# Can inverse dynamics control be regarded as a function?

I know that inverse kinematics ($p \rightarrow q$, p: desired pose of the end-effector, q: joint angles) is not a function because there might be multiple joint angle vectors q that result in the same pose p.

By inverse dynamics control I mean the mapping $(q, \dot{q}, \ddot{q}) \rightarrow u$ (u: required torques. I am not very experienced with these kind of problems. Is the mapping a function, i.e. for each triple $(q, \dot{q}, \ddot{q})$ there is a unique solution u? My intuition says it is. But I am not sure. If there is not, would it always be possible to obtain a solution by averaging two or more solutions?

• Is this inverse dynamics? i don't think so. mapping joint positions, velocities, and accelerations to the required joint torques is low level control. – Ben Feb 24 '14 at 1:48
• At least in this paper it seems like it is: robot-learning.de/pmwiki/uploads/Publications/… – alfa Feb 24 '14 at 7:15

• Ben, if you read carefully the Sciavicco's book you'll find out how the inverse dynamics is not concerned about what you said. In general, indeed, inverse dynamics problem is regarded with the solution of the dynamic equation of the manipulator $M(q)\ddot{q}+h(q,\dot{q})=u_c$ with the aim of seeking the torque $u_c$ that drives the system to make $(\ddot{q},\dot{q},q)$ follow some reference $(\ddot{q}_r,\dot{q}_r,q_r)$. In case you have non-null end-effector force $h_e$ you have to consider it in the equation. – Ugo Pattacini Dec 26 '14 at 19:26