Dynamics

Dynamics

Prerequisites

To get the most out of this Dynamics module, it is strongly recommended that you have already completed the course:

Kinematics

A solid understanding of robot configurations, coordinate transformations and velocity kinematics.
Familiarity with forward and inverse kinematics, jacobians and joint-space vs task-space representations.

Basic Mechanical Physics

Newton’s laws of motion, especially force = mass × acceleration (F = ma).
Concepts of moment of inertia, torque, and rotational motion.
Understanding of Coriolis and centrifugal forces, which play a key role in the equations of motion for moving reference frames like robotic arms.
Energy principles like kinetic energy, potential energy, and conservation of energy.

These prerequisites are essential because dynamics connects the dots between motion and its causes

General Motivation

How do robots jump over gaps, crawl under obstacles, or balance on one leg?

It’s not magic, it’s dynamics.

If kinematics is the geometry of movement, dynamics is the physics behind it. It answers questions like:

How much torque does a joint need to lift a leg while climbing a beam?
What forces act on a robot when landing after a leap?
How can a robot balance itself when slipping or being pushed?

Dynamics is the part of robotics that deals with forces, torques, mass, and motion. Whether a robot is controlled with code or trained with AI, it still needs to obey the laws of physics.

In these videos below, the robots are not just following a path, they are reacting to gravity, inertia, and impact in real time.

Studying dynamics helps you:

Predict how robots move when pushed or land from a jump
Design systems that stay stable on rough, slippery, or uneven terrain
Understand the real physics behind motion

If you want to make robots that move like athletes, adapt like animals, and react like pros, then dynamics is your next step.

_{Parkour in the Wild: ANYmal leaping over rubble and gaps. YouTube video, Mai 2025. Available at: https://www.youtube.com/watch?v=QDU_FicBPDo}

_{Boston Dynamics Atlas crawling, running, and balancing. YouTube video, April 2025. Available at: https://www.youtube.com/watch?v=I44_zbEwz_w}

Similar robotic models are available in Webots, where you can explore and analyze their implementations. These can be accessed via
File → Open Sample Robot → robots\boston_dynamics.
By interacting with the provided code, it is possible to experiment with control strategies and observe the resulting behaviors in simulation.

Course Content

Why Dynamics?

Motivation: Why kinematics is not enough

What dynamics adds: Forces, torques, and real-world behavior

Examples: Balancing, jumping, slipping, and interacting with the world

Kinematics helps us plan how a robot should move.
But what about how much force a motor needs to apply?
Or what happens when the robot lands from a jump?

That is where dynamics comes in.

Dynamics is the study of how forces and torques cause motion. It lets us:

Predict how robots react to impacts, slopes, or slippery surfaces
Simulate realistic motion for design and testing
Control how fast and how strong each joint needs to move
Understand why motion doesn’t happen exactly as planned

If kinematics gives us a map, dynamics gives us the engine that drives the robot in the real world.

Conceptual Questions

Question 1: Which statement correctly distinguishes kinematics from dynamics?

Question 2: A robot follows a planned path but slips while walking downhill. Why might this happen?

Question 3: Which robot behaviors are influenced by having a correct model of the robot and environment' dynamics?

(Part1) The Lagrangian Formulation of Dynamics

Generalized coordinates and velocities

Kinetic and potential energy of robotic systems

Euler–Lagrange equations

Example: Pendulum and simple manipulators

_{Modern Robotics, Chapter 8.1: Lagrangian Formulation of Dynamics (Part 1 of 2). YouTube video. Available at: https://www.youtube.com/watch?v=1U6y_68CjeY}

In this video, we explore the dynamics of open-chain robots and introduce two major approaches to deriving their equations of motion:

Forward Dynamics: useful for simulation. It tells us how a robot will move given known torques or forces.
Inverse Dynamics: useful in control. It tells us what torques are required to follow a desired motion.

Here two key modeling approaches are compared:

Lagrangian Formulation: a variational method based on the robot’s kinetic and potential energy. It gives us compact, system-wide equations without dealing with forces on each individual link.
Newton–Euler Formulation (see section 1.2.3.3 (part 1)): a force-based method applying Newton’s second law $ F = ma $ and torque equations link by link. It’s more direct but less elegant for complex systems.

The Power of the Lagrangian Approach

In dynamics, we often ask:

“Given a robot’s configuration and energy, what equations govern its motion?”

The Lagrangian formulation answers this by defining a single scalar value called the Lagrangian:

$$ \boxed{L(\theta, \dot{\theta}) = K(\theta, \dot{\theta}) - P(\theta)} $$

Where:

$E_K$: Kinetic energy
$P$: Potential energy
$\theta$: Generalized coordinates (e.g., joint angles)
$\dot{\theta}$: Generalized velocities

Using the Euler–Lagrange equation, we can systematically derive the equations of motion:

$$ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}_i} \right) - \frac{\partial L}{\partial \theta_i} = \tau_i $$

Where:

$\theta_i$: The $i$-th generalized coordinate
$\dot{\theta}_i$: Its velocity
$\tau_i$: Torque or generalized force applied at joint $i$

From Theory to General Dynamic Equation

By applying the Lagrangian method to full robotic systems, we arrive at the general equation of motion for manipulators:

$$ \tau = M(\theta)\ddot{\theta} + c(\theta, \dot{\theta}) + g(\theta) $$

Where:

$\tau$: Vector of joint torques
$M(\theta)$: Mass (inertia) matrix — describes how mass is distributed across the robot
$c(\theta, \dot{\theta})$: Velocity-product term — includes Coriolis and centrifugal forces
$g(\theta)$: Gravity term — torques needed to compensate for gravity

This equation is central to model-based control, trajectory planning, and robot simulation. It explains how acceleration, motion-induced forces, and gravity affect the torques at each joint.

We can use it in two ways:

Forward Dynamics::
Given: torques $\tau$
Compute: joint accelerations $\ddot{\theta}$ (→ then velocities and positions)
This is used in **simulation** to see how a robot will move when forces are applied.

$$ \ddot{\theta} = M(\theta)^{-1} \left( \tau - c(\theta, \dot{\theta}) - g(\theta) \right) $$

Inverse Dynamics:: Given: desired motion $\theta(t), \dot{\theta}(t), \ddot{\theta}(t)$
Compute: required joint torques $\tau$

This is used in control to determine the torques needed to follow a trajectory.

You can think of forward and inverse dynamics like this:

Conceptual Questions

Question 1: What does the Lagrangian represent?

Question 2: What is the advantage of using the Lagrangian method for robots with many joints?

(Part2) The Lagrangian Formulation of Dynamics

_{Modern Robotics, Chapter 8.1: Lagrangian Formulation of Dynamics (Part 2 of 2). YouTube video. Available at: https://www.youtube.com/watch?v=BjD-pL819LA}

In this video, we take a deeper look at the velocity-product terms $c(\theta, \dot{\theta})$ in the dynamic equations of motion:

$$ \tau = M(\theta)\ddot{\theta} + c(\theta, \dot{\theta}) + g(\theta) $$

Velocity-Product Terms: What Are They?

These terms arise from the nonlinear interaction of joint velocities.

Two Key Types::

Centripetal Terms:
Terms with a squared joint velocity, like $\dot{\theta}_1^2$.
They arise when one joint rotates and a mass moves in a circular path around that joint.
Coriolis Terms:
Terms with a product of two different joint velocities, like $\dot{\theta}_1 \dot{\theta}_2$.
These arise when multiple joints move simultaneously.

Example: 2R Planar Robot::

We analyze a 2R robot arm under different conditions (neglecting gravity and accelerations for clarity):

Only $\dot{\theta}_1 > 0$, $\dot{\theta}_2 = 0$
- Mass 2 moves in a circular arc around joint 1.
- It experiences centripetal acceleration $\propto \dot{\theta}_1^2$ toward joint 1.
- Joint 2 must apply positive torque; joint 1 does not need torque.
Only $\dot{\theta}_2 > 0$, $\dot{\theta}_1 = 0$
- Mass 2 circles around joint 2.
- Centripetal acceleration is proportional to $ \dot{\theta}_2^2 $.
- Only joint 2 provides torque.
Both $\dot{\theta}_1 > 0$ and $\dot{\theta}_2 > 0$
- A Coriolis acceleration appears, directed toward joint 2.
- Joint 1 must apply negative torque to keep constant speed, similar to the ice skater effect (spinning faster when arms are pulled in).

Writing the Velocity-Product Term Mathematically

Option 1: General Notation We often write the velocity-product term as:

$$ c(\theta, \dot{\theta}) = \dot{\theta}^\top \Gamma(\theta) \dot{\theta} $$

$\Gamma(\theta)$: A 3D tensor (size $n \times n \times n$) of Christoffel symbols derived from the mass matrix
$\Gamma_{i}$: an $n \times n$ matrix
$\Gamma_{i,j,k}$: the ($j,k$)th element of $\Gamma_{i}(\theta)$

Option 2: Component-wise

Christoffel symbols:

$$ \Gamma_{i,j,k} = \frac{1}{2} \left( \frac{\partial M_{i,k}}{\partial \theta_j} + \frac{\partial M_{i,j}}{\partial \theta_k} - \frac{\partial M_{j,k}}{\partial \theta_i} \right) $$

The $ij$-th component of $c(\theta, \dot{\theta})$ is:

$$ c_{ij}(\theta, \dot{\theta}) = \sum_{k=1}^n \Gamma_{i,j,k}(\theta) \dot{\theta}_k $$

Physical Intuition

Just like a changing mass affects linear momentum in simple systems, a configuration-dependent mass matrix in robots leads to extra forces (velocity-product terms).
These terms are essential for maintaining or resisting motion even when accelerations are zero.

