@pbhuang In this case, I think "pathological" just means a worst-case scenario for rendering - naive path tracing works really badly with a small directional light source + near-perfect mirror, since the probability of tracing back to the light source is small
What does "pathological" mean in this context?
Force should really be mg for this slide, meaning that acceleration would just be g, and we would not need mass in any of these equations. Motion of an object in flight should not depend on its mass.
Let me add a comment here -- in practice, it is easiest to define our polynomials p(u) for 0 <= u <= 1. What that means for this slide is that we can keep this notion of knots occurring at different times. However, it might be easier to reparameterize our f(t) where t_i <= t <= t_{i+1} in terms of parameter u, such that 0 <= u <= 1. This is pretty simple -- u is just (t-t_i) / (t_{i+1} - t_i). This reparameterization makes our polynomials easier to work with, derive, and think about. We will make the assumption of this parameterization in the class.
I'm a bit confused with this slide here. For the premultiplied (from the previous slide), isn't $$ B' = \alpha_B B$$? So the non-premultiplied equation = premultiplied equation?
Could you elaborate more on why it would be harder to sample an implicit surface like the one above?
Question! (i,j) looks like it's in the middle of the pixel ($$ f_{00} $$) but i and j are the result of a floor so should be an int (and so shouldn't be in the middle of a pixel), - is the diagram not quite right? Or am I missing something here?
Don't understand what's q bar
A and C
Yes it does
<u,v>=<v,u>
it's symmetric
yes again
Yes. It satisfies the 4 properties of a norm on Page 28
Before I replaced these slides with a few fixes, AugustSZ commented — There're 2 kinds of projections mentioned in Fundamentals of Computer Graphics: Orthographic Projection and Perspective projection. Camera Definition: position, orientation(up/look-at)
I replied — Yes, that's right. We'll have a closer look when we get to the Perspective Projection lecture on Feb 9th. We've made an assumption here that we have Perspective projection, with camera position at c (really, c=(0,0,0) given the math on the slide), up vector is y, and look-at vector is z. If we want a general camera, we'll need to loosen those restrictions later to allow arbitrary camera positions and orientations. However, you can always fall back on the rule of similar triangles. They may be oriented oddly in 3D space to begin with, but we'll see how to transform them to look like the picture above.
Orthographic projection is very simple. For orthographic projection in the picture above, v=y and u=x. Nothing more to do.
How exactly does the triangle mesh become a grid again?