
Now let us just use the lifting not the sinking to produce the energy.
Okay, now the START condition.
All water in the pipe is in the seawaterlevel position and the
piston like swimmer body must be 10 Meters under seawater level
to be able to go up to the opening of the pipe l Meter above seawater level and thus
pump 10 Meters of volume over the 1 Meter sticking out of the seawaterlevel
high pipe into an upper reservoir 1 Meter above seawaterlevel.
Okay the piston like swimmer body will be also at the START 1 Meter high and have a diameter
of 1 Meter.
Okay, how big will be the potential energy we win, if we succeed to pump this
volume of 10 Meter height of water volume of 1 Meter diameter water up:
height of water above the swimmer unit: 10 Meters= 1000 cm
Volume=Area x height,
Area=r?*Pi=2500cm*Pi= 7853,98 squarecm
Volume=Area x height=7853,98 squarecm x 1000 cm= 7853980cm? this means the water weights 7854 kg
or has a volume of 7854 Liter.
Thus we have a potential energy stored of:
Energy= 7854 Kg x 9,81 x 1 Meter height difference between seawaterlevel and reservoir = 77048 Joule or Wattseconds.
That means put into Watthours we must divide it through 3600, so
we get about 21,4 Watthours of potential energy stored 1 Meter about sealevel with
about 7854 kg= 7854 Liter of water !
Now let us see, if this would be enough energy to release the
air from the balloon swimmer body, let it sink itsself again to 10 Meters deep
inside the water and then pump air at 1 Bar pressure again down to 10 Meters
deep to inflate the ballon again to 785,4 Liter at 1 Bar pressure so it is 1 Meter high
and 1 Meter in diameter.
Then the whole cycle can start again.
Okay, now I still have to know, how we calculate the compression of the
air to get it down there and build up the balloon for 785,5 Liters at 1 Bar.
I just saw, the air outside has 1 bar of pressure, so in 10 Meters deep
we need then 2 bar of pressure. this means we need to produce
about 29 psi at 10 Meters deepth
from the 21,4 Watthours we have at the reservoir of water.
Can we do this with the amount of energy ?
hi stefan,
when you get going, you really get going. i can't even type that fast, much less think and type. i may miss a thing or two, but here's a little.
you didn't mention the weight of the balloon. if it's weight is not worth mentioning, then you only have to create enough volume to fill the 1 meter space you described. once the pressure from the compressor reaches 1 bar, the ballon will start to expand. it will stay at that pressure until the ballon can't expand any more. then if more is pumped in, the pressure will rise. but once the volume is made, no more needs to be pumped. since the balloon has (basicly) no weight, it will be on its way. once it reaches the top, with the outside pressure being reduced, the balloon , if allowed, will double in size, if not allowed to expand, the pressure will be at 2 bar. does that make sense to you?
also, it will be more productive (more potential energy from moving the water into an upper
reservoir: potential-Energy= watermass x earthaccel. g x heightdifference) if you raise the reservoir. by reducing the exit pipe diameter to 1/2 the volume starting at water level, you can raise the reservoir twice as high. according to your formula, this doubles "potential-Energy"
are you still with me or do we need more explaning (research)?
tbird
Okay, I just found this:
Formula:
P*V=n*R*T => P=n*R*T/V
Isotherm:
T=const => p*v=const
dW=F*dl with F=P*A
dW=P*A*dl
dW=P*dV
dW=n*R*T*dV/V
Integration yields:
W=n*R*T*ln(V2/V1)
W=P1*V1*ln(V2/V1)
with:
1 bar = 100 kP= 15,4 psi
1 Liter= 1/1000 m^3
hi stefan,
i just read my post and thought i might have confused you. you are right, you do need 2 bar at your compressor for your example. what you don't need is that much volume. if you reduced it by half and reduce exit tube diameter above water level, you can still get the same amount of water, but not as fast.
tbird
Quote from: tbird on August 25, 2006, 08:25:25 PM
hi stefan,
when you get going, you really get going. i can't even type that fast, much less think and type. i may miss a thing or two, but here's a little.
;) Okay, no problem. Yes,I am eager to see, if we can nail it down
also mathematically and thus prove it to every physics teacher, that
this concept just works !
Quote
you didn't mention the weight of the balloon. if it's weight is not worth mentioning, then you only have to create enough volume to fill the 1 meter space you described. once the pressure from the compressor reaches 1 bar, the ballon will start to expand. it will stay at that pressure until the ballon can't expand any more. then if more is pumped in, the pressure will rise. but once the volume is made, no more needs to be pumped. since the balloon has (basicly) no weight, it will be on its way. once it reaches the top, with the outside pressure being reduced, the balloon , if allowed, will double in size, if not allowed to expand, the pressure will be at 2 bar. does that make sense to you?
Okay, yes, the ballon will expand ,
but see it more like this:
maybe we should just use a plastic cylinder case, which has a valve at the top and is open at the button.
So it can be filled up at the top via a hose and via the compressed air, so the compressed air is filling up the
water containing plastic cylinder, which weights itsself not very much, so its own weight could
be left out of the calculation.
This plastic cylinder valve is then at the top level made open the valve,
so all aircan escape at the top and then it sinks again due to its own small
weight inside the water and sinks to 10 meters, where the valve at the top is closed
and air is put via a hose into it, so the water goes out of it and it will again have a lifting
buoyancy force.
We need 2 Bars= about 29 PSI to bring the air down there.
The volume of the air we have to bring down there depends on the height
of the water cyliner above sealevel !
If we have 1 Meter, we also must have a 1 Meter cylinder of air
under water at 10 Meters deepth.
Quote
also, it will be more productive (more potential energy from moving the water into an upper
reservoir: potential-Energy= watermass x earthaccel. g x heightdifference) if you raise the reservoir. by reducing the exit pipe diameter to 1/2 the volume starting at water level, you can raise the reservoir twice as high. according to your formula, this doubles "potential-Energy"
?
Well, Okay, maybe if we use a fountain like structure and capture the fountain water inside an upper
reservoir... But remember, the hydrostatic paradoxon requires, that the pressure weight of a water column
is only dependant from the height of the water, and if you make the water column above the seawaterlevel
higher than 1 Meter, also with smaler diameter, it needs also the height of the aircylinder under water at 10 Meters
deepth to be higher, which needs more air to compress and pump down !
Quote from: tbird on August 25, 2006, 08:36:29 PM
hi stefan,
i just read my post and thought i might have confused you. you are right, you do need 2 bar at your compressor for your example. what you don't need is that much volume. if you reduced it by half and reduce exit tube diameter above water level, you can still get the same amount of water, but not as fast.
tbird
TBird,
due to the hydrostatic paradoxon we need to keep attention how much water
we have in the tube above sealevel !
If you make this watercolumn too high, also with smaller diameter, it does not matter ,
you need more volume of air down there at 10 Meters deepth !
Reread the Hydrostatic paradoxon...
i need to respond here, but not to everthing.
QuoteThe volume of the air we have to bring down there depends on the height
of the water cyliner above sealevel !
If we have 1 Meter, we also must have a 1 Meter cylinder of air
under water at 10 Meters deepth.
this is causing you a big problem. let's think about what is going on. the reason the water goes up is there is force below. if you can move 10 pounds of water, it doesn't matter if it's in a 1 by 10 (any measurement) or 10 by 1 area.
see how you can raise the same water higher without changing volume below?
TBird,
I only agree to the water under sealevel.
The water above sealevel puts weight onto the water column
and you have to pay attention to the hydrostatic paradoxon.
Quote from: hartiberlin on August 25, 2006, 08:25:35 PM
Okay, I just found this:
Formula:
P*V=n*R*T => P=n*R*T/V
Isotherm:
T=const => p*v=const
dW=F*dl with F=P*A
dW=P*A*dl
dW=P*dV
dW=n*R*T*dV/V
Integration yields:
W=n*R*T*ln(V2/V1)
W=P1*V1*ln(V2/V1)
with:
1 bar = 100 kP= 15,4 psi
1 Liter= 1/1000 m^3
We really get a thermodynamic theory guy help us with this.
In the above formula I have missing the pressure P2.
If we take the formula:
W=P1*V1*ln(V2/V1)
W= work-energy, we have to put into compressing the air.This I don?t know yet !
P1= starting air pressure of 1 bar outdoors.
V1= volume of starting air, this I don?t know..
V2= Volume of air compressed at P2= 2 bar = 785,4 Liter
But where can I put P2 ?
If I had a formula including P2 then I could play with it,
as long raise some input energy W until I get to
P2= 2 bar and V2= 785,4 Liter.
Can anybody please help ?
Thanks.
think about it this way. what stops the piston (ballon, whatever) from rising? either it weighs more than it is displacing or in our case in the tube, a larger weight above water level than the piston is displacing. right? so it doesn't matter if it is all placed in the first meter or divided in 2 and placed on top of each other. both shapes weight the same, but 1 is only half as wide as the other. so as long as the volume of water in the tube above water level weighs less than the piston is displacing, the piston will rise. your test was a little awkward for you :( but if you could actually put a smaller tube on top at water level, the water will come out higher. like getting a flat tire on your car, the smaller the hole, the longer it takes to go flat, but it will go flat.
stefan,
i'll try another angle. if the pipe in your example were 2 meters tall (above water level) and still 1 meter in diameter, at depth, your ballon would stop going up when the displacement weight, 785,4kg, was above water level. we figured this to be 1 meter, so with the pipe 1 meter taller, all the water would be still in the pipe but 1 meter above water level, right? now in this state, one way we can make the balloon rise is to give it more volume, which displaces more weight (without increasing the balloon weight), thus giving the balloon the ablity to lift more WEIGHT, right? or we could do the oppisite. we could reduce the weight above. either way, your balloon rising depends on how much WEIGHT is above water level. as long as we keep this weight less than the lift of your balloon, we can take it as high as we want. as long as the balloon has more LIFT than the column of water above water level has WEIGHT, the balloon will rise. it doesn't matter what shape this weight is as long as it stays in column. 785.4kg is 785.4kg. we can say the rest of the water, in tube and outside, equalize each other.
i hope you can see now.
tbird
Quote from: tbird on August 25, 2006, 09:24:43 PM
think about it this way. what stops the piston (ballon, whatever) from rising? either it weighs more than it is displacing or in our case in the tube, a larger weight above water level than the piston is displacing. right? so it doesn't matter if it is all placed in the first meter or divided in 2 and placed on top of each other. both shapes weight the same, but 1 is only half as wide as the other. so as long as the volume of water in the tube above water level weighs less than the piston is displacing, the piston will rise. your test was a little awkward for you :( but if you could actually put a smaller tube on top at water level, the water will come out higher. like getting a flat tire on your car, the smaller the hole, the longer it takes to go flat, but it will go flat.
Well could be, but we have to watch the hydrostatic paradoxon
with it, as only the height plays a role what water pressure weight
is ontop the sewaterlevel.
Can you calculate your setup completely ?
How can I resolve( rewrite) the formula:
W=P1*V1*ln(V2/V1)
to:
V1= ....
Its Long time I have used mathematics..
hi stefan,
QuoteWell could be, but we have to watch the hydrostatic paradoxon
with it, as only the height plays a role what water pressure weight
is ontop the sewaterlevel.
this is not true. hydrostatic paradoxon, i think this is your label for why the balloon doesn't have to raise the total weight of the column. i can't remember hearing this term before, but did notice it in some of the reading material. what i know is the water at and below water level does is not involved in lifting the water above water line. the closest thing you can say is that it nutralizes the water above the balloon, but only to the top of the water level. the only thing that pushes the water above water level in a column is the balloon displacing that amount of water. even if the balloon didn't have positive buoyancy, it would displace it's volume above water level if contained in a column with bottom closed.
QuoteHow can I resolve( rewrite) the formula:
W=P1*V1*ln(V2/V1)
to:
V1= ....
i'm afraid i don't understand this formula. is there a drawing it refers to at the link you sent? if you can explain what each stands for, it might help.
Okay, I have now the solution.
The missing link was the gas law:
p1 x V1 = p2 x V2
So now I can calculate V1 !
V1= 2 bar x 785,4 Liter / 1 bar= 1570,8 Liters
stefan,
that's great! what does it mean in plain english?
Okay, now with this information:
P1= 1 bar
V1= 1570,8 Liter
P2 = 2 bar
V2= 785,4 Liter we can now calculate the work-energy to compress
the 1570,8 Liter from 1 bar at the top to 2 bar at 10 meters deep, which will
then have a volume of 785,4 liter :
With
W=P1*V1*ln(V2/V1)= P2*V2*ln(V2/V1)
we get
W= -30,24 Watthours.
So we need more energy to compress the air, then we got from
the reservoir 1 meter above seawaterlevel...
But now is the question, if we really need 785,4 Liter of air inside
the cylinder in 10 meters deepth as TBird also suggests ?
stefan,
i think i follow somewhat. maybe i'm just tired. better go to bed. one last thought. how much would you need if it had a 2 foot lever? 5 foot lever? 10 foot lever? ???
good night all.
tbird
Hmm, in my setup there the air-cylinder under water must be at least 1 Meter high
at 9 Meters deepth, otherwise it will not be able to lift the 1 Meter water column
ontop of the water surface, which weight 785,4 Kg or Liter.
Also if you make this diameter of this water column much smaller and have thus
less water in it, you would still need the same air volume at 9 meters deepth,
cause we have the hydrostatic paradoxon, so that 1/2 of the water diameter column,
so 50 cm in diameter , will
weight the same as if we use the full 1 Meter diameter water column ontop the seawater level.
I just tried to glue a smaller pipe onto the bigger pipe I used yeasterday and it seems to
confirm this, that the water column ontop the seawaterlevel is only this high
as the airvolume height inside the swimmer body will be.
Well, I guess I found a solution to the problem.
I assumed, that I must pump ALL the air into the cylindrical plastic-balloon
container down at 10 Meters.
This must not be !
If we define a different START condition, we need only to add some
air at 9 Meters deep to the swimmer body there being in equilibrium with the water weight column
ontop the water surface,
so the swimmer can rise and move the water again
and when have reached seawaterlevel remove some air, so the swimmer can sink again.
Then we need way less pump energy !
I will calculate this also soon, but now I have to go to bed.
Stay tuned, this is going to be exciting !
Regards, Stefan.
Well, I had still another look at it how to do it,
if we start with equilibrium at
9 Meters under sealevel= 10 Meters under the whole water column
at 2 Bar and at 785,4 Liters,
where we have equilibrium. In this position the swimmer body will stay there and
don?t move and we have a water column of 1 Meter above the seawaterlevel,
which will not yet run out of the pipe.
Then we go from 785,4 Liters to about 800 Liters
of volume in the swimmer body by pumping a bit of additional air into the swimmer body,
then we can rise all the fluid above it and only
need an energy of 0,81 Watthours for the pumping this 14,6 Liters air into the swimmer body at 2 bar !
In contrast we will then get 21 Watthours of potential water energy
from all the water being pumped up
to 1 Meter up over the seawaterlevel !
So you see, how big the overunity factor is !
It just works !
Enjoy !
Regards, Stefan.
Well the swimmer body makes a whole PV diagramm right turn circle process.
We must see at the top, how we manage it, that it sinks again.
This is now the hardest parts...
hi stefan,
it's amazing what a good nights rest will do for you.
here is how you prove to yourself a smaller diameter will make the water go higher. first let's agree on a given. if we have positive buoyance in our shuttle (balloon, piston, whatever the device is that moves the water up), we have created pressure on the water above, right? if the water can't go around our shuttle, it has to go up, right? now if we don't push the pipe too far out of the water, we will see flow above the waters surface (we proved this with test). we all agree on this now, right? here comes the proof. take your garden hose with the water running and point it straight up. this looks kinda like what our pipe would sticking out of the water, right? now cover the end slowly starting from the side going to the other side (use your thumb). did you beat me to the punch line? without changing the water pressure, what does the water coming out do? it's been my experience that the water goess farther up. remove your thumb, the water height goes down.
now what have we proved? the smaller the opening (exit diameter), the higher the water will travel. do i need to say more?
tbird
Hi stefan.
Amazing, I went to bed and you are calculating such good formulas. Congrats!
Another thing to mention: The air in 10m depth has another volume than at 0m depth. That means If we fill the baloon under water with 15liters of air at 2bar, the baloon will have MUCH more volume at 1bar at the top of sealevel...
tbird - I know what you are trying to do, but this wont work. If you squeeze your water hose (or scaling down the exit) you give force to it which has to be overcome by the water pressure. This would balance out the heigher waterflow, so you have no results from it.
Andi
hi all,
to get the most (efficiency) from our machine, we should use both the up and down strokes. after all there is energy in the fall of the shuttle too. the reason we may consider using only the up stroke, would be to make building and assemble easier (simpler). there is only so much water in the pipe we can move per energy used to compress the air. if we use the down stroke too, the pipe will fill twice. that means twice the water per cycle, per compression of the air. if the sigle stroke operation can be done in at least half the time and consume half the recompression energy, then we should continue to explore that design. this maybe possible, but before we go there, we need to get the basics of the machine (elsa) well understood. let's not throw away the 30 years of work and research John Herring has already done.
the depth we use for elsa may or may not result in more or less efficiency. if the formula stefan used last night is correct, work will be proportioned to the pressure needed. so if you only use 1/3 the depth, you only need 1/3 the pressure, but you only have 1/3 the water to do the work (if i am applying the formula right). if everything is relavent, a smaller depth would allow easier building of a self contained unit. this would eliminate the need for an ocean, lake, pond, or the like. just so many square feet (meters) in the backyard (garden).
seeing how important the needed energy to compress the shuttle is, maybe we should concentrate on this part of the machine. John Herring devotes several drawings to this job. here again, i think we should take advantage of his experience.
farther down the road we will get to the actual use of the energy we have brought up to and above ground level. how this is done is only limited to our imagination.
i'll stop here for now and let you do some studying.
if i have made a mistake in the early going, then the rest is probably wrong too. please let me know.
tbird
hi andi,
QuoteI know what you are trying to do, but this wont work. If you squeeze your water hose (or scaling down the exit) you give force to it which has to be overcome by the water pressure. This would balance out the heigher waterflow, so you have no results from it.
are you saying the water in your hose didn't go higher? if you are not saying that, then the pipe reducer does this, thus higher water delivery.
tbird
Hi Stefan,
This is a first time post for me. Just to let you know, I am a polymer research chemist with over 20 years experience in the lab. I have been following this post for several days now and find it very interesting. I also follow the idea that keeping setups as simple as possible give the best chance at success.
Look at this sketch by Mr. Herring:
http://www.icestuff.com/energy/elsa/page_36_-_780.htm
Notice that it gives a solution to shuttle compression by pushing upward with the force of the gained energy (water). To me this looks like a very real and simple solution to shuttle compression. The only problem I see could be when the shuttle is ascending. Water flowing past and opening has an ability to create a vaccuum. This might pull the lever intended for compression inward. I don't know. This would block the shuttle's ascent. Maybe a spring could keep the lever in its little alcove until water compresses it from above. Just an idea.
I also don't know if we need to worry about the calculations so much at this point. Hypothetical and real values always vary and little glitches in the workings of an apparatus tend to pop up unexpectantly. Real lab work is the only way to test for these problems. If it works, it works. If not then why and how can it be made to work?
There does seem to be a good possiblility that Mr. Herring did have a working apparatus at some point in time, but where are the pictures? In fact, where is Mr. Herring. Is he still alive? If he has worked on this concept for decades then he must have some kind of working unit.
I would love to see a picture of a shuttle he built, or anything he put together for that matter. I don't believe every sketch is just hypothetical. He must have something that works and is in existance.
Just a few thoughts I am throwing out; and I want to thank you for all you efforts.
best regards, ResinRat2
tbird, surely the water will go higher when scaling down the pipe exit, but the shuttle needs to overcome this extra force... Nothing gained from it.
ResinRat2, I tried different ways to get in contact with Mr. Herring long time ago. I don't know if he gave up his project and what he is doing now. Perhaps he was not able to build his device - at least then he could tell us where the problem is.
Andi
hi ResinRat2,
it's nice to have a person onboard with such a background. welcome!
QuoteNotice that it gives a solution to shuttle compression by pushing upward with the force of the gained energy (water). To me this looks like a very real and simple solution to shuttle compression. The only problem I see could be when the shuttle is ascending. Water flowing past and opening has an ability to create a vaccuum. This might pull the lever intended for compression inward. I don't know. This would block the shuttle's ascent. Maybe a spring could keep the lever in its little alcove until water compresses it from above. Just an idea.
of course the compression arm would be locked in place until it was needed. if not, as soon as enough water was added to leveraged water container it would overcome the weight of the arm assy. on the shuttle side of the pivot and reguardless of the "vaccuum" or whatever, it would enter the pipe. using this system, the arm should be ready to work well before the cycle is complete.
any thoughts on how to prove to these guys (doesn't look like my hose example worked) that if you reduce the diameter of the exit pipe at water level, it will be able to push the water higher?
thanks for being here.
tbird
andi,
Quotetbird, surely the water will go higher when scaling down the pipe exit, but the shuttle needs to overcome this extra force... Nothing gained from it.
what extra force? it didn't gain any weight because you moved it. it can only create as much force as it weights.
have to do a little work ow. be back in less than 2 hours.
tbird
tbird,
I agree, if you reduce the diameter of the exit pipe, you are able to lift the water higher. What I mean is, if you cylindrical scale down the pipe exit (see turbine) you would create a water jet, but therefore need more power for the shuttle and would gain nothing? Agree?
This is what I red out of your post.
Andi
andi,
NO! i don't know about a turbine how you are comparing it here, but here we have moved half of the area (= i/2 weight of water above) to another place in column with the half lift behind. now the pressure focused before on the total area is focused on half the area. to stop the piston, you have to have as much weight opposing the piston as it displaces. since it won't fit in the already occupied half, you can only go up. this doesn't change the weight, just its location. if the pressure of the shuttle doesn't change, and it won't because we didn't change the volume of the shuttle, then if we only had half the area to exit the water in, the load on the piston would be halved. this would allow it to push the 1/2 remaining water harder, thus higher.
"sometimes a potatoe is just a potatoe". let's not make this harder than it is.
tbird
andi,
i understand what you mean now. you don't have to add more power because you still have the power available that was pushing on the half that is restricted. it won't just go away. it goes to the load presented, 1/2 volume exit pipe.
tbird
All. I am again awake and TBird is right. With a constant pressure from the shuttle going up, you can get the water higher, if you squeeze the tube exit. Your garden hose example is perfect, as the pressure from the watercompany delivered to your hose input is constant !
hi stefan,
THANKS FOR COMING TO MY RESCUE!!
what's next?
Pondering in bed about it, I now see the main Problems are in sinking the shuttle again ! If we just use a balloon for the shuttle, it will not work, cause at seawaterlevel, the balloon will have a much bigger volume, due to the reduced water deepth pressure. Also then releasing some air also does not work, as it never will sink again, until you will remove ALL the air ! But then we face again the problem to have to put too much energy into pumping all the air again into 10 meters deepth !
Now the only solution I see is to build the Shuttle unit like a Cartesian diver unit.
It needs to have weight on it, so if we move some air out of it on the top at seawaterlevel, then it can sink again. Maybe it is also better to build the shuttle really this way, that it can just have 2 different volumes and switch between them. Now I have to calculate how much energy is needed to sWitch between these 2 volumes.
stefan,
did you read my earlier post today? were they not worth a comment or 2?
tbird
tbird, you are right, I see it now. The pressure is constant, so we just have to arrange the pipes diameter.
hartiberlin, thats what I'm saying. Fixed volumes will not work as the pressure is different at different sea levels. Maybe it should really be a static box with 2 volume states. Lets see, what your calculations are going to look like.
Andi
TBird, which posting do you mean ? I am only at my PDA now, cause my DSL hoster has some problems...
Well let us first look at the sinking problem and define a solution for this. I guess we really need a longer much smaller tube ontop the seawaterlevel to bring the water more up !
stefan,
first things first. read reply #119 first.
tbird
We have to differentiate between static case, where all forces compensate and the shuttle is stopped at its place and the dynamic case, where the shuttle has a higher force on it, so it moves. Only for the first static case we can use the hydrostatic paradoxon, as it limits the height we can transrt the water up. If we make the upper pipe diameter smaller and have the dynamic case, then the water will go higher. Do you agree ?
stefan,
is your last post in responce to my reply #119? if so, try again using layman terms, please.
tbird
I guess we should put the static case at 5 Meters deepth in my calculation example and add and remove volume, so it can go to 10 and 0 meters deepth, simular to an oscillation center point.
stefan,
are you talking about anything that is like the gravity mill?
tbird
Tbird, I can?t see the posting number on the PDA now, as it has a different browser surface. Will answer later, am pondering about the right parameters to get it right...
stefan,
i may not understand what you are talking about, but it sounds like you are drasticly changing the E.L.S.A. design. so much so i think it would have to be covered elsewhere. am i worng?
tbird
You are wrong TBird. Try to calculate your own example for the upward and sinking case and post it here, so we can see, if you already found the right solution. Many Thanks.
Now lets extend my example to 0 to 20 Meters movement and let put the oscillation center point at 10 Meters deepth and calculate there the static case for equilibrium first...
