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Acceleration-level contact complementarity decides whether a smooth, closed contact with zero normal velocity must carry a compressive reaction or begin to separate. At $d=0$ and $\dot{d}=0$, the normal conditions areThey do not constrain the sign of acceleration at a positive gap, and they do not resolve an impact with negative incoming normal velocity. The distinction between smooth constrained dynamics and time-stepped contact/impact models is explicit in the primary dynamics treatment.^[ref]

For a frictionless multibody model with generalized velocity $v$, invertible mass matrix $M$ and contact Jacobian $J$ defined by $\dot{d}=Jv$, writewhere $b=J{M}^{-1}f+\dot{J}v$ is the free normal acceleration and $W=J{M}^{-1}{J}^{\mathsf{T}}$. The normal reactions then solve the linear complementarity problemThe matrix couples contacts: choosing each reaction from a separate scalar formula generally fails when it changes acceleration at the other contacts. This formulation assumes the displayed velocity-coordinate kinematics and active closed contacts; additional friction forces modify the dynamics.

For an original coupled example, take free normal acceleration $b=(-3,-3{)}^{\mathsf{T}}\mathrm{m}{\mathrm{s}}^{-2}$ andThe reaction $N=(1,1{)}^{\mathsf{T}}\mathrm{N}$ gives zero normal acceleration at both contacts. Ignoring the off-diagonal coupling would instead choose $1.5\mathrm{N}$ at each; inserted into the full system, that produces $1.5\mathrm{m}{\mathrm{s}}^{-2}$ of outward acceleration at both loaded contacts and violates complementarity. Positive reaction and positive separating acceleration cannot coexist in this smooth active-contact model.

Because these are acceleration-level conditions, a numerical trajectory can still accumulate position or velocity drift. They must be paired with an integration/contact-update procedure that maintains the assumed state. During separated free flight, a negative acceleration can be perfectly admissible until the next contact event. At impact, an impulse and an appropriate velocity-level law are needed instead of applying the smooth equations across a discontinuity.

Accumulated contact-mode phase drift is the phase difference that builds over many timesteps when a numerical contact mode oscillates at a slightly incorrect frequency. A small error per step can matter after many cycles even when the amplitude remains bounded. The fast oscillations arise naturally in stiff compliant-contact models, whose numerical integration difficulty is described in the primary multibody notes.^[ref]

For a fixed undamped mode of angular frequency $\omega$, step $h$, and numerical phase advance $\theta (h\omega )$ per step, after $N$ steps the unwrapped phase error isUnwrapped phase counts complete cycles; reducing it modulo $2\pi$ would hide a missed or added cycle. This expression assumes the same fixed mode and a uniform step throughout the comparison.

For central differences, $\theta (z)=2\arcsin (z/2)$ on $0<z<2$. Expanding givesFor the trapezoidal rule, $\theta (z)=2\arctan (z/2)$ instead givesBoth are second-order phase errors at fixed duration, but their signs and leading coefficients differ. Halving $h$ approximately quarters the accumulated phase error; doubling the simulated duration approximately doubles it when the mode remains unchanged.

For an original example, choose $\omega =600rad/s$, $h=0.5\mathrm{ms}$ and $T=0.20\mathrm{s}$, so $N=400$ and $h\omega =0.3$. Central differences give $\tilde{\omega }\approx 602.273rad/s$ and a phase lead about $0.4546\mathrm{rad}$, or $26.05^{\circ }$. The relative frequency error is only about $0.379\%$, yet the phase lead has become readily visible. The leading-order estimate is $0.4500\mathrm{rad}$.

An approximate phase budget $|\Delta {\varphi }_N|\le {\varphi }_{\max }$ for central differences suggestsWith the preceding mode and duration and ${\varphi }_{\max }=0.020\mathrm{rad}$, it gives $h\lesssim 105.4\mu \mathrm{s}$. Taking $100\mu \mathrm{s}$ gives an exact phase lead about $0.01801\mathrm{rad}$. This is far below the $3.333\mathrm{ms}$ oscillator stability boundary; contact-mode timestep stability and timing accuracy impose different requirements.

The associated timing shift is approximately $-\Delta \varphi /\omega$ for a fixed phase feature of a nearly sinusoidal mode with small drift. It is not a general contact-event error bound: a changing amplitude, a grazing zero, changing tangent stiffness or an active-contact transition can invalidate that approximation. A variable-step method also needs its actual update analysis; summing uniform-step formulas without checking how successive maps compose is not automatically exact.

A proposed convergence test records a small compressed-contact oscillation over a stated duration, compares unwrapped numerical and reference phases, and repeats with half the step and twice the duration. Compare the force waveform and event times as well as the maximum penetration. Material damping changes the physical response, while numerical phase drift changes the timing of that modeled response; they should be tested separately.

