Measuring damping and friction

Key ideas

1. 01

   Two idealised frictions bracket most hinges

   Viscous friction pushes back in proportion to speed, $b\dot\theta$. Dry (Coulomb) friction pushes back with a constant magnitude $\tau_c$, always against the motion. Real bearings mix the two, and a motor adds effects of its own (idea 07).[1][2]

2. 02

   Viscous: the same fraction every cycle

   With viscous friction the peaks lie on an exponential, $|\theta_n| = \theta_0 e^{-\sigma t_n}$ (F8.03). Every cycle loses the same fraction of its amplitude, so $\ln|\text{peak}|$ against time is a straight line of slope $-\sigma$, and $b = 2I\sigma$ (F8.05). For $b = 10^{-3}$ N·m·s/rad on the running example, $\sigma$ = 0.0859 s⁻¹: the swing halves every 8.07 s, about 9.9 cycles.[1][3]

3. 03

   Dry friction: the same angle every half-swing

   With Coulomb friction every half-swing loses the same angle, $2\Delta = 2\tau_c/k$ (F8.07), so the peaks fall on a straight line on ordinary axes. Motion stops at the first peak inside the dead band $\pm\Delta$, and the pendulum rests there, off-centre. With $\tau_c = 5\times10^{-4}$ N·m, each half-swing loses 0.166°, and a 5° release stops after 30 half-swings.[2]

4. 04

   The shape of the decay is the diagnosis

   Plot $\ln|\text{peak}|$ against time. A straight line means viscous; a line that steepens as the swing dies means dry friction is present. Better, plot each half-swing peak against the one before (F8.08): it is a straight line whose slope gives $\sigma$ and whose intercept gives $\Delta$, so one fit recovers both. Put another way, an apparent damping ratio that rises as the amplitude falls is dry friction's fingerprint (F8.09).[4][5]

5. 05

   Fit the wrong model and you get a confident wrong number

   Plant $b = 10^{-3}$ and $\tau_c = 5\times10^{-4}$ together and fit a pure exponential, and the fit reports a $b$ 3.96 times too large. The number alone gives no warning, and an $R^2$ of 0.92 can pass a casual glance; the residuals arch, which the plots in idea 04 show at once. The return map, on the same data, recovers $b$ to 0.07 % and $\tau_c$ to 0.3 %.[4]

6. 06

   Light damping barely moves the frequency

   At $\zeta$ = 0.0112, the damped frequency is below $\omega_n$ by 62 parts per million (F8.02). You can usually ignore the difference, and F9.02 removes it exactly at no cost.[1]

7. 07

   A motor is a brake you may not have meant to fit

   A turning motor generates a back emf in proportion to its speed. If the winding is shorted, that emf drives a current, the current makes a torque against the motion, and the result is viscous damping, $b = K^2/R$ (F8.11). Whether a "disabled" driver shorts the winding or leaves it open depends on the driver: one common part offers both, as brake and coast. Separately, cogging acts even with no current at all.[6][7][8][9]

8. 08

   The simulator can bias the answer, slightly

   MuJoCo's default integrator treats joint damping implicitly, which lowers the simulated decay rate by a factor of about $1 - h\sigma$ (F8.10). Here $h\sigma$ = 1.7×10⁻⁴: negligible, but worth knowing before you plant heavy damping.[10][11]

Formulas

The same entries as the formula sheet. Open an entry's details for its symbols and the checks behind it.

1. Damped equation of motion, small angle

   $$I\ddot\theta + b\dot\theta + k\theta = 0,\qquad k = m g d$$
   F8.01[1][12]
   Symbols

   Symbol	Meaning	Unit
   $b$	viscous damping coefficient	N·m·s/rad
   $k$	gravity stiffness	N·m/rad

2. Decay rate, damping ratio, damped frequency

   $$\sigma = \dfrac{b}{2I},\qquad \zeta = \dfrac{b}{2\sqrt{kI}} = \dfrac{\sigma}{\omega_n},\qquad \omega_d = \omega_n\sqrt{1-\zeta^2}$$
   F8.02✓ 2/2 checked[1][12]
   Symbols and checks

