Educator Resources

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Four teaching modules for biomechanics, sport science and control courses, with learning objectives, reading paths and exercises with worked solutions.

How to Use These Modules

These modules are for instructors who teach biomechanics, sport science or control. Each one fits a single class session or a short problem set. Each module gives:

  • Learning objectives that say what a student should be able to do.
  • A reading path through pages that already exist on this site.
  • Exercises built on worked examples and interactive tools on this site.
  • Worked solutions, folded so you can hand out the exercises first.

Every number in a solution comes from one of three places: a cited page, a cited interactive tool, or a derivation shown in full on this page. Where an exercise uses round numbers that are not on another page, the solution says so and shows every step. These numbers are chosen to make the arithmetic easy to follow. They are not measurements of golfers or clubs.

Module Suggested Course Main Tool or Example
1. Orientation and Rotation Representations Biomechanics, motion capture Rotation Representation Converter
2. Constraint Torques at the Wrist Biomechanics, multibody dynamics Cardan and Torque Projection Demonstration
3. Energy Accounting in the Proximal-Distal Chain Sport science, mechanics Proximal-Distal Model Workbench
4. Control-Affine Dynamics and Counterfactuals Control, dynamical systems Worked examples in the theory pages

The modules can be taught in any order. Module 1 helps before Module 2, because the wrist model uses rotation matrices.

Reuse and Licence

You may copy, adapt and share the text of these modules, including the exercises and solutions, under the Creative Commons Attribution 4.0 International licence (CC BY 4.0). Credit “AffineDrift (affinedrift.com)”, link to the licence, and say if you changed anything. This applies to classroom handouts, course sites and commercial course material alike.

The code these modules call, such as the rotation converter and the Python functions under src/, is released under the MIT License. Figures and illustrations on the linked pages are not covered by CC BY 4.0; ask before reusing them. The full terms are in COPYRIGHT.md and LICENSE-CONTENT.

Module 1: Orientation and Rotation Representations

Motion-capture systems, simulators and papers describe the same orientation in different ways. Students who can convert between them, and who know where each one breaks, avoid a large class of analysis errors.

Learning Objectives

  • Convert one orientation between a rotation matrix, a unit quaternion, ZYX Euler angles, an axis-angle pair and a rotation vector.
  • Explain why \(q\) and \(-q\) describe the same rotation.
  • Show by hand why ZYX Euler angles lose one angle at a pitch of \(90^\circ\), and say what the converter reports there.
  • State what an orientation does not tell you, such as angular velocity or joint location.

Reading Path

  1. Rotation Representation Converter: read the Background and Mathematical Reference sections (about 20 minutes).
  2. Rotation Representations: A Cross-Reference Guide: the derivations behind each conversion (about 40 minutes).
  3. Rotations and SE(3) in The Geometry of Motion, for a longer textbook treatment (optional).

Exercises

All exercises use the converter’s conventions: right-handed active rotations, Hamilton quaternions ordered \((w,x,y,z)\), and intrinsic ZYX angles with \(R=R_z(\psi)R_y(\theta)R_x(\phi)\).

Exercise 1.1 (Quarter Turn). By hand, write the rotation matrix, quaternion, axis-angle pair, rotation vector and ZYX angles for a \(90^\circ\) rotation about \(z\). Then press the Rz(90°) preset and check every panel.

Exercise 1.2 (Two Quaternions, One Rotation). Enter the quaternion \((0,0,0,-1)\). Which quaternion does the converter report, and why is that not a different rotation?

Exercise 1.3 (The Arbitrary Preset). The preset labelled \(\hat n=[1,1,1]/\sqrt3\), \(47.3^\circ\) rotates about the diagonal axis. Before pressing it, predict the trace of \(R\), the quaternion and the rotation vector. Then press the preset and compare.

Exercise 1.4 (Gimbal Lock). Enter yaw \(\psi=30^\circ\), pitch \(\theta=90^\circ\) and roll \(\phi=10^\circ\). Read the matrix panel. Then enter \(\psi=20^\circ\), \(\theta=90^\circ\), \(\phi=0^\circ\). Explain what you see, and predict the Euler angles the converter reports if you press Use this R.

