What is the orientation of the vector of friction?

[jbriggs444]

There cannot be pure rolling given the constraint that a fixed point on the wall (to which one end of the elastic is attached) slides over the circle remaining continuously at the point of contact.

The dot B is always at the same position than the dot of contact between the circle and the red wall.

 

[A.T.]

There cannot be pure rolling ...

I'm using pure rolling just to show that this elastic band method cannot be used to determine the energy dissipated by friction.

...given the constraint that a fixed point on the wall (to which one end of the elastic is attached) slides over the circle remaining continuously at the point of contact.

I thought B is merely the contact location, constrained to be somewhere on the wall, but not necessarily fixed to the same physical wall position.

It cannot satisfy both constraints (always contact location & fixed physical wall position) in the OPs scenario either, so this interpretation doesn't make sense to me.

 

[jbriggs444]

It cannot satisfy both constraints (always contact location & fixed physical wall position) in the OPs scenario either, so this interpretation doesn't make sense to me.

It is the OP's scenario. He specified it.

The dot A is fixed on the circle. The dot B is fixed on the red wall.

 

[A.T.]

It is the OP's scenario. He specified it.

It's either a language problem or it doesn't make any sense. B cannot be the contact location and a fixed physical point of the wall, because the contact location moves along the wall.

 

[jbriggs444]

It's either a language problem or it doesn't make any sense. B cannot be the contact location and a fixed physical point of the wall, because the contact location moves along the wall.

I agree that it does not make much sense. The only way I saw reconcile it, which indeed may not have been the intent, is to have a fixed point on the wall sliding around the circle and maintaining contact.

This interpretation certainly violates the constraint in the original post that the wall is fixed to a pivot on the ground.

 

[JrK]

Thanks to both of all ! you are very very nice :)

It's either a language problem or it doesn't make any sense. B cannot be the contact location and a fixed physical point of the wall, because the contact location moves along the wall.

No, it is not a problem of language but I think with a step of integration, I change the position of the dot B at each step of the integration, B is at start (of the step of the integration) at the dot of contact between the circle and the wall and only at that time, after, the dot B is farther because it is fixed on the red wall. The dot B is not in the continuous time at the dot of contact, it is not possible. But I can increase the number of steps.

I know where is my mistake with the elastic: I didn't count the difference of the potential energy stored in the elastic (the potential energy stored from start to end of the study) and there is one ! it is (d1-d2)*F. So, like I drew, the energy lost to move the circle is recover by the elastic.

Remember, I took the idea of the elastic because I would like to find a method to count the energy because I see the friction at d2*F not d1*F. With the dot A fixed on the circle it is not like the friction could do. So, I took an elastic for each step of integration. That elastic at start has a length of 0, the two ends of that elastic is at the dot of contact, one end fixed on the circle, the other end fixed on the red wall. For each step, the elastic will increase its length of d2 not d1, or at least it is like I see that.

I resumed all here:

I think like that I will understand the distance is not d2 but d1.

I copy paste the text in the image because it seems it is not possible to download the image at its full definition:

An elastic is attached between the dot A and the dot B around the wall of the circle, always around the wall of the circle. I suppose the force F of the elastic constant when its length changes, at least for the difference of length of that example. The dot A is fixed on the circle. The dot B is fixed on the red wall. At start the length of the elastic is 0 and at final it is d2 : at start the dots A and B coincide. At each step of the numerical integration, I take a new elastic. The two ends of each new elastic is at the dot of contact between the circle and the red wall, but one end is fixed on the circle the other end is fixed on the red wall. Each elastic increases its length of d2 not d1. The force F1 is the force from the elastic to the red wall at the final of one step. The force F2 is the force from the elastic to the circle at the final of one step. The energy recovered from all the elastics is d2*F. The energy needed to move the circle is d1*F = lg*cos(a)*F. I don’t need nor recover any energy from the rotation of the red wall.

 

[A.T.]

The energy recovered from all the elastics is d2*F. The energy needed to move the circle is d1*F.

Then your elastic is not modelling friction correctly, which has to dissipate all the work done moving the circle in the original setup.