Alternative Representations

Coriolis Matrix form:
Another way to express the velocity-product term:

$$ c(\theta, \dot{\theta}) = C(\theta, \dot{\theta}) \dot{\theta} $$

Where $ C(\theta, \dot{\theta}) $ is the Coriolis matrix, assembled using Christoffel symbols.
Combined vector notation:

$$ h(\theta, \dot{\theta}) = c(\theta, \dot{\theta}) + g(\theta) $$

Often used to group all non-acceleration-dependent terms together.

SUMMARY

Velocity-product terms are quadratic in joint velocities and arise naturally in robotic motion.
They include centripetal and Coriolis effects.
They ensure that even without accelerations or external forces, internal inertial effects still generate torque requirements.
You can represent them using:
- Christoffel symbols
- A Coriolis matrix
- Or compact vector notation $ h(\theta, \dot{\theta}) $

Understanding these terms is essential for accurate simulation, control, and physical intuition in robotic systems.

Conceptual Questions

Question 1: What is a centripetal term in the context of robot dynamics?

Question 2: What makes a term a Coriolis term?

Question 3: Which of the following is true about the Christoffel symbols in robot dynamics?

Question 4: What happens to joint 1 when both $\dot{\theta}_1$ and $\dot{\theta}_2$ are positive?

Question 5: Why can $c(\theta, \dot{\theta})$ be written as $\dot{\theta}^\top \Gamma(\theta) \dot{\theta}$?

Mathematical exercises

From Practice Exercise 8.1, MODERN ROBOTICS, Practice Exercices

EXERCISE 1:

Fig 8.1 — **Figure 8.1.** RP robot moving in a vertical plane.

Figure 8.1 illustrates an RP robot moving in a vertical plane. The mass of link 1 is $m_1$ and the center of mass is a distance $L_1$ from joint 1. The scalar inertia of link 1 about an axis through the center of mass and out of the plane is $I_1$. The mass of link 2 is $m_2$, the center of mass is a distance $\theta_{2}$ from joint 1, and the scalar inertia of link 2 about its center of mass is $I_2$. Gravity g acts downward on the page.

(a) Let the location of the CoM of link $i$ be $(x_i, y_i)$ for $i=1,2$. Find $(x_i(\theta), y_i(\theta))$ and their time derivatives $(\dot{x}_i, \dot{y}_i)$.

(b) Write the potential energy of each link, $P_1$ and $P_2$, using the joint variables $\theta$.

(c) Write the kinetic energy of each link, $E_{K_1}$ and $E_{K_2}$. Use the planar rigid–body formula

$$ E_K \;=\; \tfrac{1}{2}\,m\,v^2 \;+\; \tfrac{1}{2}\,I\,\omega^2, $$

where $m$ is the mass, $v$ is the scalar linear velocity at the CoM, $\omega$ is the scalar angular velocity, and $I$ is the scalar inertia of the rigid body about its CoM.

(d) What is the Lagrangian in terms of $K_1$, $K_2$, $P_1$ and $P_2$ ?

(e) One of the terms in the Lagrangian can be expressed as

$$ \frac{1}{2}\,m_2\,\theta_2^{\,2}\,\dot{\theta}_1^{\,2}. $$

If this were the complete Lagrangian, what would the equations of motion be? Derive these by hand (no symbolic math software assistance). Indicate which of the terms in your equations are a function of $\ddot{\theta}$ , which are Coriolis terms, which are centripetal terms, and which are gravity terms, if any.

(f) Now derive the equations of motion (either by hand or using symbolic math software for assistance) for the full Lagrangian and put them in the form $$ \tau \;=\; M(\theta)\,\ddot{\theta} \;+\; c(\theta,\dot{\theta}) \;+\; g(\theta) $$

Identify which of the terms in $c(\theta,\dot{theta})$ are Coriolis and whihc are centripetal. Explain as if to someone who is unfamiliar with dynamics why these terms contribute to the joint forces and torques.

From Practice Exercise 8.3, MODERN ROBOTICS, Practice Exercices

EXERCISE 2:

The equations of motion for a particular 2R robot arm can be written $M(\theta)\,\ddot{\theta} + c(\theta,\dot{\theta}) + g(\theta) = \tau$. The Lagrangian $ \mathcal{L}(\theta,\dot{\theta}) $ for the robot can be written in components as $$ L(\theta,\dot{\theta}) = L^{1}(\theta,\dot{\theta}) + \mathcal{L}^{2}(\theta,\dot{\theta}) + \mathcal{L}^{3}(\theta,\dot{\theta}) + \cdots $$

One of these components is $\mathcal{L}^{1} = m\,\dot{\theta}_1\,\dot{\theta}_2\,\cos\theta_2$

(a) Find the joint torques $ \tau_1 $ and $ \tau_2 $ corresponding to the component $ \mathcal{L}^{1} $.

(b) (See Section 1.2.3.2 if necessary) Write the $2\times2$ mass matrix $M^{1}(\theta)$, the velocity-product vector $c^{1}(\theta,\dot{\theta})$, and the gravity vector $g^{1}(\theta)$ corresponding to $\mathcal{L}^{1}$. (Notice that $M = M^{1} + M^{2} + M^{3} + \cdots$, $c = c^{1} + c^{2} + c^{3} + \cdots$, and $g = g^{1} + g^{2} + g^{3} + \cdots$.)

Solutions

EXERCISE 1:

(a) Centers of mass and their time derivatives

Positions:

$$ x_1 = L_1\cos\theta_1,\qquad y_1 = L_1\sin\theta_1 $$

$$ x_2 = \theta_2\cos\theta_1,\qquad y_2 = \theta_2\sin\theta_1 $$

Velocities:

$$ \dot x_1 = -L_1\dot\theta_1\sin\theta_1,\qquad \dot y_1 = \;\;L_1\dot\theta_1\cos\theta_1 $$

$$ \dot x_2 = \dot\theta_2\cos\theta_1-\theta_2\dot\theta_1\sin\theta_1,\qquad \dot y_2 = \dot\theta_2\sin\theta_1+\theta_2\dot\theta_1\cos\theta_1 $$

(b) Potential energies

$$ \mathcal P_1 = m_1 g y_1 = m_1 g L_1 \sin\theta_1, \qquad \mathcal P_2 = m_2 g y_2 = m_2 g \theta_2 \sin\theta_1. $$

(c) Kinetic energies

Link 1:

$$ \mathcal E_{K_1} = \tfrac12 m_1(\dot x_1^2+\dot y_1^2) + \tfrac12 I_1 \dot\theta_1^{\,2} = \tfrac12 (I_1+m_1L_1^{\,2})\,\dot\theta_1^{\,2}. $$

Link 2:

$$ \mathcal E_{K_2} = \tfrac12 m_2(\dot x_2^2+\dot y_2^2) + \tfrac12 I_2 \dot\theta_1^{\,2} = \tfrac12\left( (I_2+m_2\theta_2^{\,2})\,\dot\theta_1^{\,2} + m_2\dot\theta_2^{\,2} \right). $$

(d) Lagrangian

$$ \mathcal L = \mathcal K_1 + \mathcal K_2 - \mathcal P_1 - \mathcal P_2. $$

(e) Equations of motion

$$ \tau_1 = 2 m_2 \theta_2 \dot\theta_1 \dot\theta_2 \;+\; m_2 \theta_2^{\,2}\ddot\theta_1, \qquad \tau_2 = -\,m_2 \theta_2\,\dot\theta_1^{\,2}. $$

In $\tau_1$, the first term is Coriolis; the second multiplies $\ddot\theta_1$ (a mass-matrix term).
$\tau_2$ is a centripetal term.

