Hi,
I have made a drawing of an idea I had the other day. If it was possible to shield water in the right part of a container so it can be air in the left side, how would a buoyancy wheel act then?
I know there is harder to push ball E into the water than it is for ball A, due to the difference in pressure. But will the buoyancy of the balls F, G and H (Or more if more balls was available) be enough to force ball E (minus force of ball A) into the water?
Vidar
B, C, D, E will 'resist' going into the water, this falls under the path of least resistance law which will always center itself out and settles into a zero kinetic potential.
If you really wanted a wheel that runs on gravity then you must also understand that Gravity is not a constant around the world, different places have different gravitational influences.
if you built a rather large ordinary wheel that borders the blue gravity area vs. a heavy gravity environment in red then one side of that wheel will be heavier than the other side causing the wheel to perpetually rotate to the path of least resistance.
on the gravity map, in the blue areas objects fall slower than objects dropped in the red area.
32ft per sec squared only counts in the area it was calculated from, it does not hold true entirely around the world, there is a spot in the Indian ocean that has the least over all ratio of gravitational influence.
after you have fumbled with this concept in your head then tell me why it won't work as planned, 'remember the path of least resistance law'
Jerry 8)
Only position E will be the counter energy with an average energy of the balls Dmax x cos45o x DepthE-A
If the spheric balls was 1 litre. The maximum area is 250 cm^2. If DeltaE-A = 1m, the maximum counterforce is approx 250N when ball E is half way in. The average force is 250N x cos45 = 177N for the whole ball to enter the water. The distance Dmax it must travel is 0,178m to be totally in the water. The total energy is then 177N x 0,178m = 31,5.
Then the three other balls F, G, H are travelling the same distance, but ball F and H are at average 45 degrees angle. What is the energy they provide?
If ball F and H are raising each approx: 0,178m x cos45 = 0,125m
In sum this is equivalent to 10N (About 1 litre) in 0,25m.
Ball G is raising approx 0,178m. The weight is 10N.
The total buoyancy energy is then (0,25m x 10N) + (0,178m x 10N) = 4,3
Ball E wins with 31,5 - 4,3 = 27,2 (something)
If we store that exsess energy in a fast spinning metal disc (Like the ones in some toys), should that energy be enough to keep the wheel to rotate in the opposite direction than I intended? ...so the balls at position F, then G, then H could provide energy each time they left position E and into the air?
I guess I'll not get a nice sleep tonight...too much thinking :)
Any thoughts @onthecuttingedge2005?
lets say you weigh 200 pounds, take one soccer ball and one pool, the object of this mission, try to get the soccer ball down to the 12 foot level and touch the bottom of the pool.
don't hold your chin out to far or the ball might smack you a good one!
Sincerely
Jerry ;)
I see your point. One h... of a black eye - but what if the wheel are turning clockwise. Ball at position E is going OUT of the water because the buoyancy of the other balls are not enough to keep it back. The question is also if the remaining balls in the water require as much energy to virtually move the 1 litre water upwards 1m....hmmmmm
I'll go to bed now....
Vidar
when it comes to gravity, you have to convert it, not use it directly.
interrupt it to be more precise.
there can be no work unless you convert it from one form to another. this is another law, the conversion of potential to kinetic energy requires some input energy.
always remember, 'all' energy is inherited as lazy, it doesn't like to do work! you have to force it to work!
think of a really lazy slave, I mean really lazy as lazy of a slave as possible, this would be your driver an electron.
Jerry
"Lazy energy" LOL ;D. Yes, you're right. Most energy is only potential - not willing to change without a push. I was however hoping that I could convert from potential to kinetic energy with my idea. I guess I've to build it. Two containers of water side by side with very flexible membranes towards each other where the balls are going in between - lubricated with some silicone oil or something. Don't hold your breath :)
Vidar
This is my stupid idea for bouyancy wheel :)
I don't think it will works but was fun to think about it :)
Quote from: Silver on December 29, 2009, 12:45:08 PM
This is my stupid idea for bouyancy wheel :)
I don't think it will works but was fun to think about it :)
I liked it! Nice drawing too. This made me think of another and similar idea:
The upper an lower roller is making sure the flexible rubber tube is keeping the air on the right side. The shape of the tube I believe is critical in order to make bouyancy more efficient - but in the end I have a bad feeling it will not work...
