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For a gas of fixed composition in hydrostatic equilibrium, with $z$ upward and downward gravitational acceleration of magnitude $g$, an adiabatic process gives the displaced element density change $\delta {\rho }_{\mathrm{element}}=\rho {\gamma }^{-1}(d\ln p/dz)\xi$. Comparing with the ambient change $(d\rho /dz)\xi$ yieldsThe ideal gas law $p=\rho \mathcal{R}T$ and $dp/dz=-\rho g$ give the title expression, where $c_p=\gamma \mathcal{R}/(\gamma -1)$ is the constant-pressure specific heat capacity. A temperature gradient less negative than the adiabatic lapse rate gives positive $N^2$; a steeper decline gives convective instability.

Apparent weight measures support rather than the gravitational force itself. In an ideal scale model, it is the magnitude $N$ of the normal force exerted on the body. A scale calibrated to display mass divides that support force by its calibration value of $g$.

Take upward as positive in a constant downward gravitational acceleration of magnitude $g$. If a body stays in contact with a horizontal scale and accelerates upward at $a$, Newton's second law gives $N-mg=ma$, hence $N=m(g+a)$. At rest, $N=mg$. In common free fall, $a=-g$, so $N=0$ despite the nonzero gravitational weight $mg$. If the formula would require $N<0$, ordinary supporting contact cannot realize it: the body loses contact instead.

Archimedes' principle states that buoyancy equals the downward gravitational weight, or weight, of the displaced fluid. In a resting fluid of constant mass density ${\rho }_f$ under uniform gravitational acceleration $g$, its magnitude isHere $V_{\mathrm{displaced}}$ is the volume from which the body excludes the fluid: the whole external volume for a fully immersed solid body, and just the immersed volume for a partly immersed one. The body's own mass does not enter this buoyancy formula. ^[ref]

A short proof outline for a fully immersed body is to imagine replacing it by exactly the fluid it displaces. The surrounding hydrostatic pressure is unchanged. That replacement fluid is at rest, so its upward pressure force must balance its downward force from gravity. The original body receives the same inward pressure forces on the same closed surface, hence the same upward buoyancy.

Equivalently, let $\Omega$ be the displaced region, $\partial \Omega$ its closed surface and ${\mathbf{e}}_z$ the upward unit direction. The divergence theorem, applied component by component to the surface integral, and hydrostatic pressure giveThis also explains why horizontal forces cancel even for an irregular shape. A spatially constant surface pressure contributes zero to the closed-surface integral. If the mass density varies with depth, replace ${\rho }_f{V}_{\mathrm{displaced}}$ by the mass ${\int }_{\Omega }{\rho }_{f}dV$ of the displaced fluid.

The closed-surface argument assumes that the fluid applies pressure over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no fluid beneath it; then the lower-face pressure force used in the proof is absent, and a supporting contact force must be treated separately. For partial immersion, the usual buoyancy formula assumes the exposed surface receives the common atmospheric pressure, neglects the much smaller buoyancy from the air, and neglects surface tension.

For the $0.10\mathrm{m}$ cube in the stipulated water, full immersion displaces $0.001{\mathrm{m}}^3$, so $F_b=1000\times 9.81\times 0.001=9.81\mathrm{N}$. If its uniform mass density is $600\mathrm{kg}{\mathrm{m}}^{-3}$, its mass is $m=600\times 0.001=0.600\mathrm{kg}$ and its downward force from gravity is $mg=5.886\mathrm{N}$. To hold it fully immersed at rest requires an additional downward forceOn release from rest it initially moves upward; its eventual floating equilibrium has a smaller displaced volume. A same-size cube of uniform mass density $1200\mathrm{kg}{\mathrm{m}}^{-3}$ has mass $1.200\mathrm{kg}$ and downward force from gravity $11.772\mathrm{N}$. Full immersion still gives $9.81\mathrm{N}$ of buoyancy, leaving an initial net downward force of $11.772-9.81=1.962\mathrm{N}$. Subsequent motion also involves drag.

