
Math and Movies (Animation at Pixar) - nazri1
https://www.youtube.com/watch?v=mX0NB9IyYpU
======
arketyp
I find it interesting that the interviewer almost calls out on the
disappointing fact that the surface smoothness is essentially just "blobized"
interpolation: "It doesn't seem very subtle." It's like when a director has
run the film through some colorizing filter to give it a desired tone and you
can tell to the point of it becoming distracting.

~~~
jcl
The interviewer was trying to figure out why the subdivision process doesn't
turn everything into a blob -- which is understandable, given that the
examples he was shown were squares turning into circles and cubes turning into
spheres. I think he's trying to understand why, for example, Geri's head
doesn't turn into a sphere.

DeRose doesn't say so explicitly, but the reason is that the position of any
given point on the infinitely subdivided surface is only determined by the
positions of a handful of points in the control mesh. That's why, for example,
the derivation for the final position of B given at the end of the video only
depends on its neighbors A and C, and not any other points.

Thus, if you want more control over a portion of a subdivision surface, you
can just use a denser control mesh in that portion of the model. You can
create arbitrarily hard edges by moving control vertices closer together. At a
microscopic scale, the edge is still smooth and blobby, but it looks sharp at
a distance.

You can also make creases in subdivision surfaces by using different
subdivision weights over the surface of the model, which may be what DeRose is
getting at when he answers with "magic numbers". But it's my understanding
that most modern CG is done using subdivision with the standard Catmull-Clark
weights over the entire model, instead relying on the density of the input
mesh to specify detail.

~~~
CyberDildonics
This is correct. Explicit/per vertex subdivision weights are basically never
used in high end CG. Geometric detail is used, since it is simple and
universal. Geometry density is no longer a significant factor in interactivity
and not the most significant factor in rendering.

------
ivan_ah
In case someone is interested in the "eigenanalysis," the recursion formula
is:

    
    
        [1/2, 1/2, 0  ] [A^n]     [A^n+1]
        [1/8, 3/4, 1/8] [B^n]  =  [B^n+1]
        [0,   1/2, 1/2] [C^n]     [C^n+1]
        \_____   _____/
               M
    

And the question is, given [A^0,B^0,C^0], find [A^∞,B^∞,C^∞], which is
equivalent to computing the infinite power of M. Waaaat? Enter the
eigendecomposition.

The eigenvalues of M are 1/4, 1/2, and 1. If you compute M^∞, the 1/4 and 1/2
"eigenspaces" will disappear, so you're left with the subspace of the
eigenvalue 1. [http://bit.ly/eigenex001](http://bit.ly/eigenex001) M^∞ = Q
_L^∞_ Q.inv(), hence the [1,4,1] appears... very cool.

Sometimes procrastinating by reading HN actually helps with your work---today
I'm working on problem sets for book 2
[http://gum.co/noBSLA](http://gum.co/noBSLA)