The center of pressure summarizes where a distributed normal pressure acts. On a planar contact region $A$ in $z=0$, take a nonnegative pressure field $p(x,y)$ and positive total normal force $N$. Its center of pressure is the pressure-weighted positionIt is a continuous weighted mean, rather than necessarily the point of largest pressure. For discrete normal forces $N_i\ge 0$ at $(x_i,y_i)$, replace the integrals by sums. The resulting convex combination places the center of pressure inside the contacts' convex hull, or support polygon.^[ref]

This definition preserves the torque as well as the resultant normal force. About the origin in the contact plane,A single vertical force $(0,0,N)$ applied at $(x_P,y_P,0)$ gives the same torque. Thus its line of action passes through the center of pressure. Conversely, force-and-torque measurements reconstruct the point throughThe origin and sign convention are part of these equations; moving the moment reference above the contact plane requires the corresponding torque translation.

For an original discrete example, four vertical reactions of $100$, $200$, $300$ and $200\mathrm{N}$ act at the successive corners $(0,0)$, $(0.40,0)$, $(0.40,0.30)$ and $(0,0.30)$ in metres. Their total normal force is $800\mathrm{N}$, with $M_x=150\mathrm{N}\mathrm{m}$ and $M_y=-200\mathrm{N}\mathrm{m}$. Hence the center of pressure is $(0.250,0.1875)\mathrm{m}$. A rigid plate on four load cells provides a simple force platform implementation: calibrate the force channels and positions, then test known vertical loads at known locations. Subtracting the unloaded plate's own force and moment removes its contribution when reconstructing the added load. This is a proposed measurement test, not experimental data.

A continuous example shows the inverse problem hidden by this useful compression. On a rectangle of length $L$ and width $B$, put $u=x/L$ and choose profiles independent of $y$:Both profiles are positive on $0\le u\le 1$. The added polynomial has zero integral and zero first weighted integral:Thus both profiles give $N=p_{0}LB$ and the same center of pressure $(0.60L,B/2)$, while their peak pressures are $1.6p_0$ and $1.9p_0$. These are pressure distributions with identical resultant force and center of pressure. For $L=0.40\mathrm{m}$, $B=0.30\mathrm{m}$ and $N=800\mathrm{N}$, both give $(0.240,0.150)\mathrm{m}$ despite different local loads. A force platform alone therefore does not identify every local pressure.

Two limits matter. If $N=0$, the normal-pressure center of pressure is undefined. At fixed $N$, an absolute moment error $\delta M_y$ changes the reconstructed $x_P$ by $-\delta M_y/N$, so fixed absolute measurement uncertainty is amplified as a contact unloads. Also, tangential contact forces can leave a residual contact torque at the center of pressure: with full resultant $\mathbf{F}$,Selecting the normal-pressure centroid does not force this yaw torque to vanish.

Under the gravity-only, resting assumptions of support polygon, the center of pressure coincides with the horizontal center of mass projection. During motion it need not. For a planar model driven only by gravity and ground contact forces, at fixed center of mass height $h$ and with constant angular momentum about that centre, force and torque balance give $x_P=x_{\mathrm{CM}}-(h/g){\ddot{x}}_{\mathrm{CM}}$. The acceleration term explains why the center of pressure is a contact-force measurement rather than a general substitute for center of mass position.^[ref]

In terrestrial robot locomotion, the center of pressure is the ground point summarizing the normal load shared by the active contacts. With contact positions $(x_i,y_i)$ and nonnegative normal reactions $N_i$, its coordinates are the weighted averages $x_P={\sum }_i{N}_i{x}_i/{\sum }_i{N}_i$ and $y_P={\sum }_i{N}_i{y}_i/{\sum }_i{N}_i$, provided the total normal force is positive. It lies in the current support polygon, but is a contact-force quantity rather than the robot's center of mass position.^[ref]

Force-sensor coordinates matter when reconstructing this point. Suppose a sensor origin lies height $h$ above a horizontal contact plane, with parallel coordinate axes. If it reports force $(F_x,F_y,F_z)$ and moment $(M_x,M_y,M_z)$, the plane point is located at $(x_P,y_P,-h)$ relative to that origin. Moment translation givesThe ground-plane transformation of a measured contact wrench must therefore precede interpreting moment/force ratios as a contact location.

For an original sensor example, take $h=0.05\mathrm{m}$, force $(30,-20,600)\mathrm{N}$, and measured roll/pitch moments $(17,-49.5)\mathrm{N}\mathrm{m}$. The reconstructed point is $(0.080,0.030)\mathrm{m}$. Ignoring the origin height would instead give $(0.0825,0.02833)\mathrm{m}$. Those millimetre differences are coordinate effects, not a change in the physical pressure distribution.