This drawing is done in paint with a small red unusable button-mouse - hence the result...
Well, look away from friction - I am only interested in if the bouyancy effect will make the big wheel spin and accelerate if no friction is present.
Vidar
Funny picture :)
Well, cause the movement happens in liquid, if it works, it will be a low rpm system.
We must count the energy needed to pump the air down and the work it can do
Many "what ifs" here, but what if the tube around the wheel is pumped up and have an even shape - except the tube is compressed at the botton due to the higher water pressure. If the tube is pumped up, and the rollers are preventing the air from evenly spread on both sides, shouldn't the tension in the tube surface at the top be greater than in the bottom? This will in that case force the upper roller to move away from the tube. The lower roller is helped by the water pressure to compress the tube and release some of the tension down there, so the roller at the bottom is forced clockwise with less force.
Then if both rollers are fixed, the wheel is forced to turn. As no water is pushed up or nown - generally no mass is moved anywhere - the only force left is the difference betwen the tension in the blown up tube between the top and the bottom.
In the top you will then have a continously repelling system that is greater than the one in the bottom. The bigger the wheel is, the greater this difference will be. The wheel is then turning clockwise.
In an opposite case, the tube can be filled with liquid and let the whole wheel work in air, so there is more pressure at the lower roller than the upper - and the wheel will turn counter clockwise.
as bad drawing as the last one LOL ;D, but I think you get the idea.
What do you think?
Vidar
Well, i think the buoyancy force will affect the whole setup not only the tube and so, no rotating force will appear. Thats why my idea has "near independent" air balloons.
Take a look at this idea for a bouyancy machine.
All you have do to is.
1. push the object down into the tub
2. switch the valves.
surely the work in should be less than the work out.
and all the lift is free.
it works because the water is locked into the tube similar to the locks
in canals that lift ships rather than dropping down the tube and water
is not liftted except for the little bit in the tub.
1. with valves closed push down into tub of water and position under the tube.
2. open bottom valve with upper valve closed.
3. when above the bottom valve and at the top valve close it and open top valve.
and then it pops up to the top.
then you have its weight that can do work.
https://www.youtube.com/watch?v=rxFXsoqbfrk (https://www.youtube.com/watch?v=rxFXsoqbfrk)
Norman
Quote from: norman6538 on July 01, 2015, 09:44:21 AM
Take a look at this idea for a bouyancy machine.
All you have do to is.
1. push the object down into the tub
2. switch the valves.
surely the work in should be less than the work out.
and all the lift is free.
it works because the water is locked into the tube similar to the locks
in canals that lift ships rather than dropping down the tube and water
is not liftted except for the little bit in the tub.
1. with valves closed push down into tub of water and position under the tube.
2. open bottom valve with upper valve closed.
3. when above the bottom valve and at the top valve close it and open top valve.
and then it pops up to the top.
then you have its weight that can do work.
https://www.youtube.com/watch?v=rxFXsoqbfrk (https://www.youtube.com/watch?v=rxFXsoqbfrk)
Norman
Looks nice!
Now you need automatic valves that is activated by the buoyant material. Only one problem; How do the buoyant material jump out of the upper reservoir and down in the lower reservoir so it can repeat the cycle?
Vidar
Thanks Vidar,
I would start by manually switching and measuring the forces to see that there is
OU potential and then go on to making the valves work automatically and then
with the power available I'd kick the block over the edge and use that power to
switch the valves and hopefully have extra left over to use elsewhere.
In some other work I have done that has OU there has to be about 250-300% to
do the switching so that it will self run. So If I used 1 foot lb to push the block
in place then it would have to have lets say 100% more to do the switching and
100% more to kick the block out and 100% more to kick the next block in place
roughly.
Norman
The question I'm struggling with now is if I have a 10 foot tall device and a 20 foot
tall device made the same will the taller one have more power available like a 10 foot dam verses a 20 foot dam of water for power.
