r*****y
发帖数: 507 | 1 突然想起一个问题，
scale, rotation, translation, squash, etc...等可以最终相乘为一个
4*4 matrix MX.
Now, given an arbitrary 4*4 transformation matrix MX, and MX(4,1:4) = [ 0 0 0
1]; Can we decompose this matrix to all possible operations, such as scaling,
rotation etc... is it unique? or for some matrix, it doens't exist feasible
solution?
Any hint/link for this? thanks. | r***l
发帖数: 36 | 2 if the 4th row is like that.
then, you can get rid of the last column and think of it as a translation,
then the problem is reduced to MX(1:3,1:3).
but it won't be unique.
if you have a rotation R, and a translation v.
then ( R 0 ) * (1 v) = (R Rv)
0 1 0 1 0 1
but (1 Rv ) * (R 0) = the same thing.
0 1 0 1
if you decide to do the translation last, and then it will be( 1 MX(1:3, 4)
)
【在 r*****y 的大作中提到】
: 突然想起一个问题，
: scale, rotation, translation, squash, etc...等可以最终相乘为一个
: 4*4 matrix MX.
: Now, given an arbitrary 4*4 transformation matrix MX, and MX(4,1:4) = [ 0 0 0
: 1]; Can we decompose this matrix to all possible operations, such as scaling,
: rotation etc... is it unique? or for some matrix, it doens't exist feasible
: solution?
: Any hint/link for this? thanks.
| r*****y
发帖数: 507 | 3 yeah, order matters and it is not unique.
)
【在 r***l 的大作中提到】
: if the 4th row is like that.
: then, you can get rid of the last column and think of it as a translation,
: then the problem is reduced to MX(1:3,1:3).
: but it won't be unique.
: if you have a rotation R, and a translation v.
: then ( R 0 ) * (1 v) = (R Rv)
: 0 1 0 1 0 1
: but (1 Rv ) * (R 0) = the same thing.
: 0 1 0 1
: if you decide to do the translation last, and then it will be( 1 MX(1:3, 4)
|
|