During motion, the contact point need not equal the horizontal projection of the center of mass. The source's constant-height, negligible-angular-momentum-rate model relates their difference to horizontal acceleration; those assumptions must be established before using that simplified relation.^[ref] A robot-foot experiment can compare known applied loads with the sensor's translated force/moment measurements. When $F_z$ approaches zero, reconstruction becomes sensitive to moment error; at zero normal load the normal-pressure point is undefined. A residual yaw couple may remain, as explained in residual contact torque at the center of pressure.

Contact-mode numerical phase error is the difference between the oscillation phase produced by a time integrator and the phase of the modeled contact dynamics. It can grow while the numerical solution remains bounded. Contact-mode timestep stability controls a different property; stable integration alone does not ensure accurate force timing or contact-event timing. The physical motivation is the fast oscillation introduced by a stiff compliant contact model.^[ref]

For a frozen undamped contact mode $\ddot{x}+{\omega }^{2}x=0$, central differences givePut $x_n=e^{in\theta }$ and use $e^{i\theta }+e^{-i\theta }=2\cos \theta$. On the stable branch $0<h\omega <2$,The numerical frequency exceeds the modeled frequency. A Taylor series givesFor the trapezoidal rule, its amplification-factor phase instead givesThat scheme preserves modal amplitude in this undamped linear setting while lagging in phase. These formulas isolate a single mode with fixed frequency and consistent initialization; general nonlinear contacts require additional analysis.

Accumulated contact-mode phase drift after a duration $T$ is $\Delta \varphi =T(\tilde{\omega }-\omega )$ at time-grid points. For an original example, let $\omega =1000rad/s$ and $T=0.10\mathrm{s}$. With $h=1\mathrm{ms}$, the central-difference frequency is about $1047.20rad/s$, producing $4.720\mathrm{rad}$ of phase lead. The step satisfies the strict stability inequality, yet the phase lead exceeds half a cycle.

Reducing $h$ to $50\mu \mathrm{s}$ gives about $0.01042\mathrm{rad}$ of lead over the same duration. At $25\mu \mathrm{s}$ it is about $0.002604\mathrm{rad}$, approximately one quarter as large, matching the leading second-order error. For a small desired phase budget ${\varphi }_{\max }$, the central-difference expansion suggestsWith ${\varphi }_{\max }=0.01\mathrm{rad}$, this estimate is about $49.0\mu \mathrm{s}$, much smaller than the $2\mathrm{ms}$ stability boundary. Check the exact frequency formula when the small-step expansion is not sufficiently accurate.

In a contact experiment or simulation, a shifted oscillation phase changes when force maxima and gap crossings are predicted. Damping, changing tangent stiffness and a contact opening can invalidate the fixed-frequency model, so the phase estimate is not by itself an event-time error certificate. A proposed convergence check compares force histories, dominant frequency and event times at successively halved steps, while keeping contact geometry and constitutive parameters fixed. Agreement in maximum penetration alone can miss a phase error that accumulates over many cycles.

Contact-mode timestep stability concerns whether a numerical integrator introduces growing errors when advancing the fast dynamics of an active penalty contact. It depends on the integration rule and the local modal frequencies, not just the size of the modeled overlap. The primary compliant-contact treatment identifies stiff contact springs as a source of numerically stiff differential equations.^[ref]

Freeze the active contacts and linearize about a compressed equilibrium. An undamped mode has $\ddot{x}+{\omega }^{2}x=0$. For step $h$, put $z=h\omega$. The Forward Euler method updates displacement and velocity with matrixwhose eigenvalues are $1\pm iz$. Their magnitudes are $\sqrt{1+z^2}>1$ whenever $z>0$. Thus the method has no nonzero stable step for indefinite integration of this undamped oscillator, although smaller steps reduce its error on a fixed finite interval.

The central-difference recurrence instead givesFor $0<z<2$ its two distinct amplification roots lie on the unit circle. At $z=2$ the repeated root is $-1$, and solutions can include $n(-1{)}^n$, so the strict bound matters. A collection of frozen undamped modes therefore requires $h{\omega }_{\max }<2$ for this oscillatory stability condition. In coupled normal-contact penalty modes, ${\omega }_{\max }$ comes from the coupled system rather than an arbitrary total-mass estimate.

Stability still allows substantial contact-mode numerical phase error. For central differences the phase per step is $\theta =2\arcsin (z/2)$, giving numerical angular frequency $\theta /h$. At $\omega =1000rad/s$ and $h=1\mathrm{ms}$, $z=1$ is stable but the numerical frequency is about $1047.2rad/s$, $4.72\%$ high. A timing-sensitive contact event can consequently drift even when oscillation amplitudes do not grow.

