 | | Re: Another matrix inverse question —Timothy Gould
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| | | (...) 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)
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| | | | | Re: Another matrix inverse question —Brian Durney
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| | | | | (...) I can see the advantages of swapping the off-diagonals for inverting the rotation(s), but that still leaves the question of how to invert translations. In a current example I'm looking at, a minifig hand, the hand is rotated 45 degrees, (...) (18 years ago, 11-Apr-07, to lugnet.cad)
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| | | | | | | Re: Another matrix inverse question (and answer?) —Brian Durney
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| | | | | | | (...) I think I just answered my own question. I applied the inverse rotation (calculated by swapping the off-diagonals) to the translation row of the original transformation and then negated it to get the translation row in the inverse. That worked (...) (18 years ago, 11-Apr-07, to lugnet.cad)
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| | | | | | | | | Re: Another matrix inverse question (and answer?) —Timothy Gould
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| | | | | | | | (...) If it's only rotations then it should make sense. Another way of passing the information is as follows (M,x) where M is the matrix and x is the translation vector the inverse of this is (Mi,-Mi x) which you can see by showing that an operation (...) (18 years ago, 11-Apr-07, to lugnet.cad)
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| | | | | | | | | | Re: Another matrix inverse question (and answer?) —Brian Durney
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| | | | | | | | (...) Yes, that makes sense. Looking back at Travis's code, I see that's basically what's happening there, except that some references to the original matrix should be references to the inverse matrix. He probably used it only to invert rotations (...) (18 years ago, 11-Apr-07, to lugnet.cad)
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| | | | | | | Re: Another matrix inverse question —Travis Cobbs
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| | | | | | (...) You might want to check your row/column order for the converted code from LDView. At least conceptually, I always invision transformation matrices like so: A B C X D E F Y G H I Z 0 0 0 1 Notice that the translation info is in the last column, (...) (18 years ago, 11-Apr-07, to lugnet.cad)
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| | | | | Re: Another matrix inverse question —Timothy Gould
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| | | | (...) 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 (...) (18 years ago, 11-Apr-07, to lugnet.cad)
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