Maths - Matrix to Axis Angle - Forum Discussion

By: Gene Gorokhovsky (geneg102) - 2008-10-30 15:48

In case of Math.PI turn, if we find the largest diagonal element, only corresponding axis dimension needs to be calculated using square root, the rest can be done from the off-center elements. If, by convention, we pick the the sign of the the elements calculated from square root as positive, rest come out automatically, simplifying logic and speeding up code. This also eliminates need for checks for negative square root checks (since any normalization jitter affects only diagonal elements close to -1, and such element cannot possibly be the largest.)

The whole PI branch becomes

 if (m.m00 >= m.m11) {
  if (m.m00 >= m.m22) {
  //00 is max diag 
   x = Math.sqrt(0.5 * (m.m00+1)); 
  y = m.m01 * 0.5 / x;
  z = m.m02 * 0.5 / x;
   } else {
  //22 is max
   z = Math.sqrt(0.5 * (m.m22+1));
  x = m.m02 * 0.5 / z;
 y = m.m12 * 0.5 / z;
  }
   } else {
  if (m.m11 >= m.m22) {
 //11 is max diag 
  y = Math.sqrt(0.5 * (m.m11+1));
  x = m.m01 * 0.5 / y;
 z = m.m12 * 0.5 / y;
  } else {
 //22 is max
  z = Math.sqrt(0.5 * (m.m22+1));
  x = m.m02 * 0.5 / z;
 y = m.m12 * 0.5 / z;
  }
 }
 
 
By: Martin Baker (martinbaker) - 2008-10-31 10:45
Gene,

Thank you very much for this. I have put a link to your message on the Matrix to Axis Angle page (in a box below the Singularities heading). I hope that is alright?

I need to do some more work on this, which may take me some time, so that I can incorporate this fully into the page.

Coincidently I had noticed that this is the method used on the matrix to Quaternion page:
https://www.euclideanspace.com/maths/geometry/rotations/conversions/matrixToQuaternion/index.htm
and so I was just starting to think along these lines myself.

Cheers,

Martin

 

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