Energy Transfer: How Power Flows Through the Kinetic Chain
A useful explanation of the golf swing must answer two different questions: what changes the motion, and where does the mechanical energy come from? Constraint forces, muscular moments and gravity all affect acceleration, but their energy accounts depend on the system boundary. A force can accelerate a body without doing work on the complete system. A joint motor can add energy while an adjacent segment loses energy. An angular-speed peak does not identify either process by itself.
This chapter develops a consistent account from whole-system work to individual segment power, then applies it to sequencing, lag and impact. The aim is to make the kinetic-chain idea testable: identify the bodies, forces, moments, storage and measurements that would support an explanation. Two explicit teaching mechanisms and six worked exercises separate mechanical identities from hypotheses about a golfer. The idealized examples establish possibilities and counterexamples; their parameter values are not measured human physiology.
Energy Fundamentals: Define the System
For a chosen collection of rigid bodies and elastic elements, write
\[ E_{\mathrm{mech}}=\sum_i K_i+V_g+U_e. \]
Here \(V_g\) is gravitational potential, \(U_e\) is recoverable elastic energy, and chemical energy is outside this mechanical account. Muscles can therefore add mechanical energy during the downswing even if the total energy of a larger thermodynamic system is conserved. Neglecting air resistance does not make an actively driven swing mechanically conservative. Active shortening, lengthening, elastic deformation and dissipation may coexist throughout the motion.
In one inertial laboratory frame, the kinetic energy of rigid segment \(i\) is
\[ K_i=\tfrac12m_i\lVert\mathbf v_{C_i}\rVert^2 +\tfrac12\boldsymbol\omega_i^T I_{C_i}\boldsymbol\omega_i. \]
The translational velocity is that of its center of mass; the angular velocity and central inertia tensor must use the same axes. A relative wrist rate is not the club’s absolute angular velocity. The scalar expression \(I\omega^2/2\) alone applies to rotation about a specified fixed axis with the appropriate inertia about that axis. A translating club generally needs both terms above. For a deforming segment this rigid-body approximation omits internal motion; whether that omission matters requires validation for the intended output.
Near the ground, with upward height \(y_i\) and constant gravitational acceleration of magnitude \(g\),
\[ V_g=\sum_i m_i g y_i+\text{constant}. \]
Changing the height origin changes no force or energy difference. It does change a ratio such as gravitational energy divided by total mechanical energy. Such a percentage is therefore not an intrinsic efficiency measure. A segment’s share of nonzero total kinetic energy is well-defined in the declared frame; at a completely stationary instant that share is \(0/0\), not zero percent.
What Is Stored, and What Is Merely a Configuration?
Kinetic energy describes motion. Elevation changes gravitational storage. A specified elastic constitutive law assigns energy to deformation. A visible angle, a constrained degree of freedom or a desired movement pattern does not by itself define an additional store. The club, shaft, tendons and body can all participate in an energy account, but including the same elastic element both inside a segment model and as a separate spring would count it twice.
In generalized coordinates \(q\), a rigid multibody model gives
\[ K=\tfrac12\dot q^T M(q)\dot q. \]
The mass matrix contains cross terms because a body’s motion depends on more than one joint rate. Those terms are required by the coordinate representation; they are not independent packets of energy traveling between anatomical parts. Physical segment energies remain the center-of-mass expressions above. Changing coordinates changes the appearance of the cross terms without changing those energies for the same physical state and observer.