Implicit schemes illustrate another tradeoff. For the Backward Euler method, the amplification factors are $(1\mp iz{)}^{-1}$ and have magnitude $1/\sqrt{1+z^2}$. At $z=1$, each step reduces the oscillatory amplitude by about $29.3\%$, despite the continuous undamped system conserving energy. The trapezoidal rule instead has factorswith phase $2\arctan (z/2)$. The same example gives $927.3rad/s$, $7.27\%$ low. Unconditional stability for this linear model does not imply correct timing at a large step.

A practical error check therefore compares both amplitude and phase while reducing $h$. For instance, a small target dimensionless step $h{\omega }_{\max }=0.1$ gives much smaller phase error than one, but the acceptable value depends on the observable and time span; it is not a universal accuracy certificate.

The frozen-mode argument has a defined scope. Contact opening changes the active stiffness, nonlinear force laws change tangent stiffness, and damping introduces additional eigenvalue scales. An integrator can satisfy the local oscillator bound while missing an intervening collision or contact release. A proposed test starts with a small compressed-contact perturbation, checks frequencies and energy behavior, then repeats across a known contact-opening event with decreasing steps. Record force histories and event times as well as penetration, so numerical damping or phase drift is not mistaken for a material response.

Coupled normal-contact penalty modes are small oscillations generated by several simultaneously compressed penalty contacts acting through shared body inertia. A displacement at one contact can accelerate another because both load the same translating or rotating bodies. Separate scalar effective mass in a normal-contact penalty mode estimates do not generally give the coupled frequencies.

Freeze a contact configuration, use independent generalized displacement perturbations $q$, positive-definite mass matrix $M$, and gap perturbations $\xi =Jq$. Let $K$ be a positive diagonal matrix of normal penalty stiffnesses. After subtracting equilibrium loads, the restoring generalized force is $-J^{\mathsf{T}}KJq$, soThe contact-Jacobian force/velocity pairing and stiff compliant-contact model come from the primary multibody formulation; the local modal reduction here follows by substitution.^[ref]

Although $AK$ need not be symmetric, it is similar to the symmetric positive-semidefinite matrix $K^{1/2}A{K}^{1/2}$. Set $z=K^{1/2}\xi$ to obtainIts positive eigenvalues are squared angular frequencies. A zero eigenvalue may describe redundant gap coordinates outside the realizable displacement subspace; it should not automatically be interpreted as an extra physical oscillation. Free generalized motions satisfying $Jq=0$ are a separate possibility.

For an original planar example, a rigid bar has vertical translation $y$, small rotation $\theta$, mass $m$, center-of-mass moment of inertia $I$, and two vertical contacts at horizontal offsets $-a,+a$. ThenWith equal contact stiffness $k$, the equal-gap mode is vertical translation and the opposite-gap mode is rocking. Their angular frequencies areTake $m=2\mathrm{kg}$, $I=0.005\mathrm{kg}{\mathrm{m}}^2$, $a=0.10\mathrm{m}$ and $k=10^{5}N/m$. Then $A$ has diagonal entries $2.5\mathrm{k}{\mathrm{g}}^{-1}$ and off-diagonal entries $-1.5\mathrm{k}{\mathrm{g}}^{-1}$. Its eigenvalues are $1$ and $4\mathrm{k}{\mathrm{g}}^{-1}$, giving $316.2$ and $632.5rad/s$. Treating the contacts independently would assign each $500rad/s$ and miss the faster rocking mode.

Contact-mode timestep stability depends on the largest active frequency and on the integrator. For the undamped central-difference recurrence, one mode has amplification equationThe distinct oscillatory roots lie on the unit circle for $0<h\omega <2$; the endpoints need separate treatment. The example therefore requires $h<2/632.5\approx 3.162\mathrm{ms}$ for this linear stability condition. Resolving the waveform accurately normally requires a smaller step. This is not a stability rule for every integration method.

The derivation applies while both contacts remain compressed and geometry is locally constant. A sufficiently large rocking amplitude can unload one contact, changing the active model. A proposed check preloads both supports, applies small symmetric and antisymmetric perturbations, and compares measured or simulated frequencies with the two predictions. Repeat with smaller amplitudes and smaller timesteps to separate linearization error, contact opening and numerical error.

The effective mass in a normal-contact penalty mode is the inertia seen by one specified relative normal displacement, rather than automatically the total mass of either body. A contact away from a center of mass can excite rotation as well as translation, changing that inertia and the normal contact penalty time scale.

Use independent generalized velocities $v$, a positive-definite mass matrix $M$, and a normal gap $g$ whose local velocity is $\dot{g}=Jv$. A scalar normal force $\lambda$ produces generalized force $J^{\mathsf{T}}\lambda$. The primary multibody formulation gives this contact-Jacobian velocity/force pairing explicitly.^[ref] At a frozen configuration, the part of gap acceleration caused by that force isIf the scalar $a=J{M}^{-1}{J}^{\mathsf{T}}$ is positive, defineThe definition uses a physical normal gap measured in metres and a matching normal force measured in newtons. Arbitrarily rescaling the gap coordinate changes the associated inertia. Constraints must first be accounted for by independent coordinates or the appropriate constrained response.

