From Lesson 1, you have seen broad families such as manipulators and mobile robots. This lesson refines that view by asking a simpler but structural question: is the robot’s base inertially fixed or dynamically moving?

A stationary-base system is attached to ground or a rigid support. Its base coordinate is constant, and only internal joints move. Typical examples: industrial arms, gantry robots, many parallel mechanisms.

A moving-base system has a base that translates and/or rotates. The base is part of the robot’s configuration and dynamics. Examples: wheeled robots, legged robots, aerial robots, mobile manipulators, humanoids.

flowchart TD
  S["Start: given a robot"] --> Q1["Is base rigidly fixed to environment?"]
  Q1 -->|yes| FB["Stationary-base system"]
  Q1 -->|no| MB["Moving-base system"]
  FB --> FB1["Internal joints define motion"]
  MB --> MB1["Base + internal joints define motion"]
  MB1 --> MB2["Dynamics are coupled"]
        

The distinction is not only geometric but energetic: in moving-base systems, momentum exchange between base and joints is unavoidable. We now formalize this.

Let the robot configuration be represented by generalized coordinates \( \mathbf{q} \in \mathbb{R}^n \). For a stationary-base manipulator, \( \mathbf{q} \) contains only joint coordinates (revolute/prismatic variables).

For a moving-base system we partition coordinates into base and internal parts: \( \mathbf{x}_b \in \mathbb{R}^m \) (base pose parameters) and \( \mathbf{q} \in \mathbb{R}^n \) (joint variables), forming the full configuration \( \mathbf{z} = \begin{bmatrix}\mathbf{x}_b \\ \mathbf{q}\end{bmatrix} \in \mathbb{R}^{m+n} \).

flowchart LR
  Z["z = [xb; q]"] --> XB["xb: base coordinates (m DOF)"]
  Z --> Q["q: joint coordinates (n DOF)"]
  XB --> EX1["translation/rotation of base"]
  Q --> EX2["internal joint motion"]
        

The configuration space is thus \( \mathcal{C}_{fixed} \subseteq \mathbb{R}^n \) for stationary-base systems and \( \mathcal{C}_{moving} \subseteq \mathbb{R}^{m+n} \) for moving-base systems. Even before dynamics, this dimensional increase changes planning and control complexity.

Because you are familiar with linear control, we write standard nonlinear robot dynamics in a control-affine form without assuming any kinematic details yet. For a stationary-base robot with generalized coordinates \( \mathbf{q} \), the Lagrangian is \( \mathcal{L}(\mathbf{q}, \dot{\mathbf{q}}) = T(\mathbf{q}, \dot{\mathbf{q}}) - V(\mathbf{q}) \), where \(T\) is kinetic and \(V\) potential energy.

Euler–Lagrange equations yield:

\[ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\mathbf{q}}}\right) - \frac{\partial \mathcal{L}}{\partial \mathbf{q}} = \boldsymbol{\tau}. \]

Under standard assumptions (rigid links, smooth energies), this can be written:

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

Key structural properties (proved in standard robotics texts, but we sketch them):

• Symmetry and positive definiteness of mass matrix: \( \mathbf{M}(\mathbf{q}) = \mathbf{M}(\mathbf{q})^\top \) and \( \mathbf{v}^\top \mathbf{M}(\mathbf{q}) \mathbf{v} > 0 \) for any nonzero \( \mathbf{v} \).
• Energy balance: in the absence of actuation and dissipation, total mechanical energy is conserved.

Proof sketch (energy conservation): Let the total energy be \( E(\mathbf{q},\dot{\mathbf{q}}) = T(\mathbf{q},\dot{\mathbf{q}}) + V(\mathbf{q}) \). Multiply the dynamics by \( \dot{\mathbf{q}}^\top \):

\[ \dot{\mathbf{q}}^\top \mathbf{M}(\mathbf{q}) \ddot{\mathbf{q}} + \dot{\mathbf{q}}^\top \mathbf{C}(\mathbf{q},\dot{\mathbf{q}})\dot{\mathbf{q}} + \dot{\mathbf{q}}^\top \mathbf{g}(\mathbf{q}) = \dot{\mathbf{q}}^\top \boldsymbol{\tau}. \]

