Natural vs Active Forces and Torques in Affine Mechanical Systems
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- Open / Medium: Critique: Double Pendulum Energy Blindness (
crit-double-pendulum-energy-blindness)
A two-link pendulum shows how an arm-like link and a club-like link push on each other. At one position and speed, changing either joint torque changes the force between the links. We separate the force with zero declared input from the added effect of that input. Both can pull on the club-like link. The split does not show whether muscles are relaxed or an action is conscious.
Natural and Active Force
We compare two parts at one instant. The natural part is the drift: the zero-input prediction, with gravity, motion effects, and any other loads kept. Current motion already reflects earlier input. The active part is the change caused by the declared torque at that same state. It can change both the force between links and the joint moment. Picture a motor-driven pendulum next to one with the motor off at the same position and speed. That holds only for one instant. Once both move, their states differ and a later comparison asks a different question.
The Double Pendulum Model
We apply this to a double pendulum standing in for arms and club.
1. Motivation
For an affine control system, we usually write \[ \dot{x} = f(x) + G(x)u, \] and think of \(f\) as the drift (what the system does “on its own”) and \(G(x)u\) as the control effect.
For mechanical systems, the same question can be asked of an observable force or wrench: at a fixed state, how much changes when the declared input changes from zero? The answer requires a model of the output, not just a label attached to a measured force.
We call the zero-input term drift and the additional term input-induced. The traditional words “natural” and “active” below refer to this declared partition. Drift is not an independent energy source, a measure of passivity, or a statement that the golfer contributes nothing. It depends on the current state, retained loads and choice of input. In particular, a joint reaction can depend on both drift and input.
This article derives the conditional force partition, specializes it to a fixed-base planar double pendulum, and checks how inter-link force power differs from joint-motor power. A manufactured numerical example and linked reproducibility checks support the mechanics; no measured golf-swing contribution percentages are inferred.
2. General Mechanical System: Natural vs Active Torques
Consider an \(n\)-DOF mechanical system with generalized coordinates \(q \in \mathbb{R}^n\), velocities \(\dot{q}\), accelerations \(\ddot{q}\). The standard joint-space dynamics (from Lagrange or Newton–Euler) can be written as
\[ M(q)\,\ddot{q} + C(q,\dot{q})\,\dot{q} + g(q) + d(q,\dot{q}) = \tau_{\text{in}} + \tau_{\text{ext}}, \tag{2.1} \]
where
- \(M(q) \in \mathbb{R}^{n \times n}\) is the inertia matrix,
- \(C(q,\dot{q})\dot{q}\) is the vector of Coriolis and centrifugal terms,
- \(g(q)\) is the gravity torque vector,
- \(d(q,\dot{q})\) collects dissipative effects (e.g., joint damping, friction),
- \(\tau_{\text{in}}=B(q)u\) is the declared applied generalized load, with \(u\in\mathbb R^m\) and \(B\in\mathbb R^{n\times m}\),
- \(\tau_{\text{ext}}\) are generalized forces from external wrenches.
Here \(\dot q\) are coordinate rates; this form assumes independent coordinates and a nonsingular mass matrix. If generalized speeds differ from \(\dot q\), their kinematic map and dual generalized loads must be included. Generalized loads are power-conjugate to their rates: rotational entries have torque units, translational entries force units. Net joint torque does not identify individual muscles or metabolic effort.
Suppose external wrenches \(\lambda\) (contact forces, etc.) act at some contact frames with Jacobian \(J_c(q)\). Then
\[ \tau_{\text{ext}} = J_c(q)^\top \lambda. \tag{2.2} \]
Substituting into (2.1),
\[ M(q)\,\ddot{q} + C(q,\dot{q})\,\dot{q} + g(q) + d(q,\dot{q}) = \tau_{\text{in}} + J_c(q)^\top \lambda. \tag{2.3} \]
2.1 Inverse-Dynamics Applied Load
Define \(h=C\dot q+g+d-J_c^\mathsf T\lambda\), where the external wrench is specified independently. Then \[ \tau_{\mathrm{ID}}=M\ddot q+h=\tau_{\mathrm{in}}=Bu. \] This is the applied generalized load, not a natural drift torque. The signed drift load is \(-h\), and the drift acceleration is \(-M^{-1}h\). Calling \(M\ddot q+h\) drift and then adding \(Bu\) counts the applied load twice.
