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### Author Topic: Sum of torque  (Read 148861 times)

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #285 on: June 01, 2016, 01:26:19 PM »
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.

I had this problem with the F3x and F3y forces because it is a double  integral. Even Wolfram is not perfect for the double integral I wrote a program with a high accuracy, and like that I found the sum of forces at 0 with the number of digit I want (Wolfram not, my program yes). But I don't need F3x and F3y in my use of the device because  the red axis is fixed. And I have the indefinite integrals for F2x, F2y, F1x and for the torques.

Do you understand the device ?

For Noether, don't forget there is an hypothesis: the potential energy don't change with time.

I corrected my integrals for the F3x and F3y, I divided by 4 but the true integral take x inside. The results are the same but the integrals are easier to calculate
« Last Edit: June 01, 2016, 06:00:51 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #286 on: June 02, 2016, 04:23:16 PM »
The indefinite integral for the force F2. I don't think this is a problem of accuracy.

And the difference of energy is more than 20 % for an angle of 1rd, with the force I use, the difference of energy is 0.25-0.203=0.046 J it is near 20%. If the force is 1000 times higher (F2=247 N) the difference is 1000*0.046=46 J. It's easy to increase the force because the springs like the balls are always in the same relative position inside the half disk. If the device turns at 10 tr/s, the power is 460 W.

@LibreEnergia: Do you understand how the device works ?

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #287 on: June 03, 2016, 11:23:02 PM »
With an extreme accuracy and with a very small angle it's possible to have directly the result without an integral, there is always a difference:

#include <stdio.h>

#include <gmp.h>
#include <mpfr.h>

#define DG 2000

int main()
{
mpfr_t   x,l,angle,c,s,r,r2,sum,dl,fx,fy,f,temp1,temp2,temp3,temp4,temp5,temp6,temp7,temp8,temp9;
mpfr_init2 (x, DG);
mpfr_init2 (l, DG);
mpfr_init2 (angle, DG);
mpfr_init2 (c, DG);
mpfr_init2 (s, DG);
mpfr_init2 (r, DG);
mpfr_init2 (r2, DG);
mpfr_init2 (dl, DG);
mpfr_init2 (sum, DG);
mpfr_init2 (fx, DG);
mpfr_init2 (fy, DG);
mpfr_init2 (f, DG);
mpfr_init2 (temp1, DG);
mpfr_init2 (temp2, DG);
mpfr_init2 (temp3, DG);
mpfr_init2 (temp4, DG);
mpfr_init2 (temp5, DG);
mpfr_init2 (temp6, DG);
mpfr_init2 (temp7, DG);
mpfr_init2 (temp8, DG);
mpfr_init2 (temp9, DG);

mpfr_set_d(x, 1e-20, MPFR_RNDD);
mpfr_set_d(l, 0.0, MPFR_RNDD);
mpfr_set_d(fx, 1.0, MPFR_RNDD);
mpfr_div_d(fx, fx, 6.0, MPFR_RNDD);
mpfr_set_d(fy, 0.182355, MPFR_RNDD);
mpfr_div(angle, fx, fy, MPFR_RNDD);
mpfr_atan(angle, angle, MPFR_RNDD);

mpfr_mul(temp2, fx, fx, MPFR_RNDD);
mpfr_mul(temp3, fy, fy, MPFR_RNDD);
mpfr_add(temp2, temp2, temp3, MPFR_RNDD);
mpfr_sqrt(f, temp2, MPFR_RNDD);

mpfr_mul_d(temp2, x, 2.0, MPFR_RNDD);
mpfr_sin(temp1, temp2, MPFR_RNDD);
mpfr_mul_d(temp1, temp1, 4.5, MPFR_RNDD);
mpfr_add_d(temp1, temp1, 4.5, MPFR_RNDD);
mpfr_sqrt(l, temp1, MPFR_RNDD);

mpfr_set_d(temp2,3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196,MPFR_RNDD);
mpfr_div_d(temp2, temp2, 4.0, MPFR_RNDD);
mpfr_add(temp2, temp2, angle, MPFR_RNDD);
mpfr_sub(angle, temp2, x, MPFR_RNDD);

mpfr_sin(s, angle, MPFR_RNDD);

mpfr_mul(temp5, l, s, MPFR_RNDD);
mpfr_mul(r2, temp5, x, MPFR_RNDD);

mpfr_sqrt_ui(temp7, 2.0, MPFR_RNDD);
mpfr_mul_d(temp7, temp7, 1.5, MPFR_RNDD);
mpfr_sub(dl, temp7, l, MPFR_RNDD);
mpfr_cos(c, angle, MPFR_RNDD);
mpfr_mul(r, dl, c, MPFR_RNDD);

mpfr_add(sum, r, r2, MPFR_RNDD);
mpfr_mul(sum, sum, f, MPFR_RNDD);

printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, l, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, angle, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, s, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, c, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, r, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, r2, MPFR_RNDD);
printf("\n");
printf("\n");
mpfr_out_str (stdout, 10, 0, sum, MPFR_RNDD);
printf("\n");
printf("\n");

//mpfr_clear (s1);

return 0;
}

#### Gabriele

• Full Member
• Posts: 248
##### Re: Sum of torque
« Reply #288 on: June 04, 2016, 04:28:40 PM »
Hello. Why don't you model your idea with WorkingModel2D... should be simple...