For a stationary-base robot, \( \dot{\mathbf{q}}^\top \mathbf{M} \ddot{\mathbf{q}} = \frac{d}{dt}T \) and \( \dot{\mathbf{q}}^\top \mathbf{g} = \frac{d}{dt}V \). Moreover, \( \dot{\mathbf{q}}^\top \mathbf{C} \dot{\mathbf{q}} = 0 \) (a standard skew-symmetry identity). Hence: \( \frac{d}{dt}(T+V) = \dot{\mathbf{q}}^\top \boldsymbol{\tau} \). If \( \boldsymbol{\tau}=\mathbf{0} \), then \( \frac{dE}{dt}=0 \), proving conservation.

For a moving-base robot, generalized coordinates are \( \mathbf{z}=\begin{bmatrix}\mathbf{x}_b \\ \mathbf{q}\end{bmatrix} \). The Lagrangian becomes \( \mathcal{L}(\mathbf{z},\dot{\mathbf{z}})=T(\mathbf{z},\dot{\mathbf{z}})-V(\mathbf{z}) \), and Euler–Lagrange gives:

\[ \mathbf{M}(\mathbf{z}) \ddot{\mathbf{z}} + \mathbf{C}(\mathbf{z},\dot{\mathbf{z}})\dot{\mathbf{z}} + \mathbf{g}(\mathbf{z}) = \mathbf{u}. \]

Partitioning reveals coupling:

\[ \begin{bmatrix} \mathbf{M}_{bb}(\mathbf{z}) & \mathbf{M}_{bq}(\mathbf{z}) \\ \mathbf{M}_{qb}(\mathbf{z}) & \mathbf{M}_{qq}(\mathbf{z}) \end{bmatrix} \begin{bmatrix} \ddot{\mathbf{x}}_b \\ \ddot{\mathbf{q}} \end{bmatrix} + \begin{bmatrix} \mathbf{h}_b(\mathbf{z},\dot{\mathbf{z}}) \\ \mathbf{h}_q(\mathbf{z},\dot{\mathbf{z}}) \end{bmatrix} = \begin{bmatrix} \mathbf{u}_b \\ \boldsymbol{\tau} \end{bmatrix}. \]

Here \( \mathbf{M}_{bq} \) and \( \mathbf{M}_{qb} \) are cross-inertia terms. If they are nonzero, joint accelerations produce base reactions, and base motion produces apparent forces in joint space.

Implication 1 (momentum exchange): If the base is unactuated (\( \mathbf{u}_b=\mathbf{0} \)), the system is often underactuated: joint torques \( \boldsymbol{\tau} \) must indirectly control some base DOF via coupling.

Implication 2 (energy structure persists): The same energy proof from Section 3 holds in full coordinates, giving: \( \frac{d}{dt}(T+V)=\dot{\mathbf{x}}_b^\top \mathbf{u}_b + \dot{\mathbf{q}}^\top \boldsymbol{\tau} \). Thus mobility does not break mechanical passivity; it expands the power ports.

Optional constraint view: Many mobile bases have velocity constraints \( \mathbf{A}(\mathbf{x}_b)\dot{\mathbf{x}}_b=\mathbf{0} \). These reduce feasible velocities, but still keep base coordinates dynamic. (We will treat such constraints carefully in later courses.)

The taxonomic split changes what designers and control engineers care about:

\[ \text{DOF gain}:\quad n \;\;\to\;\; n+m \quad\Rightarrow\quad \dim(\mathcal{C}) \text{ increases}. \]

• Stationary-base: actuation directly addresses all joint DOF. Control can often be organized joint-by-joint with only internal coupling.
• Moving-base: base mobility introduces additional DOF and cross-coupling, producing:
  • Reaction forces (joints move base).
  • Apparent forces (base motion disturbs joints).
  • Underactuation if some base DOF lack direct actuators.

A simple linear-control interpretation: around an equilibrium, the stationary-base dynamics linearize to \( \delta\ddot{\mathbf{q}} = \mathbf{A}\delta\mathbf{q} + \mathbf{B}\delta\boldsymbol{\tau} \), whereas moving-base systems yield a larger coupled linearization in \( \delta\mathbf{z} \) with potentially fewer inputs than states— a key difference for stabilizability.