At 10 Meters deepth to have equilibrium and no movement, we need to have with the same dimensions as previously posted, we need an airvolume of 785,4 Liters at 2 bar and attached to it a weight of 785,4 Kg. Then this shuttle is not moving and just hovering at 10 meters deepth. Do you agree so far ?
stefan,
i think when you get your internet back, you need to read my post and go back to http://www.icestuff.com/energy/elsa/ and re-read all pages, not just glance at them. try to get a handle on what John Herring is trying to pass along.
i don't think we need to go down that problem riddled road he already has. if we don't take advantage of his 30 year effort, shame on us!
i'll be back when you catch up. if anyone has questions about Mr. Herring's designs, i'll try to help out.
tbird
Now the pipe above seawaterlevel is still empty. Now lets fill it up with 1 meter high water and again lets calculate how much air we need, so the shuttle is not moving...
Now we must rise the volume of air to double the amount, so 1570,8 Liters at 2 bar, so the shuttle is not yet moving.
Also due to the hydrostatic paradoxon it does not matter, if we have a very small tube or a bigger one above sealevel, as only the height of 1 Meter water height counts. Agreed ?
Hi TBird, did you come up yet with a conclusive calculation yet ? The problem I see isin the calculation of the energy required to switch the shuttle into 2 different volume states. We have to calculate that via the air pressure as this also hinders any mechanical movement, if you want to do it with a mechanical see-saw torque-arm press. In this case you also have to overcome the different airpressure values, so itis just easier to do it just via pumping airpressure and where I have worked out now all the formulas !
Okay now back to the calculation. If we now fill in instead of 1570,8 Liters e.g. 1600 Liters at 2 bar, the shuttle will rise and also push the water out higher than 1 meter, cause we made the pipe smaller above sealevel. As the suttle moves up, the airvolume will also expand inside the shuttle and thus move the shuttle even faster to the top, as the waterpressure is reduced, which will make the airvolume bigger.( I still calculate it here for the balloon shuttle which can expand with attached weight) Agreed so far ?
Interesting ... found this on the web. http://www.glassbuy.com.cn/en/offer/printshowinfo.php?Id=14937
Looks like John was looking for glass tubing back in 2005. Also has a contact phone #.
At seawater level the shuttles air volume has now doubled, cause:
P1 x V1 = P2 x V2
so it is now 3200 Liters !
Now we must release almost 2415 Liters from it, so it has got still less than 785,4 liters, so the 785,4 kg weight attached to it can pull it down completely... Agreed so far ?
Now we move down to 20 meters and the airvolume will compress even further, cause we are now at 3 bar air pressure inside the shuttle ballon, due to the outside water pressure and the volume of the balloon has reduced to around 261 liters only... Agreed so far ?
hi all,
just called Mr. Herring. got his answering machine with a message from his wife. i left message telling about this forum and my phone #. hope he shows up.
tbird
fyi - Just sent John an e-mail ... also inviting him to chat ...
Okay, now we need to refill the shuttle balloon isobarly ( at the same pressure)
at 3bar with about additional 1310 Liters again so it has a total of a bit more than 1570,8 Liters,
so to be able to rise it against its own 785,4 Kg attached weight and also against the 1 Meter water colum
it has to push beyond seawater level, which is another 785,4 Kg weight.
You see this is a lot of energy we must apply for the pumping !
To pump these 1310 Liters at 3 bar we need alone 109 Watthours of energy to do this !
So, this way it does not work.
We really must use a smaller pipe diameter just directly above seawaterlevel
so there must be no water column weight to be overcome and the water must
then be sprinkled into an upper bassin via the pressure from the buoyant force,
just like a fountain.
Then we also can reduce the attached weight to the shuttle and have to pump less
air down at the 10 or 20 meter deepth.
Okay, now this step by step analysation has helped a lot to understand the
thermodynamical principle behind it and all the PV diagrams of the air which went through
these cycles.
Now to optimize the cycles we must do the following:
Have no watercolumn above seawaterlevel, so we reduce
the negative impact of the hydrostatic paradoxon.
Instead have a nozzle directly at the seawaterlevel
inside the tube?s top, so it works like a fountain
and sprinkles the water 1 Meter at least out into an upper
reservoir at 1 Meter height.
I wonder how big must then be the buoyant force on the shuttle
to get the water pushed out 1 Meter high behind the nozzle ?
Are there any forumulas to calculate this ?
2. We should try to calculate the needed energies to do this with a shuttle,
that has really only 2 different volume states and does not expand
or shrink with water deepth pressure !
So I will try to do this now.
Hi everyone,
ooandioo and tbird, thanks for the info and kind words. I've been busy getting ready for my son's B-day party and am just getting back (snuck away from the wife) to reading the posts. A great deal of activity here today.
Hope tbird's and bastonia's attempts at contacting Mr. Herring work out.
Stefan, it doesn't seem like you read my first post, but check out this link to a drawing from Mr. Herring.
http://www.icestuff.com/energy/elsa/page_36_-_780.htm
It gives a simple way to recompress the shuttle. No electronics, no timers or switches. just a lever that can be weighted down by the water above.
If you get a chance read my earlier post.
Thanks everyone for all your efforts.
ResinRat2
ResinRat2,
yes, let?s concentrate next on the recompression of the shuttle at the top.
To do this efficiently we really need to work this out with all the equatations.
I have to pause for a little, cause I have to take a bath and play with the tubes
and shuttle in it and see, how the nozzle effect turns out and how much
force it needs onto the shuttle !
recompressing , also good to understand with:
http://www.icestuff.com/energy/elsa/page_29_-_770.jpg
GP
Mr. Herring has drawn some nice graphics,
but the problem is, that you have a special air pressure at
each deepth stage inside the shuttle and a different water pressure around the shuttle,
that has to be overcome, when you want to compress or push out the shuttle volume !
This needs real energy to do so, as you also have to change the
air pressure inside the shuttle. To change the airpressure inside the shuttle
is also not less energy requireful than to do it
with just a pump.
And for the pump we already know the equatations.
If you want to use the lever torquearm-combination you also have to calculate in the
distance you have to move the other side of the lever torquearm to get the desired
pressure, so to calculate the required energy you need to calculate the
forces x distance. If you move the shuttle pistons just 10 cms to achieve the different volume,
you have to move the lever arm at the other side, maybe 50 cm to have a 5:1 force multiplication
and then the question is, if the water is high enough, so you can do this 50 cm movement ?
hi andi,
sorry to take so long to get to this post. been thinking about how to get stefan a better internet. :D
you wrote:
"hartiberlin, thats what I'm saying. Fixed volumes will not work as the pressure is different at different sea levels. Maybe it should really be a static box with 2 volume states. Lets see, what your calculations are going to look like."
can't imagine what you mean by "fixed volumes". also, what is a "static box"? does it look any thing like the shuttle in this drawing? http://www.icestuff.com/energy/elsa/
as far as having different pressures, that's ok. did you realize the shuttle in the drawing is pre-pressurised at the surface, before it ever makes it's 1st run? this pressure, when the shuttle is compressed so it will desend, will be 2 atmospheres (if we are taking it to 1 atmosphere, 10meters +or-). 30psi is a little more than that, but will make the shuttle expand to size, for sure. to compress a shuttle that is based on a one cubic foot displacement, we'll say the top area is 12 inches square (144sq.in.total), would take 30psi times the 144 square inches which equals 4,320 pounds. even if the shuttle is expanded with 15(+or-)psi at the surface, 4,320 pounds will compress it to the point where the air being compressed inside will be 30psi. since we started with 15psi, the volume will be reduced by half. back to the weight. if you use a lever to hold the container that the compression water (weight) will be put in, for every foot out, less water will be needed. if the lever was 5 feet long, the weight would only need to be 1/5th (864lbs) as much. if you took it out 10 feet, 1/10th (432lbs.). now at 10ft leverage we've reduced the amount of water needed to recompress from 7.67 cubic feet (4,320lbs.) to 0.767 cubic feet (432lbs.). since we are goin down to 10 meters (+or-) and delivering water in both directions, we should have an excess of about 65 cubic feet (over 4200 lbs). not bad for a 1 cubic foot unit, eah? using the example from this drawing http://www.icestuff.com/energy/elsa/, the height of the stored energy (water), (if you store it) could be as high as 72 inches using a 1 square inch exit pipe.
do you still want to only use the up stroke?
i have more about the shuttle, if you are not burnt out.
tbird
This is really amazing. Some times ago I found elsa on the internet, but nobody would recognize my questions to it. Now its going to be the biggest thread in this forum...
tbird, yes, I'm coming back to the original elsa design. Using a shuttle box that can have 2 different volume states will be the right decision. But, as I said some times ago, the pressure is different at different shuttle depths and it doesn't make a difference if we compress the shuttle at or above waterlevel or in 20m depth. I think first we have to find out if the gained waterpower is able to compress the shuttle. Hartiberlin is on the right way.
andi,
didn't you understand my last post? do you think my numbers are wrong?
tbird
andi,
btw,
QuoteBut, as I said some times ago, the pressure is different at different shuttle depths and it doesn't make a difference if we compress the shuttle at or above waterlevel or in 20m depth.
with this shuttle, those pressure differences are not a factor. when the shuttle expands at the bottom, it will be at it's max volume. the displacement is the same all the way up. this brings me to the "i have more about the shuttle" statement. to take advantage of the pressure left over at the top of the cycle, we could build the shuttle to accomdate more expansion. it won't deliver any more water in that cycle, but it will do it faster. so for a given time, we will have more engery (water) available. the recompression energy will stay the same too.
tbird
Hi guys,
I'm hoping to simplify things a little with the attached drawing of a shuttle. The displacement area can be full of water or air (depending on whether the shuttle is going up or going down.
Let's work with just one kilo of water and a shuttle weight of 0.5 kilos to keep things simple. We must also consider the volume of our shuttle because it will also displace that same volume of water, let's make things easy and assume it is 0.5 litres. The displacement volume, then, must be 1 litre - 1 litre to displace 1 kilo of water plus 0.5 to displace the weight of our shuttle minus 0.5 because our shuttle already displaces that much water when the displacement area is full of water. The weight and volume of our shuttle cancel each other out if they are the same. If the weight is less than the volume then it floats and if the weight is more than the volume it sinks. Ok so far?
The displacement volume will be the same at any depth but, as Steve pointed out, it requires greater air pressure to displace 1 litre of water as we go deeper because we have to displace that 1 kilo of water plus the weight of a column of water above it (just as at sea level the air pressure is approx 14.7psi - or approx 1 atmosphere - because we are measuring the weight of the column of air above it, from sea level all the way up to the top of the atmosphere.) At approximately 10m below waterlevel the weight of the column of water is the same as the weight of a column of air the height of our atmosphere!
What our shuttle will do is to contain air in the reservoir sufficient to displace 1.5 litres (i.e. 1.5 kilos, see why we love metric) at a depth of 10m That is 1 litre of water plus the 0.5 litre volume of our shuttle because we will be keeping that volume in our reservoir when we have filled the displacement area. At 10m we will require 3 litres of air in our 0.5 litre reservoir so it must be at 6 times atmospheric pressure - if my logic is correct. That should be 14.7 * 6 = 88.2psi = approx 6.2 kilos/cm2.
We can discount 1 litre of air when it comes to recompression of the reservoir because we still have that from the previous cycle. This leaves 2 litres to recompress = 2000cc. If we have a plunger with a cross section of 1 cm2 and a length of 20m then a weight of 6.2 kilos will compress the air in the cylinder to that pressure. If we only have 0.5m to compress in then we will have a cross section size of 40cm2 and require a weight of 248 kilos.
The quantity of water we can pump depends on the area of the top of our shuttle. If it is 20cm2 then 1 litre will be 50cm high in the pipe and we can therefore pump 20 litres 0.5m above the water level. Hmmm....
Is my maths wrong or did I miss something or do we have less energy than we thought?
The proposal for the valves, by the way is that V1 is used to fill the reservoir with compressed air, v2 lets the compressed air into the displacement area when the shuttle reaches the bottom and V3 lets the air out when the shuttle reaches the top.
Forgot diagram. See next message.
Oops, forgot to attach the drawing...
Quote from: tbird on August 26, 2006, 05:21:07 PM
as far as having different pressures, that's ok. did you realize the shuttle in the drawing is pre-pressurised at the surface, before it ever makes it's 1st run? this pressure, when the shuttle is compressed so it will desend, will be 2 atmospheres (if we are taking it to 1 atmosphere, 10meters +or-). 30psi is a little more than that, but will make the shuttle expand to size, for sure. to compress a shuttle that is based on a one cubic foot displacement, we'll say the top area is 12 inches square (144sq.in.total), would take 30psi times the 144 square inches which equals 4,320 pounds. even if the shuttle is expanded with 15(+or-)psi at the surface, 4,320 pounds will compress it to the point where the air being compressed inside will be 30psi. since we started with 15psi, the volume will be reduced by half. back to the weight. if you use a lever to hold the container that the compression water (weight) will be put in, for every foot out, less water will be needed. if the lever was 5 feet long, the weight would only need to be 1/5th (864lbs) as much. if you took it out 10 feet, 1/10th (432lbs.). now at 10ft leverage we've reduced the amount of water needed to recompress from 7.67 cubic feet (4,320lbs.) to 0.767 cubic feet (432lbs.). since we are goin down to 10 meters (+or-) and delivering water in both directions, we should have an excess of about 65 cubic feet (over 4200 lbs). not bad for a 1 cubic foot unit, eah? using the example from this drawing http://www.icestuff.com/energy/elsa/, the height of the stored energy (water), (if you store it) could be as high as 72 inches using a 1 square inch exit pipe.
do you still want to only use the up stroke?
i have more about the shuttle, if you are not burnt out.
tbird
Hi TBird,
I try to read your posting, but it sounds to me all pretty confusing.
First you said, you start with 2 Amtmospheres, which is about 2 bar and about 30 psi,
then later you say " since we started with 15psi", so I completely don?t understand your
setup.
Maybe it is already too late over here.
Maybe you can just post your setup in a table-form
so you include all parameters at start and at end points,
so we can exactly see, what you mean.
Many thanks.
Quote from: prajna on August 26, 2006, 08:09:43 PM
Hi guys,
I'm hoping to simplify things a little with the attached drawing of a shuttle. The displacement area can be full of water or air (depending on whether the shuttle is going up or going down.
Let's work with just one kilo of water and a shuttle weight of 0.5 kilos to keep things simple. We must also consider the volume of our shuttle because it will also displace that same volume of water, let's make things easy and assume it is 0.5 litres. The displacement volume, then, must be 1 litre - 1 litre to displace 1 kilo of water plus 0.5 to displace the weight of our shuttle minus 0.5 because our shuttle already displaces that much water when the displacement area is full of water. The weight and volume of our shuttle cancel each other out if they are the same. If the weight is less than the volume then it floats and if the weight is more than the volume it sinks. Ok so far?
Hmm, I don?t get it yet. Maybe I am too tired ?
Maybe you can explain it again in other words and make
a table, what should be what in your shuttle ?
Also is the blue color for water or for air ?
Is the shuttle open at the bottom ?
Should there water be able to get where into your shuttle and
where should the be air in it ?
Quote
At 10m we will require 3 litres of air in our 0.5 litre reservoir so it must be at 6 times atmospheric pressure - if my logic is correct. That should be 14.7 * 6 = 88.2psi = approx 6.2 kilos/cm2.
Hmm, if you have 0,5 Liter air at the top at sea level ( 1 bar) and
at 10 Meter you have 2 bar and 0,25 Liter volume, if you let it shrink.
But if you don?t let it shrink it still has internally 1 bar pressure and
0,5 Liter, then only the walls get the more pressure.
Hmm, maybe you can put all your parameters also in table form
and also state the start and end conditions, so we can look at the complete
cycle. And please sign your picture with titles where the water and the air should
be and if it will really be open at the bottom.
Also it would be good, if we would choose and agree on one example of dimensions
we can all work on the same and that not everbody jumps on different
dimensions and sizes so it gets not too confusing.
Many thanks.
Sorry the drawing was not clear, Steve. The blue is the air reservoir and is where the air is pressurised. Blue was probably not a good choice of color. The shuttle is hollow at the bottom - water goes in there (because the air escapes via valve V3) when the shuttle reaches the top and gets pumped out by opening valve V2 when the shuttle reaches the bottom.
I found it much easier to calculate things based on lifting 1 litre of water than working with the figures you have been using and hoped that you might find my examples easier to understand than being confused about the air expanding as the shuttle rises.
I don't know that a table will help. I hoped that by explaining the weight required to compress the air you might find it simpler to understand than getting your head around the formular you dug up.
I can certainly understand if you are tired: you guys have been hammering away at this thing like crazy. It took me ages to read all the posts and to figure out the physics of it!
hi prajna,
i've been out this evening for special event, so a bit late with this reply. i first have to say it was very refreshing to read your post. i like to keep things simple too (as you may have noticed in some of my posts). i think you have a good point about keeping things smaller and easier to understand. i don't have a problem reducing my stuff, but i may not be as effective with the metric stuff. hard to teach an old dog new tricks. i actual did follow you very well until you got to the 6 times stuff at the bottom of the run. you shouldn't need more than 2 bar at that depth. i think this is wrong and as a result, all things based on it there after would be wrong too. so far (maybe i shouldn't say this out loud) i think you have the best handle on the project. not to say others don't have good ideas, just that you are closest to the spirit of Mr. Herring.
having said that, at this point i should say something straight out. lets get away from suppling air at the bottom of the run! as far as that goes, forget using an external air compressor all together. my reason to say this is to deliver air to depth can never be as efficient as at the surface, if for no other reason than you need more air just to fill the pipe. i have worked with air compressors before and can say from experience, i'm glad i don't have to now. let's keep it simple, air compressors are not.
i probably should have said something sooner, but thought we all would have proved it a bad idea long ago. sorry for my bad judgement.
i think i should stop for now and do a quick reply for stefan before i hit the bed.
tbird
good evening stefan,
QuoteI try to read your posting, but it sounds to me all pretty confusing.
First you said, you start with 2 Amtmospheres, which is about 2 bar and about 30 psi,
then later you say " since we started with 15psi", so I completely don?t understand your
setup.
you must have missed where i clarified with "this pressure, when the shuttle is compressed so it will desend,". make sense now?
Quotethen later you say " since we started with 15psi",
the line just before sets the stage "even if the shuttle is expanded with 15(+or-)psi at the surface". the key is "expanded" here as compared to "compressed" before.
QuoteAlso it would be good, if we would choose and agree on one example of dimensions
we can all work on the same and that not everbody jumps on different
dimensions and sizes so it gets not too confusing.
great suggestion. of course i like mine best, but seems to be too hard for everyone else to follow, so i think prajna's example next best. there are probably other things we should agree on before we get too far along too. we can chat about it tomorrow.
i understand why you got a little lost, but you did the very best thing. telling me instantly so i could fix it instead of trying to pick it up as you went along. well done!!
bed time for me too. sleep tight everybody.
tbird
@prajna
okay, I will reply later,
but I just found a way it works in all cases ! ;)
TBird was totally right, when he said the shuttle must be precompressed !
That is the magic word ! ;)
Okay, let us start with 1/10 of the volume of my former example.
We just take a shuttle that is now 10 cm high and 1 meter diameter,
so it has 78,5 Liter volume and start at the top at seawaterlevel.
The shuttle is precompressed to 4 Bar !
Now to sink this shuttle we attach a weight of 80 Kg to it.
Now the shuttle together with the weight sinks down.
2. Now at 10 Meter deepth we let the shuttle expand itsself.
From the 4 Bar it had now at 10 Meter deepth there are also
2 bar pressure there due to the water pressure.
So the shuttle expands its volume now to double its size, so
to 157 Liter and being at 2 bar inside, as inside pressure = outside pressure= 2 bar.
Now 157 Liter means 157 Kg buoyant pressure minus the 80 Kg weight is
now 77 Kg or F= m x g = 77 Kg x 9,81 = 755,31 Newton upwards force.
This force now pushes the volume of 7854 Liter above it through a nozzle,
in the top of the pipe that is just at sealevel , so the 7854 Liter of water will be 1 Meter
above seawaterlevel, which will give us the mentioned 21,4 Watthours of potential
water energy.
3. Now at the top we must recompress the shuttle from now internally at 2 bar and
157 Liter volume ( 20 cm high at 1 Meter diameter)
back to 10 cm and 1 Meter diameter and 78,5 Liter volume at 4 bar.
This can be done again by using a pumping action, this time from the outside.
Now the formula for this is:
W= (P1-1 bar) x V1 x ln (V2 / V1)
So the energy W needed for this pumping is:
(200000 Pa - 100000Pa) x 0,157 m^3 x ln 0,5=-10882 Wattseconds / 3600= 3 Watthours !
So we only need 3 Watthours of energy to recompress the shuttle to 4 bar pressure and 78,5 Liter
volume and earn 21,4 Watthours by lifting the water up !
This is the real solution !
Many thanks to TBird to getting the idea with the precompression !
That is doing the real trick and we only need to do the compression at the top,
where it is much easier also.
In the deepth at 10 meters the shuttle can run against a rod, which will
switch its internal expansion control mechanism and the shuttle then expands itsself from
4 bar at 78,5 Liter to 2 bar at 157 Liter just by itsself !
So we now have the final solution and can seek now to optimize the nozzle and
still try to use other and better and more usefull dimensions !
Regards, Stefan.
Now we only have to find out, how high we can sprinkle pump the water through the
nozzle with 755,31 Newton upward force.
If we can get it higher than 1 Meter we will even have a higher COP as about 7 as calculated
above ! ;) Enjoy !
I am happy, I go to be now too ! ;) Have nice dreams !.
so let me see if i follow here.
after the shuttle reaches the top of the pipe it has saved plenty of water in a tank above that it can run a watermill so we can setup a circuit for the mechanical un/comprssion of the shuttle...the foam/spunge shuttle would get compressed at the top of the pipe this makes the shuttle "heavier" because it has been compressed into a "smaller size/volume" and so it sinks to the buttom of the pipe, once there it activates a siwtch type of thing and the foam/spunge shuttle expands itself and floats up again...and it starts all over :)
Hope you slept well guys.
@tbird,
Thank you for following my reasoning - what a relief.
Compressing 6 times comes from the fact that we will need 3 litres of air in the shuttle at a depth of 10m in order to achieve the boyancy we require. Since we will be storing those 3 litres in a tank with a volume of 0.5 litres we will have to compress it by a factor of 6 (0.5 =3/6). That is, in fact, the same as saying the pressure is approximately 2 atmospheres or 2 bar. Remember that there is a difference between absolute pressure and guage pressure. 1 atmosphere = 1 bar = 14.7psi (all approx). At sealevel we already have 1 atmosphere and at 10m (because we have to add the weight of the column of water to the weight of the column of air) we have 2 atmospheres. In my example we have 1.5 litres to displace at a pressure of 2 atmospheres so that is equivalent to 3 litres at sea level and I have suggested that we will compress all of that at sea level into a tank with 0.5 litres - thus 6 times.
I am sorry to lay metric on you but it makes everything easy to calculate since 1 litre of water is 1000cc and weighs 1kg. If we work with these quantities then we can easily multiply them by any value once the principle is established.
I am sorry that I didn't make it clear that I am also compressing the air at sea level. To pump air down would add 2 litres that we have to compress if our tube had an area of 1cm2 since the tube has a volume of 1 litre and it would displace 1 litre of water (assuming an ideal tube with walls of zero thickness). Actually we could get around that by having the compression pump at the bottom rather than at the top but let's keep it simple in engineering terms.
@Steve
Using my units and description we don't have to worry about gravity or newtons or pascals or watthours or W= (P1-1 bar) x V1 x ln (V2 / V1) or adding 80 kilos etc. It is easy to get confused with all of that. If you want to later calculate the potential energy in x volume of water at y height above sea level then you can do that but it is not necessary in order to determine whether or not the system will work. If, after your refreshing sleep, you forget the calculations you have done and work through my model then you should find it simple enough. If it is simple then it should be easy enough to spot any mistakes I have made.
@FreeEnergy
Yes, more or less
Here is a new diagram. I hope it helps.
Picture 1: the shuttle is at the top and we pressurise the reservoir using valve 1.
Picture 2: the shuttle is at the top and we let the air out of the displacement area by using valve 3. Now the shuttle descends.
Picture 3: the shuttle is at the bottom and we force the water out (displace it) by opening valve 2. Now the shuttle ascends in picture 4.
The volume of the shuttle including the reservoir is 1.5 litres: 1 litre in the displacement area and 0.5 litres in the reservoir. At 10m the pressure is 2 bar so the shuttle will contain 3 litres of air in the space of 1.5 litres. When the reservoir is fully pressurised it contains 3 litres in a volume of 0.5 litres and is therefore at 6 bar. When the shuttle is at the top the reservoir still contains 1 litre of air in a volume of 0.5 litres so it is at 2 bar.
Looking at this cycle we need to figure out two things: how much water will the shuttle lift and how much energy will be needed to recompress the reservoir. These I have calculated in my earlier message. I think that model is as simple as we can get.
Hi all.
Was out this weekend and am now surprised about the good ideas.
prajna, I like your shuttle design, but the same thing would happen as it happens with my baloon idea. At 10m depth, with your 6bar compressed reservoir you are able to fill the 1liter displacement area. How do you handle the air in the displacement area become double of voume while going upward and the pressure decreases (actually, you filled 3 liters air in your reservoir)?. Thats why I went away from the baloon idea.
stefan, your new calculations are allright, I think thats the way we should keep moving. A closed shuttle, able to have 2 different volume states is the solution and I think Mr. Herring had the same answer.