A short normal impulse $P$ gives $\Delta v=M^{-1}{J}^{\mathsf{T}}P$, hence $\Delta \dot{g}=P/m_{\mathrm{eff}}$. This supplies a direct interpretation: less effective mass means a larger relative normal-velocity change for the same impulse. At a fixed local configuration, a compressed linear penalty spring of stiffness $k$ gives the small-motion estimateIt omits changing contact geometry, other driving forces and damping; the source separately explains why stiff compliant contact challenges numerical integration.^[ref]

For two freely translating particles, the relative gap sees $a=1/m_1+1/m_2$, so $m_{\mathrm{eff}}=m_1{m}_2/(m_1+m_2)$. This is the reduced mass. Equal $2\mathrm{kg}$ particles give $m_{\mathrm{eff}}=1\mathrm{kg}$; replacing one by a fixed wall gives $2\mathrm{kg}$ instead.

For a planar rigid body above a fixed plane, let its mass be $m$, rotational moment of inertia about its center be $I$, and the contact normal be vertical. A contact a horizontal distance $r$ from the center hasFor $m=2\mathrm{kg}$, $I=0.020\mathrm{kg}{\mathrm{m}}^2$ and $r=0.10\mathrm{m}$, $m_{\mathrm{eff}}=1\mathrm{kg}$. At the center it is $2\mathrm{kg}$. With $k=10^{5}N/m$, the off-center angular frequency is $316.2rad/s$, compared with $223.6rad/s$ at the center. Using the full mass in both cases misses this rotational response.

Several simultaneous contacts require more than separate scalar replacements. Their matrix $A=J{M}^{-1}{J}^{\mathsf{T}}$ generally has off-diagonal terms: a force at one contact changes another contact’s acceleration. With frozen geometry and positive diagonal stiffness matrix $K$, compressed gap perturbations satisfy $\ddot{\xi }+AK\xi =0$; its nonzero modal squared frequencies are those of the symmetric matrix $K^{1/2}A{K}^{1/2}$. Coupled normal-contact penalty modes can therefore be faster than estimates treating each contact independently. A proposed verification applies a small known impulse or displacement perturbation at a stated configuration and checks relative contact motion, followed by a timestep convergence study.

For a controlled ordinary differential equation $\dot{x}=f(x,u,t)$, running cost $L(x,u,t)$ and terminal cost $\Phi (x)$, a smooth optimal value function satisfiesThis is the continuous-time Bellman equation. Split the cost into a short initial interval and the optimal continuation. Expanding the continuation by the chain rule, dividing by the interval length and taking its limit yields the equation. The optimization acts on the control at the current state and time.

For $\dot{x}=u$ and running cost $x^2+u^2$, the minimized expression is $x^2+u^2+V_{x}u$. Completing the square gives the minimizing control $u=-V_x/2$ and the equation $V_t+x^2-V_x^2/4=0$. The quadratic ansatz leads to linear-quadratic optimal control. ^[ref]

For a controlled diffusion process $dX=fdt+\sigma dW$, the corresponding equation adds $\frac{1}{2}\mathrm{tr}(\sigma {\sigma }^T{D}_x^{2}V)$ inside the infimum. A value function need not be differentiable; the viscosity solution formulation handles many such cases. A classical verification argument additionally requires the candidate feedback to be admissible and to attain the indicated infimum.

A linear complementarity problem asks for real vectors $z,w\in {\mathbb{R}}^n$ satisfyingfor a prescribed real matrix $M$ and vector $q$. The inequalities hold componentwise. Since every product $z_i{w}_i$ is nonnegative, the final equality means $z_i{w}_i=0$ for every index: each paired component has at least one zero. The linear complementarity problem is a feasibility problem; it need not come with an objective function. Its affine equation does not make the product condition linear.^[ref]

For an original two-variable example, takeThe pair $z=(1/2,0{)}^T$ and $w=(0,1/2{)}^T$ satisfies all three conditions. Guessing which member of each pair vanishes reduces the remaining equations to a candidate linear system, but each candidate must still satisfy nonnegativity. Choosing $z=0$ here gives $w=q$ and fails that check. Some data admit no solution: in one dimension, $M=-1,q=-1$ gives $w=-z-1<0$ for every $z\ge 0$.

The connection with quadratic programming explains one important tractable case. For a symmetric matrix $M$, considerThe Karush-Kuhn-Tucker conditions set its gradient equal to a nonnegative constraint multiplier $w$ and impose complementary slackness, giving precisely the displayed linear complementarity problem. If $M$ is a positive-definite matrix, the objective function is strictly convex and grows without bound as $\Vert z\Vert \to \infty$, so a unique minimizer exists. This conclusion uses that additional matrix property, not merely the word linear.