Power and the Whole-System Work Balance
Power is the rate of energy change through a specified mechanism:
\[ P=\frac{dE}{dt},\qquad W=\int_{t_0}^{t_1}P\,dt. \]
A positive input power increases the chosen mechanical account; negative input power extracts energy from it. A dissipative loss is reported here as a nonnegative quantity. For conservative storage \(V=V_g+U_e\), use
\[ M\ddot q+C(q,\dot q)\dot q+\nabla V =B(q)u+Q_{\mathrm{nc}}+J^T\lambda. \]
The motor map \(B\) converts actuator efforts \(u\) into generalized forces. \(Q_{\mathrm{nc}}\) contains the declared remaining nonconservative forces, and \(J^T\lambda\) represents an ideal constraint reaction. Do not place the same muscle, spring or contact force in two of these terms. A standard Christoffel choice of \(C\) satisfies the skew-symmetry identity for \(\dot M-2C\); an arbitrary matrix factorization of the velocity bias need not satisfy that matrix identity.
Differentiating kinetic and potential energy gives
\[ \dot K=\dot q^T M\ddot q+\tfrac12\dot q^T\dot M\dot q, \qquad \tfrac12\dot q^T\dot M\dot q=\dot q^T C\dot q, \]
\[ \dot V=\dot q^T\nabla V. \]
Consequently,
\[ \dot E_{\mathrm{mech}} =\dot q^T Bu+\dot q^T Q_{\mathrm{nc}}+\lambda^T J\dot q. \]
The cancellation is between the metric-derivative term and the velocity-bias term. It does not require \(\dot q^T C\dot q=0\), which is generally false. The general manipulator formulation and a simpler point-mass double pendulum are also developed in Tedrake’s multibody notes (Tedrake n.d.).
For a stationary ideal constraint and a decomposition into supplied power \(P_{\mathrm{in}}\) and dissipation \(D\geq0\), the integrated account is \(E(t)-E(t_0)=W_{\mathrm{in}}-\int D\,dt\). Gravity is already represented by \(V_g\). Alternatively one can track kinetic plus elastic energy and include gravitational power on the right. Adding gravitational work while also retaining its potential-energy change counts the same conversion twice. Time-dependent stiffness, an imposed spring rest angle, or a moving prescribed base requires its corresponding parameter or support work; the autonomous formula alone is then incomplete.
Signed Work, Positive Work and Metabolic Cost
For separately modeled actuator powers \(P_a\), distinguish \(W_{\mathrm{net}}=\sum_a\int P_a\,dt\) from \(W_+=\sum_a\int\max(P_a,0)\,dt\). The latter is positive work; it does not cancel energy supplied at one time against energy absorbed at another. Taking the positive part after summing powers gives a different quantity when some actuators supply energy while others absorb it. Declare which convention is used before comparing swings.
Net joint moments inferred from motion do not resolve antagonistic muscle forces, tendon storage or activation costs. A joint with zero net power can still involve active muscle and metabolic expenditure. Neither a ratio to net joint work nor a ratio to positive joint work is automatically a metabolic efficiency. Estimating that efficiency requires a separately justified chemical energy or metabolic model and a consistent measurement interval.
Constraint Power: Zero for Which Boundary?
An ideal reaction to a position constraint \(\Phi(q,t)=0\) has generalized force \(Q_c=J^T\lambda\), where \(J=\partial\Phi/\partial q\). Its actual power is
\[ P_c=Q_c^T\dot q=\lambda^T J\dot q. \]
Differentiating the constraint gives
\[ J\dot q+\Phi_t=0, \]
and therefore
\[ P_c=-\lambda^T\Phi_t. \]
It vanishes for a stationary ideal holonomic constraint. A moving support can exchange energy with the modeled bodies. Conversely, some ideal nonholonomic constraints, such as homogeneous no-slip rolling constraints, also have zero reaction power. Holonomic versus nonholonomic is not the test by itself: ideality, the admissible velocities and explicit support motion matter. Friction, tissue damping and impact losses require their own models.
