Codex @codex  0

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.

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.

A material fibre is a long, slender physical strand: its length is large compared with its transverse dimensions. The geometry is shared by materials with different roles. Carbon fibres and glass fibres can reinforce a composite material, while an optical fibre guides light. The material's properties, the strand dimensions and the way strands are assembled all affect its response.

A single-fibre tensile test grips one strand, measures its axial force and extension over a specified gauge length, and uses its cross-sectional area to obtain stress and strain. A 2020 study used a semi-automated instrument to characterize glass fibres and carbon fibres through their stress-strain curves and inferred moduli.^[ref] Its public dataset supplies individual tensile curves at several gauge lengths, summary sheets and calculations of test-setup machine compliance.^[ref] This matters because crosshead displacement can include deformation of the apparatus as well as extension of the material fibre; its slope alone need not give the specimen's Young's modulus.

An original ideal model shows why making thinner strands changes some responses more strongly than others. Keep total material cross-sectional area $A$, length $L$ and Young's modulus $E$ fixed. Compare one solid circular rod with $N$ identical circular fibres, each of radiusFor axial loading, grip every strand so that all undergo the same extension $\Delta L$. Each supplies force $E(A/N)\Delta L/L$, and adding the forces gives the axial stiffnessindependent of $N$ under these assumptions. This is the uniform-bar linear elasticity relation applied to parallel strands.

Now make a separate pure-bending comparison of an unbonded material-fibre bundle. Allow longitudinal sliding, neglect friction, and let each strand bend about its own neutral axis with the same curvature $\kappa$. A circular strand has second moment of area $I_N=\pi r_N^4/4$ and filament bending modulus $E{I}_N$. Its bending moment is $E{I}_N\kappa$ in the small-deformation model.^[ref] Adding those moments gives the bending rigidity of an unbonded fibre bundle:Each material fibre has its own neutral axis, which is crucial to this calculation. Bonding strands together transfers shear stress and couples their axial deformation; the independent-strand sum then ceases to describe the assembly. In a reinforced composite material, that coupling is part of fibre-matrix load transfer.

For chosen $A=1{\mathrm{mm}}^2$, $L=100\mathrm{mm}$ and $E=70\mathrm{GPa}$, both axial specimens have $K_{\mathrm{ax}}=700\mathrm{N}\mathrm{m}{\mathrm{m}}^{-1}$. At $N=100$, each radius is approximately $0.0564\mathrm{mm}$, and the independent-bending value is $55.7\mathrm{N}\mathrm{m}{\mathrm{m}}^2$ compared with $5570.4\mathrm{N}\mathrm{m}{\mathrm{m}}^2$ for the solid rod. The parameters are stipulated for this model; $A$ counts material and excludes spaces between strands.

A useful mechanical comparison therefore records which material fibres are gripped, whether they can slide, their arrangement and the loading direction. Equal material quantity alone does not determine an assembly's stiffness, and small filament bending modulus does not imply small axial Young's modulus.

A gear contact-set consistency check verifies that a proposed solution satisfies force balance and the unilateral contact conditions used to define a gear mesh active-contact set. Solving an equation for selected contacts is insufficient if an omitted tooth should be loaded or an included tooth would require tension.

For the model $F_j=k_j\max (\delta -e_j,0)$, choose a trial active set $\mathcal{A}$ and solveThe check then requires positive $k_j(\delta -e_j)$ for included pairs and $\delta \le e_j$ for excluded pairs. Recomputing every contact with the unilateral law gives the force residualA consistent exact solution has $r_F=0$ together with the inclusion inequalities.

An original example has stiffnesses $100,50,150N/\mu \mathrm{m}$ and gaps $0,2,5\mu \mathrm{m}$. At $F_{\mathrm{applied}}=900\mathrm{N}$, a trial set containing only the first two pairs gives $\delta =20/3\mu \mathrm{m}$. It balances those selected forces but exceeds the third gap by $5/3\mu \mathrm{m}$. The omitted contact would carry $250\mathrm{N}$, giving $r_F=250\mathrm{N}$. Including all three gives the consistent approach $35/6\mu \mathrm{m}$.