 

[Stephen Tashi]

It is very difficult to pose (not resolve) the movement by equations (math) ?

We won't know until we try!

The diagrams you have presented ignore the invention of Cartesian coordinates. They lack representations for crucial dimensions such as the radius of the circle and dimensions that establish the relation of the line of movement of the center of the circle to the point where the wall pivots.

You are attempting to discuss the problem in an abstract geometric style. Archimedes and Newton were able to think about things this way, but the modern approach is to use cartesian coordinates!

For the sake of getting advice about modern methods, you should be specific about cartesian coordindates. For example:

The wall pivots at (0,0). The radius of the circle is r. The center of the circle moves along the line (x,y_1) where y_1 is constant. .

Draw a diagram showing the above information. Try to find the coordinates of the point of contact (x_c, y_c) as functions of x and the constants.

You can get one equation by setting the distance between the center of the circle and (x_c, y_c) equal to r. You can get another equation from the condition that the vector from (x_c, y_c) to the center of the circle is perpendicular to the vector from (0,0) to (x_c,y_c). The complication in getting a solution from these two equations is that there are two possible solutions for (x_c, y_c) since there are two lines from (0,0) tangent to the circle.

I think there are clever people on the forum that can solve the above problem since it is posed in the modern style. Considering the problem in coordinate-free way is only leading to confusion.

 

[JrK]

I like a lot geometry ! Yes, a math method would be better, I tried with my program with a numerical integration (post #24), I used coordinates in the program but I don't know if my calculations are correct. I can explain the program if necessary, the program uses the first drawings: the dots, distances.

To come back to the method with the elastic, I need to place the dot A fixed not on the circle but on a needle, and the needle needs to be between the circle and the red wall. A very thin needle ! Like that the needle is always at the dot I want: the dot of contact between the circle and the red wall. The dot B is fixed on the red wall. I need only one elastic.

I give the details:

 

[Stephen Tashi]

@JrK, you're still avoiding cartesian coordinates.

I see no reason for the imaginary needle. You seem to think your coordinate free diagrams are helpful. So far, they haven't helped anybody.

 

[JrK]

In my program I used the name of the dot of the first drawing.
The (0,0) is on A0
The dot of contact between the circle and the red arm is 'i'
The center of the circle is the dot 'p'
The intersection between the horizontal line and the red arm is the dot 'q'
The height from A0 to the top of the circle is 'D'
The radius of the circle is 'R'
The length of the distance from A0 to 'i' is 'l' (small L)
I integrate from 'xinf' to 'xsup'For the dots i,p,q, I added a letter for the 'x' or the 'y' and I added an information for take in account the step: the number '1' is the step in calculation and the '2' is the last calculation.

'wf' is the energy to move the circle in translation
'wd' is the energy recovered from the friction
'wp' is to verify the vector of the force of friction is like the orientation of the red wall, must be at 0

What I need to add in the calculations ? Maybe the calculation of wd is not like I calculate because I use the dot 'i' and all people here says it is not the movement of the dot of contact that gives the energy from the friction. Maybe there is a method (an old method used in 16th) but I didn't find on internet, the name of the method could help me to find it and to apply it.

 

[A.T.]

To come back to the method with the elastic,

Here is one way to determine the slipping distance relevant for energy dissipated by friction, using an elastic, that works for your specific case (but not in general!):

Attach one end to the circle at the initial contact point, and the other end to the wall at the initial contact point. The initial length is zero, the final length is the slipping distance.

More general case:

You use two elastics, one for each body, with the ends attached to that body at the initial and final contact points. The slipping distance depends on how the elastics meet at the final contact point:

- If they meet at 180° (straight angle) you add their lengths (special case where you can use a single elastic).
- If they meet at 0° (acute angle) you subtract the shorter from the longer length.

Note that even the more general case works only for monotonic motion, not for back and forth sliding which you would have to do piece-wise. The elastics have to stay in contact with the bodies and cannot take short cuts through air. So for concave surfaces you have put them on the inside of the body, and for changing curvature sign you have to do it piece-wise.

 

[jbriggs444]

Like that the needle is always at the dot I want: the dot of contact between the circle and the red wall.

