Hi,
I have been thinking alot on how to make a gravity wheels and buoyancy wheels.
Both concepts are mostly the same deal, but the problem with "conventional" gravity- and buoyancy wheels is that they spend the same energy to move a weight to one or another part of the wheel to make it unbalanced, as what comes out of it - in fact they rotate as much clockwise as counterclockwise. Adding friction, these devices will stop.
I have already in the half baked section written about another view on gravity wheels. Well, there is not much response there lately, and I do not think the latest idea in that thread works even if I calculated positive average torque. I have no tools to make it, so it will still remain in the computer as an idea with possible flaws in the calculations.
I try to think outside the box -> The keyword is "vacuum"!
The Vacuum Controlled Buoyancy Wheel no. 1 (I just had to name it :))
Anyway:
I have an idea, a buoyancy wheel that works in air, and probably much better in water. Here is the list of my thaughts:
1. Have you tried to close the aperture of a medical squirt after the piston has beed pushed all the way in, and then pulled the piston out while keeping the aperture closed? You will then create vacuum inside the squirt. The funny thing is that vacuum is nothing, therefor it will not take more effort to pull this piston further from 1cm out to 10cm out. The force needed is the same regardless of where the piston is located inside the squirt.
The pull force is about 1 kg at 1 cm2 piston area.
2. If you have two squirts, fixing the squirts when pointing 100% away from eachother at some distance apart and glue the "head" (the part you usually use to push the piston in with) of each piston together, making the two squirt apertures 180o apart.
3. Say you have 5ml of vacuum in each squirt, each squirt has also 5ml left before the pistons pops out.
4. If you now try to move these two pistons (which is now linked together with glue) in each direction to increase space in one squirt and reducing space in the other squirt, you use no longer force to increase or decrease the vacuum volume inside them. As vacuum is nothing, the vacuum space inside each squirt is virtually an infinite space regardless of the physical volume inside. And that fact is the clue to the whole thing here - a virtual infinit space, totally nonaffected by changing in volume.
5. Vacuum weights about minus 1,4 kg pr. m3 at sea level.
6. Let us scale these squirts up big time, so we have two BIG cylinders, measuring 80cm in diameter (5000cm2), and 2 meters long. The volume inside each cylinder is 1m3.
7. A piston is placed inside each cylinder, and the air inside is pumped out. It now requires 5 tons of pull to keep a volume of vacuum for each piston. Lucky for us, both 5 tons of pull is 180o apart - and lets pray that the mechanism between the pistons can stand a total of 10 tons of pull to keep the pistons fixed with constant distance.
8. The alignment of the cylinders and the pistons is vertical. If we use just a tiny little force to slide the pistons upwards (depending on piston weight ofcourse), the cylinder at the very bottom will at last contain 1m3 of vacuum. See point 5. to figure out the buoyancy of that volume. The cylinder on the top now contains air. If there was a shaft in the middle of those cylinders, the whole thing would start to rotate if we gave it a tiny push. When it has rotated one half round, we again use a tiny force to push the pistons up again, and the wheel can continue.
9. The cylinders are 12 meters apart. 1,4 kg buoyancy will then provide a torque of 97Nm in average from bottom to top. Most of the energy used to push the pistons up, is taken from the kinetic energy in its weight on the way down in the rotation. In fact, the weight of the pistons and the remaning mechanism is of no consequence - regardless of rotating speed as well - those forces will in sum be zero in one revolution.
10. Friction is a practical thing we can deal with. It can to a sertain level be controlled and minimized - if essential, and must at least be designed to be less than the energy produced by the vacuum controlled buoyancy wheel.
Anyone who follow this idea? I believe in this idea, and cant find any flaw whatsoever which will prevent this wheel to produce energy. You can talk about forces needed to displace air, or water for that matter, but these actions is cancelled out as one part is displacing air/water while another part is consuming the same amount at the very same time. Hence it does not require more energy than defeating friction to move the pistons. As said: Friction can be minimized - if it is essential.
What will happen if we put this wheel under water? Will we get as much as 34600Nm as average torque in one revolution [(1000kg displaced water x 5m radius) / square root 2))x9,81]? It will be slow due to inertia, but it is possible make a streamlined shape outside the cylinders.
Br.
Vidar
Here is a picture of the squirt idea.