   Symbol	Meaning	Unit
   $σ$	envelope decay rate	1/s
   $ζ$	damping ratio	dimensionless
   $ω_{d}$	damped angular frequency	rad/s

   • ✓ b recovered as 2 I sigma, b = 1e-3 plantedexpected 0.001 · got 9.99863×10⁻⁴ · relative error 1.4×10⁻⁴
   • ✓ zeta from log decrement vs b / (2 sqrt(k I))expected 0.0111684 · got 0.0111684 · relative error 5.1×10⁻⁶

3. Peaks lie on an exponential envelope

   Linear viscous damping, released from rest.

   $$|\theta_n| = \theta_0\,e^{-\sigma t_n},\qquad t_n = \dfrac{n\pi}{\omega_d}$$
   F8.03✓ 2/2 checked[1][12]
   Symbols and checks

   • ✓ b recovered as 2 I sigma, b = 1e-3 plantedexpected 0.001 · got 9.99863×10⁻⁴ · relative error 1.4×10⁻⁴
   • ✓ b fitted with b = 0 planted (N m s/rad)expected 0 · got −3.78558×10⁻¹⁵ · absolute error 3.8×10⁻¹⁵the pipeline's own floor; planted values are 1e-3

4. Logarithmic decrement

   n counts half-swing peaks, as in F8.03, so peaks n and n + 2 are one full cycle apart, on the same side.

   $$\delta = \ln\dfrac{|\theta_n|}{|\theta_{n+2}|} = \sigma T_d = \dfrac{2\pi\zeta}{\sqrt{1-\zeta^2}},\qquad \zeta = \dfrac{\delta}{\sqrt{4\pi^2+\delta^2}}$$
   F8.04✓ 1/1 checked[1][3][12]
   Symbols and checks

   Symbol	Meaning	Unit
   $δ$	log decrement over one full cycle	dimensionless
   $T_{d}$	damped period	s

   • ✓ zeta from log decrement vs b / (2 sqrt(k I))expected 0.0111684 · got 0.0111684 · relative error 5.1×10⁻⁶

5. From the fit back to the coefficient

   I is the inertia about the pivot, including any rotor that turns with the pendulum.

   $$b = 2 I \sigma$$
   F8.05✓ 4/4 checked[1][12]
   Symbols and checks

   • ✓ b recovered as 2 I sigma, b = 1e-3 plantedexpected 0.001 · got 9.99863×10⁻⁴ · relative error 1.4×10⁻⁴
   • ✓ b fitted with b = 0 planted (N m s/rad)expected 0 · got −3.78558×10⁻¹⁵ · absolute error 3.8×10⁻¹⁵the pipeline's own floor; planted values are 1e-3
   • ✓ b fitted at h = 2 ms (Euler) vs 1 ms (RK4)expected 0.00100004 · got 9.99863×10⁻⁴ · relative error 1.7×10⁻⁴halving the step should not move the answer you report
   • ✓ b from the return map, both plantedexpected 0.001 · got 9.99265×10⁻⁴ · relative error 7.4×10⁻⁴

6. Inertia from the same swing

   Linear viscous model; needs k = m·g·d from a separate measurement.

   $$I = \dfrac{k}{\omega_d^2 + \sigma^2}$$
   F8.06✓ 1/1 checked[1][12]
   Symbols and checks

   • ✓ I from the same swing, k / (wd^2 + sigma^2)expected 0.0058208 · got 0.00582245 · relative error 2.8×10⁻⁴5 deg amplitude lengthens the period by ~0.05 %, so ~0.1 % on I

7. Coulomb friction, straight-line decay

   Linear stiffness. Motion stops at the first turning point where k|A| is within the static friction, so inside ±Δ when static and kinetic friction are equal (as MuJoCo models them), or ±τ_s/k when static friction τ_s is larger. The pendulum then rests off-centre.

   $$\Delta = \dfrac{\tau_c}{k},\qquad A_{n+1} = A_n - 2\Delta\ \ (\text{per half-swing})$$
   F8.07✓ 4/4 checked[2][12][13]
   Symbols and checks

   Symbol	Meaning	Unit
   $τ_{c}$	friction torque magnitude	N·m
   $Δ$	dead band half-width	rad
   $A_{n}$	n-th peak amplitude	rad