Andi
ooandioo, yes, as the shuttle rises the volume of air increases and fills the pipe below the shuttle. Because the shuttle is in a tube it simply increases the boyancy of the shuttle (since the air is now displacing more water than it did at 10m.) No problem. Indeed it is somewhat of a benefit since it increases the speed that the shuttle rises. There is always at least 1 litre of air in the reservoir. When the shuttle is rising there are always 2 litres of air in the displacement area/below the shuttle. At 10m those 2 litres take up a volume of 1 litre at 2 bar whilst at the surface those 2 litres take up a volume of 2 litres (1 litre in the displacement area and 1 litre below it) at 1 bar. No need for a closed shuttle. If it worries you that air might escape up the sides of the shuttle then just change the volume of the displacement area to 2 litres; nothing else changes except that at the bottom the air only fills half the displacement area. The pipe that the shuttle goes up is effectively a 10m displacement area with a piston on top.
Let's continue with the open bottomed shuttle because otherwise people will be worried about the mechanics of a two state cylinder rather than the considerably more fundamental problem of whether the sums add up. As I said in my previous post, I think that this model is conceptually as simple as we can get and that makes analysing the problem easier.
The only problems are 1. How much water can we pump up to what height above water level, and 2. How much energy does it take to compress 2 litres of air to a pressure of 2 bar. I think that I have answered both of those questions in my initial post but you might like to check my results.
Pranja, I am at the PDA right now and can?t yet see your graphics, will comment later to it, when I am back at the PC.
What we really need to find out now is, how high the water will jump out the nozzle ! This is now extremly important. I have not found yet a formula that states that...
This is the most important factor now for knowing the COP.
hi all,
lots of comments since i went to bed last night. looks like we might be getting somewhere, but everyone had problems with their arrangements.
first was stefan, then came prajna with basicly the same issue. the pressure value at 10m. not sure what caused this unless you forgot the weight still has displacement value that was not considered. it may not be a positive weight, but it is displacement.
Mr. Herring in most of his examples (we all should read his stuff close) makes the shuttle weight to displace 1 and a half time what it displaces. that inculdes the area for the compressed air. now all he has to do is increase the volume of the total shuttle by 2 times. at 10m the gage pressure (you might use absolute if working with refrigration, but not much else uses it) will read 1 bar (14.7 +or-). so if we have 1 bar at 10m in the shuttle, our shuttle will have the same shape as it has at the surface without any gage pressure and be netural buoyant. to make it rise, we only need a little extra expansion to make it positively buoyant. so you see, we don't need 6 or even 4 bar at that depth. we only need 1 plus the amount required to give our shuttle the strength (positive pressure) to raise the water above water level the height we want or need to do work. the 2 things that determine this height is the pressure and diameter of exit tube.
so seeing this mistake in most of the previous posts, anything figured afterwards, based on those numbers, will be wrong. this brings me to my comment last night "there are probably other things we should agree on before we get too far along too". if you make a mistake early on in your calcs, you carry that mistake the rest of the way (lots of wasted time). if we keep our post to address 1 thing at a time, they will be shorter and easier to correct.
how 'bout it, can we do that?
Hi Pranja, looks good your new Graphics. Maybe you can draw the Weight of the shuttle onto it additionally, so it gets more clear, why it sinks at all. Also maybe you can draw it this way, that you show 4 single pics which show the shuttle in each deepth position. Did you calculate yet, with how much air you have to refill your shuttle at the top and how much energy this costs ?
Thanks.
Stefan, I suggest you forget the nozzle idea or you will get caught up in static head, nozzle head, viscosity, flow rate etc. Let's keep this simple and consider a tube extending above water level. I seriously doubt that a nozzle is more efficient at raising water than a simple tube is (except that we don't need to worry about viscosity which is a very small component of the force required.)
As tbird suggested, for any given diameter of tube the pressure doubles as the diameter halves (roughly). If the pressure required to lift 1 litre of water to 1m above water level is, say, 1 pascal* in a tube of 10cm2 cross sectional area then that same pressure will lift that same litre of water 2m above water level in a tube with a 5cm2 cross sesction.
* this figure is simply for illustrative purposes. Don't use it in calculations; I haven't calculated what pressure it would actually take in this example.
hi stefan,
can't resist answering this now. i thought when you said you agreed with me about the size of the exit pipe, you had a handle on it. obviously you don't. maybe you've been looking in the wrong place for the answer. it's a hydraulic thing. for your formula, look for the relationship between vol and pressure and flow. to have the same flow if you change pipe diameter (area) you have to increase flow (at same pressure) to produce same volume delivered in same time. remember we are now talking about liquid, not gas.
don't know if that helped, but should get you down the right path.
tbird
@tbird
The reason I use absolute pressure in my calculations is that it keeps eveything simple and intuitive. At sea level there is a pressure of 14.7psi - equivalent of 1 bar - at 10m there is twice that.
That is not where your confusion arrises though. The 6 bar, or 6 times atmospherical pressure, is required to squeeze 3 litres of air into the reservoir tank that only has a capacity of 0.5 litres. To do that we must compress it to one sixth of it's volume or, to say it another way, six times its normal pressure. Does that clear that up?
Pranja and Tbird you are right, that an exit smaller pipe would be better, but I still have problems with the hydrostatic paradoxon, as this water colums above seawaterlevel is negatively influencing the buoyant force and for this reason the shuttle must be much bigger under water. So the best thing would be to have a nozzle sprinkling out the water into the upper basin.
Pranja, wouldn?t it be easier to calculate with your air reservoir in your shuttle to be not 0,5 but 1 liter ?
Stefan,
Yes I did calculate the force required in posting http://www.overunity.com/index.php/topic,570.msg11325.html#msg11325.
Again, neither tbird nor I understand what you are referring to when you say hydrostatic paradoxon do you mean 'paradox' as in http://scubageek.com/articles/wwwparad.html?
It would probably be easier to calculate things with a reservoir capacity of 1 litre. Shall I redo the diagrams and all the calculations/explainations again?
If we go with an 1 meter high exit tube instead of a nozzle at sealevel, no matter how small in diameter the exit tube is, this 1 meter height will need a much bigger volume shuttle at 10 meters down there, so this COP would be under 1 then and no overunity anymore.
Tbird and Pranja, the hydrostatic paradoxon puts a heavy burden onto our design here. To make it clear:
as in my example the 1meter height exit tube above sealevel puts a 785,4 kg weight onto the main 1 meter diameter water column, also if in the exit tube would only be 10 liter of water=10 kg ! Thus the shuttle bouyant force must be able to lift this additional 785,4 kg and not just the real 10 kg weight of the water inside the small diameter exit pipe ! think about it !
hi prajna,
stefan is hopeless when he's on his pda. your link to the 'paradox' would have cleared this up easily if he had read it.
stefan when you're on the pda, for me, i will count you out of the loop.
prajna, how many people do you know that uses an absolute gage? if we write everything factoring this in, we lose our simplicity. so for the sake of simplicity, let's keep the pressure of water at 10m at 1 bar. ok with you?
Quotehow 'bout it, can we do that?
haven't seen any responce to this question yet. anybody want to step up?
tbird
How do you reason that, Stefan?
Think of it this way: You can store 1 litre of water in a container that has an area of 1m2 and a height of just 1mm. Or you can store that same litre of water in a container that is 10cm by 10cm by 10cm, or, indeed, in one that is 5cm by 5cm by 20cm, or even one that is 1cm by 1cm by 1m.
Aah, your previous post has just arrived. Yes, we have to account for the weight of water in the tube as a weight that our shuttle has to lift.
In my example we want to lift 1 litre of water a height of, say, 1m. In our exit tube - let's say it has an area of 1cm2 - we have 1 litre (i.e. 1 kilo) of water. Now, if our shuttle has a positive boyancy of just 1.1 kilos then it will push that 1 gram (1cc) of water out the top of the 1m pipe and it will continue to do that, with an increasing rate of flow, as the shuttle rises (because the boyancy is increasing as it rises up the pipe). The total amount of water pumped is the same whether we pump it (slowly) out of a narrow tube 1m tall or quickly out of a wider tube 10cm tall.
If you are working with your figures you can still discount the weight that is causing you so much trouble by using a thin tube from water level up to where you want the water. You come adrift when you use the same diameter exit tube as the pumping tube; then, sure, you will have a lot of extra weight to lift.
Quote from: prajna on August 27, 2006, 10:28:34 AM
Again, neither tbird nor I understand what you are referring to when you say hydrostatic paradoxon do you mean 'paradox' as in http://scubageek.com/articles/wwwparad.html?
Yes, that?s it,
also have a look at this:
http://de.wikipedia.org/wiki/Hydrostatisches_Paradoxon
There you can also see the forces in the lower diagramm.
The main problem is, as the main pipe is connected to the smaller
diameter exit pipe, the hydrostatic pressure ontop the water column
is the same if you have only 10 Liters of water or 785,4 Liters of water
inside the exit pipe , the weight the water column see is always
785,4 Liters = 785,4 kg !
So the shuttle volume must be very big at 10 Meter deepth to overcome
this additional 785,4 kg and this makes the whole concept unvalid !
So only a nozzle at seawaterlevel will work in my opinion,
which sprinkles the water to an upper reservoir and avoiding
the negative impact of the hydrostatic pressure paradoxon !
http://de.wikipedia.org/wiki/Hydrostatisches_Paradoxon
Have a look at this and also watch the Quicktime movie !
Understanding the hydrostatic pressure paradoxon is a MUST before
you understand the gravity mill !
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/HydroStatic.html
Have a look at this:
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fwww.ac.wwu.edu%2F%7Evawter%2FPhysicsNet%2FTopics%2FPressure%2FPressureGifs%2FPressure08.gif&hash=5c7d3802f5fc73cfe5bba9cd9cd04484fc5fd135)
Case C is the case for the connection of our pipe into the smaller exit pipe.
The water column would see as the weight the whole big diameter of water, although there
is just this small diameter water column ontop of it !
So instead seeing the real 10 Liters= 10 Kg of water the main
pipe sees 785,4 Kg of weight, which must be additionaly lifted and which
puts a hard burden onto the volume size of our shuttle and energy
requirement !
hi stefan,
looks like you are back online again, so i'll try again.
as per your "hydrostatic pressure paradoxon" pressure at any given point is the same in all directions. so, the water has as much pressure on the bottom of the shuttle as is on top, but only to the water level. the only extra pressure is the amount of water above water level. now if your shuttle has a positive pressure created by displacing more water weight than it weighs itself, it will rise until the weigh of the water (contained in a tube) above weighs as much as the extra displacement. then it will be netural (static). like the drawings show, it can be short and fat or tall and skinny, as long as the weight is not more than the shuttle pressure can handle.
tbird
Sorry Stefan, you are confused about this. The large diameter pipe is below sea level and we do not need to consider the weight of water in that pipe we need only consider the weight of water in the small pipe that is above sealevel. Whatever volume of water that is above the sealevel we must consider as being the weight of water that our shuttle must lift.
You even did the experiment to see this effect and the whole of our effort is based on the principle that we need only account for any water we lift above the water level.
I think the confusion for you arrises in that the pressure from the water in a column 1m high will exert the same pressure per area whether the area is large or small. So a column of water 1cm by 1cm by 1m will exert a force of 1 kilo per cm2 and will weigh 1 kilo. A column of water 10cm by 10cm by 1m will also exert a force of 1 kilo per cm2 but will weigh 10 kilos. The boyancy required to lift the water in the second tube is 10 times that required to lift the water in the narower tube.
I can confirm with tbird. Never mind you are taking a 1m high, 10cm diameter pipe above sealevel or a 2m high, 5cm diameter pipe - the pressure on the shuttle beyond is is always the same.
I still think, using a pipe for transporting water in the reservoir bassin or a sprinkling noozle will have the same results, as the pressure is also the same.
hi all,
another note to remember is reguardless of the shape or size of the shuttle, the pressure we get is only the difference between what the shuttle weighs and the weight of the water it displaces. example: shuttle weights 1, 125 kg (pretty big, right?) and it displaces 1,126kg, then the positive force (pressure) is 1 kg. divided over whatever surface area the shuttle has.
i think it is easy to lose track of this fact.
tbird
ps still don't see my question addressed
Quote from: prajna on August 26, 2006, 08:09:43 PM
10m we will require 3 litres of air in our 0.5 litre reservoir so it must be at 6 times atmospheric pressure - if my logic is correct. That should be 14.7 * 6 = 88.2psi = approx 6.2 kilos/cm2.
Okay, so far.
Quote
We can discount 1 litre of air when it comes to recompression of the reservoir because we still have that from the previous cycle. This leaves 2 litres to recompress = 2000cc. If we have a plunger with a cross section of 1 cm2 and a length of 20m then a weight of 6.2 kilos will compress the air in the cylinder to that pressure. If we only have 0.5m to compress in then we will have a cross section size of 40cm2 and require a weight of 248 kilos.
Okay, that would need
W= Force x distance= 6,2 Kilo x 9,81 x 20 Meter= 0,3379 Watthours
Quote
The quantity of water we can pump depends on the area of the top of our shuttle. If it is 20cm2 then 1 litre will be 50cm high in the pipe and we can therefore pump 20 litres 0.5m above the water level. Hmmm....
Is my maths wrong or did I miss something or do we have less energy than we thought?
No, your math is not wrong, it is the 0.5 Meter exit pipe height, that destroys your design.
You only get as output
W= m x g x h = 20 Kg x 9,81 x 0,5 Meter= 98,1 Wattseconds/3600 = 0,02725 Watthours, so
you are very much underunity !
You must use a nozzle at seawaterlevel and then you can redesign your
shuttle to be in a different height ! the height should be much smaller then !
Also it could be, that just with 1 Liter it does not work yet !
Hope this helps.
Please all,also if you repeat yourself, please post always all the details
in ONE posting, so to better see your design.
Don?t leave things out, just copy and paste from old postings.
This way it is much easier to see the whole design and one does not need
to go back pages and pages...
Thanks.
Regards, Stefan.
hi andi,
QuoteI still think, using a pipe for transporting water in the reservoir bassin or a sprinkling noozle will have the same results, as the pressure is also the same.
the advantage of the pipe is you can control where the water goes. if you use leverage for recompression, it would be hard to spray it there accurately. it's cheap too.
tbird
stefan,
as long as you don't agree with the height issue, you are no use to this link. either get with the program or leave.
tbird
Quote from: tbird on August 27, 2006, 12:15:44 PM
hi andi,
QuoteI still think, using a pipe for transporting water in the reservoir bassin or a sprinkling noozle will have the same results, as the pressure is also the same.
the advantage of the pipe is you can control where the water goes. if you use leverage for recompression, it would be hard to spray it there accurately. it's cheap too.
tbird
Thats what I'm saying.
Quote from: tbird on August 27, 2006, 12:19:34 PM
stefan,
as long as you don't agree with the height issue, you are no use to this link. either get with the program or leave.
tbird
Please don't shout that way! We are still brainstorming - every idea and criticism is a good one.
Quote from: prajna on August 27, 2006, 12:00:20 PM
Sorry Stefan, you are confused about this. The large diameter pipe is below sea level and we do not need to consider the weight of water in that pipe we need only consider the weight of water in the small pipe that is above sealevel. Whatever volume of water that is above the sealevel we must consider as being the weight of water that our shuttle must lift.
You even did the experiment to see this effect and the whole of our effort is based on the principle that we need only account for any water we lift above the water level.
I think the confusion for you arrises in that the pressure from the water in a column 1m high will exert the same pressure per area whether the area is large or small. So a column of water 1cm by 1cm by 1m will exert a force of 1 kilo per cm2 and will weigh 1 kilo. A column of water 10cm by 10cm by 1m will also exert a force of 1 kilo per cm2 but will weigh 10 kilos. The boyancy required to lift the water in the second tube is 10 times that required to lift the water in the narower tube.
Hmm, you could be right, that I am confused because of the hydrostatic paradoxon,
but let me ask it this way:
If the shuttle is 1 Meter under water and has 100 Liter = 100 Kg still above it in the main pipe and
in the 1 Meter high exit pipe with much smaller diameter are 10 Liters= 10 Kg ,
so what does the shuttle see on its surface for a weight ?
is it just 110 Kg or is it 200 Kg ?
If it is 200 Kg, then we have the hydrostatic paradoxon at work.
Please let me know your view.
Thanks.
Sorry, I had a typo in my last post, I just corrected it,
I said: 100 Liter = 100 Kg in the main pipe and 10 kg= 10 Liter in the exit pipe.
Would the shuttle see 110 Kg or 200 Kg ontop of it ?
As both pipes are connected the hydrostatic paradoxon would suggest the shuttles
sees 200 Kg instead of the real 110 Kg, right ?
Yes ooandioo and tbird. Both spot on. In fact it gets even more fun since all we have to consider is the volume of water above sealevel. To pump up 1m in a 1cm2 tube we have to lift 1 kilo. That is in order to pump any water out of the top of that tube. So we know the boyancy required for our shuttle. Next we decide how much water we want to pump to that height and that will determine the cross sectional area of our shuttle; the greater the area the greater volume we pump on each cycle. If we want to pump faster over that distance then we have to increase the boyancy and the cross section of the top tube.
@ Stefan
Thanks. I think we are beginning to get a good grip on this system.
Reference your last: the shuttle will see 10kg.
I will spend some time remodelling and then post some new diagrams and a clearer explaination. It might not be too difficult to create a little html applet that you can just plonk in the values you want for each of the parameters and it will calculate the rest. Maybe I will do that.
Quote from: ooandioo on August 27, 2006, 12:21:51 PM
Quote from: tbird on August 27, 2006, 12:19:34 PM
stefan,
as long as you don't agree with the height issue, you are no use to this link. either get with the program or leave.
tbird
Please don't shout that way! We are still brainstorming - every idea and criticism is a good one.
yes, TBird, please we are just brainstorming here and please copy and past all
your ideas to a new posting with all the numbers and energy calculations in there and present it in ONE
posting, so we can see,if your design works. Don?t spread it over several posts, as it gets
hard to follow.
We also need to post some moree drawings.
I will try to paint a few drawings and scan them in, so we can have a better and easier
view for this, but today I am pretty busy with other things, so also tommorow and the
next days I will not have much time towork on it...sorry.
Quote from: prajna on August 27, 2006, 12:34:44 PM
@ Stefan
Thanks. I think we are beginning to get a good grip on this system.
Reference your last: the shuttle will see 10kg.
You mean 110 Kg ? Right ?
Okay, now I see it also, as the surrounding water around the main pipe is also up to sealevel,
the weight onto the upper shuttle surface in 1 Meter deepth is 110 Kg.
Quote
I will spend some time remodelling and then post some new diagrams and a clearer explaination. It might not be too difficult to create a little html applet that you can just plonk in the values you want for each of the parameters and it will calculate the rest. Maybe I will do that.
That would be nice, if you could do that.
Many thanks !
andi.
QuotePlease don't shout that way! We are still brainstorming - every idea and criticism is a good one
sorry about that
andi
QuoteThats what I'm saying.
glad we agree.
as i see it, there are 2 basic parts to Mr. Herrings design, the energy source and the use of the gererated energy. at this point we are looking only at the energy source, of which there are 2 major componets, the shuttle and the recompression device. since we have already put so much energy into the shuttle, i guess we should finalize that first.
in order to do that, let's agree on what size would be best to discuss on this thread. i like prajna's suggestion except for the depth of the tube. if we expect others to be interested enough to build a unit, it has to be doable. i think more people would be apt to build one if it were only like 10 feet instead of 10 meters. after all, if you can build one this size, you know enough to build one to supply whatever energy you want.
tbird
No Stefan.
QuoteI said: 100 Liter = 100 Kg in the main pipe and 10 kg= 10 Liter in the exit pipe.
Would the shuttle see 110 Kg or 200 Kg ontop of it ?
The shuttle will only see the 10 litres in the exit pipe.
@ tbird
Let's calcuate everything based on 10m depth to begin with and then when the model is validated we can change whatever we like.
To lift 1kg up 1m we have to have a displacement size of 1 litre. So long as the reservoir volume is the same as its weight then it will be boyancy neutral and doesn't need to affect our calculations (it wouldn't sink but we are adding 2 litres of compressed air to it so that will make it do so.)
I'll work on a calculator applet for these things so that you can put in any figures you like. Ok?
Okay, looking forward to your applet.
I could integrate it here into a posting, if you would send me the HTML code
via email.
Me as the admin has rights to include HTML code into the postings.
Anyway, yes , we have to find dimensions of this gravity mill,
so that we can build it also in smaller space and smaller height.
Maybe we can find parameters, where one only needs a height of 1 Meter for the main
pipe and water tank and the exit pipe is another 1 Meter, so 2 Meter height will be okay ?
Also it would be good, if one could use something like these water column bubblers:
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=4415427510 (http://partners.webmasterplan.com/click.asp?ref=284148&site=1382&subid=&type=text&tnb=32&diurl=http%3A%2F%2Fadfarm.mediaplex.com%2Fad%2Fck%2F707-3922-3266-17%3Fmpro%3Dhttp%253A%252F%252Fcgi.ebay.com%252Fws%252FeBayISAPI.dll%253FViewItem%2526item%253D4415427510)
If we can modify one of these to build such a gravity mill, it would be very easy to get these
water columns.
hi guys,
i had to take a breather. my breath of fresh air came from page 6&7 of Mr. Herrings drawings.
http://www.icestuff.com/energy/elsa/
boyle's law is what you need to study to figure out the correct pressure. see here
http://www.grc.nasa.gov/WWW/K-12/airplane/boyle.html
i attached a paint drawing (not very good) to show how a shuttle design could start (more details later). the black part at the bottom and the rest of the container can weigh 1.5 times as much as it displaces. not sure what material would work, but lots of stuff heavier than water, but don't know if anything weighes that much in that little volume. this might be a slight flaw with Mr. Herrings example. the worse result is a higher compression like you guys were trying to use or less buoyancy for a given displacement. the shuttle on the left is in the compressed mode with 2 bar (30psi+or-). that's the white part inside the shuttle. since pressure and volume are an exact ratio, if we reduce the pressure by half, the volume will double. the picture to the right shows this state. not sure if you can tell from the drawing but that is suppose to be twice what it was in the compressed state. since the outside water pressure is the same at 1 bar, the shuttle would not expand any more at this depth. if the retainers at the top of the shuttle walls were not there and the walls extended up, as the shuttle went up, the air inside would expand the volume further. but it is there so when the shuttle reaches the surface, there will still be 1 bar (15psi+or-) inside.
i modified the drawing after i attached, so not sure which will show up. if the wrong one comes, i quickly send the other.
well that didn't work at all. stefan what am i doing wrong? changed to gif and still doesn't like it.
hi all,
just emailed it to stefan. maybe he can post it.
tbird
tbird, if you double the volume then you also double the amount of water you displace and increase the bouyancy so must increase the weight to match. If you increase the pressure then you can use less weight and less volume.
Still working on the calculations.
hi prajna,
Quotetbird, if you double the volume then you also double the amount of water you displace and increase the bouyancy so must increase the weight to match. If you increase the pressure then you can use less weight and less volume.
not sure what you are driving at. maybe i confused you and you think i was talking bout the shuttle being at the top. when i spoke of doubling the volume, i was talking about when it was at the bottom. this stops it going down and makes it go up. sorry for the confusion.
tbird
hi all,
i was trying to find out how much 1 cubic foot of lead weighs. couldn't find anything i really understood, but did find the atomic weight of lead and water. lead= 207.2 water=18 does that mean for the same volume lead would be 11.5 times heavier?
anybody know if that is what it means? if so, i should check other things more suitable for building the shuttle.
tbird
Just got a mail from the inventor Mr. Herring. He stated that he visited the overunity.com forum and asked for our questions. I replied about the forums intention and asked if he was able to build a working device and if he has some photos or videos. I also invited him to join us here.
Andi.
Thanks andi,
I can't wait to see some photos from Mr. Herring.
Good job!!!!!!!!!
Hookay guys,
I have knocked up a basic calculator which you can test drive at http://declarepeace.org.uk/energy/elsa.htm (http://declarepeace.org.uk/energy/elsa.htm). It is not finished yet but you can try it out. I am getting a tad tired to do more work on it tonight. I will develop it into a proper page with diagrams and explainations later but those who have been active in this discussion should have no problem using it. Try changing a few values and let me know what you think.
Oh... the javascript doesn't work in firefox, only IE at the moment. I will sort that out too when I get a chance.
hi prajna,
that is a REALLY cool tool!! your mom would be proud of you!
i haven't converted the numbers to what i'm familiar with yet, but at first glance the weight of the shuttle seems a bit high. one thing that did impress me was how simply you have made things. EXCELLENT JOB!!!!
tbird
Pranja,
great javascript, but there the shuttle volume is 10 times too big.
Also you can also introduce the stored potential energy
in the upper reservoir via
Energy= watermass in kg x 9,81 x height in meters= Energy in Wattseconds or Joule
now if you divide it / 3600 you get energy in Watthours.
Also please include the required pumping energy which is
Energy for recompression in Wattseconds =
(P1 in Pascal - 100000 Pascal atmospherical pressure) x V1 in qubic meters x ln (Volume2 / Volume1)
If you recompress from 2times the volume back to 1time the volume, we get
ln (1/2)
"100000 Pascal atmospherical pressure" = 1 bar must be subtracted from the Pressure
P1, cause we are doing it not in a vaccuum, but at sealevel with 1 bar outdoors pressure.
also for the volume
1000 cm^3 = 1 m^3 = 1 meter x 1 meter x 1 meter = 1 qubic meter
Hope this helps.