We compare two toy systems:

1. Fixed-base 1-DOF arm: a single rotary joint with inertia \( I \).
2. Moving-base arm on a cart: cart position \( x \) plus the same joint \( q \).

The fixed-base dynamics are:

\[ I \ddot{q} + b\dot{q} + k q = \tau. \]

The moving-base coupled dynamics (simplified) are:

\[ \begin{bmatrix} M + m & m\ell\cos q \\ m\ell\cos q & I + m\ell^2 \end{bmatrix} \begin{bmatrix} \ddot{x} \\ \ddot{q} \end{bmatrix} + \begin{bmatrix} -m\ell\sin q \, \dot{q}^2 \\ m g \ell \sin q \end{bmatrix} = \begin{bmatrix} u_b \\ \tau \end{bmatrix}. \]

Note the cross term \( m\ell\cos q \) that couples cart and arm. If \( u_b=0 \), cart motion is indirectly driven by \( \tau \).

6.1 Python (NumPy/SciPy)

import numpy as np
from scipy.integrate import solve_ivp

# Parameters
I = 0.2      # arm inertia
b = 0.05     # viscous friction
k = 1.0      # linear stiffness (small-angle approx)

M = 1.0      # cart mass
m = 0.5      # arm mass lumped
ell = 0.6    # COM distance
g = 9.81

# PD controller for q
qd, qdd = 0.0, 0.0
Kp, Kd = 15.0, 3.0

def fixed_base(t, s):
    q, qdot = s
    tau = -Kp*(q-qd) - Kd*(qdot-qdd)
    qddot = (tau - b*qdot - k*q)/I
    return [qdot, qddot]

def moving_base(t, s):
    x, xdot, q, qdot = s
    tau = -Kp*(q-qd) - Kd*qdot
    ub  = 0.0  # unactuated base to show coupling

    # Mass matrix
    M11 = M + m
    M12 = m*ell*np.cos(q)
    M22 = I + m*ell**2
    Mass = np.array([[M11, M12],
                     [M12, M22]])

    # Bias terms
    h1 = -m*ell*np.sin(q)*qdot**2
    h2 =  m*g*ell*np.sin(q)

    acc = np.linalg.solve(Mass, np.array([ub - h1, tau - h2]))
    xddot, qddot = acc
    return [xdot, xddot, qdot, qddot]

# Simulate both systems
tspan = (0, 5)

sol_fixed = solve_ivp(fixed_base, tspan, [0.5, 0.0], max_step=0.01)
sol_move  = solve_ivp(moving_base, tspan, [0.0, 0.0, 0.5, 0.0], max_step=0.01)

print("Fixed-base final q =", sol_fixed.y[0,-1])
print("Moving-base final q =", sol_move.y[2,-1], "final x =", sol_move.y[0,-1])
      

6.2 C++ (Eigen for linear algebra)

#include <iostream>
#include <Eigen/Dense>
#include <cmath>

using Eigen::Matrix2d;
using Eigen::Vector2d;

struct State {
  double x, xdot, q, qdot;
};

int main() {
  // Parameters
  const double I=0.2, M=1.0, m=0.5, ell=0.6, g=9.81;
  const double Kp=15.0, Kd=3.0;

  double dt = 0.001;
  State s{0.0, 0.0, 0.5, 0.0};

  for(int i=0; i<5000; ++i){
    double tau = -Kp*s.q - Kd*s.qdot;
    double ub  = 0.0;

    // Mass matrix
    double M11 = M + m;
    double M12 = m*ell*std::cos(s.q);
    double M22 = I + m*ell*ell;

    Matrix2d Mass;
    Mass << M11, M12,
            M12, M22;

    // Bias terms
    double h1 = -m*ell*std::sin(s.q)*s.qdot*s.qdot;
    double h2 =  m*g*ell*std::sin(s.q);

    Vector2d rhs;
    rhs << ub - h1, tau - h2;

    Vector2d acc = Mass.lu().solve(rhs);
    double xddot = acc(0);
    double qddot = acc(1);