2.2 Recovering Input Channels
For full-column-rank \(B\), a consistent load gives \(u=B^+\tau_{\mathrm{ID}}\). Check the residual \(BB^+\tau_{\mathrm{ID}}-\tau_{\mathrm{ID}}\). With redundant channels, report the null-space family and physiological constraints; minimum norm does not identify muscle recruitment.
3. State-Space Drift vs Active Contribution
We can also describe the same decomposition at the state level, in the usual first-order form.
Define the state
\[ x = \begin{bmatrix} q \\ \dot{q} \end{bmatrix} \in \mathbb{R}^{2n}. \]
From (2.3), with \(\tau_{\text{in}}=B(q)u\),
\[ M(q)\ddot{q} = B(q)u + J_c(q)^\top \lambda - C(q,\dot{q})\dot{q} - g(q) - d(q,\dot{q}). \tag{3.1} \]
Therefore
\[ \ddot{q} = M(q)^{-1}\, \big( J_c(q)^\top \lambda - C(q,\dot{q})\dot{q} - g(q) - d(q,\dot{q}) \big) + M(q)^{-1} B(q)\,u. \tag{3.2} \]
Thus
\[ \dot{x} = \begin{bmatrix} \dot{q} \\ M(q)^{-1}\big( J_c(q)^\top \lambda - C(q,\dot{q})\dot{q} - g(q) - d(q,\dot{q}) \big) \end{bmatrix} + \begin{bmatrix} 0 \\ M(q)^{-1} B(q) \end{bmatrix} u. \tag{3.3} \]
Define the natural drift field
\[ f_{\text{nat}}(x,\lambda) := \begin{bmatrix} \dot{q} \\ M(q)^{-1}\big( J_c(q)^\top \lambda - C(q,\dot{q})\dot{q} - g(q) - d(q,\dot{q}) \big) \end{bmatrix}, \tag{3.4} \]
and the input field
\[ G(x) := \begin{bmatrix} 0 \\ M(q)^{-1} B(q) \end{bmatrix}. \tag{3.5} \]
Then we have the familiar affine structure:
\[ \dot{x} = f_{\text{nat}}(x,\lambda) + G(x)u. \tag{3.6} \]
For a prescribed input-independent external wrench, \(f_{\text{nat}}\) is the evolution the system would have with no active input (\(u = 0\)), subject to the declared gravity, external load, damping and state. If \(\lambda\) is an unknown constraint reaction, solve it jointly with acceleration: it generally depends on \(u\) and cannot be frozen as an independent drift load.
The input zero matters: writing \(u=u_0(x)+\delta u\) moves \(G(x)u_0(x)\) into the baseline drift without changing the total dynamics. State the reference input before interpreting the partition. A mathematical zero-input baseline also need not be physically admissible for every actuator or contact constraint.
4. Wrench-Level Decomposition: Natural vs Active Wrench
The virtual-power map \(\tau=J^\mathsf T w\) applies to the generalized load of a specified external wrench. A pseudoinverse of \(J^\mathsf T\) applied to all actuator torques does not generally recover a physical endpoint wrench: internal loads, inconsistent task loads and null-space ambiguity must be resolved.
For an identifiable reaction output, substitute the forward acceleration and solve its force balance at a fixed state. If this yields \(w(x,u)=w_d(x)+W_u(x)u\) in the declared contact mode, the same-state input contribution is \(W_u(x)u\). A forward zero-input trajectory instead changes state and cannot be subtracted later as though it were a pointwise sample. Input-dependent contact-mode switches, friction transitions or nonlinear actuator laws can prevent one affine output map from applying across the comparison.
The double pendulum below supplies an explicit example through the distal segment’s center-of-mass force balance. It does not need a wrench pseudoinverse.
A wrench must declare its reference point and frame, and its power must use the matching velocity. In an inertial frame, let \(r\) point from point \(A\) to point \(B\) on a rigid body. Then \(m_B=m_A-r\times F\) and \(v_B=v_A+\omega\times r\), giving \[ F\cdot v_A+m_A\cdot\omega=F\cdot v_B+m_B\cdot\omega. \] The total is independent of the reporting point when both transformations are made. Force power and moment power individually can change with the point; body power also depends on the observer’s motion. Comparing “pulling” and “twisting” therefore requires a common convention before any drift/input inference.
5. Planar Double Pendulum Example
We now specialize the above to a planar double pendulum (2-link planar manipulator). The model is an illustrative mechanical system, not an anatomically calibrated golfer.