There is also a short uniqueness proof. For two complementary pairs,Positive definiteness then forces $z=z^{\prime }$. This argument gives a directly checkable reason for uniqueness in the stipulated case.

In unilateral contact complementarity, a nonnegative separation pairs with a nonnegative normal contact force: an open gap has zero contact reaction, while a positive reaction requires closed contact. A time-discretized rigid body model can relate its next-step gap affinely to the unknown reaction, producing a linear complementarity problem. The particular discretization and impact law determine that affine relation; the algebraic form alone does not select the physical contact model.^[ref]

A standard finite-horizon linear-quadratic problem uses linear dynamics $\dot{x}=Ax+Bu$ and minimizeswhere $Q,Q_f$ are symmetric positive semidefinite matrices and $R$ is positive-definite. The control input is unrestricted in Euclidean space. These hypotheses, rather than the word quadratic alone, make the control minimization well posed.

The value function has the quadratic form $V(t,x)=x^{T}P(t)x$. Substitution into the Hamilton-Jacobi-Bellman equation and completion of the control square giveThus a backward matrix Riccati equation gives a linear feedback control. ^[ref]

For the scalar system $\dot{x}=u$, terminal cost zero and running cost $x^2+u^2$, the solution is $P(t)=\tanh (T-t)$ and $u_{\ast }=-\tanh (T-t)x$. Differentiation verifies $-\dot{P}=1-P^2$. Infinite-horizon formulas need additional stabilizability and detectability conditions; they do not follow from an arbitrary quadratic cost.

Local contact pressure from additional spatial measurements requires more information than total normal force and two first moments. One explicit reconstruction model partitions the contact into cells of known areas $A_i$ and centroids $(x_i,y_i)$, assuming uniform pressure $p_i$ within each cell. Writing $F_i=A_i{p}_i$ givesThese are three linear measurements of the cell loads. They need not determine every $F_i$, as the nonuniqueness argument in pressure distributions with identical resultant force and center of pressure shows. The primary centroid definition specifies precisely which force/moment information is retained.^[ref]

For an original four-cell example, use equal $0.01{\mathrm{m}}^2$ cells centered at successive positions $(-a,-a),(a,-a),(a,a),(-a,a)$ with $a=0.05\mathrm{m}$. Loads $(10,20,30,40)\mathrm{N}$ give $N=100\mathrm{N}$ and centroid $(0,0.020)\mathrm{m}$. Adding $s(1,-1,1,-1)$ to these loads changes neither the sum nor either first moment. Nonnegative loads permit $-10\le s\le 20\mathrm{N}$. The alternative $(15,15,35,35)\mathrm{N}$ therefore gives the same global measurements.

An independently calibrated measurement of the first cell load resolves this one missing degree of freedom under the four-cell model. If it reads $10\mathrm{N}$, the global moment equations recover the original other three loads. Their cell pressures are $(1,2,3,4)\mathrm{kPa}$. Measuring all four cells instead provides redundant data for checking total load and the assumed geometry.

The local sensor output must actually represent a known cell-integrated force, or have a validated response model; a point sample cannot automatically be treated as an area average. Calibration errors, cell cross-coupling and imperfect coverage become additional measurement inputs. Spatial pressure-map resolution below a sensor cell remains limited even when every cell average is known: unmeasured within-cell peaks are not recovered by simply drawing a finer image. Constitutive or smoothness assumptions can add information, but must be identified as assumptions rather than extra measurements.

The normal contact penalty time scale describes the fast dynamics introduced by a stiff normal-contact approximation. For a compressed linear penalty spring with stiffness $k$ acting on a point mass $m$, small oscillations about equilibrium satisfy $m\ddot{x}+kx=0$. Their angular frequency, reference time and period areThe reference time $t_c$ is not the full oscillation period. Rescaling time by $s=t/t_c$ gives $d^{2}x/d{s}^2+x=0$, so a numerical step is controlled by its dimensionless size $h/t_c$, not just its duration in seconds. The primary compliant-contact treatment highlights the numerical stiffness caused by such springs.^[ref]

A modeling and numerical budget interact. Limiting equilibrium overlap under a specified maximum supported load $F_{\max }$ to $d_{\max }$ requires $k\ge F_{\max }/d_{\max }$ in this linear model. For an original choice $F_{\max }=10\mathrm{N}$ and $d_{\max }=10\mu \mathrm{m}$, choose $k=10^6\mathrm{N}{\mathrm{m}}^{-1}$. At $m=0.5\mathrm{kg}$,A chosen resolution target of twenty steps per contact oscillation would set $h\le 0.2221\mathrm{ms}$. This is a resolution target, not a universal accuracy or stability theorem for every integrator.

