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Thread: building a 3d world globe

  1. #11
      jumpjack is offline
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    Quote Originally Posted by waldronate View Post
    R
    Below are examples of this process on a basic Earth map to get polar caps.
    Cool, looks like it's exactly what I need!
    But I can't install Wilbur on my office PC, as the installer requires administrator privileges?!?
    Isn't any ZIP/portable version available?

    edit:
    ...and I can't download Fractal Terrains as it appears as a "Game" site! :-(
    Last edited by jumpjack; 02-06-2009 at 04:53 AM.

  2. #12
      waldronate is online now
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    Try installing directly from the msi file rather than the exe in the WIlbur setup - it might work better.

  3. #13
      jumpjack is offline
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    Quote Originally Posted by waldronate View Post
    Try installing directly from the msi file rather than the exe in the WIlbur setup - it might work better.
    I found two better solutions:
    - I waited till I went home and I installed Wilbur on my home PC
    - I installed a virtual PC on office PC

    But now I have another difficulty: how do I know how large the saved image must be (proposed size is always 1024x1024) to "match" the sinusoidal projection which I used to obtain the petals?
    Let's suppose the starting image is 640x320 size.

  4. #14
      waldronate is online now
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    Pick any size that you want. It's the inches / degree at constant dpi when printing that's important, not the image size.

    For an arbitrary ball, the length of a pole-to-equator sinusoidal gore will be (ball diameter * pi / 4). A common 10-inch playball give a 7.85 inch gore (10 * 3.1415927 / 4).

    With a 10-inch ball, you need a sinusoidal gore that's 7.85 inches long pole to equator. If you are using an image that's 320 pixels high, then the pole-to-equator distance is half of that or 160 pixels. 160 pixels / 7.85 inches = 20.4 pixels/inch (very low resolution). To get an endcap that's 15 degrees radius and printed at the same ppi then you'll need an image that's 2/6 of the 160 pixels (15 degrees / 90 degrees = 1/6 radius * 2 = diameter) or 53.33 pixels) or 53 pixels.

    For a more reasonable 2048 height image then you're looking at 130 ppi and 341 pixel endcaps.

  5. #15
      jumpjack is offline
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    Quote Originally Posted by waldronate View Post
    Pick any size that you want. It's the inches / degree at constant dpi when printing that's important, not the image size.

    For an arbitrary ball, the length of a pole-to-equator sinusoidal gore will be (ball diameter * pi / 4)[...].
    Thanks, I think now I got the point.
    Given ball diameter D, I need a W x H map, where:
    W=2*H
    H=D*pi/2 (H/2 = C/4 = 2*pi*r/4 = D*pi/4)

    Given this map, I need, for a Lambert projection 30° wide, an image WL wide, where:
    WL=D*pi/12 (WL=C*30/360 = C/12 = 2*pi*r/12 = D*pi/12)

    I hope it's correct.
    Last edited by jumpjack; 02-10-2009 at 08:12 AM.

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  7. #17
      jumpjack is offline
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    Quote Originally Posted by RobA View Post
    Here is an inspirational video for you:

    http://videos.howstuffworks.com/scie...lobe-video.htm

    -Rob A>
    thanks, very useful.
    I need a rod.

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