Worked Solutions

With \(c=\cos 90^\circ=0\) and \(s=\sin 90^\circ=1\),

\[ R_z(90^\circ)=\begin{bmatrix}0&-1&0\\1&0&0\\0&0&1\end{bmatrix}. \]

The quaternion for angle \(\theta\) about unit axis \(\hat n\) is \((\cos\tfrac\theta2,\ \hat n\sin\tfrac\theta2)\). Here \(\cos 45^\circ=\sin 45^\circ=0.7071\), so \(q=(0.7071,\,0,\,0,\,0.7071)\). The axis-angle pair is \(\hat n=(0,0,1)\), \(\theta=90^\circ\). The rotation vector is \(\rho=\theta\hat n=(0,\,0,\,\pi/2)=(0,\,0,\,1.5708)\) rad. The ZYX angles are \(\psi=90^\circ\), \(\theta=0\), \(\phi=0\). The converter shows the same values to more digits.

The converter reports \((0,0,0,1)\). The quaternion-to-matrix formula on the converter page is quadratic in the components, so \(q\) and \(-q\) give the same matrix. Both describe a \(180^\circ\) turn about \(z\). The converter picks the sign with \(w\ge 0\) so that the reported angle lies in \([0,\pi]\). At exactly \(180^\circ\), \(w=0\), so that rule cannot decide; the converter’s largest-component extraction returns \(z=+1\). The axis sign is a free choice here: \(+z\) and \(-z\) give the same half-turn. This is a reporting convention, not a continuity rule. For a sampled swing you must choose signs so that consecutive quaternions stay close.

For any rotation, \(\operatorname{tr}R=1+2\cos\theta\). With \(\theta=47.3^\circ\), \(\cos\theta=0.6782\), so \(\operatorname{tr}R=2.3563\). The axis treats \(x\), \(y\) and \(z\) alike, so the three diagonal entries are equal: each is \(2.3563/3=0.7854\). The converter’s matrix is

\[ R=\begin{bmatrix}0.7854&-0.3170&0.5316\\0.5316&0.7854&-0.3170\\-0.3170&0.5316&0.7854\end{bmatrix}. \]

The quaternion is \(w=\cos 23.65^\circ=0.9160\) and each vector component is \(\sin 23.65^\circ/\sqrt3=0.2316\), so \(q=(0.9160,\,0.2316,\,0.2316,\,0.2316)\). The angle in radians is \(47.3\pi/180=0.8255\), and each rotation-vector component is \(0.8255/\sqrt3=0.4766\) rad.

Both entries give the same matrix. Its first row is \((0,\,-0.3420,\,0.9397)=(0,\,-\sin 20^\circ,\,\cos 20^\circ)\). To see why, multiply out the product at \(\theta=90^\circ\):

\[ R_y(90^\circ)R_x(\phi)=\begin{bmatrix}0&\sin\phi&\cos\phi\\0&\cos\phi&-\sin\phi\\-1&0&0\end{bmatrix}, \]

\[ R_z(\psi)R_y(90^\circ)R_x(\phi)=\begin{bmatrix}0&-\sin(\psi-\phi)&\cos(\psi-\phi)\\0&\cos(\psi-\phi)&\sin(\psi-\phi)\\-1&0&0\end{bmatrix}. \]

Only the difference \(\psi-\phi\) appears. Yaw \(30^\circ\) with roll \(10^\circ\) and yaw \(20^\circ\) with roll \(0^\circ\) both give \(\psi-\phi=20^\circ\). When the matrix is the input, the converter detects the lock and sets roll to zero, so it reports \(\psi=20^\circ\), \(\theta=90^\circ\), \(\phi=0^\circ\) with a gimbal-lock warning. The body can still rotate in every direction. Only this choice of three angles has stopped describing it uniquely.

Module 2: Constraint Torques at the Wrist

A two-axis wrist model can carry a torque along a direction in which it does not allow rotation. This module separates that constraint torque from actuation, energy and grip claims.

Learning Objectives

  • Find the direction of an ideal universal joint’s reaction couple and check that it does no work on the allowed motion.
  • Show with a small example that a constraint reaction can change when the actuator input changes.
  • Compute a club’s transverse inertia about the grip from point-mass and thin-rod terms.
  • Use the Cardan demonstration to tell a supported driveshaft apart from a wrist, and explain why a torque ratio above one is not an efficiency above one.
  • Derive how a constant torque error grows into an angle error.