   • ✓ mean drop per half-swing vs 2 tau_c / kexpected 0.00290418 · got 0.00290032 · relative error 0.0013
   • ✓ spread of the drops (std / mean)expected 0 · got 3.76085×10⁻⁴ · absolute error 3.8×10⁻⁴constant drop = straight line on linear axes
   • ✓ tau_c = k Delta from the interceptexpected 5×10⁻⁴ · got 4.98736×10⁻⁴ · relative error 0.0025
   • ✓ half-swings before the pendulum stopsexpected 30 · got 30 · absolute error 0motion stops at the first peak inside the dead band

8. Peak-to-peak return map (viscous and Coulomb together)

   Linear model, half-swing peaks. Fit a straight line to A_{n+1} against A_n; the slope gives σ, the intercept gives Δ.

   $$A_{n+1} = q\,A_n - \Delta\,(1+q),\qquad q = e^{-\sigma T_d/2}$$
   F8.08✓ 5/5 checked[4][12]
   Symbols and checks

   • ✓ viscous: slope vs exp(-sigma Td / 2)expected 0.96552 · got 0.965504 · relative error 1.6×10⁻⁵
   • ✓ Coulomb: intercept vs -2 Delta (rad)expected −0.00290418 · got −0.00289417 · relative error 0.0034
   • ✓ Coulomb: slope vs 1expected 1 · got 0.999911 · relative error 8.9×10⁻⁵
   • ✓ b from the return map, both plantedexpected 0.001 · got 9.99265×10⁻⁴ · relative error 7.4×10⁻⁴
   • ✓ tau_c = k Delta from the interceptexpected 5×10⁻⁴ · got 4.98736×10⁻⁴ · relative error 0.0025

9. Equivalent viscous damping of Coulomb friction

   Energy balance over one cycle (4τ_cA = πbωA²). Rises as amplitude falls, which is the signature that gives Coulomb away.

   $$\zeta_{\text{eq}}(A) = \dfrac{2\,\tau_c}{\pi\,k\,A}$$
   F8.09✓ 1/1 checked[5][12]
   Symbols and checks

   • ✓ apparent zeta of Coulomb decay vs 2 tau_c / (pi k A)expected 0.0119878 · got 0.0119903 · relative error 2×10⁻⁴

10. Bias of MuJoCo's implicit joint damping

    Semi-implicit Euler with implicit joint damping, linear oscillator. Relative bias ≈ −hσ = −hζω_n, negligible for light damping.

    $$\sigma_{\text{num}} = \dfrac{\ln(1+2h\sigma)}{2h} \approx \sigma\,(1 - h\sigma)$$
    F8.10✓ 2/2 checked[11][14]
    Symbols and checks

    • ✓ MuJoCo Euler decay rate at zeta=0.2 vs ln(1+2h sigma)/(2h)expected 1.53354 · got 1.53354 · relative error 2.7×10⁻⁶true sigma 1.53825; predicted relative bias -3.06e-03
    • ✓ b fitted at h = 2 ms (Euler) vs 1 ms (RK4)expected 0.00100004 · got 9.99863×10⁻⁴ · relative error 1.7×10⁻⁴halving the step should not move the answer you report

11. Braking from a shorted motor winding

    One winding (the brushed DC model), shorted through R, with the electrical time constant L/R much shorter than the swing. K_t = K_e follows from power balance for an ideal motor. Open-circuit windings carry no current and add no braking.

    $$b_{\text{short}} = \dfrac{K_t K_e}{R} = \dfrac{K^2}{R},\qquad K = K_t = K_e\ \text{(SI units)}$$
    F8.11✓ 1/1 checked[6][7][15][16]
    Symbols and checks

    Symbol	Meaning	Unit
    $K_{t}$	torque constant	N·m/A
    $K_{e}$	back-emf constant	V·s/rad
    $R$	total resistance around the shorted loop	Ω

    • ✓ b fitted from the coupled model vs K^2/Rexpected 0.001 · got 0.00100005 · relative error 5.2×10⁻⁵illustrative K = 0.1 N m/A, R = 10 ohm, L = 1 mH

Worked example: plant a damping, then recover it

The plant-and-recover test. Put a known damping into MuJoCo, simulate a free decay the way the bench would record it, and check that the fitting code returns what went in. Running example, released from rest at 5°, MuJoCo's default Euler integrator at $h$ = 2 ms, 40 s of data. Every number is computed by tools/robotics/verify.py.