(f) Full equations in standard form

$$ \tau = M(\theta)\,\ddot\theta + c(\theta,\dot\theta) + g(\theta), \qquad \theta=\begin{bmatrix}\theta_1\\ \theta_2\end{bmatrix}. $$

Mass matrix:

$$ M(\theta)= \begin{bmatrix} I_1 + I_2 + m_1L_1^{\,2} + m_2\theta_2^{\,2} & 0 \cr 0 & m_2 \end{bmatrix}. $$

Velocity-product (Coriolis/centripetal) vector:

$$ c(\theta,\dot{\theta})= \begin{bmatrix} 2 m_2 \theta_2 \dot\theta_1 \dot\theta_2 \cr -m_2 \theta_2 \dot\theta_1^{\,2} \end{bmatrix}. $$

Gravity vector:

$$ g(\theta)= \begin{bmatrix} (m_1 L_1 + m_2 \theta_2)\, g \cos\theta_1 \cr m_2 g \sin\theta_1 \end{bmatrix}. $$

EXERCISE 2:

(a) $$ \tau_1 = \frac{d}{dt}\left(\frac{\partial \mathcal{L}^{1}}{\partial \dot{\theta}_1}\right) - \frac{\partial \mathcal{L}^{1}}{\partial \theta_1} = \frac{d}{dt}\big(m\,\dot{\theta}_2 \cos\theta_2\big) - 0 = m\,\ddot{\theta}_2 \cos\theta_2 - m\,\dot{\theta}_2^{\,2}\sin\theta_2 . $$

$$ \tau_2 = \frac{d}{dt}\left(\frac{\partial \mathcal{L}^{1}}{\partial \dot{\theta}_2}\right) - \frac{\partial \mathcal{L}^{1}}{\partial \theta_2} = \frac{d}{dt}\big(m\,\dot{\theta}_1 \cos\theta_2\big) + m\,\dot{\theta}_1 \dot{\theta}_2 \sin\theta_2 = m\,\ddot{\theta}_1 \cos\theta_2 . $$

(b) $$ M^{1}(\theta) = \begin{bmatrix} 0 & m\cos\theta_{2} \cr m\cos\theta_{2} & 0 \end{bmatrix}, \qquad c^{1}(\theta,\dot{\theta}) = \begin{bmatrix} -m \dot\theta_{2}^{2} \sin\theta_{2} \cr 0 \end{bmatrix}, \qquad g^{1}(\theta) = \begin{bmatrix} 0 \cr 0 \end{bmatrix}. $$

Understanding the Mass Matrix

Deriving the mass matrix from kinetic energy

Properties: Symmetry, positive-definiteness, configuration dependence

Example: Two-link planar robot

In this video we focus on better understanding of the mass matrix M of theta

_{Modern Robotics, Chapter 8.1.3: Understanding the Mass Matrix. YouTube video. Available at: https://www.youtube.com/watch?v=7PFQou5l9do}

In this video, we focus on gaining a deeper intuition and mathematical understanding of the mass matrix $M(\theta)$, which appears in the robot’s kinetic energy expression and equations of motion.

From Point Mass to Robot Arm

For a point mass:

$$ E_K = \frac{1}{2} m v^2 $$

But for a robot with multiple joints, the kinetic energy is expressed in joint coordinates as:

$$ E_K = \frac{1}{2} \dot{\theta}^\top M(\theta) \dot{\theta} $$

Where:

$\dot{\theta}$: Vector of joint velocities
$M(\theta)$: Mass (inertia) matrix maps joint velocities to kinetic energy

Key Properties of the Mass Matrix

Symmetry
$$ M(\theta) = M(\theta)^\top $$
Positive Definiteness
$$ x^\top M(\theta) x > 0 \quad \text{for all } x \neq 0 $$
→ This means the kinetic energy is always positive unless the robot is at rest.
Configuration-Dependent
- $M(\theta)$ depends on the robot’s joint configuration
- For example, a stretched-out arm has a different inertia than a folded one.

Physical Interpretation

The effective mass at the robot’s end-effector changes with direction and configuration.
When you push the robot’s end-effector by hand, it doesn’t behave like a point mass —
the force and acceleration directions are not necessarily aligned.
This is because the inertia felt at the end-effector is anisotropic (depends on direction) and configuration-dependent.

SUMMARY

After this section, you should have a solid understanding of the structure of a robot’s dynamic model:

$$ \tau = M(\theta)\ddot{\theta} + c(\theta, \dot{\theta}) + g(\theta) $$

This equation is an extension of Newton’s second law:

$F = ma$ → but here, $m$ and $a$ both depend on joint velocities and accelerations,
plus additional forces are needed to balance gravity and create end-effector wrenches.

By understanding the mass matrix $M(\theta)$, you’re one step closer to simulating and controlling robot motion with accuracy and intuition.

Conceptual Questions

Question 1: What is the role of the mass matrix $M(\theta)$ in robot dynamics?

Question 2: Which of the following is not a mathematical property of the mass matrix $M(\theta)$?

Question 3: Why does not a robot's end-effector feel like a point mass when you push it by hand?

Question 4: What does this expression represent?
$E_K = \frac{1}{2} \dot{\theta}^\top M(\theta) \dot{\theta}$

Mathematical exercises

From Practice Exercise 8.2, MODERN ROBOTICS, Practice Exercices

EXERCISE 3:

Fig 8.2 — **Figure 8.2.** A 2R robot with all mass concentrated at the ends of the links

The mass matrix of the 2R robot of Figure 8.2 is

$$ M(\theta)= \begin{bmatrix} m_1 L_1^{2} + m_2\left(L_1^{2} + 2 L_1 L_2 \cos \theta_2 + L_2^{2}\right) & m_2\left(L_1 L_2 \cos \theta_2 + L_2^{2}\right) \cr m_2\left(L_1 L_2 \cos \theta_2 + L_2^{2}\right) & m_2 L_2^{2} \end{bmatrix}, $$

where each link is modeled as a point mass at the end of the link. Explain in text and/or figures why each of the entries makes sense, for example using the joint accelerations $\ddot{\theta}=(1,0)$ and $(0,1)$.

From Practice Exercise 8.4, MODERN ROBOTICS, Practice Exercices

Exercise 4:

For a given configuration $\theta$ of a two-joint robot, the mass matrix is $$ M(\theta)= \begin{bmatrix} 3 & a \cr b & 2 \end{bmatrix}, $$ which has a determinant of $6-ab$ and eigenvalues $\frac{1}{2}\left(5 \pm \sqrt{1 + 4ab}\right)$. What constraints must $a$ and $b$ satisfy for this to be a valid mass matrix?

From Practice Exercise 8.1, MODERN ROBOTICS, Practice Exercices

Exercise 5: (this exercise is a continuation of exercise 1)

Consider the configuration-dependent mass matrix $M(\theta)$ from your previous answer. When the robot is at rest (and ignoring gravity), the mass matrix can be visualized as the ellipse of joint forces/torques that are required to generate the unit circle of joint accelerations in $\ddot{\theta}$ space. As $\theta_2$ increases, how does this ellipse change? Describe it in text and provide a drawing.

Solutions

Exercise 3:

Fig 8.9 — **Figure 8.9.** The linear accelerations of the point masses of the 2R arm for joint accelerations (1,0) and (0,1).

Let $$ M = \begin{bmatrix} M_{11} & M_{12} \cr M_{21} & M_{22} \end{bmatrix}, $$ where $$M_{11}=m_1 L_1^{2}+m_2(L_1^{2}+2L_1L_2\cos\theta_2+L_2^{2})$$ $$M_{12}=M_{21}=m_2(L_1L_2\cos\theta_2+L_2^{2})$$ $$M_{22}=m_2 L_2^{2}$$

Figure 8.9 shows the linear accelerations of the masses $m_1$ and $m_2$ for joint accelerations $(1,0)$ and $(0,1)$.

The terms $M_{11}$ and $M_{22}$ are relatively easy to understand. The term $M_{11}$ is the inertia of the robot about joint 1 if joint 2 is locked. The inertia contribution due to $m_1$ is $m_1 L_1^{2}$. The distance of $m_2$ from joint 1 is $d_2=\sqrt{L_1^{2}+2L_1L_2\cos\theta_2+L_2^{2}}$ (by the law of cosines), and the inertia contribution due to $m_2$ is $m_2 d_2^{2}$.

The term $M_{22}$ is the inertia about joint 2 due to the mass $m_2$ a distance $L_2$ from the joint.

The off-diagonal term is harder to understand. But we know that if joint 1 accelerates, joint 2 has to apply a torque to keep joint 2 locked. And if joint 2 accelerates, joint 1 has to apply a torque to remain locked; otherwise, conservation of angular momentum about joint 1 would cause it to begin to rotate in a direction opposite joint 2. Using Figure 8.9 and some geometry, you could calculate the joint torque $\tau_2$ required to keep joint 2 stationary when $\dot{\theta}_1=1$, based on the moment about joint 2 generated by the line of force required to accelerate $m_2$.

EXERCISE 4:

$M(\theta)$ must be positive definite (and therefore symmetric), so $a=b$ and the eigenvalues must be positive, so $\lvert a \rvert = \lvert b \rvert < \sqrt{6}$. (The determinant $\det(M)=6-a^2$ must be positive, which gives the same condition on $\lvert a \rvert$.)

Exercise 5: (this exercise is a continuation of exercise 1)

The mass matrix $M(\theta)$ is diagonal, so the principal axes of the ellipse $M(\theta),\ddot{\theta}$ (for all $\ddot{\theta}$ satisfying $\lvert\dot{\theta}\rvert = 1$) are aligned with the $\tau_1$ and $\tau_2$ axes, and the lengths of those principal axes (the eigenvalues of $M$) are just the entries along the diagonal. As $\theta_2$ gets larger, the top-left component of $M$ gets larger. This means that larger torques at joint 1 are required to generate accelerations in the $\ddot{\theta}_1$ direction, due to the increased inertia of the robot about joint 1. Hence the ellipse gets wider in the $\tau_1$ direction. See Figure 8.7.

