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Lecture 5: Texture Mapping (33)
jacklishufan

Linear Transformations in homogenous cordinates can be represented by a 3x3 matrix with degree of freedom 8. This implies that 4 points should be able to uniquely determine a transformation. This seems contradicts the results in this figure? Moreover, the middle figure is apparently not "linear interploation" since if we look on the diagnoal from bottom left to upper right in the texture space and tilted space, the blacks and whites are clearly not mapped in a linear way. What is the explanation of this observed "non-linear" pattern in "linear interploation"?

Staffatsu-kotani

Great question!

Let's nail down what problem we are solving here. Say a square gets transformed into a trapezoid by 3×33\times3 perspective transform matrix, now the question is how to fill in that trapezoid with a checkerboard texture.

As you see in the next slide, we can cut this trapezoid diagonally, from top left to bottom right, and treat this as two triangles and do barycentric interpolation. The result is what the middle figure is showing here.

In fact, if you focus on the spacing of checkerboard blocks on the side, it is equally spaced for the middle figure, whereas it is non-uniformly spaced for the right (the square gets smaller as it gets further away). Again, slicing the middle trapezoid diagonally, the middle figure is indeed the result of linear interpolation.

When doing texture mapping for perspective projection, we have to sample texture non-uniformly, as the spacing on the side from the right figure characterizes, and that is basically the main takeaway from this set of slides.

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