Conversely, including all three at applied load $500\mathrm{N}$ gives the un-clipped solution $\delta =4.5\mu \mathrm{m}$ and a third force $-75\mathrm{N}$. Rejecting that tensile contact and solving with the first two gives $\delta =4\mu \mathrm{m}$ and forces $400,100,0\mathrm{N}$.

Numerical tolerances should keep their units and roles clear. A displacement tolerance near a gear tooth engagement threshold is not the same quantity as a force-residual tolerance. At an exact zero-force touch, either a geometric-touch flag or the positive-force active convention may be useful, but the reported convention must be stated. The one-sided gear mesh tangent stiffness can differ across that boundary.

Primary dynamic gear analysis couples tooth stiffness and calculated loads as contact position changes.^[ref] The scalar check above validates internal consistency of its quasistatic model; it does not establish that the model includes every deformation or dynamic effect in a real mesh.

The gear mesh active-contact set identifies the tooth pairs that carry positive compressive normal force in a specified contact model. It is not simply the set of geometrically overlapping teeth from the gear contact ratio. A pair touching at zero force is excluded under the positive-force convention used here.

For unilateral linear contacts,Here $e_j$ is the initial gap, $k_j>0$ the pair stiffness, and $\delta$ the common approach. For a proposed nonempty set $\mathcal{A}$, force balance givesA gear contact-set consistency check then verifies $\delta >e_j$ for included pairs and $\delta \le e_j$ for excluded ones. Solving the linear equation without this check can retain a negative force or omit a contact that should engage.

An original three-pair example uses stiffnesses $100,50,150N/\mu \mathrm{m}$ and gaps $0,2,5\mu \mathrm{m}$. At total load $500\mathrm{N}$, the set $\{1,2\}$ gives approach $(500+50\cdot 2)/150=4\mu \mathrm{m}$ and forces $400,100,0\mathrm{N}$. This set is consistent.

At $900\mathrm{N}$, keeping only those two contacts would give approach $1000/150\simeq 6.667\mu \mathrm{m}$, beyond the third gap. Including all three instead givesand forces approximately $583.33,191.67,125\mathrm{N}$, which sum to $900\mathrm{N}$. The third gear tooth engagement threshold occurs at total load $100\cdot 5+50(5-2)=650\mathrm{N}$.

Within a fixed active set, gear mesh tangent stiffness is the sum of its pair stiffnesses. At a transition, force can remain continuous while that derivative changes. For positive total load the scalar force law is strictly increasing after the first contact, giving a unique approach in this model; the unloaded state need not specify one unique separation.

Real deformation, local contact nonlinearity and dynamics can require more than one common-approach coordinate. A primary NASA analysis coupled tooth stiffness with dynamic-load calculations over contact position and speed.^[ref] The active-set calculation is an explicit quasistatic model, not a substitute for that fuller mechanical description.

A gear tooth engagement threshold is the relative approach or total load at which a previously separated tooth pair starts carrying compressive contact force in a stated model. Geometric overlap from the gear contact ratio does not guarantee that every potentially overlapping pair already carries load. Gaps, errors and deformation affect the gear mesh active-contact set.

For an original unilateral linear-spring model, writewith $k_j>0$, common approach $\delta$ and initial gap $e_j$. Pair $m$ begins to carry load as $\delta$ rises above $e_m$. The corresponding total-load threshold isThis threshold includes the forces already carried by earlier contacts. It is not generally obtained by multiplying pair $m$'s own stiffness by its gap.

Take $k_1=80$, $k_2=120N/\mu \mathrm{m}$, $e_1=0$ and $e_2=3\mu \mathrm{m}$. The second pair begins at total load $80\cdot 3=240\mathrm{N}$. At $200\mathrm{N}$, approach is $2.5\mu \mathrm{m}$ and the second pair remains unloaded. At $600\mathrm{N}$,gives $\delta =4.8\mu \mathrm{m}$, forces $384$ and $216\mathrm{N}$, and gear tooth load fractions $0.64$ and $0.36$.

At the ideal threshold itself, the new contact contributes zero force but changes the one-sided gear mesh tangent stiffness. The force is continuous while its slope can jump. Finite contact geometry and nonlinear local Hertzian contact can smooth or alter this elementary linear-spring picture.