Increasing the stiffness by a factor of 100 reduces the same-load overlap by 100 but reduces these times by only ten. Keeping the old step then gives roughly two samples per new oscillation instead of twenty. The parent penalty approximation of unilateral contact also shows why the Forward Euler method's undamped-oscillator instability cannot be cured merely by calling the step small.

For a multibody contact, the effective mass in a normal-contact penalty mode replaces an arbitrary total body mass. Additional damping and nonlinear force laws can introduce other time scales. A convergence study should therefore check force histories and contact events as well as maximum overlap; smaller geometric penetration alone does not establish a more accurate simulation.

A penalty approximation of unilateral contact allows a small modeled overlap and uses a stiff restoring force instead of enforcing an exact rigid nonpenetration constraint. With signed gap $d$, positive in separation, a simple nonadhesive law isAn open or just-touching contact has zero force in this spring law; a sustained compressive load requires negative $d$. Thus it approximates unilateral contact rather than satisfying its rigid gap condition exactly. Tedrake's primary compliant-contact treatment uses stiff springs and damping to resist penetration and explains the resulting numerical stiffness.^[ref]

For an original gravity-supported example, a $2\mathrm{kg}$ mass with no other vertical load has $N=mg=19.62\mathrm{N}$. With $k=2\times 10^5\mathrm{N}{\mathrm{m}}^{-1}$, its static overlap is $mg/k=98.1\mu \mathrm{m}$. Increasing $k$ by a factor of 100 reduces that overlap to $0.981\mu \mathrm{m}$, but also makes the contact dynamics faster.

While the contact stays compressed, linearization about equilibrium givesThe normal contact penalty time scale is therefore proportional to $\sqrt{m/k}$. For the first stiffness above, ${\omega }_c\approx 316.2\mathrm{rad}{\mathrm{s}}^{-1}$; after increasing stiffness it is ten times larger. At fixed numerical step size, a smaller overlap error need not mean a more accurate trajectory.

For example, the Forward Euler method applied to this undamped oscillator has amplification factors $1\pm i{\omega }_c\Delta t$, whose magnitudes exceed one for every positive step. Merely shrinking the step does not make that method asymptotically stable for indefinite undamped oscillation, although it can reduce error on a finite interval. A suitable integration method, damping model and convergence check must be chosen. The source notes specifically identify stiff differential equations as a challenge for compliant contact simulation.^[ref]

A penalty stiffness is a chosen model parameter, not automatically the actual material stiffness. Impact dissipation and tangential friction still need their own constitutive choices; a purely elastic normal spring does not imply a perfectly inelastic collision.

Pressure distributions with identical resultant force and center of pressure demonstrate that a pressure centroid is not a reconstruction of a pressure field. A normal-load measurement and its two first spatial moments constrain onlyA perturbation $\delta p$ preserving all three integrals leaves the resultant and center of pressure unchanged. It is physically admissible only when the perturbed field remains nonnegative under the nonadhesive pressure model. The polynomial profiles in the parent article give one smooth example; the following construction shows that even a peak-pressure bound is missing.

Take a rectangular region of length $L$ and width $B$, centered about its midpoint, and fix a normal load $N>0$. A uniform field has pressure $N/(LB)$ and centroid $(L/2,B/2)$. Instead place the entire load uniformly on a smaller centered square of side $a\le \min (L,B)$, with zero pressure outside that square. Its pressure isSymmetry fixes the same centroid, and integration still gives total force $N$. Yet its peak is $N/a^2$, which has no upper bound as $a$ shrinks. Smooth symmetric concentrated profiles can provide the same conclusion if a continuous pressure field is required.

For an original numerical example, let $L=0.40\mathrm{m}$, $B=0.20\mathrm{m}$, and $N=800\mathrm{N}$. Uniform pressure is $10\mathrm{kPa}$. Concentrating the same load on a centered $0.05\mathrm{m}$ square gives $320\mathrm{kPa}$ while preserving the measured point $(0.20,0.10)\mathrm{m}$. Both satisfy the stated nonnegative-field moment constraints.

Consequently a force/moment sensor cannot infer local peaks from the centroid alone. Local contact pressure from additional spatial measurements needs more data or justified material/contact assumptions, such as a constitutive model and known geometry. The source's normal-force centroid definition explains exactly which resultant information was retained.^[ref] The examples are inverse-problem constructions; they are not claims that arbitrary pressure maps can be realized by the same elastic surfaces.

A residual contact torque at the center of pressure is the moment left after translating a planar contact's resultant force to its normal-pressure centroid. Let the resultant wrench at an origin in $z=0$ be force $F=(F_x,F_y,N)$ and moment $M=(M_x,M_y,M_z)$, with $N>0$. Put $r_P=(x_P,y_P,0)$, where $x_P=-M_y/N$ and $y_P=M_x/N$. Translation givesThe two horizontal moment components vanish, but the yaw component need not. The zero-moment-point construction for flat robot contact removes moments perpendicular to the surface normal; it does not assert that every component is zero.^[ref]

A useful invariant makes the limitation exact. Since $F\cdot (r_P\times F)=0$,If this scalar is nonzero, no change of the moment origin can eliminate the entire moment vector. Replacing the whole contact by only one force at the center of pressure would discard that information.