Fig 8.7 — **Figure 8.7.** The mass matrix $M(\theta)$ represented as the ellipse of joint forces and torques corresponding to a unit circle of joint accelerations $\ddot{\theta}$ (when gravity and the joint velocities are zero).

(Part1) Dynamics of a Single Rigid Body

Newton-Euler equations for a free rigid body

Linear and angular momentum

Inertia tensor and spatial representation

_{Modern Robotics, Chapter 8.2: Dynamics of a Single Rigid Body (Part 1 of 2). YouTube video. Available at: https://www.youtube.com/watch?v=7PFQou5l9do}

This section introduces the Newton–Euler method for modeling the dynamics of a rigid body, the foundation for building full robot dynamics. Unlike the Lagrangian approach, which uses energy, the Newton–Euler formulation is built directly from:

Force = mass × acceleration
Torque = moment of inertia × angular acceleration

Rigid Body Basics

A rigid body is modeled as a collection of point masses rigidly connected.
A body frame {b} is fixed at the center of mass (COM) of the body.
The twist of the body in the {b} frame is denoted by:

$$ V_b = \begin{bmatrix} \omega_b \\ v_b \end{bmatrix} $$

Where:
- $\omega_b$: Angular velocity of the body (in {b})
- $v_b$: Linear velocity of the COM (in {b})

Kinematics of the Point Masses

For the $i$-th point mass:

Position relative to the COM: $p_i$
Velocity:
$$ \dot{p}_i = v_b + \omega_b \times p_i $$
Acceleration (after differentiating):
$$ \ddot{p}_i = \dot{v}_b + \dot{\omega}_b \times p_i + \omega_b \times (v_b + \omega_b \times p_i) $$

Newton–Euler Dynamics

Using $f_i = m_i \ddot{p}_i$ and summing over all point masses:

Total Force: $$ f_b = m \left( \dot{v}_b + \omega_b \times v_b \right) $$

Total Moment: $$ m_b = I_b \dot{\omega}_b + \omega_b \times (I_b \omega_b) $$

Where:

$I_b$: Inertia matrix of the rigid body (in {b})
The full wrench applied on the body is:
$$ F_b = \begin{bmatrix} m_b \\ f_b \end{bmatrix} $$

Inertia Matrix

The inertia matrix $I_b \in \mathbb{R}^{3 \times 3}$ describes how mass is distributed around the center of mass.

Symmetric and positive definite
Diagonal terms (e.g., $ I_{xx} $) are moments of inertia
Off-diagonal terms (e.g., $ I_{xy} $) are products of inertia

Computation (for point masses):

$$ I_b = -\sum_i m_i [p_i]_\times^2 $$

$[p_i]_\times$: Skew-symmetric matrix of vector $p_i$

From Discrete to Continuous Mass

When replacing point masses with a continuous mass density $\rho(x,y,z)$,
the summation becomes an integral:

$$ I_b = \int_{\text{Body}} \rho(x,y,z) [p]_\times^2 \, dV $$

Kinetic Energy of a Rigid Body

Rotational kinetic energy in frame {b}:

$$ E_K = \frac{1}{2} \omega_b^\top I_b \omega_b $$

Just like joint-space kinetic energy used the mass matrix, rotational energy uses the inertia matrix.

Principal Axes and Diagonalization

The principal axes of inertia are the eigenvectors of $I_b$ (called $R_{bp}$)
The principal moments of inertia are the eigenvalues of $I_b$
In the {p} frame (aligned with principal axes), $I_p$ is diagonal:

$$ I_p = R_{bp}^\top I_b R_{bp} $$
Using the principal axes simplifies the rotational equations of motion.

SUMMARY

Linear equation: $$ \boxed{f_b = m (\dot{v}_b + \omega_b \times v_b)} $$
Rotational equation: $$ \boxed{m_b = I_b \dot{\omega}_b + \omega_b \times (I_b \omega_b)} $$

These equations form the building blocks for modeling full robot dynamics using recursive Newton–Euler algorithms in later sections.

Conceptual Questions

Question 1: What physical quantity does the inertia matrix $I_b$ describe?

Question 2: What does the twist $V_b$ of a rigid body represent?

Question 3: What is true about the inertia matrix $I_b$?

Question 4: What is the benefit of aligning the body frame with the principal axes of inertia?

Question 5: The rotational equation $m_b = I_b \dot{\omega}_b + \omega_b \times (I_b \omega_b)$ includes what kind of terms?

(Part2) Dynamics of a Single Rigid Body – Spatial Inertia and Equations of Motion

_{Modern Robotics, Chapter 8.2: Dynamics of a Single Rigid Body (Part 2 of 2). YouTube video. Available at: https://www.youtube.com/watch?v=2rUWVdslaI4}

In this chapter, we extend the Newton-Euler formulation using spatial vector algebra to derive compact dynamic equations for a single rigid body.

Key Concepts

We represent rigid body motion with a spatial twist:

$$ V_b = \begin{bmatrix} \omega_b \\ v_b \end{bmatrix} $$

We define the spatial inertia matrix $G_b$ in the body frame {b} as:

$$ G_b = \begin{bmatrix} I_b & 0 \cr 0 & m \cdot I_{3 \times 3} \end{bmatrix} $$

Where:

$I_b$: Rotational inertia matrix about the center of mass
$m$: Total mass of the rigid body
$I_{3 \times 3}$: Identity matrix

Kinetic Energy

The kinetic energy of the rigid body becomes:

$$ E_K = \frac{1}{2} V_b^T G_b V_b $$

Lie Bracket for Twists

The Lie bracket of two spatial twists $V_1$ and $V_2$ is an acceleration measuring how motion along these twist change if the body follows the other twist (here it measures how motion along the twist $V_2$ would change if the body follows the twist $V_1$):

$$ [V_1, V_2] = \text{ad}_{V_1} V_2 $$

The little adjoint operator is:

$$ \text{ad}_V = \begin{bmatrix} [\omega] & 0 \cr [v] & [\omega] \end{bmatrix} $$

Where $[\cdot]$ represents the skew-symmetric matrix (cross-product bracket).

Equations of Motion in Frame {b}::

The spatial Newton-Euler equation becomes:

$$ F_b = G_b \dot{V_b} - \text{ad}_{V_b}^T G_b V_b $$

Where:

$F_b$: Wrench (torque + force)
$V_b$: Spatial twist
$\dot{V}_b$: Spatial acceleration
$\text{ad}_{V_b}^T G_b V_b$: Velocity-product term

This equation generalizes the classic rigid body dynamics with:

Twist instead of angular velocity
Spatial inertia instead of scalar inertia
Lie bracket instead of cross product

Frame Change to {a}::

Given a transform $T_{ba}$ from frame {b} to {a}, the spatial inertia transforms as:

$$ G_a = \mathrm{Ad}(T_{ba})^{\mathrm{T}} \, G_b \, \mathrm{Ad}(T_{ba}) $$

And the equation of motion becomes:

$$ F_a = G_a \dot{V_a} - \text{ad}_{V_a}^T G_a V_a $$

Inverse Dynamics (compute wrench from motion):

$$ F_b = G_b \dot{V_b} - \text{ad}_{V_b}^T G_b V_b $$

Forward Dynamics (compute acceleration from wrench):

$$ \dot{V_b} = G_b^{-1} \left( F_b + \text{ad}_{V_b}^T G_b V_b \right) $$

SUMMARY

The spatial inertia matrix $G_b$ combines mass and rotational inertia into a single $6 \times 6$ matrix.
Spatial dynamics equations use twists, wrenches, and the Lie bracket to unify motion representation.
This lays the foundation for the recursive Newton-Euler algorithm for full robot manipulators.

Conceptual Questions

Question 1: Why is the spatial inertia matrix $G_b$ a 6×6 matrix?

Question 2: What does the Lie bracket $[V_1, V_2]$ represent in spatial dynamics?

Question 3: Why do we sometimes prefer to use the principal axes of inertia when modeling a rigid body?

Question 4: In spatial dynamics, what does the velocity-product term $\text{ad}_{V_b}^T G_b V_b$ represent?

Mathematical exercises

From Practice Exercise 8.6, MODERN ROBOTICS, Practice Exercices

EXERCISE 6:

Fig 8.3 — **Figure 8.3.** A ring of radius 3.

Figure 8.3 shows a ring in the $\hat{y}_b$–$\hat{z}_b$ plane (the $\hat{x}_b$ coordinate of each point on the ring is zero). The radius of the ring is $3$ (all mass is a distance $3$ from the $\hat{x}_b$-axis). The mass of the ring is $10$, and the mass is uniformly distributed around the ring. Write the spatial inertia matrix $G_b$. All entries should be numerical, no symbols or math.

From Practice Exercise 8.10, MODERN ROBOTICS, Practice Exercices

EXERCISE 7: Consider the object in Figure 8.5 that consists of a cube and sphere that are rigidly attached. The ${c}$ and ${s}$ frames of each component are aligned with the principal axes and located at the center of mass. The $z$-axes of both frames are colinear. Given that the body inertia of a sphere is $I_s=\left(\frac{2mr^2}{5}\right) I_{3\times3},$ the body inertia of a cube is $I_c=\left(\frac{ml^2}{6}\right) I_{3\times3},$ $r=1$, $l=2$, the cube has mass $2$, and the sphere has mass $1$, solve for the spatial inertia matrix $G_b$ for the object.

