Hi guys I don't know if this was my idea or I saw it somewhere but it does'nt matter it's easy to do this.
Quote from: guruji on February 18, 2011, 08:15:37 AM
Hi guys I don't know if this was my idea or I saw it somewhere but it does'nt matter it's easy to do this.
good Morning
How are the angles kept in the same position?
Thanks
Bizzy
Quote from: Bizzy on February 18, 2011, 08:20:02 AM
good Morning
How are the angles kept in the same position?
Thanks
Bizzy
Good Morning again
Or let me rephrase that , how do you keep the angles in the same direction as it is turning?
Thanks
Bizzy
I think he is saying that at the bottom of the L it is a little heavier so gravity keeps it at 90 degrees. I am going to have to take this under review because at first glance, I thought no way it could work....but now....I am not saying it will but...it is an approach I have never seen before. hey, you never know. That is what experiments are for....and he is right....very simple and not hard to experiment with this design.
Bill
Quote from: Pirate88179 on February 18, 2011, 08:23:33 AM
I think he is saying that at the bottom of the L it is a little heavier so gravity keeps it at 90 degrees. I am going to have to take this under review because at first glance, I thought no way it could work....but now....I am not saying it will but...it is an approach I have never seen before. hey, you never know. That is what experiments are for....and he is right....very simple and not hard to experiment with this design.
Bill
Hi Bill
That was my first thought that the bottom would be heavier. The only problem is that the bottom L would be at full extension while it is at the bottom. So I am not sure if the other L's would have enough weight to move that. However like you said it is a very simple method and would be easy to try. When I was younger, much younger I would tinker with such devices. Unfortunately I didn't keep good records back then of my work So I don't recall if I tried this or not. So I now I'll have to try this when I get some more time
Bizzy
As I see it the key would be to keep it out of equilibrium. That is why you would have an L and not just the weighted bottom. So in looking at this, I now wonder if it would work better if the horizontal part of the L was two or three or more time longer than the veritcal wieghted part. This would keep it out of equilibrium on the right/extended side. and give teh machine some direction.
While on the left /retracked side the horizontal arm would not be extended.
Just a thought.
Bizzy
Bizzy:
I still think there is something wrong with this but, I can't put my finger on it. I think you might have hit on it by saying that if the horizontal part of the "L" was heavy enough to create an over balance, then it would dip down and not stick out at all. But, all it takes is a few grams heavier sticking out on the downside and a few grams less on the up side and those L's would appear to do that so.....to quote Einstein..."I don't know". The only way to know is to try.
Bill
Ok you two are both in trouble with my wife becuase I can't stop thinking about this and taht means more experiments ....
I am also thinking that instead of an even number of Ls there should be an odd number of Ls to keep it out of equilibrium.
Bizzy
My apologies to your wife.
I still think this will find equilibrium but without trying it, I can't explain how. I have a great bike wheel mounted on a pedestal for such experiments that I have saved. Decent low friction (mostly) bearings and good balance. With a dime taped on it it will move from 12:00 around to a little past 11:00 (CW) with just letting it go. Of course, everything I have tried makes it go only to 10....then 7 and then 5, ha ha. It was better before I started. I discovered a lot of different rotation dampers while doing so, ha ha.
Bill
Hi Bill
You are right the equilibrium is still staying in teh back of my head. The only way to find out is to actually try it. I already told my wife I need to get some more part for another testing table. She said I can after we finish bottle more wine this weekend...
I am attaching a crude diagram of an idea I had many years ago but work and my ex wife prevented me from exploring it. But now that my head is spinning again(no pun inteaded) I will have to try it. I am attaching acopy in case yo also wise to try it. This is my first attcahment on this site, so I apologize if it is not clear.
Bizzy
To explain this. The Ls on mine are tubes with capes on each end. There are heavy metal balls in each tube. At 3 oclock the ball rolls down the tube and impacts the endcap on the bottom of the vertical tube which gives is force to move the others. at 6 oclock the ball rolls backwards towards the bend in the L.
At 9 oclock the ball is rolling towards the center. I forgot to draw a small sping at the inside portion of the L to prevent any unwanted force on the left side of the wheel. at 12 oclock the ball is resting at the bottom of the L.
Bizzy
Even if the vertical stem is heavier then the horizontal stem, if the union point is a pivoting point, here is what will happen.
1) The heavier vertical stem holds all its weight in its limited diameter.
2) Anything going outwards from the vertical stem is the horizontal stem.
3) Since the vertical is balanced, 100% of the horizontal will be considered as excess or off balance to the vertical, pushing the vertical to the right somewhat.
4) This means when the weight is to the right of the wheel it will be pushing the vertical to the left of the pivot point, but when the weight is on the left side of the wheel, it will also push the vertical again to the left somewhat, thus canceling any momentum potential.
Any modification of the weight ratios between the vertical and horizontal stems will just change the overall angle of the weight, but the distribution will be the same.
So in reality, the drawing is impossible.
But I love these mind teasers because always at first glance your mind wants to work it through as intended.
wattsup
wattsup:
I can not disagree with your analysis.
Bill
Quote from: wattsup on February 18, 2011, 09:26:57 AM
Even if the vertical stem is heavier then the horizontal stem, if the union point is a pivoting point, here is what will happen.
1) The heavier vertical stem holds all its weight in its limited diameter.
2) Anything going outwards from the vertical stem is the horizontal stem.
3) Since the vertical is balanced, 100% of the horizontal will be considered as excess or off balance to the vertical, pushing the vertical to the right somewhat.
4) This means when the weight is to the right of the wheel it will be pushing the vertical to the left of the pivot point, but when the weight is on the left side of the wheel, it will also push the vertical again to the left somewhat, thus canceling any momentum potential.
Any modification of the weight ratios between the vertical and horizontal stems will just change the overall angle of the weight, but the distribution will be the same.
So in reality, the drawing is impossible.
But I love these mind teasers because always at first glance your mind wants to work it through as intended.
wattsup
Good morning
You are right that is the equilibrium portion that was bugging us earlier.
Thanks
Bizzy
Hi Bill
another dumb techinal question if you don't mind. I attached a PDF file successfully and it can be accessed other it is not showing up on the thread like gurji's original drawing.How can I make that happen?
