How about:

#declare Initial_Position=<5,3,0>;
#declare Radius_Increase=5;
#declare Revolvements=1;
#declare
Cam_Location=vrotate(Initial_Position+x*clock*Radius_Increase,y*360*Revolvem
ents*clock);

This would get you a path beginning at Initial_Position,
looking from positive x axis, and encircle the origin while
increasing the distance (but not the height) to the object.

To increase the height as well, you'd have to add another
+y*clock*Some_Value before the "," of the vrotate (look
that one up in the Docs). For more rotations, increase
Revolvements.
If you want to begin from in front of the object (I think
left-handed, -z is front), exchanget the Initial_Position
with <0,3,-5> and the "+x" in vrotate with "-z".

When you dissect that, you should understand the concept.

Regards,
Tim

-- 
Tim Nikias v2.0
Homepage: http://www.digitaltwilight.de/no_lights

> Hello Friends
>  I have been part of the news gruop for some time as a silent reader but
> this is my first post. I have not been able to find a satisfactory answer
> to a technical problem I'm having within previous posts in the newsgroup.
>  Can any math wizz out there tell me what the formula would look like that
> would describe the motion of a camera that looked at one spot in space
> continually as it gradually began to move in a circular and expanding
> spiral, steadily gaining speed and steadily rising above what it is
looking
> at?
>  If this motion could be visualized by a single curved line, it would
> probably appear to be a very symetrical and conical tornado.
>  Also; how would a mathematical formula discribing the above motion be
> worded in POVray?
>
>                                     Thanx
>
>
>


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