Sum of torque

[EOW]

1/ I'm sure of the values of the forces because I found the sum of forces at 0. I need only the forces on the red wall and on the black center because the device turns around the red center
2/ The device don't move in straight direction, it's only a rotation around the red center. The black center turns too around the red center even the  black arm turns around the green center because the length of the black arm increases during the rotation
3/ The springs never lost any potential energy because all the springs are attached in the red center and the device turns around the red center.
4/ I'm sure about the torque on the red wall, it is 0.25
5/ I'm sure about the forces on the black center: x: 1/6 y:0.182
6/ Like I drawn the device can turn because I turned it with an angle of 30°, I need only to adjust the length of the black arm
7/ The last thing to be sure is the energy won by the black arm:
   I'm sure of:
     7.1/ If the device turns of X degrees then the black arm turns of X/2 degrees, I calculated and verify with the geometry
   I need to take in account:
     7.2/ The torque on the black arm, and take care about the direction of the force. The value of the force is always : sqrt( 0.166² + 0.182² ) because the springs don't change their lengths
     7.3/ I need to increase the length of the black arm, so I need an energy for that.

So, my calculation is:

---------------------- ok from 0 to pi/2 ---------------------------
Fx=1/6
Fy=integrate( 0.5 - 1/sqrt( 0.5*cos(x) + (1.5 + 0.5*sin(x) ) )*sin(x)*0.5 from -pi/2 to pi/2 = 0.182355
L=sqrt( (1.5*cos(2*x))^2+(1.5+1.5*sin(2*x))^2) // I need to have 2x inside trigo functions because when I integrate I do from 0 to A/2
F=sqrt( Fx^2 + Fy^2 )
B=atan( Fx / Fy )
Energy recover from black arm = Integrate( F * L * sin( pi/4 -x + B ) from 0 to A / 2
Energy needed to increase the length of the black arm = Integrate( F * L ) )* cos( pi/4 -x + B ) from 0 to A / 2
Energy lost from red arm = A*integrate( ( 0.5 – 1/(2-x) ) * (x-2) ) from 0 to 1 = 0.25 * A

At final the energy won is :

Integrate( F * L * ( sin( atan( C ) + B ) -  cos( atan(C) + B ) ) from 0 to A/2 - A*integrate( ( 0.5 – 1/(2-x) ) * (x-2) ) from 0 to 1

Example, A=0.1 rad sum at : 0.005829 J

With the force 1000 times higher (182 N) the device gives 0.1J and if it turn at 1000 tr/s it will give 100 W



I will verify today again but I think my calculations are correct.

[EOW]

I checked my equations, all seems fine, there is a difference of torque. I compute with 500 digits and I find the same value than Wolfram, so the values of the integrals are correct and from 0 to 0.1 rd the value 0.0494171 it is not 0.5.

Datas:

A: the angle of rotation of the red arm, the black arm turns of A/2
Fx=1/6
Fy=integrate( 0.5 - 1/sqrt( 0.5*cos(x) + (1.5 + 0.5*sin(x) ) )*sin(x)*0.5 from -pi/2 to pi/2 = 0.182355
L=sqrt( (1.5*cos(2*x))^2+(1.5+1.5*sin(2*x))^2) // I need to have 2x inside trigo functions because when I integrate I do from 0 to A/2
F=sqrt( Fx^2 + Fy^2 )
B=atan( Fx / Fy )
Energy recover from black arm = Integrate( F * L * sin( pi/4 -x + B ) from 0 to A / 2
Energy needed to increase the length of the black arm = Integrate( F * L ) )* cos( pi/4 -x + B ) from 0 to A / 2
Energy lost from red arm = A*integrate( ( 0.5 – 1/(2-x) ) * (x-2) ) from 0 to 1 = 0.25 * A

At final the energy won is :

Integrate( F * L * ( sin( pi/4 - x + B ) -  cos( pi/4 - x + B ) ) from 0 to A/2 - A*integrate( ( 0.5 – 1/(2-x) ) * (x-2) ) from 0 to 1

I try with an attraction with the law 1/d and another law and it's the same, there is a difference. The final integral is logical. The difference come from sin(x)-cos(x). The integral from 0 to 2pi is 0 but from 0 to 0.1 for example the sum is not at 0.

The max is gave from x=0.7629 to pi+0.7629

Nobody ?

[EOW]

Note the start of the integration, it's the vertical not the horizontal. And the direction is clockwise. Axis x is horizontal and axis y is vertical.

I found where come from the difference, it is :

integrate sqrt( (1/6)^2+0.182355^2 )*(sqrt( (1.5*cos(2*x))^2+(1.5+1.5*sin(2*x))^2)-1.5*sqrt(2))*(sin( pi/4-x+0.7404789456 )-cos(pi/4-x+0.7404789456))  dx from x=0 to 0.05

it's exactly that. This work come from the length of the arm increases, at start it is 1.5*sqrt(2), but the arm increases when the red arm turns so this extra length give an extra energy.

The energy won is the extra length of the black arm * the force (constant) * sin (angle) and the sinus is 0.99 at 0.05 - extra length of the black arm * the force (constant) * cos (angle)  and the cos is 0.094 the difference is here

The energy recovered is 0.2 J for a turn, so with a force 1000 times higher (180N) and a rotation of 1000 tr/s the power is 200 kW. The rotation can be high because there is no mass.

[EOW]

For me it's logical but maybe not for all: the device is not composed of one body. There are 2 parts:

1/ The red wall with its red arm connected to the red center.
2/ The half circle and its black center of rotation. The black arm is connected to the black center.

If I let the device like that, the red wall rotates to the counterclockwise and the half circle clockwise, and sure, the balls move closer to the springs, and the springs lost their potential energy. But it's not like that I want to use the device. I have a motor on the red center that force the red wall to rotate and follow in the clockwise direction the half circle. Like balls are very small (in theory imagine balls like molecules of water) I need to add gaskets. So, I calculated the energy I need to give to the motor and I calculated the energy I recover from the black arm. The black arm needs to change its length because the are 2 centers of rotation.

It's possible to replace the springs by magnets or electrostatic force or any hydraulic device.  It's possible to use a law of attraction different of 1/d², like (1/d^x) with x any real number. It's possible to use another shape than a circle like an ellipse. Here, it's a 2D device, but it's possible to have a 3D device.

Here the force of pressure is very small (0.5) but it's easy to have a high pressure to increase the power of the device.

[EOW]

Maybe I made a mistake about the force F2y, it is (maybe) 0.14709. My program at start was good for the forces F3x and F3y.The result change but the sum of torque is not 0. In fact, this is not very important because it is possible to change the law of attraction and to find a vector with a good value.

With 500 binary digits and 3e6 steps the result is 0.025034 and even I change a little the values this don't change the difference enough to have the sum of energy at 0.

I got it, it's 0.14780 not 0.182 for F2x. It's nothing the result it's the same, there is a difference to have the sum of energy not at 0.

And don't forget, the force is small and I integrated for a small angle, if I integrate from 0 to 0.5 rd I have 0.219, it's far from 0.25

Someone can confirm ?