Fig 8.5 — **Figure 8.5.** Rigid object consisting of a sphere and cube.

Solution

EXERCISE 6:

$$ G_b \;=\; \begin{bmatrix} 90 & 0 & 0 & 0 & 0 & 0 \cr 0 & 45 & 0 & 0 & 0 & 0 \cr 0 & 0 & 45 & 0 & 0 & 0 \cr 0 & 0 & 0 & 10 & 0 & 0 \cr 0 & 0 & 0 & 0 & 10 & 0 \cr 0 & 0 & 0 & 0 & 0 & 10 \end{bmatrix}, $$

where the inertia about the $\hat{x}_b$-axis is $mR^2 = 90$ since $m = 10$ and $R = 3$. The inertia about the $\hat{y}_b$ and $\hat{z}_b$ axes is $\tfrac{1}{2} m R^2$ (derive this formula from the integral or look it up online).

EXERCISE 7:

$$ COM_s = (0,0,3), \quad COM_c = (0,0,1). $$

$$ COM_b = \frac{COM_s\,m_s + COM_c\,m_c}{2}. $$

$$ q_c = COM_b - COM_c \quad,\quad q_s = COM_b - COM_s $$

$$ I_1 = I_c + m_c \left(q_c^{T} q_c\, I - q_c q_c^{T}\right) $$

$$ I_2 = I_s + m_s \left(q_s^{T} q_s\, I - q_s q_s^{T}\right) $$

$$ I_b = I_1 + I_2 = \operatorname{diag}(6.48,\, 6.48,\, 1.73) $$

$$ G_b = \operatorname{diag}(6.48,\, 6.48,\, 1.73,\, 3,\, 3,\, 3) $$

Inverse Dynamics with Newton-Euler

Recursive Newton-Euler algorithm

Computing joint torques given a motion

Efficiency advantages for real-time control

_{Modern Robotics, Chapter 8.3: Newton-Euler inverse Dynamics. YouTube video. Available at: https://www.youtube.com/watch?v=ZASVKAlegfQ}

In this chapter, we apply the inverse dynamics of a rigid body to compute the joint torques for an open-chain robot using the Newton–Euler algorithm.

We consider an n-link robot with:

Frame {0} fixed in the world
Frames {1} to {n} at each link’s center of mass
Frame {n+1} at the end-effector

We define:

$V_i$: Twist of link $i$, expressed in frame ${i}$
$\tau_i$: Joint torque applied at joint $i$
$A_i$: Joint screw axis of joint $i$ (expressed in frame ${i}$)
$T_{i,i-1}$: Transformation from frame ${i-1}$ to ${i}$
$F_{n+1} = F_{\text{tip}}$: External wrench applied by the end-effector
$\dot{V}_0$: Base acceleration to model gravity

Step 1: Forward Iteration (from base to end-effector)

Compute twists and accelerations:

1.Compute transformation
$$ T_{i,i-1} = e^{-A_i \theta_i} M_{i,i-1} $$

2.Compute twist
$$ V_{i} = Ad_{T_{i,i-1}} \, V_{i-1} + A_i \dot{\theta}_i $$

3.Compute acceleration
$$ \dot{V_i} = Ad_{T_{i,i-1}} \, \dot{V_{i-1}} + \text{ad}_{V_b} \, A_i \, \dot{\theta}_i + A_i \, \ddot{\theta}_i $$

Where:

$Ad_{T_{i,i-1}}$ is the Adjoint transformation from frame ${i-1}$ to ${i}$
$\text{ad}_V$ is the Lie bracket operator (captures velocity-product terms)

Step 2: Backward Iteration (from end-effector to base)

Compute wrenches and torques:

4.Compute wrench
$$ F_i = Ad_{T_{i+1,i}}^T F_{i+1} + G_i \dot{V_i} - \text{ad}_{V_i}^T G_i V_i $$

Where:

$G_i$: Spatial inertia matrix of link $i$

5.Compute joint torque
$$ \tau_i = F_i^T A_i $$

Only the component of $F_i$ along $A_i$ must be supplied by the joint actuator.

Final Output::

After forward and backward passes, we obtain:

Joint torques $\tau = [\tau_1, \dots, \tau_n]^T$ required for:
- Joint positions $\theta$
- Velocities $\dot{\theta}$
- Accelerations $\ddot{\theta}$
- End-effector wrench $F_{\text{tip}}$

SUMMARY

No differentiation is required
Recursive algorithm → computationally efficient
Suitable for model-based robot control

Conceptual Questions

Question 1: What is the main goal of the Newton-Euler inverse dynamics algorithm?

Question 2: What does the forward iteration of the Newton-Euler algorithm compute?

Question 3: How is joint torque $\tau_i$ computed from the wrench $F_i$?

Question 4: Why is the Newton-Euler algorithm considered efficient?

Mathematical exercises

From Practice Exercise 8.5, MODERN ROBOTICS, Practice Exercices

EXERCISE 8:

Link $i$ of an open-chain robot has two frames attached to it, a frame ${b}$ at its center of mass and a frame ${a}$ on the axis of joint $i$, a revolute joint, that drives the link. In the frame ${a}$, the screw axis of the revolute joint is expressed as $S$. In the backward iterations of Newton–Euler inverse dynamics, it was determined that the wrench $F_b$ (expressed in ${b}$) must be applied to the link. What joint torque $\tau_i$ must be applied at joint $i$, in terms of $F_b$, $S$, and the frames ${a}$ and ${b}$?

From Practice Exercise 8.11, MODERN ROBOTICS, Practice Exercices

EXERCISE 9:

Fig 8.6 — **Figure 8.6.** 2D quadcopter and attached pendulum

You are teaching Newton–Euler inverse dynamics, and you are using the 2R robot from the beginning of section 1.2.3.8 (see also Figure 8.6) as an example. Each link has length $L_i$ and the mass of each link is $m_i$, concentrated at a point at the end of the link. You already know the correct dynamics from the Lagrangian derivation. Now you will show how to arrive at the same answer using Newton–Euler. Go through the method step by step, showing intermediate results if it is helpful.

(a) Give $M_i$, $M_{i-1,i}$, $A_i$, $S_i$, $g$, $\mathcal{G}_i$, $\mathcal{V}_0$, $\dot{\mathcal{V}}_0$.

You can assume the frame ${3}$ is coincident with ${2}$ and $\mathcal{F}_{\text{tip}}$ is zero.

(b) Forward iteration: First calculate the transformation, twist, and twist derivative for link 1, then calculate them for link 2.

(c) Backward iteration: First calculate $\mathcal{F}_2$ and $\tau_2$, then calculate $\mathcal{F}_1$ and $\tau_1$. Confirm that your final result agrees with the result in the notes.

Solutions

EXERCISE 8:

Take the dot product of the wrench with the screw axis after they’ve been expressed in the same frame; e.g., in the frame ${b}$:

$$ \tau_i = F_b^{\top} \mathrm{Ad}_{T_ba} S . $$

Exercise 9:

(a) Observe the drawing. Find the transformation matrix $\mathcal{M}_i \in SE(3)$ for each link. $\mathcal{M}_i$ is the transformation from the base frame ${0}$ to the frame ${i}$, which is attached to the center of mass of the $i$-th link, when the robot is in its home configuration.

$$ \mathcal{M}_{1} = \begin{bmatrix} 1 & 0 & 0 & L_1 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \end{bmatrix}, \qquad \mathcal{M}_12 = \begin{bmatrix} 1 & 0 & 0 & L_1 + L_2 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \end{bmatrix}. $$

$\mathcal{M}_{12} \in SE(3)$ is the transformation matrix from the frame ${1}$ (attached to center of mass of link 1) to the frame ${2}$ (attached to the center of mass of link 2), when the arm is in its home configuration.