The mass per unit horizontal area above a level in an atmosphere. With constant gravitational acceleration, hydrostatic balance gives $m=(p_s-p_{\mathrm{top}})/g$. This relation connects an observed surface pressure to the total reservoir mass without assuming an isothermal temperature profile.

A laboratory gravity gradiometry experiment used two vertically separated clouds of chemical atoms in vacuum, launched into free fall. Shared laser beams applied three pulses ($\pi /2,\pi ,\pi /2$), splitting and recombining matter waves. Balanced photodiodes measured probe-light absorption to infer internal-state populations; scanning the last pulse's wave phase produced an interference fringe.^[ref]

The difference of fringe wave phases obeys $\delta \varphi ={\mathbf{k}}_{\mathrm{eff}}\cdot \Delta \mathbf{g}{\tau }^2$, with effective wave vector ${\mathbf{k}}_{\mathrm{eff}}$ and pulse spacing $\tau$: shared reference motion cancels, leaving differential gravitational acceleration. Eight lead bricks, each approximately $12.5\mathrm{kg}$, were placed symmetrically around the lower chamber, about $0.2\mathrm{m}$ from the fountain apex. The measured signal was $(8.1\pm 2.1)\times 10^{-9}g$, versus a calculated $8.2\times 10^{-9}g$, where $g$ denotes terrestrial gravitational acceleration.^[ref]

Controls deliberately shook the shared reference reflector and compared actively isolated and unisolated operation. Tilting tests exposed sensitivity to Coriolis acceleration and centrifugal acceleration. Thus shared translation is rejectable, while rotation and finite measurement uncertainty remain limitations. The brick signal detects spatially varying gravity; it does not establish a precision value of the gravitational constant.^[ref]

In the flat-surface, constant-gravitational acceleration model, a coasting projectile conserves horizontal velocity. If it arrives at the top of its trajectory with horizontal speed $v_o$, its earlier velocity at angle $\theta$ above the horizontal has vertical component $v_o\tan \theta$. Its remaining height gain is therefore $v_o^2{\tan }^2\theta /(2g)$. The displayed burnout height follows. This is a local projectile motion approximation, not an exact Kepler orbit. Computing $v_o=\sqrt{g(R_E+h_o)}$ assumes constant gravitational magnitude also at the orbit; computing $v_o=\sqrt{g_0{R}_E^2/(R_E+h_o)}$ instead assumes an inverse-square law with surface value $g_0$. Those assumptions must be distinguished.

Buoyancy is the resultant upward force exerted by a resting fluid on an immersed body in a downward gravitational field. The fluid pushes inward, perpendicular to each part of the body's surface. Because hydrostatic pressure increases with depth, the upward push from below exceeds the downward push from above. Opposite horizontal forces cancel; buoyancy is the remaining vertical force. A sinking body still receives this upward support. ^[ref]

More precisely, pressure $p$ on a small surface of area $dA$ with outward unit normal vector $\mathbf{n}$ exertsThe minus sign says the force points into the body. On a cube, the top force points down, the bottom force points up, and the side forces cancel in pairs because opposite sides have the same pressure at each depth. This is a surface traction; gravity on the body itself is a separate body force. ^[ref]

For a computed example, stipulate resting water with constant mass density ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$ and uniform gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. A fully immersed cube has side $L=0.10\mathrm{m}$, top depth $h_t=0.20\mathrm{m}$ and bottom depth $h_b=0.30\mathrm{m}$. Each horizontal face has area $A=L^2=0.010{\mathrm{m}}^2$, and its volume is $V=L^3=0.001{\mathrm{m}}^3$. Its top and bottom gauge pressures areThe corresponding forces relative to the common surface pressure are $p_{g,t}A=19.62\mathrm{N}$ downward and $p_{g,b}A=29.43\mathrm{N}$ upward. Therefore the net buoyancy isUsing absolute pressures adds the same $p_{\mathrm{surface}}A$ to both face forces, leaving their difference unchanged. Zero gauge pressure at the surface does not mean zero absolute pressure. Increasing both depths by the same amount also leaves buoyancy unchanged in this constant-mass density model.