The pistons can be moved from side to side without using more energy than the friction - friction can be controlled, and reduced to a minimum by using an approperiate method. This can then get heavier on one side without using energy - as consumption of air or water at the back of one squirt is the same a the other squirt displace air or water.
Imaging if we scale this up, and fix the squirts to a rotating disc, and let the pistons inside be fixed ofset the axis of the disc so the pistons can move from side to side during rotation, in order to allways make more emty space, thus more buoyancy, at the left or right side of the wheel.
Vidar
Interesting idea. Your logic seems to comply with mine. So far I can't see something wrong.
@Vidar
Here is a design I posted awhile back that uses fluid and vacuum principles.
AB the difference is that he's not using mass transfer. I'm still tinkering with Vidar's idea trying to either find convince myself it's possible or not.
@ Vidar
Just found a video that is not 100 % similar to your idea but deals with vacuum: http://www.youtube.com/watch?v=yJL72ZT2vJo&feature=related.
It is a vacuum engine. It seems to be different from the flame sucker who needs energy (= heat).
Thanks for you picture and for your idea.
Best wishes
Alana
Quote from: Low-Q on November 20, 2008, 07:30:35 PM
1. Have you tried to close the aperture of a medical squirt after the piston has beed pushed all the way in, and then pulled the piston out while keeping the aperture closed? You will then create vacuum inside the squirt. The funny thing is that vacuum is nothing, therefor it will not take more effort to pull this piston further from 1cm out to 10cm out. The force needed is the same regardless of where the piston is located inside the squirt.
The pull force is about 1 kg at 1 cm2 piston area.
2. If you have two squirts, fixing the squirts when pointing 100% away from eachother at some distance apart and glue the "head" (the part you usually use to push the piston in with) of each piston together, making the two squirt apertures 180o apart.
3. Say you have 5ml of vacuum in each squirt, each squirt has also 5ml left before the pistons pops out.
4. If you now try to move these two pistons (which is now linked together with glue) in each direction to increase space in one squirt and reducing space in the other squirt, you use no longer force to increase or decrease the vacuum volume inside them. As vacuum is nothing, the vacuum space inside each squirt is virtually an infinite space regardless of the physical volume inside. And that fact is the clue to the whole thing here - a virtual infinit space, totally nonaffected by changing in volume.
5. Vacuum weights about minus 1,4 kg pr. m3 at sea level.
6. Let us scale these squirts up big time, so we have two BIG cylinders, measuring 80cm in diameter (5000cm2), and 2 meters long. The volume inside each cylinder is 1m3.
7. A piston is placed inside each cylinder, and the air inside is pumped out. It now requires 5 tons of pull to keep a volume of vacuum for each piston. Lucky for us, both 5 tons of pull is 180o apart - and lets pray that the mechanism between the pistons can stand a total of 10 tons of pull to keep the pistons fixed with constant distance.
8. The alignment of the cylinders and the pistons is vertical. If we use just a tiny little force to slide the pistons upwards (depending on piston weight ofcourse), the cylinder at the very bottom will at last contain 1m3 of vacuum. See point 5. to figure out the buoyancy of that volume. The cylinder on the top now contains air. If there was a shaft in the middle of those cylinders, the whole thing would start to rotate if we gave it a tiny push. When it has rotated one half round, we again use a tiny force to push the pistons up again, and the wheel can continue.
9. The cylinders are 12 meters apart. 1,4 kg buoyancy will then provide a torque of 97Nm in average from bottom to top. Most of the energy used to push the pistons up, is taken from the kinetic energy in its weight on the way down in the rotation. In fact, the weight of the pistons and the remaning mechanism is of no consequence - regardless of rotating speed as well - those forces will in sum be zero in one revolution.
10. Friction is a practical thing we can deal with. It can to a sertain level be controlled and minimized - if essential, and must at least be designed to be less than the energy produced by the vacuum controlled buoyancy wheel.
Anyone who follow this idea? I believe in this idea, and cant find any flaw whatsoever which will prevent this wheel to produce energy. You can talk about forces needed to displace air, or water for that matter, but these actions is cancelled out as one part is displacing air/water while another part is consuming the same amount at the very same time. Hence it does not require more energy than defeating friction to move the pistons. As said: Friction can be minimized - if it is essential.
What will happen if we put this wheel under water? Will we get as much as 34600Nm as average torque in one revolution [(1000kg displaced water x 5m radius) / square root 2))x9,81]? It will be slow due to inertia, but it is possible make a streamlined shape outside the cylinders.
