Lagrangian Reference for Control-Affine Multibody Dynamics
What the Lagrangian Description Does
Lagrangian mechanics derives motion from kinetic and potential functions, the chosen coordinates, and the forces and constraints that complete the model. Knowing only total energy at two instants is not enough to recover a swing. Different configurations and velocities can have the same energy, and a force can redirect motion without doing instantaneous work.
For a golfer, the useful connection is between preparation and later response. Earlier loading changes joint configuration, velocity and shaft deformation. Those quantities affect both the motion that continues with zero declared drive and the acceleration produced by a new drive. Lagrange’s equations organize that coupling; they do not establish which muscles supplied a load or which coaching instruction will improve delivery.
This reference derives a smooth mechanical model first, then states what changes for constraints, moving supports and actuator memory. Its manufactured examples verify bookkeeping, not a fitted golfer. The normative notation defines the site’s drift, input, ZTCF family and instantaneous ZVCF terminology.
Coordinates, Velocities, and the Mechanical State
Let \(z=(q,\eta)\in\mathbb R^N\) be a local coordinate chart, with rigid-body coordinates \(q\) and retained shaft-deformation coordinates \(\eta\). Here \(v=\dot z\). Coordinates need not all be angles: their conjugate generalized forces must be paired with the corresponding rates to compute power.
A global unconstrained Euclidean chart is not available for every configuration space. For example, a unit quaternion uses four numbers for orientation but only three independent angular-velocity components. Implementations often use \(\dot q=N(q)\nu\), where \(\nu\) denotes generalized velocities that need not be coordinate derivatives and can have a different dimension from \(q\). The simpler \(\dot z=v\) derivation below applies in its declared local chart; inserting quaternion components into it as four independent rotation coordinates is not valid. Singular charts, massless coordinates and redundant constraints also require treatment before asserting an invertible mass matrix.
The state \((z,v)\) is complete only for a model whose future is determined by those variables and its prescribed inputs. Activation, viscoelastic memory, controller state or contact history may require additional state. A photograph of a pose does not specify that complete state.
Energy Functions and Flexible Coordinates
For time-independent kinematics in an inertial observer, assume a regular mechanical model with
\[ T(z,v)=\tfrac12v^\mathsf T M(z)v,\qquad \mathcal L(z,v)=T(z,v)-V(z). \]
The inertia matrix is symmetric and positive definite when every nonzero velocity in these independent coordinates has positive kinetic energy. This is an assumption to check, not a remedy for a redundant or singular parameterization. Prescribed support motion can make the reduced Lagrangian explicitly time-dependent and introduce terms linear in velocity; it need not have precisely the autonomous form above.
For a flexible shaft, use the geometry of the deformed system:
\[ V(z)=V_{\mathrm{grav}}(q,\eta)+V_{\mathrm{el}}(q,\eta),\qquad M(z)=\begin{bmatrix}M_{qq}&M_{q\eta}\\M_{\eta q}&M_{\eta\eta}\end{bmatrix}. \]
Deflection can change elevation as well as elastic strain, so gravitational potential need not depend on \(q\) alone. A small-deformation approximation may use \(V_{\mathrm{el}}=\tfrac12\eta^\mathsf T K(q)\eta\) about a specified reference. Check the mode shapes, boundary conditions, retained modal inertia and range of validity. Off-diagonal inertia couples rigid motion and deformation.
Do not add an unexplained second “geometric energy” on top of a strain-energy model that already contains the same effect. Tangent stiffness comes from linearizing the declared force and constraint equations about a loaded state. Prestress can contribute geometric stiffness; it is not automatically a separate energy supply, nor is every geometric-stiffness effect a centrifugal effect.
Count Every Generalized Force Once
The forced Euler–Lagrange equations are
\[ \frac{d}{dt}\frac{\partial\mathcal L}{\partial v} -\frac{\partial\mathcal L}{\partial z}=Q. \]
For time-independent coordinate kinematics, virtual work defines \(\delta W=Q^\mathsf T\delta z\), and the associated power is \(Q^\mathsf T v\). A wrench must be reduced about a declared point and expressed in a compatible frame before its Jacobian transpose maps it into \(Q\). Net generalized load does not identify a unique muscle-force or bilateral-hand-force allocation.
Choose the following disjoint bookkeeping convention:
\[ Q=\bar B(z)u-D(z)v+Q_r(z,v). \]
Here \(u\) is the declared applied force or torque input, \(\bar B\) maps that input into generalized loads, \(-Dv\) is the modeled viscous damping, and \(Q_r\) contains other retained generalized-force laws. A load already represented by \(V\) or \(-Dv\) is excluded from \(Q_r\). An independently prescribed disturbance needs its own declared argument or channel; it is not silently identified with the input removed by a counterfactual.