5.1 Configuration and Parameters
Let
- \(q = \begin{bmatrix} q_1 \\ q_2 \end{bmatrix}\) where
- \(q_1\) = angle of link 1 from the downward vertical,
- \(q_2\) = relative angle of link 2 from link 1.
- Link lengths: \(l_1, l_2\).
- Center-of-mass distances along each link: \(c_1, c_2\).
- Masses: \(m_1, m_2\).
- Inertia about COM (out of plane): \(I_1, I_2\).
- Gravity: \(g > 0\).
Use a fixed inertial plane with horizontal \(x\) positive right and vertical \(y\) positive upward. Both angles increase counterclockwise; the distal absolute angle is \(q_1+q_2\). The base is fixed, the two links are rigid, and their common pin joint transmits force without slip. \(\tau_1=u_1\) acts on link 1 against the fixed base; \(\tau_2=u_2\) acts positively on link 2 and equally oppositely on link 1. Thus the generalized input is \((u_1,u_2)\) in these relative coordinates. Lengths are in metres, masses in kilograms, centroidal inertias in \(\mathrm{kg\,m^2}\), and torques in \(\mathrm{N\,m}\).
We assume no explicit damping (for simplicity), i.e. \(d(q,\dot{q}) = 0\), and no external endpoint load (\(J_c^\top\lambda = 0\)). The fixed-base reaction has zero power because its application point is stationary; the internal pin force remains present in each link balance.
Then the equations of motion take the canonical form
\[ M(q)\ddot{q} + C(q,\dot{q})\,\dot{q} + g(q) = \tau_{\text{in}}, \tag{5.1} \]
with \(\tau_{\text{in}} = \begin{bmatrix} \tau_1 \\ \tau_2 \end{bmatrix}\) the joint torques.
5.2 Inertia Matrix \(M(q)\)
A standard planar 2-link manipulator model gives:
\[ \begin{aligned} M_{11}(q) &= I_1 + I_2 + m_1 c_1^2 + m_2 \big( l_1^2 + c_2^2 + 2 l_1 c_2 \cos q_2 \big), \\ M_{12}(q) &= I_2 + m_2\big( c_2^2 + l_1 c_2 \cos q_2 \big), \\ M_{21}(q) &= M_{12}(q), \\ M_{22}(q) &= I_2 + m_2 c_2^2. \end{aligned} \tag{5.2} \]
So
\[ M(q) = \begin{bmatrix} M_{11}(q) & M_{12}(q) \\ M_{21}(q) & M_{22}(q) \end{bmatrix}. \]
5.3 Coriolis/Centrifugal Terms \(C(q,\dot{q})\dot{q}\)
There are various equivalent parameterizations. A commonly used one defines
\[ \gamma(q) = - m_2 l_1 c_2 \sin q_2. \tag{5.3} \]
Then the product \(C(q,\dot{q})\dot{q}\) can be written as
\[ C(q,\dot{q})\dot{q} = \begin{bmatrix} \gamma(q)\dot{q}_2(2\dot{q}_1 + \dot{q}_2) \\ - \gamma(q)\dot{q}_1^2 \end{bmatrix}. \tag{5.4} \]
You can think of this as \(C(q,\dot{q})\dot{q}\) without ever writing out \(C\) explicitly as a matrix.
5.4 Gravity Term \(g(q)\)
Assuming angles are measured from the downward vertical for simplicity, the gravity torques are
\[ \begin{aligned} g_1(q) &= (m_1 c_1 + m_2 l_1)\,g\,\sin q_1 + m_2 c_2 g \sin(q_1 + q_2), \\ g_2(q) &= m_2 c_2 g \sin(q_1 + q_2). \end{aligned} \tag{5.5} \]
So
\[ g(q) = \begin{bmatrix} g_1(q) \\ g_2(q) \end{bmatrix}. \]
5.5 Applied Torque and Drift Acceleration
For the fully actuated example \(B=\operatorname{diag}(1,1)\) (the identity map, distinct from the distal inertia \(I_2\)), \[ \begin{aligned} u&=M(q)\ddot q+C(q,\dot q)\dot q+g(q),\\ \ddot q_d&=-M(q)^{-1}[C(q,\dot q)\dot q+g(q)]. \end{aligned} \] Inverse dynamics evaluated on \(\ddot q_d\) returns zero applied torque. The resultant acceleration balance is \(M\ddot q=-C\dot q-g+u\).
6. Energy Transfer Decomposition
The force partition becomes useful when combined with power flow. A large joint force need not deliver large power: its projection onto the velocity of the shared joint point determines its force power. Torque inputs can change that force through the coupled accelerations, even when the distal torque is zero.