Br.
Vidar
1: Actually, the vacuum you will be able to pull in this manner will still contain a lot of air. And if your proposition were true, then commercial vacuum pumps wouldn't need to run for hours just to get those last few microns of air out of a chamber...but they do. SO it does in fact take more work to pull that plunger further out--because you are pulling a greater vacuum.
2,3: OK
4: See #1. Your idealization is incorrect because your vacuum isn't perfect. You will in fact be doing work as you move the pistons back and forth, against friction and against the residual air (you will be heating the residual air, some of this heat will escape, the process is lossy).
5: Totally wrong. Vacuum weighs nothing, not negative. Buoyancy is different, but here you say "weighs" and that's just wrong.
6, 7, 8: OK
9: the key is that you realize: "In fact, the weight of the pistons and the remaning mechanism is of no consequence - regardless of rotating speed as well - those forces will in sum be zero in one revolution."
10: That's right. Refer to #9 above, and you will see that if you reduce friction to ZERO, the forces will sum to zero. The only way you can get continuous rotation is to reduce friction to BELOW zero. Which is the same as providing an external source of driving power.
Your conclusion:
"You can talk about forces needed to displace air, or water for that matter, but these actions is cancelled out as one part is displacing air/water while another part is consuming the same amount at the very same time. Hence it does not require more energy than defeating friction to move the pistons. As said: Friction can be minimized - if it is essential."
That's totally right. You just need to reduce friction to less than zero, and then you've got it. Of course putting the apparatus under water, while increasing buoyancy (and increasing the force required -- you have to lift all that water now, not just the pistons) also increases the friction, greatly, from viscous drag. So by going into the water you have an increased torque developed, an increased torque required (these exactly balance) and greatly increased frictional drag.
Quote from: AB Hammer on November 21, 2008, 12:25:52 PM
@Vidar
Here is a design I posted awhile back that uses fluid and vacuum principles.
I think the problem with that design is that the brass piston is using gravity to push the oil upwards. Which means that the wheel is forced backwards by the piston, and forward by the oil. I cannot decide in what direction this wheel is going. Maybe i haven't looked close enough.
Edit: I looked at it once more. I cant for sure say if it works or not, it is interesting though. It should be possible to calculate if you have dimentions, oil density, brass piston weight etc.
Br.
Vidar
Everyone. I have thinked a bit more.
Btw. vacuum has no weight, but in air the weight is negative. Weight is relative. Even if the vacuum isn't perfect, it will still work. I have tried those squirts. It takes 1,5 kilo pull from the first millimeter to the last 10cm. So it is sufficient linearity of that vacuum.
Here is a "small" part of what idea I came up with. it is similar to what I already have with a few differences. I started in MS Word to take som e notes, so here it is "copy and paste":
Perpetial motion without defying the laws of physics.
The power of vacuum in an environment not meant for it.
Vacuum is a state in physics which is known as an empty space, with no mass, with no gravity. Here on earth this state is not natural as there is an atmosphere which surrounds us with a certain pressure â€" named atmospheric pressure. Due to this fact, if vacuum could exist in that environment, it would disappear in an instant. It is however possible to create vacuum in sealed boxes by pumping out the air inside it.
My concept:
This is a buoyancy wheel, but do not stop reading here. It is a common and accepted knowledge that buoyancy wheels doesn’t work. That is because they are constructed to work in liquid. That simply doesn’t work. Liquid will have different pressure at bottom and the surface of a tank filled with liquid, but liquid cannot be compressed to a denser liquid, therefore density remains the same in both places regardless of pressure. That means that the extra pressure at the bottom will make an air bubble smaller there and therefore contain less volume for buoyancy.
Using gas instead of water:
The alternative is to put a buoyancy wheel inside a box of heavy gas. Buoyancy of air in a heavier gas is not as much as air in water. A heavy gas such as Sulfur Hexafluoride, is 6 times heavier than air and is the heaviest gas compound known to man. An air bubble will at the bottom of a deep tank filled with Sulfur Hexafluoride bee compressed a bit due to the weight of the heavy gas (Like what’s happening in water), so the volume of air will decrease. The good thing is that the gas at the bottom is denser, so even if the air bubble is smaller, the denser gas will compensate for it so buoyancy of that air bubble will be maintained in any position in the gas tank.