With \(D=D^\mathsf T\succeq0\), the optional Rayleigh function \(\mathcal R=\tfrac12v^\mathsf T Dv\) gives \(Q_d=-\partial\mathcal R/\partial v=-Dv\). Writing damping directly in \(Q\) is equally legitimate. Rayleigh’s construction is a useful representation for suitable force laws, not a prerequisite for including every dissipative force. Dry friction or hysteresis can need a nonsmooth law or internal state; they do not become conservative because they appear in Lagrange’s equations. See the distinction between general external forces and the Rayleigh subclass in the author’s abstract on dissipation.
Manipulator Form and Its Energy Check
Expanding the derivatives and moving the same damping term to the left gives
\[ M(z)\dot v+c(z,v)+\nabla_zV(z)+D(z)v=\bar B(z)u+Q_r(z,v). \]
There is no additional copy of damping on the right. Define \(h=c+\nabla_zV+Dv-Q_r\), so the equation is \(M\dot v+h=\bar Bu\). The physical conservative generalized load is \(-\nabla_zV\); the positive potential gradient is a left-hand load term, not its signed force on the system.
For the kinetic energy above, one construction of the velocity-product vector is
\[ c_i=\sum_{j,k}\Gamma_{ijk}v_jv_k,\qquad \Gamma_{ijk}=\tfrac12\left( \frac{\partial M_{ij}}{\partial z_k} +\frac{\partial M_{ik}}{\partial z_j} -\frac{\partial M_{jk}}{\partial z_i}\right). \]
Writing \(c=C(z,v)v\) does not make \(C\) unique. The Christoffel choice \(C_{ij}=\sum_k\Gamma_{ijk}v_k\) satisfies \(C+C^\mathsf T=\dot M\). Thus \(\dot M-2C\) is skew-symmetric and gives \(v^\mathsf Tc=\tfrac12v^\mathsf T\dot Mv\). This identity is why the inertial terms fit the energy balance; treating Coriolis terms as an additional external energy source would break it. The Modern Robotics velocity-product derivation provides the standard construction, while MIT’s multibody notes cover the broader manipulator and constrained forms.
For \(E=T+V\) and the time-independent model,
\[ \dot E=v^\mathsf T\bar Bu+v^\mathsf TQ_r-v^\mathsf TDv. \]
This is mechanical power. It is not metabolic power, and zero mechanical power does not establish zero muscle effort. Elastic release changes the partition between \(T\) and \(V\); it is not another external input to the same whole-system energy balance.
As a small check, take one angular coordinate with inertia 2 kg m², stiffness 3 N m/rad and damping 0.2 N m s/rad. At angle 0.2 rad, rate 1 rad/s and zero applied torque, acceleration is \((-0.6-0.2)/2=-0.4\) rad/s². The mechanical energy rate is \(2(1)(-0.4)+3(0.2)(1)=-0.2\) W, exactly the damping loss. Counting the damping twice would predict a different motion and loss.
What Control-Affine Structure Establishes
For \(x_m=(z,v)\), invertibility of \(M\) and input-independent \(M,h,\bar B\) give
\[ \dot x_m=f(x_m)+G(x_m)u,\qquad f=\begin{bmatrix}v\\-M^{-1}h\end{bmatrix},\quad G=\begin{bmatrix}0\\M^{-1}\bar B\end{bmatrix}. \]
The inverse denotes a linear solve; numerical implementations should solve the system rather than form a dense inverse unnecessarily. The model may be very nonlinear in state while remaining affine in this particular input.
At one fixed complete state, changing \(u\) by \(\Delta u\) changes the vector field by \(G(x_m)\Delta u\). Doubling \(u\) doubles its input contribution, not the drift, total acceleration, subsequent trajectory, clubhead speed or energy. Two forward branches generally develop different states and therefore different \(f\) and \(G\). The relation is not a superposition theorem for whole swings or for neural effort.
The drift also depends on the declared zero of the input. Under a new convention \(u=u_0(x_m)+\widetilde u\), the same physical dynamics have drift \(f+Gu_0\) and input \(G\widetilde u\). Gravity compensation illustrates why a physical source and a chosen input convention must not be confused. State coordinates and output projections can change the displayed components without changing the motion.
Flexibility Changes the Input Response Too
Even when a shaft mode has no direct actuator, its acceleration can respond to joint torque through \(M^{-1}\). Consider two angular coordinates with manufactured SI coefficients
\[ M=\begin{bmatrix}2&0.4\\0.4&0.8\end{bmatrix},\qquad \bar B=\begin{bmatrix}1\\0\end{bmatrix},\qquad M^{-1}\bar B=\begin{bmatrix}5/9\\-5/18\end{bmatrix}. \]
The entries of \(M\) have units kg m²; \(\bar B\) is dimensionless, while \(M^{-1}\bar B\) has units (rad/s²)/(N m). The second coordinate has a zero row in the direct load map and a nonzero row in the acceleration map. It would therefore be wrong to say that adding shaft dynamics can change only drift. A changed inertia or constraint model changes how input is transmitted as well.
