5 ms·

I understand what you're saying, but they're still very magical to me. They are taking something that is a non-linear operation in R3 (like a translation) and m

by geokon 9y ago

I understand what you're saying, but they're still very magical to me. They are taking something that is a non-linear operation in R3 (like a translation) and making it linear by using a trick and giving it an extra dimension. This is somehow very unsettling to me. (what is the set of operations that can be linearized like this?) I wish I "got it" better

Sharlin 9y ago
The trick is that a translation of a point (a1, ..., an) is essentially a shear of a point (a1, ..., an, 1) - a linear operation in R^(n+1). But because the points that are of interest lie on a (hyper)plane that is not a vector space - it does not intersect the origin - a shear looks like a nonlinear operation after projection to R^n. ^ |---a | | --+---------> | ^ | ,---a |/ / --+---------> |
- Sharlin 9y ago
  And you can see how points on the a_(n+1) = 0 plane ("infinitely far away" in projective terms) are invariant under this operation - so they behave like pure direction vectors and can be used as normals and transformed with the same matrix as points!
- UnquietTinkerer 9y ago
  This explanation really helped, thank you.
deleted 9y ago
[deleted]
contravariant 9y ago
The translation part is not that complex, you're basically just introducing an extra coordinate 'w' which is always 1, so you can write x -> x + w, which is linear, rather than x -> x + 1, which is not. Things start to get weird when you note that this works even when w isn't 1. If you then decide that e.g. (x,y,w) and (x/w, y/w, 1) should be the same point, you can do some rather interesting things. Although, if you identify (x:y:w) with all the points on the line (lx:ly:lw) (for any l) then suddenly (x/w, y/w, 1) is just the point where this intersects the plane w=1, which isn't too hard to visualize. And if you think about it this also kind of explains why this coordinate system can be used to project a 3D scene onto a 2D surface with correct perspective.

by geokon 9y ago

Sharlin 9y ago

The trick is that a translation of a point (a1, ..., an) is essentially a shear of a point (a1, ..., an, 1) - a linear operation in R^(n+1). But because the points that are of interest lie on a (hyper)plane that is not a vector space - it does not intersect the origin - a shear looks like a nonlinear operation after projection to R^n. ^ |---a | | --+---------> | ^ | ,---a |/ / --+---------> |

Sharlin 9y ago
And you can see how points on the a_(n+1) = 0 plane ("infinitely far away" in projective terms) are invariant under this operation - so they behave like pure direction vectors and can be used as normals and transformed with the same matrix as points!
UnquietTinkerer 9y ago
This explanation really helped, thank you.

deleted 9y ago

[deleted]

contravariant 9y ago

The translation part is not that complex, you're basically just introducing an extra coordinate 'w' which is always 1, so you can write x -> x + w, which is linear, rather than x -> x + 1, which is not. Things start to get weird when you note that this works even when w isn't 1. If you then decide that e.g. (x,y,w) and (x/w, y/w, 1) should be the same point, you can do some rather interesting things. Although, if you identify (x:y:w) with all the points on the line (lx:ly:lw) (for any l) then suddenly (x/w, y/w, 1) is just the point where this intersects the plane w=1, which isn't too hard to visualize. And if you think about it this also kind of explains why this coordinate system can be used to project a 3D scene onto a 2D surface with correct perspective.