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Re: Mathematica and 3D surface.

• To: mathgroup at smc.vnet.net
• Subject: [mg23138] Re: Mathematica and 3D surface.
• From: Martin Zacho <zac at mip.sdu.dk>
• Date: Thu, 20 Apr 2000 03:20:56 -0400 (EDT)
• Organization: UNI-C
• References: <8djlnk$7qi@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

Quite nice to make a reply to ones own problem :o)


> http://www.mip.sdu.dk/~zac/temp/bucket.jpg ).

Was the original data set. I then transformed the points from cartesian
coordinates (where I couldn't make an Interpolation on each cross section)
to cylindrical coordinates (where I was able to make an Interpolation). Then
I was able to make a nice uniform grid of points which could be used by
ListSurfacePlot3D (from the Graphics package). The result can be viewed at:

http://www.mip.sdu.dk/~zac/temp/bucket1.jpg

But then again... the original problem is still valid because this method
uses a special property of the data set (the cylindrical nature). A more
general method would still be appreciated :o)

Thanks for the input from all of you (via e-mail).

M'

• Follow-Ups:
  • Re: Re: Mathematica and 3D surface.
    • From: Hartmut Wolf <hwolf@debis.com>