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In this case: It's a theoretical study, just for find my error. No gravity. No buoyancy. All volumes are constant. I suppose no friction. Outside, the pressure of air is constant at $P$ (atmospheric pressure for example at 1 bar). A basic device is composed of an arm and a circle. There is a gas inside each circle at a constant pressure. The black arm can turn around the blue axis. The circle is fixed to the arm. The circle doesn't turn around itself. I have drawn 2 basic devices but it possible to have $N$ devices: in the third image, I have drawn 16 basic devices. Walls between circles can be removed when it is necessary: red in the drawing. Two gaskets not drawn prevent the gas to escape. The system suppresses wall like that a part of each circle receives a torque on it. Like all circles don't turn around the same center, the torque is not the same. The pressure inside a circle can be $2P$ or $\frac{P}{2}$ and the pressure is constant because the volumes are constant.

There is a difference of torque:

enter image description here

The sum of torque of this basic device at $time=0$ is $F_1d_1\cos\theta-F_2d_2\cos\theta$. Like the distance $d1$ is greater than $d2$, the torque is higher counterclockwise from $time=0$ to $time=1$. So, the sum of the torques is not 0, the system gives an energy.

The device turns from $time=0$ to $time=2$.

enter image description here

The device turns from $time=0$ to $time=0.5$

enter image description here

At $time=1$, the pressure inside circles must change. This step doesn't need energy (in theory) because all volumes are constant. With a device with 16 basic devices I can, for example, exchange 2 circles with $2P$ with 2 circles with $\frac{P}{2}$ if the angular velocity of the all device is low.

It's possible to use another method for exchange the pressure: use an external device. The temperature will increase when the external device changes the pressure inside the circles, so it's necessary to give an energy to its external device but this energy is at the higher temperature. I don't lose any energy in this step with an external device.

Even in a short time (without changing the pressure inside circles) the device can't have a net torque on it from $time=7$ to $time=1$.

The third drawing shows a device with 16 circles, 2 circles at each time, all 16 arms turn counterclockwise at the same angular velocity. Two circles work together. At the exact position: $time=1$ $time=3$ $time=5$ and $time=7$ there is no work because there isn't a difference of torque.

enter image description here

Inside circles, the pressure is $2P$ from $time=7$ to $time=1$, from $time=3$ to $time=5$. Inside circles, the pressure is $P/2$ from $time=1$ to $time=3$, from $time=5$ to $time=7$.

Gaskets rotate like the following image shows:

enter image description here

Maybe the device lost an energy with the gaskets but I don't see how.

So my question is: what wrong, where I lost energy for compensate the energy I won in the difference of torques ?

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  • $\begingroup$ The explanation of the supposed workings of the machine are quite difficult to follow. Your final statement seems to suggest that "the lost energy is compensated", a device moving indefinitely without friction or another process extracting energy is not surprising (correct me if I understood something wrong). $\endgroup$ – Sebastian Riese May 31 '15 at 16:34
  • $\begingroup$ Like there is a net torque (d1>d2) on each basic device, the angular velocity increase more and more, the device win kinetic energy. $\endgroup$ – Sx7 May 31 '15 at 16:38
  • $\begingroup$ No matter how complex the scheme, they all fall over when they meet the laws of thermodynamics. It is impossible to prove that an "over unity" device works just using theory. If you want to convince people it is possible, build one and openly demonstrate it $\endgroup$ – user56903 May 31 '15 at 16:39
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    $\begingroup$ Aside from analyzing the mechanics of the device (assuming no friction seems unreasonable) there is the question- What's maintaining the pressure ? $\endgroup$ – paisanco May 31 '15 at 17:22
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    $\begingroup$ Moderators - you already know that perpetual motion machines don't work. Why wasn't this question "filtered"? $\endgroup$ – David White Apr 1 '16 at 0:02
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I'm having trouble understanding how this machine works (does the torque come from buoyancy? If so, you need gravity.), but the following statement is false.

At time=1, the pressure inside circles must changed. This step don't need energy (in theory) because all volumes are constant.

In order to change the gas pressure inside a container, you must either change the amount of gas or change the temperature of the gas, both of which require work or energy. For example, to increase the pressure from $P/2$ to $2P$, you have to force more air into the container (requiring work), or you must raise the temperature of the container (requiring heat input).

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  • $\begingroup$ I think I understood my error, there is no torque if radius of circles are the same. If I use a liquid inside circles, how centrifugal forces are compensated ? If the device turn very quickly, enough to be higher than the speed of sound in the liquid, the pressure dont' have time to be the same in each circles. $\endgroup$ – Sx7 Jun 1 '15 at 15:09
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    $\begingroup$ @Sx7: You could save yourself a lot of wasted time by not struggling to invent perpetual motion machines by adding more and more complications. If you don't understand something, it usually isn't helpful to add complexity. Also, I suspect more advances have been made by reducing complex problems to simple ones than vice versa. $\endgroup$ – RedGrittyBrick Oct 13 '15 at 14:22
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Your most obvious problem is:

"I suppose no friction."

This is an assumption that is never physically true and is a leading reason that perpetual motion machines, in general, never work.

Even orbiting planets experience a tiny amount of friction from tidal forces on bodies that are never truly perfectly rigid that converts the kinetic energy of angular momentum into heat or other forms of internal motion and reduces angular momentum ever so slightly over time.

I also suspect that another main reason is that even in the absence of friction in the machinery (which is impossible) the Second Law of Thermodynamics in the fluid is in play, although it is a bit difficult to follow what is going on in the contraption to do a proper analysis of it.

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you are only considering the torque on the device without the air inside. But the mass of air is moving within the device so it needs to be taken into the system on which you compute the torque. The torque on the whole systeme (device+air) is zero as expected. Action/Reaction if you want. Or no external force on the system.

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