Find $\mathcal{M}_{2}$ by observing the drawing or by using the equation:

$$\mathcal{M}_{12}=\mathcal{M}_1^{-1}\mathcal{M}_2$$.

$$ \mathcal{M}_{2} = \begin{bmatrix} 1 & 0 & 0 & L_2 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \end{bmatrix}. $$

From observing the drawing, obtain the screw-axis $S_i$ for each joint, expressed in the space-frame:

$$ S_1 = [\,0,\;0,\;1,\;0,\;0,\;0\,]^\top,\qquad S_2 = [\,0,\;0,\;1,\;0,\;-L_1,\;0\,]^\top . $$

$A_i$ is the twist-vector for joint $i$ expressed in the frame ${i}$ when the arm is in its home configuration ($\theta_i=0$). For a simple 2R arm it can be obtained by observing the spatial velocity of frame ${i}$ when rotating about joint $i$ from the home configuration. Alternatively one may use the equation

$$ A_i = \mathrm{Ad}_{\mathcal{M}_i^{-1}}\,S_i . $$

$$ A_1 = [\,0,\;0,\;1,\;0,\;L_1,\;0\,]^\top,\qquad A_2 = [\,0,\;0,\;1,\;0,\;L_2,\;0\,]^\top . $$

Define the gravity vector $g=[\,0,\;g,\;0\,]^\top$ with $g<0$. Define the spatial inertia matrix $G_i$ for each link $i$, expressed in the frame ${i}$. In this case we assume the mass is concentrated as a point mass:

$$ G_1= \begin{bmatrix} 0&0&0&0&0&0\cr 0&0&0&0&0&0\cr 0&0&0&0&0&0\cr 0&0&0&m_1&0&0\cr 0&0&0&0&m_1&0\cr 0&0&0&0&0&m_1 \end{bmatrix}, \qquad G_2= \begin{bmatrix} 0&0&0&0&0&0\cr 0&0&0&0&0&0\cr 0&0&0&0&0&0\cr 0&0&0&m_2&0&0\cr 0&0&0&0&m_2&0\cr 0&0&0&0&0&m_2 \end{bmatrix}. $$

The base is fixed to the ground. It therefore has no velocity. It is however subject to gravity. The gravity vector $g$ needs to be incorporated in $\dot{\mathcal{V}}_0$.

$$ \mathcal{V}_0=[\,0,\;0,\;0,\;0,\;0,\;0\,]^\top,\qquad \dot{\mathcal{V}}_0=[\,0,\;0,\;0,\;0,\;g,\;0\,]^\top . $$

(b) During the forward iteration of Newton–Euler inverse dynamics, we obtain the states and accelerations of the frames attached to each link. Because the velocity and acceleration of each link is influenced by those of its predecessors, we start our calculations at the base and incrementally move out-board until the states and accelerations for each link have been obtained. As a convention we will express velocities $\mathcal{V}_i$ and accelerations $\dot{\mathcal{V}}_i$ for each link $i$ in the frame ${i}$, which is attached to the center of mass of the respective link.

Link 1 states and acceleration:

We now calculate the transformation $T_{01}$ from link 1’s predecessor (frame ${0}$) to itself (frame ${1}$). The equation $T_{01}=\mathcal{M}_1 e^{[A_1]\theta_1}$ takes $\mathcal{M}_1$ (the transformation from the base frame ${0}$ to the frame ${1}$ when the robot is in its home configuration $\theta_1=\theta_2=0$) as a reference point, and incorporates twists (exponential coordinates $A_1\theta_1$) about joint 1 to find the transformation from frame ${0}$ to frame ${1}$ for any given $\theta_1$.

$$ T_{01}= \begin{bmatrix} \cos\theta_1 & -\sin\theta_1 & 0 & L_1\cos\theta_1\cr \sin\theta_1 & \phantom{-}\cos\theta_1 & 0 & L_1\sin\theta_1\cr 0&0&1&0\cr 0&0&0&1 \end{bmatrix}. $$

Calculate the absolute velocity $\mathcal{V}_1$ of the frame ${1}$ expressed in frame ${1}$.

$$ \mathcal{V}_1=[\,0,\;0,\;\dot{\theta}_1,\;0,\;L_1\dot{\theta}_1,\;0\,]^\top . $$

Calculate the absolute acceleration $\dot{\mathcal{V}}_1$ of the frame ${1}$ expressed in frame ${1}$.

$$ \dot{\mathcal{V}}_1=[\,0,\;0,\;\dot{\theta}_1,\;g\sin(\theta_1),\;g\cos(\theta_1)+L_1\dot{\theta}_1,\;0\,]^\top . $$

Link 2 states and acceleration:

We now calculate the transformation $T_{12}$ from link ${2}$’s predecessor (frame ${1}$) to itself (frame ${2}$).

The equation:

$$ T_{12}=\mathcal{M}_{12} e^{[A_2]\theta_2} $$

takes $\mathcal{M}_{12}$ (the transformation from frame ${1}$ to frame ${2}$ when the robot is in its home configuration $\theta_1=\theta_2=0$) as a reference point, and incorporates twists (exponential coordinates $A_2\theta_2$) about joint 2 to find the transformation from frame ${1}$ to frame ${2}$ for any given $\theta_2$.

$$ T_{12}= \begin{bmatrix} \cos\theta_2 & -\sin\theta_2 & 0 & L_2\cos\theta_2 \cr \sin\theta_2 & \phantom{-}\cos\theta_2 & 0 & L_2\sin\theta_2 \cr 0&0&1&0 \cr 0&0&0&1 \end{bmatrix}. $$

Calculate the absolute velocity $\mathcal{V}_2$ of the frame ${2}$ expressed in frame ${2}$.

1.2.3.5: Forward Dynamics of Open Chains

Computing motion from applied torques

Articulated body algorithm (brief overview)

Simulation pipelines

_{Modern Robotics, Chapter 8.5: Forward Dynamics of Open Chains. YouTube video. Available at: https://www.youtube.com/watch?v=L8zpJOxDbh4}

Forward dynamics answers the question:

Given joint torques $\tau$, what are the resulting joint accelerations $\ddot{\theta}$ ?

It complements the Newton–Euler inverse-dynamics algorithm derived in the previous chapter.

Re-using Inverse Dynamics

Bias torque (zero-acceleration call)
Set $\ddot{\theta}=0$ and run inverse dynamics once.
The result is

$$ \tau_{\text{bias}} \;=\; c(\theta,\dot{\theta}) \;+\; g(\theta) \;+\; J^{\top}(\theta) F_{\text{tip}} $$

capturing Coriolis, gravity, and any end-effector wrench terms.
Mass-matrix construction
Call inverse dynamics n times (one per joint):
- For call $i$, set $\ddot{\theta}$ = 1, all other $\ddot{\theta}$ = 0 , and set gravity, $F_{\text{tip}}$, and $\dot{\theta}$ to zero.
- The returned torque vector is column $i$ of the mass matrix $M(\theta)$
- After n calls, assemble all columns to obtain the full symmetric matrix $M(\theta)$.
Solve for accelerations

$$ M(\theta)\,\ddot{\theta} \;=\; \tau \;-\; \tau_{\text{bias}} $$

Use any efficient linear solver to obtain $\ddot{\theta}$.

Simulation Loop

At each simulation time-step:

Compute $\ddot{\theta}$ via forward dynamics.
Integrate $\ddot{\theta}$ → update $\dot{\theta}$ and $\theta$ (Euler, Runge–Kutta, etc.).
Repeat.

Example: A 6-R arm dropped with $\tau = 0$.
With no friction, total mechanical energy (kinetic + potential) should remain nearly constant, an easy check on simulation accuracy.

SUMMARY

Forward dynamics: torques → accelerations.
Inverse dynamics: accelerations → torques.
Needs only n + 1 inverse-dynamics calls (one bias term + n columns of $M$).
Energy conservation is a quick simulator sanity-check.
Forward dynamics underpins trajectory simulation, controller testing, and physics-based animation.

Conceptual Questions

Question 1: What is the primary purpose of the forward-dynamics algorithm for an open-chain robot?

Question 2: How many inverse-dynamics calls are required to build the mass matrix $M(\theta)$ for an n-DOF robot?

Question 3: The “bias torque” $\tau_{\text{bias}}$ (obtained with all $\ddot{\theta}=0$) contains which terms?

Question 4: When constructing column i of $M(\theta)$, which quantity is set to 1 while others are zero?

Question 5: In a simulation with zero torques and zero friction, why should the robot’s total mechanical energy remain nearly constant?

1.2.3.6: Task-Space Dynamics

Mapping dynamics from joint space to end-effector (operational) space

Operational space inertia matrix

Force control and impedance concepts

_{Modern Robotics, Chapter 8.6: Task-Space Dynamics. YouTube video. Available at: https://www.youtube.com/watch?v=iQa01aFgf8U}

So far we have described robot dynamics in joint space (joint motions, forces, and torques).
We can express the same dynamics directly in task space, the space of end-effector twists and wrenches.

Velocity and Acceleration in Task Space

End-effector twist
$$ V = J(\theta)\, \dot{\theta} $$
If the Jacobian $J(\theta)$ is invertible, differentiate once: $$ \dot{V} = J(\theta)\, \ddot{\theta} + \dot{J}(\theta)\, \dot{\theta} $$
Solve for $(\dot{\theta}, \ddot{\theta})$ in terms of $(V, \dot{V})$ and substitute into the joint-space dynamics.

Task-Space Dynamic Equation

The resulting wrench equation becomes

$$ \boxed{F_{\text{task-space dynamics}} \;=\; \Lambda(\theta)\, \dot{V} \;+\; \eta(\theta, V) \;+\; F_{\text{tip}}} $$

where

$\Lambda(\theta)$: task-space mass matrix (a mapping from end-effector acceleration to wrench)
$$ \Lambda(\theta) \;=\; J^{-T} M(\theta) J^{-1} $$
$\eta(\theta, V)$: task-space bias wrench (Coriolis + gravity terms expressed at the end effector)
$$ \eta(\theta, V) \;=\; J^{-T} h(\theta, J^{-1} V)- \Lambda(\theta) \dot{J}(\theta) J^{-1} V $$
$F_{\text{tip}}$: any additional wrench applied directly by (or to) the tool.