1. 01

   Predict before you fit

   For $b = 10^{-3}$ N·m·s/rad, from F8.02 and F8.04: $\zeta$ = 0.0112, and a log decrement of 0.0702 per cycle.

   $$\sigma = \dfrac{b}{2I} = \dfrac{10^{-3}}{2 \times 0.0058208} = 0.08590\ \text{s}^{-1}$$
2. 02

   Find the turning points

   A peak is where the velocity changes sign. Fit a parabola through the three samples around each sign change to place the peak between samples. Forty seconds gives 97 half-swing peaks, from 5° down to 0.17°.

3. 03

   Fit a straight line to the logarithm

   Least squares on $\ln|\theta_n|$ against $t_n$; the slope is $-\sigma$ (F8.03). Then $b = 2I\sigma$ (F8.05).

   $$\sigma_{\text{fit}} = 0.085887\ \text{s}^{-1} \;\Rightarrow\; b = 2I\sigma_{\text{fit}} = 9.9986\times 10^{-4}\ \text{N m s/rad} \quad (-0.014\,\%)$$
4. 04

   Run the null test

   Repeat with nothing planted. The pipeline reports $b$ = −3.8×10⁻¹⁵ N·m·s/rad against a planted $10^{-3}$: on clean data, the fitting code adds no damping of its own. That says nothing yet about noise or encoder steps on bench data; to test those, add them to the simulated angle and fit again.

5. 05

   Halve the step

   Refit at 1 ms with Euler and with RK4. In units of $10^{-3}$ N·m·s/rad, the fitted $b$ is 0.99986 at the default step, 0.99995 at 1 ms and 1.0000 with RK4: the default differs from RK4 by 0.017 %. At this damping the integrator is not part of the answer. If halving $h$ moves your number by more than your tolerance, it is.

6. 06

   Swap in dry friction

   Plant $\tau_c = 5\times10^{-4}$ N·m instead. Then $\Delta = \tau_c/k$ = 0.0832°, so every half-swing should lose 0.166° and a 5° release should stop after 30 half-swings (F8.07). MuJoCo agrees. The first 11 drops vary by 0.04 % (standard deviation over mean), their mean differs from the prediction by −0.13 %, and the swing stops after the predicted number of half-swings, at 12.3 s against 12.25 s for that many half-periods. The last half-swing runs slow because MuJoCo's friction constraint is soft.

7. 07

   Both at once

   Plant both and fit each half-swing peak against the one before (F8.08), using the 18 pairs well above the dead band. The slope gives $\sigma$ and the intercept gives $\Delta$. Fit the same data with a pure exponential and you get the wrong answer, and only the residuals tell you so.

   Recovered from the same 40 s of simulated decay

   Quantity	Planted	Exponential fit	Return-map fit
   $b$, $10^{-3}$ N·m·s/rad	1	3.96	0.9993
   $\tau_c$, $10^{-4}$ N·m	5	not modelled	4.987

Fit the model whose shape matches the data. Prove the fit on a simulation where you know the answer, prove it reports nothing when there is nothing, and check that the answer does not move when the timestep does. Then, and only then, fit the bench.

Best practice

• Plant, then recover, before you fit anything real.

  A pipeline that cannot return a value you put into a simulation will not return the truth from a bench. It costs one extra run.

• Run the null test.

  Fit a decay with nothing planted. Whatever comes back is the floor of your method on clean data; here it is −3.8×10⁻¹⁵ N·m·s/rad. Then add realistic noise and encoder steps and repeat.

• Find peaks from velocity sign changes, refined with a parabola.

  The sample nearest a peak can be up to half a sample period away from it. A parabola through the three samples around the sign change places the peak between them.

• Look at the shape before you fit a number.