Let \(\vec F_{21}\) be the force on link 2 from link 1, applied at their shared joint, and let \(\tau_{21}=u_2\) be the couple on link 2. With uniform gravity included in \(E_2\), and no other distal loads, \[ \begin{aligned} E_2&=\tfrac12 m_2\|\vec v_{c2}\|^2+\tfrac12 I_2\omega_2^2\\ &\quad +m_2g y_{c2},\\ \omega_2&=\dot q_1+\dot q_2. \end{aligned} \] The COM-to-joint lever is \(-c_2 e(q_1+q_2)\), where \(e(\theta)=(\sin\theta,-\cos\theta)\). Newton’s law and the COM moment balance give the following energy rate. Crucially, the force multiplies joint-point velocity, not COM velocity without its corresponding moment correction:
\[ \dot{E}_2 = \vec{F}_{21} \cdot \vec{v}_{\text{joint}} + \tau_{21} \, \omega_2. \]
To see why the joint-point velocity appears, write \(r\) for the COM-to-joint lever. The COM moment balance is \[ I_2\dot\omega_2=(r\times\vec F_{21})_z+u_2. \] Differentiating COM kinetic energy and gravitational potential cancels the gravity-work terms. Adding rotational kinetic energy gives \[ \begin{aligned} \dot E_2&=\vec F_{21}\cdot\vec v_{c2} +(r\times\vec F_{21})_z\omega_2\\ &\quad+u_2\omega_2\\ &=\vec F_{21}\cdot\vec v_{\text{joint}}+u_2\omega_2, \end{aligned} \] because \(\vec v_{\text{joint}}=\vec v_{c2}+\boldsymbol\omega_2\times r\), with \(\boldsymbol\omega_2\) directed out of the plane. This identifies the point and sign of each power term.
We calculate the joint constraint force \(\vec{F}_{21}\) from Newton’s law applied to the COM of link 2:
\[ \vec{F}_{21} = m_2 (\vec{a}_{c2} - \vec{g}). \]
In the same inertial axes, \(\vec r_{c2}=l_1e(q_1)+c_2e(q_1+q_2)\). Its translational COM Jacobian (distinct from the contact Jacobian \(J_c\)) is \[ \begin{aligned} J_{c2}&=\frac{\partial\vec r_{c2}}{\partial q} =\begin{bmatrix}j_1&j_2\end{bmatrix},\\ j_1&=l_1t(q_1)+c_2t(q_1+q_2),\\ j_2&=c_2t(q_1+q_2). \end{aligned} \] Here \(t(\theta)=(\cos\theta,\sin\theta)\). The curvature acceleration is \[ \dot J_{c2}\dot q=-l_1\dot q_1^2e(q_1)-c_2(\dot q_1+\dot q_2)^2e(q_1+q_2). \] The column vectors and their components use the frame declared in Section 5.1; \(\vec g=(0,-g)\).
Using the affine decomposition of acceleration from (3.2), \[ \ddot{q} = \ddot{q}_{\text{nat}} + M(q)^{-1} B(q) u, \] the linear acceleration of the COM \(\vec{a}_{c2}\) decomposes as:
\[ \begin{aligned} \vec a_{c2}&=J_{c2}\ddot q+\dot J_{c2}\dot q\\ &=\vec a_{\text{nat}}+\vec a_{\text{act}},\\ \vec a_{\text{nat}}&=J_{c2}\ddot q_{\text{nat}}+\dot J_{c2}\dot q,\\ \vec a_{\text{act}}&=J_{c2}M(q)^{-1}B(q)u. \end{aligned} \]
Thus, this identifiable joint force splits into a same-state drift and declared-input contribution:
\[ \begin{aligned} \vec F_{\text{nat}}&=m_2(\vec a_{\text{nat}}-\vec g),\\ \vec F_{\text{act}}&=m_2\vec a_{\text{act}}. \end{aligned} \]
Depending on the actuator model, the joint torque \(\tau_{21}\) also splits (e.g., into passive spring/damper \(\tau_{\text{nat}}\) and active motor \(\tau_{\text{act}}\)). For a purely torque-driven joint, \(\tau_{21} = \tau_{\text{act}} = u_2\).
