Subject:
|
Re: Another matrix inverse question
|
Newsgroups:
|
lugnet.cad
|
Date:
|
Wed, 11 Apr 2007 12:11:01 GMT
|
Viewed:
|
1953 times
|
| |
|
|
In lugnet.cad, Timothy Gould wrote:
> In lugnet.cad, Brian Durney wrote:
> > Thanks for the information, everyone. Although the test case I posted was
> > arbitrary, I did some other tests with matrices that were used for rotations and
> > translations. The upper-left 3x3 seemed to be OK, but the bottom row had
> > problems. I'm guessing that the inversion routine wasn't written to include
> > those.
> >
> > I wanted the inverse so that I could move a part back to the origin and do
> > rotations in the local coordinate space. I think that what I will do is keep an
> > inverse matrix with the transformation matrix, and every time I do a
> > transformation (like rotate 45 degrees around the X axis) I will do the inverse
> > (-45 degrees) and apply it to the inverse matrix.
> >
> > I suppose another possibility is to transform the origin and/or axis of rotation
> > and then use quaternions to specify rotation around a point other than the
> > origin, but it seems like that would have more opportunities for mistakes,
> > considering my limited experience with quaternions.
> >
> > Thanks again for the help.
> > Brian Durney
>
> Quarternions are, IMO, a bad way to do what matrices do well.
>
> For a rotation you don't need to (and shouldn't) calculate the inverse by a
> routine. Just transpose (swap the off-diagonals) and you have the inverse.
> There's no point running unneccessay numerical risks.
>
> Tim
Som further clarification... I was back in the 3x3 matrix world when I wrote
this: the true inverse of a rotation + translate in psuedo-4x4 notation is as
follow
[ R11 R12 R13 x
R21 R22 R23 y
R31 R32 R33 z
0 0 0 1 ]
goes to
[ R11 R21 R31 xp
R12 R22 R32 yp
R13 R23 R33 zp
0 0 0 1 ]
where xp = -(R11 x + R21 y + R31 z), yp = -(R12 x + R22 y + R32 z), zp = -(R13 x
+ R23 y + R33 z)
Tim
|
|
Message is in Reply To:
 | | Re: Another matrix inverse question
|
| (...) Quarternions are, IMO, a bad way to do what matrices do well. For a rotation you don't need to (and shouldn't) calculate the inverse by a routine. Just transpose (swap the off-diagonals) and you have the inverse. There's no point running (...) (18 years ago, 10-Apr-07, to lugnet.cad)
|
12 Messages in This Thread:
 
 
- Entire Thread on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
|
|
|
|