  Plot $\ln|\text{peak}|$ against time, and each peak against the one before. Straight on the log plot is viscous; a return map with slope near 1 is dry friction; the return map handles a mixture.[4]

• Halve the timestep and refit.

  If the fitted value moves by more than your tolerance, the integrator is part of your answer. Here halving the step moved $b$ by 0.0086 %, and switching to RK4 by 0.017 %.[11]

• Measure passive damping with the motor windings open, then in each electrical state you will run.

  An open winding carries no current, so it adds no braking. A shorted one adds $K^2/R$, which can be as large as the bearing friction you are trying to measure. Know which state your driver is in when it says "off".[6][8]

• Use a long record, and stop fitting before the noise floor.

  Light damping needs time: here the swing takes 8.07 s to halve. Drop peaks only a few sensor counts high; their logarithms are dominated by the sensor's resolution.

Questions

Should I fit the energy or the peaks?

The peaks. They are the turning points of the motion and need neither the inertia nor a velocity estimate. Energy needs $\dot\theta$, which on the bench means differentiating a measured angle and amplifying its noise.

What does MuJoCo call these?

Viscous damping is the joint's damping; dry friction is frictionloss. MuJoCo's documentation uses "Coulomb friction" for friction that scales with normal force, as in contacts, and calls load-independent joint friction "friction loss". It models friction loss as a constraint rather than a force law. One consequence: a joint that should have stuck can creep. In the dry friction run above, the stopped pendulum drifted from 0.010° at 20 s to 0.0029° at 40 s. That is the solver, not physics.[10][14][17]

How long a record do I need?

Long enough for the swing to shrink by a factor you can measure well. With $\zeta$ = 0.0112, the amplitude halves every 9.9 cycles; 40 s spans 5.0 halvings.[1]

Is the log decrement the same thing as $\sigma$?

Nearly. $\delta$ is the drop in $\ln|\theta|$ over one full cycle, so $\delta = \sigma T_d$ (F8.04). Prefer $\sigma$ from a fit over many cycles to $\delta$ from two adjacent peaks: one ratio of two measured peaks carries the noise of both.[3]

Why do I need the inertia to get $b$?

Because the decay rate is $b/(2I)$: the swing reveals the ratio, not $b$ alone. A heavier pendulum on the same bearing decays more slowly. Measure $I$ from the period (F9.02), rotor included, and use that.[1]

Could the integrator fake a decay?

Some integrators can: one with numerical damping shrinks a swing that has nothing planted. That is what the null test is for. Module 07 explains which integrators do it and why.

My bench decay is a straight line on ordinary axes. Is that bad?

It means dry friction dominates. Get $\tau_c$ from the drop per half-swing (F8.07), use the return map if there is a viscous part too, and model it in MuJoCo as frictionloss, not as damping.[2][10]

Watch

• Estimation of natural frequencies and damping ratios from measured response: the logarithmic decrement (opens in a new tab)[18]

  J. Kim Vandiver · MIT OpenCourseWare, 2.003SC Engineering Dynamics, Fall 2011 (a chapter of Lecture 19)

  Estimating natural frequency and damping ratio from a measured decay: the lecture-hall version of this page's worked example.

  listed

• Lecture 20: Linear System Modeling a Single Degree of Freedom Oscillator (opens in a new tab)[19]

  J. Kim Vandiver · MIT OpenCourseWare, 2.003SC Engineering Dynamics, Fall 2011

  Damped free response, and the experimental determination of damping ratio.

  listed

• Modern Robotics, Chapter 8.9: Actuation, Gearing, and Friction (opens in a new tab)[20]

  Kevin M. Lynch · Northwestern University, Modern Robotics video series (YouTube)

  Motors, gearing and friction models, from the book this guide follows.

  listed

Read

• University Physics Volume 1, §15.5 Damped Oscillations (opens in a new tab)[1]

  William Moebs, Samuel J. Ling, Jeff Sanny · OpenStax, 2016. CC BY 4.0

  Damped oscillations, including the under-, critically and over-damped cases.

  read

• Free Vibration of Damped System (contd.), Module 4 Lecture 3 (lecture notes, PDF) (opens in a new tab)[2]

  NPTEL (National Programme on Technology Enhanced Learning, India), course 101105022

