Sum of torque

[EOW]

I found an error in my device, if I reduce the size of the semi disk the force decreases in the same time. I tried to find a good simulator on Linux but they are difficult to use.

I'm trying to find where is my error in this case with 2 wheels.

1/ The two wheels move to the right
2/ One wheel rotates clockwise, the other counterclockwise
3/ The pressure is at 0.01 P everywhere except in the square shape delimited by the green/red walls
4/ The pressure is at P inside the shape delimited by the green/red walls
5/ Altitude of the wheels don't change
6/ There is no gravity
7/ The green V shape is free to rotate around the orange axis
8/ The orange arm is free to rotate around the black axis
9/ The red V shape is free to rotate around the blue axis
10/ The black arm is free to rotate around the pink axis
11/ Like there is a difference of pressure, the green V shape has a torque on the orange axis, this torque is 1
12/ Like there is a difference of pressure, the red V shape has a torque on the blue axis, this torque is 1
13/ I add a spring between the green V shape and the red V shape (not at the axes), to cancel the torque on each V shape, I need a force of 1/sqrt(2) because the distance is sqrt(2) and the torque is at 1
14/ So, now with the spring, each V shape don't want to rotate, they keep their relative position
15/ The force from the difference of pressure on the V shape on the orange axis is f11, this force =1
16/ The force from the difference of pressure on the V shape on the orange axis is f10, this force =1
17/ Each wheel receives a torque of 1-sqrt(2)*R,  like each wheel turn they can give an energy
18/ All volumes are constant, the gas don't lost any potential energy (I hope...)

I have only a problem with the wall of the circle between the V shapes but it's possible to take wheels just side by side at start, like the straight velocities are the same, the wheels are always side by side, an example:

The equation of the cycloid with a radius of the circle of 1:

x=a-sin(a)
y=1-cos(a)

But be careful that the reference is not the same for the left wheel and the right wheel.

For the spring at start:
Left wheel, the point at start is (0,0)
Right wheel, the point at start is (2,0)
For the spring at final, after 0.01 rd for each wheel:
Left wheel, the point at final is (0.02,50e-6)
Right wheel, the point at final is (2+1.666e-7,50e-6)

The difference of length is 0.02

The force needed to compensate the torque is 1/sqrt(2)

The energy for the spring is 1.414 e-2

The force from the V shape is 2 so the force on each axis is 2-1/sqrt(2). Note I take a big V shape, the radius of the wheel is 1 and a side of the V shape is sqrt(2).

The energy from the wheels is 2*1*(2-1/2)*0.01 = 2.58 e-2

There is a difference

The forces:

F1 and F4 come from the spring
F1 gives F2 to the orange axis and F2 gives F3 to the center of the left wheel
F4 gives F5 to the blue axis and F5 gives F6 to the center of the right wheel
F10 come from the pressure on the red V shape
F11 come from the pressure on the green V shape

[dieter]

Wikipedia has a long list of open source simulations:

https://en.m.wikipedia.org/wiki/Physics_engine

Sorry, I don't see where the error is.

[EOW]

Thanks for the link, if I don't find my error I will simulate the device.

With the following example and an angle of 0.001 rd:

spring lost 0.0010005
wheels win 0.001

In this example I can find near the sum of energy at 0. But not others examples.

It seems before pi/4 the wheels win more energy than the springs lost and after it is the contrary. I calculated for pi/8 (near like the first case I drawn in a last message). I found 2.7e-4 for the springs and 7.8e-7 for the wheels (the force on the axis is near 0)

[EOW]

I took another example:

Radius of the wheels = R = 1
β = pi/6 rd
δ = 0.01 rd (the small angle I calculate the energy)

The length of one V shape's wall is R/2*√2, so the force is  R/2*√2 = 1/√2 for each wall (I consider the pressure outside at 0 to simplify), like forces are at 90° the force from the difference of pressure is 1 N

The torque is 2*R/4*√2 = 1/√2 Nm

The force Fs for the spring is 1/√2 /(2*sin(β)) = 1/√2 N

The force on the orange or the blue axis is 1-1/√2 = 0.2929 N
The energy for the spring is 2*(cos(β)-cos(β+δ))*Fs = 7.13 e-3 J
The energy for the wheels is δ*2*Fa*Rf = 2.93 e-3 J

Maybe I can take Fb in another position and the red object could be not half square (an angle different of 90°)

I can replace the spring by 2 stems (telescopic stems like hydraulic cylinders) like that I don't lost any energy.

If I cancel the torque from the center I lost sqrt(2)*angle but I win angle/2

[EOW]

I was wrong about the torque with a constant pressure.

If I use small balls attracted from the center or use a fluid (liquid is better, centrifugal forces are higher), the force and the torques are not the same.