Thanks
Bizzy
Quote from: Bizzy on February 18, 2011, 09:56:07 AM
Hi Bill
another dumb techinal question if you don't mind. I attached a PDF file successfully and it can be accessed other it is not showing up on the thread like gurji's original drawing.How can I make that happen?
Thanks
Bizzy
I am sorry but I really have no idea. I have never tried that.
Bill
Hi Bill
Ok. I'll try it as a a jpg
Thanks
Bizzy
@Bizzy
Something is wrong with your jpg file. Even though it ends with jpg, my editor says it "can't determine type". Seems like you saved in some funny format.
wattsup
Quote from: Bizzy on February 18, 2011, 10:18:16 AM
Ok. I'll try it as a a jpg
Forgive me for interupting here but maybe I can help. Simply renaming a file does not convert that file to another type. Your document is still a pdf even though you renamed it as jpg. Many freeware utilities can do this for you. I like PrintKey2000. Open your pdf in a reader - launch Printkey2000 and capture the area you want in your jpg and save that capture as a jpg. When you attach this saved file to your post, it will display correctly. Hope this helps!
Quote from: b_rads on February 18, 2011, 11:49:28 AM
Forgive me for interupting here but maybe I can help. Simply renaming a file does not convert that file to another type. Your document is still a pdf even though you renamed it as jpg. Many freeware utilities can do this for you. I like PrintKey2000. Open your pdf in a reader - launch Printkey2000 and capture the area you want in your jpg and save that capture as a jpg. When you attach this saved file to your post, it will display correctly. Hope this helps!
Hi Brad
Ok lets try it
Bizzy
Quote from: b_rads on February 18, 2011, 11:49:28 AM
Forgive me for interupting here but maybe I can help. Simply renaming a file does not convert that file to another type. Your document is still a pdf even though you renamed it as jpg. Many freeware utilities can do this for you. I like PrintKey2000. Open your pdf in a reader - launch Printkey2000 and capture the area you want in your jpg and save that capture as a jpg. When you attach this saved file to your post, it will display correctly. Hope this helps!
Hi Brad\
Thanks great advise!!!
Bizzy
Quote from: b_rads on February 18, 2011, 11:49:28 AM
Forgive me for interupting here but maybe I can help. Simply renaming a file does not convert that file to another type. Your document is still a pdf even though you renamed it as jpg. Many freeware utilities can do this for you. I like PrintKey2000. Open your pdf in a reader - launch Printkey2000 and capture the area you want in your jpg and save that capture as a jpg. When you attach this saved file to your post, it will display correctly. Hope this helps!
You are right about the pdf format. So I just downloaded the file as is. Changed the extension to pdf and opened it. No problem.
I grabbed so the image is here.
But again this will not work. The top is dead center so it cancels itself. The extended side to the right equals the left and bottom being on the other side.
wattsup
Quote from: wattsup on February 18, 2011, 12:03:30 PM
You are right about the pdf format. So I just downloaded the file as is. Changed the extension to pdf and opened it. No problem.
I grabbed so the image is here.
But again this will not work. The top is dead center so it cancels itself. The extended side to the right equals the left and bottom being on the other side.
wattsup
How about if the tubes were tilted down at a slight angle which would have the metal balls rolling down sooner?
Thanks
Bizzy
Quote from: Bizzy on February 18, 2011, 12:06:06 PM
How about if the tubes were tilted down at a slight angle which would have the metal balls rolling down sooner?
Thanks
Bizzy
Here is what I meant by angling the tubes more downward.
Let me know what you think
Bizzy
@Bizzy
No matter how you change the angles, one side will enjoy it, and, the other will hate it. As beneficial as it can be to one side, it will be equally detrimental to the other side.
Try and draw your idea with a real drawing program and once it is done, make a copy and then take that copy and rotate is about 2 degrees. Then draw some straight lines through the center of each shifting weight where they are at their new position. You will eventually see at which degrees the device just fails.
We have all been through this exercise many times.
From what I have learned of these wheels, you would require movements that create at least 5-10 different events for a wheel to have any decent "chance" of turning on its own. That means what? If your wheel has 10 equal sections, then each takes up 36 degrees and in that 36 degrees, at each 3.6 to 6 degrees something has to shift in favor of the direction of rotation. The more events you have, the smaller the shifts that are required to maintain rotation, and the more chance you have of capitalizing on momentum, weight shift, etc.
The less sections you have, the less chances of getting it going. But technically speaking, even if you had 20 sections or more, you have to know in advance that what you are trying is next to impossible, but hey, when did the impossible stop us.
wattsup
Wattsup maybe you're right regarding those L's cause of it's centre of gravity but what if the horrizontal be a little more heavier so that it tilts a bit down? ;) Maybe this works.
Regarding Bizzy idea that wheel works too cause it would be like a bessler but with tubes out like in this diagram
Thinking again on those L's yes Wattsup is right cause it always creating balance the L.
Quote from: guruji on February 18, 2011, 04:06:33 PM
Wattsup maybe you're right regarding those L's cause of it's centre of gravity but what if the horrizontal be a little more heavier so that it tilts a bit down? ;) Maybe this works.
Regarding Bizzy idea that wheel works too cause it would be like a bessler but with tubes out like in this diagram
guruji & wattsup
I truely appreciate your critique and contructive critisism. I think when I comes down to a working unit it will be a coupling of several ideas.
As I mentioned beforemy wife said we can go shopping this weekend once our wine is bottled so I plan on assembling another test table. I understand it may not work as well as I hope, but is my hope that I can get enough of an understanding to combine this with other ideas I have to get a working unit. I think for this test I will start with two tubes as guruji shows. to see how it reacts. Then add tubes and or adjust tubes to prevent equilibrium.
Again thanks and I'll let you know what progress I make.
Bizzy
I would advise you against it because we have all put in so much time on such wheels and if we can save you some time, then great.
The wheel itself has to be perfectly balanced to start with. Just that can kill any attempt even before you start.
Now take the last drawing. One would look at this and expect it to work. But now turn that wheel so the left tube is at 0 degrees and the right tube is at 180 degree. One is at the top with the ball fallen down dead center, where is the other ball? It is now to the left side of the wheel and working against any further rotation. On such extension ideas, once the angle reaches the bottom at 180 degrees, it passes to the left side sooner then the top can get pass to the right side.
