Hi,
I have an idea on how a Buoyancy wheel might work. What we already know is that these wheels does not work, but I have been thinking a little, and tried to imagine what would happen if just the half wheel was under water.
If you take the picture below as an example. (This is a drawing I made a long time ago but I have modified it a bit).
The vacuum tubes is altered in volume, and a rod is fixed on each oposite piston which seal the vacuum. To move this rod back and forth does not require energy other than friction. So it is easy to increase emty space on one side and reduce it on the oposite side. The initial idea was to produce more buoyancy on one side than the oposite side. The idea was further to alow the greater buoyancy on one side to force the wheel to turn around.
The problem is that the weight of water this wheel is trying to push from the bottom of the wheel to the top, requires as much energy as the energy from the buoyancy. So it won't work.
So I had another idea. I filled the water tank half full. So the wheel does not have to move water upwards, beyond the wheels axis, and to the pistons on the top.
Instead, with the half full tank, the water is just moved sideways, or is just tumbeling around in the tank going nowhere. Moving mass sideways does not require energy! The pistons in the bottom of the wheel is moving upwards on both left and right side, so the forces acting on the pistons cancels out. The best had been if the pistons was as hig or higher than water level (not shown), so no force from the water is acting at the pistons at all.
Then you have a wheel where the weight of the mass rotating is practically less on the bottom than the top - all the time. And the lowest weight is on one side of the wheel, and the greater weight is on the other side - all the time.
So left I have only one force to care about, and that is the buoyancy force on the right side which is forcing the wheel to spin counter clockwise.
However, I have a feeling that there is something I have missed. So PLEASE anyone, can you try to find out why this new wheel does not work.
All replies is welcome - but please try to stay objective :)
Br.
Vidar
It does take energy to move mass sideways.
Quote from: brian334 on March 04, 2009, 11:45:11 AM
It does take energy to move mass sideways.
Thanks @brian334, I will think on that. My thought was that it only requires energy to acellerate mass in order to move it from A to B, but the kinetic energy is going back when deaccelerating before that mass stops at point B. So no net work is done.
Br.
Vidar
When things move there is friction, you need to add energy to the system to overcome friction.
Quote from: brian334 on March 04, 2009, 12:41:44 PM
When things move there is friction, you need to add energy to the system to overcome friction.
Yes, you're right, but friction is a variable factor that isn't directly related to mass, size, torque, or force etc, so it is controllable as it can be minimized or changed regardless of potential energy found in torque or forces.
I mean, if the friction is all that is to make the water in the tank above moving around, that friction can be changed or controlled. I have taken into consideration that there will be friction, and inertia through water, and inside bearings etc., but what I am looking for is if there is a directional force or torque that works more in one direction than the other - regardless of friction. If it is, the torque will potentially start this wheel even with friction in it.
Do you have any opinion about the design by the way - even if its so there is friction?
Br.
Vidar
It takes energy to make something move, what energy makes your wheel turn?
What energy moves the pistons?
What energy creates the vacuum?
What energy moves the water?
What energy overcomes the friction of all the moving parts?
Quote from: brian334 on March 04, 2009, 01:12:20 PM
It takes energy to make something move, what energy makes your wheel turn?
What energy moves the pistons?
What energy creates the vacuum?
What energy moves the water?
What energy overcomes the friction of all the moving parts?
I'll try to answer as good as I can :):
You need force to make something move, but as it moves it develops kinetic energy. From there the energy is utilized - not provided to the system.
In one complete round, the energy used to move the pistons is zero + friction.
Vacuum does not create energy. Vacuum is used to make the movement of the pistons easier to alter the volume inside the cylinders. If it was air inside, it would require much force to compress the air in one cylinder and decompress air in the oposite cylinder. With vacuum the force on the surface of the piston will be the same regardless of volume inside. With a rod that controls the two oposite pistons, the net force to alter the volume is therfor zero.
The water is moved around due to rotation of the wheel and movement of the pistons. However, the water content is constant, and the water inside the tank isn't actually going anywhere. The inertia in the water is slowing down the system, but if there is potential force or torque in the wheel it will start spinnig - maybe a little slow.