Reading Path

  1. Constraint Torques at the Wrist: read through “The Grip Angle Hypothesis” and “The Cardan Driveshaft Analogy” (about 60 minutes).
  2. The Null Space of the Constraint Jacobian: the motion a constraint still allows (about 30 minutes).
  3. Screw Theory Reference: wrenches, twists and reciprocity (optional).
  4. Cardan and Torque Projection Demonstration: the interactive tool used in Exercises 2.3 and 2.4.

Exercises

Exercise 2.1 (Reaction Direction). Model the wrist as \(R=R_x(\phi)R_y(\psi)\) with allowed rotation directions \(\mathbf a=\mathbf e_x\) and \(\mathbf b=R_x(\phi)\mathbf e_y\) in the forearm frame. At \(\phi=30^\circ\), \(\psi=20^\circ\), find the unit reaction direction \(\mathbf n_c\) in the forearm frame and in the hand frame. Check that it does no work on the allowed motion.

Exercise 2.2 (A Reaction That Depends on the Input). Take \(M=\begin{bmatrix}2&0&0.5\\0&2&0\\0.5&0&2\end{bmatrix}\), the constraint \(A=(0,0,1)\), inputs \(B=(\mathbf e_x,\mathbf e_y)\) and no bias. Apply the input \((u,0)\). Find \(\ddot q_x\) and the multiplier \(\lambda\) two ways: by substitution, and with the wrist page’s formula \(\partial\lambda/\partial u=-(AM^{-1}A^T)^{-1}AM^{-1}B\).

Exercise 2.3 (The Cardan Demonstration). Open the demonstration and set Demo Angle to \(30^\circ\). Read the Angular Velocity Ratio and Torque Transmission Ratio in Model Information at Input Shaft Phase \(0^\circ\) and \(60^\circ\). Derive both readings, and explain why a torque ratio above one does not mean the joint creates energy.

Exercise 2.4 (Grip-Point Inertia). With the demonstration’s default club properties (200 g head at 0.85 m, 100 g shaft of length 1.00 m), the tool reports \(I_\alpha\). Derive it. Then explain why the tool’s \(I_\gamma=0.5\,I_\alpha\) is not a measured club property.

Exercise 2.5 (How a Torque Error Grows). A constant torque error \(\eta\) with standard deviation \(\sigma_\tau\) acts on a single inertia \(I\) for time \(T\), starting from zero angle and velocity error. Derive the angle standard deviation \(\sigma_\theta\). Evaluate it for \(\sigma_\tau=0.15\) N m, \(T=0.010\) s and \(I=0.005\) or \(0.00015\) kg m², the illustrative values on the wrist page.

Worked Solutions

The ideal reaction couple must do no work on any allowed rotation, so it is perpendicular to both \(\mathbf a\) and \(\mathbf b\): \(\mathbf n_c=\mathbf a\times\mathbf b=R_x(\phi)\mathbf e_z=(0,\,-\sin\phi,\,\cos\phi)\). At \(\phi=30^\circ\) this is \((0,\,-0.500000,\,0.866025)\) in the forearm frame.

In the hand frame, \({}^H\mathbf n_c=R^T\mathbf n_c=R_y(\psi)^TR_x(\phi)^TR_x(\phi)\mathbf e_z=R_y(\psi)^T\mathbf e_z=(-\sin\psi,\,0,\,\cos\psi)\). At \(\psi=20^\circ\) this is \((-0.342020,\,0,\,0.939693)\). Both match the wrist page.

Check: \(\mathbf n_c\cdot\mathbf a=0\) because the first component is zero, and \(\mathbf n_c\cdot\mathbf b=-\sin\phi\cos\phi+\cos\phi\sin\phi=0\), with \(\mathbf b=(0,\cos\phi,\sin\phi)\). So the couple does no work on any allowed relative rotation \(\mathbf a\dot\phi+\mathbf b\dot\psi\). The direction moves with the wrist: it is neither a fixed forearm axis nor a fixed hand axis.

By substitution. The constraint forces \(\ddot q_z=0\). The equations \(M\ddot q=Bu+A^T\lambda\) read

\[ 2\ddot q_x+0.5\ddot q_z=u,\qquad 2\ddot q_y=0,\qquad 0.5\ddot q_x+2\ddot q_z=\lambda. \]

With \(\ddot q_z=0\), the first gives \(\ddot q_x=u/2\), and the third gives \(\lambda=0.5\cdot u/2\), so \(\lambda=u/4\).