Important: $\Lambda$ and $\eta$ are functions of joint positions $\theta$, not directly of the task-space pose $X$, because multiple joint configurations can map to the same end-effector pose.

Why Task-Space Dynamics?

Intuitive control: force/impedance control at the tool tip
Decoupling: $\Lambda(\theta)$ acts like an “apparent mass” felt at the end effector
Planning & compliance: specify motions and forces directly where the robot interacts with its environment

Task-space dynamics provides a powerful framework for advanced control strategies such as operational-space control, impedance control, and hybrid position/force control.

Conceptual Questions

Question 1: In task-space dynamics, what does the matrix $\Lambda(\theta)$ represent?

Question 2: Which term appear on the right-hand side (total force balance needed to achieve the motion in task space) of the task-space equation $F_{\text{ee}} = \Lambda(\theta)\dot{V} + \eta(\theta,V) + F_{\text{tip}}$?

Question 3: Why are $\Lambda$ and $\eta$ expressed as functions of joint positions $\theta$ rather than the end-effector pose $X$?

Question 4: Which condition must hold to express $\dot{V}$ directly in terms of $\ddot{\theta}$ and $\theta$?

Question 5: For a planar 2-DOF end effector, let: $ J(\theta)=\begin{bmatrix}1&0\cr 0&2\end{bmatrix}, M(\theta)=\begin{bmatrix}2&0\cr 0&8\end{bmatrix}. $ The task-space mass matrix is defined by $\Lambda(\theta)=J(\theta)^{-T}\,M(\theta)\,J(\theta)^{-1}$. What is $\Lambda(\theta)$?

Mathematical exercises

From Practice Exercise 8.1.(h), MODERN ROBOTICS, Practice Exercices

EXERCISE 10: (this exercise is a continuation of exercise 1)

(h) Now visualize the configuration-dependent end-effector mass matrix $\Lambda(\theta)$, where the “end-effector” is considered to be at the point $(x_2, y_2)$, the location of the center of mass of the second link. For a unit circle of accelerations $(\ddot{x}_2,\ddot{y}_2)$, consider the ellipse of linear forces that are required to be applied at the end-effector to realize these accelerations. How does the orientation of this ellipse change as $\theta_1$ changes? How does the shape change as $\theta_2$ increases from zero to infinity when $\theta_1=0$? Provide a drawing for the case $\theta_1=0$. If you have access to symbolic computation software (e.g., Mathematica), you can use the Jacobian $J(\theta)$ satisfying

$$ \begin{bmatrix} \dot{x}_2 \cr \dot{y}_2 \end{bmatrix} = J(\theta)\,\dot{\theta}\, . $$

to calculate $\Lambda(\theta) = J^{-T}(\theta),M(\theta),J^{-1}(\theta)$ for the case $\theta_1 = 0$. If you do not have access to symbolic computation software, you can plug in numerical values for $I_1$, $I_2$, $m_1$, $m_2$, and $L_1$ (make them all equal to $1$, for example) to say something about how $\Lambda$ changes (and therefore how the ellipse changes) as $\theta_2$ goes from zero to infinity while $\theta_1 = 0$.

From Practice Exercise 8.8, MODERN ROBOTICS, Practice Exercices

EXERCISE 11:

Consider the four equivalent forms of dynamics shown below:

$$ \tau \;=\; M(\theta)\,\ddot{\theta} \;+\; h(\theta,\dot{\theta}) \;+\; J^{T}(\theta)\,F_{\text{tip}} \tag{8.1} $$

$$ \tau \;=\; M(\theta)\,\ddot{\theta} \;+\; c(\theta,\dot{\theta}) \;+\; g(\theta) \;+\; J^{T}(\theta)\,F_{\text{tip}} \tag{8.2} $$

$$ \tau \;=\; M(\theta)\,\ddot{\theta} \;+\; C(\theta,\dot{\theta})\,\dot{\theta} \;+\; g(\theta) \;+\; J^{T}(\theta)\,F_{\text{tip}} \tag{8.3} $$

$$ \tau \;=\; M(\theta)\,\ddot{\theta} \;+\; \dot{\theta}^{T}\Gamma(\theta)\dot{\theta} \;+\; g(\theta) \;+\; J^{T}(\theta)\,F_{\text{tip}} \tag{8.4} $$

(a) List the variables common to all of the equations, what they represent, their dimension, how they are derived, and any constraints they must always follow or properties they must satisfy.

(b) For the unique variables in each of the equations, describe what they represent and provide the dimension.

Solutions

EXERCISE 8: (this exercise is a continuation of exercise 1)

The Jacobian relating joint velocities $\dot{\theta}$ to the velocity of the end-effector $(\dot{x}_2,\dot{y}_2)$ is $$ J(\theta)= \begin{bmatrix} -\theta_2 \sin\theta_1 & \ \cos\theta_1 \cr \ \theta_2 \cos\theta_1 & \ \sin\theta_1 \end{bmatrix}. $$

and the end-effector mass matrix is $$ \Lambda(\theta)=J^{-T} M J^{-1}. $$

We are interested in the ellipse $\Lambda(\theta)[\ddot{x}_2\ \ddot{y}_2]^T$ (in the $(f_x,f_y)$ space) when the end-effector acceleration is a unit vector.
The orientation of this ellipse rotates with $\theta_1$, so we can just consider the case for a particular constant $\theta_1$, i.e., $\theta_1=0$ (the end-effector is at $(x_2,y_2)=(\theta_2,0)$). In this case, a force applied to the end-effector in the $f_x$ direction acts to extend or retract joint 2 while a force in the $f_y$ direction acts to rotate the robot about joint 1.

Evaluating $\Lambda(\theta)$ with $\theta_1=0$, we get the diagonal matrix $$ \begin{bmatrix} m_2 & 0\cr 0 & \dfrac{(I_1+I_2+m_1 L_1^{2}+m_2 \theta_2^{2})}{\theta_2^{2}} \end{bmatrix} \;=\; \begin{bmatrix} m_2 & 0\cr 0 & \dfrac{(k+m_2 \theta_2^{2})}{\theta_2^{2}} \end{bmatrix}, $$ where $k$ is a positive constant. Since the matrix is diagonal, the principal axes of the ellipse $\Lambda(\theta)[\ddot{x}_2\ \ddot{y}_2]^T$ (where the end-effector acceleration is a unit vector) are aligned with the $f_x$ and $f_y$ axes and the lengths of the principal components are the entries along the diagonal.

The apparent mass at the end-effector in the radial ($x$) direction is $m_2$, i.e., it is independent of $\theta_2$. The apparent mass in the tangential ($y$) direction depends on $\theta_2$, however. As $\theta_2$ approaches zero from above, the bottom-right component of $\Lambda$ approaches infinity. This means large $f_y$ forces are needed to accelerate the tip in the $y$ direction. This is because the torque about joint 1 provided by a force $f_y$ through the end-effector tends to zero as the end-effector approaches joint 1, and therefore $f_y$ must become large to generate the angular acceleration of the inertia about joint 1 needed to generate a modest $\ddot{y}_2$ acceleration. Accordingly, the principal axis of the end-effector mass ellipse in the $f_y$ direction becomes large (Figure 8.8). As $\theta_2$ approaches infinity, the bottom-right element of $\Lambda$ drops to $m_2$, and the end-effector mass matrix ellipse approaches a circle: the end-effector feels like a mass $m_2$ in every direction.

Fig 8.8 — **Figure 8.8.**The end-effector mass matrix $\Lambda(\theta)$ represented as the ellipse of forces that must be applied to the end-effector to create a circle of accelerations $(\ddot{x}_2,\\ \ddot{y}_2)$. As $\theta_2$ goes to infinity, the magnitude of the force required to generate a unit acceleration $(0,\\ \ddot{y}_2)$ approaches $m_2$, i.e., the robot feels like a point mass with mass $m_2$.

EXERCISE 9:

(a)

$\tau$: the torque or force at each of the joints represented by the generalized coordinates. Dimensions are $n \times 1$ array.
$M(\theta)$: the configuration dependent mass matrix. Dimensions are $n \times n$ matrix. $M$ must be symmetric and positive definite.
$\theta$: The generalized coordinates for the joints. Dimensions are $n \times 1$ array.
$\ddot{\theta}$: The acceleration of the joints represented by the generalized coordinates. Dimensions are $n \times 1$ array.
$J(\theta)$: The Jacobian (depends on configuration $\theta$). Dimensions are $n \times n$ matrix.
$\mathbf{F}_{\text{tip}}$: The force applied at the tip of the robot. Dimensions are $n \times 1$ array.

(b)

8.1: The most general representation, and $h(\theta,\dot{\theta})$ is an $n \times 1$ array that contains the centripetal, coriolis, and gravity terms.
8.2: $c(\theta,\dot{\theta})$ is an $n \times 1$ array that contains the centripetal and coriolis terms, and $g(\theta)$ is an $n \times 1$ array that contains the gravity terms.
8.3: $C(\theta,\dot{\theta})$ is the $n \times n$ Coriolis matrix.
8.4: $\Gamma(\theta)$ is the $n \times n \times n$ Christoffel matrix. Emphasizes that the Coriolis and centripetal (velocity product) terms are quadratic in the velocity and that $\Gamma$ depends only on $\theta$.