    // Euler integration
    s.x    += dt*s.xdot;
    s.xdot += dt*xddot;
    s.q    += dt*s.qdot;
    s.qdot += dt*qddot;
  }

  std::cout << "Final q=" << s.q << " Final x=" << s.x << std::endl;
  return 0;
}
      

6.3 Java (simple RK4 integrator)

public class MovingBaseArmSim {

    static class State {
        double x, xdot, q, qdot;
        State(double x, double xdot, double q, double qdot){
            this.x=x; this.xdot=xdot; this.q=q; this.qdot=qdot;
        }
        State copy(){ return new State(x, xdot, q, qdot); }
    }

    // Parameters
    static double I=0.2, M=1.0, m=0.5, ell=0.6, g=9.81;
    static double Kp=15.0, Kd=3.0;

    static State f(State s){
        double tau = -Kp*s.q - Kd*s.qdot;
        double ub  = 0.0;

        double M11 = M + m;
        double M12 = m*ell*Math.cos(s.q);
        double M22 = I + m*ell*ell;

        double det = M11*M22 - M12*M12;

        double h1 = -m*ell*Math.sin(s.q)*s.qdot*s.qdot;
        double h2 =  m*g*ell*Math.sin(s.q);

        double rhs1 = ub - h1;
        double rhs2 = tau - h2;

        // Solve 2x2 system manually
        double xddot = ( M22*rhs1 - M12*rhs2)/det;
        double qddot = (-M12*rhs1 + M11*rhs2)/det;

        return new State(s.xdot, xddot, s.qdot, qddot);
    }

    static State rk4Step(State s, double dt){
        State k1 = f(s);
        State s2 = new State(s.x + 0.5*dt*k1.x, s.xdot + 0.5*dt*k1.xdot,
                             s.q + 0.5*dt*k1.q, s.qdot + 0.5*dt*k1.qdot);
        State k2 = f(s2);
        State s3 = new State(s.x + 0.5*dt*k2.x, s.xdot + 0.5*dt*k2.xdot,
                             s.q + 0.5*dt*k2.q, s.qdot + 0.5*dt*k2.qdot);
        State k3 = f(s3);
        State s4 = new State(s.x + dt*k3.x, s.xdot + dt*k3.xdot,
                             s.q + dt*k3.q, s.qdot + dt*k3.qdot);
        State k4 = f(s4);

        return new State(
            s.x    + dt*(k1.x    + 2*k2.x    + 2*k3.x    + k4.x)/6.0,
            s.xdot + dt*(k1.xdot + 2*k2.xdot + 2*k3.xdot + k4.xdot)/6.0,
            s.q    + dt*(k1.q    + 2*k2.q    + 2*k3.q    + k4.q)/6.0,
            s.qdot + dt*(k1.qdot + 2*k2.qdot + 2*k3.qdot + k4.qdot)/6.0
        );
    }

    public static void main(String[] args){
        State s = new State(0,0,0.5,0);
        double dt=0.001;
        for(int i=0;i<5000;i++){
            s = rk4Step(s, dt);
        }
        System.out.println("Final q=" + s.q + " Final x=" + s.x);
    }
}
      

6.4 MATLAB / Simulink-compatible script

% Moving-base arm on cart (ode45)
I=0.2; M=1.0; m=0.5; ell=0.6; g=9.81;
Kp=15; Kd=3;

f = @(t,s) dyn(t,s,I,M,m,ell,g,Kp,Kd);

tspan = [0 5];
s0 = [0;0;0.5;0]; % [x; xdot; q; qdot]
sol = ode45(f, tspan, s0);

xf = sol.y(1,end);
qf = sol.y(3,end);
disp(['Final q=', num2str(qf), ' Final x=', num2str(xf)]);

function ds = dyn(~, s, I, M, m, ell, g, Kp, Kd)
  x=s(1); xdot=s(2); q=s(3); qdot=s(4);

  tau = -Kp*q - Kd*qdot;
  ub = 0; % unactuated base

  M11 = M + m;
  M12 = m*ell*cos(q);
  M22 = I + m*ell^2;

  Mass = [M11 M12; M12 M22];

  h1 = -m*ell*sin(q)*qdot^2;
  h2 =  m*g*ell*sin(q);

  acc = Mass \ [ub - h1; tau - h2];
  xddot = acc(1);
  qddot = acc(2);

  ds = [xdot; xddot; qdot; qddot];
end
      

Library note: in later chapters you will see how toolboxes (e.g., ROS, MATLAB Robotics System Toolbox) encode these structures automatically. For now, the example emphasizes the coupling term and additional base state.

