Peter Popov wrote:

>//try this
>
>#declare Pos=array[11]
>
>#declare i=0;
>#while (i<11) #declare Pos[i]=<i,0,0>; #declare i=i+1; #end
>
>#declare i=1;
>#while (i<11)
>  #declare Pos[i]=vrotate((Pos[i]-Pos[i-1]),<0,0,10*clock>)+Pos[i-1];
>  #declare i=i+1;
>#end

No, I'm afraid that's not what I'm looking for. What I expected to see is 
the spheres arranged in a arc, about one quarter of a circle.

mmm, all spheres end up on a circle, the length of the arcsegment between 
two spheres is known (1 in this case). The angle between the two vectors 

circle will always go through the origin, where the first sphere is.
With this, I think the radius and centerposition of the circle can be 
calculated. Once this circle is known, take the distance frome any sphere 
on the x-axis to the origin. Take an arcsegment of the circle with the same 
length and put the sphere there.
Would it work? I'll sleep over this and try if I can figure it out 
tomorrow.

Thanks Peter for your effort.

Ingo

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