Why Energy Can Cross an Ideal Joint
Consider an arm and a club connected at hinge \(H\), with coincident material-point velocities \(\mathbf v_H\) in the ideal attachment. Let \(\mathbf F\) act on the club at \(H\), with \(-\mathbf F\) on the arm. Then
\[ P_{F,\mathrm{club}}=\mathbf F\cdot\mathbf v_H, \qquad P_{F,\mathrm{arm}}=-\mathbf F\cdot\mathbf v_H. \]
The pair sums to zero while either term may be nonzero. This is energy crossing an internal boundary, with equal and opposite entries in the two ledgers. Equality of attachment-point velocities enables the cancellation; it does not prevent transfer. Substituting the two centers’ different velocities for the contact-point velocity would calculate the wrong contact power.
In a planar model let a hinge motor apply moment \(n\) to the club and \(-n\) to the arm. If gravity is included in each segment’s potential, the club balance is
\[ \dot E_{\mathrm{club}}=\mathbf F\cdot\mathbf v_H+n\omega_{\mathrm{club}}, \qquad P_{\mathrm{hinge\ motor}}=n(\omega_{\mathrm{club}}-\omega_{\mathrm{arm}}). \]
The moment’s power on the club uses its absolute angular velocity, whereas the motor’s net mechanical power uses relative angular velocity. Their difference is the equal-and-opposite moment’s power on the arm. An ideal unactuated hinge transmits attachment force while allowing relative rotation; setting its motor moment to zero does not disconnect the club.
On narrow screens, scroll the figure or open it at full size.
These balances do not select a universal direction of transfer or maximize club energy. Geometry, motion and applied efforts determine the signed power. The opposite signs of the interface entries establish accounting consistency; they do not identify what would happen if the golfer deliberately changed an effort or movement. That requires a new dynamically consistent trajectory.
Reporting Point, Observer and Ground Contact
A wrench’s power is \(\mathbf F\cdot\mathbf v_O+\mathbf n_O\cdot\boldsymbol\omega\) when both the moment and velocity refer to the same rigid body’s reporting point \(O\). Moving that point changes the moment and linear velocity together; the power is unchanged. Changing the inertial observer by a constant velocity changes individual kinetic energies and force powers consistently. A segment energy fraction is therefore an explicitly frame-dependent diagnostic, not an observer-independent property of the swing.
For an ideal fixed no-slip foot contact, local contact velocity is zero and the contact force does no work there, although it can change the golfer’s momentum. Multiplying ground-reaction force by whole-body center-of-mass velocity produces a center-of-mass power quantity, not necessarily actual foot-contact power. Deforming footwear, slipping contacts and moving ground alter the boundary. Gravity, air and the ball can also supply external forces; ground reaction is not the only external interaction in a complete swing and collision.
Energy Partition During a Declared Passive Motion
Before interpreting a golfer’s downswing, test the transfer claim on a mechanism whose assumptions and trajectory are known. Take two uniform rods, each of mass 1 kg and length 1 m, a fixed base, ideal hinges, zero gravity and no motors. Absolute angles are measured from vertically downward. Initially the inner rod has angle zero, the outer rod angle \(-\pi/2\), and both absolute angular rates are 2 rad/s. Their kinetic energies are \(2/3\) J and \(8/3\) J, respectively. Total energy is \(10/3\) J.
The initial hinge-force power into the outer rod is \(+1\) W. Reflecting the outer angle to \(+\pi/2\), without changing either angular rate, gives \(-1\) W. Both states have the same segment kinetic energies. The transfer direction therefore cannot be inferred from the energies or speed ordering alone. The force and attachment-point velocity matter.
Numerical integration of the first state gives the following rounded outer-rod energies at 0, 0.5, 1, 1.5 and 2 s: 2.6667, 3.3327, 2.9780, 2.5007 and 2.3421 J. At all five times the total remains \(10/3\) J within integration error. The outer fraction rises from 80 percent to about 99.98 percent at 0.5 s and subsequently falls. The signed power changes direction again later. There is no active braking and no dissipative sink in this model: a decrease of outer-rod energy is an increase of inner-rod energy.
On narrow screens, scroll the figure or open it at full size.
