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### Author Topic: Sum of torque  (Read 139766 times)

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #270 on: November 14, 2015, 11:16:09 PM »
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

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #271 on: November 15, 2015, 04:46:34 PM »
Like that the red center don't have any torque from torus.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #272 on: May 23, 2016, 07:56:22 PM »
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.
« Last Edit: May 24, 2016, 07:44:41 AM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #273 on: May 24, 2016, 08:07:30 AM »
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.

« Last Edit: May 24, 2016, 05:12:15 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #274 on: May 25, 2016, 08:33:28 AM »
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.
« Last Edit: May 25, 2016, 04:28:48 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #275 on: May 26, 2016, 06:30:06 AM »
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.
« Last Edit: May 26, 2016, 06:08:11 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #276 on: May 26, 2016, 08:00:37 PM »
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 ?
« Last Edit: May 27, 2016, 06:59:53 AM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #277 on: May 27, 2016, 10:22:48 AM »
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.
« Last Edit: May 27, 2016, 09:46:11 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #278 on: May 28, 2016, 08:50:21 PM »
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.

« Last Edit: May 28, 2016, 11:19:24 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #279 on: May 29, 2016, 01:09:34 PM »
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 ?
« Last Edit: May 29, 2016, 10:24:54 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #280 on: May 30, 2016, 11:33:50 AM »
I think the integrals are good, I had a problem of coef in my program and it is ok now. Like that I have the sum of forces at 0 and without the green center I have the sum of torques at 0 too.

But with the green center I can have a difference in the sum of energy.

I have the definite integral so it can't be a problem of accuracy

« Last Edit: May 30, 2016, 11:28:40 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #281 on: May 31, 2016, 07:33:48 AM »
I drawn an image to show how is the black axis, without the balls and the red wall to watch details. Spoke are not in the same plan like that they don't interact with balls. The semicircle can only turn around the black axis and the black axis is fixed to the black arm. The semicircle can only rotate around itself but there is no torque on it (because it is a part of a circle and the forces of pressure are perpendicular to the surface). The semicircle is in an unstable position but in theory there is no torque on it and in practice it's possible to correct with an external device the position.

I don't wrote but I show an image with a rotation of the device, but note:

The red wall is free to rotate around the fixed red axis, a motor force the red wall to rotate clockwise and follow the half disk
The black arm (telescopic) is free to rotate around the fixed green axis, it has the force F2 at end (black dot). The black arm rotates around the green center and increases its length
the semicircle is free to rotate around the black dot but it don't rotate around it because there is no torque
The trajectory of the black dot is a circle with a radius of 1.5 and the center is the red axis.
I used balls inside the half circle but it's possible to use another shape like ellipse.

The black axis is not connected with the red arm.
« Last Edit: May 31, 2016, 04:28:17 PM by EOW »

#### dieter

• Hero Member
• Posts: 938
##### Re: Sum of torque
« Reply #282 on: May 31, 2016, 01:59:34 PM »
I see you're very creative. As usual this is way over my head ^^ nevertheless: keep up the good work!

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #283 on: May 31, 2016, 05:35:44 PM »
Thanks. Maybe someone could understand what I'm trying to explain

#### LibreEnergia

• Sr. Member
• Posts: 332
##### Re: Sum of torque
« Reply #284 on: June 01, 2016, 10:07:44 AM »
Thanks. Maybe someone could understand what I'm trying to explain

Noether's Theorem tells us that for every differential symmetry there is a corresponding conservation law.  In my opinion if you are finding anything other than zero change in energy while integrating over a path that starts and ends in the same physical positions then your analysis is wrong.

In this case 1, The motion is continuously differentiable, ie it has no inflection points, and it does represent a mathematical  'symmetry' , in this case a rotation,  so it meets the conditions for which Noether's theorem applies.

In my opinion you will find the error in analyzing the way you are calculating the integrals numerically, in that small errors introduced when summing numbers containing large differences in numerator and denominator give rise to an error in the totals.