Sum of torque

[EOW]

With the wheel in rotation the energy is :

Energy lost by the moment: √3*R*δ*F = 0.001*√3*1*0.45 = 7.79e-4 J
Energy win by the center: √3*R*δ*F = 7.79e-4 J
Energy needed for the force F2: ( √ ( (θ-sinθ)² + (1-cos(θ))²) - √ ( ((θ+δ)-sin(θ+δ))² + (1-cos(θ+δ))²) ) *2*F = 4.27e-4 * 2 *F = 3.84 e-4 J

[EOW]

With a gas under pressure inside the triangle, the sum of energy is not 0:

L = 1 m =  thickness of the device
P = 1 Pa = pressure of the gas
R = 1 m = radius of the wheel

The force on the red arm is √3 N so F2 = √3 N
The torque on the red arm around the red axis is 1.5 Nm

Energy studied for an angle of δ = 0.001 rd

I apply the force F2 from the ground, I lost an energy because the length increases. The force F2 come from a spring for example. I consider there is no acceleration, I can use a big mass for the wheel of use an external device to keep constant the velocity.

Energy lost by the moment on the red axis: √3*R*δ  = √3e-3 J
Energy win by the center in translation: √3*R*δ = √3e-3 J
Energy needed for the force F2: ( √ ( (θ-sinθ)² + (1-cos(θ))²) - √ ( ((θ+δ)-sin(θ+δ))² + (1-cos(θ+δ))²) ) *2*F = 4.27e-4 * √3 = 7.39 e-4 J

[EOW]

I don't need the balls and the springs. I can use the pressure of a gas.

The wheel rotates and turns before I studied the sum of energy.

There are 4 bodies:
A: the wheel + 2 red walls, the center of the wheel receives the black force, the outer circle of the wheel receives the red forces
B: the black wall, can turn around the red axis (i), receives the gray forces
C: the spring, gives F2 to the black arm and -F2 to the ground
D: the ground, it is fixed, receives the green force

Radius of the wheel = 1 m
All forces at sqrt(3) N
P = 1 Pa

F1 come from the difference of pressure of the gas on the black wall
F2 come from the spring to the black arm
-F2 come from the spring to the wall
Fi come from the black arm on the axis
-Fi come from the axis to the black arm
Fo come from Fi to the center of the wheel
-Fo come from the center to the outer circle
F4 come from the difference of pressure of the gas on the red walls

I studied the sum of energy for a small angle δ like  0.0001 rd for example

3 walls (2 red+1black) shape a equilateral triangle, there is a gas under pressure at 2P inside the triangle. The surface of the triangle is always constant, the gas never lost its potential energy. Outside the triangle there is the pressure P.

The wheel moves to the right and rotates clockwise like a wheel of a bike, BUT THERE IS NO FRICTION BETWEEN THE GROUND AND THE WHEEL.

Like I want the surface of the triangle constant, I need the spring and the force F2.

There is a gasket between the black arm and the red wheel but I don't drawn it.

The mass is in the center of the wheel and in the outer circle of the wheel.

The center of the wheel (move to the right) win the energy the moment on the wheel lost (the moment is counterclockwise and the wheel rotates clockwise) : +1.5Rδ-1.5Rδ = 1.5 e-3 -1.5e-3 = 0 J
The spring lost 7.4e-4 J
The force F4 works very few compared to others, it is  √3Rδ²/2 = 8.6e-7 J

The sum of energy is not at 0, in this case the device destroy an energy

[EOW]

I drawn forces with less length .

First: destroy an energy
Second: create an energy

I don't drawn them before but there are spokes on the wheel

[EOW]

In fact, I don't need pressure nor the black and red walls, just a bike and a spring. I drawn only the wheel not the bike. There is NO friction between the ground and the wheel.

The force Fo gives the same energy than the torque -Fo/Fw on the wheel lost. But the spring lost an energy.

The energy won by the force Fo to the bike is lost by the torque -Fo/Fw on the wheel but the spring lost an energy. The sum of energy is not at 0. The center win FRδ. The wheel lost FRδ. If F=1 N and R=1m, the energy lost/win is δ. For δ=0.001 rd the energy win/lost is 0.001 J.

The length increases: dl = ( √ ( (θ-sinθ)² + (1-cos(θ))²) - √ ( ((θ+δ)-sin(θ+δ))² + (1-cos(θ+δ))²) ) with θ at 300° The energy for the force F2 is at dl*F2 = 4.27e-4 * 1 = 4.27 e-4 J
The spring lost 4.27e-4 J

Maybe I can attach the spring with a wheel bigger (trochoid)  like that the spring can win/lost more.