Vacuum instead of air:
Altering a volume of air in a sealed box is difficult. The process of increasing and decreasing such volume is not linear, and requires different force to double or halve the volume. It requires 4 times more force to halve than it takes to double it. That said you use twice as long distance to double the volume as it take to halve the volume, which means it takes 2 times more energy to halve.
If the box is “filled†with vacuum, it is used the same force to decrease or increase the volume. In addition the force is always pointing in the same direction when halve or double the volume, as opposed to an air filled box.
With vacuum it is also easier to counterforce as it is a nearly linear function regardless of volume. Vacuum is virtually an infinite space, therefore.
The construction of the vacuum buoyancy wheel:
This is made simple â€" very simple. I have two tubes made of aluminum. Both tubes have the same length, and the same diameter. They are both polished inside. There is one sealed end with a tire valve mounted backwards on both tubes. The other end is open. Into this open end, a piston is pressed in, and presses the air inside the tube out of the tire valve. The tire valve will close when we pull the piston back again. This creates vacuum inside the tube. It is important that the piston fits perfectly into the tube to prevent leakage.
So, I have two tubes. Each tube is fixed on a solid aluminum profile with the tire valves pointing 180 degrees apart. The pistons are then linked together, so one piston is using force from its vacuum to counterforce the vacuum in the other tube. The distance between the tubes and length of the piston links, makes the volume of vacuum to the half in each tube.
Now as the pistons are linked, there is no net force that acts on the system. Anyone is now, without much force, free to move the pistons in each direction in order to decrease vacuum volume in one tube, and increase it in the other tube.
Further, these tubes have an axis right in the middle between them, right through the aluminum profile they are fixed to, now called “main axisâ€. The pistons are fixed to an axis off centre the tubes axis. This will force the pistons to alter direction in proportion to the tubes, when one starts spinning the tubes around on its main axis. So the volume in each tube will go from almost nothing to maximum during one round, and practically without spending force to do it. Compare it with moving an object from side to side in thin air.
The axis of the pistons are fixed at the left side of the main axis â€" this offset horizontally, left or right, is important. So there will always be most volume in the tube at the right side of the wheel in this case.
Let’s see how it works in practice:
In air, the tube with most volume, at the right side, is lighter than the tube on the opposite side, with almost no vacuum volume. In fact all parts of the right side have always more vacuum volume than the left side. So vacuum is used to remove mass on the right side.
When the tubes are aligned vertically, there is a half and equal amount of volume in them. At horizontal alignment, there is full volume in the right tube, and almost no volume in the left.
Now this wheel is placed in a box of heavy gas. The parts of the wheel that is nearest the bottom will have slightly more pressure than the one on the top. Accordingly the lowest parts of the wheel will have greater buoyancy as the gas is denser at the bottom and the volume is locked by the link between the pistons and its axis. What I say is that there is no net force that will prevent this wheel to rotate at any position. There will always be a torque in one direction.
As you understand, the buoyancy of a small volume of vacuum in air isn’t much, as 1 liter of air weighs about 1,4 grams, the buoyancy of 1 liter vacuum is therefore 1,4 grams. So we need to scale up the wheel a great deal, increase diameter to increase torque, use a heavier gas to increase torque, and increase volume to increase torque. Say we make a 1m^3 tubes a few meters apart. Fill a tank with Sulfur Hexafluoride and put a multiple vacuum tube wheel in it.
Ok here's a further addition. I ended up having to use a sealed container (black part on drawing). Which has water between the pistons/plates. the idea as Low-q mentioned is that buoyancy will pull the vacuum part of the container up by a very big force. When it has rotated 180° the pistons need to be pushed up. This is not free as Low-q mentions. But my argument is that the energy of the rotation is far above the energy needed to push the liquid 1 meter up. According to my primarily calculations that is...
1) finding out how much energy it takes to lift the mass of liquid between the pistons up 1m
2)finding out how much energy the buoyancy provides due it's torque
And comparing 1 and 2 I found out that using my dimensions I have 1.5x more energy from the rotation than lifting the pistons 1m up. I could have made a mistake and could be wrong. But please take not this is pure conceptual. Meaning every part is considered mass less. The liquid is considered non-viscous, that is when a object moves through it there's no resistive force in essence it can move freely. The last one is important in calculating the energy from the rotation. Pretty much assuming that the wheel has no resistance during rotation.
More will follow.
