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This seems like a pretty elementary math/graphics question, but for some reason I can't seem to wrap my head around it.

What I've got is four line segments, outlining a quad. Each vertex on each line segment has a known color value. For simplicity, let's assume that each line segment has 100 vertices (aka 100 known color points). An example might look something like this:

What I need to do is, render a solid quad with every internal pixel colored, based on the colors of the points on the outline. The most obvious way to do this (without specifying a vertex for every 100*100 color point) would be to first generate a 100x100 bitmap with each interpolated color value, then apply this as a texture to the quad. I'm just having difficulty figuring out how to calculate those internal color values.

It seems like it would be a matter of bilinear interpolation, but since I'm not trying to find values within four known corner points but at the intersection of a "+" of known color points, I keep getting confused. My math is a bit rusty :P

As an example, how might I approach calculating the color value of the point at (50,50) - in the middle of the quad - with known color points at (0,50), (50,0), (50,100), and (100,50)? Is this even bilinear interpolation, or is it something else entirely?

Thanks in advance!

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J23
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1 Answers1

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This looks like a boundary value problem with Dirichet conditions (that is, the values are specified at the boundary). You can't solve this using bilinear interpolation because it will usually turn out (if you have more than four input points), that the pixels adjacent to your edge pixels won't have a color continuous with their immediate neighbor.

The main thing you need to solve this is an equation that's reasonably smooth and that always gives a color value close to it's neighbor at the boundary, and there are multiple options. The obvious thing to use is Laplace's equation, which is basically like pinning a rubber sheet to the value of each color channel at the boundary and then letting it relax. Solving Laplace's isn't trivial because you need to simulate the rubber sheet for each new set of boundary conditions, but it's very common, so you could look for solvers or examples in almost any language.

tom10
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  • Well, the colors along the edges *are* more or less continuous - i.e. there won't be any particularly sharp changes. The data points are actually soil density measurements obtained by pushing a radar in lines over the ground; we want to "complete" the picture by interpolating between the values explicitly collected (the lines of colors). Previously we just had passes going along one axis, so LERPing between the parallel lines was easy. Now we're going across the other axis to add a bit more detail. Bilinear interpolation wouldn't be an acceptable approximation in this case? – J23 Mar 24 '12 at 19:27
  • I was considering a situation like this when I wrote my answer. In general, the answer is no. Specifically, if you lertp between two parallel edges and get the perpendicular ones back, then yes, I guess you can lerp, but that won't usually turn out to be the case. In general, if you need the fill to be continuous with all four edges, then you need to take all four edges into account equally, and the Laplace solution (or other boundary value soln) does this. Also, since this is a physical system, Laplace is easy to justify and explain... it's just the simplest boundary value system. – tom10 Mar 24 '12 at 20:08
  • Also, "the colors along the edges are more or less continuous" isn't the issue I'm referring to. Instead I mean, consider the top row of values that you've measured. Now how to you fill in the row right below this? This row needs to be smooth with the top row, but also smooth at the endpoints which are the tops of the vertical data points. I don't see a way to guarantee all of these conditions are met (as well as all of the other sides) unless you specifically build it into the solution, and bilinear interpolation doesn't do this. – tom10 Mar 24 '12 at 20:11
  • Ah, yeah that makes sense. Sounds like this will indeed end up being the solution - thanks so much for your thoughtful (and quick) feedback! :) – J23 Mar 26 '12 at 15:50