🚀 go-pugleaf

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.
(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.

Lee

On 23 Aug 1998 16:12:24 GMT, ferre@irisa.fr (Sebastien Ferre) wrote:

>In article <35DFD6E7.F2A5DA30@xs4all.nl>, Remco de Korte <remcodek@xs4all.nl> writes:
>> Matthias Weiss wrote:
>
>> When the radius is variable as well as the angle between the two lines
>> connecting the center of the circle with both points the length of the
>> arc can be a constant.
>> I was just suggesting circles because I think it's the most
>> straigtforward solution. It could also be a parabole or something like
>> that but I'm afraid that'll get far to complicated for what I want and
>> besides you'd have more then one (actually two) solutions.
>> (With a circle you'd have only two - if any).
>> 
>
>If I have well understood, you want to find the radius or, better, the equation of the circle that match the tube joining both points ?
>
>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)  
>
>So R is defined by:  R = d/(2.sin(L/R))  (1)
>and A is defined by: A = 2.L.sin(A)/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
>
>