The Power Transfer then decomposes into:
Natural Power Transfer: The signed zero-input contribution evaluated at the same state and with the declared retained loads. \[ P_{\text{nat}} = \vec{F}_{\text{nat}} \cdot \vec{v}_{\text{joint}} + \tau_{\text{nat}} \omega_2. \]
Active Power Transfer: The signed change in link power due to the declared input at that same state. \[ P_{\text{act}} = \vec{F}_{\text{act}} \cdot \vec{v}_{\text{joint}} + \tau_{\text{act}} \omega_2. \]
This is a decomposition by declared input dependence, not a pulling-versus-twisting split: both drift and input can contribute to the joint force, and the input also supplies joint moment. Here \(\omega_2=\dot q_1+\dot q_2\) is the distal absolute angular velocity. The distal moment power \(u_2\omega_2\) differs from the actuator power \(u_2\dot q_2\) delivered to the two-link system; the equal-and-opposite moment on link 1 supplies the difference.
This model includes kinetic and gravitational energy but omits shaft strain energy and its dissipation. A flexible model needs an extended energy balance, deformation state and validated constitutive assumptions. Stored strain energy can return later; the rigid comparison alone establishes neither how much returns nor a speed advantage. Its omission does not justify a universal claim about delayed-release capacity.
6.1 Energy Balance in the ZTCF Family Trajectory
The Zero Torque Counterfactual (ZTCF) sets \(u = 0\), leaving only drift dynamics \(\dot{x} = f(x)\). For a conservative mechanical system (no dissipation), the total mechanical energy is conserved along the ZTCF:
\[ \frac{d}{dt}\left(T(q, \dot{q}) + V(q)\right)\bigg|_{\text{ZTCF}} = 0. \]
This provides a useful sanity check for this fixed-base conservative model. Nonclosure can indicate inconsistent equations, sign conventions, constraint treatment or numerical error; it does not uniquely identify the cause. Moving supports, time-dependent parameters or other external work invalidate the stated zero-power assumptions. In practice, dissipation (joint damping, air resistance) breaks strict energy conservation; the ZTCF trajectory then follows the conservative drift augmented by passive dissipative forces.
With identical initial energy, no dissipation, no other external work and fixed parameters, the energy difference equals applied-input work along the driven trajectory. With damping, changing impedance or other external work, include differences in those contributions along both trajectories.
6.2 Segment Transfer and Whole-System Work
The joint force contributes \(\vec F_{21}\cdot\vec v_{\text{joint}}\) to link 2 and its negative to link 1, because both share the same joint-point velocity. These terms redistribute energy and cancel in the sum. With \(\omega_1=\dot q_1\), the joint couples instead give \[ \begin{aligned} u_2\omega_2-u_2\omega_1&=u_2\dot q_2,\\ \dot E_1+\dot E_2&=u_1\dot q_1+u_2\dot q_2. \end{aligned} \] Thus neither zero distal torque nor positive distal force power proves zero motor input. At an instant with \(\dot q_2=0\), the joint motor has zero net pair power even though its equal-and-opposite couples can transfer energy between links rotating together. This instantaneous condition does not lock the joint or require \(\ddot q_2=0\).
6.3 A Manufactured Same-State Example
Take \(m_1=2.5\), \(m_2=0.4\) kg; \(l_1=0.35\), \(c_1=0.175\), \(c_2=0.5\) m; \(I_1=0.025\), \(I_2=0.03\ \mathrm{kg\,m^2}\); and \(g=9.81\ \mathrm{m/s^2}\). At \(q=(0.6,-0.8)\) rad and \(\dot q=(1.7,-2.3)\) rad/s, solve (5.1) separately for each input without advancing the state.
- With \(u=(0,0)\ \mathrm{N\,m}\), the two coordinate accelerations are approximately −25.838742 and 39.646836 rad/s². The force components are \(F_{21,x}\approx-0.493169\) N and \(F_{21,y}\approx1.737295\) N. The force supplies \(0.341483\) W to link 2 while whole-system energy is constant.
- With \(u=(2,0)\ \mathrm{N\,m}\), the accelerations are approximately −10.717772 and 18.853237 rad/s². The force components are \(F_{21,x}\approx0.142103\) N and \(F_{21,y}\approx3.158003\) N. Distal force power is \(1.130753\) W: the same-state drift contribution remains \(0.341483\) W and the proximal-input contribution is \(0.789270\) W. Distal moment power is zero, but the proximal motor supplies \(3.4\) W to the complete system.
- With \(u=(0,2)\ \mathrm{N\,m}\), distal moment power is \(2(-0.6)=-1.2\) W, whereas motor pair power is \(2(-2.3)=-4.6\) W. A positive torque can remove energy when its conjugate rate is negative.