After parametrization of the formulas I came to the conclusion. That the work W_rot due to the rotation of the structure and the work W_w due to the water displacement has a certain fixed relationship. I really think I did something wrong since this is just too simple to have missed for centuries :P.
Anyways the relation in my case having the vacuum cylinder always volume being V= pi m^3. That is a diameter of 1m and height of 1m. I found out no matter how deep you go (thinking it'll increase the torque if I had the volume at some deeper level) there's always the same relation between the two work formulas.
Namely W_rot/W_w = r / (r - 0.25) ...r being the distance from the center to the center of the vacuum. As you can see W_r (the energy from the rotation) is bigger than the energy needed to displace the rod with pistons.
From Simanek's site: http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm (http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm)
"Consider a body of mass m and volume V that is less dense than water. If it is pushed underwater and released, it moves upward. As it moves upward from position A to position B, water must move downward. The volume V at position A which the body occupied must now be replaced with water, and the water that was in the space it now occupies at position B must be displaced. The net effect is that as the volume V of the body moves from position A to position B, an equal volume V of water moves from position B to position A.
(I couldn't get the table to format correclty so it's omitted--TK.)
Whenever a mass moves in a gravitational field, work may be done on the body by the field or on the source of the field by the body. That work is of size Fh where F is the size of the force exerted on the body by the field and h is the distance moved vertically. When a body moves up in the field (opposite to the gravitational force) it does work on the source of the field. When the body moves down in the field (in the direction of the gravitational force) the source of the field field does work on the body.
We tabulate the force and work on the body (mass m) and on the equal volume of water (mass M). Since the body is lighter than water, m < M. The net work done on the body moving up in the liquid is (M-m)gh = (Mg-mg)h. We identify Mg as the size of the Buoyant force, and mg as the force due to gravity acting on the body of mass m.
Mgh is the work done by gravity on the water during the displacement downward. -mgh is the work done on the lighter body being displaced upward. "
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fwww.lhup.edu%2F%257Edsimanek%2Fmuseum%2Fthemes%2Fpiston2c.gif&hash=f2bb6663fefd5fdad1a56ee7efcc17b628393a34)
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fwww.lhup.edu%2F%257Edsimanek%2Fmuseum%2Fhydraul3.gif&hash=c4baa60a72f5b8d315eba4aad0d8fb028cd77b2d)
In broli's figure 1, a 2 meter by 2 meter cylinder of water falls one meter. This is all the energy there is in the system, and this is where the buoyancy comes from. In figure 2, that same cylinder of water must be raised back up. If you think that the rotating path makes a difference in the energy, then yes, you have made a mistake in your calculations or assumptions!!
Quote from: broli on November 21, 2008, 04:53:55 PM
Ok here's a further addition. I ended up having to use a sealed container (black part on drawing). Which has water between the pistons/plates. the idea as Low-q mentioned is that buoyancy will pull the vacuum part of the container up by a very big force. When it has rotated 180° the pistons need to be pushed up. This is not free as Low-q mentions. But my argument is that the energy of the rotation is far above the energy needed to push the liquid 1 meter up. According to my primarily calculations that is...
1) finding out how much energy it takes to lift the mass of liquid between the pistons up 1m
2)finding out how much energy the buoyancy provides due it's torque
And comparing 1 and 2 I found out that using my dimensions I have 1.5x more energy from the rotation than lifting the pistons 1m up. I could have made a mistake and could be wrong. But please take not this is pure conceptual. Meaning every part is considered mass less. The liquid is considered non-viscous, that is when a object moves through it there's no resistive force in essence it can move freely. The last one is important in calculating the energy from the rotation. Pretty much assuming that the wheel has no resistance during rotation.
More will follow.
I think the problem is that you cannot use liquid. Liquid is not compressable so at the bottom the density of the liquid is not higher due to higher pressure. So the pressure acting on the piston at the bottom, will be greater than the top, and the piston cannot move upwards. Since the volume doesn't change from <
oto 360
o, the buoyancy will be the same at all levels of depth, but the pressure on the piston wont.
In a heavy gas the buoyancy will be greater the deeper the constant volume is present, or at least the buoyancy at greater depths will increase parallell to the increasing pressure acting on the piston at the bottom. So those forces cancels out. Left you have the a greater buoyncy at the right side of the wheel - which i also believe will be sufficient to defeat friction.
Br.