Anybody got a bead on that concept?
There have been some other buoyancy devices around in the last few years but
I could not understand the basics enough to believe in them.
Norman
Quote from: Low-Q on July 01, 2015, 01:59:15 PM
Looks nice!
Now you need automatic valves that is activated by the buoyant material. Only one problem; How do the buoyant material jump out of the upper reservoir and down in the lower reservoir so it can repeat the cycle?
Vidar
its at the top of the water, because its floating.
so you only have to move it a small amount, then you have available all of the energy of the mass as it falls back to the bottom.
some kind of pulley system could be generating energy with the mass of several floats, and some of that can be used to pick up the floats out of the water.
that can be automated easily.
have a wheel with hooks on it, that picks them up and transfers them to the pulley belt. then at the bottom, you have some sort of mechanism to reload them into the tube.
switches inside the tube can be activated by the floats to open and close the locks.
If we put the buoyant object beneath the rock, it will float like a dock
channels at the top and bottom can be used to move the rocks to and from the tube
to the loading areas where they enter and exit the pulleys.
assuming all objects are the same physical size::
a 1 Ton rock displaces 3/4Ton of water
the buoyant object (Styrofoam?) weighs 0.01 Ton, but also displaces 3/4 Ton of water
each object has an upwards force from 3/4 ton of water they are displacing (1 & 1/2 Tons of water!!)
so gravity is only affecting the (now) 1/4Ton rock and 20 lbs of Styrofoam which leaves 0.49 Tons for upwards force.
that's obviously too much buoyancy for a practical system, but if you were able to get it into the tube the rock will have no problem raising itself as high as you can build your tube....
Also, with the locks, the water pressure is not the full force of water above,
but only what is in the lowest lock
if the buoyant force and the total mass of rock and float are dissimilar enough only to make it slightly buoyant
the input energy can be minimized and easily compensated for by height of tube.
someone builds this..... may just win that prize :-\
Energy input takes place 4 times::
1) E = mgh1
2) E = mgh2
3) E = dgh3 (d is the displacement)
4) E = energy to control valves in the tube (unknown?)
note that this valve energy will be a constant value based on water pressure and volume of the water system.
This is completely separated from the energy value of the mass cycle. we change the mass and buoyancy, the valve control energy does not change.
Energy out: E = mgh`
Sm0key2, could you detail your formula a little clearer and where each takes place.
ie m, I assume to be mass
h, I assume to be height
d ??? displacement
as I see it there are 6 work inputs
work 1. the object is inserted into the water displacing water and moved to the tube.
work 2. lowest valve is opened
work 3. object rises to valve 2 and valve 1 is closed
work 4. lowest valve is closed
work 5. upper valve is opend
work 6. object kicked off the water to do some work out
Hello everyone, I am from France, I offer my concepte, if you are interested in her by her and chat with me, thank you for the questions http://overunity.com/15847/total-mouvement-autonome-france/msg454472/#new
Quote from: fredenergie on July 02, 2015, 09:10:48 AM
Hello everyone, I am from France, I offer my concepte, if you are interested in her by her and chat with me, thank you for the questions http://overunity.com/15847/total-mouvement-autonome-france/msg454472/#new (http://overunity.com/15847/total-mouvement-autonome-france/msg454472/#new)
why are you spamming ?
sorry no abituer the forum, how I can remove about thank you ???
Quote from: sm0ky2 on July 01, 2015, 08:08:06 PM
someone builds this..... may just win that prize :-\
And the winnwer is: https://www.youtube.com/watch?v=UGPEOlQ2_uA
I think this is very close to your drawings except for the ejection system of the spheres which occurs by the thrust of the other floating spheres, and for the system of reintegration of the sphere in the riser tube, which is covered by the acceleration in the fall . The balls seem to slow the fall as they pass through the square boxes, maybe there are magnets in the ball and in those boxes there are coils that capture the energy and slow down the fall
Just an idea:
Let there be plenty of buoyant objects in that tube. They cannot be so buoyant that they cannot submerge after a free fall from the top of the tube or reservoir. Imagine ice cubes.