Problem 1 (DOF Counting): Consider (i) a 6-DOF industrial arm bolted to the floor, (ii) the same arm mounted on a wheeled platform with planar motion (2 translations + 1 rotation). Compute the total configuration dimension for each case.

Solution: (i) Stationary-base: only joints move, so \( n=6 \). Thus \( \dim(\mathcal{C}_{fixed})=6 \). (ii) Moving-base: base adds \( m=3 \) planar DOF. Thus \( \dim(\mathcal{C}_{moving})=n+m=6+3=9 \).

Problem 2 (Mass Matrix Symmetry): Show that for any rigid-body robot (fixed or moving base), the mass matrix derived from kinetic energy is symmetric and positive definite.

Solution: Kinetic energy of a rigid-body system can always be written as a quadratic form in generalized velocities: \( T(\mathbf{z},\dot{\mathbf{z}})=\tfrac{1}{2}\dot{\mathbf{z}}^\top \mathbf{M}(\mathbf{z}) \dot{\mathbf{z}} \). Since \(T\) is a scalar and quadratic in \( \dot{\mathbf{z}} \), the matrix of second partial derivatives satisfies \( \mathbf{M}_{ij}=\frac{\partial^2 T}{\partial \dot{z}_i \partial \dot{z}_j} =\mathbf{M}_{ji} \), hence symmetry. Moreover, for any nonzero velocity \( \mathbf{v} \), physical kinetic energy obeys \( T = \tfrac{1}{2}\mathbf{v}^\top\mathbf{M}\mathbf{v} > 0 \), implying positive definiteness.

Problem 3 (Coupled Lagrangian): For the moving-base arm-on-cart example in Section 6, assume kinetic energy \( T=\tfrac{1}{2}(M+m)\dot{x}^2 + \tfrac{1}{2}(I+m\ell^2)\dot{q}^2 + m\ell\cos q \,\dot{x}\dot{q} \) and potential energy \( V=m g \ell (1-\cos q) \). Derive the equations of motion.

Solution: Form \( \mathcal{L}=T-V \). Compute partial derivatives: \( \frac{\partial T}{\partial \dot{x}}=(M+m)\dot{x}+m\ell\cos q\,\dot{q} \), so \( \frac{d}{dt}\frac{\partial T}{\partial \dot{x}}=(M+m)\ddot{x} + m\ell(-\sin q\,\dot{q})\dot{q} + m\ell\cos q\,\ddot{q} \). Also \( \frac{\partial T}{\partial x}=0 \), \( \frac{\partial V}{\partial x}=0 \). Thus:

\[ (M+m)\ddot{x} + m\ell\cos q\,\ddot{q} - m\ell\sin q\,\dot{q}^2 = u_b. \]

\[ m\ell\cos q\,\ddot{x} + (I+m\ell^2)\ddot{q} + m g \ell \sin q = \tau. \]

Problem 4 (Underactuation Check): In the same cart–arm system, suppose \( u_b=0 \). Show that the number of inputs is less than the number of DOF. What does this imply for control?

Solution: The configuration is \( [x,q]^\top \), hence \( m+n=2 \) DOF. With \( u_b=0 \), only one input \( \tau \) remains. Therefore the system has 2 DOF but 1 actuator → underactuated. Control must exploit coupling: joint torques indirectly influence base motion. Not every trajectory is feasible with arbitrary accuracy, and stabilization requires careful design (a theme for later courses).

Stationary-base robots have configurations described solely by internal joints, leading to dynamics in \( \mathbf{q} \). Moving-base robots expand configuration to \( \mathbf{z}=[\mathbf{x}_b;\mathbf{q}] \), producing block-structured mass matrices with cross-coupling. This coupling yields momentum exchange and often underactuation, fundamentally affecting taxonomy, design, and control complexity.