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Re: texture mapping/memory leak?

  • To: mathgroup at
  • Subject: [mg86794] Re: texture mapping/memory leak?
  • From: michael.p.croucher at
  • Date: Fri, 21 Mar 2008 01:53:33 -0500 (EST)
  • References: <frt5i1$t28$>

On 20 Mar, 08:00, "Fred Klingener" <gigabitbuc... at> wrote:
> Ref: Mathematica 6.0.2, WinXP, mid-range Intel dual core, 1GB RAM.
> Lately, I've been trying to work out texture mapping with Mathematica.
> By 'texture mapping' I mean transforming and applying the contents of
> an image file onto the surface of a 3D object.
> I'll describe the approach I've taken, the results so far, then try to pose
> some questions. The example problem I'll use is a more-or-less standard
> one - mapping a Mercator global map onto a sphere.
> I copied a map image from Google Images ("earth map Mercator"), picking a
> small thumbnail, maybe 100 or 130 pixels wide, and using it to define a
> variable
> map = First[*paste image here*]
> map is a Raster, in my case specified as a List of 3-Lists interpreted as
> RGBColors. After a long time foundering around, I discovered that the
> specification for the Option Mesh-> has to be:
> grid = Take[Dimensions[map[[1]], 2] - 1
> This gives a pixel count of the form {width, height}.
> Testing with a rectangular mapping:
> RegionPlot[-2<=x<=2&&-1<=y<=1,{x,-3,3},{y,-2,2}
> ,MeshStyle->None
> ,Mesh->Reverse[grid]
> ,MeshShading->Map[RGBColor,map[[1]],{2}]]
> seems to do what I want after, for completely inscrutable reasons, I Reverse
> the Mesh Option spec to represent {height, width}.
> Then  I tried a sphere, added a red axis line, and finally got this:
> grid = Take[Dimensions[map[[1]]], 2]-1;
> Timing[Graphics3D[{
> First[SphericalPlot3D[
> 1.0,
> ,{\[Theta],0,Pi}
> ,{\[CurlyPhi],0,2 Pi}
> ,Mesh->Reverse[grid]
> ,MeshStyle->None
> ,MeshFunctions->{#5 &,-#4&}
> ,MeshShading->Map[RGBColor,map[[1]],{2}]
> ]]
> ,Red,Thick,Line[{{0,0,-1.5},{0,0,1.5}}]}
> ,ImageSize->{400,400}
> ]]
> and the result is at least recognizable. If you choose an image that's not a
> true Mercator, Greenland and South America might extend to the poles -
> insignificant details. At least North, South, East, and West are right.
> To get this to work required a vast amount of thrashing, not helped at all
> by SphericalPlot3D's parameterization inverting that advanced by Mathworld.
> I can't even dope out whether SP3D is internally consistent, considering the
> odd form required for the MeshFunctions. The (/[CurlyPhi]) is the azimuth,
> the second in the SP3D parameter list, but first in the return order (#4
> before the #5 (/[Theta]) elevation.)
> I can't quite make sense of the minus sign in front of the #4 either. Even
> though \[Theta] runs the direction opposite to latitude, its direction is
> consistent with the direction of x in the 2D example, which seemed to work
> ok.
> Converting #5 to latitude is evidently not necessary, because the image rows
> count from the top.
> I'm left, though, with a pile of imponderables.
> 1. Why so slow? As written, Timing[] in the code fragment above reports
> around 1.5 seconds, but the actual meat clock elapsed time is more like a
> minute. How come?
> 2. Is there a pattern I can't discern in the way the coordinate defnitions
> relate to the MeshFunctions and the orientation of the MeshShading.
> 3. Are there any reported memory leakage problems associated with these
> calculations? When I work with MeshShading, my page file usage (as reported
> in my Windows Task Manager) gradually climbs to a point where the system
> hangs. If I put it into a Manipulate, I've had to pull the power and the
> battery.
> 4. Is there a faster or otherwise preferred way to do this?
> TIA,
> Fred Klingener

Hi Fred

I recently did something like this while mucking around with 'Easter
Eggs.'  No idea if its any better or worse than what you have done but
you may find it useful - Essentially I make use of color function to
take care of the mapping.  Its very hackish and I made no attempt to
optimise it or anything but the details can be found here


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