Robotics Kinematics and Dynamics/Serial Manipulator Position Kinematics

Forward position kinematics

The forward position kinematics problem can be stated as follows: given the different joint angles, what is the the position of the end-effector? With the previous sections in mind, the answer is rather simple: construct the different transformation matrices and combine them in the right way, the result being $\underset{ee}{\overset{}{bs}}T$, where $\{bs\}$ is the base frame of the robot manipulator.

Solution

Suppose the mutual orientation matrices between adjacent links are known. (As the fixed parameters of each link are known, and the joint angles are a given to the problem, these can be calculated.) The transformation that relates the last and first frames in a serial manipulator arm, and thus, the solution to the forward kinematics problem, is then given by (the axes are moving, and so the compound homogeneous transformation matrix is found by premultiplying the individual transformation matrices):

$\underset{ee}{\overset{}{bs}}T=$ $\underset{N}{\overset{}{0}}T=$ $\underset{1}{\overset{}{0}}T$ $\underset{2}{\overset{}{1}}T$ $\dots$ $\underset{N-1}{\overset{}{N-2}}T$ $\underset{N}{\overset{}{N-1}}T$

Examples

The Planar Three-link Manipulator

The equations below use 3 × 3 pose matrices, as this is just a 2-dimensional case.

FIGURE

The pose of the first link, relative to the reference frame, is given by (recall the elementary rotation about the z-axis from the previous section):

$T_1({\theta }_1)=\left(\begin{array}{ccc}\cos ({\theta }_1)& -\sin ({\theta }_1)& 0\\ \sin ({\theta }_1)& \cos ({\theta }_1)& 0\\ 0& 0& 1\end{array}\right)$

The pose of the second link, relative to the first link, is given by:

$T_2({\theta }_2)=\left(\begin{array}{ccc}\cos ({\theta }_2)& -\sin ({\theta }_2)& (1-\cos {\theta }_2)l_1\\ \sin ({\theta }_2)& \cos ({\theta }_2)& -\sin {\theta }_2{l}_1\\ 0& 0& 1\end{array}\right)$

This corresponds to a rotation by an angle ${\theta }_2$ and a translation by a distance $l_1$, where $l_1$ is the length of the first link.

The pose of the third link, relative to the second link, is given by:

$T_3({\theta }_3)=\left(\begin{array}{ccc}\cos ({\theta }_3)& -\sin ({\theta }_3)& (1-\cos {\theta }_3)(l_1+l_2)\\ \sin ({\theta }_3)& \cos ({\theta }_3)& -\sin {\theta }_3(l_1+l_2)\\ 0& 0& 1\end{array}\right)$

The solution to the forward kinematics problem is then:

$T_3({\theta }_3)=\left(\begin{array}{ccc}\cos ({\theta }_1+{\theta }_2+{\theta }_3)& -\sin ({\theta }_1+{\theta }_2+{\theta }_3)& a\\ \sin ({\theta }_1+{\theta }_2+{\theta }_3)& \cos ({\theta }_1+{\theta }_2+{\theta }_3)& b\\ 0& 0& 1\end{array}\right)$

Here, $a=l_1\cos ({\theta }_1)+l_2\cos ({\theta }_1+{\theta }_2)-(l_1+l_2)\cos ({\theta }_1+{\theta }_2+{\theta }_3)$, en $b=l_1\sin ({\theta }_1)+l_2\sin ({\theta }_1+{\theta }_2)-(l_1+l_2)\sin ({\theta }_1+{\theta }_2+{\theta }_3)$.

As the begin position (${\theta }_i=0$) of the end effector is given by $(l_1+l_2+l_3,0)$, the coordinates, after applying a certain value for ${\theta }_i$, are found through:

$\left(\begin{array}{c}x\\ y\\ 1\end{array}\right)=T\left(\begin{array}{c}{l}_1+l_2+l_3\\ 0\\ 1\end{array}\right)$

The resulting kinematic equations are:

$\begin{array}{c}x=l_1\cos {\theta }_1+l_2\cos ({\theta }_1+{\theta }_2)+l_3\cos ({\theta }_1+{\theta }_2+{\theta }_3)\\ y=l_1\sin {\theta }_1+l_2\sin ({\theta }_1+{\theta }_2)+l_3\sin ({\theta }_1+{\theta }_2+{\theta }_3)\end{array}$

Reverse position kinematics

The reverse kinematics problem is the opposite of the forward kinematics problem and can be summarized as follows: given the desired position of the end effector, what combinations of the joint angles can be used to achieve this position?

Two types of solutions can be considered: a closed-form solution and a numerical solution. Closed-form or analytical solutions are sets of equations that fully describe the connection between the end-effector position and the joint angles. Numerical solutions are found through the use of numerical algorithms, and can exist even when no closed-form solution is available. There may also be multiple solutions, or not a solution at all.