A guide will guide them back into the tube in the moment they are submerged and still in motion. A torpedo shaped object will be good for reducing water resistance.
Now, let all the buoyant objects in the tube use their combined buoyancy to push one object at the time out of the tube. While that pushed object leaves the tube, it must fall down, submerge, be guided back into the tube. Now there is enough buoyant objects in the tube to push next one out of the tube.
The valves are in theory an effortless operation. At least one cannot take the energy they require to open and close into consideration. This energy requirement has nothing to do with the buoyant system.
We need only two valves because the tube must maintain the water column all the time.
I have not made my calculus on this one, but nature usually don't give a shit about what we believe or conclude. So what's left is to do the practical experiment.
This far, I am just confused.
EDIT: I did't notice the video link above when I posted this. It appears to be the same idea... ;D
Vidar
Another idea that is similar but don't need buoyancy:
Valved control the water level, and keeps the water to one side.
Displaced water is the working medium. Torque of the hub depends on the displaced water, not the weight of the spheres.
Looking forward to another sleepless night...because I cannot figure out at the moment why this wouldn't work "perpetual".
I will definitely work for a while, until water is leveled out - or something...
I have had a hard working week, very tired, but I got it now; When the sphere enter the water at the bottom, water level rise at the bottom.
When the sphere exit the water on top, water level descent at the top.
As the cycle continue, the water level will finally equalize...
So, no, it will not be OU. :-\
Quote from: fredenergie on July 02, 2015, 12:22:30 PM
sorry no abituer the forum, how I can remove about thank you ???
ask help to the admin of this forum
Quote from: tagor on July 03, 2015, 01:02:06 AM
ask help to the admin of this forum
ok thank you Tagor
I really want to test this buoyancy principle in a simple way. The key will be to have
2 low friction valves. For each cycle they have to open and close so that means
4 valve movements. If those 4 valve movements take less work than the floated
object then the device will be OU. And as said before each additional vertical foot
is that much more work available so it seems to me it can be done if the tube is
tall enough.
A friend suggested for the water valve a plate between 2 orings would slide in and
out and would not displace much water. or just a ball valve which probably is
a ball with a hole through it in between 2 orings/seals.
In the video you can see the boy push and pull the a knob that opens and closes the
valve so it is not a ball valve but like the plate described above.
But the water displacement puzzles me. if you push a buoyant object down into
the tub of water it will make the water level rise. Then when you slide that
object under the tube which is filled with water also will the tub water level
drop? And likewise when a submerged object is removed the water level
will drop - so at the top of the tube will the water lever drop when removed?
I probably will just have to make it and ignore the valve work requirement losses
and then adjust the height to match those losses.
Norman
in an open system, the water will be displaced, and transferred to the surface level. it will reflect as a rise in water level at the top of the tank and tube. However, in this set-up, an open system will result in all the water from the tube flowing down into the lower basin.
Because the tube remains sealed at either end, water does not leave the tube. What occurs is a change in pressure, equal to the buoyant force x volume displaced. when the float and object leave the top of the tube, the pressure drops again as the "hole" is filled with water. the lower end of the tube is sealed when this takes place so the water does not fall down, a small amount will be transferred into and out of the tube to the upper basin, but this is negligible and recycled each time an object floats upwards.
If properly designed and executed, there should be no loss in water height. ( very small amounts of water may travel on the surface of the objects as they leave the system, but compared to the volume displaced, this is very small)
theoretically, the energy value for opening and closing the valves can be negated, or treated as 0.
even further, the work-function of the buoyant force as it travels up the water column, can be converted to perform this task, without affecting the potential gravitational energy of the mass once it reaches the top.
buoyancy, by default, is an overunity process. Once the mass has been removed from the fluid medium in which it is buoyant, the object has gained a gravitational potential proportional to its' change in height.
This is the power source that operates every hydroelectric power station on earth.
The same power source that brings divers and their gear to the top of the ocean from great depths using only the tiny energy of a CO2 cartridge.
- which is but a fraction of the energy that would be consumed if you lifted the diver with a mechanical force over distance.