I would suggest you study and study and study as much as possible before you decide to spend money and time on an actual build.
wattsup
Quote from: wattsup on February 18, 2011, 04:24:40 PM
I would advise you against it because we have all put in so much time on such wheels and if we can save you some time, then great.
The wheel itself has to be perfectly balanced to start with. Just that can kill any attempt even before you start.
Now take the last drawing. One would look at this and expect it to work. But now turn that wheel so the left tube is at 0 degrees and the right tube is at 180 degree. One is at the top with the ball fallen down dead center, where is the other ball? It is now to the left side of the wheel and working against any further rotation. On such extension ideas, once the angle reaches the bottom at 180 degrees, it passes to the left side sooner then the top can get pass to the right side.
I would suggest you study and study and study as much as possible before you decide to spend money and time on an actual build.
wattsup
Yes you're right regarding 180 deg but if there are many around the wheel other compensate for that as bessler's.
Quote from: guruji on February 18, 2011, 04:35:32 PM
Yes you're right regarding 180 deg but if there are many around the wheel other compensate for that as bessler's.
Good Morning
How many do you think it would take to compenstae for teh others and keep it pout of equilibrium?
Thanks
Bizzy
Quote from: Bizzy on February 19, 2011, 06:51:11 AM
Good Morning
How many do you think it would take to compenstae for teh others and keep it pout of equilibrium?
Thanks
Bizzy
Hi Bizzy this guy did eight around for the bessler wheel as shown so eight would work maybe even less.
http://www.youtube.com/watch?v=HqR60Z8jIm8
These simulations would help too:
http://www.youtube.com/watch?v=Bq0P6XDKrnY&feature=related
Quote from: guruji on February 19, 2011, 07:22:54 AM
Hi Bizzy this guy did eight around for the bessler wheel as shown so eight would work maybe even less.
http://www.youtube.com/watch?v=HqR60Z8jIm8
These simulations would help too:
http://www.youtube.com/watch?v=Bq0P6XDKrnY&feature=related
Those were fasinating. Originally I thought about using just one large metal ball but never considered serveral smaller ones.
Thanks
Bizzy
Another idea on same principle wheel what do you think guys?
@Guruji,
That last one I couldn't figure out... but the first design you posted got me wondering... probably won't be a self runner, but I'll have to find some time and play with it, just to see how it behaves, so to speak.
Made a sketch of what I was thinking about... Man, it looks so close... ;D
Thanks for sharing
http://img442.imageshack.us/img442/8540/33389627.jpg
spiralout
No matter how you shape your arms. It only counts at it point of hang.
Alan
Quote from: spiralout on February 21, 2011, 05:10:13 PM
@Guruji,
That last one I couldn't figure out... but the first design you posted got me wondering... probably won't be a self runner, but I'll have to find some time and play with it, just to see how it behaves, so to speak.
Made a sketch of what I was thinking about... Man, it looks so close... ;D
Thanks for sharing
http://img442.imageshack.us/img442/8540/33389627.jpg
In Picture 33389627.jpg the hanging weights cant' hang like the picture. Becasue in the L shape hanging weight , the large weight's moment is zero (becasue of the distance from the centre of axis is zero). the small weight has a moment. so it wont hang like shown in the picture. It can hang like ^ this
@bizzy
have you thought of incorporating your idea with the Milosevic principle of dual oscillation? maybe with a little spring help, you could get it to throw some weight around!
keep thinking man
sam
@
Quote from: supersam on February 22, 2011, 12:39:58 AM
Good morning
I am unfamiliar with that principle can you elaborate?
Thanks
Bizzy
@bizzy
have you thought of incorporating your idea with the Milosevic principle of dual oscillation? maybe with a little spring help, you could get it to throw some weight around!
keep thinking man
sam
@
here is a thought.
you need to have a component in your wheel that picks up the weight creating an eccentric load, or else it is not off-balance. the blue lines I added represent a shelf/platform to take-on/engage the weight during a phase of the rotation. the radius of this blue line will have to be slightly less than the length of the arm. there will certainly be a horizontal vector from the load back to the pin when the weight engages the blue platform, but it should be small. some derivative of this idea might create the off-balance condition you seek.
Quoteyou need to have a component in your wheel that picks up the weight creating an eccentric load, or else it is not off-balance.
Agreed. For these types of wheels to work negative feedback (ensured by some kind of a ramp) is a must. Finding a way to ensure that with the least friction the biggest drag in this whole research.
Quote from: mdmiller on February 22, 2011, 10:58:25 AM
here is a thought.
you need to have a component in your wheel that picks up the weight creating an eccentric load, or else it is not off-balance. the blue lines I added represent a shelf/platform to take-on/engage the weight during a phase of the rotation. the radius of this blue line will have to be slightly less than the length of the arm. there will certainly be a horizontal vector from the load back to the pin when the weight engages the blue platform, but it should be small. some derivative of this idea might create the off-balance condition you seek.
HI md
This looks similar to guruji's original idea, which means it would also have the same issues. It looks like that at some point the system would reach equilibrium. Do you have a way to prevent that?
Thanks
Bizzy
The first wheel I think whatsup was right although mdmiller is creating me a bit of doubt now again if this works or not. The second one I think it works for sure. The second one is on same principle but weights this time are hanged with a moving belt to every small wheel around to the fixed centre of the big wheel. When big whell turns all wheels around turn to same position so that weight remain on one side. There's not much to understand this. If toothed belts or sprockets of bicycles are used would be much better to cancel friction.
Thanks
I went through various wheel builds several years ago and had the issues/problems understood to a point. Now I'm fuzzy. Someone please tell me if the attached image is balanced or unbalanced. The red dot is loosely pinned. The green dot is fixed. I'm thinking I went through this type of build in the past, and it assumed equilibrium in this position, but today it seems unbalanced. Please set me straight. Thanks !
Hi mdmiller that image seems balanced. What is on the left is on the right.
The one you've posted about the L's seems to work if the horrizontal rods are long and skips the wheel centre as you did. One has to build this too to see regarding the L Gravity wheel.
Thanks
@ All, thanks for the eye openers.