At this point I'm unsure. Will the pistons try to move water upwards? Look at the pistons to the left. They are not moving the water upwards as the wheel is going downwards. What about right side. Is the piston easier to pull out as the water is gradually more shallow? There is a few things here I'm not getting right.
The torque from the buoyancy effect are suppose to overcome all these frictions. If the net force or torque is mostly working in one direction, and isn't cancelled out during one revolution, left you have a potential energy waiting to be released by leting the wheel go. Friction only brakes the movement a bit, so friction isn't an oposite equivalent to the force created by the buoyancy.
Bottom line: I'm ONLY looking for the net torque of the whole system during one complete revolution. If the sum of all forces in one complete revolution is positive or negative, the wheel will start spinning. As I mentioned, friction is a variable and can be controlled, and isn't allways so significant that is stops any motor completely. Conventional motors works because of torque, not because of zero friction. The energy is torque times revolutions pr. time. So if the wheel will spin by help of positive or negative torque, I'll probably be a dead man ;D
Br.
Vidar
You say,
You need force to make something move, but as it moves it develops kinetic energy. From there the energy is utilized - not provided to the system.
Two questions,
1. What is the force that makes the kinetic energy?
2. If your machine works why would you be a dead man?
Quote from: brian334 on March 04, 2009, 05:02:13 PM
You say,
You need force to make something move, but as it moves it develops kinetic energy. From there the energy is utilized - not provided to the system.
Two questions,
1. What is the force that makes the kinetic energy?
2. If your machine works why would you be a dead man?
1. The hope is to get more torque from the left side than the right side of the wheel. There is more displaced water on the left side than on the right side. This sum of torque other than 0 will make the force to spin the wheel. A spinning or moving mass is kinetic energy. The question is if that is the case here...
2. Because the oil industry is not the industry to fu#*?k with.
Vidar
I have been brain dead. If both cylinders is in air, the force to pull the oposite pistons back and forth will be zero. But if one of the cylinders is under water there will be greater force on that piston inside, so the piston is forced by weight of water to decrease the volume inside that cylinder. That will stop the wheel. I have to think a little more, on how it could be possible to make a counter force for that. I have been thinking on a magnet which is shaped and placed in a way that it gradually attracting a piece of steel on the rod or in the pistons which perfectly balanced the weight of water pressing on the piston. I'll be back with more drawings.
Br.
Vidar
Here is a drawing of the new idea. I have made this drawing very roughly, and the proportions of the magnets isn't correct, but it should give you a hint of the idea, and how it works.
The idea:
The weight of the water is a force that want to move the pistons downwards as soon as they are under water. This force is a perfect match to the forces provided by the buoyancy effect, and the wheel will not turn.
So I have added magnets on the pistons and a stator magnet which is shaped to counterforce equal to the force that is working on the piston surface under water plus the weight of the magnets and pistons. The coloured arrows indicates equal force for same colour - just an example to show you what I mean by equalizing forces.
So now there isn't any force from the weight of water which will prevent the pistons to move upwards when the wheel is turned counter clockwise.
I could ofcourse alternatively used an iron disc or something inside the rotor magnets to make the same counterforce. So try to imagine this alternative :)
My claim (maybe not true, but I'll try):
As the magnetic counterforce doesnt influence on the interaction between the water and the vacuum (buoyancy effect), or the non magnetic water, or the air, or the cylinders, or the vacuum inside the cylinders. There is then only one working force left - That is the buoyancy force that is greater on the right side than the left side. There isn't any other forces that will counterforce the rotation. The wheel is now allways lightest on the bottoms right side - and therfor allways heavier on the top left side. The consevation of energy is conserved only if the wheel starts spinning just by itself without help. Because it takes less energy to move the pistons upwards counterclockwise than clockwise.
How cool is that?
Any comments folks? As previously I still want to know what I'm doing wrong (Friction isn't an issue, I only want to know about the forces). As I am still very sceptical to perpetual motion. So bring it on, but please be objective, and please try to substantiate, or at least explain in detail, your claims. :) :)
br.