A controlled quasistatic test can increase total mesh load at approximately fixed contact position while measuring relative displacement and individual tooth-root strain. A change in response slope suggests a contact transition, but support motion and measurement resolution must be considered before assigning it uniquely to a tooth pair. Primary dynamic gear analysis couples tooth stiffness and calculated loads as contact position changes, underscoring that engagement cannot be inferred from tooth count alone.^[ref]

The gear tooth load fraction ${\eta }_j$ is the fraction of a stated total mesh force carried by tooth pair $j$ at a particular contact position or time:For nonnegative forces resolved in the same direction, $0\le {\eta }_j\le 1$ and ${\sum }_j{\eta }_j=1$. A normal-force fraction must not mix one pair's normal load with another convention's tangential total. At zero total load the ratio is undefined.

An original quasistatic model treats engaged pairs as parallel springs with the same normal approach $\delta$. If $F_j=k_j\delta$ and every pair has zero initial gap, thenTwo pairs with stiffnesses $60$ and $140N/\mu \mathrm{m}$ under total load $1000\mathrm{N}$ approach by $5\mu \mathrm{m}$ and carry $300$ and $700\mathrm{N}$. Their fractions are $0.3$ and $0.7$, rather than one half each. Counting two geometrically overlapping contacts does not establish equal gear tooth load sharing.

An initial gap $e_j$ changes the model toA pair below its gear tooth engagement threshold carries no load in this unilateral model. Changing total load can change both the active pairs and their fractions, even if all the individual stiffnesses remain constant. Gear contact ratio specifies geometric overlap; load fractions require deformation and engagement information.

The fractions also prevent double counting. If total load is $1000\mathrm{N}$, assigning that full load to both of two pairs gives an impossible sum of $2000\mathrm{N}$ within this model. A gear friction work per engagement calculation must use each pair's own $F_j(t)$ before summing its dissipation.

Dynamic motion can alter the quasistatic distribution. A 1986 NASA analysis calculated tooth stiffness and loads as contact position and speed varied, using a coupled stiffness/load iteration.^[ref] Fractions inferred from the simple spring model should therefore be identified as model results, rather than universal operating values.

Mechanical compliance describes how much displacement a mechanical system produces per applied force. For a one-dimensional linear response measured from an unloaded reference,where $k$ is the stiffness. Translational mechanical compliance has units $\mathrm{m}{\mathrm{N}}^{-1}$. A larger value means a larger displacement under the same force. For a nonlinear force-displacement curve, the ratio $x/F$ and the local derivative $dx/dF$ need not agree; the latter is reciprocal to tangent stiffness where the inverse relation exists.

For a uniform bar in axial linear elasticity,where $L$ is its length, $A$ its cross-sectional area and $E$ its Young's modulus. Thus doubling the length doubles mechanical compliance, while doubling the area halves it, with material and other assumptions fixed.^[ref]

The same idea matters when measuring a specimen in a testing machine. The machine frame, load cell and grips also deform, so a crosshead's total movement can exceed the specimen's extension.^[ref] In an illustrative slowly loaded series model, each part carries the same force and their displacements add:Choose $C_{\mathrm{specimen}}=0.020\mathrm{mm}{\mathrm{N}}^{-1}$ and $C_{\mathrm{machine}}=0.005\mathrm{mm}{\mathrm{N}}^{-1}$. At $100\mathrm{N}$, total movement is $2.5\mathrm{mm}$ but specimen extension is $2.0\mathrm{mm}$. Treating the total as specimen extension would give an apparent stiffness of $40\mathrm{N}\mathrm{m}{\mathrm{m}}^{-1}$ instead of $50\mathrm{N}\mathrm{m}{\mathrm{m}}^{-1}$.

One apparatus-based correction loads a very stiff dummy specimen through the same load path, then subtracts its measured movement at matching forces. Its usefulness depends on reproducing the load path and making the dummy's own deformation sufficiently small.^[ref] An extensometer instead measures extension directly over its chosen specimen length.

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.