@mdmiller, could the missing component be a couple of simple stoppers? I'm not
completely sure about their current placement, but the idea should be clear.
http://img573.imageshack.us/img573/4971/gw1k.jpg
btw, how do I embed the picture into the post? just by attaching?
Quote from: spiralout on February 22, 2011, 05:05:02 PM
@ All, thanks for the eye openers.
@mdmiller, could the missing component be a couple of simple stoppers? I'm not
completely sure about their current placement, but the idea should be clear.
http://img573.imageshack.us/img573/4971/gw1k.jpg
btw, how do I embed the picture into the post? just by attaching?
You need to look at this string of mine from 08. It might help you.
http://www.overunity.com/index.php?topic=5226.0
Alan
Quote from: AB Hammer on February 22, 2011, 06:01:55 PM
You need to look at this string of mine from 08. It might help you.
Well it didn't... any other suggestions, officer gravity?
Quote from: spiralout on February 22, 2011, 06:21:06 PM
Well it didn't... any other suggestions, officer gravity?
Find another approach. ;)
Look back at old strings and you will find many of the same things over and over again. Anyone who has tried the gravity wheels have done things that have been tried before. It is unavoidable. To find a new approach is where you can find what you are trying to achieve. ;)
Alan
PS WOW I have never been called "officer gravity" before. LOL
Posted a vid on my last gravity wheel idea.
http://www.youtube.com/watch?v=MaOjUcPDbVU
Gravity wheels are ALL about potential energy. If the weights are going as much upwards as it is going downwards, there is no point what so ever to try making a gravity wheel. All parts in a gravity wheel are ALWAYS limited within a given radius. This is the very reason why gravitywheels, regardless of its design, will never work.
Try find something else to do. Wind power, sun power, power from heat exchange. Gravity wheels will always be a dead end.
Vidar
No gravity wheels do work if one finds the way to always unbalance one side from the other.
gravity wheels ? think half wheel only way to unbalance the process, the idea is to be lifting less weight than going down. here is a idea
Protech good idea it seems to work did you try to build this?
Thanks
Hi Protech,
Maybe replace the blocks with heavy washers? This would reduce the friction of having to flip them around corners, they will just roll over the ends.
Quote from: DreamThinkBuild on March 28, 2011, 04:28:51 PM
Hi Protech,
Maybe replace the blocks with heavy washers? This would reduce the friction of having to flip them around corners, they will just roll over the ends.
the idea of the blocks with the slot are to provide a weight drop and as the shafts are all turning there is no drag. Just a rollover till it drops , But keep the idea,s coming my Idea is like a half wheel it all looks good on paper
Quote from: guruji on March 28, 2011, 03:28:15 PM
Protech good idea it seems to work did you try to build this?
Thanks
not yet no build,
I realized that even if it were to function well the amount of torque it would produce would be minimal like gravity wheels, unless it was the size of a 3 story building.
HOWEVER I was Serious enough to buy a milling machine though .... then I had another idea of a catch and release pendulum this is in the works. sure takes a lot of time though.
Hi Protech,
QuoteThe idea of the blocks with the slot are to provide a weight drop and as the shafts are all turning there is no drag. Just a rollover till it drops.
Ok, I see. Your design is very interesting, worth testing to see if working or not.
protech
Nice thought pattern. You will learn a lot from this build.
Alan
please note the vector component of the force mg (weight = mass X gravity) in the slanding rope is mg cos theta , the perpendicular rope the force is mg..
so 17 X mg cos theta = 7 x mg
please measure the angle and check
please see the attachment
Wow. I loved all your posts. Gravity wheels don't work but I really enjoyed reading this. Very amusing.
Hi Quantum you're very convinced that gravity wheels don't work. Everything is possible in life :)
Hi Quantam,
Gravity wheels may work or may not. If some one start to go in a wrong path, if we know that it is a wrong path it is our duty to inform them that it is a wrong path. Accept or not acceept is there decission.
I belive gravity wheels are possible. From physics i have some valid derived theories that shows this is possible. But with that we cannot say this is possible. a working model is the solution. I am also trying on that.
May be i can get it or maynot. so if you have a belief that gravity wheels will work , keep the momentum and go on. sure you will get a solution oneday.
Nixon
I said gravity wheels "don't work" in the sense that no working gravity wheel exists (that can be replicated independently) and so far no such device has been shown to work. That appears to be the state of play concerning experiments and working models (to the best of my knowledge).
Turning to the theory behind gravity wheels, resourceful and wonderful people such as many on this specific forum try to build these devices, and earn admiration from the rest of us for their efforts.
But my theoretical objection to gravity wheels is this. When considering a system, one must consider all of the forces and torques acting on each part of the system. Even if the forces (including rotational forces or torques due to gravity acting on one part of the system) are unbalanced on one side of it, other forces also exert torques on balls or other objects attached to the gravity wheel. Tension in the chains or ropes joining the balls together have to be taken into account as well.
When all these forces are considered, it is the case that the net torque in any part of the system is zero, despite a powerful counter-intuitive sense that this cannot be so.
When you add to this the fact that the net forces and net torques on surrounding objects like balls and pulleys are also zero, this has the effect that the wheel can never move by itself.
I remain open to persuasion (unlike many so called scientists who get angry when topics they consider closed are raised) but this is my current understanding.
Good luck with the gravity wheels. I really hope you prove me wrong :)
Quantumangles: Hey at least your mind is still open somewhat, I have to give you 4 stars for that! Have you studied Bessler at all?
@ Dr
Sorry for the delay responding. I moved house. Just had my phone connected up again.
In answer to your question...no. I have not studied Bessler.
But I will most certainly check out his work.
Thanks for the reference to Bessler. I am sure it will be interesting.
;D
Gravity wheels does not work for one simple reason: The weights is ALLWAYS present within a fixed difference in altitude.
The weights alone are for sure overbalancing, but not the energy. That means there is a factor which adds up in zero, inwhich the weights must be multiplied with in order to get the energy. You can guess what factor that would be. ;)
Vidar
Quote from: Low-Q on August 02, 2011, 02:27:06 PM
Gravity wheels does not work for one simple reason: The weights is ALLWAYS present within a fixed difference in altitude.