Vidar
If that idea would work it would do so without the water. The water adds nothing except drag.
Think about it.
Hans von Lieven
Hans
And LIFT?[buoyancy]
Chet
Pulling a piston against a vacuum requires a force greater than 1kg/cm2 Where is that coming from?
Hans von Lieven
Quote from: hansvonlieven on March 05, 2009, 05:23:22 PM
Pulling a piston against a vacuum requires a force greater than 1kg/cm2 Where is that coming from?
Hans von Lieven
That force is taken from the oposite piston. There is a link or rod between two oposite pistons. So the force to pull vacuum is cancelled out. No net force is done to pull the pistons as there is vacuum inside the cylinders regardless of its volume.
I have tested the theory in practice by pulling out the piston in a squirt when the nozzle is tight. Started from the very bottom, the force measured is the same if I pull it out 1cm, or 7cm. So the pull force from a vacuum volume much less on the left side is in fact the very same as the pull force on the vacuum volume at the right side. So, those oposite pull forces cancelles out to zero. Rest is friction, but we can deal with that. So, with the magnet configuration, we can play with altering the volumes in the pistons as we whish without lifting one single finger. And left is pure buoyancy force, forcing the mass in the wheel to move. Drag through water will just slow down the system, but torque will still be there to move it around. At least I hope, but I'm still very sceptical.
I might however be very wrong about details here, but I do know for sure how the vacuum cylinders work :)
Br.
Vidar
Vidar
LOTS of variables[opportunities] in your system
I am drawn like the moth to the flame on these ,your's more so .One thing you just said was you found the resistance on the vacuum was not linear to distance?
Chet
The force on 1cm2 is 1 kilo regardless of the vacuum volume, but when one of the pistons is under water, the water weight will push the piston down. So magnetism is the counter force for that weight.
@ Vidar,
Why have the vacuum there at all, why not vent the cylinders?
Where is the advantage?
Hans von Lieven
Quote from: hansvonlieven on March 06, 2009, 02:13:48 AM
@ Vidar,
Why have the vacuum there at all, why not vent the cylinders?
Where is the advantage?
Hans von Lieven
I have been thinking on the same idea. Thank you for reminding me :)
Also when I look at the drawing, I should only have a straight bar magnet at the bottom and the same on the top, with the same pole pointing up on both. This will reduce cogging when the rotor magnet is approaching and leaving the bottom magnet, as the top magnet will attract the rotor magnet as much as the bottom magnet will push it away.
One more thing: If the counterforce from the magnets is great enough, we can maybe fill the tank all the way up with water.
Br.
Vidar
Quote from: ramset on March 05, 2009, 06:13:36 PM
Vidar
LOTS of variables[opportunities] in your system
I am drawn like the moth to the flame on these ,your's more so .One thing you just said was you found the resistance on the vacuum was not linear to distance?
Chet
Hi,
My last reply was made with my cellphone, so here I will explain further.
It could seem to be some misunderstandings in my explanations earlier. I meant that th vacuum volume is less on the right side, not the force that pulls the vacuum.
And there isn.t that much variables. You have:
Buoyancy
Gravity
Magnetism
Gravity and magnetism cancells out the weight of the water pushing on the pistons.
Magnetism cannot affect the weight of water. So that buoyancy force is the only force left.
Force can again acellerate mass untill friction limits the speed. So there you go - A spinning wheel.
But I will again mention that I'm rather sceptical to this. I'm just trying to make ideas I cannot understand fully. It is my form of brain workout :)
[off-topic]I hope you fly AROUND the fire, as moth tends to burn up rather quickly when flying OVER or THROUGH fire ;D.[/off-topic]
Br.
Vidar
hi Low-Q,
your design has a lot of points to chat about. since it is changing rapidly, let's address what is looking at us now.
QuoteGravity and magnetism cancells out the weight of the water pushing on the pistons.
