We get it. We use ad blockers too. But GameDev.net displays them so we can continue to be a great platform for you.
Please whitelist GameDev.net and our advertisers.
Also consider a GDNet+ Pro subscription to remove all ads from GameDev.net.
Subscribe to GameDev.net's newsletters to receive the latest updates and exclusive content.
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
Posted 29 January 2013 - 01:33 PM
Posted 29 January 2013 - 02:03 PM
I don't understand how there could be a more direct measurement, unless you're looking for an equation.
According to the definition from wolfram, an affine transformation is one that "preserves collinearity ... and ratios of distances."
Thus if you're scaling everything, the distance will scale with the lines as well. Thus new h = old h * scale amount. Otherwise the h will stay the same.
Hope this helps,
Selenaut
Posted 29 January 2013 - 02:17 PM
The distance can be seen as the length of a direction vector d. The vector can be transformed as well, and the length of the vector after transformation is the distance of the lines after transformation.
Let d := |d| be the original length.
Uniform scaling S * d =: d' will cause d' = s * d, that's right. However, non-uniform scaling isn't that simple, because the direction of d plays a role. Moreover, if the transformation in question is a composite one w.r.t. the well known primitive transformations, then S is perhaps applied in a space where the direction of d isn't the same as those it is computed initially.
I think that for the most general case the transformation must be applied.
Posted 29 January 2013 - 03:07 PM
Affine transformation is just another name for a matrix with non uniform scale, rotation, and translation baked in (more or less). I didn't mean to be too obtuse when forming the question
Anyway, I came up with: h' = h * length(S*(b - a)) / (length(b - a)) after a bit of algebra. Can someone confirm/refute that? I took the method I described and worked it out algebraically. But intuitively I'm surprised I'm using b-a and not a vector perpindicular to b-a or something along those lines.
Posted 29 January 2013 - 04:15 PM
h' = h * length(S*(b - a)) / (length(b - a))
Edited by Ravyne, 29 January 2013 - 04:16 PM.
throw table_exception("(ノ ゜Д゜)ノ ︵ ┻━┻");
Posted 29 January 2013 - 04:20 PM
Posted 29 January 2013 - 05:04 PM
Let points X and Y be arbitrary closest point pairs on the two lines. That is, X lies on the first line, and Y lies on the second line, and the distance between those two points is the same as the distance of the two lines themselves, i.e. in the original picture, h = |X-Y|.
Let H := X-Y =: (x,y,z,0). That is, H is a direction vector between the two points and it's length |H| = h is the distance between the two lines.
Let A be the affine map in question. Since A is affine, it is representible by a matrix M, using operation A(v) = M*v.
Then the sought new distance is
h' = |A(H)| = |M*H|
If the affine map A does contain shear and therefore M is not guaranteed to be orthonormal, then one can't do much better than to compute the matrix multiplication above.
If the affine map A does not contain any shearing, the formula does simplify: For ease of notation, fix ourselves to a 3D space (although 2D or other dimensions are equivalent). We can decompose the matrix M to a form M = N*S, where S is a diagonal matrix S=diag(sx,sy,sz,1) and N consists of column vectors vx,vy,vz,t, where vx, vy and vz are normalized and t is a translation component. That is, decompose the scaling part out of matrix M to matrix S.
Since A does not contain any shearing, the matrix N is orthonormal (only rotates, and potentially mirrors and its determinant is +/-1), and the new distance is
h' = |M*H| = |N*S*H| = |S*H| = |(sx*x,sy*y,sz*z)|
Posted 29 January 2013 - 06:15 PM
@clb: I don't think you can do h' = |M*H| because M*H is not guaranteed to be a shortest path between the two parallel lines after they're transformed by M.
That is, just because H is perpendicular to both lines before transformation doesn't mean it's perpendicular to both lines after transformation. Consider the case of a sheering of a square in to a parallelogram. One of those funny properties of affine transformations: closest point pairs on parallel lines aren't preserved.
Posted 30 January 2013 - 01:23 PM
Numsgil: You are absolutely correct.
For the case of no shear, it should hold. For the case with shear, I think it's possible to arrive to a closed expression by deriving the distance between the lines under the operation by the affine map, but that's indeed a more detailed examination.
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
GameDev.net™, the GameDev.net logo, and GDNet™ are trademarks of GameDev.net, LLC.