The weights alone are for sure overbalancing, but not the energy. That means there is a factor which adds up in zero, inwhich the weights must be multiplied with in order to get the energy. You can guess what factor that would be. ;)
Vidar
If there are two weights and one is overbalanced than gravity wheel works. The puzzle is how can one put that overbalanced weight in a way that it move by gravity for perpetium.
Quote from: guruji on August 02, 2011, 02:42:42 PM
If there are two weights and one is overbalanced than gravity wheel works.
No. The problem is not how to overbalance, but how to achieve extra potential energy from a system which never change its potential energy level. A weight put on a wheel, or just use an elevator, or your hand, to move up and down, is the same thing as long the weight (regardless of its shape and alignment) is returned to its origin for each cycle. You can walk ten meters, then walk ten meters back. How far is the net sum of distance you walked? It's zero. No change. However, you have SPENT much energy to fight friction while moving your mass 20 meters.
In a gravity field, there is one thing that matters; up and down. What the wheel does to a weight sideways, or in what pattern the weight is traveling before it is returned to the same spot, does not change anything. Gravity wheels are conservative as the gravity itself. The factors in total is not overbalancing - only the weights. So therfor no work can be carried out.
QuoteThe puzzle is how can one put that overbalanced weight in a way that it move by gravity for perpetium.
The gravity can only do work if the field change. That means Earth has to gain its weight, then loose it, then gain weight again. Or, the weight on the wheel itself must become heavier on one side of the wheel, and lighter on the other side. This will never happen, so we do have to look at alternatives to gravity powered wheels. Because gravity, and mass is conservative - doesn't change. No change means no potential difference which possibly could be converted into useful energy. Potential energy IS that change we need, but there is no change in gravity or mass to harvest energy from.
This is very, very simple, but most people just don't get it.
However, there is one thing that might work. If gravity can be used together with the Earth momentum as it spins around its own axis, you might achieve something interesting. But that does not mean energy from nothing, but energy from the Earth incredible kinetic energy - assisted by gravity/heavy objects.
Vidar
@ Vidar: If you are 100% sure gravity wheels will not work, you should be able to prove this with Math, I am still waiting for your secrete formula, that will prove your point! ;)
Not allowing for friction:
One plus minus one equals zero.
@Dr. You also know that 1-1=0. You do the math too. Take 10 Newton times (1m - 1m) you get 0 joule. No matte what number you multiply with zero you'll always get zero. There is your proof. If you have another equation please let me look at it.
Interesting to see the discussion unfold.
I confess I think gravity wheels cannot work because net forces (emphasis added) are always zero.
However, consider fluids.
If you have a dense fluid (for example seawater of density 1020kg/m3) and you wish to recirculate a less dense working fluid (say castor oil or dry-cleaning fluid of density circa 860 kg/m3), in this case less dense working fluid should rise to the top of the seawater tank.
Working fluid can then fall onto the cups of an impulse turbine via gravity before rising to the top of the seawater tank due to lower density.
The tricky part is getting working fluid to re-enter the base of the seawater tank.
I concluded air pressure (via an air compressor) and a one way valve would be fairly efficient.
Pressure relief valves also prevent P1V1=P2V2 equalisation.
So when one says gravity is a conservative force (which of course it is) what precisely does this mean in practical terms (when one should be able to cause a fluid to strike a turbine via gravity, and then 'float it' up through a dense substrate so it can fall onto the turbine all over again?
I did some maths in a post concerning this (recirculating fluid turbine).
It is on topic because the turbine is a 'wheel' and the machine works using gravity.
It is off topic because the maths is complicated for the uninitiated.
Please let me know your views ;D
Quote from: quantumtangles on August 03, 2011, 07:21:28 AM
Interesting to see the discussion unfold.
I confess I think gravity wheels cannot work because net forces (emphasis added) are always zero.
However, consider fluids.
If you have a dense fluid (for example seawater of density 1020kg/m3) and you wish to recirculate a less dense working fluid (say castor oil or dry-cleaning fluid of density circa 860 kg/m3), in this case less dense working fluid should rise to the top of the seawater tank.
Working fluid can then fall onto the cups of an impulse turbine via gravity before rising to the top of the seawater tank due to lower density.
The tricky part is getting working fluid to re-enter the base of the seawater tank.
I concluded air pressure (via an air compressor) and a one way valve would be fairly efficient.
Pressure relief valves also prevent P1V1=P2V2 equalisation.
So when one says gravity is a conservative force (which of course it is) what precisely does this mean in practical terms (when one should be able to cause a fluid to strike a turbine via gravity, and then 'float it' up through a dense substrate so it can fall onto the turbine all over again?
I did some maths in a post concerning this (recirculating fluid turbine).
It is on topic because the turbine is a 'wheel' and the machine works using gravity.
It is off topic because the maths is complicated for the uninitiated.
Please let me know your views ;D
You are talking about a buoyancy effect. For oil to enter the bottom of a tank filled with water, you must spend the energy required to displace the same volume in water, minus the displacement of same volume of oil. If you displace 1 litre of water (1kg) at 10m depth, with 1 litre of oil (0.8kg), the required energy to do this operation is 1kg - 0.8kg * 9.81 * 10m = 19.62 Joule. The very same energy you can take out of the oil on its way up to the surface.
It wont work. You need to supply energy/heat to alter the fluids density on demand. That would be the only way to make a working gravity wheel. Heating one side, and cool the other side. Then the temperature will displace the weight you need in order to achieve rotation.
Vidar
@Vidar: You are not taking into account weight transfer, which is completely different than movement! Take a 10 ft. dia. wheel, at the 12 oclock position fasten a pivot point and hang a 5 ft. lever. at the end of the lever fasten a 5lb. weight. Where does the wheel see the weight? You see the weight in the center of the wheel! The wheel see,s the weight at the 12 oclock position! I am absolutely convinced that Bessler had a working gravity driven wheel, and I think he found a way to transfer the weight rather than move it. ::)
Quote from: Dr on August 03, 2011, 08:53:09 AM
@Vidar: You are not taking into account weight transfer, which is completely different than movement! Take a 10 ft. dia. wheel, at the 12 oclock position fasten a pivot point and hang a 5 ft. lever. at the end of the lever fasten a 5lb. weight. Where does the wheel see the weight? You see the weight in the center of the wheel! The wheel see,s the weight at the 12 oclock position! I am absolutely convinced that Bessler had a working gravity driven wheel, and I think he found a way to transfer the weight rather than move it. ::)
What equation prooves that this wheel will selfrun just by placing a lever on it?