By the formula. The \(x\)–\(z\) block of \(M\) has determinant \(2\cdot2-0.5^2=3.75\), so \(AM^{-1}A^T=2/3.75=0.5333\) and the \(x\) entry of \(AM^{-1}B\) is \(-0.5/3.75=-0.1333\). The \(y\) entry is zero. Then \(\partial\lambda/\partial u=-(-0.1333)/0.5333=0.25\), which again gives \(\lambda=u/4\).

A changed input changes the reaction at the same state. This disproves the claim that every constraint load is drift that the input cannot affect. It does not show that any desired reaction can be chosen freely.

The wrist page gives the supported-shaft speed ratio \(\dot\chi/\dot\varphi=\cos\delta/(1-\sin^2\delta\sin^2\varphi)\), and the demonstration computes exactly this. With \(\delta=30^\circ\), \(\cos\delta=0.866025\) and \(\sin^2\delta=0.25\).

  • At phase \(0^\circ\), \(\sin\varphi=0\), so the speed ratio is \(0.8660\) and the torque ratio, its reciprocal, is \(1.1547\).
  • At phase \(60^\circ\), \(\sin^2\varphi=0.75\), so the denominator is \(1-0.25\cdot0.75=0.8125\). The speed ratio is \(0.866025/0.8125=1.0659\) and the torque ratio is \(0.9382\).

For a lossless joint with fixed supports, input power equals output power: \(\tau_{\rm in}\dot\varphi=\tau_{\rm out}\dot\chi\). Torque goes up exactly when speed goes down, so the product of the two ratios is one. This is a driveshaft held in two bearings. It is not the freely moving wrist, and the demonstration’s phase is not a wrist angle.

About the grip, a point head contributes \(mr^2\) and a thin uniform shaft contributes \(\tfrac13mL^2\):

\[ I_\alpha=0.200(0.85)^2+\tfrac13(0.100)(1.00)^2=0.1445+0.0333=0.177833\ \mathrm{kg\,m^2}, \]

which the tool shows as \(0.1778\) kg m². This matches the wrist page. The tool then sets \(I_\gamma=0.5\,I_\alpha\), and its source labels this an illustrative ratio. A real shaft-axis inertia needs the shaft radius, the head’s own inertia and its offset from the shaft. As the wrist page says, an assumed ratio does not become a measured property by appearing in a simulator.

With constant \(\eta\), the angular acceleration is \(\eta/I\). Integrating twice from zero gives \(\delta\omega(T)=T\eta/I\) and \(\delta\theta(T)=T^2\eta/(2I)\). The standard deviation scales the same way: \(\sigma_\theta=T^2\sigma_\tau/(2I)\).

  • With \(I=0.005\): \(\sigma_\theta=(0.010)^2(0.15)/(2\cdot0.005)=0.0015\) rad.
  • With \(I=0.00015\): \(\sigma_\theta=(0.010)^2(0.15)/(2\cdot0.00015)=0.0500\) rad.

The ratio is \(0.005/0.00015=33.333\), as in the table on the wrist page. Both inertias are illustrative and unmeasured. The result shows how sensitivity scales with inertia under these assumptions. It is not a clubface error or a grip comparison.

Module 3: Energy Accounting in the Proximal-Distal Chain

Coaches and textbooks often say energy “flows” from the body to the club. This module teaches students to check that kind of claim with a clear system boundary, a consistent model and a stated falsifier.

Learning Objectives

  • Show how two mass models of the same club give different kinetic energies for the same motion.
  • Compute the net force and moment from two hand forces, and recognize a pure couple.
  • Tell an instantaneous counterfactual apart from a forward one, and say what result would count against a stated mechanism.

Reading Path

  1. How a Golf Swing Carries Energy: the visual narrative (about 30 minutes).
  2. Explore the Proximal–Distal Models: read “What Would Count Against the Mechanism?” and “Read Work and Energy With the Same Model” (about 40 minutes). Students who install the workbench can run the guided experiments it lists.
  3. Proximal-to-Distal Energy Transfer: the technical summary (optional).