The only time buoyancy systems are in thermodynamic equilibrium, is when you cycle the system by forcing the buoyant object back down through the fluid medium, without changing its' buoyant state. - this is a specific case when Energy In = Energy Out. And both sides of the cycle can be treated as a potential gradient. ALL other buoyant systems, are OU by nature.
I think that the water level drops because the spheres steal the space of the water when them rises. In the system shown in the video there is a water displacer in the upper side that reintegrates the water ( i think ). The water level will drop and must be reintegrate, this is the only problem to solve or the cost to be payed. we need to understand if energy expenditure is greater than that obtainable (not hard to believe that it is more)
Thanks guys for the excellent water level discussion. I'm pondering them all.
I was thinking that if the water level drops with each cycle then where does it go?
It seems like when the floating object rises into a given section the level will rise
then fall - that would make the most sense to me....
Norman
Thank you guys. This sheds some light in the issue.
has anyone done the experimentation on this?
take a look at this illustration, and I will walk through each step of the cycle, as I gently insert my foot into my mouth.....
1) As the object is placed into the lower basin, water is displaced, rising its' water level by that amount. lets call it X
2) When loch1 is opened, there is a negative pressure head, keeping the water pushed up inside the water column. As the object floats up into the "load chamber", the water that is displaced by the object is also pushed into the lower basin, but at the same time the water level in the lower basin drops by X, so the lower basin is still +X
3) loch1 is closed, loch2 opened. now the water displaced by the object is transferred from the "float chamber" to the "load chamber". an amount of X
4) loch 2 is closed, and loch 3 is opened. the object displaces water in the upper basin, and at the same time, that amount of water (X) is transferred into the "float chamber", so there is no noticeable change in water height of the upper basin.
until the object is removed. Then we have mass (Y), which is less than displacement (X) moved to the top. and displaced mass (X) moved to the bottom.
5) the difference between the mass, over height, is equal to the buoyant force over distance for the height of the column.
So,. I retract my earlier assumptions, and now state, verifiably, that yes, water is being moved down the column by the displaced buoyant mass.
the energy would be "free" if the source of water came from an already elevated point, such as the top of a waterfall, or the pressure from your local water tower to fill the column. But the obvious assumption would be that the mass of the elevated water would possess more potential energy than the buoyant mass you float to the same height.
For instance, micro-hydro turbines inline with your sink faucet to extract energy from your city's water pressure?? there are places in your house where you don't need the full pressure, just the water that comes out. so extracting the pressure as electrical energy won't affect the way you live. just lower your energy costs.
we might have to take this project back to the lab and figure out a way to keep the water level up, or use a different source of fluid medium that is replenished as we displace it.
My version of the water levels and movements.
When the floating object moves from the tub into the bottom section of the
tube it will displace that water in the tube down into the tub and
raise the tub water level a second time....
Then the lower valve is closed sealing it off from the tube to the tub.
and the upper valve is opened and the floating object will rise
and push the water above to below the floating object changing nothing.
there is no pressure change and no water level changes until the floating object pops above the top water level and is removed...then that level will drop....
and in each cycle it will drop more.... and that should be a problem.
So the net is the tub will rise and eventually overflow and the tube will
drop its level and need refilling.
Does this make sense now?
Norman
Quote from: norman6538 on July 08, 2015, 04:15:48 PM
Thanks guys for the excellent water level discussion. I'm pondering them all.
I was thinking that if the water level drops with each cycle then where does it go?
It seems like when the floating object rises into a given section the level will rise
then fall - that would make the most sense to me....
Norman
Norman,
Water level at the bottom reservoir will rise when the object submerge. When the first valve opens, and the object goes into next reservoir, and the valve behind it close, the water level at the bottom reservoir will remain at the new hight. Now the object will bring with it its displaced water, but when the object leaves the top reservoir, that object will not displace water anymore, and water level of the top reservoir will drop to a lower level, and remain this new level. Then the cycle repeats. Result is increasing level in the bottom reservoir, and reduced level in the top reservoir. Finally the water level is the same, and the process stops.