Vidar
@ Vidar
That is most interesting and helpful.
In other words there is no benefit to using less dense working fluids.
I suspected this and said so in earlier posts about the invention.
I only used the example of less dense working fluid to answer critics who claimed working fluid had to be "lifted" to the top of tank A.
I always preferred the idea of using seawater as working fluid (due to its greater density).
However, what I have had difficulty calculating is the amount of energy required to force one cubic metre of seawater per second (density 1020kg/m3) into the base of a 30m high cylinder full of seawater.
one cubic metre per second is drawn from the top of the vessel at the same time as one cubic metre per second is crammed into the base of it.
You calculated required energy in joules, which makes sense.
But somewhere I recollect warnings about trying to convert joules to watts (I need to know the value in watts).
How would you go about calculating the precise energy in watts required to force one cubic metre of water per second into the base of a 30m high cylinder full of seawater? (mass out at the top equals mass in at the base).
Would the removal of 1m3/s from the top of the tank create a suction effect that would minimise the energy required to draw water into the base of the tank? Would adherence caused by covalent bonds in the water molecules 'glue' them together en bloc?
The pressure at the base of the 30m high cylinder is 300,186 Pascals gauge pressure (1020kg/m3 x 30m x 9.81 m/s/s) + Patmos of 101325 Pascals = 401,511 Pascals = 401.511kPa
I use compressed air in vessel B to force seawater back into the base of vessel A. But the amount of energy required is critical.
I took the direct approach of ensuring that the pressure in vessel B always exceeded the pressure at the base of vessel A (in other words I did the maths to ensure vessel B always maintained pressure of 500kPa).
So it is not a buoyancy question (as the working fluid is seawater just as the rest of the fluid in tank A is also seawater).
Rather it is a question about the energy required to cram one cubic metre per second of seawater into the base of a 30m high cylinder, a cylinder which is already full of seawater (note that one cubic metre per second of seawater is simultaneously being drawn out of the top of tank A by a pump assisted siphon so that mass in equals mass out).
Many thanks Vidar. I am getting closer to knowing the truth of the matter. if you have any thoughts whatever on how to calculate the energy in watts to cram this cubic metre of water per second into the base of cylinder A I would much appreciate them.
1 Joule is the same as 1W in a period of one second. In one hour you get 3600 Joule of energy at 1W/s. So the term Watt is not energy alone. Watt is energy only over a given period of time. So Joule should be the term to use in any context. Hope that clears it up:). To lift 9810N 30 meters require approx 30kJ of energy. Or 30kW/s. You need 30kW to lift 1m^3 of water 30 meters up pr second.
@ LowQ
Many thanks.
If it really only takes 30kW to move one cubic metre of water up 30m, then my machine should work.
The same cubic metre of water falling 25m onto an impulse turbine generates about 200kW of electricity.
Here is the maths for electrical output.
Pwatts = h(25m) x g (9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW
So it seems we spend 30kW and get output of 225kW.
Interesting.
Quote from: quantumtangles on August 03, 2011, 03:03:50 PM
@ LowQ
Many thanks.
If it really only takes 30kW to move one cubic metre of water up 30m, then my machine should work.
The same cubic metre of water falling 25m onto an impulse turbine generates about 200kW of electricity.
Here is the maths for electrical output.
Pwatts = h(25m) x g (9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW
So it seems we spend 30kW and get output of 225kW.
Interesting.
I believe the input would be 30 kw per second. I think the time has to be factored in.
Bill
Just have in mind that Watt isn't energy unless you apply time. My guess is that the water falling 25 meters will for sure generate 225kW, but for how long? You talk about impulse turbine! That is exactly where the flaw is. You have a short burst of 225kW.
That is why I said that mesuring everything in Joules is better. Watt alone does not show how much energy there is because you miss time in the equation. Sorry. It will most probably not work. Do a recalculation in Joules. That water falling 25 meters creates less energy than it takes to push it up 30m.
This is btw a very good example why gravitywheels works inside the inventors mind, but does not work in real world. What do the inventor miss? Time!
Vidar
@ LowQ & Pirate
Many thanks. Of course you are correct about time.
However the equation for calculating power output in watts uses SI units.
Acceleration due to gravity (9.81 m/s/s) is time based (per second per second).
The mass flow rate of one cubic meter per second is precisely 1020 kg/s.
The only units that are not time related are height (or head) in metres (25m) and the unitless fraction representing turbine efficiency (always a figure between zero and one for any system, and always between 0.69 and 0.95 for Pelton impulse turbines). I somewhat generously allowed 0.9 for my turbine because larger turbines have better efficiency and a 0.9m diameter turbine is enormous by Pelton turbine standards).
So I completely agree that time is the critical factor. However I respectfully suggest that time is already 'factored in' if one uses SI units.
There is a way of cross checking the electrical output result obtained from the standard hydroelectric flow/head equation, and that is by using the Pmech equation. In order to deploy the Pmech equation, one must first calculate RPM, which in turn requires one to calculate teh force in Newtons that is applied to the buckets of the turbine.
Beginning with the force equation, F= m*a
F = 1020kg/s x 9.81 m/s/s
F = 10006.2 Newtons (enormous force).
But velocity will be greater than 9.81 m/s/s.
Bernoulli's equation gives us the velocity of the water jet applied to the turbine.
P = ½ r . V2
P = Pressure (401,536 Pa)
rho = density (1020kg/m3)
V = velocity (m/s) mystery value
401536 = ½ 1020 . V2
401536 = 510 . V2
V2 = 401,536 / 510
V = 28 m/s
Force of water Jet (Fjet)
1 m3/s of seawater (1020kg/m3) represents a flow rate of 1020kg/s.