Looking forward for the complete Javascript.
Many thanks.
hi prajna,
me again. been playing with your neat program. i have a thought or 2. the first thing you do is "head of water in the feed tube". this weight is nice to know, but i think just as (if not more) important would be to know what the max diameter of that pipe could be for a given height. and maybe if given a pipe diamter, what max height could be.
shortly after you get to the shuttle. it's a bit unclear if you are referring to expanded or compressed state. don't know if you read my previous post
Quotei attached a paint drawing (not very good) to show how a shuttle design could start (more details later). the black part at the bottom and the rest of the container can weigh 1.5 times as much as it displaces. not sure what material would work, but lots of stuff heavier than water, but don't know if anything weighes that much in that little volume. this might be a slight flaw with Mr. Herrings example. the worse result is a higher compression like you guys were trying to use or less buoyancy for a given displacement. the shuttle on the left is in the compressed mode with 2 bar (30psi+or-). that's the white part inside the shuttle. since pressure and volume are an exact ratio, if we reduce the pressure by half, the volume will double. the picture to the right shows this state. not sure if you can tell from the drawing but that is suppose to be twice what it was in the compressed state. since the outside water pressure is the same at 1 bar, the shuttle would not expand any more at this depth. if the retainers at the top of the shuttle walls were not there and the walls extended up, as the shuttle went up, the air inside would expand the volume further. but it is there so when the shuttle reaches the surface, there will still be 1 bar (15psi+or-) inside.
Quotei was trying to find out how much 1 cubic foot of lead weighs. couldn't find anything i really understood, but did find the atomic weight of lead and water. lead= 207.2 water=18 does that mean for the same volume lead would be 11.5 times heavier?
and you may need to wait to see the drawing, but if we made lead (if i really did figure it right) the standard (least volume for weight, common) to use for weight, then we could get a better picture of the shuttle. Mr. Herring's drawings would lead you to beleave (as i did for awhile) the paper thin housing could provide this weight. not quite true to life. don't know if i put this part right. hope you get my drift.
from your program "We know that a depth of 10m will compress the air 2 times its volume at the surface.
height / 10 + 1". not sure if this is worded right.
that's it, just some thoughts.
i have to say again GREAT JOB!!!
tbird
QuoteAlso please include the required pumping energy which is
Energy for recompression in Wattseconds =
(P1 in Pascal - 100000 Pascal atmospherical pressure) x V1 in qubic meters x ln (Volume2 / Volume1)
oh yes, good idea. i'd prefer knowing the weight needed to recompress (helps those who use leverage to do this).
tbird
QuoteIf you recompress from 2times the volume back to 1time the volume, we get
ln (1/2)
stefan what does the "ln " mean here?
tbird
Here is the picture from TBird attached.
Many thanks.
ln means "Logarithmus naturalis" and is the inverted function of
e^x
So if you want to to know x then you would
do ln (e^x) = x
Look on the windows calculator program, it is there under
scientific setup.
Hi Pranja,
if you will work on the javascript,
maybe you can include also the weights in pounds and
the height in inches and the pressure in psi,
so our US friends, who don?t use the metric system
can also easier see their unit results.
Many thanks for this great work !
Regards, Stefan.
hi stefan,
thanks for posting my picture. any idea why it wouldn't work for me? and thanks for the explantion (still don't know what all that means). if i ever need to use it, i'll call you. :D
Quotei was trying to find out how much 1 cubic foot of lead weighs. couldn't find anything i really understood, but did find the atomic weight of lead and water. lead= 207.2 water=18 does that mean for the same volume lead would be 11.5 times heavier
you are pretty good with this stuff. am i right? if so, then that bottom space would need to be about 15% of the volume. if that's the case, the compression would need to be a little over 2 bar at 10m so it would expand more than twice the size. hmmmm....
tbird
Quote from: hartiberlin on August 27, 2006, 12:23:58 AM
@prajna
okay, I will reply later,
but I just found a way it works in all cases ! ;)
TBird was totally right, when he said the shuttle must be precompressed !
That is the magic word ! ;)
Okay, let us start with 1/10 of the volume of my former example.
We just take a shuttle that is now 10 cm high and 1 meter diameter,
so it has 78,5 Liter volume and start at the top at seawaterlevel.
The shuttle is precompressed to 4 Bar !
Now to sink this shuttle we attach a weight of 80 Kg to it.
Now the shuttle together with the weight sinks down.
2. Now at 10 Meter deepth we let the shuttle expand itsself.
From the 4 Bar it had now at 10 Meter deepth there are also
2 bar pressure there due to the water pressure.
So the shuttle expands its volume now to double its size, so
to 157 Liter and being at 2 bar inside, as inside pressure = outside pressure= 2 bar.
Now 157 Liter means 157 Kg buoyant pressure minus the 80 Kg weight is
now 77 Kg or F= m x g = 77 Kg x 9,81 = 755,31 Newton upwards force.
This force now pushes the volume of 7854 Liter above it through a nozzle,
in the top of the pipe that is just at sealevel , so the 7854 Liter of water will be 1 Meter
above seawaterlevel, which will give us the mentioned 21,4 Watthours of potential
water energy.
3. Now at the top we must recompress the shuttle from now internally at 2 bar and
157 Liter volume ( 20 cm high at 1 Meter diameter)
back to 10 cm and 1 Meter diameter and 78,5 Liter volume at 4 bar.
This can be done again by using a pumping action, this time from the outside.
Now the formula for this is:
W= (P1-1 bar) x V1 x ln (V2 / V1)
So the energy W needed for this pumping is:
(200000 Pa - 100000Pa) x 0,157 m^3 x ln 0,5=-10882 Wattseconds / 3600= 3 Watthours !
So we only need 3 Watthours of energy to recompress the shuttle to 4 bar pressure and 78,5 Liter
volume and earn 21,4 Watthours by lifting the water up !
This is the real solution !
Many thanks to TBird to getting the idea with the precompression !
That is doing the real trick and we only need to do the compression at the top,
where it is much easier also.
In the deepth at 10 meters the shuttle can run against a rod, which will
switch its internal expansion control mechanism and the shuttle then expands itsself from
4 bar at 78,5 Liter to 2 bar at 157 Liter just by itsself !
So we now have the final solution and can seek now to optimize the nozzle and
still try to use other and better and more usefull dimensions !
Regards, Stefan.
Hi All,
I recalculated this now for to see, if also a mechanical lever action can recompress
this shuttle from 2 bar back to 4 bar ( 400 000 Pascal) pressure and thus shrink the
volume again from 20 cm to 10 cm, so difference 10 cm= 0,1 Meter.
We can use then an external pressure of 4 bar to do this.
As we can also do it via a lever we must see, what force
4 bar means.
This can be calculated this way:
P= F / A ( means: Pressure= Force / Area )
thus
F= P x A ( means: Force= Pressure x Area)
Energy = F x s = P x A x s ( means Energy= Pressure x Area x distance)
So:
400 000 Pa x 0,785398 m^2 x 0,1 meter =8,7 Watthours
This 8.7 Watthours is bigger than the upper only 3 Watthours,
but this is logical, cause we apply already at the start 4 bar
where the shuttle only has 2 bar.
So at the start we could only apply 2,1 bar and so on rising,
until the 4 bar are reached, which would save us some energy
and if you do it with a syringe compression via air pressure,
you would only need the above 3 Watthours, so
doing it with a lever-torquearm will waste some energy.
But anyway, it is still overunity ! ;)
COP = 21,4 / 8,7 =2,46
Enjoy !
Regards. Stefan.
Quote from: tbird on August 28, 2006, 12:07:38 AM
hi stefan,
thanks for posting my picture. any idea why it wouldn't work for me? and thanks for the explantion (still don't know what all that means). if i ever need to use it, i'll call you. :D
You must click the Reply button,
do not use the Quick-Reply ,
then you can also attach a picture file.
Quotei was trying to find out how much 1 cubic foot of lead weighs. couldn't find anything i really understood, but did find the atomic weight of lead and water. lead= 207.2 water=18 does that mean for the same volume lead would be 11.5 times heavier
Yes, 11.5 times heavier ! That?s a lot ! Now you see, why they use it for fishing
as a weight...
Did somebody else already try, how high
the water can be pressed through a small diameter
exit pipe ?
In my experiment, where I only had a pretty slippery
shuttle, the pressure from the buoyant force was not very big
and thus it did not go very high in a smaller pipe at the top....
I wonder, if the shuttle must not be this high itsself
as high as the exit pipe will be ?
Maybe I just need a tighter fitted shuttle to see how high I can bring
up the water in the exit pipe and I also have to get a much longer main
pipe to play with it....
TBird, what did you see in your experiments ?
Did you try already a smaller exit pipe ontop the main pipe ?
Regards, Stefan.
Hi all.
No reply from Mr. Herring yet.
prajna, very good and clearly laid out job. I think we now should discuss the way the shuttle should look like. As stefan allready mentioned, compressing the shuttle via leverage will cost more energy then via air pressure - i agree with that.
Perhaps we should come back to pranjas design.
Andi.
Hello all
Yesterday I tried the same experiment as you Stefan.
This time with a higher tube (1m High) from the boxing of a Barbiefamily from my daughter and a plastic Cokebottle as a swimmer. I tried both with seal and without, even with valve like function on the upper and lower side of the tube. Tomorrow I will post a photo from my setup. In all cases there was no water lifted up! The water level was always the same. Nor a mm difference.
So 2day I was reviewing your video and saw the difference between yours and my experiment.
I attached a drawing and I hope you all will understand what I?m talking about.
Picture No. 3 is the most important one. It shows that your swimmer body lifted also the tube up, perhaps caused by fiction of the water between the tube and the swimerbody, and that your swimmerbody didn?t reach his whole level as in the startposition.
I assume that the weight of water you "lifted" plus the weight of the tube is equal to bouyancy force of the swimmerbody. So in that moment the setup is in balance. But now we have to understand why the water stands there "lifted". Will it stands there forever?? I think that it won?t!
Stefan - please try again your experiment and wait a few minutes after the water is lifted. As result I expect that it will flow out between the tube and the body, as long you have no seal there. Or that the tube will slide slowly down and let the water flow out at the upper border of the tube.
If I do my homeworks right noone reported that the tube in the E.L.S.A.setup is a moving part. In the most graphics it is a fixed part, or am I wrong? Otherwise the shuttle have to lift up the hole watercolum and tube against gravity.
This is very important for your experiment Stefan, because when you fix your tube in anykind on the ground, you will not see the same result. Please try again and you will see I?m right.
by
2Tiger
http://www.grc.nasa.gov/WWW/K-12/WindTunnel/Activities/buoy_Archimedes.html
Congratulation Stefan!
Now I am sure that the E.L.S.A. device won?t work as you assume at the beginning of this thread.
But your device will work if you add some things. ;) ;D
If you let your device lift up with water and tube and try to hold the tube in this upper position (look at my drawing).
Then you have only to open the valve in order to unbalance the system. Some water will flow out and the lowered weight will let the swimmerboddy raise up till he reaches his upper position (floating).
Again CONGRATULATION accidently you have find the right setup.
The best thing in your setup is that it doesn?t have to be a large tube, because the lifting work begins right under the watersurface, the depth will equal to the hight of the lifted watercolum.
Good job!
Now you have to do some maths to calculate if it is still OU.
By
2Tiger
Hi 2tiger.
I think you misunderstood something about elsa. Your second drawing is allright, but that was the idea from the beginning. Imagine a large sea or pool and one long pipe going 10m under water and 1m above the water. The pipe is surely fixed in this position.
The transporting shuttle is able to freely rise up from 10m under water in the pipe and transport all of the water above it in an upper basin, 1m above sealevel if it has the righ boyancy.
We are now researching some math about how we are going to compress the shuttle, or the air inside the shuttle, so it can increase its volume 10m under sealevel.
Andi.
Glad you like it guys. There is a lot to answer - mostly because the page is not complete yet.
@Stefan - yes I will be adding recompression calculations, COP and metric to imperial conversion.
@tbird - the shuttle weight is calculated as that weight required to sink the shuttle (i.e. balance the volume of the shuttle.) The volume of the shuttle can be set by changing the pumping cylinder internal diameter or the shuttle height (the most logical thing to do.) Yes, I have based everything on an ideal shuttle - that is a cylinder that has infinitely thin walls. There are many improvements that can be made such as to choose material for the shuttle so that we can calculate the internal volume of the shuttle e.g. if the shuttle is made of lead the walls will be
Xmm thick and the capacity will be
Ycc less than the displacement volume. We can add a friction calculation to tell us how much energy will be consumed by the seal between the pumping cylinder walls and the shuttle. We can add viscosity calculations for the water in the header tube. etc.
Quotethe first thing you do is "head of water in the feed tube". this weight is nice to know, but i think just as (if not more) important would be to know what the max diameter of that pipe could be for a given height. and maybe if given a pipe diamter, what max height could be.
shortly after you get to the shuttle. it's a bit unclear if you are referring to expanded or compressed state.
The max diameter can be anything you like. Increasing the diameter (or the height) increases the static head (i.e. the weight of water in the header pipe) and thus the amount of water your shuttle must lift.
When discussing the shuttle volume it might be better to think of it as the capacity of the shuttle. My shuttle never expands or compresses, rather it is the air that it contains which is compressed at the top of the cycle
into the shuttle (which is simply a closed cylinder). At the bottom of the cycle air is released from the bottom of the shuttle (think of it as an upside-down hollow piston) into the pumping cylinder until the pressure is equal inside the shuttle and below it. At this point the air under the shuttle will provide lift to the head. We can ingnore the air inside the shuttle at this point since it is the shuttle's displacement (or physical volume) being equal to its weight (because the calculator sets the weight equal to its displacement) that makes the shuttle bouyancy neutral. In fact the weight of the shuttle needs to be slightly greater than its displacement so that it will sink and the air under the shuttle needs to be slightly more than the static head so that it will 1. lift the head
above the top of the header tube and 2. lift the slightly unbouyant shuttle. I will make a note to include these factors in the calculator. I hope this all makes sense.
Quotefrom your program "We know that a depth of 10m will compress the air 2 times its volume at the surface.
height / 10 + 1". not sure if this is worded right.
I can't have worded it right or it would have been crystal clear to you. :) If a balloon contains 1 litre of air at sealevel then at 10m depth it will be compressed to a volume of 0.5 litres. I should probably have said "We know that a depth of 10m will compress the air to half of the volume it occupies at the surface."
Thanks for all the compliments guys. I will continue working on it.
good morning,
i know how tempting and easy it is to take something new you just bought out of the box and try to use it instantly without reading the operating instructions only to find out you weren't as clever as we thought and then go hunting for the little piece of paper that came with your new purchase so you can make it work. everybody is guilty of this. i believe this is why the elsa never really got off the ground before. the instructions (Mr. Herring's drawings) aren't as clear as they could be, but if you take the time to really study them, you will understand how it works and be able to build a working model.
maybe a rewrite of his drawings will be needed. for sure if you don't read them, you may eventually end up with something that works but at an extra expense of time and frustation (possibly money too).
i have in previous post presented his design of various parts and methods, but seems to be like water on a duck's back. since you insist on writting your own instructions, and sometimes physics, i will just be on the side and if i can throw in an obvious correction, i will. good luck
tbird
Quote from: prajna on August 27, 2006, 06:00:52 AM
Here is a new diagram. I hope it helps.
Picture 1: the shuttle is at the top and we pressurise the reservoir using valve 1.
Picture 2: the shuttle is at the top and we let the air out of the displacement area by using valve 3. Now the shuttle descends.
Picture 3: the shuttle is at the bottom and we force the water out (displace it) by opening valve 2. Now the shuttle ascends in picture 4.
The volume of the shuttle including the reservoir is 1.5 litres: 1 litre in the displacement area and 0.5 litres in the reservoir. At 10m the pressure is 2 bar so the shuttle will contain 3 litres of air in the space of 1.5 litres. When the reservoir is fully pressurised it contains 3 litres in a volume of 0.5 litres and is therefore at 6 bar. When the shuttle is at the top the reservoir still contains 1 litre of air in a volume of 0.5 litres so it is at 2 bar.
Looking at this cycle we need to figure out two things: how much water will the shuttle lift and how much energy will be needed to recompress the reservoir. These I have calculated in my earlier message. I think that model is as simple as we can get.
prajna, if this is the shuttle you plan to use, i think you have a problem. steps 1 & 2 cancel each other. if you put air in and then let it out so it will sink, you won't have air for step 3. maybe you just left out another step that would leave you with enough compress air.
i thank you again for your efforts with the program. it will be very useful, even if you have read the instructions (i think you have, more so than anyone else)
tbird.
tbird,
my approach is to get a rough idea of how something works and then work all of the details out for myself. That means to ignore the manual completely and, though that might sound dumb, it means that I end up with a deep understanding of how something works - much deeper than if I had simply read about it.
I've updated the calculator a little. Nothing significant changed but I have added a picture to make things a little clearer.
http://declarepeace.org.uk/energy/elsa.htm (http://declarepeace.org.uk/energy/elsa.htm)
Quotemaybe you just left out another step that would leave you with enough compress air.
I think I straightened this out in a recent posting:
QuoteAt the bottom of the cycle air is released from the bottom of the shuttle (think of it as an upside-down hollow piston) into the pumping cylinder until the pressure is equal inside the shuttle and below it.
Aah, I see your confusion. The displacement area is below the shuttle (not the reservoir inside the shuttle). I let the air out of the displacement area but don't let out the air I have just pressurised into the reservoir. Is that clearer?
Quote from: prajna on August 28, 2006, 09:13:20 AM
tbird,
my approach is to get a rough idea of how something works and then work all of the details out for myself. That means to ignore the manual completely and, though that might sound dumb, it means that I end up with a deep understanding of how something works - much deeper than if I had simply read about it.
I've updated the calculator a little. Nothing significant changed but I have added a picture to make things a little clearer.
http://declarepeace.org.uk/energy/elsa.htm (http://declarepeace.org.uk/energy/elsa.htm)
Quotemaybe you just left out another step that would leave you with enough compress air.
I think I straightened this out in a recent posting:
QuoteAt the bottom of the cycle air is released from the bottom of the shuttle (think of it as an upside-down hollow piston) into the pumping cylinder until the pressure is equal inside the shuttle and below it.
Aah, I see your confusion. The displacement area is below the shuttle (not the reservoir inside the shuttle). I let the air out of the displacement area but don't let out the air I have just pressurised into the reservoir. Is that clearer?
had a short look again at your program. with 15cm (+or-6inches) diameter and 30cm (+or-12inches) high (sounds like less than 1 cubic foot (64.7lbs)) the result is 53.01 kg. doesn't that sound heavy to you? maybe you need to use internal radius instead of internal diameter.
so there is a small passage outside the shuttle (still inside the tube) to valve 3?
your shuttle does get pretty close to elsa design. if i see your shuttle right, the bottom of the tube will be open? to allow the shuttle to descend without moving water up threw exit pipe?
tbird
Quote from: prajna on August 28, 2006, 09:13:20 AM
Aah, I see your confusion. The displacement area is below the shuttle (not the reservoir inside the shuttle). I let the air out of the displacement area but don't let out the air I have just pressurised into the reservoir. Is that clearer?
Ok, now its much clearer and actually I think its the best design we currently have.
Guys, for the calculator the shuttle design is even simpler than the previous one. Take a look at the picture on my calculator site. It is just a hollow piston into which we can compress air. At the bottom of the stroke the air is released from the piston (via a valve, presumably) underneath the piston. This means that the compressed air that was inside the piston is now partly inside the piston and partly underneath it. But now it has expanded from the volume of the piston (actually slightly less because we have to take into account the thickness of the piston's walls so its internal capacity is slightly less than its displacement) to the volume it will occupy at the depth we release it (10m in my default example).
Does this all make sense?
Quotehad a short look again at your program. with 15cm (+or-6inches) diameter and 30cm (+or-12inches) high (sounds like less than 1 cubic foot (64.7lbs)) the result is 53.01 kg. doesn't that sound heavy to you? maybe you need to use internal radius instead of internal diameter.
It is heavy, tbird. That is because you have a much taller shuttle than you need. What kind of head are you lifting? If it were 1kg (2.2 pounds, I think) then you would need a much smaller shuttle. Tell me the head sizes as well as the pump sizes and I will tell you a reasonable shuttle height.
How much does one cubic foot of water weigh?
Allright, the idea is very good - if we are able to seal the shuttle sides so that no air is able to escape.
We can do that, as per my previous design, once all the figures have been calculated but it is simpler to consider it as a simple piston for now.
Hi Pranja, good work with your calculator.
Well, one thing:
So we can pump 77754.42 litres on each stroke
using Ãâ,¬ r2 * (height of the pumping tube - height of the shuttle)
So now your are subtracting the height of the shuttle ?
Is that with the displacement area full of air or without it ?
Maybe you make better the main tube x meter longer when you
ask for the deepth, so the tube will be automatically 20 cm longer than
10 Meter, if one chooses 10 Meter deepth...
Then it will not be too confusing...
2. Also it would be better, if you draw your picture this way, that
you attach some open buttom case for the air displacement area,
otherwise one can not easily see, that air is expanded from the shuttle
down below the shuttle into the displacement area.
Many thanks for your hard work.
Stefan. I have changed it again. Now it calculates the pumped volume by subtracting the volume of the total air required to lift the shuttle and the head from the volume of the pumping tube. To save questions, this is the volume that the air will occupy at the bottom of the pumping cycle including the displacement of the shuttle.
The height of that air in the pumping cylinder will become important when we need to figure out how close to the bottom of the pumping tube the shuttle can go so I will also include the height figure in the results. It also tells us the height of the shuttle including a skirt to contain the expanded air (which you suggested I show on the drawing).
Quote from: prajna on August 28, 2006, 10:29:10 AM
Quotehad a short look again at your program. with 15cm (+or-6inches) diameter and 30cm (+or-12inches) high (sounds like less than 1 cubic foot (64.7lbs)) the result is 53.01 kg. doesn't that sound heavy to you? maybe you need to use internal radius instead of internal diameter.
It is heavy, tbird. That is because you have a much taller shuttle than you need. What kind of head are you lifting? If it were 1kg (2.2 pounds, I think) then you would need a much smaller shuttle. Tell me the head sizes as well as the pump sizes and I will tell you a reasonable shuttle height.
How much does one cubic foot of water weigh?
unless i really don't know my metric values, 15cm diameter by 30cm high is only about (not even, 6x6square has more area than a 6"diameter circle) 25% as big as 1 cubic foot. so 25% of 64.7 pounds (16.175 pounds or 7.35kg) doesn't sound like 53.01 kg (116.6 pounds). what am i missing?
Quoteso there is a small passage outside the shuttle (still inside the tube) to valve 3?
your shuttle does get pretty close to elsa design. if i see your shuttle right, the bottom of the tube will be open? to allow the shuttle to descend without moving water up threw exit pipe?
didn't see an answer to this part of my post. should i quack?
tbird,
a cylinder of diameter 15cm and a height of 30cm has a volume of 53.01 litres
53.01 litres = 1.87203048 cubic feet so that is where your error is.
Unless I have my decimal point in the wrong place. I'll go check.
I have added a detail picture of the shuttle so that might answer your second question. Don't quack :)
Yes, decimal point in the wrong place. Sorry tbird the volume should be 5.3 litres = 0.187167734 cubic feet. I'll fix my calculator. Thanks for pointing it out.
I'd better check my pumping capacity too.
phew! I think I have decimal points in all the right places now. Can someone please check by calculating the volumes for themselves and comparing the results to my calculator.
Oh, and I have fixed it so that the javascript runs in DOM compliant browsers like Firefox too.
Time to add some compression calculations I think...
http://declarepeace.org.uk/energy/elsa.htm (http://declarepeace.org.uk/energy/elsa.htm) to save you having to find the link.
Announce, Announce...
ELSACALC now calculates the recompression requirements and excess water. Read it 'n weep ('cause we never worked on this earlier).
Anyone care to check my calculations?
http://declarepeace.org.uk/energy/elsa.htm (http://declarepeace.org.uk/energy/elsa.htm)
Hi guys,
I think the ELSA system is the first system which can be calculated mathematically
to run and to have overunity.
Now I thought a lot in the last days, how to best build such a unit, but I must say, it is
very complicated with all the mechanics involved.
So I am thinking now about how to do it much more easily.
One way could be to just rotate the main water case by 180 degrees,
when the water has been pumped up and over the center of gravity
of the whole unit, so it can tilt by 180 degrees and the cycle can begin again.
Then you also don?t need to compress the shuttle anymore.
This is also not so easy cause you have to see how to get more water weight
over the center of gravity of the whole unit..
A second new idea is to just use a slow rotating water cylinder with
a fixed volume shuttle in it and every time the water cylinder has turned
by 180 degrees the shuttle with buoyant force is again at the ground of the cylinder
and will rise up again.
Think about a Plastic coke bottle which is filled with water to the top.
Then put a ping-pong ball in it and screw the lid tight.
Then rotate the coke bottle and you see, that the ping-pong ball
rises every 180 degrees from the buttom to the top
inside the coke bottle, if you continue to rotate the bottle.
Now if you have the a bigger and longer watercylinder and
have a bigger shuttle than a ping-pong ball you have real good
bouyant forces, which can pull via a thread on a generator axis
in the watercylinder, so you could generate quite some power
this way and couple it to a flywheel that keeps
the watercylinder rotating !