Magnetism cannot affect the weight of water. So that buoyancy force is the only force left.
if we start with cylinders "g" and "c", we find them in a position they can not be. since they are in the same surroundings, the pistons can not be in different positions in the cylinders. the only thing that might make a very small offset is the magnets force on cylinder "c". as we know, the force will be greatly reduced with distance, so at the surface, there won't be much push.
now let's look at cylinders "b" and "f". at this point in rotation, you are getting closer to the magnet, so the force is getting stronger. i think it will be so strong that it will prevent "f" from getting vertical. like all the magnet motors being developed, you now have a "sticky spot" and a buoyancy on the wrong side.
if you add another magnet above, you are giving it another "sticky spot". this one is because of the attraction. the magnet/piston won't want to leave.
i think if you work out these "sticky spots", you could have a working magnet motor that so many others have tried to build, without luck.
i'll stop here for now. if we can work out these problems, we can then address the next.
good luck!
tom
Quote from: ramset on March 05, 2009, 05:16:52 PM
Hans
And LIFT?[buoyancy]
Chet
Yes, bouyancy... on both sides. "Lift" on one side is "resistance" on the other.
Quote from: tbird on March 06, 2009, 09:56:52 AM
hi Low-Q,
your design has a lot of points to chat about. since it is changing rapidly, let's address what is looking at us now.
if we start with cylinders "g" and "c", we find them in a position they can not be. since they are in the same surroundings, the pistons can not be in different positions in the cylinders. the only thing that might make a very small offset is the magnets force on cylinder "c". as we know, the force will be greatly reduced with distance, so at the surface, there won't be much push.
now let's look at cylinders "b" and "f". at this point in rotation, you are getting closer to the magnet, so the force is getting stronger. i think it will be so strong that it will prevent "f" from getting vertical. like all the magnet motors being developed, you now have a "sticky spot" and a buoyancy on the wrong side.
if you add another magnet above, you are giving it another "sticky spot". this one is because of the attraction. the magnet/piston won't want to leave.
i think if you work out these "sticky spots", you could have a working magnet motor that so many others have tried to build, without luck.
i'll stop here for now. if we can work out these problems, we can then address the next.
good luck!
tom
Hi,
I have made a new drawing. It's bigger and better. The surroundings are the same for all pistons and cylinders now, as water, or a havy liquid matter, is filled all the way up. There is a vent between the oposite cylinders to let the air go back and forth.
Let me explain the magnets in this drawing:
The bottom stator magnet is with a sertain force preventing the piston to pass through. This is however just temporary, as it is later cancelled out with the same force on the other side. It is called cogging. The important thing is then to reduce this cogging so much so the buoyancy force will be strong enough to make that magnet pass through the repelling forces at the bottom.
That is done by adding an perfectly oposite magnetic force at the top. As the top rotor magnet is approaching the stator magnet on top, it will feel greater and greater attraction, but the repelling forces at the bottom is doing the very oposite, and cancels out the forces at the top - and visa versa. So eventually there is no cogging left to prevent the buoyancy force from moving the wheel.
Initially the magnets was introduced to equalize the force needed to move the mass of water upwards from the lowest to highest point of the wheel. And as the water isn't affected by magnetism, and therfor still is forced downwards by gravity, it will continously try to lift the empty volume at the right side without using the same force to lift the water upwards - which is preventing "conventional" buoyancy wheels from working in the first place.
Br.
Vidar
Quote from: erickdt on March 06, 2009, 12:09:53 PM
Yes, bouyancy... on both sides. "Lift" on one side is "resistance" on the other.
Can you please explain what you mean with resistance, and lift on both sides?
Br.
Vidar
Quote from: Low-Q on March 06, 2009, 12:33:55 PM
Can you please explain what you mean with resistance, and lift on both sides?
Br.
Vidar
I'm saying what you're calling lift on one side will work against you with an equal ammount of force on the other side (in your original design).
Quote from: erickdt on March 06, 2009, 12:54:55 PM
I'm saying what you're calling lift on one side will work against you with an equal ammount of force on the other side (in your original design).
Maybe it does. I'm very sceptical to free energy, so you're probably right, but I cannot see how a big "bubble" has the same buoyancy force as a small "bubble".
