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26 Nov 2024 19:44:30 EST (-0500)
  "bending" an isosurface (Message 1 to 2 of 2)  
From: H E  Day
Subject: "bending" an isosurface
Date: 24 Oct 1999 16:33:45
Message: <38136C3F.DF6C67C7@teleport.com>
Just curious, but does anyone here have any clue as to go about bending
an isosurface?
Preferably at a single point. I've already thought of CC's object
bender, but it is too slow for what I want to do.
Thanks in advance!

H.E. Day


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From: Thomas Willhalm
Subject: Re: "bending" an isosurface
Date: 25 Oct 1999 04:47:38
Message: <qqmd7u34yja.fsf@goldach.fmi.uni-konstanz.de>
"H.E. Day" <mil### [at] teleportcom> writes:

> Just curious, but does anyone here have any clue as to go about bending
> an isosurface?
> Preferably at a single point. I've already thought of CC's object
> bender, but it is too slow for what I want to do.

I'm not sure whether I've completely understood what you want. Anyway,
this might be the solution:

Spherical coordinates may be what you're looking for. Instead of Cartesian
coordinates (x, y, and z) you use:
r     the distance from the origin
phi   Project your point to the xy-plane. Then phi is the angle between
      the x axis and the line through the origin and the projected point.
theta the angle between the z-axis and the line through the origin and
      your point.
I suggest, that you make a drawing to help you imaging the situation.
It should be clear that
r     is in [0,infinity]
phi   is in [0,360[ degrees
theta is in [0,180] degrees

Your isosurface is defined by a equation f(x,y,z) = 0. So, what I think
that you want is to map the Cartesian coordinates to spherical coordinates
and put them in your function f. This can be achieved by:
r = sqrt(x*x + y*y + z*z)
phi = arctan (y/x)
theta = arctan ( sqrt(x*x + y*y) / z )

It's up to you to ensure that the resulting surface is smooth.

I hope that my limited English does not prevent you from understanding 
my posting.

Thomas

-- 
http://thomas.willhalm.de/
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