F = 1020kg/s x 28 m/s/s
= 28560 Newtons
Fjet Momentum Change Calculation
Turbine speed may not exceed 50% of water jet speed
Vjet = 28 m/s
Vrunner = 14 m/s
Delta Mom = mass flow rate x Delta V
Delta Mom = mass flow rate x (Vjet â€" Vrunner)
Delta Mom = 1020kg/s x (28 m/s â€" 14 m/s)
Delta Mom = 14280 N
Fjet = 14280N.
Turbine RPM
The RPM figure for the turbine is based on runner velocity
of 14 m/s (half of water jet velocity*).
First we need to know the circumference of the turbine.
Diameter = 0.9m
radius = 0.45m
2.pi.r = 2.827433388m circumference
Vrunner = 14 m/s
RPS = 4.951487 revolutions per second x 60
= 297 RPM
Power Output
Applying (297 RPM) and also (Fjet = 14280 Newtons) to the Pmech equation:
Pmech = Fjet x Njet x pi x flowrate x RPM x 0.9 x 0.87m / 60
= 14280N x 1(jet) x pi x 1m3/s x 297RPM x 0.9(eff) x 0.87m / 60
= 173Kw
So far as I am aware, this is 173kW per second (it is not 173kW per year, or per hour or per month).
All of the units which lead to the result of the calculation in the Pmech equation are in seconds or are based on seconds (apart from revolutions per minute but you will note that I divided by 60 to obtain the result, thus converting the RPM into revolutions per second. The SI unit for angular velocity is radians per second but the equation I used amounts to precisely the same thing again because of the division by 60.
So I think the output is somewhere between 173 and 225 kW per second.
Please let me know your views on this.
* Note on turbine velocity being half of water jet velocity
x = vb / vj
x = ratio
vb = Cup velocity at pitch circle diameter of turbine
vj = Jet velocity
F = mb. vj . (1-x) (1+ z.cos g)
h = mb . (vj . vj) . x . (1-x) . (1+z.cosg) / ½ . mb (vj . vj)
P = F . vb = mb . vj . (1-x) . (1+z.cos g) . vb = mb . vj . x . (1-x) . (1+z.cos g)
dh / dx = 2(1-2x). (1+z.cos g) = 0
x = 0.5
h = system efficiency as a unit-less fraction between zero and one
F = force of water on cups (N)
mb = mass flow rate into cup (kg/s)
vj = Jet velocity (m/s)
vb = runner tangential velocity at pitch circle diameter (m/s)
z = efficiency factor for flow in buckets (unit-less fraction between zero and 1)
g = angle of sides of cups
x = speed ratio of vj to vb
Quote from: quantumtangles on August 03, 2011, 03:03:50 PM
@ LowQ
Many thanks.
If it really only takes 30kW to move one cubic metre of water up 30m, then my machine should work.
The same cubic metre of water falling 25m onto an impulse turbine generates about 200kW of electricity.
Here is the maths for electrical output.
Pwatts = h(25m) x g (9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW
So it seems we spend 30kW and get output of 225kW.
Interesting.
You might like to try the math again ?
Vidar's logic always makes good sense to me - in this case I think he made a slight typo - understandable as it's just hit his radar & he is actually explaining thoughtfully why gravity is conservative for another example [applies here as well].
Keeping it simple:
Problem A: Energy required to raise 1 m^3 of water [say 1000kg] 30 meters N.B. we can think of this two ways [they are equivalent].
1. what is the volume [diameter of cylinder] of the vessel & therefore how much displacement volume occurs injecting 1 m^3 - this will see a rise in water level h - so we need to know the entire tanks mass x h to find the increase in Potential Energy in Joules.
2. we don't need to know area & volume & tank contents mass - all we need to know is that 1 m^3 will be raised 30 meters - this will see a rise in Potential Energy in Joules.
i.e. Pe = mgh => 1000kg x 9.80665N x 30 m = 294,200 Joules or 294 KJ's [per sec].
This is how much energy in Joules [Work Done] must be supplied to force the water to gain Potential Energy.
Problem B: What energy [Work Done] can we get out of the descending water mass.
Ans: Potential Energy Joules at 100% efficiency will convert to an equivalent Kinetic Energy in Joules i.e. energy of position is converted into an equivalent energy of motion.
Therefore Pe at start is mgh => 1000kg x 9.80665N x 25m = 245,200 Joules or 245 KJ's [per sec]
However, we all know that there are frictional losses in water ducting & flow systems, head loss etc - added to that it that engines & pumps are generally inefficient [electric engine perhaps 80% then deduct further efficiency for pump losses, lets say engine & pump 70% efficient].
Therefore Kinetic Energy from gravity is 245KJ's per sec x 70% =
171.5 KJ's per sec useable Output.
But we needed an Input of 294KJ's per sec.
Clearly this simple analysis shows IMO that there is a shortfall in Joules of energy per sec to make this theory viable & the reality self sustaining.
EDIT:
Quote from: quantumtanglesSo I think the output is somewhere between 173 and 225 kW per second.
This seems to agree closely with the 171.5 to 245 KJ's per sec from the simple math approach.
@ Fletcher
That is extremely helpful. I have been hoping for months for some feedback on the maths.
The problem I have tried to solve is that of making hydroelectricity viable using recirculating water (without the need for naturally occurring falling water).
I have known from the outset that lifting water was not viable. My initial sums indicated that 4.5 times more energy would be required to lift water up than could be derived from it falling onto a turbine.
But yes, 300kW is about right if the system were supernaturally efficient (for output of 173kW).
It makes perfect sense that input of at very least 294kW would be required to obtain electrical output of 173kw to 225kW (if water has to be lifted 30m upwards).
But what if air pressure were used to force recirculated water sideways back into the base of tank A? I thought in April 2011 that that might work and I am still unsure about it.
Put bluntly, the pressure at the base of tank A (arising from the height of the column of seawater and its density) is only about 4 atmospheres (about 401,000 Pascals).
Generally speaking, fluid always moves from areas of high pressure to areas of low pressure.
If all I have to do is ensure tailgate water in tank B (after striking the turbine) is pressurised to in excess of this, then the tailgate water will force its way through a one way valve sideways and back into tank A.
All that is needed is an air compressor powerful enough to pressurise the standing tailgate water in tank B to over 400kPa. Surely (I said to myself) such an air compressor would not need to consume 300kW or more?