Vidar
hi Low-Q,
let's look at 1 thing you didn't address in your last design. water pressure at depth.
to start, looking at assembly "ae", you gave the magnet force, of the 2 stators combined, = to 1 gf (gravity force). so doesn't this make it neutral when influenced by both magnets? won't this allow the higher water pressure at the deeper end ("e") to simply push the piston down to the bottom of the cylinder? since the air has a passage to "a", it won't offer ant resistance to the piston and at 1/2gf i don't think the bottom gravity compensation stator magnet would be enough resistance to prevent the piston from going to the bottom of the cylinder.
even if it didn't go all the way, you would still end up with more buoyancy at the top ("a") than the bottom ("e").
would it work if the magnet strengths were increased to 1 gf each? would that increase the cogging too much?
QuoteThat is done by adding an perfectly oposite magnetic force at the top. As the top rotor magnet is approaching the stator magnet on top, it will feel greater and greater attraction, but the repelling forces at the bottom is doing the very oposite, and cancels out the forces at the top - and visa versa. So eventually there is no cogging left to prevent the buoyancy force from moving the wheel.
one point here. these don't cancel. they help each other. both are trying to raise the piston assemblies. they work together. being farther from the bottom magnet helps get the piston past there, but hurts getting its other end past the attractive magnet at the top.
if you can work this out, we can chat more about other things in the design.
tom
Quote from: tbird on March 06, 2009, 02:14:04 PM
hi Low-Q,
let's look at 1 thing you didn't address in your last design. water pressure at depth.
to start, looking at assembly "ae", you gave the magnet force, of the 2 stators combined, = to 1 gf (gravity force). so doesn't this make it neutral when influenced by both magnets? won't this allow the higher water pressure at the deeper end ("e") to simply push the piston down to the bottom of the cylinder? since the air has a passage to "a", it won't offer ant resistance to the piston and at 1/2gf i don't think the bottom gravity compensation stator magnet would be enough resistance to prevent the piston from going to the bottom of the cylinder.
even if it didn't go all the way, you would still end up with more buoyancy at the top ("a") than the bottom ("e").
would it work if the magnet strengths were increased to 1 gf each? would that increase the cogging too much?
one point here. these don't cancel. they help each other. both are trying to raise the piston assemblies. they work together. being farther from the bottom magnet helps get the piston past there, but hurts getting its other end past the attractive magnet at the top.
if you can work this out, we can chat more about other things in the design.
tom
Thanks for your input. It helps me seeing things from another angle.
The fg is the total gravital force acting on both A and E pistons. Equals 1. Therfor fm is 2 x 1/2 fg. The higher pressure deeper down is among the total force fg. A and E is also linked, but also fixed so it is not possible to lift or lower the pistons if we stop the wheel by hand at point AE. So the magnets must lift AE with the same force as the total weight of the pressure acting on the pistons and the weight of the pistons itself. As long buoyancy of B< F, C<G, D<H, and the force to lift water from EFGH to ABCD is taken care of by the force of magnetism, it will allways be unbalanced.
Regarding the buoyancy force of a given volume, it is the same in any depths. The volume in A and E are the same. That means the buoyancy force is equal. The pressure on the piston E is greater than piston A, but that doesnt change the level of the pistons in AE. I believe that "conventional" buoyancy wheels does have a smaller volume in the bottom because the force of the water weight will compress the volume at the bottom more than on the top - so the wheel will allways be heaviest at the bottom and stops rotation.
When it comes to the magnets, I cannot see what you think. It is right that these two magnets are helping each other, but that is much of the whole point. Two magnets will make a more uniform magnetic field through the whole revolution of the wheel, than only one magnet would do. The force working on the stator magnet, is with two magnets a push-pull system. The closer to the magnets the rotormagnet is going, the closer to point AE you go, the more the magnets will push and pull the rotor magnets upwards. The magnets are maybe not correctly shown, but with right shape and flux distribution, we can discuss further. And remember that the rotor magnets are not fixed at the cylinders, but the moving pistons which feels the water pressure.
I will take a closer look into your reply above, and try to figure out more about this, and where the flaw is hiding. I have a bad feeling that there is an obvious flaw somwhere, but I simply cannot fint it anywhere.....
