Why isn't it the way that the smaller t is, the more black the color at x is? From the equation that calculates the color at x, it seems that the computation is opposite as in the picture of the triangle.
joel
Yes, I believe there's a typo. The color should be computed as t * [0 0 1] + (1-t) * [0 0 0]
keenan
blue = large value (1)
black = small value (0)
The barycentric coordinate for vertex c should get bigger as we approach c, so that when we're at c the triple of barycentric coords is equal to (0,0,1).
keenan
Ah, right. Yes, @joel is right - I didn't notice the expression for the color itself.
Why isn't it the way that the smaller t is, the more black the color at x is? From the equation that calculates the color at x, it seems that the computation is opposite as in the picture of the triangle.
Yes, I believe there's a typo. The color should be computed as t * [0 0 1] + (1-t) * [0 0 0]
blue = large value (1) black = small value (0)
The barycentric coordinate for vertex c should get bigger as we approach c, so that when we're at c the triple of barycentric coords is equal to (0,0,1).
Ah, right. Yes, @joel is right - I didn't notice the expression for the color itself.