The examples use arbitrary mechanical parameters and states, not a measured golfer or a calibrated swing phase. They demonstrate why force-versus-couple power and drift-versus-input power are different decompositions. They establish no general ranking of those contributions over a downswing.
Theoretical Synthesis
The corrected force balance separates same-state reaction-force contributions; it does not identify intent or biological cost. The state at which the partition is evaluated encodes previous actuation and external interactions.
Time-varying stiffness need not be energetically passive. For \(V=\tfrac12 k(t)e^2\) at fixed reference, \(\dot V=ke\dot e+\tfrac12\dot k e^2\): changing stiffness can add or remove energy even at zero displacement velocity. A frozen-stiffness ZTCF therefore answers a different question from a controlled-impedance experiment.
A rigid shaft omits elastic storage and distributed wave propagation. A flexible shaft can exchange stored energy with the rest of the system; this is not by itself the physiological stretch-shortening cycle. Test modal convergence and the complete work balance before assigning a speed benefit to recoil.
Suggested Cross-References
- Proximal-to-Distal Energy Transfer: the proposed timing comparison and interface-power accounting for this two-link model.
- Theory Part 3: classification of the retained drift terms.
- Force-Mobility Matrices: why a static Jacobian transpose alone cannot determine dynamic acceleration or reaction forces.
Relation to Published Work
MIT’s Underactuated Robotics notes explain manipulator dynamics and constrained multibody systems. The finite-inertia equations and COM balances used here are derived and checked for the explicitly declared model.
Robertson and Winter (1980) describe segment energy accounting in a sagittal walking model. That work motivates the interface-power bookkeeping; it is not golf-specific evidence for drift dominance.
MacKenzie, McCourt and Champoux (2020) estimate grip force and couple contributions to driver energy in 76 right-handed golfers. Such a force/couple split does not identify the zero-input and input-induced parts of either contribution. Their between-golfer associations also do not establish that instructing an individual to change a contribution will produce the corresponding speed gain.
7. Python Implementation of the Double Pendulum Model
The repository’s two-link mechanics reproducer implements the mass matrix, velocity bias and gravity used in the example. The distal-force and power checks calculate COM kinematics, recover the joint force and compare its interface power against an independent directional derivative of distal energy. They also check the COM moment balance and the numerical values above. The retained chapter checks cover COM kinetic energy, potential gradients and whole-system input power.
These are reproducibility checks for the stated mechanical identities, not a Streamlit application, an inverse-dynamics fit to golfer measurements or a validated prediction of performance.
7.1 Imports and Parameters
The reproducer declares the parameter values in Section 6.3. In a repository checkout with the documented test dependencies installed, run:
python -m pytest tests/test_drift_wrench_review.py tests/test_double_pendulum_chapter_rigor.pyThe power tests use a central directional difference of the independently computed COM energy, with an explicit absolute numerical tolerance. This checks the instantaneous rate; it does not integrate a zero-torque trajectory or establish long-horizon solver accuracy. Any forward comparison must specify its initial state, input history, numerical convergence and complete work balance separately.
Citation
@online{olson2025,
author = {Olson, Dieter},
title = {Natural Vs {Active} {Forces} and {Torques} in {Affine}
{Mechanical} {Systems}},
date = {2025-11-28},
url = {https://affinedrift.com/articles/drift-components-wrench-double-pendulum.html},
langid = {en}
}
The following are questions about the model and its interpretation, not quotations or an attributed scientific consensus:
The "Active" Definition vs. Impedance Control
What does zero input mean if activation changes stiffness or damping? If $u$ denotes net joint torque, muscle contributions are aggregated into that input. If the model instead retains spring or damping loads outside $u$, the zero-input prediction retains those loads. Activation-dependent parameters require an explicit state/input model; they cannot simply be labelled passive.
The "Energy Bank" Fallacy
How can an instantaneous partition account for previously stored energy? A flexible shaft can store strain energy and return some of it later, subject to damping and coupling to the rest of the system. Whether its elastic force belongs to drift depends on the declared model. The force partition alone does not identify who supplied the stored energy or whether its return increases clubhead speed.
Planar vs. Spatial Realities
Can a planar model explain clubface closure? It has no independent spatial face-orientation coordinate. A spatial model needs consistent axes, angular momentum, moments and hand constraints; this planar calculation cannot assign the size or sign of those omitted effects.