Br.
Vidar
hi Low-Q,
i'm confused.
QuoteA and E is also linked, but also fixed so it is not possible to lift or lower the pistons if we stop the wheel by hand at point AE.
you didn't really mean this (fixed), did you?
tom
Quote from: tbird on March 06, 2009, 07:07:50 PM
hi Low-Q,
i'm confused.
you didn't really mean this (fixed), did you?
tom
It's not locked into position so it can't move at all. But if you see that small circle, which is the path for center of the rods, the rods are fixed in that path. When the piston rods are following that path, the pistons will alter direction inside the cylinders when the wheel is rotating.
That means If I lock the wheel, you cannot move the pistons back and forth. If I release the wheel, and you move the pistons AE, BF, CG, or DH back and forth in the cyinders, you will also force the wheel to rotate.
Also if you rotate the wheel by hand, the pistons are forced to alter position - regardless of the magnets are present of course (These are present just to cancel the gravity force working on the pistons, by its own weight and water pressure, in any position.
Br.
Vidar
hi Low-Q,
so there are 4 small disk geared to the main shaft that keeps everything relevant?
i'll have to think about that a bit.
tom
hi Low-Q,
does the wheel rotate 360 degrees? if so, how do you keep the joints on one side only?
tom
Quote from: tbird on March 07, 2009, 07:15:39 AM
hi Low-Q,
does the wheel rotate 360 degrees? if so, how do you keep the joints on one side only?
tom
Let's say that all four joints are fixed to each rod. Each rod with its piston and the belonging cylinders are apart in depth of the picture but tilted 45 degrees from eachother - as you see it on the picture. So there is four separate pairs of pistons and cylinders. The torque close to the wheel are suppose to move the pistons by following the path of the joints.
In order to move the pistons upwards the joint is following the paths right side. This action need force as long the piston is going upwards because of water pressure and weight of the pistons. This force is gravity, but is perfectly, or close to perfectly, counterforced by the magnet configuration.
In the position around C-G there is no uward force provided by the magnets as the rotor magnets are tilted 90 degrees on the magnetic field from the stator magnets. In the particular magnet configuration it does not take energy to tilt the magnets from 0 degrees in A-E to 90 degrees in C-G. So in that last position the joints want to follow the path downwards, or counter clockwise, by gravity - just as I want.
When the joints continue to go counter clockwise, the downward forces from water and weight of pistons is gradually counterforced by upward lift from the magnet configuration. This doesnt require energy either as explained earlier. And remember, there is no sticky spot to fight against. The magnets does not provide work to the system at all. They are in place for counterforce gravity ONLY in the desired positions of the pistons - when they are suppose to lift water upwards. So there is
no counterforce working clockwise on the joints in position A-E - not counter clockwise either. The buoyancy force of the bigger volume from E to A via D, C, and B, is the only force that is left to affect the wheel. So what is the result? More torque counter clockwise than clockwise.
I my theory at least.... I put in the picture again so it is easier to follow the explanation.
We should meet some place so it is easier to explane - where do you live? Send PM if interested :)
Br.
Vidar
Quote from: tbird on March 07, 2009, 08:55:53 AM
hi
you have presented a design i have never seen before. i suppose it is used elsewhere, but i have never taken anything apart that used it. it will take me some time to imagine the different forces at different positions.
just to be clear, the joint is attached to a guide (the small circle) and has to follow that path, right? if so, it could go backwards if it wanted to?
you can continue in the thread or answer here, it's up to you.
tom
ps are the pistons light enough to float or will they sink on their own?
That joint is free to move back and forth unless there is a one-way bearing in place.
I have now also seen the flaw I have been searching for in that particular design. When the pistons are in position D-G there is no torque at the point of the joint. However, both the poles is forced to the right in D-G. As the joint now is on the top og the path, it will be forced backwards rotation.
Further, the force is to the left for B-F, but the joint is now on the bottom of the path. I both cases the joints want to go in the oposite direction.
Case closed untill I try another twist on that design :)
br.
Vidar