Hi. I've a question . Are the matrices C and V the extrinsics and intrinsics equivalent in Computer Graphics? C transfers to an arbitrary coordinate system, V performs perspective projection? Also, isn't [x,y,z,w] a homogeneous matrix?
31:50 _[...] I'm going to subtract P from both sides_ Somehow I - P = Q1 + t(Q2 - Q1) - P is the same as I - P = (Q1-P) + t((Q1-P) - (Q2-P)) Does this have something to do with the fact that _I_ is a point and _I - P_ is a vector?
Instead of saying you subtract P from both sides, you could also describe this step as subtracting P from each point to get the vector from P to that point. It should become more clear at 43:54, when using the origin (0,0,0) as P
Camera Transform: 0:35 - 13:00
Image Space & Screen Space: 13:00
Alternative equation of plane: 22:00
Convex Polygon: 41:40
20:40 Clipping starts
36:50 How to clip a 2d polygon with a plane (have to know how to clip a segment, seen just previously)
Whoa first time I try to understand from a course, and I got it ! good explanation, good smiles
Hi. I've a question . Are the matrices C and V the extrinsics and intrinsics equivalent in Computer Graphics? C transfers to an arbitrary coordinate system, V performs perspective projection? Also, isn't [x,y,z,w] a homogeneous matrix?
Divide by W is the new multiply by zero
31:50 _[...] I'm going to subtract P from both sides_
Somehow
I - P = Q1 + t(Q2 - Q1) - P
is the same as
I - P = (Q1-P) + t((Q1-P) - (Q2-P))
Does this have something to do with the fact that _I_ is a point and _I - P_ is a vector?
+hanzo2001 (Q2 - P) - (Q1 - P) = Q2 - P - Q1 + P = Q2 - Q1
Instead of saying you subtract P from both sides, you could also describe this step as subtracting P from each point to get the vector from P to that point. It should become more clear at 43:54, when using the origin (0,0,0) as P
pls listen 43:15 in 25% speed, it killed me.