Use \(K=\operatorname{diag}(3,5)\), \(D=\operatorname{diag}(0.2,0.1)\), \(z=(0.2,-0.1)\) rad, \(v=(1,2)\) rad/s and \(u=4\) N m, with \(K\) in N m/rad, \(D\) in N m s/rad, \(Q_r=0\) and \(V=\tfrac12z^\mathsf TKz\). The acceleration is \((61/36,-17/36)\) rad/s². Applied power is 4 W, damping loss is 0.6 W and \(\dot E=3.4\) W. These numbers test the coupling and the energy balance together.
A Neural Command Is Not an Applied Torque
Suppose a scalar applied torque \(a\) has illustrative dynamics \(\dot a=(u_c-a)/T_a\). The mechanical equation uses \(a\), while command \(u_c\) enters the equation for \(a\). With state \((z,v,a)\) the system can still be affine in \(u_c\), but its input column acts in the actuator-state block. It is not the same \(G\) as the direct-torque model above.
If \(a(0)=6\) N m, \(T_a=30\) ms and \(u_c=0\), the torque after one time constant is \(6/e\) N m. A free inertia of 2 kg m² still accelerates at \(3/e\) rad/s². Directly removing applied torque would instead give zero acceleration in that otherwise unloaded example.
Activation-dependent stiffness, force–length and force–velocity effects, saturation, delay and recruitment can change the input law or require additional state. An input bound on an affine model does not itself remove affineness; passing an upstream command through saturation need not be globally affine in that command. State explicitly which level is being modeled before interpreting “zero control” or an effort increase.
Constraints, Contact, and Moving Supports
Solve Reactions With the Acceleration
For ideal, stationary, bilateral constraints \(\Phi(z)=0\), let \(J=\partial\Phi/\partial z\). Compatible motion satisfies \(Jv=0\) and \(J\dot v+\dot Jv=0\). In coordinates independent before these added constraints,
\[ \begin{bmatrix}M&-J^\mathsf T\\J&0\end{bmatrix} \begin{bmatrix}\dot v\\\lambda\end{bmatrix} = \begin{bmatrix}\bar Bu-h\\-\dot Jv\end{bmatrix}. \]
With \(M\succ0\) and independent constraint rows, the acceleration and multiplier are uniquely determined. Holding that smooth mode fixed, the constrained input-to-acceleration map is
\[ \begin{aligned} G_{v,c}={}&M^{-1}\bar B\\ &-M^{-1}J^\mathsf T(JM^{-1}J^\mathsf T)^{-1}JM^{-1}\bar B. \end{aligned} \]
Thus reactions depend on the input too. A counterfactual branch must solve its own reactions; retaining the original reaction forces after changing drive is a different model. Dependent constraints, unidentified internal forces or incompatible initial conditions need additional analysis rather than an unqualified inverse.
For a manufactured example, take \(M=\operatorname{diag}(2,3)\), \(J=[1,-1]\), \(\bar B=(1,0)^\mathsf T\) and a 10 N m input with \(h=0\). Both angular accelerations are 2 rad/s² and \(\lambda=-6\) N m. At rates \((0.4,0.4)\) rad/s, the two reaction-power terms are -2.4 and +2.4 W. They cancel for the complete constrained system while transferring power between its coordinates. The constrained acceleration map \(G_{v,c}\) is \((0.2,0.2)^\mathsf T\), whereas the unconstrained acceleration map \(M^{-1}\bar B\) is \((0.5,0)^\mathsf T\), both in (rad/s²)/(N m).
Nonholonomic Constraints Are Not Excluded
Ideal velocity constraints such as rolling without lateral slip can be modeled with Lagrange–d’Alembert equations and their reaction forces. They need not be integrable into position constraints. This is an established Lagrangian formulation, as described in the authors’ nonholonomic-mechanics abstract. Simply substituting a nonintegrable velocity restriction into an unconstrained Hamilton action is not generally the same variational problem.
For a shoe and ground, choose a contact law appropriate to the question: compliance, sticking, sliding, separation and rolling are different assumptions. Unilateral contact, frictional transitions and impacts can produce mode changes, set-valued forces or velocity resets. One smooth affine vector field does not automatically cover all of those regimes. Check feasibility, contact-law validity and mode transitions alongside the integration residuals.