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #289 on: June 04, 2016, 05:42:55 PM »
Thanks for the name of the software but it's not possible to simulate N balls I tried with Comsol.

I found my error it was in the integral about the length of the arm.

Another idea: take 3 semicircles. One like before, radius of 0.5 and 2 smaller with a radius of 0.25. There is no straight wall like before. I do the same with balls and springs with the law 1/d^2 (or 1/d^3). I place the device on an object in rotation.  Each semicircle can turn with its arm (color).

The torque from the 3 semicircles device is 0 on the green dot but the object turns and it can receive a difference.

I calculated the 3 forces (radial), I have:

Tangential forces :
F1=1/28 (black)
F2=1/6 (red)
F3=15/100 (green)

The value are exact it is indefinite integrals.

The torque at 1.5 is 1/28*1.75-1/6*1.5+15/100*1.25 = 0
But at 10 it is 1/28*10.75-1/6*10.5+15/100*10.25 = 0.1904

Don't remember the semicircles don't receive any torque around themselves from pressure of balls because the pressure is perpendicular to the surface (center of the circle). I want the device keeps its shape because walls are 3 bodies, so I need to apply a torque on the center, for example I give a torque on the red center to compensate the 1/6 torque. I need an energy but I can recover more from 2 others.

Note the main object will receive a torque on it.
« Last Edit: June 04, 2016, 11:01:57 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #290 on: June 05, 2016, 02:47:32 PM »
I verified my tangential forces, they are correct.

Note, it's very important do not block the semicircle on the main object. Each semicircle must give it's force on the gray axis to have the sum of forces at 0 on the gray axis. So the device must accelerate more and more, it wins only a potential energy. This energy can be recover later not in same time.

And it's pretty good with the Neother's theorem

I added an image with the forces:

F1: force from pressure on the black semicircle to the black center
F2: force from pressure on the red semicircle to the red center
F3: force from pressure on the green semicircle to the green center

Like each semicircle can turn around the gray axis, forces on the red/green/black axis are reported to the gray axis
F1' : from F1
F2' : from F2
F3' : from F3

The springs gives a force on the gray axis:
Fs1 : for attract all balls of the bigger semidisk
Fs2: for attract all balls of the smaller semidisks

I added the external force Fext because I want the red semicircle follows the others semicircles.

The sum of forces on the gray axis is 0 (x or y)

At d=0 the torque is 0 but with d different of 0 there is a counterclockwise torque.

I don't need radial forces, but tangential forces are:

F1t = 1/28
F2t = 1/6
F3t = 3/20
Fs1t = 0.02648
Fs2t = 0.007433
Fext = 1/6

Tangential force on the gray axis  = 0

3 images to show each axis alone.

It's not only a sum of torque different of 0 but when the forces F1, F2, F3 apply their force on axis, these forces are applied to the gray axis, so the sum of force is at start (tangential forces):

F1=+1/28
F2=-1/6
F3=+3/20
On gray axis=-0.019

The sum is at 0 but the forces F1-F2+F3 go to the gray center so the sum is:
On gray axis: -0.019 +1/28-1/6+3/20 = 0
But the forces F1-F2+F3 is not 0 it is 0.019

----------------------------------------------------------------------------------------------------------------------------

If I come back with only one semicircle with straight red line (look before) it's strange too, the sum of force on the gray axis is 0 but the forces are not the same so at distance the sum of torque is not 0. If d=10 then the red torque is integrate((0.5-1/(2-x))*(x-10) dx from x=0 to 1) = 1.79 but the black dot torque is 9.5 * 1/6 = 1.583 there is a difference. There is no force on the red axis.

---------------------------------------------------------------------------------------------------------------------------

The last device, the torque is -0.25+0.0568 +1/6*1.5-1/6*0.5 different of 0
« Last Edit: June 06, 2016, 01:16:31 AM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #291 on: June 06, 2016, 12:01:45 PM »
Note: axes don't interact with balls ! Each basic heart shape is composed with 3 bodies (3 semicircles). Axes are in another plane.