For an original two-contact example, put equal normal loads of $100\mathrm{N}$ at $(+0.10,0)$ and $(-0.10,0)$ metres. Add tangential forces $(0,+15,0)\mathrm{N}$ at the positive position and $(0,-15,0)\mathrm{N}$ at the negative position. The total force is $(0,0,200)\mathrm{N}$, the pressure centroid is the origin, and the opposing tangential forces give a pure yaw couple of $3\mathrm{N}\mathrm{m}$. A friction coefficient of at least $0.15$ permits the stipulated tangential magnitudes at these loads.

A normal-load-only reconstruction would identify the correct centroid while missing the couple. A sensor measuring all force and moment components can distinguish the two situations. The origin, surface orientation and axis signs must be retained when comparing measurements. Near zero normal load, the centroid reconstruction itself becomes singular, independently of whether a tangential moment remains.

Russ Tedrake authored Underactuated Robotics by Russ Tedrake, developing mathematical methods for controlling systems with fewer independent actuators than degrees of freedom. Its treatment of the linear-quadratic optimal control derives feedback gains from a Riccati equation. This connects a local dynamical model and a quadratic performance objective with an explicit stabilizing control law under appropriate assumptions.^[ref]

Unilateral contact permits bodies to separate while preventing them from passing through each other. In an ideal rigid, nonadhesive model, the supporting normal contact force can push but cannot pull. It is therefore different from a permanently attached support: an upward pull may lift a body away instead of producing a downward support reaction.

Let $d$ be the signed normal gap, positive when the bodies are separated, and $N$ the outward compressive normal contact force. The nonpenetration constraint and absence of adhesion give the unilateral contact complementarity conditionsAn open gap requires $N=0$, while $N>0$ requires a closed gap. The pair $d=N=0$ is also allowed: touching does not by itself imply a loaded contact.^[ref] Tangential friction requires an additional relation such as Coulomb friction; it is not specified by the normal-gap conditions.

The equation of motion is still needed to determine the reaction. Consider a block of mass $m$ on a fixed horizontal floor, initially at rest at $d=0$, with an upward applied force $u$ through its center of mass. With upward positive, Newton's second law isIf the contact remains closed, $\ddot{d}=0$ would require $N=mg-u$. That is feasible for $u\le mg$. For $u>mg$, a negative normal reaction would violate unilateral contact, so the contact opens with $N=0$ and $\ddot{d}=(u-mg)/m>0$. At a closed contact with zero normal velocity, the appropriate acceleration-level contact complementarity iswhich yieldsThese acceleration conditions are being applied at a closed, initially stationary contact; they are not a rule forbidding downward acceleration during separated free flight.

For an original numerical example, let $m=2\mathrm{kg}$ and $g=9.81m/{\mathrm{s}}^2$. Pulling upward with $u=5\mathrm{N}$ leaves a normal reaction of $14.62\mathrm{N}$. At $u=25\mathrm{N}$, the reaction is zero and the initial upward acceleration is $2.69m/{\mathrm{s}}^2$. At $u=19.62\mathrm{N}$, both the reaction and the initial acceleration vanish, while touching remains possible.

A simple apparatus test places the block on a calibrated load cell and applies the upward pull through a spring balance, keeping the pull vertical through the center of mass to avoid an extra overturning torque. The support reading should decrease with the pull while contact persists, then vanish at lift-off. Recording the gap as well distinguishes an unloaded touch from a genuinely open contact. Real support elastic deformation and adhesion must be accounted for before treating that apparatus as the ideal rigid, nonadhesive model.

A penalty approximation of unilateral contact replaces perfect rigidity by a chosen stiff response, for exampleHere a small negative gap represents an allowed modeled overlap relative to undeformed reference surfaces. It is not an exact solution of $d\ge 0$. For the $14.62\mathrm{N}$ supported load above and chosen penalty stiffness $k=10^{5}N/m$, equilibrium requires $d=-0.1462\mathrm{mm}$. Increasing $k$ reduces that overlap at the same load. The penalty method and the ideal constraint are different approximations of the contact response.

Finally, impact requires a rule for the velocity change. The position-level inequalities alone do not decide whether an approaching body sticks or rebounds. A coefficient of restitution or another specified impact law determines that behavior, and the impulsive contact force is summarized by its impulse. Tedrake's particular time-stepping formulation identifies its block collisions as perfectly inelastic; that is a feature of the stated model, not a consequence of the three gap/reaction inequalities alone.^[ref]