# Position and orientation of local coordinate system relative to another local coordinate system (both described in world coord. system)

This question might have been answered before, but I couldn't find one that will best describe my problem, or before I got confused.

The following image depicts three coordinate systems: a world coordinate system, W, and two local coordinate systems A and B described in the world coordinate system in terms of their position, T, and orientation in quaternion format, Q (all relative to W):

Coordinate systems in 3D space

• $$T_A$$ - position of the origin of coordinate system A, in the world coordinate system, W.
• $$T_B$$ - position of the origin of coordinate system B, in the world coordinate system, W.
• $$Q_A$$ - orientation of coordinate system A, in (with reference to) the world coordinate system, W (in quaternions).
• $$Q_B$$ - orientation of coordinate system B, in the world coordinate system, W (in quaternions).
• $$t_{BA}$$ - position of the origin of coordinate system B, inside (with reference to) coordinate system A.
• $$q_{BA}$$ - orientation of coordinate system B, described in terms of (with reference to) coordinate system A.

I would like to get $$t_{BA}$$ and $$q_{BA}$$ (with reference to coordinate system A)? Or basically, the position and orientation of coordinate system B from an observer's perspective standing at (and inside) coordinate system A.

Any help would be appreciated!

This is how you could find out that $$t_{BA} = T_B - T_A$$ and $$q_{BA}=(Q_A)^{-1}Q_B$$:

Let $$\mathbf{x_P^W}= (x^W_P,y^W_P,z^W_P)$$ be the coordinates $$(x^W_P,y^W_P,z^W_P)$$ or the position $$\mathbf{x_P^W}$$ of any point $$P$$ measured along the $$x$$-,$$y$$- and $$z$$-axis of the world coordinate system $$W$$. If the $$x$$-,$$y$$- and $$z$$- coordinates or the positions of the origins $$0_A$$ and $$0_B$$ of the coordinate systems $$A$$ and $$B$$ with respect to the world coordinate system $$W$$ are $$\mathbf{x_{0_A}^W} = (x_{0_A}^W,y_{0_A}^W,z_{0_A}^W)$$ and $$\mathbf{x_{0_B}^W} = (x_{0_B}^W,y_{0_B}^W,z_{0_B}^W)$$, then the position of the arbitrary point $$P$$ with respect to $$A$$ and $$B$$ can be calculated as (just vector addition in $$A$$ or in $$B$$)

$$\mathbf{x_P^A}= -\mathbf{x_{0_A}^W}+\mathbf{x_P^W}=(x^W_P-x_{0_A}^W,y^W_P-y_{0_A}^W,z^W_P-z_{0_A}^W)$$

$$\mathbf{x_P^B}= -\mathbf{x_{0_B}^W}+\mathbf{x_P^W}=(x^W_P-x_{0_B}^W,y^W_P-y_{0_B}^W,z^W_P-z_{0_B}^W)$$

Now fix this point $$P$$ and make it equal the origin $$0_B$$ of the coordinate system $$B$$, let $$\mathbf{x_P^W}=\mathbf{x_{0_B}^W}=(x_{0_B}^W,y_{0_B}^W,z_{0_B}^W)$$, then the equations above will become

$$\mathbf{x_{0_B}^A}= -\mathbf{x_{0_A}^W}+\mathbf{x_{0_B}^W}=(x_{0_B}^W-x_{0_A}^W,y_{0_B}^W-y_{0_A}^W,z_{0_B}^W-z_{0_A}^W)$$

$$\mathbf{x_{0_B}^B}= -\mathbf{x_{0_B}^W}+\mathbf{x_{0_B}^W}=(x_{0_B}^W-x_{0_B}^W,y_{0_B}^W-y_{0_B}^W,z_{0_B}^W-z_{0_B}^W)$$

The first equation is saying: the $$x$$-,$$y$$- and $$z$$-coordinates of the origin $$0_B$$ of the coordinate system $$B$$ measured along the $$x$$-,$$y$$- and $$z$$-axis of the coordinate system $$A$$ are $$(x_{0_B}^W-x_{0_A}^W,y_{0_B}^W-y_{0_A}^W,z_{0_B}^W-z_{0_A}^W)$$

The second equation is saying: the $$x$$-,$$y$$- and $$z$$-coordinates of the origin $$0_B$$ of the coordinate system $$B$$ measured along the $$x$$-,$$y$$- and $$z$$-axis of the coordinate system $$B$$ are $$(0,0,0)$$

Look here and here and here and here for some information about orientation. I just found this powerpoint presentation part 1 of 2 and powerpoint presentation part 2 of 2 (see page 19) which is stating that:

If $$Q_A^W$$ is quaternion for orientation of reference frame $$A$$ relative to some base reference frame $$W$$ and $$Q_B^W$$ is quaternion for orientation of another reference frame $$B$$ relative to that same base reference frame $$W$$, then $$Q^{A}_{B}=(Q_A)^{-1}Q_B$$ is a quaternion for orientation of reference frame $$B$$ relative to reference frame $$A$$ where $$Q_A^{-1}$$ is inverse of $$Q_A$$ (if quaternions are unit quaternions, then $$Q^{-1}=Q^{*}$$ (conjugate)).

• Thank you very much for all the effort and extra reading material you've put into the answer. I appreciate it! Those powerpoints on quaternions are great. – Beard Jul 25 '19 at 20:58