I guess this also could work. I will
calculate the energy you would win per 180 degrees lift.
Regards, Stefan.
Quote from: prajna on August 28, 2006, 05:23:45 PM
Announce, Announce...
ELSACALC now calculates the recompression requirements and excess water. Read it 'n weep ('cause we never worked on this earlier).
Anyone care to check my calculations?
http://declarepeace.org.uk/energy/elsa.htm (http://declarepeace.org.uk/energy/elsa.htm)
at the beginning you say "To compress 1 litre of air to twice its volume we need to double the weight on top of it." twice sounds like just the opposite of what you mean. that is just a "being tired" mistake. i understand.
QuoteWe can compress the air in a cylinder. The smaller the diameter of the cylinder the less weight we need on our piston but we will have to compress it over a greater distance.
i believe this is wrong. it may work with liquid, but not gas. if you apply 1kg of weight, that is all the pressure you will gain divided by the area it is applied to. to gain the whole 1kg., you would have to apply that amount to each square unit you are using.
i'll try to find a website to confirm. hope i'm wrong, but think not.
tbird
stefan, nice idea - but how will you arrange the whole system (especially the header tube and main water feeding)?
Can you explain more?
Andi.
Yes, tbird, you caught me out with back to front explainations again. 'twice' should read 'half'.
If I have a cylinder full of air it has a weight of 1kg/cm2 acting on the top of it (the atmospheric pressure). If I add 1kg/cm2 to the top of it then it will compress it to half its volume. It is now at twice atmospheric pressure. Double the weight, double the pressure, halve the volume.
Height comes into it when we need to store a certain volume in a particular diameter cylinder. I managed to confuse myself there I think. I'll try to clarify my thinking on the page. The sums should be correct now though, don't you think?
Quote from: prajna on August 28, 2006, 07:38:49 PM
Yes, tbird, you caught me out with back to front explainations again. 'twice' should read 'half'.
If I have a cylinder full of air it has a weight of 1kg/cm2 acting on the top of it (the atmospheric pressure). If I add 1kg/cm2 to the top of it then it will compress it to half its volume. It is now at twice atmospheric pressure. Double the weight, double the pressure, halve the volume.
Height comes into it when we need to store a certain volume in a particular diameter cylinder. I managed to confuse myself there I think. I'll try to clarify my thinking on the page. The sums should be correct now though, don't you think?
i still have a problem with you using absolute pressure. if anybody should use it, it would be a scuba diver, but they don't. consider this; if you have an empty container and put a lid on it, how much pressure is inside? if you use an average gage, say like you use to check the pressure in your car tires, it will read 0. now if you can apply 1 bar to the container, you could take it to 10m (we'll say that is 1 atmosphere) deep without it being crushed in because it exerts more pressure inside than is being applied outside at any depth less than 10m. if you try to take it deeper, it will start to collapse. now at 10m if want to expand the volume, we would need twice the pressure. that's how we get to the 2nd bar needed to expand the volume to twice at 10m. compressing the air will change the weigh, but i doubt we could read it on our scales. to get the container to depth, we have to build i the weight to the container. a 10kg container with 1 bar of pressure inside will float or sink the same as if it had 2 bar of pressure. if we allow the container with 2 bar to expand the air volume to twice, then the pressure in both containers would be the same, but the buoyance would be different.
in your example, to read 2 bar, you would have to use an absolute gage. i don't have one and it won't expand at 10m. sorry.
tbird
confuses the hell out of me too tbird. 1 bar = 14.503861 psi = 0.9872 atmosphere = 1.02 kg/cm2 (which means I should fix my 14.7 psi etc on my site; dunno where I got that figure from).
What is the pressure in a container if I double the volume of air in it? If you are going by gage pressure then the pressure is 0 psi or 0 bar before I double it. What is 2 times 0? Yep, that is why I use absolute pressure because then my unpressurised container has 1 bar or 1 atmosphere or 14.5 psi in it and when I double the pressure it becomes 2 bars or 2 atmosphere or 29 psi.
Don't use my 'absolute gage' when you inflate your tires and I won't use your tire gage when I am calculating pressures. :)
Quote from: prajna on August 28, 2006, 08:29:40 PM
confuses the hell out of me too tbird. 1 bar = 14.503861 psi = 1.01295 atmosphere = 1.03325 kg/cm2 (which means I should fix my 14.7 psi etc on my site; dunno where I got that figure from).
What is the pressure in a container if I double the volume of air in it? If you are going by gage pressure then the pressure is 0 psi or 0 bar before I double it. What is 2 times 0? Yep, that is why I use absolute pressure because then my unpressurised container has 1 bar or 1 atmosphere or 14.5 psi in it and when I double the pressure it becomes 2 bars or 2 atmosphere or 29 psi.
Don't use my 'absolute gage' when you inflate your tires and I won't use your tire gage when I am calculating pressures. :)
i see where you go wrong. if you double the volume of air, you are not doubling the pressure, you are using the pressure to double the volume.
tbird
i'll go further now. if you have 0, like you say there is nothing to double. once you have a pressure, the ratio kicks in.
btw the 14.7 is the accepted value at least in the states. it really does change with the weather. how much water is in air? depends on the weather. the more water there is, the heavier it is. so pressure changes. some days atmosphere is heavier than others.
QuoteI see where you go wrong. if you double the volume of air, you are not doubling the pressure, you are using the pressure to double the volume.
If I have a rigid container of a given size and I double the quantity of air in it then the pressure inside will be twice what it was. Isn't that so? Twice the air = twice the pressure. If I take a sealed container and squeeze it to half of its size then the pressure inside will double.
in absolute value, you are right. the only problem is your 1st bar will be only half enough at 10m to keep your container displacing its shape. it will take another whole bar just to be static. to get it to expand to twice, you need to put 2 more of your absolute bars for a now total of 4 bar. see how confussing it becomes.
tbird, if I have a box 10cm x 10cm x 10cm it contains 1 litre of air at sea level.
In absolute terms the air inside the box exerts 1kg/cm2 on the walls of the box but that is exactly balanced by the air outside the box which exerts 1kg/cm2 outside the walls of the box. Your tire gauge measures its pressure at 0 bar.
If I compress another litre of air into it then it contains 2 litres which exerts a pressure of 2kg/cm2 inside the walls of the box at sea level which is balanced by the air outside exerting 1kg/cm2 on the walls of the box. Your tire gauge reads 1 bar.
When I take that box down to 10m below sealevel the water is exerting 2kg/cm2 on the outside of the box which is equal to the pressure exerted by the air inside the box. Your tire gauge will still read 1 bar because the pressure inside the box has not changed. It is only the pressure outside the box that has changed. It is a sealed box and the air inside the box couldn't care less what the pressure is outside the box.
right! the box is static. 2 bar outside, 2 bar inside. it will not expand.
i suggest you take a survey to see which pressure (gage or absolute) our friends would like to use.
bed time. see ya tomorrow.
tbird
At sealevel we have 1 bar=14,5 psi pressure. That is right. These are the accepted values.
Your tire measuring gauge is subtracting this 1 bar from its reading ! So if you put 2 bar=29 psi into your tire as displayed on the gauge, it really has already 3 bar inside it !
As we have one bar more pressure 10 meters deep than at sealevel, it is okay, that we add just this one bar at the top to our shuttle. Then it can expand in 10 meters deepth to double its size.
Aah, I think it dawns on me now. 2 bar inside and 2 bar outside means no air will come out at 10m depth so I have to put 4 litres of air into the box if I want to have 2 litres of air inside it (at 2 bar) and 2 litres outside it (also at 2 bar) displacing 1 litre of water. Ok, I will check the calculator tomorrow and see if I have accounted for pressures correctly. Thanks tbird. Sleep well.
Today I did lots of drawings to see and calculate, if it is possible to shift the center of gravity for the whole water case over a symmetrical axis, so that it could tilt and turn by 180 degrees.....but I have found so far no solution. The maximum was, that it just turns once by 180 degrees, but then the center of gravity is even deeper down and it can not shift anymore the center of gravity over its axis for the next turn...too bad...
Pranja, Tbird is right, we need at 10 meters down 4 bar absolutely to expand the shuttle twice ! At sealevel we would just need 2 bar, but down at 10 meter we need twice as much !
It seems we really need a 2 part system to get this device to work as we have so far worked out here. With 2 parts I mean the basic watercase, the pipes and the shuttle and the second part I mean the upper reservoir with its shuttle recompression mechanics and waterwheel for energy extraction.
Ahh, I guess I just found a solution to circumvent the recompression of the shuttle by turning the watercase by 180 degrees, after the shuttle has lifted all the water above it !
Imagine, the basic watercase with the 2 (now indeed 3) tubes has a symmetrical axis. now the shuttle just delivers the water as before to the upper reservoir ! After this is done, the exit pipe is closed and pulled off from the upper reservoir and the water up there is used via the generated torque of a waterwheel to rotate the whole watercase with the shuttle and mainpipe ! this way the shuttle will then be again down at 10 meter, when the whole watercase has turned by 180 degrees ! Great idea !
You then have 3 tubes. One main tube, where the shuttle slides up and down in and 2 exit tubes, one at each side. There must be valves at the exit tubes and the lower one is always closed. This way the shuttle has a fixed volume and does not need to be changed ! Also there needs to be no recompression device ! Major breakthrough ! we only need some more passive valves, so the water can go its way it want to go and the main water tank must be put on a turnable axis and the upper water reservoir must stay in its place as before. The pumped up water will be used to turn the main watercase and additionally a generator for the output. Enjoy. Regards, Stefan.
Think about it, the shuttle is at the top of the main pipe, now the whole watercase will turn by 180 degrees. This means the shuttle is now again at the buttom of the watercase and can again lift the water above up via the next exitpipe into the upper reservoir. now this cycle repeats and repeats again. The upper reservoir water must also be returned into the lower watercase. This can be done at the right moment after it has driven a waterwheel to generate output power.
Stefan,
Just a thought. Once the main tube begins to rotate, the shuttle may begin its ascent once it goes past 90 degrees of rotation. Maybe some type of lock may be needed to hold the shuttle in place until it reaches a full 180 degrees of rotation, otherwise valuable energy is lost.
This is getting more complicated. I like simple, but then again, I am simple minded.
Nice inovative idea though.
Ok guys, I couldn't sleep so I have designed a simple system for recompressing the shuttle and releasing the air when it reaches the bottom of the pumping tube. It assumes that you use some of that lovely excess energy you have stored to run a compressor that will supply compressed air to the shuttle. Now it is just a matter of designing what will happen above water level.
It's on my website http://declarepeace.org.uk/energy/elsa.htm (http://declarepeace.org.uk/energy/elsa.htm). Enjoy.
Hi prajna, the idea would work if we are able to completely seal up the shuttle from the pipe sides. Do you think, that can be made that easy?
stefan, I also thinked about your idea this night. We could also do it that way (see image).
Perhaps we can use the compressed shuttle as it is, without the need of recompression, and simple rotate the whole apparatus INSIDE a water basin.
I'm going to make some calculations about it.
Mr Herring replied again. He says that currently he is not able to join our forum or submit images because of his old computer. In 6 weeks he is going to buy a new one. He attached an image with some personal informations, its the same as we know from http://www.theverylastpageoftheinternet.com/forsale/plans/elsa/elsa700.htm.
I again invited him to join us here - lets see what happens.
Andi
Yes Andi, this way it could also be done, but then you need a much bigger main watercase, but if you have a garden with a pool that will be okay, so the main tube could rotate in it.
Ooandioo,
QuoteHi prajna, the idea would work if we are able to completely seal up the shuttle from the pipe sides. Do you think, that can be made that easy?
Sure, why not? We do this in internal combustion engines all the time. A two stroke engine even uses ports rather than valves. So this machine is familiar technology using this design. Any engineer 'skilled in the art' ... as they say. :)
Perhaps I should have drawn in the 'piston rings'.
prajna - you are completely right. I'm amazed we have finally discussed the elsa and found out such interesting things.
How will we go in now?
Andi
Well, I have just fixed some of the pressure calculations in ELSACALC and I am tidying up some of the text so that it is easier to understand. Someone needs to design the energy extraction system above water level. If we generate electricity then we can use a standard compressor to recompress the shuttle. The actuator and sensors need to be designed or discovered too.
I will also be adding energy conversion and COP calculations to ELSACALC. Then I will track down a physicist or two to check my calculations. Then we can all say 'fuck George Bush and the rest of the Oiligarchy.' (Actually, I've been saying that for years but it will be nice to demonstrate it rather than simply say it.)
Hello Prajna and Others!
Many things are better than todays energy "constelation". Even ELSA.
So I'll contribute with this Free Energy H2O2 generator in spray bottle. Ha Ha ha
...actually it can be done. Its even much much better than any todays energy system we use.
wizkycho
Ok, I have added power calculations to ELSACALC and, if my calculations are correct, I might not be going off the grid just yet.
Have a look at http://declarepeace.org.uk/energy/elsa.htm (http://declarepeace.org.uk/energy/elsa.htm)
hi prajna,
thank you for your kind words, but they are not necessary. your actions are quite satisfying!
now for the reason i'm posting these days, obvious mistakes.
you didn't state (anywhere that i saw) which pressure you were using, gage or absolute. looking at your numbers,
QuoteSo the pressure inside our shuttle will be 4.41 bar
i assume absolute.
after coming up with that answer, you then say
QuoteSo we will have to compress 4 times our displacement volume into the shuttle i.e. 43.28 litres
[2 * compression * displacement volume]
So the pressure inside our shuttle will be 4.41 bar
[total compressed volume / shuttle volume]
So we need a weight of 432.8 kg times our compression ratio of 3:1 to recompress the shuttle.
hmmm..... is this obvious? i didn't do the math, is this just a typing mistake? if not i'll explain in next post why 3:1 is wrong.
i have to say, just because i think you guys are not really working on the elsa, i will try not to be a negative. in fact, i will help you get the device and numbers as accurate as possible so when you do get around to working on it, you will see the real differences.
this may help; in the default example of prajna's program, he uses
"First we need to decide to what height we will pump the water:
1 (meters)
Then choose the internal diameter of our header tube:
3.57 (centimeters)"
i don't think 1 meter is the max height the shuttle can lift the water.
QuoteWe know that the shuttle volume is 9.82 litres
[Ãâ,¬ r2 * height]
Therefore our shuttle will have to weigh at least 9.82 kg in order to sink since it displaces that weight of water
if you reduce this diameter, you can gain height (i believe that is part of the formula you used to determine stored energy).
also,
QuoteSo we need a weight of 432.8 kg times our compression ratio of 3:1 to recompress the shuttle.
which equals 1298.4 kg
Subtracting that from the total volume of water pumped gives us an excess of 654.28 litres stored at 1 m after we have recompressed the shuttle.
i'm not trying to point out the ratio error, but the fact there is no mention of how you used the compression weight. would it be possible to use that energy (weight) again? maybe the water has to travel the full distance to develope the energy. this is not my best subject. but,... if you did use leverage (hydraulics is leverage as well as a pole) of 10 feet, you would only need 1/10th the weight. if you put 90% of the weight you figured to use back in the excess column, how would that look to you?
one last thought. what if you could get twice the water stored for the same compression energy?
fyi, there is no value for the kwh you converted to from the kilojoules.
tbird.
@wizkycho
nice idea !
@pranja
how do you calculate the output power now ?
Via the gas laws or via the mechanical lever pressure ?
The output seems to be pretty low now...
Also we need to check how high we really can pump the
water through the small exit tube with just a 2 or 5 cm high shuttle...
I did not yet have time to get better tubes and a shuttle that fits in tightly.
Maybe 2tiger with the "boxing of a Barbiefamily" can check into it ?
Many thanks.
Well, tbird, it looks like I don't handle pressure well. :) Going back to the 10cm x 10cm x 10cm box we were discussing in an earlier message... it has a capacity of 1 litre. If I compress 3 more litres of air into it then it now contains 4 litres of air. What does the pressure guage read?
As far as the shuttle is concerned we need to take the original 1 litre of air into account because it is a factor in the calculation of how much water will be displaced at 10m depth. The total 4 litres will expand to displace 2 litres of water. Now, suppose I used gauge pressure in my calculation: the calculation would be wrong. Balanced against that we have the problem that for anyone who frequently works with gauge pressure, they will look at the figure of 4.41 bar and think 'Man, that can't be right.' Which is your reaction, understandably. For anyone who is not so familiar with the difference between gauge pressure and absolute pressure it is more intuitive (I think) to comprehend the calculations if they are expressed as absolute pressure. Perhaps I am wrong or perhaps I should just indicate that the pressure is absolute.
Where I calculated the pressure I have included the formular I used:
QuoteSo the pressure inside our shuttle will be 4.41 bar
[total compressed volume / shuttle volume]
So anyone who is concerned should be able to see what I did.
As far as the compression ratio is concerned, how should I calculate it. I am not sure why I chose to calculate it this way. Perhaps you can help here.
As to ELSA, I haven't studied it. I did take a look at John's pages but only sufficiently to get a rough idea of what he was up to and did all the rest from 1st principles (which, as I explained to you in an earlier post, is my preferred way of working.) If this system is wildly different than ELSA then perhaps we could refer to it as 'ELSA inspired.'
The default value of 1m for the header pipe was just an example. You have probably already discovered that you can set it to whatever you like. 1m by 3.57cm gives a head of 1 litre and that is the only reason I set those values as defaults. My preferred height is 5m.
By the way, in case you haven't already figured out, all of the values that are in
bold type in the text are calculated (or copied from calculations further up the page). Perhaps I should point that out on the page too.
Quoteif you reduce this diameter, you can gain height
Yes, of course. You can change any of the values from their defaults and then click the 'calculate' buttons to recalculate the results. That is the whole point of the application.
Quote'm not trying to point out the ratio error, but the fact there is no mention of how you used the compression weight. would it be possible to use that energy (weight) again? maybe the water has to travel the full distance to develope the energy.
Certainly we can compress more efficiently by using some fraction of the water we currently use for compressing via a lever. I am building this thing up bit by bit, as you have seen, and I wanted to start with the simple case. We can add a calculation for that as well. It will simply require adding an input box for the lever-arm lenght and the appropriate calculations.
Quoteone last thought. what if you could get twice the water stored for the same compression energy?
It would look very much like the lever system we have just discussed. :)
Quotefyi, there is no value for the kwh you converted to from the kilojoules.
There is on my screen:
QuoteThat might sound like a lot but it represents a stored potential energy of 6.41 kilojoules which is approximately 0.002 kwh
It should display as 0 if it works out at less than 0.0009 kwh.
Thanks again for your careful evaluation, tbird.
@stefan
I use the mechanical pressure (mgh)
prajna,
i'm going to start a little out of order.
QuoteBalanced against that we have the problem that for anyone who frequently works with gauge pressure, they will look at the figure of 4.41 bar and think 'Man, that can't be right.' Which is your reaction, understandably.
you misunderstood my meaning. i do think the figure is right. it's the ratio that bothers me. maybe i'm wrong, but i have always been under the impression if you have something that is 4 times something, then the ratio is 4 to 1, not 3 to 1, right? have i lived under the wrong impression all my life?
QuoteNow, suppose I used gauge pressure in my calculation: the calculation would be wrong.
the first thing i want to say is why? i guess i know why, but can't the calculation be rewritten? the only thing you would have to do is divide by 2 at the end. in your example the 4.41 would become 2.205, which would be right. you could kinda compare it to why we don't have to worry about the water in the pipe. it all takes care of itself. you will always need a greater difference inside the container than it is in to get it to expand. since we are always (maybe we shouldn't) want to double or halve our volume, simply determine the gage pressure at the depth (this will make it static) you want, then to expand just double that amount. works every time.
QuoteFor anyone who is not so familiar with the difference between gauge pressure and absolute pressure it is more intuitive (I think) to comprehend the calculations if they are expressed as absolute pressure. Perhaps I am wrong or perhaps I should just indicate that the pressure is absolute.
the best way i can commit is comparing metric to what i grew up with (and visa versa for stefan). i really struggle getting thru the math in metric, not to mention to visualize without converting is impossible. if you have never been exposed to a gage (never checked your tire pressure, never been scuba diving, etc.) then it absolute will be your standard. you can see where i'm going. gage is easiest for me (and i know about absolute). one more question you can ask yourself. why isn't absolute pressure used more?
QuoteThe default value of 1m for the header pipe was just an example. You have probably already discovered that you can set it to whatever you like. 1m by 3.57cm gives a head of 1 litre and that is the only reason I set those values as defaults. My preferred height is 5m.
this comes back to one of my orignal suggestions (quack quack). if you build a unit, what do you know? not much. the fewer details you have to provide, the better for the user. the first thing you might know is the size of the pipe available. you might have (could be required) a size of the shuttle in mind to be used. remember, the first brick hasn't been laid (so to say). why not give the user the max height, using as a standard 10% of shuttle pipe size, or something reasonable, automaticly. this would be better than guessing about the height. the user could still enter other sizes, but i think most people would rather see the highest energy practical first. don't get me wrong, if you left it as is, it would still be a wonderful tool.
QuoteIt would look very much like the lever system we have just discussed.
no. it would only have to use both strokes. water delivery just doubled. if time is a factor (we haven't ever addressed that), then we need to find out which one does the best job. if not, twice is better with only one recompression.
speaking of time. when you figure energy stored, simply raising the delivery height increases the energy stored. wouldn't time have to be considered somewhere? if we have a working unit and decided to raise the tank twice as high, this would take twice as long (if shuttle stayed the same) to deliver the same amount of water. soooo..... am i missing something? any thoughts?
enough for now.
tbird
tbird,
Quoteratio is 4 to 1, not 3 to 1, right?
Yes, absolutely. Must've been tired and hung up over that original 1 atmosphere or something, I'll fix it. It knocks on to the following calculations too.
Quotethe only thing you would have to do is divide by 2 at the end.
Surely just subtract 1 at the end (or am I tired and confused this morning too?) Whatever the absolute pressure is in the container the gauge pressure will read 1 atmosphere less because the gauge is zeroed to 1 atmosphere, yes? Scuba diving? I've done it twice: once with a friend who taught me a handsignal underwater that nearly had me coughing up my regulator because it looked so funny and once when I was in the army where they seemed more concerned at beasting you through physical exercises to prove just how 'hard' Royal Engineers divers are! Neither seemed, to my poor recall, to have spent much energy on familiarising me on the niceties of the diffference between absolute pressure and gauge pressure. As to tire pressures, most people read the pressure chart in their handbook or the label on the door post and fill the tire until it reads that pressure. Oh dear. I think I will have to print the pressure as gauge pressure and indicate, for poor, confused characters like myself that the absolute pressure is 1 bar more than the gauge pressure.
It sounds as if you are now confused between neutral pressure at 10m and neutral pressure at sea level. The pressure required to double the displacement at any particular depth will be twice the neutral pressure at that depth and the gauge pressure will be 1 bar less. Are we there yet?
Quotewhy not give the user the max height, using as a standard 10% of shuttle pipe size, or something reasonable, automaticly.
Because I want to allow the user to experiment with the effect of changing the height of the head. All of the inputs are there so that the user can experiment with the effect that changing their values has on the various results. It seemed logical to me that one of the first things you might want to decide is 'how high can I pump the water?' and the head height is the main determinant for that.
The user doesn't have to provide a value for this, since it is a default. I understand what you are saying though: it would be nice to have the most efficient system specified as the defaults but calculating the most efficient system is not a trivial task. Perhaps later I will change some of the default values.
Quoteit would only have to use both strokes. water delivery just doubled.
Yes, the original ELSA. I spent some time looking at John's chaotic drawings last night and that is certainly the way to go. Designing this system has gone a long way towards helping me understand all of the factors involved so that I could prove to myself (and others, of course) that the physics really do work. When this version is complete and correct I will do another calculator based on the 'true' ELSA - hopefully with rather less effort than this has taken since I am now pretty familiar with what is involved. I would very much like to come up with some elegant way to recompress the shuttle for ELSA though. All of the rest of John's design is nice and simple but the recompression methods are all a bit ugly and Heath-Robinson. Perhaps that is why he has had difficulty introducing the concept to the world.
Quoteraising the delivery height increases the energy stored. wouldn't time have to be considered somewhere?
Certainly. At a later date I would like to introduce fluid dynamics to the mix and friction so that we can calculate the stroke speed. That will allow proper output values to be calculated and we will know if you really can power your house with one of these in your yard.
Ok guys, I think that is as far as I will go with this design of ELSACALC (unless tbird wants me to add metric to imperial conversion in this version). Next I will work on John's double acting ELSA so that we can calculate the output from that.
Hi all.
New message from John Herring:
QuoteYes you may post if you wish.
Plesae includ my e-mail address, gravity_machine@yahoo.com in your posting`s
I am going in the hospital tomorrow and hope to be home by Monday.
sorry no time today !
Sincerely
John
QuoteJust real Productions wrote:
Hello Mr. Herring.
Thanks for your reply. We are looking forward, discussing and working with you. Maybe you would like to join our foru m board at
http: //www.overunity.com/index.php/topic,570.0.html.
Thank you for the image ? we already know it from the internet.
I?m looking forward, hearing from you. Kind regards, Andreas Lutz.
PS: May I post your replies in th e overunity.com forum?
Hi all
Please take a look at this graphics. It is a simulation from the software Interactive Physics.
The marbles (1cm diameter) are representing the water. The orange one is representing the shuttle (50% more weight).
