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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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)
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