Archimedes' principle turns this calculation into a rule for bodies of any shape. It explains floating equilibrium and lets us infer an immersed body's volume from the difference between its force from gravity and the force needed to support it. In particular, buoyancy depends on the displaced volume and the surrounding mass density, whereas the body's own mass density decides how its force from gravity compares with that support.

The capillary length measures when gravity and surface tension are comparably important in shaping a liquid interface. For a liquid of uniform mass density $\rho$, beneath a gas whose mass density is negligible, with uniform gravitational acceleration $g>0$ and constant surface tension $\gamma >0$,A deformation of height $a$ and horizontal length scale $L$ has hydrostatic pressure variation of order $\rho ga$ and small-slope capillary pressure of order $\gamma a/L^2$. Equating these gives $L\sim {\ell }_c$: surface tension smooths small features, whereas gravity matters for broad ones. For two resting fluids, replace $\rho$ by their positive mass density difference $\Delta \rho$ in this static balance. This change alone does not give the two-fluid wave dynamics, because both fluids can have inertia. ^[ref]

With the stipulated illustrative parameters $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the capillary length is ${\ell }_c=2.71\mathrm{mm}$. These parameters define the ideal calculations below; they are not a measurement of a particular sample.

For a sinusoid with wavenumber $k>0$, the relevant horizontal scale is $L=1/k$, while its wavelength is $\lambda =2\pi /k$. Thus the Bond number for this scale and the reciprocal restoring-term ratio areEquality is at $k_c=1/{\ell }_c$, hence ${\lambda }_c=2\pi {\ell }_c=17.02\mathrm{mm}$ for these parameters. Using $L=\lambda$ instead produces a different numerical Bond number; state the length convention. The capillary length itself is not the equality wavelength. This distinction helps choose whether a measured ripple should be modeled as a surface gravity wave, a nearly pure capillary wave, or a gravity-capillary wave. ^[ref]

A catenary is the static shape of an ideal flexible, inextensible chain hanging under its own gravitational weight in a uniform gravitational field, with uniform linear mass density along its arc length. With the lowest point above $x=0$, one convenient vertical origin givesHere $a$ has units of length. The supports and the chain's arc length determine the scale and position; horizontal and vertical shifts give the general catenary. ^[ref]

For $a=1$, the three points at $x=-1,0,1$ have heights $\cosh 1\approx 1.543081$, $1$, and $\cosh 1\approx 1.543081$. If heights should instead be measured from the lowest point, subtract $a$:The same three heights are then approximately $0.543081$, $0$, and $0.543081$. The shift changes the vertical origin and preserves the catenary shape.

The scale also measures tension: if $H>0$ is its constant horizontal component, $\mu$ is the linear mass density, and $g$ is the magnitude of gravitational acceleration, then $a=H/(\mu g)$. A short force balance explains this. For the portion from the lowest point to $x\ge 0$, with arc length $s(x)$, vertical tension balances its gravitational weight and horizontal tension givesApplying differentiation yields $a{y}^{\prime \prime }=\sqrt{1+y^{\prime 2}}$. The proposed hyperbolic cosine solves this differential equation, because $y^{\prime }=\sinh (x/a)$ and $a{y}^{\prime \prime }=\cosh (x/a)=\sqrt{1+{\sinh }^2(x/a)}$. This is the mechanism behind the shape, under the stated ideal chain assumptions.

The same identity simplifies the arc length between symmetric supports at $x=\pm b$:For $a=b=1$, $L\approx 2.350402$, exceeding the horizontal span of two. A larger $a$ at fixed $b$ gives a shallower sag: the Taylor series gives $a(\cosh (x/a)-1)=x^2/(2a)+x^4/(24a^3)+\cdots$. Thus a parabola is a local approximation, while the exact ideal chain shape is a catenary. The useful connection is that the same hyperbolic Pythagorean identity controls its coordinates, force balance, and arc length.

