POV-Ray : Newsgroups : povray.advanced-users : circles in circles : Re: circles in circles Server Time
30 Jul 2024 14:17:24 EDT (-0400)
  Re: circles in circles  
From: Ron Parker
Date: 26 May 1999 13:21:54
Message: <374c1fa2.0@news.povray.org>
On Wed, 26 May 1999 16:50:50 +0200, Ph Gibone wrote:
>I would say that the relation is :
>
>sin(pi / n) = r/(R-r)
>
>where n is the number of small circles, r the radius of the small circles
>and R the radius of the big one.
>
>for n=6 this gives r = R/3 which is true !

I get this too (the argument to sin is in radians).  Here's the reasoning:

The centers of the smaller circles are the vertices of a regular n-gon.  
The distance between centers is 2*r, but it is also the length of the side 
of a regular n-gon of radius R-r. That length is 2*(R-r)*sin(pi/n).  So, 
2*r=2*(R-r)*sin(pi/n).

Solving for r I get:

r = R*(sin(pi/n)/(1+sin(pi/n)))

Note, however, that there is a more efficient way to pack 7 wires than the 
one shown in the image: pack six around the outside and put the seventh in 
the center.  The seventh wire won't be twisted, but you'll be able to use 
the larger wire radius of R/3.  This leads to a much hairier problem, but 
if you never look at the end of the cable nobody will ever know anyway.

Depending on how you're doing the twist, you may find that this value for 
r is too big.  If you're using helices with a minor radius of r and a 
major radius of R-r, a cross-section of the cable will contain something 
resembling ellipses with a minor radius of r and a major radius, s, that 
depends on the amount of twist you're applying (s/r increases as you twist 
more.)  The minor radius of each ellipse will lie on a radius of the cable.
As a result, the equation is now much, much uglier.  After this point I 
don't trust my math much but I'll make an attempt:

I think that the distance between centers is now something like
2*(r+sqrt(s^2-r^2)*cos(pi/n)).

I think the relationship between s and r is something like 
s^2 = r^2+(2*pi*(R-r)*r/w)^2), where w is the "wavelength" of a single 
twist.  So s^2-r^2 is (2*pi*(R-r)*r/w)^2 and we get

r*(1+(2*pi*(R-r)/w)*cos(pi/n))=(R-r)*sin(pi/n).

Notice that as w approaches infinity, corresponding to zero twist, this 
approaches the equation we had before.  As w approaches zero, 
corresponding to an infinite "amount of twist", r approaches zero as 
well.  Somebody else can try to solve this mess for r.  :)


Post a reply to this message

Copyright 2003-2023 Persistence of Vision Raytracer Pty. Ltd.