Manipulator Equations of Motion

Manipulator Equations of Motion

For any mechanical system whose kinetic energy is a quadratic form in joint velocities, $T=\frac{1}{2}{\dot{q}}^{\top }M(q)\dot{q}$ — which holds for every robot built from rigid links — the Euler–Lagrange equations collapse into the standard manipulator form:
${\rm Y}=\underset{\text{inertia}}{\underbrace{M(q)\ddot{q}}}+\underset{\text{velocity terms}}{\underbrace{C(q,\dot{q})\dot{q}}}+\underset{\text{gravity \& springs}}{\underbrace{G(q)}},$
where $q\in {\mathbb{R}}^n$ is the joint vector, ${\rm Y}\in {\mathbb{R}}^n$ is the joint torque/force vector, and the three coefficient objects are:

• $M(q)\in {\mathbb{R}}^{n\times n}$ — the system inertia matrix, symmetric and positive definite. It captures the aggregate mass and rotational inertia of all links and depends on configuration $q$ but not velocity. For a serial chain with body Jacobians $J_i^b(q)$,
  $M(q)={\sum }_{i=1}^n{J}_i^b(q{)}^{\top }{\mathcal{M}}_i{J}_i^b(q),$
  where ${\mathcal{M}}_i$ is the $6\times 6$ generalized inertia of link $i$ from 6.2. Kinetic Energy of a Rigid Body.
• $C(q,\dot{q})\in {\mathbb{R}}^{n\times n}$ — the Coriolis matrix, capturing all velocity-dependent fictitious forces. Diagonal contributions (${\dot{q}}_i^2$) are centrifugal; off-diagonal cross terms (${\dot{q}}_i{\dot{q}}_j$, $i\ne j$) are Coriolis proper. The standard formula uses Christoffel symbols of the first kind:
  $C_{ij}(q,\dot{q})={\sum }_{k=1}^n\frac{1}{2}\left(\frac{\partial M_{ij}}{\partial q_k}+\frac{\partial M_{ik}}{\partial q_j}-\frac{\partial M_{jk}}{\partial q_i}\right){\dot{q}}_k.$
• $G(q)\in {\mathbb{R}}^n$ — the gravity vector, derived purely from potential energy: $G(q)=\partial V/\partial q$. It captures gravity and conservative spring forces and depends on $q$ but not $\dot{q}$.

The standard form is the workhorse of robot dynamics and control: it cleanly separates the three physical contributions, and each piece has a different role in computed-torque control, gravity compensation, and trajectory planning.

$M(q)$ is not the same as ${\mathcal{M}}_i$ $M(q)\in {\mathbb{R}}^{n\times n}$ is the system inertia matrix in joint space, aggregating all links via Jacobians. ${\mathcal{M}}_i\in {\mathbb{R}}^{6\times 6}$ is the generalized inertia of a single link in body-twist space. They have different sizes, units, and physical meanings — confusing them is a major source of derivation errors.

Intuition

Every Lagrangian-derived equation of motion for a manipulator factors into the same three pieces: how hard it is to accelerate the joints (the inertia term), what velocity-induced forces the system experiences as it moves (the Coriolis/centrifugal term), and what static torque is needed just to hold a configuration against gravity and springs (the gravity term). Reading ${\rm Y}=M\ddot{q}+C\dot{q}+G$ left to right is reading off these three contributions one at a time.

The factorization is also what makes the equations useful for control. Computed-torque control commands
${{\rm Y}}_{\text{cmd}}=M(q){\ddot{q}}_{\text{des}}+C(q,\dot{q})\dot{q}+G(q),$
which, substituted into the dynamics, gives $\ddot{q}={\ddot{q}}_{\text{des}}$ — perfect tracking, modulo modeling error. None of this works without isolating $M$, $C$, and $G$ explicitly.

The Coriolis term being velocity-dependent is the part students most often miss. Calling it “fictitious” is misleading: $C(q,\dot{q})\dot{q}$ is mandatory, not optional. It is the dynamical signature of describing a multi-body rotating mechanism in non-inertial generalized coordinates, and dropping it produces a dynamically incorrect model that fails at any non-trivial joint speed.

Properties

System Inertia Matrix $M(q)$

(Symmetric Positive-Definite) $M(q)=M(q{)}^{\top }\succ 0$ for any physical configuration.