Vidar
I have to point out that I did make a mistake in my calculation. The 1.5x came out of the radius which was really 1m so the end result should have been 1x in other words no gain in output. I realized the mistake when I was in my bed trying to sleep ::).
Here is a new and better picture of my idea.
The idea is that the torque from the buoyancy in cylinder B, C, and D should defeat the differential force that exist between 2-6, 4-8 and 3-7 and still have torque to spear.
I will point out that this wheel is not going to be tested in water, but in a heavy gas. If the gas have the same atmospheric pressure and density at the bottom as on the top, will it take energy to displace gas at the bottom and consume the same amount at the top?
And the question is: Will it make any difference in using gas instead of water?
Here is an example of a heavy gas that can be used:
http://www.youtube.com/watch?v=lpWjR8r_Uvc (http://www.youtube.com/watch?v=lpWjR8r_Uvc)
Here is the model. The green numbered arrows, represent the force that acts on the pistons.
That linkage would be mechanically feasible for one pair of pistons, as you showed in your earlier animation, but with several like here, the linkage is impossible. It requires parts to interpenetrate, at the axis.
Now, ignoring that, the substitution of a heavy gas to eliminate the pressure gradient doesn't change anything, because the pressure gradient in water isn't what prevented the submerged wheel from operating in the first place. It's the fact that, in all these configurations, as broli has discovered, the Work Out equals the Work In, once you have found all the masses that are being moved and account for them.
Keep thinking, though.
Quote from: TinselKoala on November 22, 2008, 11:33:42 AM
That linkage would be mechanically feasible for one pair of pistons, as you showed in your earlier animation, but with several like here, the linkage is impossible. It requires parts to interpenetrate, at the axis.
Now, ignoring that, the substitution of a heavy gas to eliminate the pressure gradient doesn't change anything, because the pressure gradient in water isn't what prevented the submerged wheel from operating in the first place. It's the fact that, in all these configurations, as broli has discovered, the Work Out equals the Work In, once you have found all the masses that are being moved and account for them.
Keep thinking, though.
I have been blind on the thaught that a gas (maybe water) in the same gas (or water) is relatively weightless, thus it isn't more work done by moving it upwards than sideways. The second thing that made me blind, was the thaught of a volatile gas (maybe water) which didn't "care" where the gas (maybe water) came from to fill up the cylinders rear side as the vacuum volume are decreasing gradually from the rightmost part to the leftmost part of the wheel. If we make a bubble of vacuum in this tank of heavy gas, I have been thinking of a net force working on the linkages due to vacuum in both cylinders and the possible difference in atmospheric pressure on the upper and lower pistons.
Do you have any comments on these thaughts?
Br.
Vidar
What happens if we attach a bubble or a springload right in the center on the rods?
This bubble contains a volume equivalent to the mass of displaced water + pressure differance + weight of the rod and pistons.
If spring loaded the force upwards will be adjusted accordingly.
This bubble or spring load is following the path of the linkages. The buoyancy of this volume, or force of the spring load, will equalize the force needed to displace water, the pressure difference, and the weight of the rod and pistons, regardless of angle of the system.
This bubble or spring load have only one task: To equalize the factors that stop this wheel from spinning.
I have attached a picture to show you how it can be solved with a weight. Now it will never require force or energy to lift the water upwards. So left there will be only the torque from the buoyancy from the empty space on the right side.
Did I miss anything?
Br.
Vidar
Further we can put this model in a wheel:
There is some small text in the picture.
Hi, T-Koala,
Quote from: TinselKoala on November 22, 2008, 11:33:42 AM
That linkage would be mechanically feasible for one pair of pistons, as you showed in your earlier animation, but with several like here, the linkage is impossible. It requires parts to interpenetrate, at the axis.
I think the "interpenetrating parts" are technically rather easy doable (axle bypass pipes). The main technical obstacles are still (not mentioning a conservative principles...) a
really high friction with vacuum-tight pistons, a very bad force parallelogram (look at the Force vectors in certain areas - a piston-push circular offset...).... etc...
Quote
Now, ignoring that, the substitution of a heavy gas to eliminate the pressure gradient doesn't change anything, because the pressure gradient in water isn't what prevented the submerged wheel from operating in the first place. It's the fact that, in all these configurations, as broli has discovered, the Work Out equals the Work In, once you have found all the masses that are being moved and account for them.
Keep thinking, though.