This is what has haunted me since April 2011. The thought that an air compressor can indeed side step the need to lift water 30m upwards (by pushing water 'sideways' into the base of tank A at the same time as the pump assisted siphon draws an equal volume of water up and away from the surface of tank A).
Very interesting and helpful analysis. Many thanks but please let me know what you make of the sideways pressure idea.
I think I remember coming across your thread about this idea - you were looking for a very select bunch of rather exclusively qualified people to comment IIRC - perhaps you could point me back to it & I could take a look at any drawings you have done, time willing - might save some time & also not hijack this thread.
Pushing water sideways [horizontally] thru a one-way valve takes the same amount of energy to enter the tank as lifting the water thru the height of the tank - this may not be what you meant & I don't want to go off on too many tangents - but ... hydraulic pressure is force x area - water pressure increases with depth in a linear manner [that'll be where you get your 4 atmosphere's at 30 meters etc] - liquids are also virtually incompressible so this linear relationship is constant & the density doesn't effectively change with depth - the next thing is that water pressure always acts normally [i.e. at right angles] to surfaces - IOW's it has no 'sheering moment' - that means that the water pressure at 30 meters will be the same on any surface at that depth - that is, whether it be the bottom of the tank, the side walls, or any other face or direction water pressure can apply itself to.
This may not be relevant to what you are proposing but it probably does illustrate that 'pushing water sideways' is no easier [Work Done in Joules] than entering the tank from above etc.
Mass that is moving sideways, where gravity is the key force, you will in best case not gain or spend energy on moving a mass horizontally. In worst case, which will be the practical outcome, is to spend energy and loose energy output by any movement of mass in any direction.
The energy output of your system will regardless of complexity be at most 25/30 of the energy input. Even if there was a 30m fall, your output would not be higher than the input. If that was the case you shouldn't have to rise the water at all. Just take energy out of it while it stands still in the same spot all the time. How plausible is that?
Vidar
Thanks all. I hope this was on topic as the turbine is a wheel and gravity is involved.
The initial idea evolved with a series of posts under Recirculating Fluid Turbine Invention. Been driving me mad for months. Just want certainty.
I'm about to go off & start reading his thread/topic Vidar.
Before having read & understood it [so I don't really know yet] I think he is suggesting that compressing air & using that to drive a piston which pushes a volume of water into the bottom of a tank possibly takes less energy in Joules [Work Done] than other mechanical means - it would have to be way less as we've all pointed out - I think he is also suggesting an advantage may be gained from a siphon effect of some sort ?
Perhaps there is some advantage [small though it might be if it exists] in using compressed air in terms of the Carnot Cycle & Adiabatic heating & Isothermal Cooling of the tank of compressed air - I'll expand these thoughts on the other thread perhaps - any advantage might be able to be engineered into the situation quantum is proposing - who knows if it has a place, will need to read his logic first.
@ Vidar & Fletcher et al
I really appreciate your input. Most of all I really need your help.
Quote from: fletcher on August 04, 2011, 06:02:43 PM
I'm about to go off & start reading his thread/topic Vidar.
Before having read & understood it [so I don't really know yet] I think he is suggesting that compressing air & using that to drive a piston which pushes a volume of water into the bottom of a tank possibly takes less energy in Joules [Work Done] than other mechanical means - it would have to be way less as we've all pointed out - I think he is also suggesting an advantage may be gained from a siphon effect of some sort ?
Perhaps there is some advantage [small though it might be if it exists] in using compressed air in terms of the Carnot Cycle & Adiabatic heating & Isothermal Cooling of the tank of compressed air - I'll expand these thoughts on the other thread perhaps - any advantage might be able to be engineered into the situation quantum is proposing - who knows if it has a place, will need to read his logic first.
Ofcourse, input of energy will make this work, but that energy is going through a process with loss, so one can only hope for an energy output which is less than the energy you put in. A siphon will definitely not provide the extra energy. Try to emty the ocean with a siphon. That is impossible because the ocean is the bottom level on this planet (except a few areas which is held dry with water pumps).
The gravity wheel, or other types of gravity "powered" devices is a dead end, but gravity can be handy in combination with other factors such as waterfalls, as waterfalls is created by the sun, ocean, mountains and gravity.
Vidar
@ Vidar
Your example of trying to empty the ocean with a siphon is somewhat different. Functional siphons require tubes (leading up to the siphon crest from the water source) that are shorter than the tubes leading down from the siphon crest to discharge the water.
Any attempt to place a siphon in the ocean would result in the 'long end' of the siphon being under water and therefore it could not possibly work.
By contrast (in the system under review) the long end of the siphon tube is below the level of the water intake without actually being 'under water'. Accordingly the siphon in question should work.
One of the key questions (re the proposed system as a whole) is whether it can work if 'unlimited' energy is available to circulate the water. I say this because initially I feared P1V1=P2V2 equalisation would prevent circulation regardless how much energy was expended.
If recirculation is possible with unlimited energy (which seems to be the position) the second key question is: how much energy is required.
Calculating power output is trivial to mathematicians and physicists. But calculating required power input is not so straightforward.
Certainly any idea of lifting water 30m is doomed.
Even though 'water lifting' is path dependent (methods of lifting may vary from the oestentatiously inefficient to the slightly more efficient) it certainly cannot work. Power expenditure will be in the order of 4.5 times power output.
But by pressurising tank B to more than the base of tank A, water should flow from B into the base of tank A.
But then the problem moves further round the circuit.
If tank B is under high pressure, how can the upper siphon work? How can low pressure water from A enter the high pressure environment of Tank B?
At first it seems tank B has to be both a high pressure area (to evacuate tailgate water after it has struck the turbine) and also a low pressure area (to allow the siphon to keep flowing).
The solution is to have two distinct pressure areas inside tank B.
The main body of tank B is always kept at higher pressure than the base of tank A. Fine. But there is also a second much smaller very high pressure area inside the exit nozzle of the siphon, that keeps it flowing because the air compressor drives siphon water out.
It all boils down to expenditure versus output (bearing in mind the turbine will have to fight against high air pressure (air of greater density) which will reduce RPM and therefore output.
I find the issues fascinating and much appreciate your views on the way to debunking or establishing the concept.
Kind regards and thanks,