I take the basic shape I drawn before (like a heart shape). But there is nothing inside the heart shape. I imagine I have a lot of sizes, bigger, smaller, like that I can full all the blue disk with basics heart shapes. Each basic heart shape gives a clockwise torque. It's possible to reduce the size of the heart shape so small that it can full all the blue disk. So at final the springs attracted balls from the white center with the law 1/d^2 (or another law) and there are very few balls if balls are very small. In fact, I need only one thickness of ball and if the size of ball is near 0, the volume attracted by springs is near 0, so I can have no torque on the white center from springs.

In fact, I can dot like before and place balls inside the heart shapes IF I take a lot of heart shapes with different sizes, like that I attract all around the device balls everywhere. And like that I'm sure of my results. The torque is counterclockwise in this  case.

I plot the difference of torque, it's logical that it is always in the same direction but like that I'm sure all basic heart shapes give a clockwise torque (nothing in it).
« Last Edit: June 06, 2016, 10:46:56 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #292 on: June 07, 2016, 06:21:01 AM »
I drawn the side view: balls don't touch the axes.

Note that each semicircle receives the forces only from the balls not from another semicircle.

I need to give to the red semicircle a torque in the good direction but I recover more from 2 others semicircles.

My integral at start was good. For example, the big circle (main device) is at 20, so I used 1/20 in the equation. I plot the result of the integral with plot3d, function of 'd' and function of 'r'. 'd' is the distance from the white center (main device) and 'r' is the radius of the semicircle. A basic heart shape is composed of 3 semicircles, 2 with the radius 'r' and one with the radius '2r' like you can see in the equation of the integral.

#### dieter

• Hero Member
• Posts: 938
##### Re: Sum of torque
« Reply #293 on: June 07, 2016, 07:31:48 AM »
It's just amazing how productive you are. Maybe you should really try it in some physics simulation software, as Gabrielle suggested.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #294 on: June 07, 2016, 08:28:45 AM »
I don't have a simulator for now. Maybe it's possible to simulate only one heart shape to show if I'm right with the torque, someone ?

I think it's not a problem of sum of torque, but sum of forces. The sum of torques is always at 0 but the sum of tangential forces of 3 dots (green, red, black) is not 0, with one shape the springs compensate this difference but if I place a lot of sizes I don't have anything to attract (very few balls) so the main device can be a sector of a disk full with heart shapes. The forces from pressure on the walls 1 and 2 are canceled by the buoyancy force that each heart shape has on it. But the sum of tangential forces from the difference of green/red/black forces can't be canceled.
« Last Edit: June 07, 2016, 12:51:36 PM by EOW »

#### Gabriele

• Full Member
• Posts: 248
##### Re: Sum of torque
« Reply #295 on: June 07, 2016, 12:01:51 PM »
I think any solid on which we apply a an equal force for each point of its surface,will obtain only equilibrium

#### Gabriele

• Full Member
• Posts: 248
##### Re: Sum of torque
« Reply #296 on: June 07, 2016, 12:04:10 PM »
i dont know

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #297 on: June 07, 2016, 01:09:54 PM »
My forces are correct for one device, I found the sum of forces at 0 and the sum of torques at 0 too. The difference of tangential force from the green/red/black force is canceled by the springs because the shape is asymmetrical. But I can take a sector of disk full with heart shapes with different sizes, in theory I can full the device and like that there is very few balls inside the sector and the springs can't compensate the sum of tangential forces. The buoyancy force is compensated by the forces on the straight lines (1) and (2). I calculated the indefinite integral and I plotted in 3d with the parameter 'd' and 'r' to show there is always the difference in the same direction. Have you any idea what can cancel the tangential forces ?

And I can take any shape, like half disk (straight line and center of the semicircle), or another shape, I need only to fill the sector with all the basic shape with different sizes. The sum of torque is 0 but not the sum of forces.

I need to take small semidisks, very small to prevent the asymmetrical forces from fa/fb. For example, if the sector has a radius of 10 I take for the semidisks a radius from 1e-10 to 1e-6 not more.

I calculated the difference of force when I reduce the size of the semidisk, if I divide by 2 the radius of the disk, the force is divided by 4 but like the surface is multiply by 4 I don't lost anything in the force:

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

#define N 100

int main()
{

double sum=0, sum2=0,r,d=2,l;
int i;

l=1.0;
r=l/(double)N/2.0;

for(i=0;i<N;i++)
{
l=1+1/(double)N*(double)i;
sum+=r*(-sqrt(l*l+4.0*r*l+4.0*r*r)/(r*l+r*r)+l/(r*l+r*r)+2.0/d);
//   printf("\nl=%f",l);

sum2+=-log(l+2*r)+log(l)+2*r/d;
}

printf("\nr=%f\ns=%f\ns2=%f\ndiff=%f\n",r,sum,sum2,sum-sum2);

return 0;
}

And for a big precision, it's possible to see the force is divided always by 4 when the radius is divided by 2 but the surface is 4 times higher:
#include <stdio.h>
#include <math.h>