This idea (gravity/buoyancy) using a "vacuum"/volume in a dense gas environment has some advantages and many disadvantages comparing to a "common" (Air/water) buoyancy combinations.... Of course, both concepts should be "less effective" than a classical overbalanced wheel (weights in a normal atmosphere, sorry, gravity)...
The buoyancy difference between the gasses/fluids should define the theoretical "buoyancy difference" leading to a (possible) overbalance and potentially work performed. Mass in a gravity field has a better potential difference in relation to any buoyancy concept so far, so it's a way to go.... ;)
How many buoyancy wheels has been built or designed based on vacuum?
Do they only exist on paper? At least mine does...
Is vacuum, as a non-material, a part of thermodynamics?
Wouldnt vacuum be the absolute zero-energy level? Wouldn't it allways be a potential difference in an environment full of matter?
Vidar
Some, sort of:
From the Simanek site:
1833 [No. 6510] Barthelemy Richard Comte de Predaval of Leicester Place, London, Engineer. Complicated mess of water, turning cylinder, drum, pistons, friction plates, springs, etc. It acts "by a joint power derived from the buoyancy of a body in fluids, and the weight of a body in vacuo." (A joint stock company was based on this patent.) Long description in Dircks (1861) pp. 420-427, with four drawings. A rotating drum is in a cylinder. Carefully machined "seals" divide the space between drum and cylinder into a left portion, filled with water, and the right portion in a vacuum. Buoyancy was expected to keep it turning. The inventor describes a simple experiment with a half-drum intended to demonstrate the ability of the buoyant force to turn the half-drum.
"Wouldnt vacuum be the absolute zero-energy level? Wouldn't it allways be a potential difference in an environment full of matter?"
Pretty much. That's why it's so hard to find a real vacuum. For example, I'm working with a vacuum chamber about the size of a washing machine. Say 200 liters capacity or so. Call one atmosphere of pressure 1 million microns. It takes a mechanical roughing pump about an hour to get the chamber down to 1000 microns, then another hour to get to 100 microns, then another hour to get to 10 microns. Then I start the turbo-molecular pump. It takes 10 minutes to speed up, but then it fairly quickly gets the chamber down to below 1 micron (or 10e-3 Torr). After a couple hours, we are down to the mid 10e-6 Torr, and I can turn on the high-voltage ion pumps, which are totally electrical and have no moving parts. If things are looking good, I can valve off and shut down the mechanical pumps at this stage. By evening the chamber may be down to 10e-8 Torr. But there is STILL air in there, and other stuff too. At these pressures, almost everything has a detectable vapor pressure. So when I fire up the residual gas analyzer I might be able to detect water, air gases, crud from lubricants and seals, even some metals, you name it. And that's in a medium-hard laboratory vacuum, far better than anything you are going to be able to achieve by pulling on plungers.
In fact, any vacuum that's in direct contact with water can't be lower pressure than the vapor pressure of water at that temperature.
Thanks for the information above :)
Can't you start with an initial vacuum made by pulling a piston out? If you use CNC made mechanism, you start with no volume, at least a micro-volume. Then you pull back the piston by a great force. Let's say you manage to get 1 micrometer average space between the piston and the bottom of the cylinder. Then you pull the piston out to one meter. Shouldn't you get 1micro bar in an instant? Then you can continue to pump out the rest of it. You will save alot of time I guess.
With my biggest squirt, I fill some of the volume at the bottom with vaseline to prevent space when I press the piston to the bottom. Then I seal the aperture, and I visually inspect the bottom to see there is no bubbles in there. Then I pull back the piston. A scale show me that I use the same force of pull all the way - about 2,7kg - according to the area of the cylinder too. That rubber piston will in fact not leak. I pulled it out for a few minutes. When I released it, the piston went back to the bottom, and no visible bubbles after a few minutes of vacuum. I was quite surprised it worked that well.
Of course, if you have 200 liters, the great area of the piston would be hard to pull out...
br.
Vidar
Hi,
I understand now that I have to use my hand and energy to lift up the weight in water displaced by the cylinders in order to have a rotating buoyancy wheel. So I now replaced that hand with hollow pistons that have the amount of volume required to lift itself up, making it easy for the pistons to move that water upwards. There is no limit how much volume it can be that lifts the pistons upwards. And still it will be possible to have the very same buoyancy difference between left and right.
Where is the flaw? It must be a flaw, so where is it?
br.
Vidar
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