These times are integration times for this declared mechanism, not five measured phases of a golf swing. A realistic model needs measured or justified segment properties, gravity, multiple joints, applied efforts and boundary conditions. A sequence of arbitrarily selected joint rates is not a simulated trajectory; it cannot establish a missing-work budget or a universal club-energy percentage.
Sequential Speed Peaks and Their Interpretation
The phrase summation of speed can obscure three different statements. Velocities combine vectorially through kinematics. Powers add algebraically in a declared energy balance. Peak times are features of a time series. None of these implies that scalar peak angular speeds should simply be added, or that the observed ordering proves a one-way energy cascade.
For an isolated rotor about a fixed axis, equal assigned energy gives
\[ \omega=\sqrt{2K/I}. \]
For example, separate rotors with inertias 2, 0.5 and 0.1 kg m\(^2\), each assigned 100 J, have angular speeds 10, 20 and approximately 44.72 rad/s. This algebra does not describe three connected segments at successive swing phases. Their energies need not be equal; their centers translate; their physical inertia tensors and the full joint mass matrix have different roles. A smaller scalar inertia alone does not guarantee a later or higher speed peak in the chain.
An especially useful counterexample uses the same two unit rods, now aligned vertically downward with both rates 2 rad/s, zero gravity, a base moment of \(-1\) N m and zero hinge moment. The inner and outer absolute angular accelerations are \(-12/7\) and \(18/7\) rad/s\(^2\). Yet their kinetic-energy rates are \(-8/7\) and \(-6/7\) W. The outer rod accelerates rotationally while losing energy because its translational energy falls more rapidly. Total power is \(-2\) W, exactly the base moment times its angular velocity.
A proximal segment slowing while the club speeds up can accompany transfer, active work, gravity conversion, elastic release or combinations of them. It does not show that the proximal segment has exhausted its energy. Testing a sequencing explanation requires synchronized physical segment energies, interface powers and actuator powers, together with a model that reproduces the measured motion. Prescribing the observed speed peaks in a model and then claiming that the model explains those peaks would be circular.
Storage, Active Work and a Balanced Swing Ledger
The following five-state account is deliberately hypothetical. It illustrates bookkeeping, not a typical golfer. Start from rest at the chosen reference height with no elastic deformation. The successive stored energies, in joules, are given by the three rows
\[ \begin{aligned} K &: (0,\ 0,\ 80,\ 115,\ 115),\\ V_g &: (0,\ 15,\ 5,\ 0,\ 0),\\ U_e &: (0,\ 10,\ 2,\ 0,\ 0). \end{aligned} \]
Their sums are 0, 25, 87, 115 and 115 J. Choose cumulative active mechanical work of 0, 25, 90, 120 and 120 J and cumulative dissipative loss of 0, 0, 3, 5 and 5 J. Every state satisfies stored energy equals initial energy plus supplied work minus losses. Between the second and third states, stored energy increases 62 J while active work increases 65 J and dissipation increases 3 J. Falling gravitational and elastic storage explains part of the kinetic-energy rise; it does not replace the active-work contribution.
The model could instead start at the top with its already stored 25 J. Then that amount is initial energy, while only subsequent work belongs in the downswing input. Adding both the initial store and all work previously used to create it would double count the backswing. A claim that impact energy exceeds downswing muscular work can be consistent with energy conservation, but it requires an explicit initial store and other inputs rather than an efficiency greater than one inferred from incomplete accounting.
