🚀 go-pugleaf

In article <35ea85fb.90925333@newshost.unx.sas.com>, leball@unx.sas.com (Lee D. Ballard) writes:
> If I understand Matthias's initial question correctly, I believe there
> are quite a few erroneous assumptions with Sebastien's method.  The
> first is with this statement:
> >Let d be the distance between the two points, L the length of the tube, R the radius of the circle and A the angle between the two lines connecting the center of the circle with both points.
> >
> >We have      L = A.R            
> >   and       sin(A) = d/(2.R)  
> (1)  First, we are assuming that the curve (i.e. the "tube") between
> the two points takes on the shape of an arc of a circle.  I claim that
> it could take on infinitely more shapes than simply the arc of a
> circle.  For instance, the curve's shape can be that of a sin curve, a
> cosine curve, a parabola, or something far more complex.

You're right, but Matthias was speaking of an arc of a circle...

> (2)  Assuming the curve is restricted to that of an arc of a circle, I
> claim that, in general, sin(A) != d/(2*R).  Take the case where A is
> 60 degrees (or Pi/3).  The triangle that is formed by the center of
> the circle and the two points is then an equilateral triangle (i.e.
> d=R).  Therefore, sin(A) = sqrt(3)/2, NOT d/(2*R) = 1/2.

Sorry, I made a mistake with the A angle. Actually, we have  sin(A/2) = d/(2*R).


So R is defined by:  R = d/(2*sin(L/(2*R)))  (1)
and A is defined by: A = 2*L*sin(A/2)/d      (2)

With the radius and the coordinates of both points, it's easy to find the equation of the circle. The only problem is that (1) and (2) are reflexive. But I think it's possible to solve them by giving any initial value to R or A and by applying (1) or (2)  until R or A values don't move too much. Just try it...

Sebastien Ferre