Constrained Dynamics

Dealing with closed kinematic chains and contact

Constraint forces and Lagrange multipliers

Applications in legged robots and grasping

_{Modern Robotics, Chapter 8.7: Constrained Dynamics. YouTube video. Available at: https://www.youtube.com/watch?v=E6Yp6DwJh24}

Real robots often operate under constraints:

Loop-closure constraints (parallel robots, humanoid feet on ground, gripped objects)
Non-holonomic constraints (wheeled vehicles)
Surface-contact constraints (e.g., an end-effector sliding on a whiteboard)

Describing the Constraints

Configuration constraints
$$ b(\theta(t))=0 $$
Velocity (Pfaffian) constraints
$$ \dot{b}=A(\theta)\,\dot{\theta}=0 \quad\text{with }A\in\mathbb{R}^{k\times n} $$
- $k$ independent constraints on $n$ joints
Work-less assumption: forces that enforce the constraints do no mechanical work.

Splitting Joint Torques

Joint torques are decomposed into
$$ \tau = \tau_{\text{motion}} + \tau_{\text{con}} $$ where

$\tau_{\text{motion}}$ drives the robot,
$\tau_{\text{con}} = A^{T}\lambda$ enforces the constraints (no work).

Here $\lambda\in\mathbb{R}^{k}$ are Lagrange multipliers.

Constrained Equations of Motion

Unconstrained dynamics:
$$ M(\theta)\,\ddot{\theta} + h(\theta,\dot{\theta}) = \tau $$

Add constraints ⇒ solve the coupled system $$ \tau = M(\theta)\,\ddot{\theta} + h(\theta,\dot{\theta})+ A(\theta)^{\top}\lambda, $$

$$ A(\theta)\,\dot{\theta} = 0 \quad\rightarrow\quad A(\theta)\,\ddot{\theta} + \dot A(\theta,\dot{\theta})\,\dot{\theta} = 0 $$

These give $n+k$ equations in $n+k$ unknowns: the $k$ multipliers $lambda$ and either the $n$ joint accelerations $\ddot{\theta}$ (for constrained forward dynamics, with $\tau$ given) or the $n$ joint torques $\tau$ (for constrained inverse dynamics, with $\ddot{\theta}$ given).

Projection Matrix

Define an $n\times n$ projector
$$ P(\theta)=I - A^{T}\bigl(A\,M^{-1}A^{T}\bigr)^{-1}A\,M^{-1} $$

rank$(P)=n-k$
projects any vector onto the constraint-free subspace

Constrained Inverse Dynamics

Project the dynamics to eliminate $\lambda$:

$$ P\,\tau = P\,[M(\theta)\,\ddot{\theta} + h(\theta,\dot{\theta})] $$

Solving steps:

Insert desired $(\theta,\dot{\theta},\ddot{\theta})$ → compute motion-producing torque $P\tau$.
Any additional torque of the form $A^{\top}\lambda$ can be added without altering motion (pure constraint force).

SUMMARY

Handles closed-loop mechanisms, wheeled bases, surface contacts
Forms the basis for hybrid motion–force control (Chapter 11)
Separates motion control (projected torques) from force control (constraint torques)

Constraint dynamics lets us model and control robots that must simultaneously move and push/pull against their environment.

Conceptual Questions

Question 1: What is the purpose of the projection matrix $P(\theta)$ in constrained dynamics?

Question 2: Constraint torques are expressed as $\tau_{\text{con}} = A^{\!\top}\lambda$. What does the vector $\lambda$ represent?

Question 3: The velocity constraint $A(\theta)\dot{\theta}=0$ is called a Pfaffian constraint when it is:

Question 4: Why do we assume constraint forces are “workless” in this formulation?

Question 5: In constrained inverse dynamics, what additional torque can we add without affecting the robot’s motion?

Actuation, Gearing & Friction

Modeling motors and gear ratios

Viscous and Coulomb friction

Backdrivability and actuator limitations

_{Modern Robotics, Chapter 8.9: Actuation, Gearing & Friction. YouTube video. Available at: https://www.youtube.com/watch?v=w1kYLT3pETc}

Robot dynamics so far assumed “ideal” joint torques.
Real joints use actuators + transmissions whose own dynamics can dominate the system.

Direct-Drive vs. Geared Actuation

Direct-drive robots: motor outputs torque directly at the joint without any gearing –> uncommon (motors too fast, too little torque).
Practical robots add a gearhead or other transmission (belts, chains, etc.) to trade speed for torque.

Anatomy of a Typical Geared Motor

Rotor (spins with the motor shaft)
Stator (motor housing)
Gearhead with ratio $G>1$
- Speed ↓ by factor $G$ , torque ↑ by factor $G$ (less when losses included).
Encoder measures motor (and thus joint) position.

Apparent Rotor Inertia

Rotor spins $G$ times faster than the link.
Rotor kinetic energy
$$ E_{K,\text{rotor}}=\tfrac12 \, I_{\text{rotor}}\,(G\,\dot{\theta})^{2} \;=\;\tfrac12\,(G^{2}I_{\text{rotor}})\,\dot{\theta}^{2} $$
Apparent inertia about the joint axis is $G^{2}I_{\text{rotor}}$.
- Even a small real inertia becomes significant when $G$ is large (common ratios: 10 – 100+).

Effect on the Mass Matrix

Example 2-R arm:
- Gear ratio 10 → moderate increase in diagonal entries.
- Gear ratio 100 → diagonal terms dominate; off-diagonal coupling & configuration dependence shrink.
High gearing ⇒ robot behaves like $n$ nearly independent joints.

Modified Newton–Euler Algorithm

Treat the rotor as part of link $i$ (mass & enhanced inertia) and the stator as part of link $i-1$.
Recursive inverse-dynamics still applies, but motor torque must accelerate both rotor & link.
Controller often commands motor current ∝ required torque.

Joint Friction

Gearheads introduce significant friction (often grows with $G$).
Add simple friction models (Coulomb, viscous, etc.) to torque commands for better simulation and control accuracy.

SUMMARY

High gear ratios amplify rotor inertia (${\sim}G^{2}$), reshape the mass matrix, reduce dynamic coupling, and demand friction modeling.
Accurate actuator/transmission models are essential for realistic simulation and torque-based control.

Conceptual Questions

Question 1: Why are direct-drive joints (no gearing) uncommon in most industrial robots?

Question 2: If the gear ratio is $G$, the apparent rotor inertia reflected to the joint scales as ...

Question 3: As gear ratios become very large, what generally happens to the off-diagonal elements of the mass matrix $M(\theta)$?

Question 4: With very high gear ratios, the robot’s joint dynamics increasingly resemble...

Question 5: When adding friction to the joint-torque model, which statement is generally true?

Mathematical Development Questions

For additional practice and deeper derivations, see the PDF: Modern Robotics – Practice Exercises. The dynamics problems begin on page 55, and the solutions are on page 62.

Additional exam-level exercises covering other chapters of the book Modern Robotics and their solutions are also available here:

Credits

This course page was created by Shujiro Shobayashi, MSc in Robotics at EPFL, under supervision of Aude Billard, and funded by IEEE RAS and EPFL. We thank Kevin Lynch for agreeing to use material of his course.

Additional Resources

Books

Modern Robotics: Mechanics, Planning, and Control,” by Kevin Lynch and Frank Park, Cambridge University Press 2017.
Springer Handbook of Robotics (Chapter 3. Dynamics)

Videos

Contents shared by Prof. Kevin Lynch, Professor of Mechanical Engineering at Northwestern University.

Exercices

Modern Robotics — Practice Exercises (PDF) (Dec 6, 2018).
Supplemental to Modern Robotics: Mechanics, Planning, and Control (Cambridge University Press, 2017).
Contributions: Tito Fernandez, Kevin M. Lynch, Huan Weng, Zack Woodruff.
Dynamics exercises start on p. 55; solutions on p. 62.
https://hades.mech.northwestern.edu/images/e/ef/MR_practice_exercises.pdf

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Dynamics

Prerequisites

General Motivation

Course Content

Why Dynamics?

(Part1) The Lagrangian Formulation of Dynamics

The Power of the Lagrangian Approach

From Theory to General Dynamic Equation

(Part2) The Lagrangian Formulation of Dynamics

Velocity-Product Terms: What Are They?

Writing the Velocity-Product Term Mathematically

SUMMARY

Understanding the Mass Matrix

From Point Mass to Robot Arm

Key Properties of the Mass Matrix

Physical Interpretation

SUMMARY

(Part1) Dynamics of a Single Rigid Body

Rigid Body Basics

Kinematics of the Point Masses

Newton–Euler Dynamics

Inertia Matrix

From Discrete to Continuous Mass

Kinetic Energy of a Rigid Body

Principal Axes and Diagonalization

SUMMARY

(Part2) Dynamics of a Single Rigid Body – Spatial Inertia and Equations of Motion

Key Concepts

SUMMARY

Inverse Dynamics with Newton-Euler

SUMMARY

1.2.3.5: Forward Dynamics of Open Chains