Impact Is a New Boundary and Time Interval
For an isolated one-dimensional head–ball collision, assume head mass \(m_h=0.2\) kg, ball mass \(m_b=0.04593\) kg, initially stationary ball, head speed \(u_h=45\) m/s and restitution \(e=0.8\). These are teaching assumptions, including the effective isolated head mass; a shaft-coupled, spinning or oblique impact requires additional dynamics. Momentum conservation and relative restitution give
\[ v_h=\frac{m_h-e m_b}{m_h+m_b}u_h, \qquad v_b=\frac{(1+e)m_h}{m_h+m_b}u_h. \]
The final speeds are approximately 29.8724 and 65.8724 m/s. Initial head energy is 202.5 J; residual head energy is 89.2360 J; ball energy is 99.6491 J; the mechanical loss in this collision is 13.6148 J. Ball energy, residual head energy and collision loss sum to the initial energy. Residual head motion is not collision dissipation, and swing-phase air or tissue losses are not added again to this impact account.
With reduced mass \(\mu=m_hm_b/(m_h+m_b)\), the loss is \(\mu(1-e^2)u_h^2/2\). The ratio of ball energy to pre-impact head energy is \(m_hm_b(1+e)^2/(m_h+m_b)^2\), approximately 49.21 percent here. This is an impact energy-transfer ratio for the declared model, not a measured swing efficiency. A halving of a body’s speed quarters its kinetic energy at fixed mass; it does not halve it. These distinctions are essential when interpreting residual club speed, ball launch and claims about efficient energy delivery.
Lag, Elasticity and Release Timing
A lag angle is a relative configuration. An ideal freely rotating hinge has no elastic potential associated with that angle. If a torsional spring is explicitly included, with stiffness \(k\) and rest angle \(q_{20}\), then
\[ U_e=\tfrac12 k(q_2-q_{20})^2, \qquad \tau_e=-k(q_2-q_{20}). \]
At a 0.5 rad displacement, stiffness zero stores zero joules; stiffness 40 N m/rad stores 5 J. The angle alone cannot distinguish these cases. A deforming shaft or tendon needs its own strain coordinate and constitutive law; its energy is not automatically a function of the visible wrist angle.
If the rest angle or stiffness changes, the explicit parameter contribution is \(\partial U_e/\partial t=\dot k(q_2-q_{20})^2/2-k(q_2-q_{20})\dot q_{20}\). That term represents work associated with changing the modeled spring, rather than free passive release. A viscous hinge damper with coefficient \(d\geq0\) dissipates \(d\dot q_2^2\). Active moment, elastic moment and damping moment must be separated if the question concerns their individual contributions.
Holding a relative angle fixed is a different idealization from storing energy in a spring. A stationary ideal lock can transmit forces and moments with zero net constraint power. Removing such a lock need not change velocity or kinetic energy instantaneously, although subsequent dynamics change. Engaging a lock when the relative rate is nonzero generally requires an impulsive transition and a specified loss or actuator-work model. Releasing a rotational lock leaves the translational attachment intact; it does not remove the hinge force.
A Fair Release-Timing Comparison
Compare strategies from the same full initial state and with declared actuator limits, allowed controls and delivery event. The event should include relevant club position, face orientation, path and contact conditions, rather than only the largest angular speed anywhere in the trajectory. Report terminal kinetic and elastic energy, signed and positive work, losses and residual motion. Equal commanded torque histories do not generally imply equal work, because the trajectories and rates may differ.
If one strategy begins with 90 J and another with 140 J, their final speeds cannot isolate the effect of timing. If terminal energies are 60 and 150 J for the same isolated fixed-axis rotor, the speed ratio is \(\sqrt{150/60}\), approximately 1.581, not two. For a translating multibody club, even that ratio requires checking the relationship between the reported speed and total kinetic energy. There is no universal release delay or speed multiplier established by these examples.
Setting one motor effort to zero at a frozen state is a useful instantaneous acceleration comparison. It is not the complete trajectory produced by a different release strategy. Passive tissue forces, gravity, constraints and other motors can remain active. This is where energy analysis connects to counterfactual dynamics and control: the intervention, retained terms, starting state and event must all be specified before assigning a causal benefit.
