Full playlist: • Computer Graphics (CMU... Course information: 15462.courses.cs.cmu.edu/
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Пікірлер: 16
@ricardomarino85542 ай бұрын
For the question in 43:09: It's true that if the point x get's closer to x_i, its value get's closer to f_i, since for phi_i the ratio becomes one (the triangles overlap) but for the other ones the ratio becomes zero. However, it doesn't do it in the same way of the other one (the 2d previous linear interpolation). For instance, consider the point just in the middle of the triangle. In the previous linear interpolation the value of phi_i, phi_j and phi_k was the same 0.5 (if the triangle is equilateral), but for the interpolation with the areas it's 0.3 for phi_i, phi_j and phi_k, since the area of each subtriangle is a third of the bigger one.
@rupeshmandke3 жыл бұрын
These lectures are like a gold mine for computer graphics enthusiasts ... thanks a ton Keenan.
@Karuska22ps
3 жыл бұрын
did this course help you a lot?
@GrandomamaBulletto2 жыл бұрын
Solution for 01:06:53: We get the same result. Intuitively: If we blend four colors on a painters palette, the order in which we mix them doesn't matter, so neither should it here. Mathematically: Proof by expanding out horizontal-first interpolation and factorizing by s and (1-s) (1-t) * ( (1-s) * f00 + s * f10) + t * ( (1-s) * f01 + s * f11) Expand by multiplication = (1-t) * (1-s) * f00 + (1-t) * s * f10 + t * (1-s) * f01 + t * s * f11 Group differently = (1-t) * (1-s) * f00 + t * (1-s) * f01 + t * s * f11 + (1-t) * s * f10 Factorize out (1-s) and s = (1-s) ( (1-t) * f00 + t * f01) + s * ( (1-t) * f10 + t * f11) Which is first horizontal interpolation and then vertical. qed.
@user-pl4df6le1g2 жыл бұрын
00:44:50: should that be cross-product (for example $ji \times jx$) instead of dot product?
@tomasdougan18122 жыл бұрын
Hello, where can I find the article you reference @49:05? Thank you
@mathiaz9432 жыл бұрын
11:20 why do we have a negative w-vector in the R-matrix? Does it not make this a left-handed coord. system?
@praveensanap Жыл бұрын
@Keenan Crane Why don't we apply texture maps before perspective projection?
@GuillermoValleCosmos3 жыл бұрын
the simple "divide by z" matrix in 26:00 does a perspective projection, not just the transformation into clip coordinates right?
@keenancrane
3 жыл бұрын
Right. You have one matrix that both transforms you into clip coordinates, and performs the transformation that (after a homogeneous divide) will determine what your projection looks like (perspective, orthographic, etc.).
@GuillermoValleCosmos
3 жыл бұрын
@@keenancrane what I mean is that the bottom matrix keeps information about depth, after the homogeneous divide (NDC are a volume) while the top one doesn't (everything lies on a plane after divide). The element in the fourth column seems to be what causes this difference
@animeshkarnewar33 жыл бұрын
30:04 correction: orthographic projection instead of orthogonal projection.
@mohamedhashem19383 жыл бұрын
1:04:29 i don't understand why we subtract 0.5 from the u and v before we round them down ? why not just round them down to get the right texel ?
@mat9300
3 жыл бұрын
A sample point is located in between four texels' centers. By subtracting 0.5 and rounding down we find coordinates of the lower-left texel. Without subtracting first we wouldn't know in which direction to interpolate. I think that the diagram is slightly wrong, (i, j) point should correspond to a texel's lower-left corner and not its center.
@sahhaf12347 ай бұрын
@49:00 where the heck is "the useful article down below"?
@sharkbonebroth
6 ай бұрын
I believe its Perspective-Correct Interpolation by Low Kok Lim. Should pop up on google when you search it.
Пікірлер: 16
For the question in 43:09: It's true that if the point x get's closer to x_i, its value get's closer to f_i, since for phi_i the ratio becomes one (the triangles overlap) but for the other ones the ratio becomes zero. However, it doesn't do it in the same way of the other one (the 2d previous linear interpolation). For instance, consider the point just in the middle of the triangle. In the previous linear interpolation the value of phi_i, phi_j and phi_k was the same 0.5 (if the triangle is equilateral), but for the interpolation with the areas it's 0.3 for phi_i, phi_j and phi_k, since the area of each subtriangle is a third of the bigger one.
These lectures are like a gold mine for computer graphics enthusiasts ... thanks a ton Keenan.
@Karuska22ps
3 жыл бұрын
did this course help you a lot?
Solution for 01:06:53: We get the same result. Intuitively: If we blend four colors on a painters palette, the order in which we mix them doesn't matter, so neither should it here. Mathematically: Proof by expanding out horizontal-first interpolation and factorizing by s and (1-s) (1-t) * ( (1-s) * f00 + s * f10) + t * ( (1-s) * f01 + s * f11) Expand by multiplication = (1-t) * (1-s) * f00 + (1-t) * s * f10 + t * (1-s) * f01 + t * s * f11 Group differently = (1-t) * (1-s) * f00 + t * (1-s) * f01 + t * s * f11 + (1-t) * s * f10 Factorize out (1-s) and s = (1-s) ( (1-t) * f00 + t * f01) + s * ( (1-t) * f10 + t * f11) Which is first horizontal interpolation and then vertical. qed.
00:44:50: should that be cross-product (for example $ji \times jx$) instead of dot product?
Hello, where can I find the article you reference @49:05? Thank you
11:20 why do we have a negative w-vector in the R-matrix? Does it not make this a left-handed coord. system?
@Keenan Crane Why don't we apply texture maps before perspective projection?
the simple "divide by z" matrix in 26:00 does a perspective projection, not just the transformation into clip coordinates right?
@keenancrane
3 жыл бұрын
Right. You have one matrix that both transforms you into clip coordinates, and performs the transformation that (after a homogeneous divide) will determine what your projection looks like (perspective, orthographic, etc.).
@GuillermoValleCosmos
3 жыл бұрын
@@keenancrane what I mean is that the bottom matrix keeps information about depth, after the homogeneous divide (NDC are a volume) while the top one doesn't (everything lies on a plane after divide). The element in the fourth column seems to be what causes this difference
30:04 correction: orthographic projection instead of orthogonal projection.
1:04:29 i don't understand why we subtract 0.5 from the u and v before we round them down ? why not just round them down to get the right texel ?
@mat9300
3 жыл бұрын
A sample point is located in between four texels' centers. By subtracting 0.5 and rounding down we find coordinates of the lower-left texel. Without subtracting first we wouldn't know in which direction to interpolate. I think that the diagram is slightly wrong, (i, j) point should correspond to a texel's lower-left corner and not its center.
@49:00 where the heck is "the useful article down below"?
@sharkbonebroth
6 ай бұрын
I believe its Perspective-Correct Interpolation by Low Kok Lim. Should pop up on google when you search it.