Symmetry follows from $M(q)={\sum }_i{J}_i^{b\top }{\mathcal{M}}_i{J}_i^b$ and the symmetry of each ${\mathcal{M}}_i$. Positive definiteness follows from the kinetic energy $T=\frac{1}{2}{\dot{q}}^{\top }M\dot{q}$ being strictly positive whenever $\dot{q}\ne 0$ for a body of positive mass. Consequence: $M$ is always invertible, so the forward-dynamics map $\ddot{q}=M^{-1}({\rm Y}-C\dot{q}-G)$ is well-defined.

(Configuration Dependence Only) $M(q)$ depends on $q$ alone, not on $\dot{q}$.

The $\dot{q}$-dependence of $T$ is entirely in the explicit ${\dot{q}}^{\top }M(q)\dot{q}$ factor, so $M$ inherits no velocity dependence. This is what makes the standard form linear in $\ddot{q}$.

(Read Off From Kinetic Energy) $M_{ij}={\partial }^{2}T/\partial {\dot{q}}_i\partial {\dot{q}}_j$.

After computing $T$ in the Lagrangian procedure, extract $M$ directly by inspection — no need to invoke Euler–Lagrange first. Coefficients of ${\dot{q}}_i^2$ give diagonal entries (with a factor of 2); coefficients of ${\dot{q}}_i{\dot{q}}_j$ ($i\ne j$) give off-diagonal entries.

Coriolis Matrix $C(q,\dot{q})$

(Christoffel Construction) Once $M(q)$ is in hand, $C$ is built directly from partial derivatives of $M$ via

$C_{ij}(q,\dot{q})={\sum }_{k=1}^n\frac{1}{2}\left({\partial }_{q_k}{M}_{ij}+{\partial }_{q_j}{M}_{ik}-{\partial }_{q_i}{M}_{jk}\right){\dot{q}}_k.$
No return to the full Euler–Lagrange machinery is required. This particular construction (out of several valid choices) yields the skew-symmetry property below.

(Skew-Symmetry of $\dot{M}-2C$) With Christoffel-defined $C$, the matrix $N(q,\dot{q}):=\dot{M}(q)-2C(q,\dot{q})$ is skew-symmetric, i.e. $N+N^{\top }=0$.

Equivalently, ${\dot{q}}^{\top }N\dot{q}=0$ for every $\dot{q}$. This is the dynamical statement of energy conservation when no torque is applied: it forces $\frac{d}{dt}(\frac{1}{2}{\dot{q}}^{\top }M\dot{q})={\dot{q}}^{\top }({\rm Y}-G)$ — power in equals power out. Lyapunov-based controllers (PD-plus-gravity, passivity-based) rely on this property.

(Centrifugal vs. Coriolis Terms) Inside $C(q,\dot{q})\dot{q}$, ${\dot{q}}_i^2$ contributions are centrifugal; ${\dot{q}}_i{\dot{q}}_j$ ($i\ne j$) contributions are Coriolis proper.

A single-joint arm spinning about a fixed axis has only centrifugal terms. Multi-joint arms with simultaneous joint motion produce mixed Coriolis cross terms. Both arise mechanically from the same Christoffel construction.

(Quadratic Velocity Scaling) $\Vert C(q,\dot{q})\dot{q}\Vert$ scales as $\Vert \dot{q}{\Vert }^2$.

At low speeds, the Coriolis term is quadratic-small and can sometimes be ignored for back-of-envelope analysis. At design speeds it cannot be ignored — this is why high-speed manipulators need careful Coriolis modeling and slow assembly arms can sometimes get away without it.

(Constant $M$ ⇒ $C=0$) If $M(q)$ does not depend on $q$, the Christoffel formula gives $C\equiv 0$.

For Cartesian (prismatic-only) mechanisms moving in a horizontal plane, $M$ is a constant matrix, $C\equiv 0$, and $G\equiv 0$ — the dynamics reduce to ${\rm Y}=M\ddot{q}$. This is the simplest possible manipulator dynamics.

Gravity Vector $G(q)$

(Gradient of Potential) $G(q)=\partial V/\partial q$, where $V$ sums gravitational $m_{i}g{h}_i$ over links and elastic $\frac{1}{2}{k}_j{x}_j^2$ over springs.

No derivatives of $T$ are needed to compute $G$. Unlike $C$, the gravity term is linear in nothing — it is purely configuration-dependent.