As you can see the 50 % more weight on the right side is only able to lift the blue marble on the left side half his height - 5 mm. If the marbles were water, the water will rise 5 mm too. But if you want the water to overcome the 5 mm so that it can flow out, you have to choose a tube that is 4 mm high OR you have to make the shuttle 50%+10% = 60% heavier (by same volume of water)!
So the relation between the weight of the shuttle and the max. lift is 10 % more weight (by same volume of water) for 1mm lift up. 1cm (marble diameter) = 100%, 2cm =200% and so on. And remember you have to lift this weight with your "balloon" only by doubling the volume of the shuttle.
Now take a look at the lower graphic the ratio between both tubes is 1:1.
What is the ratio in the ELSA design? Let us say 1:5 (lift tube : shuttle piston tube).
So if we don?t want to violate the hydrostatic paradoxon that stefan pointed out in his earlier posts, there have to be on the shuttle side, 5 times more downforce (for example press the water 1 mm down) in order to lift the water in the lift-tube 1 mm up.
Now we know the relation 10% more weight (shuttle) for one 1mm lift. But when the ration between the tubes is
1:5 we will need 5 times 10% more weight (shuttle) for lift the water 1 mm up.
Stefan - did you try what i write in my last post?
From the beginning you have been right with your assumptions, that this won?t work.
If you now are still believing that ELSA works, I will send you my "barbyfamilytube" for free (without shiping costs).
I live near by Bad Segeberg (Karl May) in Germany.
Next i will try to analyse the upstroke of ELSA. But the downstroke definitive won?t work in that manner.
CU
2Tiger
2Tiger,
Water at sealevel has a force of 14.5psi or 1.02kg acting on it. If I have a 'U' tube such as the one in your drawing that has an internal diameter of 3.57cm it will hold 1 litre/m of tube. Let's assume that there is enough water in the tube so that the bend at the bottom is under water. If I then pour 1 litre of water into the tube then the water level in the tube will rise 50cm. Equally, if I put a 1kg weight in one leg the water level will raise 50cm (since 1 litre of water weighs 1kg). Is that what you were trying to say?
What will happen if there is just 40cm of height of tube over the level of the water and I put a weight of 1kg in one leg? The water will flow out of the other leg (because we know that 1kg will push the water up 50cm) and the weight will continue to sink, pushing even more water out of the tube until it gets to the bend at the bottom. It will have pushed nearly half of the water out of the tube (considerably more than half a kg of water) even though the weight I put in is only 1kg. This is how ELSA works when the weight is going down.
Hello prajna
QuoteWater at sealevel has a force of 14.5psi or 1.02kg acting on it. If I have a 'U' tube such as the one in your drawing that has an internal diameter of 3.57cm it will hold 1 litre/m of tube. Let's assume that there is enough water in the tube so that the bend at the bottom is under water. If I then pour 1 litre of water into the tube then the water level in the tube will rise 50cm. Equally, if I put a 1kg weight in one leg the water level will raise 50cm (since 1 litre of water weighs 1kg). Is that what you were trying to say?
Yes this is right. But only if the U tube has two legs of the same diameter. As the ratio between the legs is 1:2 you need to pour 2 liter, if you want the water to rise 50 cm.
Quote
What will happen if there is just 40cm of height of tube over the level of the water and I put a weight of 1kg in one leg?
Yes - this is what I wrote. Remember (look at the picture 1 Marble 1cm in diameter):
Quote
As you can see the 50 % more weight on the right side is only able to lift the blue marble on the left side half his height - 5 mm. If the marbles were water, the water will rise 5 mm too. But if you want the water to overcome the 5 mm so that it can flow out, you have to choose a tube that is 4 mm (over sealevel) high OR you have to make the shuttle 50%+10% = 60% heavier (by same volume of water)!
Quote
The water will flow out of the other leg (because we know that 1kg will push the water up 50cm) and the weight will continue to sink, pushing even more water out of the tube until it gets to the bend at the bottom. It will have pushed nearly half of the water out of the tube (considerably more than half a kg of water) even though the weight I put in is only 1kg. This is how ELSA works when the weight is going down.
Yes - this is the conclusion of that, but only in a 1:1 ration between the both legs of the U tube!
And therefor you will need 2kg to overcome 100 cm, 3kg to overcome 150cm and each time subtracting 10 cm of the tube over waterlevel, so that the water can flow out!!
In the case of a 1:2 ratio, you will need 4 kg to overcome 100 cm, 6 kg to overcome 150cm.
In the case of a 1:5 ratio, you will need 10 kg to overcome 100 cm, 15 kg to overcome 150cm.
...and so on...
And stefan here is the answer to your question, how high will the water rise over sealevel.
I do the math for you.
Imagine a U-tube 100 cm under sealevel, one leg of U-tube over sealevel and diameter ration between the legs 1cm:1cm.
Now if the shuttle is 50% heavier than the water by same volume, the water will rise 0,4 cm.
In a 1:5 U tube just 0,008 cm.
Cu
2Tiger
Hi All, Maybe this should be under another topic: but here is another water gravity wheel.
http://www.raska.info/tocak2/Invention.pdf#search+%22vukosavljevic%20perpetuum%22
Regards,
Keytronic
hi prajna,
QuoteQuote
the only thing you would have to do is divide by 2 at the end.
Surely just subtract 1 at the end (or am I tired and confused this morning too?) Whatever the absolute pressure is in the container the gauge pressure will read 1 atmosphere less because the gauge is zeroed to 1 atmosphere, yes?
i have to say, i'm sorry. i didn't verify my statement "works every time". this is another example of how easy it is to get confussed with absolute. to check your formula, i changed the pumping tube depth to 20m. the answer is right. it comes in at 6 atmospheres. in order to get gauge pressure, if you subract 1, it would be wrong. it would be 2 times the 1 because you had to double it as well as the water pressure. so instead of dividing by 2, subtract 2. not being a program writer myself, i can't tell you the best way to write this formula. but to get to the weight needed, all you need to know is the pressure and area the force needs to be applied to. assuming we only want to double volume from static state at depth.
Quote from: prajna on August 30, 2006, 03:16:17 AM
Ok guys, I think that is as far as I will go with this design of ELSACALC (unless tbird wants me to add metric to imperial conversion in this version). Next I will work on John's double acting ELSA so that we can calculate the output from that.
it would be a wonderful addition for me (probably Mr. Herring too). it's your program, you have to do the work (i no it takes a bunch), so it's up to you.
QuoteQuote
why not give the user the max height, using as a standard 10% of shuttle pipe size, or something reasonable, automaticly.
Because I want to allow the user to experiment with the effect of changing the height of the head. All of the inputs are there so that the user can experiment with the effect that changing their values has on the various results. It seemed logical to me that one of the first things you might want to decide is 'how high can I pump the water?' and the head height is the main determinant for that.
The user doesn't have to provide a value for this, since it is a default. I understand what you are saying though: it would be nice to have the most efficient system specified as the defaults but calculating the most efficient system is not a trivial task. Perhaps later I will change some of the default values.
can't you do both? from just above " It seemed logical to me that one of the first things you might want to decide is 'how high can I pump the water?' ". imagine an head pipe the size (diameter) of a toothpick with 50kg of pressure behind it. how high is that, max? no, because you can always divide that into a smaller size. not practical. thus my suggestion of a percentage. i didn't mean to limit what the user could use, just start there and still be able to change the value. basicly the same as you are doing, but with a footnote like "this value is x% of pumping pipe diameter". it would be your default.
QuoteI would very much like to come up with some elegant way to recompress the shuttle for ELSA though. All of the rest of John's design is nice and simple but the recompression methods are all a bit ugly and Heath-Robinson. Perhaps that is why he has had difficulty introducing the concept to the world.
i agree. just be careful not to get too cute. like you said something about an electric control for.... having lived on a boat for as long as i have, you learn to keep your electric devices as far from water as possible. you only mix them if there is absolutely no other way (if you don't want to be doing without that device from time to time until you can repair the water damage). john had an idea (shows in drawing #47) that might be option for those electric things you talked about. simple magnets. don't think the drawing covers all you wanted to do, but you're a clever guy, keep going down that path.
QuoteQuote
raising the delivery height increases the energy stored. wouldn't time have to be considered somewhere?
Certainly. At a later date I would like to introduce fluid dynamics to the mix and friction so that we can calculate the stroke speed. That will allow proper output values to be calculated and we will know if you really can power your house with one of these in your yard.
when you get there, my post will probably slow down. they may only consist of questions.
keep up the good work!
tbird
2tiger,
you must be overlooking (at least) the fact we don't use the same size diameter pipe above water. true it's slower delivery, but it all will go there except what is left in the head pipe. since our examples (so far) are based on 10m depth, there's plenty of water lifted to do work.
tbird
Hello Tbird
It doesn?t matter if it is 10 m or 100 m depth, because the watercolums (without shuttle) are canceling its weight eatch other out thrugh the hydrosttic pressure. The only thing that takes effect (unbalance) here is the shuttle with 50% more weight (comparing with same volume of water). As result you?re still lifting half of the shuttle height.
Is the shuttle 1m high then you can lift 50 cm over seawater, but the tube has to have a little bit less (49cm) so the water can flow out.
If you choose a lower diameter for the tube over sealevel, let us say the half diameter of the tube under water, then you will have to double the pressure by making the shuttle two times heavier.
ONLY in that case you were able to double the height over sealevel.
That is the hydrostatic paradoxon!
By
2Tiger
Tbird
I forgot something.
This will only work by a ration beween "lifting tube" and "shuttlepistontube" from 1:1.
Do the math and see what happen, if you have a ration from 1:5.
By
2Tiger
Hi all
I attache some calculations to previews posts, so that you can proof this.
You can change al diameters of all tubes, height of the shuttle and the height of the whole device. The lifted water will always have half volume/height of the shuttle.
You can only lift the water to a higher level when you increase the weight of the shuttle, but the you need more bouyancy force to lift the shuttle back to the top.
Tell me if I?m wrong.
CU
2Tiger
tbird,
From: http://www.challengers101.com/Pressure.html (http://www.challengers101.com/Pressure.html)
"Absolute pressure = Gauge pressure + Atmospheric pressure"
Thus Gauge pressure = Absolute pressure - Atmospheric pressure. If I have the box open the gauge reads 0 bar. If I close the box and double the pressure inside it then the absolute pressure is 2 bar and the gauge reads 1 bar. If I double the pressure again then the absolute pressure is 4 bar and the the gauge reads 3 bar. If I double the pressure again then the absolute pressure is 8 bar and the gauge reads 7 bar. Yes? So to get gauge pressure from absolute pressure I always subtract 1 from the absolute pressure. If not then I am still lost over this.
I'll add imperial units to the page then, tbird. It's a chunk of work but I guess it will make it a lot easier for those who live in imperialistic countries. :)
I didn't suggest electric control. Sensors were mentioned with reference to my design but my preferred approach would be hydraulic. I think I have something in mind for recompression. I have been working on a design and also a page to explain the principles in really simple terms. I have drawn the pictures and I just have to add the explaination, then I will put it all up on my site.
2tiger,
be careful with giving definitive statements here you can't really back up.
QuoteIt doesn?t matter if it is 10 m or 100 m depth, because the watercolums (without shuttle) are canceling its weight eatch other out thrugh the hydrosttic pressure. The only thing that takes effect (unbalance) here is the shuttle with 50% more weight (comparing with same volume of water). As result you?re still lifting half of the shuttle height.
Is the shuttle 1m high then you can lift 50 cm over seawater, but the tube has to have a little bit less (49cm) so the water can flow out.
you are doing ok except for 1 minor detail. you should use weight instead of height in this sentence, "As result you?re still lifting half of the shuttle height." they are proportionate, but weight is more appropriate. i think this is where you start going wrong.
QuoteIf you choose a lower diameter for the tube over sealevel, let us say the half diameter of the tube under water, then you will have to double the pressure by making the shuttle two times heavier.
ONLY in that case you were able to double the height over sealevel.
That is the hydrostatic paradoxon!
i have covered this extensively in other post, but here goes again. the doubling the pressure is not needed. that amount of pressure (created by the weight difference) will pickup a given WEIGHT. it doesn't matter if it is contained in an area of x by y or y by x. it exerts the same pressure. the reason it would stop lifting is because the WEIGHT above water level equals the amount of WEIGHT the shuttle can lift. the hydrostatic paradoxon is helpful by providing pressure to the back side of the shuttle to lift the water that is above the shuttle, but only to the surface. this doesn't apply to the water in the head pipe because hydrostatic paradoxon is based on pressure differences. this makes them want to be level and equal. in our case the water outside the pumping pipe has already neutralized the water above the shuttle, but only to the surface. thus the only weight or pressure the head water can apply to the shuttle is what it WEIGHS. if the shuttle has more lift, the water will go up.
Quote from: 2tiger on August 30, 2006, 09:17:52 AM
Tbird
I forgot something.
This will only work by a ration beween "lifting tube" and "shuttlepistontube" from 1:1.
Do the math and see what happen, if you have a ration from 1:5.
By
2Tiger
the ratio you refer to should be comparing the head pipe volume to the pumping pipe volume. the pressure area (the part that applies the force) of the shuttle will be close to this. it depends on how well it fits (could have been a big problem with your test). the other figure is more reliable. i hope you understand now.
prajna,
i'm sure you will blush when the light turns on.
the pressure you figure for the depth includes 1 bar of atmosphere at the surface. now if you double that (to allow for expansion at depth), you have 2 of those units. that amount doesn't change if you go deeper because in your pressure calc you only add 1 bar for the above atmosphere. so whatever figure you double, it still only comes out 2 for the atmoshere pressure. but it is 2 not 1.
QuoteI'll add imperial units to the page then, tbird. It's a chunk of work but I guess it will make it a lot easier for those who live in imperialistic countries.
i'm a happy camper ;D
QuoteI didn't suggest electric control. Sensors were mentioned with reference to my design but my preferred approach would be hydraulic. I think I have something in mind for recompression. I have been working on a design and also a page to explain the principles in really simple terms. I have drawn the pictures and I just have to add the explaination, then I will put it all up on my site.
can't hardly wait to see! did the magnets give you any thoughts. i thought of a refrigator door seal. magnetic and seal well.
tbird
@2tiger,
your calculation is a bit hard to understand:
1. What case do you want to calculate,
a)the lifting of the shuttle and pumping up the water
or
b) the downward movement of the shuttle ?
2. You seem to forget, that it is done in big watercase, where the
outer water must be calculated in and which has a big weight.
This is also the reason, why so much water is lifted,
if you look at it, if you disconnect the exit pipe and just look
how much water is lifted to the same height of the seawaterlevel
via the shuttle.
Maybe we should just calculate it all with the hydrostatic pressures,
once with disconnecting the exit pipe and one time with
connceting the exit pipe to the main tube
and only look at the case, that the shuttle
is deep inside the water and will lift the water up
above it.
Regards, Stefan.
@pranja,
please only do the calculations with absolute pressure,
otherwise it is getting too confusing, don?t use the gauge
reading with subtracting 1 bar... it is just only good for
inflating your car tires !
@2tiger,
where is exactly your 1 cm^2 and 5 cm^5 area ?
Seems you have drawn this in the same spot...
This is confusing...
Sorry tbird, this is like wading through toffee...
Quote
So we will need to displace 4.93 litres of water to lift the head and the shuttle. We will call this the displacement volume.
...
We know that the pressure at 10m depth is 2 times the pressure at water level.
[(height in meters / 10) + 1]
So we will have to compress 4 times our displacement volume into the shuttle
i.e. 19.72 litres
[2 * max pressure * displacement volume]
So the pressure inside our shuttle will be 4.02 bar (remember that most pressure gauges are set to read 0 at sealevel.)
[(total compressed volume / shuttle volume) - 1]
Where do I go wrong in the above and why?
Hi guys,
I had another look at:
http://en.wikipedia.org/wiki/Hydrostatic_pressure
and it seems 2tiger is half way right.
As the buoyancy is only calculated by hydrostatic pressure differences
onto the the upper and lower surface of our shuttle,
only the heights of the water colums above and below it will
give the buoyancy onto the shuttle.
So if the shuttle is with its lower surface at 10 meters deep,
it has a hydrostatic pressure there of
p= rho x g x 10 meter = 100 000 Pascal
and the upper surface has
p= rho x g x 9 meter = 90 000 Pascal,
if our shuttle is 1 meter high and we have no exit pipe,
but the water above the shuttle is only at seawater level.
(assume g= 10 m/s^2 for easier calculation)
Now this is a buoyancy force of 10 000 Newton
if the shuttle has an area of 1 m^2.
Now, if we add water above the shuttle water column via
an exit tube, only the additional height of the water column counts
and not the amount of water, this means, not the weight,
but only the height counts !
Also if the exit pipe is only 1 cm^2 in diameter, if it is also 1 Meter
high, we have no more buoyancy for onto our shuttle,
as now the hydrostatic pressure oonto the shuttle?s upper
surface is the same as the the hydrostatic pressure on the
lower surface of the shuttle= 100 000 Pascal and the shuttle
does not move !
So this is the hydrostatic paradoxon at work making our device
less efficient, as we can only pump the water as high, as the shuttle
itsself is high !
This is too bad.
So the hydrostatic paradoxon limits the height,
the water can be pumped up.
I have to recalculate , if we can find a relationship,
where it still works, also maybe by using both the
up- and downgoing of the shuttle-cycles
for the water transfer energy output.
I guess we have to look for a different case, where we can use
the hydrostatic paradoxon positively and not negatively as in this
case.
Regards, Stefan.
Maybe we still can use the hydrostatic paradoxon
positively in our case, if we apply
low pressure ( some vaccuum) in the exit pipe.
As there is only very small volume, it also
needs low power to create a vacuum over there
and thus suck the water higher up.
Also the normal 1 bar air pressure at sealevel will help then
to press the water up via the exit pipe.
Quote from: prajna on August 30, 2006, 12:09:16 PM
Sorry tbird, this is like wading through toffee...
Quote
So we will need to displace 4.93 litres of water to lift the head and the shuttle. We will call this the displacement volume.
...
We know that the pressure at 10m depth is 2 times the pressure at water level.
[(height in meters / 10) + 1]
So we will have to compress 4 times our displacement volume into the shuttle
i.e. 19.72 litres
[2 * max pressure * displacement volume]
So the pressure inside our shuttle will be 4.02 bar (remember that most pressure gauges are set to read 0 at sealevel.)
[(total compressed volume / shuttle volume) - 1]
Where do I go wrong in the above and why?
the pressure in your shuttle at 10m that you compressed at surface has a net force of what? we would figure, 4 inside minus 2 (that includes your absolute) outside equals 2. one of which is your surface bar. if you used the net value, you could subtract 1 (at any level). once the shuttle is allowed to expand you will be left with a net 2 bar (1 of which is still your surface bar). without the surface bar, your GAUGE pressure will be 1. so 1 bar expanded, 2 bar compressed. when you change a formula, you have to change it everywhere it is a factor. if you started with gauge pressure where the absolute bar is not considered, we would see at 10m the pressure is 1 bar. so to expand a volume 2 times, you will have to have 2 bar. net pressure then is; 2 bar inside minus outside pressure of 1 bar gauge equals 1 bar net, expanded pressure.
i'm running out of ways to say it.
did that work for you? i'll keep trying until we get it right.
tbird
stefan,
in your example to prove this wrong, you are only appling this to the area the size of the head pipe (1 cm^2 in diameter). do you really think this can stop the shuttle with many more time the area pushing up? did you forget your hydraulics?
i have a hard time with your metric examples, as i have said before. if you give me an inch example, i'll be more exact.
tbird
TBird,
just do the experiment to attach a smaller diameter exit pipe
at the top of the main tube and see, how high the water will be in it,
if you let the shuttle go up under water inside the main tube.
In my case the water in the exit pipe was only as high as the shuttle
was itsself high, but okay, I had not so thight tubes and
some leakage, but I already saw the hydrostatic paradoxon
in my experiments to some extent.
Also the buoyance formulas predict you can only bring the water as high in the exit tube
as high as the shuttle is itsself high !
"Too bad to burst our bubble over here..."
All,
if you look at it, that the main water at sealevel and the
maintube with connected exit tube are a "U" shaped tube,
the water column?s heights in it must be equal, also if the volumes
of the 2 legs are much different.
As the shuttle is the partitioning wall between the both
legs, only the shuttle height can be added to one
water column, if the other has much more weight.
stefan,
if you had "not so thight tubes and
some leakage", how could you expect it to work?
if i build a unit to prove this to you, what prize do i get?
tbird
just read your 2nd post. stop it! you are not making any sense. i will try to build something this afternoon and vedio it. if i can't make it work, i'll leave. if i can,.........
Quotei'm running out of ways to say it.
Sorry to try your patience, tbird.
Can we try approaching it from another angle. If I have 2 kg to lift from a depth of 10m I could attach a balloon to the weight and blow it up with air. How many litres of air would I have to put into the balloon at 10m to make the weight rise? I figure that I would need to have just over 4 litres in the balloon. If I were to compress those 4 litres of air into my 1 litre box then what would the pressure be inside the box? I figure the compression ratio would be 4:1 and the pressure would be 4 bar absolute or 3 bar gauge. Am I anywhere close?
Quoteif i can't make it work, i'll leave.
Please don't leave tbird. I'd go crazy trying to deal with the pressure. :)
prajna,
don't worry, i have time for you.
if you have 2 U.S. coins that add up to 55 cents and one of them is not a 5 cent coin, what 2 coins would you have?????
given; U.S. makes 1 cent, 5cent, 10 cent, 25 cent, 50 cent, and 1 dollar coins.
if you think this is impossible, think again.........the answer is a 50 cent and 5 cent coin. you say...but there is not suppose to be a 5 cent coin! that's right, the 50 cent coin is NOT a 5 cent coin.
sometimes the obvious is too obvious. let's assign our pressures a color. the 1st bar is the atmosphere at sea level. this will be yellow and each atmosphere (bar) of water will be blue. we won't concern ourselves with weight at this time. we know we are trying to create a nice positive lift. so, if the numbers demand 4 bar (absolute gauge)to be compressed in the container to go down, 2 bar will be blue and 2 bar will be yellow. you say how can that be??? well we start with 1 yellow bar and add a blue bar, that would be what was needed to make the container static at 10m. but we want it to be able to double it's volume so we have to multiple this by 2, so 2 times 1 yellow is 2 yellow and 2 times 1 blue is 2 blue. you know this has to be true because when you read the expanded pressure at 10m, you will see 1 yellow bar and 1 blue bar, because you are still reading absolute pressure. if it where 2 blue bars, you would be reading gauge pressure. now if we go to 20m we would need 1 yellow bar and 2 blue bar (we didn't add atmosphere pressure (it still weighs the same) we added 1 bar water pressure). so to make it double volume, we multiple by 2. this gives us 2 yellow and 4 blue. and so on and so no. the atmosphere pressure always being the same (2 yellow). the thing that is increasing is the water pressure not the atmosphere, but the absolute gauge always has to account for it.
so it is easier to just work with the blue bar.
maybe it would be better to write a formula that deals with the pressure rather than the volume. we know that 2 times the pressure equals half the volume and half the pressure equals 2 times the volume. the main reason we want to know this pressure is to figure how much weight we need to recompress. the pumping pipe diameter will give us the area to apply the pressure to, so if we take the pressure time the area, we have the weight. it is easy enough to figure how much water will be delivered just by knowing how much is available. the shuttle size is really only important to determine how high and how fast the water will be delivered.
hope i didn't bore you to tears.
tbird
ps i was out for awhile buying stuff for test.
tbird,
Quotethe thing that is increasing is the water pressure not the atmosphere, but the absolute gauge always has to account for it.
Absolutely (if you'll pardon the pun), and the absolute gauge always does. We can think of it as a gauge that is 'zeroed' at 1 bar when the tire gauge is zeroed at zero. When we go down to 10m they both change by the same amount but the absolute gauge always reads 1 bar or 14.5 psi or 1 atmosphere more than the tire gauge. To indicate gauge pressure, having done my calculations in absolute terms I simply subtract 1 from whatever pressure I am converting from absolute to gauge pressure. Isn't this back where I started from?
If I am talking volumes then it is easy because I can simply add them or double them and nobody gets confused. Likewise with weights. Ho boy!
I am writing an 'Introduction to ELSA' and in that I state it like this:
QuoteLet's say that our shuttle has a weight of 8kg, radius of 14.27cm and a height of 25cm. We can calculate the volume (and therefore the displacement) of the shuttle using the formular Ãâ,¬ r2 h:
3.1415926535897932384626433832795 * 14.27 * 14.27 * 25 = 15999.7
To convert that to litres we divide by 1,000 = 15.9997; that's 16 litres, near as damn it.
If we were to compress the shuttle to half of its length then it would have a displacement of 8 litres and the pressure of the air inside would be at 2kg/cm2. Normal air pressure is taken to be approximately 1kg/cm2.
I'm looking forward to seeing the results of your experiments.
Quote from: tbird on August 30, 2006, 01:53:48 PM
stefan,
if you had "not so thight tubes and
some leakage", how could you expect it to work?
if i build a unit to prove this to you, what prize do i get?
tbird
just read your 2nd post. stop it! you are not making any sense. i will try to build something this afternoon and vedio it. if i can't make it work, i'll leave. if i can,.........