So this "needle" is not like a physical pin that pierces the elastic at a particular point. It is merely an indicator (like a "needle" on a speedometer) marking the contact point.

Like any other position marker, this indicator has no physical significance. It does not "do" anything. It is merely a convenience for the purposes of accounting.

It seems that you would like to use this marker to separate the elastic into a "circle piece" and a "wall piece". The idea would be that the stretching or shrinking of the "circle piece" corresponds to the work done on the circle by the force of kinetic friction. However, this is not correct.

To gauge the work done on the circle, the relevant place on the elastic is the place (if any) where it is stationary. This is the place where the elastic material does not move past a stationary line at right angles to the contacting surfaces. This place will not, in general, be at the point of contact.

It is worth noting that this point on the elastic changes with your choice of reference frame. Similarly, the work done on the circle changes with your choice of reference frame. Neither energy nor work are invariant quantities.

 

[JrK]

The needle is a real object. It is there to be near the dot of contact, I would like to be at the exact position of the dot of contact but it is possible only in theory. I suppose all the objects rigid and without any friction. Remember I done that because I don't know another method to calculate that example with the friction between the objects. I cannot place the image in the message, a bug ?

 

[A.T.]

Remember I done that because I don't know another method to calculate that example with the friction between the objects.

I gave you a method using integrals and I gave you a method using elastics. The needles seem needless here.

 

[JrK]

You changed your message, I didn't see it at start. Ok, I will try with your method.

The needle is useful, I can suppose the 2 bodies rigid and a small diameter for the needle, for example some atoms, like that the needle is near the dot of contact. I have well the distance I though with the friction. I have only one elastic and in straight shape (not curved). I calculate with my program the energy from F3 and F4 and it is the same:

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

int main()
{
long double pi=M_PI;

long int N=1000000;
long double R=.1;
long double D=1.;
long double xinf=1.;
long double xsup=xinf/tanl(44/180.0*pi);

long double px1=0,py1=0,a1=0,px2=0,py2=0,a2=0,dl=0,wd=0,wf=0,ix1=0;
long double iy1=0,xi2=0,yi2=0,qx1=0,qy1=0,s=0,xl1=0,yl1=0,xl2=0,yl2=0;
long double sx=0,sy=0,qx2=0,qy2=0,xi1sav=0,yi1sav=0,xsav=0,suma=0,l=0,asav=0;
long double xf=0,yf=0,wp=0,af=0,sumdx=0,dlrc=0,dlpc=0;

long int i;
qy1=D;

for(i=0;i<N;i++)
{
  //a1=(long double)(asup/180.0*pi-ainf/180.0*pi)/N*i;
  qx1=xinf+(xsup-xinf)/N*i;
  a1=atanl(qy1/qx1); if(i==0) a2=a1;
  s=R/tanl(a1/2.);
  px1=qx1-s;
  py1=qy1-R;
  ix1=px1+R*sinl(a1);
  iy1=py1-R*cosl(a1);
  if(i<1) {xi1sav=ix1;yi1sav=iy1;xsav=px1;asav=a1;}
  //printf("\npx1=%Lf , py1=%Lf , a=%Lf , ix1=%Lf , iy1=%Lf , a1=%Lf",px1,py1,180/pi*atanl((fabsl(iy1-yi2)-fabsl(py1-py2))/(fabsl(ix1-xi2)-fabsl(px1-px2))),ix1,iy1,180/pi*a1);
  suma+=a1;

  if(i>0)
  {
   sumdx+=(fabsl(px1-px2)-fabsl(ix1-xi2))*cosl(a1);

   l=(sqrtl(ix1*ix1+iy1*iy1)+sqrtl(xi2*xi2+yi2*yi2))/2.;
   xf=xi2+l*sinl(a1)*fabsl(a2-a1);
   yf=yi2-l*cosl(a1)*fabsl(a2-a1);
   af=fabsl(atanl((iy1-yf)/(ix1-xf)));
   wf+=fabsl(px2-px1)*cosl(af);
   wd+=sqrtl((ix1-xf)*(ix1-xf)+(iy1-yf)*(iy1-yf))*cosl(a1-af);
   //printf("\nl=%Lf , xf=%Lf , yf=%Lf , af=%Lf , diffa=%Lf",l,xf,yf,180/pi*af,180/pi*(a1-af));
   wp+=fabsl(sinl(a1-af))*l*fabsl(a2-a1);
   dlpc+=(px2-px1)*sinl(a1)*1000.;
   dlrc+=l*(a2-a1)*1000.;

