Sum of torque

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Another position,the arm turned of 45° the squares too

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I change the gas every 90°.

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This device don't create energy, just use the energy from the temperature of a gas (pressure). The arm turns at w clockwise. The disk don't turn around itself. The pneumatic cylinder has a spring inside and has the pressure P everywhere except between the cylinder and the disk. The disk decreases its angular velocity in the arm reference and the cylinder compress the spring. The temperature of the gas decreases. I drawn the device at the time the pneumatic cylinder must work, exactly at this time, it's a transcient action.

Second image: all the device turns clockwise around the red center. The square can turn around the blue center. The device don't receive a torque but the square receives one. The square receives the forces Fa and Fb and Fc, these forces give F2 on the blue center. Inside the red circle there is no pressure.

The torque on the red circle from F1 is:

C1=(R1+R2/2)*R2*P  (I suppose the depth is 1)

The torque on the red circle from F2 from the blue center is:

C2=R1*R2*P+(R1+(R2²-2R1²)/2/R1)R1+ (R1-(R1+(R2²-2R1²)/2/R1))*R1

The square in the contrary receives a torque of:

C3=(R1+(R2²-2R1²)/2/R1)R2*P

Numerical application:

R1=10
R2=1
C1=10.5
C2=10.50012
C3=0.05

If the red circle accelerates more and more, the square keeps its relative position with the red circle, so the square is always like I drawn. I need to give an energy to turn the red circle (the device) but I win the energy of the rotation of the square around itself.

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I can use one or 2 half torus. Outside the pressure is near 0 and inside the grey container there is 1bar. The arm turns clockwise at w. At start the half torus don't turn around itself so it turn at -w in the arm reference.

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Shows the image