Sum of torque

[EOW]

If the green (interior) torus is fixed to the red walls the solution doesn't works. I need to get the torque from F1 in the center of the torus, for that I need to have the force from pressure from the red walls. So when I attract the green part of the torus from the green line, I give only one clockwise torque ,  but if the green torus and the red walls are the same piece the torque on the torus is twice.

With the green torus and red walls the same piece:

- torque from F1 =  17
- torque from torus = -11.6*2
- torque from the container = 6

So the sum is at 0

With the green torus not the same piece that red walls:

- torque from F1 =  17
- torque from torus = -11.6
- torque from the container = 6

The sum is not 0

To keep constant the length of the springs I can use a motor on the blue axis, this motor needs an energy but less than the device can recover. I adjust the torque of the motor to have the length of the springs constant.

[EOW]

Like that the red center don't have any torque from torus.

[EOW]

I calculated the sum of forces on this device and I don't find 0. I used small balls attracted by springs like before.  Friction is low. The law of attraction is in 1/d² not 1/d. There are a lot of balls inside the area (or the volume). Balls give pressure like water inside a recipient under gravity but the law need to be in 1/d² (like gravity) or another law 1/d³ for example.

The device is in a unstable position but like I drawn the sum of forces is not 0. I don't try to rotate the main black arm, just have the sum of forces on the gray device, and I don't find 0.

Datas:

Radius of the circle : 0.5
Length d = 1
Pressure at top : 0
Pressure at bot : 0.5

RED WALL:
Force on axis x from red wall:
integrate( 0.5-1/(2-x) dx from 0 to 1 = 0.193

Torque on the black arm from the red wall:
integrate( (0.5-1/(2-x))*(x-2) dx from 0 to 1 = 0.25

BLUE WALL
Force on axis x from the black axis:
0.5*integrate( 0.5-1 / ( sqrt( (0.5*cosx)² +(1.5+0.5*sinx)² ) ) * cosx dx from -pi/2 to pi/2 = 0.1666
Force on axis y from the black axis:
0.5*integrate( 0.5-1 / ( sqrt( (0.5*cosx)² +(1.5+0.5*sinx)² ) ) * sinx dx from -pi/2 to pi/2 = 0.364
Torque on the black axis from the force:
1.5*0.1666 = 0.25

The torques cancel themselves so the half circle don't give any energy.

The force from the springs on the red axis:
In x : 0.0674
In y : 0.129

The sum of forces on x: -0.2+0.1666+0.0674 = 0.034 MAYBE if I divide 0.0674 by 2, I have 0.034 but not exactly, it is 0.03371 so the sum is not 0 too, the result 0.2 and 1.6666666 is known, the result 0.03371 is very precise, the sum is 0.00037
The sum of forces on y: 0.364-0.129 = 0.235, here if I divide by 2 or 4 it's worse..., the sum is not 0

I can't have 0 in x and y axis in the same time. I'll integrate to find the best result.

Below, an example for calculate the sum of springs in the red center, I changed the result because I tried to find 0 :

#include <stdio.h>
#include <math.h>

int main()
{
double N=2000;
int i, j, c=0;
double s1=0, s2=0 ,x,y1,y2;
for(i=0;i<N;i++)
{
  x=(double)i/N*0.5;
  for(j=0;j<N;j++)
  {
   y1=+(double)j/N*0.5;
   y2=-(double)j/N*0.5;
   if ((x*x+y1*y1)<=0.25)
   {
    s1+=fabs(cos(atan((y1+1.5)/x))/(x*x+(y1+1.5)*(y1+1.5)));
    s1+=fabs(cos(atan((y2+1.5)/x))/(x*x+(y2+1.5)*(y2+1.5)));
    s2+=fabs(sin(atan((y1+1.5)/x))/(x*x+(y1+1.5)*(y1+1.5)));
    s2-=fabs(sin(atan((y2+1.5)/x))/(x*x+(y2+1.5)*(y2+1.5)));
    c++;
   }
  }
}
printf("\nc=%i\n",c);
printf("s1=%lf\ns2=%lf\n",s1/c,s2/c);
return 0;
}

In the i7.png image, I drawn the red line side by side the black arm, the red wall don't have a link with the black center, and the red forces is reported to the red axis, sure.