Hi TBird,
if you can show us, that your water in the exit pipe goes higher than the height
of the shuttle volume, I will wire you 10 US$ via PayPal , so you can buy yourself
a bottle of champain or a sixpack and have a party ! ;)
hi prajna,
did you get the answer to my quiz? of course the first time i heard it, i failed.
one question and we'll put this to bed. according to your default example, if you took your shuttle to pressurize it at the gas station, what pressure would you fill to, according to their gauge?
houston, i think we have a problem. i'm not sure about this, but the amount of weight to recompress you come up with is different (default example) than what i figured. being a simple arkie, i just figured the square cm of the pumping pipe diameter, then muliplied it times the pressure needed. your example being in metric, i couldn't follow how you came up with your number. the big reason i have to wonder about the way i did it is if you made the diameter of the pipe smaller, the weight would be less. i kinda said that was ok, it would just travel further. the thing that is on my side (in my head) is if i use psi (i know what this means), that's pounds per square inch. since the only place we have to apply weight is the top of the shuttle, i figured this area would have to have the ratio (whatever it is) times (in inches) 14.5 (or there abouts) per square inch. i think i used 1kg for each square cm in your example. houston, do we have a problem?
i think your 'Introduction to ELSA' is pretty cool. i didn't see mentioned the type of pressure (g or a) you were using. might be helpful.
i read through your description of the system again and have a suggestion. at the bottom of the pipe where you expand the shuttle, make the pipe so it looks like a donut is around it, but without the inside wall. this way if for some reason the shuttle spins, it won't miss the transfer pipe. for that matter you could have more ports for a faster transfer and wouldn't have to work the valve at the top. now if you use a magnet arrangement at the top to hold it inplace while it fills, this would elimate another operation of the valve. to make sure it didn't release too soon, the ports for intake and exhaust could be different sizes. leaving just one thing it had to do, allow water to fillin behind shuttle's descent. this could be controlled with a check valve. now no need at all for slector valve. now everything would work automatic with less to go wrong. just a thought.
tbird.
Hi Stefan
Quote@2tiger,
your calculation is a bit hard to understand:
1. What case do you want to calculate,
a)the lifting of the shuttle and pumping up the water
or
b) the downward movement of the shuttle ?
I was analysing only the down stroke of ELSA. I mean the downward movement of the shuttle and pumping water up throug smaller tube.
Quote
2. You seem to forget, that it is done in big watercase, where the
outer water must be calculated in and which has a big weight.
This is also the reason, why so much water is lifted,
if you look at it, if you disconnect the exit pipe and just look
how much water is lifted to the same height of the seawaterlevel
via the shuttle.
Do you mean to disconnect the exit pipe so that it is on sealevel? In that case the shuttle will sink and all the water of the "pistontube" will flow out right over sealevel. But then you have no usefull work, as energy is W=F x distance (lifting height) and distance is cero, then you get 0 Nm of energy.
Quote
where is exactly your 1 cm^2 and 5 cm^5 area ?
Seems you have drawn this in the same spot...
This is confusing...
I mean the area of the cross-section of the tube, so 1 cm^2 = 0,564 cm radius. The tube will have a diameter from 1,128 cm.
Quote
So this is the hydrostatic paradoxon at work making our device
less efficient, as we can only pump the water as high, as the shuttle
itsself is high !
This is wrong. If the shuttle is 50% heavier than te same volume of water, then you will lift the water on the other side ONLY HALF a height of the shuttle.
As prajna pointed out before, in an U-tube (ration 1:1, dimension 1liter per 1m height) you have to pour 1 liter water to increase the waterlevel on both sides 50 cm.
In U-tube (ration 1:2) you will have to pour 2 liters to reach the same height.
Stefan, if have not understood my calculations or my drawings compairing water with marbles, here I have another example. This times with boxes/cubes and without pressure and water. Just by using the lever rules.
In drawing No. 2 you see on the balance 0,1 N versus 0,20 N. To be conform with the hydrostatic paradoxon I moved the axis of the balance in a ratio 2:1 (NOT 1:2), so that it is balance out. You have to put twice weight on the shorter lever to lift the same amount on the longer one.
Now look at drawing No. 3. There you have on the balance 0,1 N versus 0,21 N (plus 50% weight of the shuttle).
On the right side you have 0,01 N more than on the left side. But because of the lever rules, you will only lift 0,005 N on the left side.
Remember 1 box/cube = 0,01N =1cm^3 there for 0,005 N = 0,5 cm^3 ist a HALF box/cube.
Volume = area x height => height = volume / area,
so height = 0,5cm^3 / 1 cm^2
= 0,5 cm
= half the height of the shuttle
And you will not be able to lift the water higher since you don?t change the weight of the shuttle.
Do you still want me to test this out with my "barbiefamilytube" ?
SeeU
2Tiger
Yes, tbird. I did get it and I did understand why. I must admit that I did have to look twice.
I will change the calculator to read:
QuoteSo the pressure inside our shuttle will be 4.41 kg/cm2 (air pressure at sealevel is taken to be approximately 1kg/cm2.)
[total compressed volume / shuttle volume]
I think that is unambiguous since I state what I consider normal air pressure to be as well as the shuttle pressure.
If I take the compressed shuttle to the garage the tire gauge will read 3.41 kg/cm
2 or 3.41 bar or 49.4578686 psi (according to google's wonderful convert function.)
Quotebeing a simple arkie, i just figured the square cm of the pumping pipe diameter
Did you figure the suare cm by taking
half the diameter and multiplying that by
pi or by multiplying it by itself? I hope the former since
we are dealing with the area of a circle rather than the area of a square.
The way that I figure the pressure is to take the combined weight of the shuttle and the head, convert that to a volume and divide it by the volume of the shuttle. That figure is multiplied by the pressure at whatever depth you selected (10m so 2 times in the default example). That tells me the neutral bouyancy pressure at the required depth. I then double that pressure so that the air will expand to twice its volume at the required depth. Previously I indicated that pressure as gauge pressure by subtracting 1 bar but now I indicate it as you see above.
Gotta go out so I'll continue this later....
@tbird
If you want to win the bottle of champaigne that Stefan offered you, make the shuttle 110% heavier than the same volume of water. Volume of water for example 1000 cm^3 = 1kg = 10 N, then the shuttle must have 2,1 kg = 21 N of weight.
;)
bye
2Tiger
Hi 2tiger and Tbird, I meant to have the shuttle itsself only to have a few grams of weight, so it could be avoided to have any impact. So only the volume of water it is displacing counts. In my view you can still only get the water up over sealevel with an exit tube as high as the shuttle height is. Maybe we then really have to go via a nozzle at sealevel to get it higher or use a vacuum inside the upper reservoir to suck it up.
Hi Tbird, did you already setup your experiment or are you still needing to get parts for it ? Many thanks.
Hmmm - you?re try to lift the water to a higher level via a nozzel?
This seems to me like a syringe.
Please let me compare both setups (with and without a nozzle).
Let us say the pipe has cross-section area from 1cm^2 and it is a little bit over sealevel - 5 cm.
Let us assume that the hydrostatic pressure (in the pipe full of water) is for example 2N/cm^2 equal 0,2 bar.
How high will the water rise?
F=area x pressure => 1 cm^2 x 2 N/cm^2 = 2 N => 200 cm^3 water
h= volume / area => 200 cm^3 / 1 cm^2 = 200 cm (over sealevel)
Now we put a nozzle at sealevel height. The nozzle is ten times smaller than the pipe so cross-section area is
0,1 cm^2. Pressure is the same.
F=area x pressure => 0,1 cm^2 x 2 N/cm^2 = 0,2 N => 20 cm^3 water
h= volume / area => 20 cm^3 / 0,1 cm^2 = 200 cm (over sealevel)
That?s paradox, isn?t it ?
2Tiger
Quote from: hartiberlin on August 31, 2006, 08:21:36 AM
Hi Tbird, did you already setup your experiment or are you still needing to get parts for it ? Many thanks.
bought what i think i need, but have some building (parts) to do.
QuoteMaybe we then really have to go via a nozzle at sealevel to get it higher or use a vacuum inside the upper reservoir to suck it up.
why do you think this will make the water go higher? will you attach a pump to the nozzle?
2tiger
why do you always use the static state of the liquid (not always, some times you don't use liquid)? doesn't the positive pressure that causes the movement count?
tbird
Tbird
Quotewhy do you always use the static state of the liquid (not always, some times you don't use liquid)? doesn't the positive pressure that causes the movement count?
But where is the positive pressure? As long the shuttle has 50% more weight than same volume of water and the exit pipe is half of the shuttle height over surface, then you have the system in balance = static. If you want to get a positive pressure to do some useful work, than you have to make the shuttle more than 50% heavier in order to lift water more than the HALF of the shuttle height OR cut a mm from the top of the exit pipe.
So if you want to lift the water more than the height of the shuttle, you have to make him more than 100% heavier. But don?t forget the shuttle is then down on the ground and you have to lift him again by ONLY doubling its volume.
2Tiger
Quote from: 2tiger on August 31, 2006, 09:47:21 AM
Tbird
Quotewhy do you always use the static state of the liquid (not always, some times you don't use liquid)? doesn't the positive pressure that causes the movement count?
But where is the positive pressure? As long the shuttle has 50% more weight than same volume of water and the exit pipe is half of the shuttle height over surface, then you have the system in balance = static. If you want to get a positive pressure to do some useful work, than you have to make the shuttle more than 50% heavier in order to lift water more than the HALF of the shuttle height OR cut a mm from the top of the exit pipe.
So if you want to lift the water more than the height of the shuttle, you have to make him more than 100% heavier. But don?t forget the shuttle is then down on the ground and you have to lift him again by ONLY doubling its volume.
2Tiger
i don't have too much time right now (i'm building a test project), but if i understand you, if the exit pipe is the same height as equals the lift (different topic) but only half the volume, it can stop the force twice its size? sounds like OU in itself since i don't see any leverage.
tbird
@2tiger,
you always calculate the case, where the shuttle goes down to its
own weight.
We are discussing always the other case over here,
where the shuttle itsself has no weight ( only a few grams for the plastic case that can be
overlooked) and pushes the whole water column above it up due to its buoyance forces.
Please only calculate for this case.
Also you use only the pressure forumula P= F/A for calculation.
If you have a nozzle, then you need to calculate in water streaming and
water velocities in it and before it , etc.. and also the exit profile of the nozzle counts in,
only then you can calculate how high the water will rise.
So please calculate to have a good nozzle directly at sealevel= 0 Meter
and have a shuttle weight= 0 Kg and have for instance a shuttle volume
of 10 Liter = 10 Kg and have an area of 100 cm^2 surface area of the shuttle
at its top and also the same at the bottom and have the
shuttle started at 10 Meters deepth to go up.
So the shuttle is 1 Meter high and has 100 Liters of water above it.
the main pipe will then have a diameter of 11,28 cm that is about 4,429 inches.
So how high will the water rise up after the nozzle output ?
Thanks.
Regards, Stefan.
P.S: With the calculation of the hydrostatic pressure paradoxon
to my opinion, if you take any diameter 1 Meter high exit tube instead
of the nozzle ,then the shuttle would be in equilibrium forces and would not
rise and stay in 10 Meters deepth.
Quote from: tbird on August 31, 2006, 09:20:15 AM
why do you think this will make the water go higher? will you attach a pump to the nozzle?
Well, maybe if will use a small diameter exit tube instead of a nozzle,
we might try to evacuate the upper water reservoir to 1/2 bar pressure,
before starting the experiment and see, if this needs less energy, than we
gain by sucking this way the water up maybe also a few meter more ?
I have to calculate this with the gas pressure laws and see, if this will
bring in any gain ?
Well, I had another thought onto how to use the Hydrostatic paradoxon
not negatively as in the Gravity Mill but positively.
Well, if you look again at the Cartesian diver:
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2F2%2F2b%2FCartesischer_Taucher_animiert.gif%2F180px-Cartesischer_Taucher_animiert.gif&hash=7e92e226e94af6ef99d945923467919f72ceada5)
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fen%2Fthumb%2Fc%2Fcf%2FCartdivr.jpg%2F180px-Cartdivr.jpg&hash=9ba35fec0805acfcff62b6ef51e9530bbdf5002b)
Then you can see, that the upper opening of the bottle,
where the blue cap is, is pretty small.
If we need now to pressure the bottle to sink the diver
we could just place a weight onto the opening of the bottle at the water surface.
The smaller the opening of the bottle is, the smaller must be the weight to
get the same pressure inside the water as:
P= F / A and
F= mass x 9,81
So if we make the opening of the bottle very small, say only 0.1 cm^2 we can use
a small weight to get a huge water pressure inside the water bottle.
So if we want to sink and rise the diver, which is connected to a generator for output
generation, we only need to lift a small weight up and down a few Millimeters
and thus do for the input only a very small lifting work of the weight that puts
pressure onto the water.
With the Cartesian diver we just convert a STATIC Weight force
into a moving work force=
usefull energy-work output !
What do you think ?
Regards, Stefan,
hi guys,
in for lunch and had to ask. did you guys lose sight of the fact the elsa is just a pump?
Quote from: hartiberlin on August 31, 2006, 12:28:17 PM
With the Cartesian diver we just convert a STATIC Weight force
into a moving work force=
usefull energy-work output !
What do you think ?
Regards, Stefan,
Isn't the deep the diver can sink in relation to the pressure you apply on the top of the bottle?
I think this will net out.
Andi.
Quote[If I take the compressed shuttle to the garage the tire gauge will read 3.41 kg/cm2 or 3.41 bar or 49.4578686 psi (according to google's wonderful convert function.)
/quote]
hi prajna, i'm a bit concerned about continuing to try to explain to you why you are still doing this wrong. i'm worried i will offend you and you'll be put off the project all together. my hope is you will figure out on your own so i won't have to sound negative any more. i would suggest you figure this same example using guage pressure and see if it comes out the same pressure. if it doesn't, you'll know something is wrong. if you can get it to equal, explain to me.
QuoteDid you figure the suare cm by taking half the diameter and multiplying that by pi or by multiplying it by itself? I hope the former since
we are dealing with the area of a circle rather than the area of a square.
i believe i did that part right. our answers were within 10 square cm (i used rounded pi). the numbers are; 25 x 25 x 3.14 = 1962.5 square cm. this then would be mulitied by 1kg (i think that's what you say 1 square cm has at sea level) times 3.41, the number of bar you need at 10m (just for arguments sakes we'll say it's right). this gives us this; 1962.5 x 1 x 3.41 = 6692.125kg. your answer was 1731.2 kg. did i miss something because of the metric system?
if i know your heart is in the right place (i think yours is), you can be pretty ruff with me. i do have thick skin when i know you are trying to help me and not just be mean. if anything in that last post didn't set right, don't worry about telling me.
if we keep up the struggle, we'll get there.
all parts for test are made. just have to wait on the glue to cure and clean them up. hopefully test tomorrow pm.
tbird
QuoteQuote
Did you figure the suare cm by taking half the diameter and multiplying that by pi or by multiplying it by itself? I hope the former since
we are dealing with the area of a circle rather than the area of a square.
i believe i did that part right. our answers were within 10 square cm (i used rounded pi). the numbers are; 25 x 25 x 3.14 = 1962.5 square cm. this then would be mulitied by 1kg (i think that's what you say 1 square cm has at sea level) times 3.41, the number of bar you need at 10m (just for arguments sakes we'll say it's right). this gives us this; 1962.5 x 1 x 3.41 = 6692.125kg. your answer was 1731.2 kg. did i miss something because of the metric system?
hi prajna, just had a light go on in my head. maybe we both are right. when i did the calc, i used the size of the shuttle with only 5cm height. this only requires a compression stroke of 2.5cm. you use external compressing and feed it to the shuttle. the size of the compression container could have a smaller compression area, but would need a longer stroke. is that right? it's been a long time since i actually worked with air compressors and the like.
tbird
tbird,
Quotei'm worried i will offend you and you'll be put off the project all together.
No chance. I'll be put off the project only when it is proved to my satisfaction that it doesn't work. Today I proved to my satisfaction that it does work. I have posted the proof at http://declarepeace.org.uk/elsa/ if you would care to check my work. It does tell me that you need both strokes, up and down, and that you need to use a leverage to get it to work. I haven't used any pressure units except kg/cm
2. :) Sorry it is in metric but all the units are easy and I will add an example in imperial as soon as I get a chance.
I will have a think about calculating using gauge pressure and the light may dawn. Thanks. Oh, and my calculator has been set on hexidecimal rather than decimal for days too! Duh! :)
Quotesize of the compression container could have a smaller compression area, but would need a longer stroke.
Yes, absolutely. It is like leverages: if you have a 2:1 ratio you apply twice the force by moving half the weight over twice the distance.
hi prajna,
you silver tongued orator!! great job!! this is what Mr. Herring has needed from day one. to put all that on paper so anyone with half a brain can understand it is nothing short of genius!! great job! great job! if it doesn't fly now, no one really wants free energy.
I DECLARE YOU KING OF THIS THREAD!
how may i serve you?
tbird
phew!
Quotehow may i serve you?
well actually, how about converting it to imperial for me. You don't need all the blurb, just 1' x 1' x 3' shuttle, 32.808399' pumping cylinder etc. i.e. just choose some sensible sizes in imperial units that work. Then I could add it all to the page and others who are not blessed with a familiarity with metric would have an intuitive approach to understanding it.
I am about to completely rewrite ELSACALC based on a double-acting pump with leverage. And sure, having worked out the intro page I have seen that your suggestion re choosing just the shuttle size and pumping depth is sensible. They are certainly the first things that one would need to consider.
Thanks for the accolades. I kinda feel that if Mr Herring had presented such a page to the patent office (even the US one) they could hardly have failed to grant him a patent, regardless of their position on perpetual motion.
Quotewell actually, how about converting it to imperial for me.
you mean any place i find a metric number, convert it (adjust size as logical) to imp and label it? be happy to.
QuoteI am about to completely rewrite ELSACALC based on a double-acting pump with leverage. And sure, having worked out the intro page I have seen that your suggestion re choosing just the shuttle size and pumping depth is sensible. They are certainly the first things that one would need to consider.
if it's anywhere near what you have alredy set up, it'll be wonderful.
QuoteI kinda feel that if Mr Herring had presented such a page to the patent office (even the US one) they could hardly have failed to grant him a patent, regardless of their position on perpetual motion.
i think when he gets back to his internet and reads your essay, his heart will be lifted.
i better get busy, unless you want to chat about something.
tbird
Quotethink when he gets back to his internet and reads your essay, his heart will be lifted.
Nothing I like better than lifting hearts.
Quotei better get busy, unless you want to chat about something.
Please do. Me too.
Hi Pranja,
wouldn?t your new 10 x 10 cm shuttle not pump 98 Liters when it goes down ?
Why do you say 90 Liters ?
When you set it into the water in compressed state it has 1 Liter volume and
if it goes down and decompresses, it has 1,5 Liter displacement,
so it would lift down 99 Liters and lift up 98,5 or 95,5, dending on how deep you let it go.
(if you use the upper surface or the lower surface of the shuttle to count the deepth)
Also you assume, that you can press with just 0,05 Kg of weight water through a small
pipe to 40 cm over the sealevel , when the shuttle goes down ?
This is not very realistic.
Also it depends if the hydrostatic paradoxon can really be neglected in this
experiment.
If your shuttle is only 30 cm high at the top, I think it can only press the
water 30 cm high in the header tube and when it is at 10 Meters, it can only
press the water 15 cm high inside the header tube due to the
hydrostatic paradoxon.?
So let?s see, what TBird will report with his setup.
Maybe TBird you can already post a picture of your setup,
not yet in water,how it looks like, if you still need to wait until the glue dries ?
Many thanks.
QuoteouldnÃ,´t your new 10 x 10 cm shuttle not pump 98 Liters when it goes down ?
Think about it for a bit longer, Stefan. 1. If the shuttle is in the tube then it occupies at least 1 litre of the space in the tube. 2. The shuttle is always in the tube. 3. When the shuttle reaches the bottom of the tube it is still compressed and only occupies 1 litre. 4. When the shuttle decompresses there are exactly 99 litres above it.
Ahh, you have pointed out one error I have made though; it does pump 99 litres in each direction and not 90 (my brain was a bit slow.) I'll correct the page.
With all due respect, Stefan, the hydrostatic paradox is just confusing you. Just let it go because it is not important in understanding this system and has already been taken into account. It tells us that the water level is equal across containers that are connected. You really do not need to be concerned that any aspect of it will change the calculations I have included in this example. The height of the shuttle has absolutely no bearing on the height to which it will lift water in this system. The ONLY things that matter in this respect are the displacement, the weight and the cross sectional area of the header tube. Sorry, but that is it.
Pranja,
I hope it is like you say, so lets see, what the experiment of
TBird will show.
I will now try to calculate a good version of the Cartesian diver.
Hi prajna
QuoteThe height of the shuttle has absolutely no bearing on the height to which it will lift water in this system. The ONLY things that matter in this respect are the displacement, the weight and the cross sectional area of the header tube. Sorry, but that is it.
Well when the height of the shuttle has absolutely no bearing on the height to which it will lift water in this system, how could it be the displacement? Displacement caused by what? By the shuttle perhaps? What weight? Of the displaced water?
The only thing I see here that can displace water is the shuttle. As the desplacement is a product of height x area than the height of the shuttle is directly in ralation to the desplacement and also the "weight" (if you mean the displaced water!).
So for me it is hard to understand this sentence.
Please reply.
2Tiger
Quote from: 2tiger on September 01, 2006, 03:19:19 AM
Hi prajna
QuoteThe height of the shuttle has absolutely no bearing on the height to which it will lift water in this system. The ONLY things that matter in this respect are the displacement, the weight and the cross sectional area of the header tube. Sorry, but that is it.
Well when the height of the shuttle has absolutely no bearing on the height to which it will lift water in this system, how could it be the displacement? Displacement caused by what? By the shuttle perhaps? What weight? Of the displaced water?
The only thing I see here that can displace water is the shuttle. As the desplacement is a product of height x area than the height of the shuttle is directly in ralation to the desplacement and also the "weight" (if you mean the displaced water!).
So for me it is hard to understand this sentence.
Please reply.
2Tiger
displacement could be 10cm by 10cm by 100cm or 10cm by 100cm by 10cm. obvious the height is different, but displacement the same.
tbird
QuoteAs the desplacement is a product of height x area than the height of the shuttle is directly in ralation to the desplacement and also the "weight" (if you mean the displaced water!).
2tiger, that statement also applies to area.
@TBird, any news about your experiment ?
Did you try to setit already up ?
Does anybody interested in gravity and water buoyance have a comment to
my other thread ?
See:
http://www.overunity.com/index.php/topic,1469.0.html
Thanks.
I'm too busy with this to check your calculations there, Stefan. My gut feeling is that the physics will be much the same with a diver system as with ELSA. Parts of it may be easier to impliment and other parts more difficult. We shal have to see as it develops.
By the way, I have updated my ELSA page. It seems to produce more energy than I had anticipated.
http://declarepeace.org.uk/elsa/ (http://declarepeace.org.uk/elsa/)
Quote@TBird, any news about your experiment ?
Did you try to setit already up ?
hi stefan, i have been working all afternoon (just quit) on the test. sure am glad i don't have to work for a living. nobody in their right mind would pay me for the work i have done on the setup. to prove my point, piston i made was too big to fit the pipe. i couldn't believe it. it's plastic pipe, so sanding it down to fit is very slow. as soon as a little friction (sanding) builds up, the pipe gets soft and the sandpaper gets stuck on it. not all is my fault. the pipe is not perfect either. looks good, but with a close fit like i have, the imperfections show up. if tomorrow morning things don't look up, i'll build a different design (try to incorporate styrofoam). nobody hates this delay more than me, but i'm not giving up.
tbird
Quote from: prajna on September 01, 2006, 06:31:26 PM
I'm too busy with this to check your calculations there, Stefan. My gut feeling is that the physics will be much the same with a diver system as with ELSA. Parts of it may be easier to impliment and other parts more difficult. We shal have to see as it develops.
By the way, I have updated my ELSA page. It seems to produce more energy than I had anticipated.
http://declarepeace.org.uk/elsa/ (http://declarepeace.org.uk/elsa/)
Hi Pranja, good work,
but the problem I see is, if we really get the water 2 Meters up.
If you look at your old drawing with the header pipe above the main tube
you can see, that the upwards forces are "compensated" at the top of the main tube
and there only push against the main tube walls !
Only the small area of the exit header pipe will get the upwards
pressure of the lifting water...
so it is clear that the hydrostatic paradoxon is at work and
we can only lift the water as high in the exit header tube as high
as the shuttle itsself is high...
So you will never get 2 Meters of height !
ONLY IF YOU MAKE THE MAIN TUBE STATIC !
But what happens, if you close the main tube and make
the main tube able to move upwards together with the shuttle ?
Then we are are to move all water upwards over the shuttle I guess
as the sealevel airpressure can not come into the main tube and
we are able to lift it all up until we compensate the air pressure
with the water weight.
Make an experiment and use a cylinder that is closed at the top and
fill it with water and then turn it by 180 degrees and put in in a big water bassin,
so that no air will come into the cylinder.
Now how far can you pulk up the cylinder, until all the water will flow
out of the cylinder ?
If you do this right, it could probably be a few meters and you could
design the cylinder this way, that if you open the top,
most of the water could then flow out into
a higher reservoir, if you let air again come into the lifted cylinder.
Maybe this is a better method to get the water up ?
Regards, Stefan.
Hmm, I just tried a 180 degrees turned 1,5 Liter coke bottle full of water
to put this way upside down into a water reservoir.