  Coulomb damping worked through, with the linear decay and the dead band.

  read

• Engineering Vibration, 4th edition (opens in a new tab)[12]

  Daniel J. Inman · Pearson, 2014. ISBN 978-0-273-78521-7 (international edition)

  A standard engineering-vibration textbook: free and damped response, Coulomb damping and the log decrement in full.

  listed

• MuJoCo Computation chapter, Numerical integration (3.17.0) (opens in a new tab)[14]

  Google DeepMind

  How MuJoCo integrates damping and models friction loss, from the people who wrote it.

  read

Sources

Numbered in order of first use on this page. read: checked against the claim. listed: exists and covers the topic, wording not checked. derived: worked out here and checked numerically.

1. [1]

   University Physics Volume 1, §15.5 Damped Oscillations (opens in a new tab)

   William Moebs, Samuel J. Ling, Jeff Sanny · OpenStax, 2016. CC BY 4.0

   Eq. 15.24 (underdamped solution, exponential envelope), Eq. 15.26 (damped frequency); under-, critically and over-damped conditions

   Exponential envelope e^(−bt/2m), damped frequency, damping regimes. (Translational form; the rotational form swaps m→I, k→mgd.)

   read
2. [2]

   Free Vibration of Damped System (contd.), Module 4 Lecture 3 (lecture notes, PDF) (opens in a new tab)

   NPTEL (National Programme on Technology Enhanced Learning, India), course 101105022

   Coulomb damping loses 2F/k per half cycle and 4F/k per cycle (linear decay). The notes take static friction as larger than kinetic.

   read
3. [3]

   Logarithmic decrement (opens in a new tab)

   Wikipedia

   Definition (natural log of the ratio of successive peaks, or of peaks n periods apart) and when it applies (underdamped; precision falls as ζ rises past about 0.5). The page's equations did not render when checked, so the ζ relations are derived here from OpenStax Eqs. 15.24 and 15.26.

   read
4. [4]

   Half-swing return map with viscous and Coulomb damping (linear stiffness)

   F8.08.

   derived
5. [5]

   Equivalent viscous damping of Coulomb friction from energy per cycle

   F8.09.

   derived
6. [6]

   University Physics Volume 2, §13.6 Electric Generators and Back Emf (opens in a new tab)

   Samuel J. Ling, Jeff Sanny, William Moebs · OpenStax, 2016. CC BY 4.0

   Back emf and Fig. 13.32

   A turning motor generates a back emf that is zero at rest and rises in proportion to the motor's angular velocity.

   read
7. [7]

   University Physics Volume 2, §11.5 Force and Torque on a Current Loop (opens in a new tab)

   Samuel J. Ling, Jeff Sanny, William Moebs · OpenStax, 2016. CC BY 4.0

   Eqs. 11.19, 11.21 (μ = NIA) and 11.22 (τ = μ × B)

   Torque on a coil in a magnetic field is proportional to the current through it.

   read
8. [8]

   DRV8833 Dual H-Bridge Motor Driver, data sheet (SLVSAR1E) (opens in a new tab)

   Texas Instruments, 2011, revised 2015

   §7.3.2 Bridge Control and Decay Modes, Table 1 (H-Bridge Logic); §7.3.5.3 Undervoltage Lockout

   A driver’s two “off” states: coast (outputs high-impedance; in fast decay the recirculation current flows through the body diodes) and brake (both outputs low, which shorts the winding). Below the undervoltage threshold (VM < 2.5 V) all circuitry in the device is disabled.

   read
9. [9]

   Cogging torque (opens in a new tab)

   Wikipedia

   Cogging is the interaction between the rotor magnets and the stator slots; it is a detent (no-current) torque, especially prominent at low speed.

   read
10. [10]

    MuJoCo XML Reference (3.17.0) (opens in a new tab)

    Google DeepMind

    option-timestep (default 0.002 s; RK4 costs about four times per step); option-gravity (default 0 0 −9.81); option-integrator; option-flag-energy; option-flag-eulerdamp (implicit joint damping in Euler, on by default); body-joint-armature (reflected inertia, gear ratio squared; positive armature improves stability); actuator-general-armature (rotor inertia, scaled by gear squared); body-joint-damping (Euler treats joint damping implicitly); body-joint-frictionloss; compiler-inertiafromgeom; body-geom-density (default 1000 kg/m³, used only when mass is not given)

