On 19 Nov 1999 17:52:36 -0500, ing### [at] homenl (ingo) wrote:

>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

I got your intentions wrong. This should do what you want, as
described in your original post.

#declare Pos=array[11]

#declare i=0;
#while (i<11) #declare Pos[i]=<i,0,0>; #declare i=i+1; #end

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

If it doesn't, I'll look at it tomorrow.


Peter Popov
ICQ: 15002700

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