No air inside the coke bottle.
If you pull out the bottle up to the height, that the opening
is still a bit under water and no air comes into the bottle,
then you have about 1,5 Lieter of water above the water surface,
but the bottle weights about 1,5 Kg then...
So our shuttle would only push so much weight up, what volume
it displaces, so that does also not work.
So, it the shuttle would be 1 Meter high and its own
weight would be almost zero then we could just push the whole
main tube also only 1 Meter high above sealevel. So that is also no option...
Hmm, I just had a good idea, how to explain the negative impact of the hydrostatic
paradoxon in our case.
The whole ELSA setup works as a hydraulic press in reverse !
In a closed water container the pressure onto the walls is
all over case the same !
So that is also the case in the hydraulic press.
There we have for the pressure in every point onto the walls,
if we use the formula:
P1 = F1 / A1 = P2 = F2 / A2 so we can write:
F1/A1 = F2/A2
which means:
Force1 / Area1 = Force2 / Area2
So if we divide both sides by Area2 we get:
(Force1 x Area2) / Area1 = Force 2
Now we use in the ELSA main tube for instance a 1m^2
surface shuttle, that generates for instance 100 Newton upwards force.
Now we want to look, what force it generates onto the
1 cm^2 upper exit header pipe area.
So we have:
Area1=10000cm^2 = 1 m^2
Area2= 1 cm^2
Force1= 100 Newton
So if we put this in the upper forumula we have:
Force2= 100 Newton x 1 cm^2 / 10000 cm^2= 1/100 Newton.
So we have only a force of 1/100 Newton acting onto our upper
surface of the exit tube.
As F = mass x g
we get
mass= F / g= 1 /100 Netwon x 9,81= 0,0981 Kg
we can therefore only put 0,0981 Liter of water into the exit header tube
to compensate this force, If we put more water into the exit tube,
the shuttle will be pressed down again !
So you see, this is a very small amount of water, that can be lifted
and our ELSA system works backwards as a hydraulic press unfortunately
and the hydrostatic paradoxon is at full work here... too bad...
The only thing, which might still work is to calculate,
how much energy one would need to evacuate the exit tube and
the upper reservoir and see how much energy one would need to do this to pump
the air out there to around 0,5 bar pressure only and thus suck up all the water
via the pressure difference. Maybe one needs less pumping energy to evacuate the
upper reservoir than one gains in potential water energy ?
Then we maybe also need no shuttle and just can use the always available air outdoor
pressure of 1 bar to do the work all in all...
Regards, Stefan.
P.S. As 0,0981 Liter that is about 0,1 Liter of water
has about 100 cm^3 of volume, in a 1cm^2 area exit
header tube we would have a height then of:
100 cm^3 / 1 cm^2 = 100 cm = 1 Meter.
Hmm, this is quite a lot as I just see,
now it depends on how big the shuttle must be
at 100 Newton positive buoyancy force.
Regards, Stefan.
Stefan,
Head height = ((max displacement - weight) + (weight - min displacement)) / 2. Let's put some figures in and see if this works: max displacement = 1.5 litres, min displacement = 1 litre, weight = 1.25kg so at the top the pressure on the header tube = (1.25 - 1) = 0.25kg/cm2 and at the bottom the pressure on the header tube = (1.5 - 1.25) = 0.25kg/cm2. To get the average pressure we calculate (0.25 + 0.25) / 2 = 0.25kg/cm2.
We can ignore the pressure at the sides of the tube because both the tube and the shuttle are rigid (if not then we have a problem because the shuttle will stick or water will leak around it).
The hydrostatic paradox has bitten you again: In calculating the static head we need account for only the pressure on top of the shuttle and underneath the shuttle. Indeed this is precisely what the hydrostatic paradox is about. We need only take account of a column of water.
The height of the shuttle has absolutely no effect on the head because the shuttle is always submerged. If we are talking about floating bodies then the aspect ratio might come into it (haven't given it a great deal of thought, probably it doesn't even effect it then.) The shuttle can be 30cm x 10cm x 10cm or 10cm x 10cm x 30cm but so long as it is submerged then it will always displace exactly displacement - weight of water. If you think you know better then think better. I believe you are getting confused by this paradox thing because once a body comes to the surface it must support the weight of any part that is above the surface. That doesn't come into play in our system. That is why I said in a much earlier message "The hydrostatic paradox doesn't apply in this system", or at least it doesn't as far as the floating bodies part is concerned and that is the part which is causing you so much confusion.
hi all,
i'm not sure how to start, so i'll just get right to the answer you've all been waiting for.
it doesn't work. i was wrong and i owe stefan and 2tiger, in particular, a huge apology. I'M SORRY!! i feel like the kind of guy i was fussing about from time to time. making definitive statements without having the definitive science to back me up. in my heart, i believed it. i wasn't trying to just be disruptive.
the definitive answer comes from the formulas pertaining to pumps, head pressure in particular.
http://www.engineeringtoolbox.com/pump-head-pressure-d_663.html
this site gives all the needed formulas to prove the point. not to mention my test that failed. as a result i came back to the computer to find the truth. armed with these formulas and a bit of time, i saw the light. to me, the small exit pipe (when filled with the amout of pressure [water] it is exposed to) serves as a cork to what is otherwise already closed.
true to my word, this will be my last post, unless popular opinion thinks i should continue.
good luck to all!! :'(
tbird
Quote from: tbird on September 02, 2006, 02:51:15 PM
hi all,
i'm not sure how to start, so i'll just get right to the answer you've all been waiting for.
it doesn't work. i was wrong and i owe stefan and 2tiger, in particular, a huge apology. I'M SORRY!! i feel like the kind of guy i was fussing about from time to time. making definitive statements without having the definitive science to back me up. in my heart, i believed it. i wasn't trying to just be disruptive.
the definitive answer comes from the formulas pertaining to pumps, head pressure in particular.
http://www.engineeringtoolbox.com/pump-head-pressure-d_663.html
this site gives all the needed formulas to prove the point. not to mention my test that failed. as a result i came back to the computer to find the truth. armed with these formulas and a bit of time, i saw the light. to me, the small exit pipe (when filled with the amout of pressure [water] it is exposed to) serves as a cork to what is otherwise already closed.
true to my word, this will be my last post, unless popular opinion thinks i should continue.
good luck to all!! :'(
tbird
:'(
I concur with tbird. I have finally discovered the error in my calculations that made it appear to work. If you would like the difinitive answer then you can see it at http://declarepeace.org.uk/elsa/ (http://declarepeace.org.uk/elsa/). Make sure that you read all the way down to the end of the main text (past the output power calculations). I declare ELSA a non-runner, unless anybody would like to find fault in my detailed analysis.
Thanks Stefan, Prajna, tbird, for all your efforts and time.
I was going to order the clear pvc pipe and fittings to start my own experiments on Friday, but thankfully I was delayed and now you all convinced me this is just another pipe dream (pun).
So we must assume that Mr. Herring never had a working model, and he is just another talker. He never toured with a working unit either. No wonder we never heard of him before. Sadly I don't understand why people do this again and again to try and fool us all.
Again thank you all. The logic, calculations, and discussion were very interesting. You saved me the money I was about to outlay on the piping too.
Good luck everyone.
I think Mr Herring was only trying to fool us because he had fooled himself. It is very easy to do. I sure managed to fool myself several times into thinking this would work; with tiny errors in my calculations proving first that it would and then that it wouldn't and then that it would again. Unless you have another pair of eyes to check your work it is very easy to miss small mistakes in logic or a decimal point in the wrong place.
No problem; roll on the next project.
Hi Tbird
As I lost my first girlfriend, my mother said to me:
"Don?t cry because it?s over, better smile, thinking of the good time you have had together."
Well this is not quite the same, but it will help you to go on with your researches.
Thanks a lot for the good discussion.
Bye
2Tiger
Hi guys, I've just come up with a way around the problem with ELSA. Take a look at http://www.DeclarePeace.org.uk/elsa/buoyeng.htm (http://www.declarepeace.org.uk/elsa/buoyeng.htm) and let me have your comments.
Prajna,
Now that looks very interesting. I need to admit that I had not given up on the ELSA idea, and was brainstorming on Stephan's Cartesian diver variation. I wanted to simplify the shuttle and get rid of the compression problem but I just couldn't figure a way to get the shuttle to descend. I love the way you put in the pressure piston idea. I could see using a the water collected above to drive a variation on leverage to get enough pressure to press on the piston.
Nice innovation, Elsa may not be dead yet.
QuoteI could see using a the water collected above to drive a variation on leverage to get enough pressure to press on the piston
That's the plan.
The whole thing is significantly simpler, in engineering terms, than ELSA.
Just noticed that this new design reduces the quantity of water pumped on each cycle. So it won't work. Oh well, back to the drawing board.
Hi Pranja,
yes, we should use the massive volume the ELSA was able to
lift above the shuttle, but just use it in a different way, like
pumping it still under water into a pipe, where it turns a generator
or would wind up a spring or something else, so one could later recompress
the ELSA shuttle.
As long as you keep the water below the sealevel, it still has great
potential to lift really a lot of water above the shuttle...
So this must just be used differently I guess.
Let?s see, if you turn a gear via a pump at sealevel and
gear it up, so you can lift an external weight, how much
weight can be lifted how high ?
Regards, Stefan.
Stefan,
Now that sounds like great "out of the box" type thinking. Whoever said the water had to be pumped above sea-level? The moving water needs to be harnessed for energy. Enough to compress the shuttle, or a variation of compression in Prajna's cartesian diver type design. The compression mechanism would need to be above sea level but not necessarily the energy collecting one.
Prajna's idea of the magnetic piston, cartesian-diver shuttle may still have potential. It does simplify the shuttle design.
here is an idea !
We never have used hydrogens lightweight and hardcompresibility properties.
I suggest shuttle that is when went is opened fills up with water and goes down to bottom of
device.
At the bottom of device (or some natural area... sea,lake) there is effifcient (only 80%) electrolyzer who will fill the shuttle
with lightweight H2O2 gass. Shuttle than climbes up and when reaches top point of device we take H2O2.
So energy should be extracted from
heavy shuttle going down (metal shuttle filled with watter),
light shuttle (filled with H2O2) going up,
and H2O2 extraction from shuttle when it reaches the top.
wizkycho
@ Stefan and ResinRat2,
There is a difference between pumping volume and pumping pressure. In the ELSA example we have great pumping volume and tiny pumping pressure. In the Cartesian Pump example the situation is reversed. The head height is simply a measure of pumping pressure. You can think of it as an analogy of electricity. In the first example we have large amperage and low voltage, in the second we have big volts and low amps.
@ wizkycho,
I have added calculations of the total power generated in each cycle and it is disappointingly low compared to the energy required for recompression. I don't know the energy that would be required for electrolysis but I imagine it would be greater than the energy generated by the cycle. If you could calculate the energy required to produce 3 litres of H2O2 and the energy contained in that H2O2 then we might be able to evaluate your suggestion.
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fwww.besslerwheel.com%2Fforum%2Ffiles%2Fmt108.jpg&hash=a31d9cfffafbfd16773ac57fb11478fc813fd339)
Quote from: prajna on September 06, 2006, 04:42:48 AM
@ Stefan and ResinRat2,
There is a difference between pumping volume and pumping pressure. In the ELSA example we have great pumping volume and tiny pumping pressure. In the Cartesian Pump example the situation is reversed. The head height is simply a measure of pumping pressure. You can think of it as an analogy of electricity. In the first example we have large amperage and low voltage, in the second we have big volts and low amps.
Hi Pranja,
we could also have a SMALLER exit header tube going sideways under the water already.
So then we will have no problem to pump all the water through this small exit header tube and
thus have there also a lot of pressure inside this smaller diameter pipe.
This way we can avoid the hydrostatic paradoxon.
We can do this easily
by having the exit header tube turn sideways under the sealevel already, so
the buoyancy force of the shuttle can always press the whole water above it
into the small pipe, which drives a water pump or a syringe like piston inside the exit tube.
If we go with a syringe like piston, we can have a look at it,
with how much force it moves and then calculate via multiplication
of the distance s it moves the output energy W= F x s .
Now if we pull up with this syringe piston a weight,
we could use this later to recompress the shuttle
again.
Will this work ?
Stefan,
At the end of my Cartesian Pump (http://www.DeclarePeace.org.uk/elsa/buoyeng.htm (http://www.declarepeace.org.uk/elsa/buoyeng.htm)) page I have calculated the energy required to recompress the shuttle and the energy generated by the sinking/buoyancy pumping cycle. You can ignore all of the mechanics of how the pumping energy is used and just compare it to the recompression energy requirement. It doesn't come close. So it makes no difference what scheme you come up with, it will never produce enough energy to recompress the shuttle (or compression chamber or whatever else you choose to use.)
I have even added a calculator that lets you change various parameters and then calculates the results (though I have yet to add a recompression section to it.)
Hi Pranja,
for what case is the calculation there ?
I think,
if we stick with the original ELSA,
but just move the water inside a pipe sideways at seawaterlevel and
never go over seawaterlevel with it and
have a syringe piston in the exit tube, so the
piston will pull up a weight, which then
could recompress via leverage the shuttle,
it will indeed work.
When I have more time I will calculate it all through
with an example.
We just have to avoid the hydrostatic paradoxon only
and thus pump the water sideways out into an exit pipe
below seawaterlevel.
Then we will benefit from all the water that is pumped up
via the shuttle.
Stefan,
The calculation is for any case - up, down or sideways. The energy generated by a weight and that generated by buoyancy added together is possible to calculate. That is what I did. This is the only energy that can be generated by the system regardless of whether it is an ELSA configuration or a Cartesian Pump configuration or even some strange configuration you might invent some time in the future. This is the only energy in the system regardless of how you store or use it, whether you use it to pump a large volume of water a small distance or a small volume a large distance. Even using some form of leverage it will not be enough to recompress the shuttle.
Pranja,
maybe you can clarify, what your value
l to r mean ?
How did you calculate them and
why do you give as an energy the pressure
unit Pascal, which is indeed no energy unit.
Pascal is Newton / Meter and Energy is
Newton x Meters !
I don?t understand your calculations there.
Thanks.
Sorry Stefan, typo. I meant joules. I'll fix it. The letters used in the formulae refer to the input values and result values, which have letters next to them: so 'l' represents negative buoyancy and 'r' represents total energy, 'a' would represent shuttle diameter, etc.
Okay All,
now let me recalculate my old example and see if it works
with a horinzontal exit pipe at sealevel :
1. We have a shuttle, that has 1 Meter diameter and
an water column of 10 meters.
The shuttle is 10 cm high and 1 meter diameter,
so it has 78,5 Liter volume and starts at the top at seawaterlevel.
The shuttle is precompressed to 4 Bar !
Now to sink this shuttle we attach a weight of 80 Kg to it.
Now the shuttle together with the weight sinks down.
2. Now at 10 Meter deepth we let the shuttle expand itsself.
From the 4 Bar it had now at 10 Meter deepth there are also
2 bar pressure there due to the water pressure.
So the shuttle expands its volume now to double its size, so
to 157 Liter and being at 2 bar inside, as inside pressure = outside pressure= 2 bar.
Now 157 Liter means 157 Kg buoyant pressure minus the 80 Kg weight is
now 77 Kg or F= m x g = 77 Kg x 9,81 = 755,31 Newton upwards force.
This force now pushes the volume of 7854 Liter of water above the shuttle
through a 90 degrees right turn output pipe,
where a piston is located in it
so as the water is moved up, at the top at seawaterlevel ( 0 meters)
the water goes into a horizontal pipe which has a smaller diameter.
Inside the exit pipe is a piston that is pushed out of the pipe to the side
at seawaterlevel and
the piston is pulling a string(thread) which will
lift a weight up over seawaterlevel.
Now let us calculate, with which force a weight could be lifted this way:
Let us just make the exit pipe area just halfed in diameter, that means diameter= 50 cm,
which would mean a 4 times smaller surface area and a 4 times longer pipe to hold
all the 7854 Liter of water which is pumped into it. So the exit pipe
will be 40 meters long horizontally at seawaterlevel.
As we have now the condition of a hydraulic press, where
F1/F2= A1/A2 and with A2= 4 x A1
we find that
F1= A1 x F2 / A2 = 1/4 x F1= 755,31 Newton / 4 = 188,83 Newton
So we get out of it a force of 188,83 Newton over a distance of 40 Meters
which is an energy of around 7553 Newtonmeters= 7553 Wattseconds / 3600= 2,1 Watthours.
So you see, this is the same energy you would get, if you multiply the
force x distance from our water column the shuttle itsself lifts,
so Energy= 755,31 Newton x 10 Meter = 7553 Wattseconds / 3600= 2,1 Watthours.
3. Now at the top we must recompress the shuttle from now internally at 2 bar and
157 Liter volume ( 20 cm high at 1 Meter diameter)
back to 10 cm and 1 Meter diameter and 78,5 Liter volume at 4 bar.
This can be done again by using a pumping action, this time from the outside.
Now the formula for this is:
W= (P1-1 bar) x V1 x ln (V2 / V1)
So the energy W needed for this pumping is:
(200000 Pa - 100000Pa) x 0,157 m^3 x ln 0,5=-10882 Wattseconds / 3600= 3 Watthours !
So you see, that we need 3 Watthours to compress the shuttle, but only
get about 2.1 Wattshours from our output pipe piston.
So this is only 70 % efficient, so way underunity.
Well, it seems this does not work, as also other
dimensions have the same problems.
So with this I put the ELSA principle to rest.
I will still try to see, if I can get the Cartesian
diver principle to work somehow.
Regards, Stefan.
you know i think instead of compressing/recompressing the shuttle maybe its better to just lift some kind of small weight that attaches itself (do to some kind of locking mechanism) to the shuttle at the water surface to make the shuttle sink, then once down it dettaches itself (do to some kind of unlocking mechanism) from the weight.
then the process repeats itself over and over again :)
i think this has been said im not sure.
the water that the shuttle pushs up and saves to a container is the energy needed to lift the weith to the shuttle so it will sink.
Here's an idea I have been thinking about for a few days now and cannot see why it cannot work, I am not even sure where the additional energy comes from:
You have a cylinder 1m diameter, 2m high with a piston weighing 10,000Kg.
The cylinder weighs 1000Kg.
They are connected to a pulley and counterweight of 10,500Kg.
With the top of the piston sitting level with the water surface the piston is 2m below the surface.
The piston is allowed to sink onto the piston expelling air via a tube at the surface.
The whole assembly is allowed to sink to 9m (bottom of cylinder).
Then the piston is released and the weight pulls in air from the surface and overcomes the pressure at 10m depth by its own weight.
The whole assembly is now bouyant and can travel back to the surface with the force of 2,000Kg - difference in counterweight = 1500Kg.
When the piston and weight reaches the surface it can start the process over again.
I have racked my brains, but I cannot see where the flaw in this is.
Maybe I missed something.
The only big loss that can occur is if you have to move the very heavy piston upwards using force, but I have avoided this and still managed to get air in and out of the piston at the required points.
There are various locks required to hold parts in place during the piston move stages.
If I cannot find out where the flaw lays then I may end up building a small model to prove the idea.
Regards
Rob
Kingrs,
Sorry, but I am having a hard time picturing what you are describing. Perhaps you could post some type of drawing showing what you have in mind. A little detail on what you mean by a counterweight/pulley system.
We appreciate your input.
Hi Resinrat,
As with everything, you start to draw it and then suddenly it hits you.
I got to step 4 and realized that it cannot work.
At step 4 you need to release the main piston weight which is energy you have lost and cannot get back during the ascent back to the surface. So what will happen is the cable for the counter weight will need to get longer each time. Shame, I thought it was too good to be true, that rules out any gravity mill I think.
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fi100.photobucket.com%2Falbums%2Fm25%2Fkingrs%2Fgrav1.jpg&hash=a571fab28e79ac6b8530c755c02c883ede43a0e5)
Regards
Rob
kingrs,
Thanks for going through the trouble of drawing it up anyway, even though it is a failure.
One thing I learned in Research: all failures need to be recorded and analyzed WHY they failed. That way we all know what NOT to do.
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2F2%2F2b%2FCartesischer_Taucher_animiert.gif%2F180px-Cartesischer_Taucher_animiert.gif&hash=7e92e226e94af6ef99d945923467919f72ceada5)
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fen%2Fthumb%2Fc%2Fcf%2FCartdivr.jpg%2F180px-Cartdivr.jpg&hash=9ba35fec0805acfcff62b6ef51e9530bbdf5002b)
can't this be used in a geothermal way? doesn't temperature have to do with pressure?
maybe you can have a tube going from the inside of your house's basement to the outside roof. i dont know i could be wrong.
This is a bit off topic to the Gravity mill, but worth looking at.
Ocean Gliders have been around for a while. They work basicly by compressing air and sinking, then decompress the air and float.
here is a link
http://spray.ucsd.edu/
There are lots of variations to this model and some are designed to stay in the ocean indefinately.
They are powered by batteries and recharged using different methods including heat diferentiation, drag generator, etc.
They require a fair amount of energy to compress the air. But they only have to sink or float.
The link above has good diagram of how it works.
Simply the point here is if an object can sink and rise for long enough you can use that force to generate electricity to compress and decompress the air. The sinking and floating are free. Only at the top and bottom of travel do you need to add force.
ar
Quote from: arbus on February 26, 2007, 09:13:34 PM
This is a bit off topic to the Gravity mill, but worth looking at.
Ocean Gliders have been around for a while. They work basicly by compressing air and sinking, then decompress the air and float.
here is a link
http://spray.ucsd.edu/
There are lots of variations to this model and some are designed to stay in the ocean indefinately.
They are powered by batteries and recharged using different methods including heat diferentiation, drag generator, etc.
They require a fair amount of energy to compress the air. But they only have to sink or float.
The link above has good diagram of how it works.
Simply the point here is if an object can sink and rise for long enough you can use that force to generate electricity to compress and decompress the air. The sinking and floating are free. Only at the top and bottom of travel do you need to add force.
ar
i dont know but i can be wrong...the energy that it takes to compress and de-compress the float is greater than the free ride that you get. the sinking and floating is free, but the energy gained is not enough to compress/de-compress.
The original Gravity Mill was a float that was compressed at the top of the stroke and sank to the bottom, displacing water in both directions. The problem is in the force required to collapse the float.
Why not use a compound pulley within the float to compress the float. The pulley would be powered by a water wheel. The release mechanism would be a tension release set to activate when the float was at its maximum weight at the top of the stroke and the cable would freewheel to the bottom of the stroke. When the float fully inflates the tension on the cable will be minimal and the retrieving mechanism would engage and reel the cable to repeat the cycle.
The latching mechanism to keep the float compressed should be inside the diameter of the float cylinder. This is a slight modification of the original release idea, but using the same release idea concept.
William66tell
Quote from: william66tell on February 27, 2007, 11:09:49 PM
The original Gravity Mill was a float that was compressed at the top of the stroke and sank to the bottom, displacing water in both directions. The problem is in the force required to collapse the float.
Why not use a compound pulley within the float to compress the float. The pulley would be powered by a water wheel. The release mechanism would be a tension release set to activate when the float was at its maximum weight at the top of the stroke and the cable would freewheel to the bottom of the stroke. When the float fully inflates the tension on the cable will be minimal and the retrieving mechanism would engage and reel the cable to repeat the cycle.
The latching mechanism to keep the float compressed should be inside the diameter of the float cylinder. This is a slight modification of the original release idea, but using the same release idea concept.
William66tell
could you please draw a picture for us for better understanding?
@free energy
Regarding the ocean glider.
You are right about the energy it takes to compress and decompress the air. and of course the deeper you go the more pressure you have, the more energy needed. Maybe using a gas that liquifies close to ocean temperatures or an already partially compressed gas would work better. i.e. small compression turns if from gas to solid(liquid). Decompression would require only opening a valve(if the pressure inside is greter then outside).
One other thought. is looking at whales. They use Wax in their head to alter their bouyancy. They solidify the wax to sink and heat it up to make it liquid to float. This is not going to make you sink and float very fast but works for whales :P
Ar
The enclosed schematic is showing only the functional design. It has not addressed the float latching or the re-engaging of the clutch.
Once the compression of the float is achieved, the details of the design to latch the float and engage the clutch become minor.
The diameter of the output pipe and the weight of the float is very important as to how high the water will be lifted. One pound of force will lift one square inch of water 27 inches.
william66tell
Here is a thought. If you could bypass the issue of compressing the piston, it would be a lot simpler. I have attached a diagram of my basic idea. It would loop. If you notice the right side has hole to allow water to enter and leave the tube.
The one thing that would make this work, and I'm not sure if it would be available, is a water wheel at the bottom, that is air tight, As the water enters the bottom of the tube, the boyancy of the float pushes the wheel up thus staging the next float, entering the water stream and up the column. At the top of the colum, the float is pushed around the bend (probably by the two or three floats that follow. The float then falls down the tube to the bottom, ready to be re-entered into the water filled tube. This bypasses the issue of air compression all together.
I have no idea if there is any efficient water tight water type wheel that would keep the left tube dry, but I thought it was a relatively novel approach that would warrent posting.
Here is a little something to help understand the situation. It is a way to raise water, but not over-unity.
Would working for several HUNDRED years be perpetual motion? NO, but it is amazing.
http://en.wikipedia.org/wiki/Hydraulic_Ram
http://www.lhup.edu/%7Edsimanek/museum/unwork.htm#buoy4