Hand–eye calibration problem

Given the equation AX=ZB, it is possible to decompose the equation into a purely rotational and translational part; methods utilizing this are referred to as separable methods. Where R_A represents a 3×3 rotation matrix and t_A a 3×1 translation vector, the equation can be broken into two parts:[4]

R_AR_X=R_ZR_B

R_At_X+t_A=R_Zt_B+t_Z

The second equation becomes linear if R_Z is known. As such, the most frequent approach is to solve for R_x and R_z using the first equation, then using R_z to solve for the variables in the second equation. Rotation is represented using quaternions, allowing for a linear solution to be found. While separable methods are useful, any error in the estimation for the rotation matrices is compounded when being applied to the translation vector.[5] Other solutions avoid this problem.

Simultaneous solutions are based on solving for both X and Z at the same time (rather than basing the solution of one part off of the other as in separable solutions), propagation of error is significantly reduced.[6] By formulating the matrices as dual quaternions, it is possible to get a linear equation by which X is solvable in a linear format.[5] An alternative way applies the least-squares method to the Kronecker product of the matrices A⊗B. As confirmed by experimental results, simultaneous solutions have less error than separable quaternion solutions.[6]

Iterative solutions are another method used to solve the problem of error propagation. One example of an iterative solution is a program based on minimizing ||AX−XB||. As the program iterates, it will converge on a solution to X independent to the initial robot orientation of R_B. Solutions can also be two-step iterative processes, and like simultaneous solutions can also decompose the equations into dual quaternions.[7] However, while iterative solutions to the problem are generally simultaneous and accurate, they can be computationally taxing to carry out and may not always converge on the optimal solution.[5]

The matrix equation AX=XB, where X is unknown, has an infinitive number of solutions that can be easily studied by a geometrical approach.[8] To find X it is necessary to consider a simultaneous set of 2 equations A₁X=XB₁ and A₂X=XB₂; the matrices A₁, A₂, B₁, B₂ have to be dermined by experiments to be performed in an optimized way. [9]

${\displaystyle {\begin{bmatrix}p_{b}\\1\end{bmatrix}}={\begin{bmatrix}R_{b}&T_{b}\\0&1\end{bmatrix}}\centerdot {\begin{bmatrix}R_{s}&T_{s}\\0&1\end{bmatrix}}\centerdot {\begin{bmatrix}p_{s}\\1\end{bmatrix}}}$

where ${\displaystyle p_{b}}$ represents the unknown coordinate of the point ${\displaystyle p}$ in the robot base system, ${\displaystyle R_{b},T_{b}}$ represent the known relationship between the robot base system and end-effector, ${\displaystyle R_{s},T_{s}}$ are the unknown relationship between the end-effector and the scanner, and ${\displaystyle p_{s}}$ is the known coordinate of the point ${\displaystyle p}$ in the local scanner system. Methods are as follows,

• Straight edges

There is a method using straight edges for hand-eye calibration.[10]