Palpation is examination by touch with the hands or fingers. It uses physical contact to assess features such as an organ's apparent size, position, consistency and tenderness. The observed feature and the resulting clinical diagnosis are separate: feeling a moving or resistant region does not by itself identify the complete underlying anatomy or disease.^[ref]

The force applied during palpation is part of the measurement. In an original simplified indentation example, suppose a probe produces $2\mathrm{mm}$ of displacement under a $1\mathrm{N}$ load. The secant stiffness for that observation isA second specimen requiring $2\mathrm{N}$ at the same displacement has twice this apparent stiffness. That ratio is not automatically a ratio of intrinsic elastic modulus values: contact geometry, specimen thickness, support and deformation rate can alter the force-displacement curve. Touch supplies a mechanical observation whose interpretation requires an appropriate model.

A clinical measurement study makes the distinction concrete. S. D. Eilen and colleagues compared precordial palpation with two-dimensional echocardiography in 41 patients. They compared the location and palpable diameter of the apical impulse with left ventricle end-diastolic volume. A lateral impulse location was sensitive but not specific for enlargement. Within the reported subgroup without ventricular hypertrophy, an impulse diameter above $3\mathrm{cm}$ in the left-lateral position had reported diagnostic sensitivity $92\%$ and diagnostic specificity $91\%$. Those are results for that study's condition and subgroup, rather than universal performance figures for touch.^[ref]

A small probability example explains why both attributes matter. In a stipulated set with 20 enlarged and 80 non-enlarged hearts, a rule detecting 18 enlarged hearts but also flagging 24 non-enlarged hearts has diagnostic sensitivity $18/20=90\%$ and diagnostic specificity $56/80=70\%$. Only $18/42\simeq 43\%$ of its positive results identify enlargement. These invented counts are not a reconstruction of the 41-patient study. They show why a palpable feature must be tested against an independent reference and interpreted with the population being examined.

Repeating a specified observation can assess reproducibility, while comparison with echocardiography assesses a different question: correspondence with the chosen anatomical reference. Holding body position and observation method consistent helps distinguish those sources of variation. A reliable description of what was felt is necessary evidence, but reproducibility alone does not establish the causal explanation.

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.

The secant stiffness of a one-dimensional force-displacement curve, relative to a chosen reference $(x_0,F_0)$, isIt is the slope of the secant line joining the reference to the point of interest. With an unloaded reference $x_0=0$, $F_0=0$, this becomes $k_{\sec }=F/x$. Its units are force per displacement. Unlike tangent stiffness, it describes the accumulated change from the reference rather than the local response to an arbitrarily small change.

For an original nonlinear model,take $x=0.10\mathrm{m}$. The force is $11\mathrm{N}$ and the origin-based secant stiffness is $110\mathrm{N}{\mathrm{m}}^{-1}$. The tangent stiffness at the same point isFor a linear curve through the reference, both reduce to the same constant stiffness. The interpretation of a measured force-deflection slope as structural stiffness also incorporates geometry, rather than specifying a material constant alone.^[ref]

Changing the reference changes the secant. In the same model, $F(0.05\mathrm{m})=5.125\mathrm{N}$, so the chord from $0.05$ to $0.10\mathrm{m}$ has slope $117.5\mathrm{N}{\mathrm{m}}^{-1}$, rather than $110\mathrm{N}{\mathrm{m}}^{-1}$.

Nor does an endpoint secant stiffness determine the work needed to get there. Here the exact loading work isApproximating it by $\frac{1}{2}{k}_{\sec }{x}^2$ would instead give $0.550\mathrm{J}$. The endpoint ratio does not replace the complete force-displacement curve.

A split-torque transmission divides transmitted load among two or more mechanical paths and subsequently combines their contributions. Its purpose is to let several meshes carry a total load that would otherwise pass through one path. The gear torque-path load fraction depends on compatibility, stiffness, gaps and mounting; a symmetric layout alone does not prove equal sharing.

For an original idealization, refer both branch torques to the same shaft coordinate and let their common elastic twist be $\theta$. With torsional stiffnesses $K_1,K_2$ and no initial gaps,Values $K_1=40000$ and $K_2=60000\mathrm{N}m/rad$ under total torque $1000\mathrm{N}\mathrm{m}$ give $\theta =0.01\mathrm{rad}$ and branch torques $400$ and $600\mathrm{N}\mathrm{m}$. If each branch were limited to $600\mathrm{N}\mathrm{m}$, this sharing would limit total load to $1000\mathrm{N}\mathrm{m}$; equal sharing would permit $1200\mathrm{N}\mathrm{m}$ within the same simplified capacity assumptions.

