Sum of torque

[EOW]

Maybe with the attraction from springs like that. I can change the shape for find the good forces: the force on C2 is like the yellow force. Here I have 2 areas with 2 different gravities.

F1, F2, F3 come from the springs, note that like there is a curve (torus) the forces from springs are lower than the forces from pressure so I need to add F7, F8 and F9
F4, F5, F6 come from the forces of pressure
F7, F8, F9 come from the forces of pressure, gravity2 is small compared to gravity1

Dotted arrows are the reported forces on C2 or for construct the sum of forces. The torus turns counterclockwise. Don't forget, the torus accelerates more and more for follow the acceleration of the Object. The angular position between the Object and the torus is ALWAYS the same.

[EOW]

The torus and the square object turn counterclockwise. The torus turns around C1. C1 is fixed to the ground. The square object turns around C2. C2 is fixed on the torus. The square object if free to turn on the torus. Consider the mass of the springs and the mass of the balls like zero, like that there is no problem with centrifugal forces, or like the position of each ball don't move in reference of the torus it's possible to cancel with an external arm for each ball the centrifugal force.

First image: the square object without the torus for understand how forces are on it. The are two areas, first Attraction1 where the pressure is higher than the other area Attraction2. The springs attrack more and more and give more and more pressure at "bottom" (bottom of the image). The forces F1, F2, F3 com from the springs. The forces F4, F5, F6 come from the pressure of the first area. The forces F7, F8, F9 come from the pressure of the second area. The sum of forces F1+F2+F3 is lower than the sum F4+F5+F6 because the shape of the torus is curved. I drawn small red forces on springs to look how springs attrack, I drawn only 3 springs but there is one spring for one ball. And the balls are very small.

Second image: it's the same but now with the torus. Look at the sum of forces F1+F2+F3+F4+F5+F6+F7+F8+F9 on the center C2: the yellow force. The yellow force don't give a torque on the torus. So I can accelerate like I want the torus  and after recover all this energy. Even, I can set F7+F8+F9 higher for have a counterclockwise torque on it, I lost a part of a torque on the square object but I win a bigger torque because the distance with C1 is higher.

Third image: I drawn the device for 4 positions. The angle between the square object and the torus is always the same. The square object want to turn counterclockwise, the square object accelerate more and more BUT I accelerate with an external motor the torus like that the square object keep its position. The springs don't move inside the square object so the potential energy of the springs is contant. I noted the letter 'a' for look at the position.

I can't recover the energy in the same time, I need to accelerate the torus and recover the energy after. I recover the potential energy from the square object only, it turns more and more.

The device is unstable, it's necessary to control very well the angular acceleration of the torus. And the center C2 must be with friction as lower as possible.

[EOW]

I try to explain like I drawn the forces:

Image1: Water with gravity, there are force from pressure at left and at right and forces at bottom (the weight). I did not drawn all forces, just the first, the middle and the last, because there are a lot of forces from pressure.
Image2: I replace gravity+water with springs+balls, like that I have the same forces from pressure at right and at left, sure there is no "weight" F1+F2+F3 cancelled by -F1-F2-F3. I use compressible balls or I change the arrangment like that I have a pressure on the side walls. Springs attrack exactly like gravity or with another law. I did not drawn all springs, but there is one spring for each ball.
Image3: The forces on the square object only (not forces on the torus). There are 2 differents areas where springs don't attrack with the same force, like 2 different gravities. I have forces from pressure at left and at right. I have only the up forces from the springs. I don't have the down forces because there are on the torus.
Image4: Now I drawn the forces on the torus too. There only a point of connection between the torus and the square object, it is the center C2. So, the sum of forces on the square object will be on the center C2. The sum of forces is drawn by the yellow force. The goal is to cancel the torque on the torus, and have the yellow force like I drawn. F4+F5+F6>F1+F2+F3 because there is a curve so it's necessary to have the forces F7+F8+F9.
Image5: Details of forces
Image6: The device
Image7: The device at 4 positions, look at the point 'a' it rotates like the torus. The angular velocity of the torus is exactly the same than the square object.

There is a torque on the square object so it will accelerate more and more, if you let the square object alone on C2 the device will turn a little and springs will lost their potential energy. But if the square object is place on the torus, and it the torus is accelerate more and more EXACTLY like the square object in this case the angle between the square object and the torus is always the same. There is no torque on the torus, look at the yellow force so I can accelerate it without lost an energy, I can recover the energy I give to the torus later.

Cycle: just accelerate the torus for have the same angular velocity of the square object. It necessary to accelerate more and more the torus and never stop until you can.
Recover energy from the rotation of the square object.

I think it's possible to have a better efficiency with a torque on the torus from F7+F8+F9 because when F7+F8+F9 increase it don't decreases the forces F1+F2+F3 but only F4+F5+F6, I lost a part of torque on the square object but I win the torque on the torus and like the radius is higher the efficiency is higher. Or reduce the distance between C1 and C2 like that it could be easier to turn the torus.

[EOW]

I noted:

'c' the lenght of the square
'R' the external radius of the torus
'd' the distance c1c2 x or y axis
Values: c=3 and R=6.5, d=4.34

I consider the pressure like the height of the balls like I can choose with springs.

**************************************************************************CALCULATION*********************************************************************

******************************************Springs**************************************
Start integration x:
s1=d-c/2=2.84


Middle integration x:
sm=d=4.34

End integration x:
s2=d+c/2=5.84

Height:
H=d+c/2=5.84

Integration1:
[tex]\frac{1}{c/2}\int_{s1}^{sm} (H-\sqrt(R^2-x^2))(d-x) dx[/tex]

Integration2:
[tex]\frac{1}{c/2}\int_{sm}^{s2} (H-\sqrt(R^2-x^2))(d-x) dx[/tex]

Values: -0.20+1.63

Sum of forces:
[tex]\frac{1}{c}\int_s1^s2 (H-\sqrt(R^2-x^2)) dx[/tex]

Value: 1.15

******************************************Right side**************************************

[tex]\frac{1}{c/2}\int_{0}^{c/2} x(-x+\frac{c}{2}) dx [/tex]

[tex]\frac{1}{c/2}\int_{c/2}^{c} x(x-\frac{c}{2}) dx [/tex]

Values: -0.37+1.87

Sum of forces:
Value: c/2=1.5

******************************************Left side**************************************

I want cancel sum of forces on C2, so I need Fx: 1.5-1.15=0.35

Torque:

Value: -0.35*4.34=-1.5

*******************************************Results**************************************

Torque on the object: -0.20+1.63-0.37+1.87-1.5=1.42

Torque on the support: 0

[EOW]

The red part is full with small compressible balls. All springs are attached on C2 and each spring attrack a ball. The force F1 come from the springs. The support (black) turns around C1, C1 is fixed to the ground. The red object turns around C2, C2is fixed on the support.

The support and the object turn at the same angular velocity.

The force F1 don't give a torque to the support. The force F2 don't give a torque to the support. The force F2 give a torque to the object.