The energy we put in by filling up the tube with water, is the energy we get out from the buoyant objects.
Vidar.
Quote from: Low-Q on July 09, 2015, 01:59:42 PM
Norman,
Water level at the bottom reservoir will rise when the object submerge. When the first valve opens, and the object goes into next reservoir, and the valve behind it close, the water level at the bottom reservoir will remain at the new hight. Now the object will bring with it its displaced water, but when the object leaves the top reservoir, that object will not displace water anymore, and water level of the top reservoir will drop to a lower level, and remain this new level. Then the cycle repeats. Result is increasing level in the bottom reservoir, and reduced level in the top reservoir. Finally the water level is the same, and the process stops.
The energy we put in by filling up the tube with water, is the energy we get out from the buoyant objects.
Thanks Vidar for your excellent reply. I made one mistake in my earlier post -
...the tub water level will rise a second time. (wrong cause the object
leaving reduces the water level but is simultaneously replaced from the tube.)
But Vidar consider this. Everything you say is correct with respect to buoyancy but
net the potential energy of gravity from the lifted object. Every foot it rises beyond the work in is free except for replacing the water in the tube. So, imagine a 100 foot tall tube would give quite a bit of extra gravity power unless the power to replace
the water is too great.
The original pyramid idea did not consider the water replacment in the tube.
Norman
Norman,
If the tube is 100ft tall you have to supply a respective amount of water the same hight in advance. So the net energy from lifting mass and water will combined end up in zero.
It is only the weight of displaced water that is the 'active medium' here. The weight of the objects are not interesting because it is only the difference in buoyancy in water vs. buoyancy in air that is interesting. The difference here is therfor the weight of displaced water, and only this.
The objects can be made of osmium or hydrogen. Doesnt matter. And for each object you rise 100ft you must supply the displaced water.
Vidar
you guys are getting close to the answer.
allow me to add some clarity.
the object displaces the water equal to its spacial volume. regardless of the number of locks, or chambers it travels to, when it reaches the top, an equal volume of its' displacement will have moved to the bottom.
giving you a net = 0 potential energy in gravity.
The weight of the displaced water at the top of the column (E = mgh) is exactly what can be lifted by the buoyant force. Now, there are losses due to friction, water resistance, etc. that make the actual value of PE slightly less than the displaced mass over the height.
Now - you CAN lower the mass of the buoyant object to much less than the displaced water mass that is traveling down.
However -
the EXACT amount of energy between the potential energy of the water mass at the height, and the potential energy of the buoyant mass at the same height,
will represent itself in Buoyant Force over Distance !!!!!!! <------
all in all, Energy in = Energy out minus losses.
the math is long and convoluted, but at the end of the day
i must conclude that this type of buoyant system cannot be overunity.
That is not to say we cannot extract buoyant force in another way, but this type of cyclical system, all energy can be accounted for.
possible buoyancy motor idea for discussion
This is based on the following video.
https://www.youtube.com/watch?v=rxFXsoqbfrk (https://www.youtube.com/watch?v=rxFXsoqbfrk)
and was discussed somewhere in overunity.com but I lost track
of where that was.
Given a 3 lb weight that barely floats. The volume of the object is enough
that it can be pushed down and submerged by 1 ft lb. which means the object
displaces less than 3 lb of water and floats.
Given that the 3 lb weight is submerged into the bottom of a sealed 10 ft tube
of water it will rise to the top.
Then when removed from the water with maybe 3 ft lbs of work
it will have 30 ft lbs of potential work.
We then subtract the work in of 1 ft lb plus 3 ft lbs from 30 ft lbs we have
26 usable ft. lbs of work....
This will work because in water the 3lb weight does not have its full weight
because of buoyancy whereas in air out of water it has the full 3lbs of weight.
This would not work if the float had little weight like foam because
there would be few ft. lbs of potential energy when out of the water
and it would take more ft lbs of work to submerge it than it weighs due to the
large volume and low weight.
Of coarse I have not considered the valve changing work required.
I tested the idea with a plastic slide film container and several nuts from
a bolt. With 2 nuts it floats and with 3 it sinks. So in a sealed tube it would rise
many feet giving many "foot nuts" of work.
