4
$\begingroup$

This is the follow-up question from here: Minimum requirements to uniquely represent a 3D object in space

Assume I have 3 original points in a 3D object (in 3D space) as A1=<xa,ya,za>, A2=<xa,ya,za>, and A3=<xa,ya,za> (assume we have all the requirements mentioned in our previous question). The 3D object is moved and rotated in the 3D space, and the new destination points become B1=<xb,yb,zb>, B2=<xb,yb,zb>, and B3=<xb,yb,zb>.

What is the formula for the transformation matrix? Basically, I need a matrix that if applied to all points of the origin object, I get the displaced object.

$\endgroup$
  • 1
    $\begingroup$ For near minimal you store a vector for the translation and a quaternion for the rotation, so 7 elems to store. So the matrix is simply the quaternion to rotation matrix + translation. $\endgroup$ – MB Reynolds Jun 15 '17 at 20:05
  • $\begingroup$ Yes, that I know. But can you post how that rotation matrix really looks like? The rows and columns. $\endgroup$ – Tina J Jun 15 '17 at 21:25
2
$\begingroup$

Skimming the math. Starting with a quaternion $Q=w+\left(x,y,z\right)$ then we can rotate $\mathbf{v}$ by:

$$\mathbf{v}' = Q\mathbf{v}Q^{-1}$$

and if $Q$ is unit magnitude this reduces to:

$$\mathbf{v}' = Q\mathbf{v}Q^*$$

To create a matrix we need to apply the rotation to the basis set to form our three equations:

$$ \mathbf{x} = \left(1,0,0\right) \\ \mathbf{y} = \left(0,1,0\right) \\ \mathbf{z} = \left(0,0,1\right) \\ $$

which expanded and reduced gives:

$$ \mathbf{x}' = Q\mathbf{x}Q^{-1} = \left(1 - 2 \left(y^2+z^2\right), 2\left(xy+wz\right), 2\left(xz-wy\right) \right) \\ \mathbf{y}' = Q\mathbf{y}Q^{-1} = \left(2 \left(xy-wz\right), 1 - 2\left(x^2+z^2\right), 2\left(wx+yz\right) \right) \\ \mathbf{z}' = Q\mathbf{z}Q^{-1} = \left(2 \left(wy+xz\right), 2\left(yz-wx\right), 1 - 2\left(x^2+y^2\right) \right) $$

Sticking to the math convention of column vectors, then we shove the three equations into the first three columns and to add a translation by $\left(t_x,t_y,t_z\right)$ we shove that into the last column giving:

$$ { \left( \begin{array}{ccc} 1 - 2 \left(y^2+z^2\right) & 2\left(xy-wz\right) & 2\left(xz+wy\right) & t_x\\ 2 \left(xy+wz\right) & 1 - 2\left(x^2+z^2\right) & 2\left(wx-yz\right) & t_y \\ 2 \left(wy-xz\right) & 2\left(yz+wx\right) & 1 - 2\left(x^2+y^2\right) & t_z \\ 0 & 0 & 0 & 1 \end{array} \right) } $$

$\endgroup$
  • $\begingroup$ Thanks. What was Q*? $\endgroup$ – Tina J Jun 16 '17 at 14:34
  • 1
    $\begingroup$ It's the conjugate which is the same as inverse when unit magnitude. $\endgroup$ – MB Reynolds Jun 16 '17 at 16:12
  • 1
    $\begingroup$ There was a seemly deleted comment along the lines of "not useful to have the matrix". This is true, you work with the rotation translation pair..and when transforming points you don't actually create a matrix (don't need to at least). The matrix is to show the sub-expressions that need to be computed. And this is assuming that the average number of points is (somewhere) greater than one. Otherwise you're better off performing the rotation directly from the quaternion expression. $\endgroup$ – MB Reynolds Jun 17 '17 at 18:15
  • 1
    $\begingroup$ It's more efficient to convert to a matrix instead of using quaternion directly if you transform more than one vector with the same quaternion. Also for more compact representation you don't need to store w since you are dealing with unit quaternions: w=sqrt(1-x^2-y^2-z^2). $\endgroup$ – JarkkoL Jun 18 '17 at 4:25
  • 1
    $\begingroup$ I was attempting to say the matrix doesn't need to be stored..it's on the stack. Dropping the scalar doesn't work very well: marc-b-reynolds.github.io/quaternions/2017/05/02/… $\endgroup$ – MB Reynolds Jun 18 '17 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.