  }
  px2=px1;  py2=py1;  qx2=qx1;  qy2=qy1;  xi2=ix1;  yi2=iy1;  xl2=xl1;  yl2=yl1;  a2=a1;

}
printf("\nN=%ld , R=%Lf , D=%Lf , xinf=%Lf , xsup=%Lf",N,R,D,xinf,xsup);
printf("\ndlp=%Lf , dli=%Lf, dlpc=%Lf , dlrc=%Lf",px1-xsav,ix1-xi1sav,dlpc,dlrc);
printf("\nsuma=%Lf , cos=%Lf , wf=%Lf , wd=%Lf , wp=%Lf , diff=%Lf , eff=%Lf , ddx=%Lf , sumdx=%Lf\n",(asav-a1)*180/pi,cosl(suma/N),wf,wd,wp,wf-wd-wp,(wd+wp)/wf, fabsl(px1-xsav)-fabsl(ix1-xi1sav),sumdx);
return 0;
}

Edit: @A.T. : the first method you gave for my specific case is the method I used before and I realized it is not like the friction. At least, I don't see it like the friction because for me the force of friction moves only of d2.

@jbriggs444: no, the needle is not an indicator, it is a real needle (without friction), but with a theoretical diameter very small, some atoms, to be very near to the dot of contact. Maybe in theory, I can take the diameter of the needle of 1 atom. I used the needle to block the force of the elastic with the two lateral walls (circle and red wall) like my images show in the message #74. I drew the forces F2 from the elastic and the forces F3 and F4 to the lateral walls. And I calculate the energy from F3 and F4, they cancel themselves. So, at final there is only the energy stored in the elastic and the energy to move the circle. And the distance moved by the needle is well d2 like I could compare with my first message.

 

[A.T.]

Edit: @A.T. : the first method you gave for my specific case is the method I used before and I realized it is not like the friction.

It gives you the distance of slippage, relevant for energy dissipated by friction

At least, I don't see it like the friction because for me the force of friction moves only of d2.

Wrong.

You should look at your problem from the rest frame of the wall. It is easier to understand relative motion, if one object is static. In the frame of the wall the circle is doing back-spinning , like the wheel of a car sliding forward on ice, while spinning the wheels in reverse gear. This increases the distance of slippage, which here is the sum of the contact point movements relative to each body.

In other cases, if the wheels were spinning forwards you would have to take the difference. In pure rolling that difference becomes zero.

 

[jbriggs444]

@jbriggs444: no, the needle is not an indicator, it is a real needle (without friction), but with a theoretical diameter very small, some atoms, to be very near to the dot of contact

But what does the needle do?

Does it fasten the elastic to the wall?
Does it fasten the elastic to the circle?
Does it pierce all three and thereby fasten the wall to the circle?

If it does not do any of the three, it might as well not exist. If it does one of the three, just say so. [You cannot fasten the needle to the point of contact because the point of contact is not a physical entity]

 

[JrK]

But what does the needle do?

I didn't says but if I use the needle I need to control in position the red arm. I use an hydraulic cylinder to control the position of the circle and I count the energy that the hydraulic cylinder consumes. I have an electric motor that controls the position of the red wall, so its rotation, like the hydraulic cylinder I count all the energy needed/recovered from it.

The elastic pulls the needle closer to the dot B.

The needle is between the two walls and it cannot move closer to the dot B because the distance between the two walls reduces more and more at the dot of contact. At the dot of contact the distance between the two walls is 0, so with a diameter of the needle greater than 0 the needle is blocked. The elastic pulls the needle but the two walls prevent it to move, because I control the position of the circle and the position of the red wall. The needle (in theory) is at the dot of contact.