#include <gmp.h>
#include <mpfr.h>

#define DG 400
#define N 1000

int main()
{

mpfr_t   s1,s2,r,d,l,t1,t2,t3,t4,diff;
mpfr_init2 (s1, DG);
mpfr_init2 (s2, DG);
mpfr_init2 (r, DG);
mpfr_init2 (d, DG);
mpfr_init2 (l, DG);
mpfr_init2 (t1, DG);
mpfr_init2 (t2, DG);
mpfr_init2 (t3, DG);
mpfr_init2 (t4, DG);
mpfr_init2 (diff, DG);

mpfr_set_d(s1, 0.0, MPFR_RNDD);
mpfr_set_d(s2, 0.0, MPFR_RNDD);
mpfr_set_d(d, 2.0, MPFR_RNDD);
mpfr_set_d(l, 1.0, MPFR_RNDD);

mpfr_div_d(t1, l, N, MPFR_RNDD);
mpfr_div_d(r, t1, 2.0, MPFR_RNDD);

//long double sum=0.0, sum2=0.0,r,d=2.0,l;
int i;

//l=1.0;
//r=l/(long double)N/2.0;

for(i=0;i<N;i++)
{
//l=1+1.0/(long double)N*(long double)i;
mpfr_set_d(t1, 1.0, MPFR_RNDD);
mpfr_div_d(t1, t1, N, MPFR_RNDD);
mpfr_mul_d(t1, t1, i, MPFR_RNDD);
mpfr_add_d(l, t1, 1.0, MPFR_RNDD);

//sum+=r*(-sqrtl(l*l+4.0*r*l+4.0*r*r)/(r*l+r*r)+l/(r*l+r*r)+2.0/d);
//   printf("\nl=%f",l);
mpfr_mul(t2, r, r, MPFR_RNDD);
mpfr_mul(t1, r, l, MPFR_RNDD);
mpfr_add(t2, t1, t2, MPFR_RNDD);

mpfr_mul_d(t1, r, 4.0, MPFR_RNDD);
mpfr_mul(t3, t1, r, MPFR_RNDD);

mpfr_mul_d(t1, r, 4.0, MPFR_RNDD);
mpfr_mul(t4, t1, l, MPFR_RNDD);

mpfr_mul(t1, l, l, MPFR_RNDD);

mpfr_add(t1, t1, t4, MPFR_RNDD);
mpfr_add(t1, t1, t3, MPFR_RNDD);
mpfr_sqrt(t1, t1, MPFR_RNDD);
mpfr_div(t1, t1, t2, MPFR_RNDD);

mpfr_div(t3, l, t2, MPFR_RNDD);

mpfr_d_div(t4, 2.0, d, MPFR_RNDD);

mpfr_add(t2, t3, t4, MPFR_RNDD);
mpfr_sub(t1, t2, t1, MPFR_RNDD);
mpfr_mul(t1, t1, r, MPFR_RNDD);
mpfr_add(s1, s1, t1, MPFR_RNDD);

//sum2+=-logl(l+2*r)+logl(l)+2*r/d;
mpfr_mul_d(t1, r, 2.0, MPFR_RNDD);
mpfr_add(t1, t1, l, MPFR_RNDD);
mpfr_log(t1, t1, MPFR_RNDD);
mpfr_sub(s2, s2, t1, MPFR_RNDD);

mpfr_log(t1, l, MPFR_RNDD);
mpfr_add(s2, s2, t1, MPFR_RNDD);

mpfr_div(t1, r, d, MPFR_RNDD);
mpfr_mul_d(t1, t1, 2.0, MPFR_RNDD);
mpfr_add(s2, s2, t1, MPFR_RNDD);

}

mpfr_sub(diff, s1, s2, MPFR_RNDD);

//printf("\nr=%Lf\ns=%Lf\ns2=%Lf\ndiff=%Lf\n",r,sum,sum2,sum-sum2);
printf("\n");
mpfr_out_str (stdout, 10, 0, s1, MPFR_RNDD);
printf("\n");
mpfr_out_str (stdout, 10, 0, s2, MPFR_RNDD);
printf("\n");
mpfr_out_str (stdout, 10, 0, diff, MPFR_RNDD);
printf("\n");

return 0;
}
« Last Edit: June 07, 2016, 11:22:20 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #298 on: June 08, 2016, 10:40:55 AM »
Maybe like that it's easier to the sum of forces: there are the forces from the springs and there are the forces from the axis, all at left.
« Last Edit: June 08, 2016, 05:56:03 PM by EOW »

#### dieter

• Hero Member
• Posts: 938
##### Re: Sum of torque
« Reply #299 on: June 13, 2016, 10:52:33 PM »
Have you found a physics simulation software yet?

Do you believe a physics simulation software could accurately and correctly simulate it?

Maybe the linux community has some opensource tools for free.