Advanced 10 min 3 of 3

Dynamics: why control is harder than kinematics

Inertia depends on configuration, and the coupling terms mean no joint moves alone.

What this repairs

Assuming each joint can be controlled as an independent second-order system with a fixed inertia.

Assumes

The Jacobian

The equation of motion for any rigid manipulator is

$M(\mathbf{q})\,\ddot{\mathbf{q}} + C(\mathbf{q}, \dot{\mathbf{q}})\,\dot{\mathbf{q}} + \mathbf{g}(\mathbf{q}) = \boldsymbol{\tau}.$

Four terms, and three of them depend on where the arm is:

$M(\mathbf{q})$ — the inertia matrix. Symmetric, positive definite, and a function of configuration. An extended arm has far more inertia about the base than a folded one.
$C(\mathbf{q}, \dot{\mathbf{q}})$ — Coriolis and centrifugal terms, quadratic in velocity. These are what make a fast arm behave unlike a slow one.
$\mathbf{g}(\mathbf{q})$ — gravity, which is a configuration-dependent torque and usually the largest single term at low speed.
$\boldsymbol{\tau}$ — what the motors supply.

Insight The off-diagonal entries of M are the real problem

$M$ is not diagonal. Accelerating joint 1 exerts a torque on joint 2 through the mechanical coupling, and the size of that coupling depends on the elbow angle. So an independent PID per joint is controlling a system that is not independent, and it copes by treating the coupling as a disturbance to be rejected — which works at low speed and degrades exactly as the arm gets faster, i.e. as the term it is ignoring grows quadratically.

Computed-torque control confronts it: measure $\mathbf{q}, \dot{\mathbf{q}}$ , evaluate the model, and command $\boldsymbol{\tau} = M(\mathbf{q})\mathbf{a}_{\text{desired}} + C\dot{\mathbf{q}} + \mathbf{g}$ . The nonlinearities cancel and what remains is a linear system per joint. It works to the extent your model is right, which is why parameter identification is a whole subfield.

Inertia of the two-link arm

$M_{11} = m_1 l_1^2 + m_2\left(l_1^2 + 2 l_1 l_2 \cos q_2 + l_2^2\right)$

The effective inertia at joint 1. It depends on q₂ and on nothing else — the elbow angle changes how far the outer mass sits from the shoulder axis.
$M_{12} = M_{21} = m_2\left(l_1 l_2 \cos q_2 + l_2^2\right)$

The coupling. Maximal with the arm straight (cos q₂ = 1), zero at q₂ = ±90°, negative when folded back.
$M_{22} = m_2 l_2^2$

Joint 2 sees a constant inertia: the outer link’s geometry relative to its own axis never changes. The asymmetry between M₁₁ and M₂₂ is why the two joints of the same arm need different gains.

Check yourself

A tuned arm controller oscillates when the arm is extended but is stable when folded. Why?

Show answer

Because the effective inertia at the shoulder changed — $M_{11}$ is largest when $\cos q_2 = 1$ , i.e. extended. Gains tuned for the folded configuration are, at full extension, being applied to a plant with several times the inertia, which lowers the effective damping ratio and produces overshoot and ringing. The principled fixes are gain scheduling on configuration or computed-torque control; the usual field fix is detuning for the worst case, which gives up performance everywhere else.