Two-Link Dynamics and Physical Energy Transfer
To derive the coupling explicitly, take a fixed planar base, inner-link length \(L_1\), masses \(m_1,m_2\), center distances \(c_1,c_2\) from each proximal hinge and central planar inertias \(I_1,I_2\). Define relative coordinates \(q_1=\theta_1\) and \(q_2=\theta_2-\theta_1\), where the absolute angles \(\theta_i\) are measured from downward. Let \(u=\dot q_1\), \(w=\dot q_2\) and \(a=I_1+m_1c_1^2+m_2L_1^2\), \(b=I_2+m_2c_2^2\), \(d=m_2L_1c_2\). Directly differentiating the center positions gives
\[ K=\tfrac12a u^2+\tfrac12b(u+w)^2+d\cos q_2\,u(u+w). \]
The club rotates at \(u+w\), not \(w\). The inner contribution is \((I_1+m_1c_1^2)u^2/2\); the rest belongs to the translating and rotating outer body. Hence the physical outer-body energy includes the cross term and an \(m_2L_1^2u^2/2\) contribution. Assigning only \(bw^2/2\) to the club would omit its motion induced by the inner link.
Euler–Lagrange equations with base and relative-hinge motor efforts \(\tau=(\tau_1,\tau_2)^T\) give \(M\ddot q+h+G=\tau\), where
\[ M=\begin{bmatrix} a+b+2d\cos q_2&b+d\cos q_2\\ b+d\cos q_2&b \end{bmatrix},\qquad h=\begin{bmatrix}-d\sin q_2(2uw+w^2)\\d\sin q_2u^2\end{bmatrix}. \]
Both velocity-bias components matter. For upward gravitational potential,
\[ G=\begin{bmatrix} (m_1c_1+m_2L_1)g\sin q_1+m_2c_2g\sin(q_1+q_2)\\ m_2c_2g\sin(q_1+q_2) \end{bmatrix}. \]
Multiplying by the rates and retaining the metric derivative recovers
\[ \dot E=\tau_1u+\tau_2w. \]
In particular,
\[ \dot q^T h=-d\sin q_2\,uw(u+w) =\tfrac12\dot q^T\dot M\dot q. \]
This scalar can be nonzero. It is canceled in the total energy derivative, not identified as a separate physical input or equated to the club’s energy rate. A term called Coriolis power in one coordinate decomposition must not be promoted into an invariant segment-transfer measurement.
To compute the actual hinge-force transfer, let \(\mathbf a_{C_2}\) be the outer center acceleration and \(\mathbf g=(0,-g)^T\). With no other outer-body force, Newton’s equation gives
\[ \mathbf F=m_2(\mathbf a_{C_2}-\mathbf g), \qquad P_{F,2}=\mathbf F\cdot\mathbf v_H. \]
The physical body rates follow from center translation, rotation and gravity:
\[ \dot E_2=P_{F,2}+\tau_2(u+w),\qquad \dot E_1=\tau_1u-P_{F,2}-\tau_2u. \]
Their sum agrees with the generalized balance. This independent Newton-force calculation is the check that a coordinate-based interpretation needs. In three dimensions replace scalar moments and rates by compatible moment/angular velocity pairings; in a two-arm closed chain include both attachment wrenches and avoid duplicating the shared club energy.
Reproducible Calculation
The shared rod_energy_ledger routine computes center energies and their derivatives from RodPair dynamics. Its state uses absolute angles and rates; motor inputs remain base and relative-hinge moments. For uniform rods the central inertias are \(m_iL_i^2/12\) and \(c_i=L_i/2\). Run this example from the repository root:
import numpy as np
from src.tools.energy_ledger_examples import rod_energy_ledger
from src.tools.lagrangian_examples import RodPair
model = RodPair(gravity=0)
state = np.array([0, -np.pi / 2, 2, 2])
ledger = rod_energy_ledger(model, state, np.zeros(2))
np.testing.assert_allclose(ledger.kinetic, [2 / 3, 8 / 3])
np.testing.assert_allclose(ledger.mechanical_rates, [-1, 1])
assert np.isclose(ledger.force_power, 1)
assert np.isclose(ledger.actuator_power, 0)
braking = rod_energy_ledger(
model, np.array([0, 0, 2, 2]), np.array([-1, 0])
)
np.testing.assert_allclose(braking.mechanical_rates, [-8 / 7, -6 / 7])The calculation has been independently checked against directional derivatives of each physical body’s energy, total generalized motor power and the force power at the hinge. Conservation alone would be insufficient verification: two incorrectly assigned segment rates could cancel in the total. Conversely, a small integration drift is a numerical diagnostic, not evidence of unmodeled muscular work or physical dissipation.
