clipka wrote:
> Maybe because you're not perfectly sure what point you should be rotating it
> about?
> 
> From your formula I gather that you are rotating about <1/2,.866/2,0>. However,
> the described equilateral triangle's centroid (i.e. the "center of mass" so to
> speak) is at <1/2,.866/3,0>.
> 
> 
> 
Your right! Thanks for the help. I was trying to rotate from the center 
of the box contanting the triangle.
I now use the more percise .5*tan(radian(30)) for the y adjust.

Also I had the y componet added to the whole function:
Fnc=function(x,y,z,Ang){Fna((x-.50)*cos(Ang)+(y-.433)*sin(Ang)+.50,
                      -(x-.50)*sin(Ang)+(y-.433)*cos(Ang),1)+.433};
                                                         ^  ^^^^^
                                                         |    This
                                                     Should be here

But now that's taken care of, I get a nice little cracks. (picture attached)

Have fun!

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