(Static Hold Torque) Setting $\dot{q}=\ddot{q}=0$ gives ${\rm Y}=G(q)$ — the static torque to hold the configuration against gravity.

Plotting $G(q)$ over the workspace gives the static “gravity load” map; this is the first thing to check when sizing motors or designing gravity-compensation control.

Exam & Homework Hacks

(Term-Size Triage) The three terms scale very differently with $\dot{q}$:

${\rm Y}\approx G(q)$ at rest; ${\rm Y}\approx M(q)\ddot{q}+G(q)$ at low speed; all three matter at high speed. Use this hierarchy to decide which terms to compute first.

(Dimensions Cheat Sheet) For a system with $n$ DoF: $M$ is $n\times n$, $C$ is $n\times n$, $G$ is $n\times 1$, ${\rm Y}$ is $n\times 1$, $\dot{q}$ and $\ddot{q}$ are $n\times 1$.

Quick dimension checks catch most algebra errors. If $C\dot{q}$ ends up the wrong size, the most common bug is a misindexed Christoffel sum.

(Sign Check on $G$ at Extremes) For a planar arm at $\theta =0$ (horizontal) gravity load is maximal; at $\theta =\pi /2$ (vertical) it should be zero or sign-flipped.

A quick limiting-case check on $G$ catches sign errors in the height function inside $V$.

(Verify Skew-Symmetry of $\dot{M}-2C$) Compute $\dot{M}-2C$ symbolically and check the diagonal is zero and off-diagonals satisfy $a_{ij}=-a_{ji}$.

This is a cheap, powerful sanity check on the Christoffel computation — and an exam favorite.

(Positive-Definiteness Check on $M$) For an $n\times n$ symmetric $M$, verify all leading principal minors are strictly positive.

For a 2-link arm $M=\left[\begin{array}{cc}a+2b\cos {\theta }_2& b\cos {\theta }_2+d\\ b\cos {\theta }_2+d& d\end{array}\right]$, both diagonal entries and $\det M$ should be strictly positive over the workspace.

(Recognize the Forms) $T$ that contains ${\dot{\theta }}_i{\dot{\theta }}_j$ cross products ⇒ $M$ has off-diagonal entries ⇒ Coriolis cross terms exist.

$T$ that contains explicit $\cos {\theta }_j$ or $\sin {\theta }_j$ multipliers ⇒ $M$ depends on $q$ ⇒ Christoffel derivatives are nonzero ⇒ $C\ne 0$.

Derivation

Standard form from $L=\frac{1}{2}{\dot{q}}^{\top }M(q)\dot{q}-V(q)$

For a system of rigid bodies, $T(q,\dot{q})=\frac{1}{2}{\dot{q}}^{\top }M(q)\dot{q}$, so $L=\frac{1}{2}{\dot{q}}^{\top }M(q)\dot{q}-V(q)$. Compute the two derivatives that enter Euler–Lagrange. First,
$\frac{\partial L}{\partial {\dot{q}}_i}={\sum }_j{M}_{ij}(q){\dot{q}}_j,\text{so}\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_i}\right)={\sum }_j{M}_{ij}{\ddot{q}}_j+{\sum }_{j,k}\frac{\partial M_{ij}}{\partial q_k}{\dot{q}}_k{\dot{q}}_j.$
Second,
$\frac{\partial L}{\partial q_i}=\frac{1}{2}{\sum }_{j,k}\frac{\partial M_{jk}}{\partial q_i}{\dot{q}}_j{\dot{q}}_k-\frac{\partial V}{\partial q_i}.$
Subtracting and identifying $G_i=\partial V/\partial q_i$,
${{\rm Y}}_i={\sum }_j{M}_{ij}{\ddot{q}}_j+{\sum }_{j,k}\left(\frac{\partial M_{ij}}{\partial q_k}-\frac{1}{2}\frac{\partial M_{jk}}{\partial q_i}\right){\dot{q}}_j{\dot{q}}_k+G_i.$
Symmetrizing the velocity term in $j\leftrightarrow k$ produces ${{\rm Y}}_i={\sum }_j{M}_{ij}{\ddot{q}}_j+{\sum }_j{C}_{ij}{\dot{q}}_j+G_i$, the row form of ${\rm Y}=M\ddot{q}+C\dot{q}+G$. The Christoffel formula for $C_{ij}$ falls out of the symmetrization step.