Evidence, Measurement and an Investigable Golf-Swing Argument
Nesbit and Monika Serrano’s work-and-power study used four selected amateur golfers and a selected swing from each, with model-derived mechanical quantities (Nesbit and Serrano 2005). It supports examining coordinated body and club work within that study’s model. It does not establish a population-wide energy percentage, directly measure metabolic efficiency or identify the causal effect of an instruction to delay release. Those questions need additional evidence.
To investigate an energy-transfer explanation, first define the outcome: clubhead delivery speed and orientation, ball launch, repeatability or effort. Then choose the body and time boundaries. Estimate segment properties and synchronized motion; state coordinate, frame and joint-moment conventions. Use the inverse-dynamics chapter and constraint-force chapter to construct physically compatible forces. Close the whole-system ledger and both sides of each important interface before interpreting a sequence of peaks. The complete-swing chapter and synthesis chapter connect these local mechanisms to finite-time delivery and competing task objectives.
Power estimates amplify synchronization and differentiation errors. To first order, the error in a wrench power includes
\[ \delta P\simeq\delta\mathbf F\cdot\mathbf v +\mathbf F\cdot\delta\mathbf v +\delta\mathbf n\cdot\boldsymbol\omega +\mathbf n\cdot\delta\boldsymbol\omega. \]
These errors can be correlated; uncertainty in integrated work also depends on timing, filtering, interval endpoints and their covariance. Compare a reported transfer difference with that uncertainty, test plausible segment properties, and inspect energy residuals without treating residual magnitude alone as a diagnosis of the missing mechanism. Force-plate, motion-capture and club measurements must share time and frame conventions.
The resulting argument is interconnected but specific: configuration and inertia determine how efforts and constraints affect acceleration; actual velocities determine their power; constitutive behavior determines storage and loss; control changes the trajectory and delivery event; observations constrain which of those explanations can be distinguished. A convincing coaching claim requires a controlled intervention or a validated counterfactual model with uncertainty, not just a mechanically plausible story. The mechanics makes better questions possible without prescribing a universal hip-braking or lag strategy.
- Proximal-to-Distal Energy Transfer in the Golf Swing — Power ports, segment work, and sequential energy flow
- Rotation Representations Reference — Relating joint rates to task-space kinetic energy
Chapter Exercises: Energy Transfer
Derive the mechanical-energy balance from the manipulator equation. Explain why setting the scalar Coriolis contraction to zero is unnecessary. Then derive the power of an ideal constraint with explicit time dependence.
At an arm–club hinge, force power into the club is 12 W. The base motor moment is 7 N m, hinge motor moment 3 N m, and arm and club absolute angular rates are 2 and 5 rad/s. Find both body mechanical-energy rates, hinge-motor power and total motor power. Assume all other interactions are conservative.
Check the five-state storage ledger between the second and third states. If the second state is the beginning of the downswing, explain which initial energy and work interval belong in its balance. Can a video lag angle determine the initial elastic store?
For the stated isolated impact, derive the speeds, energy-transfer ratio and loss from momentum and restitution. Explain why residual head energy and swing losses cannot be relabeled as collision dissipation.