 

[jbriggs444]

I didn't says but if I use the needle I need to control in position the red arm. I use an hydraulic cylinder to control the position of the circle and I count the energy that the hydraulic cylinder consumes. I have an electric motor that controls the position of the red wall, so its rotation, like the hydraulic cylinder I count all the energy needed/recovered from it.

The elastic pulls the needle closer to the dot B.

The needle is between the two walls and it cannot move closer to the dot B because the distance between the two walls reduces more and more at the dot of contact. At the dot of contact the distance between the two walls is 0, so with a diameter of the needle greater than 0 the needle is blocked. The elastic pulls the needle but the two walls prevent it to move, because I control the position of the circle and the position of the red wall. The needle (in theory) is at the dot of contact.

So if I understand it properly, the needle is oriented so that it is perpendicular to the sheet of paper on which the wall and the circle are portrayed. It is acting like a watermelon seed, being squeezed away from the point of contact. Is that correct?

 

[JrK]

Is that correct?

Yes.

 

[jbriggs444]

Yes.

Let me restate the scenario in my own words and you can tell me whether you agree.

This means that you must have a very large squeezing force on the needle and must have a very large force holding the circle against the wall. That is fine. One could imagine arrangements to accomplish this without any use of energy.

The elastic band is attached at one end to the needle and at the other end to the wall at the starting contact point of circle with wall. This point of attachment of elastic to wall is point B. Point B is fixed to the wall. We might consider a label "B" painted on the wall here.

The initial contact point of circle with wall is labelled point A. Point A moves with the circle. We might consider a label "A" painted onto the circle here. Nothing is attached at point A.

The wall has a stationary and frictionless pivot point down and to the right where it is anchored to the ground. The wall is massless and infinitely thin. The wall is rotating clockwise, restrained by a motor or generator that controls the rate of rotation.

The circle is driven up and to the right by a hydraulic cylinder that pushes on its center and which somehow braces the circle against rotation. The anchor mechanism for this hydraulic cylinder has not been specified. The details of the anchoring mechanism are important if one is examining the work done on or by the hydraulic cylinder. But for now we can consider it as an arbitrary external force applied at the center of the circle and directed up and to the right and as an arbitrary external clockwise torque on the circle.

I want to present a simple schematic picture of the arrangement after the circle has been displaced some distance up and to the right and after the wall has rotated through some angle clockwise just for the purposes of clarifying the scenario. Not for the purposes of analyzing it.

Do we agree that this is an accurate description of the scenario currently under discussion?

 

[JrK]

1/ The dot A is the dot where the needle is. Why do you want to add a dot fixed on the circle ? Note: in the program I used the dot 'i' (intersection).

2/ I don't understand your force F on the drawing you drew. The elastic pulls. The circle moves in HORIZONTAL translation to the right, not up and right.

3/ Yes, I didn't speak a lot about the hydraulic cylinder. The cylinder is fixed on the ground. The circle is glue to the cylinder: the circle cannot rotate around itself. Is the dot of attachment of the cylinder important ? Yes, it could be an arbitrary force applies on the center of the circle.

I added a pdf file with all the drawings. It is better than an image. Please, look at the drawing at left, I drew two positions of the circle and the wall.

Note: in the program lowercase 'a' is the angle of the red wall relatively to the horizontal.

 

[Stephen Tashi]

I added a pdf file with all the drawings.

Those sorts of drawings have not been helpful.
I repeat my suggestion that you use Cartesian coordinates.

Let's write the path of the point of contact $(x_c,y_c)$ between the circle and the "wall" as a function of the length $x$ of the piston.