The Cavendish experiment makes the attraction predicted by Newton's law of universal gravitation measurable between laboratory masses. Two small test masses hang at the ends of a horizontal rod in a Cavendish torsion balance; nearby larger source masses exert a tiny torque. A thin suspension fiber supplies an opposing torque as it twists. Measuring its twist and calibrating its torsion constant lets an experimenter infer the gravitational force and, in a modern analysis, the gravitational constant. ^[ref]

The beautiful feature is that the two horizontal forces can cancel as vectors while their torques add. The instrument converts weak attraction into a rotation, with sensitivity set by the fiber's torsion constant. Earth's much larger downward gravitational weight is supported by the suspension rather than supplying the intended torque about its vertical axis. A laboratory measurement can therefore probe the same Newtonian gravitational field that governs the two-body problem.

Henry Cavendish's original paper was Experiments to determine the Density of the Earth, published in Philosophical Transactions, volume 88 (1798), pp. 469–526, and read on 21 June 1798. On p. 469 he credits John Michell with the apparatus idea and explains that Michell died before using it; the instrument reached Cavendish through Francis John Hyde Wollaston. Cavendish's stated aim was Earth's mean mass density. Calling the experiment a determination of the gravitational constant is a modern interpretation, not Cavendish's own formulation of his result. ^[ref]

The connection between Earth's mean mass density and the gravitational constant follows in a spherical, nonrotating model of Earth. With Earth mass $M_E$, radius $R_E$, and surface gravitational acceleration $g$, the spherical shell theorem and Newton's law of universal gravitation giveThus a local comparison of gravitational attractions, together with knowledge of Earth's size and surface gravitational acceleration, reaches a property of the entire planet. These are modern equations explaining the connection, not equations quoted from Cavendish's paper.

For a worked example, use an explicitly simplified two-near-pair model. Each test point mass is $m$, each source point mass is $M$, and each lever arm is $L$, so the rod's full length is $2L$. Neglect the rod's mass. At zero twist put the test masses at $(-L,0)$ and $(L,0)$, and the sources at $(-L,-r)$ and $(L,r)$. Here $r$ is the near-pair center separation. Ignore each source's attraction on the test mass at the opposite end. Evaluate the two near-pair forces at zero twist and treat their torque as constant for a small twist, requiring $L|\theta |\ll r$. This is a teaching model, not a precision model of Cavendish's actual apparatus.

By Newton's law of universal gravitation, each retained force has magnitude $F=GmM/r^2$. The left force points down and the right force points up. Using torque about the suspension axis, both contributions are counterclockwise when viewed from above:The net retained horizontal force is zero, yet the net torque is positive. Moving the sources to $(-L,r)$ and $(L,-r)$ reverses both forces and hence the torque. In a real apparatus, the opposite source also attracts each test mass; that correction is explicitly included in the university derivation. ^[ref]

For illustrative parameters, take $m=0.010\mathrm{kg}$, $M=1.0\mathrm{kg}$, $r=0.050\mathrm{m}$, $L=0.10\mathrm{m}$, and torsion constant $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$. Use $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, the central value in NIST's 2022 CODATA table; that table gives standard uncertainty $0.00015\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. This is an input for the example, not a measurement made with these assumed parameters. ^[ref]This sub-nanonewton force produces a measurable twist in the Cavendish torsion balance. The displayed digits make the toy calculation reproducible; they do not describe experimental precision.

The centre of buoyancy is the point through which the resultant hydrostatic buoyant force acts. In a uniform fluid under uniform gravitational acceleration, it is the volume centroid of the displaced fluid. For a floating body its position generally changes when the body tilts, because the displaced volume changes shape.

The coefficient of kinetic friction is the dimensionless quantity ${\mu }_k=|f_k|/N$ for sliding contact with positive normal force $N$ in the ideal kinetic friction model. It characterizes the pair of contacting surfaces and their condition. A constant ${\mu }_k$ means the model neglects dependence on contact slip speed; unlike the coefficient of static friction, it specifies a sliding force magnitude rather than a no-slip upper bound.