Derive \(M\), \(h\) and \(G\) for the two-link relative chart. For the aligned unit-rod braking case, compute both absolute angular accelerations and both body-energy rates. Explain how the outer rod can gain angular speed while losing energy.
Design a comparison of two release strategies that could distinguish active work, elastic release and interface transfer. State the common initial conditions, actuation constraints, delivery event, measured quantities and uncertainties. Explain why equal torques or a sequence of speed peaks is insufficient evidence of equal effort or one-way transfer.
Worked Answers
Differentiate \(K=\dot q^TM\dot q/2\), substitute the equation of motion, and add \(\dot V=\dot q^T\nabla V\). The terms \(\dot q^T\dot M\dot q/2\) and \(\dot q^TC\dot q\) cancel by the compatible Christoffel identity, leaving actuator, nonconservative and constraint powers. From \(J\dot q+\Phi_t=0\), constraint power is \(-\lambda^T\Phi_t\); only the stationary ideal case guarantees zero through this argument. A moving support must remain in the work account.
Club moment power is \(3(5)=15\) W, so its energy rate is \(12+15=27\) W. The arm receives base power \(7(2)=14\) W and loses 12 W through the force and \(3(2)=6\) W through the hinge moment: its rate is \(-4\) W. Hinge-motor power is \(3(5-2)=9\) W; total motor power is \(14+9=23\) W. The body rates sum to the same 23 W. Equating club moment power to motor power would miss the arm’s opposite moment entry.
Stored energy rises from 25 to 87 J, a 62 J increase. Active work over that interval is \(90-25=65\) J and loss is 3 J. Kinetic energy rises 80 J, supplied by 65 J active work plus 10 J gravitational release plus 8 J elastic release, minus 3 J loss. Starting at the second state means 25 J initial storage and work counted only afterward. Lag alone supplies neither a spring stiffness nor its rest deformation; an ideal hinge can have that angle with zero elastic energy.
Solve \(m_hu_h=m_hv_h+m_bv_b\) and \(v_b-v_h=e u_h\). The stated speeds follow. Divide \(m_bv_b^2/2\) by \(m_hu_h^2/2\) to obtain the ratio, about 0.492094. Subtract both final kinetic energies from the initial energy to obtain \(\mu(1-e^2)u_h^2/2\), about 13.6148 J. Residual head energy is still mechanical energy, and earlier swing dissipation occurred outside the collision interval.
Differentiate the kinetic-energy quadratic twice with respect to rates to obtain \(M\); Euler–Lagrange differentiation gives both components of \(h\). Differentiate gravitational potential to obtain both entries of \(G\). In the aligned unit-rod case the absolute-coordinate mass matrix is \(\left[\begin{smallmatrix}4/3&1/2\\1/2&1/3\end{smallmatrix}\right]\), with zero instantaneous velocity bias and gravity. Solving against \((-1,0)^T\) gives \((-12/7,18/7)^T\). Inner energy rate is \((1/3)(2)(-12/7)=-8/7\) W; total motor power is \(-2\) W, so outer energy rate is \(-6/7\) W. Independent center and hinge-power calculations give the same result. Increasing rotational energy is outweighed by decreasing translational energy; angular speed alone does not measure total outer-body energy.
Use identical positions, velocities and internal elastic states, with the same stated motor capacities and support conditions. Define release as a specific control or constraint change, including any switching impulse or parameter work. Evaluate at the same physical delivery event and report its orientation and position as well as speed. Integrate separate active, elastic, dissipative and interface powers; compare signed work, positive work, residual energy and uncertainty across strategies. Equal torque commands can yield different rates and therefore different work. Peak order alone leaves those accounts unidentified. Repeat the comparison under plausible measurement and model variations before attributing a robust benefit to timing.
References
Citation
@online{untitled,
author = {},
url = {https://affinedrift.com/articles/The_Physics_of_Golf/quarto/ch10_energy_transfer.html},
langid = {en}
}