Exercises

Exercise 3.1 (Two Models of One Club). One part of the workbench puts the shaft mass \(m_2\) at the club tip with the head mass \(m_c\). Another part treats the shaft as a uniform rod. With the arm still and the club turning at \(\omega\), write the club’s kinetic energy in each model. Evaluate both for \(m_2=3\) kg, \(m_c=1\) kg, \(L_2=1\) m and \(\omega=2\) rad/s, the test values on the workbench page.

Exercise 3.2 (Two Hands, One Couple). The hands are 0.1 m apart along the shaft, so \(\mathbf d=\mathbf r_R-\mathbf r_L=(0,0,0.1)\) m. The right hand pushes on the club with \(\mathbf f_R=(10,0,0)\) N and the left hand with \(\mathbf f_L=(-10,0,0)\) N. There are no contact moments. Find the net force and the moment about the midpoint. Does the moment change if you take it about the right hand instead? These round numbers are chosen for this exercise only.

Exercise 3.3 (What Would Count Against It?). A student turns off the wrist torque at one instant, sees no change in the club’s acceleration at that instant, and concludes that the wrist does nothing. Using the workbench page, explain what is wrong with this test and describe a better one.

Worked Solutions

With the arm still, the club spins about the grip. A point mass at distance \(L_2\) has speed \(L_2\omega\), and a uniform rod has inertia \(\tfrac13m_2L_2^2\) about its end. So

\[ T_{\mathrm{tip}}=\tfrac12(m_2+m_c)L_2^2\omega^2,\qquad T_{\mathrm{rod}}=\tfrac12(m_2/3+m_c)L_2^2\omega^2. \]

With the given values, \(T_{\mathrm{tip}}=\tfrac12(4)(1)(4)=8\) J and \(T_{\mathrm{rod}}=\tfrac12(2)(1)(4)=4\) J. The two parts of the same tool disagree by a factor of two on the same state, as the workbench page warns. Any energy-transfer comparison must fix this mismatch first.

The net force is \(\mathbf F=\mathbf f_R+\mathbf f_L=(0,0,0)\). The moment about the midpoint, from the workbench page, is

\[ \mathbf M_{\mathrm{mid}}=\frac{\mathbf d}{2}\times(\mathbf f_R-\mathbf f_L) =(0,0,0.05)\times(20,0,0)=(0,1,0)\ \mathrm{N\,m}. \]

The cross product uses \((a\times b)_y=a_zb_x-a_xb_z=0.05\cdot20-0=1\); the other two components are zero. Because the net force is zero, this is a pure couple, and its moment is the same about every point. About the right hand, only the left force has a moment arm: \((\mathbf r_L-\mathbf r_R)\times\mathbf f_L=(0,0,-0.1)\times(-10,0,0)=(0,1,0)\) N m. The hands turn the club without pushing it anywhere.

The workbench page gives three reasons. First, an instantaneous change only measures \(G(x)(u_1-u_0)\) at one state. Symmetry, cancellation or the choice of output can make that change zero even when the torque matters later. Second, the later effect of removing the torque includes the changed state, so it needs a forward counterfactual that starts from the same cut state and runs on. Third, the test did not say in advance what result it expected.

A better test states, before running it, the observable (for example club speed at a defined event), the time window, what stays the same (stored elastic energy, damping, contact) and the size of effect expected. It then runs the matched forward counterfactual, a negative control and a step-size check. A failure of that stated, nonzero prediction would count against the mechanism.

Module 4: Control-Affine Dynamics and Counterfactuals

The theory pages on this site each include a worked example whose numbers come from code in the repository. This module turns four of them into a problem set for a control or dynamical-systems course.

Learning Objectives

  • Evaluate the drift of a pendulum and linearize it at a state.
  • Show that the width of a reachable set does not depend on a constant drift, even when the Drift-Control Ratio (DCR) is large.
  • Explain why the response of a control-affine system is affine in its input, and check the superposition identity with numbers.
  • Distinguish a single pointwise drift sample from a forward zero-torque trajectory.

Reading Path

  1. Part 4: Worked Pendulum Dynamics (about 40 minutes).
  2. Drift, Control Capacity and Golf-Swing Correction: read the worked example on reachable width (about 30 minutes).
  3. Superposition in Affine Control Systems: read the worked example on the double pendulum (about 30 minutes).
  4. Zero-Torque Counterfactual (ZTCF) Family: read the worked example on a pointwise sample (about 30 minutes).