The distance from $(0,0)$ to $(x_c,y_c)$ is $d = \sqrt{ x^2 + y_h^2 - r^2}$

$\angle \alpha = \arctan{ \frac{r}{d}}$
$\angle \theta = \arctan{ \frac{y_h}{x}}$
$\angle \phi = \angle \theta - \angle \alpha$

$x_c = d \cos{ \phi}$
$y_c = d \sin { \phi}$

Apparently, we are to assume that a force $F = (F_x, F_y)$ of constant magnitude $|F|$ acts on the circle at $(x_c, y_c)$ in the direction from that point toward the origin. The $x$-component $F_x$ of that force acts to oppose the piston.
$F_x = -|F| \cos{\phi}$

The work done by $F$ on the piston as it lengthens from $x = a$ to $x = b$ is computed by doing the line integral representing a force $F_x(x)$ acting on the path $(x_c(x), y_c(x))$ from $x = a$ to $x = b$.

A numerical approximation for the answer should be a numerical approximation to that line integral.

 

[JrK]

Notice this is for the example with the friction not the needle (for others users that could read too).

I wrote the numerical integration:

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

int main()
{
 int i,N=1000000;
 long double xc=0,yc=0,x_last=0,d=0,alpha=0,theta=0,phi=0,F=1.,Fx=0,r=0.1,yh=0.9,w=0,x=0,pi=M_PI;
 long double a=0.758579;
 long double b=0.788021;
 
 w=0;
 x=a;
 x_last=a;
 for(i=0;i<N;i++)
 {
  x=a+(b-a)/N*i;
  d=sqrtl(x*x+yh*yh-r*r);
  alpha=atanl(r/d);
  theta=atanl(yh/x);
  phi=theta-alpha;
  xc=d*cosl(phi);
  yc=d*sinl(phi);
  Fx=F*cosl(phi);
  w+=Fx*(x_last-x);
  x_last=x;
 }
    
 printf("\nw=%Lf",w);   
  
}

It gives the same work I found for the wall (with the friction). You didn't give the formula for the energy from the friction, do you have it ?

 

[Stephen Tashi]

You didn't give the formula for the energy from the friction, do you have it ?

How do you define "the energy from friction"? Isn't it the same as the work done by friction on the circle?

 

[JrK]

How do you define "the energy from friction"? Isn't it the same as the work done by friction on the circle?

I see less friction than the energy from the displacement of the circle and I would like to have calculations to understand where is my mistake. My program computes the energy from friction but it uses the dot of contact. So, I found the method with the needle + the elastic. In that method, I have the same result I thought with the friction: the elastic won d2*F but the energy needed to move the circle is d1*F.

 

[Stephen Tashi]

I see less friction than the energy from the displacement of the circle

How do you define "the energy from the displacement of the circle"?

In order to displace the circle, the piston exerts a force against the circle that is equal and opposite the force of friction on the circle - provided we are allowed to move the circle without accelerating it. So why do you think that the "energy from the displacement of the circle" is different than the work done by friction upon the circle?

 

[A.T.]

I see less friction than the energy from the displacement of the circle and I would like to have calculations to understand where is my mistake.

Your mistake is right here:

My program computes the energy from friction but it uses the dot of contact.

Why this approach is wrong is demonstrated here:
https://www.physicsforums.com/threa...of-the-vector-of-friction.987415/post-6330146

 

[JrK]

I'm agree it is a mistake to see that slip, even I see it. I replied by private message at Stephen to not overflow that thread. I found the method with the needle + elastic to have the SAME dot compared to the friction. So, I would like to understand the mistake in the example with the needle + elastic BUT WITHOUT FRICTION because the length won by the elastic is well d2 not d1, so the energy won by the elastic is d2*F and the energy needed to move the circle is d1*F. I calculated the energy needed/given by the forces F3 and F4 and the sum is at 0. So maybe a problem with an angle ?

The example, the circle moves in horizontal translation, the red wall rotates around A0, I drew the start and final position, the circle and the red wall are controlled in position with a cylinder and a motor for example. There are 4 objects: the circle and the wall, the elastic and the needle:

Enlargement of the previous image:

Enlargement of the area where there is the elastic, I drew the start and final position, the elastic is taken by the dots A and B, the dot B is fixed on the red wall, the dot A is fixed on the needle, the needle is near the dot of contact between the circle and the red wall:

To watch the forces of the elastic, I drew the start and final position, at start the elastic has a length at 0, and at final its length is d2:

To watch the needle (the needle is perpendicularly to the screen), I drew the final position, the needle is pulled by the elastic but the lateral walls prevent it to move closer the dot B:

Enlargement, I drew the final position, to watch the forces from the needle to the lateral walls (circle and red wall):

The program :

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

int main()
{
long double pi=M_PI;

long int N=100000000L;
long double R=.1;
long double D=1.;
long double xinf=1.;
long double xsup=xinf/tanl(44.99/180.0*pi);

long double px1=0,py1=0,a1=0,px2=0,py2=0,a2=0,dl=0,wd=0,wf=0,ix1=0;
long double iy1=0,ix2=0,yi2=0,qx1=0,qy1=0,s=0,xl1=0,yl1=0,xl2=0,yl2=0;
long double sx=0,sy=0,qx2=0,qy2=0,xi1sav=0,yi1sav=0,xsav=0,suma=0,l=0,asav=0;
long double xf=0,yf=0,wp=0,af=0,sumdx=0,dlrc=0,dlpc=0;

long int i;
qy1=D;

for(i=0;i<N;i++)
{
  //a1=(long double)(asup/180.0*pi-ainf/180.0*pi)/N*i;
  qx1=xinf+(xsup-xinf)/N*i;
  a1=atanl(qy1/qx1); if(i==0) a2=a1;
  s=R/tanl(a1/2.);
  px1=qx1-s;
  py1=qy1-R;
  if(i==0 || i==N-1) printf("\npx1=%Lf",px1);
  ix1=px1+R*sinl(a1);
  iy1=py1-R*cosl(a1);
  if(i<1) {xi1sav=ix1;yi1sav=iy1;xsav=px1;asav=a1;}
  //printf("\npx1=%Lf , py1=%Lf , a=%Lf , ix1=%Lf , iy1=%Lf , a1=%Lf",px1,py1,180/pi*atanl((fabsl(iy1-yi2)-fabsl(py1-py2))/(fabsl(ix1-ix2)-fabsl(px1-px2))),ix1,iy1,180/pi*a1);
  suma+=a1;

  if(i>0)
  {
   sumdx+=(fabsl(px1-px2)-fabsl(ix1-ix2))*cosl(a1);

   l=(sqrtl(ix1*ix1+iy1*iy1)+sqrtl(ix2*ix2+yi2*yi2))/2.;
   xf=ix2+l*sinl(a1)*fabsl(a2-a1);
   yf=yi2-l*cosl(a1)*fabsl(a2-a1);
   af=fabsl(atanl((iy1-yf)/(ix1-xf)));
   wf+=fabsl(px2-px1)*cosl(af);
   wd+=sqrtl((ix1-xf)*(ix1-xf)+(iy1-yf)*(iy1-yf))*cosl(a1-af);
   //printf("\nl=%Lf , xf=%Lf , yf=%Lf , af=%Lf , diffa=%Lf",l,xf,yf,180/pi*af,180/pi*(a1-af));
   wp+=fabsl(sinl(a1-af))*l*fabsl(a2-a1);
   dlpc+=(px2-px1)*sinl(a1);
   dlrc+=l*fabsl(a2-a1);

  }
  px2=px1;  py2=py1;  qx2=qx1;  qy2=qy1;  ix2=ix1;  yi2=iy1;  xl2=xl1;  yl2=yl1;  a2=a1;

}
printf("\nN=%ld , R=%Lf , D=%Lf , xinf=%Lf , xsup=%Lf",N,R,D,xinf,xsup);
printf("\ndlp=%Lf , dli=%Lf, dlpc=%.18Lf , dlrc=%.18Lf, diff=%.18Lf",px1-xsav,ix1-xi1sav,dlpc,dlrc,(dlpc-dlrc));
printf("\nsuma=%Lf , cos=%Lf , wf=%Lf , wd=%Lf , wp=%Lf , diff=%Lf , eff=%Lf , ddx=%Lf , sumdx=%Lf\n",(asav-a1)*180/pi,cosl(suma/N),wf,wd,wp,wf-wd-wp,(wd+wp)/wf, fabsl(px1-xsav)-fabsl(ix1-xi1sav),sumdx);
return 0;
}