Examples

Two-link planar arm — extracting $M$, $C$, $G$

Continuing the two-link arm example, the kinetic energy after collecting terms is
$T=\frac{1}{2}[\alpha +2\beta \cos {\theta }_2]{\dot{\theta }}_1^2+\frac{1}{2}\delta {\dot{\theta }}_2^2+[\delta +\beta \cos {\theta }_2]{\dot{\theta }}_1{\dot{\theta }}_2,$
with $\alpha =(m_1+m_2){\ell }_1^2+m_2{\ell }_2^2$, $\beta =m_2{\ell }_1{\ell }_2$, $\delta =m_2{\ell }_2^2$.

System inertia matrix $M(q)$ Read off $M_{ij}={\partial }^{2}T/\partial {\dot{\theta }}_i\partial {\dot{\theta }}_j$: $M(q)=\left[\begin{array}{cc}\alpha +2\beta \cos {\theta }_2& \delta +\beta \cos {\theta }_2\\ \delta +\beta \cos {\theta }_2& \delta \end{array}\right].$

Gravity vector $G(q)$ From $V=(m_1+m_2)g{\ell }_1\sin {\theta }_1+m_{2}g{\ell }_2\sin ({\theta }_1+{\theta }_2)$: $G(q)=\left[\begin{array}{c}(m_1+m_2)g{\ell }_1\cos {\theta }_1+m_{2}g{\ell }_2\cos ({\theta }_1+{\theta }_2)\\ m_{2}g{\ell }_2\cos ({\theta }_1+{\theta }_2)\end{array}\right].$

Coriolis matrix $C(q,\dot{q})$ via Christoffel Only $\partial M/\partial {\theta }_2$ is nonzero ($\partial M_{11}/\partial {\theta }_2=-2\beta \sin {\theta }_2$, $\partial M_{12}/\partial {\theta }_2=-\beta \sin {\theta }_2$). Plugging into the Christoffel formula: $C(q,\dot{q})=\left[\begin{array}{cc}-\beta \sin {\theta }_2{\dot{\theta }}_2& -\beta \sin {\theta }_2({\dot{\theta }}_1+{\dot{\theta }}_2)\\ \beta \sin {\theta }_2{\dot{\theta }}_1& 0\end{array}\right].$

The Coriolis term $C\dot{q}$ contains both centrifugal ($-\beta \sin {\theta }_2{\dot{\theta }}_2^2$) and Coriolis-proper ($-2\beta \sin {\theta }_2{\dot{\theta }}_1{\dot{\theta }}_2$) contributions. Verifying $\dot{M}-2C$ is skew-symmetric is a useful check on the algebra.

Static gravity compensation

A 2-link arm at rest at $({\theta }_1,{\theta }_2)=(\pi /2,0)$ — first link straight up, second link extending the same direction. With $\dot{q}=\ddot{q}=0$ the standard form collapses to ${\rm Y}=G(q)$:
$G(\pi /2,0)=\left[\begin{array}{c}(m_1+m_2)g{\ell }_1\cos (\pi /2)+m_{2}g{\ell }_2\cos (\pi /2)\\ m_{2}g{\ell }_2\cos (\pi /2)\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right].$
Both joint torques are zero — the arm is in a vertical, balanced configuration where gravity contributes no moment about either joint (modulo stability, a separate question).

Cartesian gantry — degenerate case

A two-axis Cartesian gantry has prismatic joints $q=(x,y)$ moving frictionlessly in a horizontal plane. Let $m_x$ be the total mass moved by the $x$-axis and $m_y$ the mass riding on the $x$ stage. Then
$T=\frac{1}{2}(m_x+m_y){\dot{x}}^2+\frac{1}{2}{m}_y{\dot{y}}^2,$
$M=\left[\begin{array}{cc}{m}_x+m_y& 0\\ 0& m_y\end{array}\right],C\equiv 0,G\equiv 0.$
$M$ is constant (no $q$-dependence) and gravity is absent for purely horizontal motion, so the dynamics reduce to the decoupled linear system $F_x=(m_x+m_y)\ddot{x}$, $F_y=m_y\ddot{y}$. The simplest manipulator dynamics possible — and the textbook illustration of why Coriolis effects vanish whenever inertia is configuration-independent.