Le 11/09/2013 00:31, galenwolfe nous fit lire :
> I forgot to add:
> 
> how do I find the new circumference created by the initial row of threads? Thus
> allowing me to continue the thread 'weaving' d # of times?
> 

From looking at
> http://www.wikihow.com/Make-a-Dreamcatcher

it seems that, starting with a 8 attach points, the thread go from N to
N+2 until it cross the line of N-1 to N+1 at which point it goes to N+1.

8 regular points on a circle are every pi/4 (or 45°). Going from N to
N+2, the line is also at 45° and stop after pi/8 from the center.

Therefore, the ratio between external radius and internal radius (for
the next iteration, turned by pi/8 (22.5°)) is 1:sqrt(10)/4

(distance from center of inflexion point is sqrt( (sqrt(2)/2)^2 +
(sqrt(2)/4)^2), angle is pi/8+k*pi/4, when the external points are at 1
at angle of k*pi/4 )

(ratio, aka scale factor for povray, is about 1 to 0.79056941504209483299 )

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