    MuJoCo defaults, energy definitions, armature, damping and friction-loss behaviour. Read from the 3.17.0 source of the docs (doc/XMLreference.rst).

    read
11. [11]

    Decay-rate bias of semi-implicit Euler with implicitly treated joint damping

    F8.10. Matched MuJoCo 3.17 to 3×10⁻⁶ at ζ = 0.2.

    derived
12. [12]

    Engineering Vibration, 4th edition (opens in a new tab)

    Daniel J. Inman · Pearson, 2014. ISBN 978-0-273-78521-7 (international edition)

    Standard reference for viscous damping, logarithmic decrement and Coulomb damping. Listed for further reading; every formula that cites it also cites a source checked on this page.

    listed
13. [13]

    Where dry friction stops a swing (static moment balance at a turning point)

    The dead band in F8.07, checked by counting half-swings in MuJoCo.

    derived
14. [14]

    MuJoCo Computation chapter, Numerical integration (3.17.0) (opens in a new tab)

    Google DeepMind

    explicit, semi-implicit and implicit-in-velocity Euler; D matrix; implicitfast drops Coriolis and centripetal terms; discrete integrator; §Friction loss (dry friction modelled as a constraint whose effect is to drive the joint velocity to zero); §Reproducibility; §Numerical integration, RK4 and the tip “Choosing timestep and integrator”

    The exact update equations of every MuJoCo integrator, and how friction loss is modelled. The docs recommend implicitfast for most models (similar cost to Euler, more stable), keep Euler for compatibility, and call RK4 best for energy-conserving or nearly conserving systems; they report RK4 qualitatively better than single-step methods on an energy-conserving example even at a quarter of the timestep, and single-step methods with implicit velocity-dependent forces significantly more stable than RK4 when those forces are large.

    read
15. [15]

    University Physics Volume 2, §9.4 Ohm's Law (opens in a new tab)

    Samuel J. Ling, Jeff Sanny, William Moebs · OpenStax, 2016. CC BY 4.0

    Eq. 9.11 (V = IR)

    Ohm's law, used for the current a back-emf drives through a shorted winding.

    read
16. [16]

    Braking from a shorted winding (back emf, Ohm’s law, torque proportional to current)

    F8.11, checked against an independent simulation of the coupled pendulum and RL circuit.

    derived
17. [17]

    MuJoCo Modeling chapter, solver parameters and friction (3.17.0) (opens in a new tab)

    Google DeepMind

    Solver parameters: solref default "0.02 1" (time constant, damping ratio 1: critically damped); timeconst at least twice the timestep (enforced unless refsafe is off); Restitution (zero damping in the direct format); friction loss is first-order, purely damped, no exact stick (slow slippage)

    Constraints are soft mass-spring-dampers, critically damped by default; restitution needs the direct format with zero damping, and even then energy only fluctuates around its initial value. Why simulated friction loss creeps instead of sticking.

    read
18. [18]

    Estimation of natural frequencies and damping ratios from measured response: the logarithmic decrement (opens in a new tab)

    J. Kim Vandiver · MIT OpenCourseWare, 2.003SC Engineering Dynamics, Fall 2011 (a chapter of Lecture 19)

    Lecture chapter on estimating natural frequency and damping ratio from a measured free decay.

    listed
19. [19]

    Lecture 20: Linear System Modeling a Single Degree of Freedom Oscillator (opens in a new tab)

    J. Kim Vandiver · MIT OpenCourseWare, 2.003SC Engineering Dynamics, Fall 2011

    Lecture video on damped free response and on determining the damping ratio by experiment (topics confirmed from the lecture page's description).

    listed
20. [20]

    Modern Robotics, Chapter 8.9: Actuation, Gearing, and Friction (opens in a new tab)

    Kevin M. Lynch · Northwestern University, Modern Robotics video series (YouTube)

    The book’s own video for §8.9: motors, gearing, apparent (reflected) inertia and friction models. Linked from the book site’s chapter 8 video list.

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