Exercises

Exercise 4.1 (Pendulum Drift and Linearization). For a rigid pendulum \(\dot x=(\dot\theta,\ -g\sin\theta/L+u/(mL^2))\) with \(m=1\) kg, \(L=1\) m, \(g=9.81\) m/s², evaluate the drift at \(\theta=0.5\) rad, \(\dot\theta=0\), \(u=0\), and find the linearization \(A\) at that state.

Exercise 4.2 (Reachable Width). For \(\dot x=d+u\) with \(x(0)=0\), \(|u(t)|\le1\) and horizon \(T=1\), find the set of reachable \(x(1)\) for \(d=0\) and for \(d=100\). What does this say about using the size of \(d\) relative to \(u\) as a measure of control?

Exercise 4.3 (Superposition). The superposition page defines \(\Phi(\tau)=M(q)^{-1}(\tau-C(q,\dot q)\dot q)\) for a double pendulum and reports \(\Phi(0)\), \(\Phi(\tau_1)\) and \(\Phi(\tau_2)\) for \(\tau_1=(2,0)\) and \(\tau_2=(0,1)\) N m. Use only those three results to predict \(\Phi(\tau_1+\tau_2)\), then compare with the page. Why can this identity never fail?

Exercise 4.4 (One Drift Sample). Reproduce the zero-torque page’s pointwise sample for the default three-link model at rest, using the code below. Explain why this sample alone does not predict the club’s path with the torques removed.

import numpy as np
from src.affine_control.golf_model import GolfModel

GolfModel().drift_acceleration(np.zeros(3), np.zeros(3))

Worked Solutions

With \(\sin 0.5=0.479426\) and \(\cos 0.5=0.877583\), the drift is \(\dot x=(0,\ -9.81\cdot0.479426)=(0,\ -4.7032)\), in rad/s and rad/s². The Jacobian of \(-g\sin\theta/L\) with respect to \(\theta\) is \(-g\cos\theta/L\), so

\[ A=\begin{bmatrix}0&1\\-9.81\cdot0.877583&0\end{bmatrix}=\begin{bmatrix}0&1\\-8.6091&0\end{bmatrix},\qquad B=\begin{bmatrix}0\\1\end{bmatrix}. \]

These match the worked example on Part 4, which computes them with SimplePendulum in src/tangent_models/examples.py.

Integrate: \(x(1)=d\cdot1+\int_0^1u\,dt\). The integral can take any value from \(-1\) to \(1\). So the reachable set is \([-1,1]\) for \(d=0\) and \([99,101]\) for \(d=100\). Both have width 2. A large drift moves the set but does not shrink it, so the control still has the same authority. A ratio of drift size to control size rose from 0 to 100 without any change in what the control can do. This matches the worked example on the Drift-Control Ratio page.

From the superposition page, \(\Phi(0)=(-0.7709,\,2.5089)\), \(\Phi(\tau_1)=(1.8448,\,-1.1359)\) and \(\Phi(\tau_2)=(-2.5933,\,9.0483)\). Then

\[ \Phi(\tau_1)+\Phi(\tau_2)-\Phi(0)=(1.8448-2.5933+0.7709,\ -1.1359+9.0483-2.5089)=(0.0224,\ 5.4035), \]

which is the page’s value for \(\Phi(\tau_1+\tau_2)\). The identity cannot fail because \(\Phi(\tau)=M^{-1}\tau-M^{-1}C\dot q\) is a fixed matrix times \(\tau\) plus a fixed vector, at a fixed state. Subtracting \(\Phi(0)\) removes the fixed vector once, which is why it appears with a minus sign. The numbers illustrate the theorem; they do not test it.

The call returns about \((-31.40,\ 29.98,\ 1.57)\) rad/s², the value on the zero-torque page. At rest the velocity terms are zero, so this is the acceleration that gravity alone would cause at that one state. It is a single sample of the drift \(f(x)\). The forward zero-torque trajectory starts from a state and then evolves under the drift alone, so the state, and the drift, change at every later instant. Predicting that path requires integrating the equations forward; one sample cannot do it.

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  title = {Educator {Resources}},
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  langid = {en}
}
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