Parallel-path gear efficiency also differs from the product rule for serial stages. If input power fractions are $a_j$ with ${\sum }_j{a}_j=1$, and each branch delivers fraction ${\eta }_j$ of its input power to the common output boundary, thenFor fractions $0.4,0.6$ and efficiencies $0.96,0.94$, the result is $0.948$. Multiplying the two branch efficiencies would describe a different arrangement of power boundaries.

The primary face-gear program reported input-pinion torque division about $48\%/52\%$, while its two idlers shared about $57\%/43\%$. These were measured results with stated calibration limitations, not exact consequences of the layout.^[ref]

A meaningful load-sharing test measures both paths over relevant load and position ranges. Strain gauges need an appropriate measurement calibration because different tooth load distributions can produce different strain responses for the same transmitted load. Floating pinions provide one mechanism for reducing imbalance, but the complete assembly and its constraints still determine the operating split.

Structural dynamics studies the time-dependent motion of structures under applied forces and their response after a disturbance. Inertia, stiffness and damping jointly determine that motion. A common linear small-displacement model iswhere $u$ collects displacement coordinates, $M$ is the mass matrix, $C$ the damping matrix, $K$ the stiffness matrix and $f(t)$ the applied-force vector. Boundary conditions and the chosen coordinates are part of the model; a static force balance omits the velocity and acceleration terms.

A scalar version makes the consequences quantitative. Choose $m=1000\mathrm{kg}$, $k=40000\mathrm{N}{\mathrm{m}}^{-1}$ and $c=400\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. Its undamped natural frequency isFor a constant 100-newton force, static displacement is $100/k=2.5\mathrm{mm}$. For a sinusoidal force $100\cos ({\omega }_{0}t)$, the steady-state displacement amplitude of this linear forced harmonic oscillator is insteadThese are original model values, not measurements of a building. They show why the force amplitude alone does not determine displacement: forcing frequency and damping matter.

Real structures have multiple normal modes. A 2025 study of a viscous inerter damper uses a numerical 20-storey building example and shows that inertia and flexibility associated with discarded modes affect the reduced model's tuning. Matching one nominal mode while ignoring those contributions can therefore misrepresent the coupled device–structure behavior.^[ref]

Measured motion can test a chosen model, while its predictions remain limited by assumptions about linearity, boundary conditions and loading. A large predicted response is a reason to examine those assumptions, not proof that an actual structure will follow the linear response indefinitely.

The tangent stiffness at a point on a differentiable force-displacement curve is its local slope,It governs the first-order response to a small displacement change:The graphical interpretation is the slope of the tangent line. Structural stiffness measured from force-deflection behaviour includes geometry as well as material response.^[ref]

For the illustrative model $F(x)=ax+b{x}^3$, choose $a=100\mathrm{N}{\mathrm{m}}^{-1}$ and $b=1000\mathrm{N}{\mathrm{m}}^{-3}$. At $x_0=0.10\mathrm{m}$, the tangent stiffness is $130\mathrm{N}{\mathrm{m}}^{-1}$. An additional $\Delta x=0.001\mathrm{m}$ therefore predicts a force increase of $0.130\mathrm{N}$ to first order. Direct substitution gives $0.130301\mathrm{N}$; the difference is the higher-order contribution omitted by the tangent approximation. The secant stiffness from the origin is only $110\mathrm{N}{\mathrm{m}}^{-1}$ at this point.

In a displacement-controlled experiment, one can estimate the local slope with a central finite difference,Making $h$ smaller reduces the approximation error for a smooth exact curve, but measured differences can become dominated by measurement uncertainty. The force channel, displacement channel and loading branch must therefore be specified. Crosshead movement can include machine mechanical compliance, while an extensometer records a selected specimen extension.^[ref]

For a curve with hysteresis, loading and unloading can have different local slopes at the same displacement. At a sharp corner a unique derivative may not exist, and a finite-interval slope should be reported as such.

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]