Do you see any flaws in this idea? Do you have any suggestions?
My former assessment of the pyramid idea was discussed here
http://overunity.com/8539/buoyancygravity-wheel-another-approach/msg455188/#msg455188
Norman
The example is very straight forward.
As long as the water level does not significantly drop during the raise of the floated weights (which the valves are supposed to prevent) there is a surplus on the energy side. The tube can be as long as you like therefore the energy needed to push the weight under water to get it into the tube is negligible.
The energy to operate the valves is only friction related. The valve which is opened or closed does not be under stress from the upper or lower side which would cause friction caused by water pressure. The valve could be constructed as such that it only slices a cookie of water sidewards, so the overall water level does not depend on whether a valve is closed or not.
The valves are essential I think for functioning.
As long as the water level does not significantly drop during the raise of the floated weights...
And this is the point: each 'floated weight' displaces water . No energy surplus
Thank you TheCell. I am working on testing the water levels. That is critical.
I will post the results.
Norman
Quote from: TheCell on February 01, 2016, 01:28:33 PM
The example is very straight forward.
As long as the water level does not significantly drop during the raise of the floated weights (which the valves are supposed to prevent) there is a surplus on the energy side. The tube can be as long as you like therefore the energy needed to push the weight under water to get it into the tube is negligible.
The energy to operate the valves is only friction related. The valve which is opened or closed does not be under stress from the upper or lower side which would cause friction caused by water pressure. The valve could be constructed as such that it only slices a cookie of water sidewards, so the overall water level does not depend on whether a valve is closed or not.
The valves are essential I think for functioning.
As long as the water level does not significantly drop during the raise of the floated weights...
And this is the point: each 'floated weight' displaces water . No energy surplus
By using some simple tube and a small eye drop bottle I have seen how
There will be water creep down from the top tube.
This is how it occurs.
When the submerged object is moved
into the first tube with its top closed water will be displaced by the
submerged object below that part of the tube which is the tub. Then when
the bottom of the middle tube is closed and connected to the top tube and the submerged object is moved up water from the top tube will drop down and replace the submerged object. And so we have water creep down.
There was someone purporting to use some displacers that might compensate
for this. ie. If that displaced water was moved off to the side or up 1 foot instead
of down then we could prevent the water creep down.
I used several props to show me what happens. 1. a M&M candy tube. 2. A small
eye drop bottle as a displacer. 3. A 1 qt container. I filled item 1, the tube with water
and then turned item 3 upside down to cover it tight . Then flipped the container
right side up and filled it with water and lifted item 1 up and pushed item 2 into the tube displacing the water and adding it to item 3. Then pushed item 1 down to
seal against item 3 and flipped it upside down draining the water. Then pulled
item 2 out and you can see how much water was displaced into item 3 by item 2.
So this leads me to this question
I have a 1 qt displacer object.
If I displace one quart of water into the upper tube a foot above
the valve so that it can later be dropped down below the valve.
How much work does it take to do that?
If that work is less then the potential ft lbs of the rising submerged object
then we can do this thing....
The limit will be the slowly rising submerged object and the lifted distance.
Norman
as Archimedes was trying to teach us,
a submerged mass displaces its' mass in water.
its the same mass of water that is dropping down, as the mass you are raising through buoyancy.
e=mgh
there is no gain
what makes the pyramid theory important, is that water can be moved one bucket at a time.
we don't have to lift 100 tons of water all at once.
but 100 tons of water, carried bit by bit to the top, can lift the 100 ton stone.
Quote from: sm0ky2 on February 01, 2016, 10:40:29 PM
Thanks Smokey, I missed the point below....
A similar lesson from the ants and the Chinese - many small pieces will surely add up.
Norman
what makes the pyramid theory important, is that water can be moved one bucket at a time.
we don't have to lift 100 tons of water all at once.
but 100 tons of water, carried bit by bit to the top, can lift the 100 ton stone.
norman6538
That is a very interesting approach and thank you for posting it. Can it be converted? Well the gates seem to be the real problem.
Alan