The sum of forces in x and in y axis is not 0. It's possible to move the device in any direction.  I drawn spring but it's possible to use any device for have the attraction like magnets, electrostatic of hydraulic.

At start, I calculated for the torus device, in this last device, there is nothing in the half disk. The balls are all around the torus, attracted by springs from the red center, center of the torus. The torus is stable, it will move in straight line with a force  and turn in the same time.

I drawn half disk but it's possible to use another shapes.

[EOW]

The forces from red and blue walls are correct. Note, the half blue disk can turn around the black center, sure the half disk don't turn around the black center. The black arm don't turn because the sum of torque is 0 but it is in a unstable position.

I need to divide by 4 for my program because I integrate a square 0.5 by 0.5

My program:

x: 0.0337
y: 0.232

Double integration:

x:0.0308 or 0.0249 if I divide only by pi
y:0.282 or 0.228 if I divide only by pi

I think I need to divide by pi because the length of the arc is 0.5*pi but I divide by 0.5 due to the first integral, so the result seems near from the program:

integration:

x: 0.0249 near 0.033
y: 0.2288 near 0.232

But the sum of forces is not 0

For cancel the forces in x, need to have 0.0264
For cancel the forces in y, need to have 0.182, I'm far with 0.23

With the lib MPFR and a precision of 200 bits (62 digits in base 10), the sum at x is -0.193+0.166666+0.0337 =0.007 and in the y axis, the sum is 0.182 - 0.232 = 0.05 the result converge, even I change NN, the result don't change for 7 first digits.

Double integration with double integration gives 0.078 in x axis, I think I need to multiply by the area of the disk (mean) so it is 0.392*0.0785 = 0.0308 and the result is not 0.193 but 0.197, note that the program and Wolfram find near the same result

With the y axis the difference is 0.282 - 0.182 = 0.1

I think Wolfram don't integrate with 200 bits precision so I think my program is more accurate.

Centrifugal forces can do the same.

[EOW]

I think I need to divide by 4 the integrals not by pi, like that the sum of forces are at 0.

Now, for increase the torque, I can change the position of the center for the black arm, the length of the black arm must be variable (telescopic). I take the green center at 1.5 in horizontal, like that I have a radius of 2.121.

X: the force in axis x, it is 1/6
Y: the force in axis y, it is 0.182
A: the small angle to rotate, like 0.1°
L: the length of the black arm, it is 2.12

The energy from the black arm is sqrt( X^2 + Y^2 ) * A / 2 * L * sin(45°+atan( X/Y )= 0.2617*A and I need to take in account the length of the black arm change, so the length increase, I lost energy. This energy is dL*F, dL is the difference from start to end:

dL = L - sqr( ( L/sqrt(2)*cos(A) ) ^2 + ( (1+sin(A)*L/sqrt(2) ) ^2 )
F = sqrt( X^2 + Y^2 ) * cos(45+atan(X/Y))

If A = 0.1, dL = 1.85e-3 and F=0.011 so the energy lost is 20e-6 it is very small compare to 0.02617 (A=0.1)

sqrt( X^2 + Y^2 ) * A / 2 * L * sin(45°+atan( X/Y ) - ( L - sqr( ( L*sqrt(2)*cos(A) ) ^2 + ( (1+sin(A)*L*sqrt(2) ) ^2 ) ) * (  sqrt( X^2 + Y^2 ) * cos(45+atan(X/Y)) ) - 0.25 * A

At final, the energy is 0.02617-20e-6-0.025 = 0.001053 J for an angle of 0.1°
Like the angle can be very small, I don't take the effect of the direction at final, I consider the direction of force and arm is the same than at start.

I added true value of the integral because all value depend of the angle and it changes all the time

Note the result is low but the force on the black center is 0.182 N it is small. With an angle of 5° and a force of 200 N the energy recover is 58 J and the device can turn at a very high angular velocity. Note I don't need mass, it's only shapes with attraction, like that I don't have any problem from the centrifugal force, it's possible to have a high angular velocity.