Moving Boundaries Can Supply Power at Zero Drive
For a time-dependent constraint \(\Phi(z,t)=0\), compatible velocity satisfies \(Jv=-\Phi_t\). Differentiating this condition gives
\[ J\dot v=-\dot Jv-\frac{d\Phi_t}{dt} =-\dot Jv-J_t v-\Phi_{tt}. \]
Here \(\dot J\) is the total derivative along the motion and \(J_t\) is the partial time derivative. This replaces the stationary constraint’s second-row right-hand side in the block solve. With reaction \(J^\mathsf T\lambda\), its power is
\[ P_{\mathrm{constraint}}=v^\mathsf TJ^\mathsf T\lambda=-\lambda^\mathsf T\Phi_t. \]
It need not vanish. For a 2 kg mass moving vertically upward at 0.3 m/s with upward acceleration 0.4 m/s² and \(g=9.81\) m/s², the supporting force is 20.42 N. It supplies 6.126 W even with no separate actuator on the mass. Resetting that mass’s velocity to zero would violate the prescribed motion.
A prescribed hand path or moving base can therefore be an external power source in a reduced club model. Report that boundary work. If prescribed motion makes the mechanical model explicitly time-dependent, write \(f(x_m,t)\) or augment the state with a clock. The latter does not turn supplied boundary energy into a free internal resource. Once coordinates are eliminated, use the resulting time-dependent energy identity rather than silently reusing the autonomous balance above.
Counterfactuals With an Explicit Intervention
Distinguish Pointwise Evaluation From Forward Motion
The site uses pointwise ZTCF for evaluating the zero-applied-control vector field at an achieved complete state. A stitched trace of such evaluations follows the achieved states as its sampling locations; it is not an integrated zero-input motion. A forward ZTCF integrates from one specified state with the declared applied drive removed. A branched ZTCF starts such a forward motion at an achieved state and compares its future with the original future.
Retain the specified elastic loading, damping, constraints, other internal state and boundary laws, and identify exactly which input channel is removed. With actuator memory, setting command to zero need not implement applied-torque removal. A zero-input branch can contain energy stored by earlier active work and power supplied by a retained moving boundary. Calling it drift does not identify a biologically inactive golfer.
Give the Zero-Velocity Evaluation Its Correct Sign
For the unconstrained smooth model, or after valid reduction to independent coordinates, the instantaneous ZVCF acceleration is
\[ a_{\mathrm{ZVCF}}(z)=-M(z)^{-1}h(z,0),\qquad u=0. \]
If constraints remain explicit, solve their acceleration/reaction system at the reset state instead of using this unconstrained inverse. Reset all declared generalized velocities and velocity-like internal variables; retain configuration, non-velocity memory and frozen parameters. Check that the reset is admissible under the retained constraints. If it is not, do not report the resulting incompatible state as a physical release. In particular, when \(\Phi_t\ne0\), setting all velocities to zero cannot satisfy \(Jv=-\Phi_t\): that ZVCF reset is undefined within the retained constrained model. Freezing the boundary or resetting only relative velocities defines a different intervention that must be named and derived separately.
Only under additional assumptions does \(h(z,0)=\nabla_zV\). Quadratic velocity products and smooth viscous damping vanish at zero velocity, but retained internal loads, friction laws or prescribed boundary effects need not reduce to that potential gradient. A friction law can be set-valued at rest.
When reporting a generalized-force representation, its signed drift load is \(-h(z,0)\). The formal holding load is \(+h(z,0)\), and the ZVCF itself is an acceleration. In the coupled example, the signed drift load is \((-0.6,0.5)\) N m and acceleration is \((-17/36,31/36)\) rad/s². The holding load \((0.6,-0.5)\) N m cannot be supplied by the single first-coordinate actuator. A mathematical holding load is not automatically an executable control policy.
A Useful Golf Interpretation
Preparation, current state and current drive are intertwined. Earlier work can load a shaft and alter configuration; configuration and retained deformation change inertia, potential gradients and constraint geometry; current inputs act through that changed system. Measure the resulting power and endpoint energy without counting the same elastic release or constraint transfer twice.
A complete mechanical model still needs identified parameters, measured boundary conditions, and evidence that its approximations fit the task. It does not infer muscle recruitment, neural intent, metabolic efficiency or a coaching benefit from the affine form. Those questions need additional models and observations.
Energy functions describe how swing motion and loading depend on geometry. The energy total alone cannot give the full motion. Forces, constraints and the starting state still matter.
Preparation Changes the Next Response
Earlier loading changes position, speed and shaft bend. The same later torque can then produce a different acceleration. A loaded shaft can return stored energy. It does not create a new supply of energy.
A Local Input Rule Has Limits
At one fixed state, doubling a declared applied torque doubles its share of the instantaneous acceleration. It need not double the total acceleration or the clubhead speed. It also does not show that twice the mental or muscular effort gives twice the torque.
Citation
@online{olson2025,
author = {Olson, Dieter},
title = {Lagrangian {Reference} for {Control-Affine} {Multibody}
{Dynamics}},
date = {2025-11-28},
url = {https://affinedrift.com/articles/lagrangian-reference.html},
langid = {en}
}