The ratio compares the tangential kinetic friction magnitude with the perpendicular normal force:Here $\mathbf{w}$ is the tangential relative velocity defining contact slip. Both numerator and denominator have the same physical dimension, so the coefficient of kinetic friction has no units. It belongs to a contact pair under specified conditions, rather than to one material alone. The coefficient of static friction instead describes the threshold $|f_s{|}_{\max }/N$ before contact slip; a pull required to start motion cannot be substituted for a sliding measurement.^[ref]

As an original model example, stipulate ${\mu }_k=0.30$ and gravitational acceleration $g=9.8m/{\mathrm{s}}^2$. An already sliding, nonrotating block of total mass $m=2\mathrm{kg}$ is pulled horizontally across a stationary horizontal surface at constant velocity. With no other vertical force, Newton's second law gives $N=mg=19.6\mathrm{N}$. Zero horizontal acceleration then requires a pulling force $F=|f_k|={\mu }_{k}mg=5.88\mathrm{N}$. Doubling the total mass to $4\mathrm{kg}$ gives $N=39.2\mathrm{N}$ and $F=11.76\mathrm{N}$. The force doubles while the coefficient of kinetic friction stays $0.30$ by assumption. These numbers are model predictions, not observations or a calibration for a particular material pair.

A constant coefficient of kinetic friction is useful because one ratio can summarize many loads, as tested by load dependence of kinetic friction. It is an empirical approximation over a stated range of normal forces, speeds and surface conditions. It implies no universal independence from speed or apparent contact area: a change of face or sliding regime can change the result. Temperature, wear and lubrication are further conditions to report. Newton's second law still applies when the constant-coefficient approximation fails.^[ref]

A conical pendulum is a pendulum whose bob executes uniform horizontal circular motion beneath a fixed pivot, with its string at a constant angle $\theta$ to the downward vertical. The string sweeps a right circular conical surface; the bob follows its circular rim, without any physical cone supporting it. The ideal model has a point mass $m>0$, a taut, massless, inextensible string of length $\ell >0$, uniform gravitational acceleration of magnitude $g>0$, and negligible drag force and pivot friction. A nonzero circular motion requires $0<\theta <\pi /2$. This geometry and the corresponding forces are described in Richard Fitzpatrick, The conical pendulum, equations 262–267.

The right triangle formed by the pivot, the orbit centre, and the bob gives $r=\ell \sin \theta$ and $h=\ell \cos \theta$. Here $r$ is the orbit radius and $h$ is the vertical drop from the pivot to the bob's horizontal plane, not the bob's height above a table. Write $T$ for tension, $P$ for the oscillation period of one revolution, and $\omega =2\pi /P$ for the angular speed; then the bob's speed is $v=\omega r$. A side view compresses the horizontal circle to a line segment; the top view reveals the circle and its tangent velocity.

The two actual forces on the bob are tension toward the pivot and gravitational weight vertically downward. The vertical acceleration is zero, while the horizontal centripetal acceleration is $v^2/r={\omega }^{2}r$. Thus Newton's second law givesThe inward net force is the horizontal component of tension, rather than an extra interaction to add to these two forces. Constant speed still requires acceleration because the velocity changes direction. ^[ref]

Dividing the two component equations and using $r=\ell \sin \theta$ gives the useful collection of predictionsIf $\varphi$ is the azimuthal angle, then $\dot{\varphi }=\pm \omega$; either direction of rotation satisfies ${\dot{\varphi }}^2\cos \theta =g/\ell$. The mass cancels from the angular speed and oscillation period, but not from the tension. The motion is a relative equilibrium: in a rotating frame following the bob at angular speed $\omega$, its position is fixed. This is a concrete way to see how rotational symmetry can turn an apparently moving solution into a stationary one.

A damped pendulum is a simple pendulum whose mechanical energy is removed by a resisting torque. Here the model is undriven, with a fixed pivot, a point mass $m>0$ at distance $\ell >0$, uniform gravitational acceleration $g>0$, and resisting torque $-bd\theta /dt$ with constant $b>0$. The angle $\theta$ is measured from the downward vertical in radians. An ideal viscous rotary dashpot realizes this torque law; tangential linear drag $-\alpha v$ on the bob gives the same equation with $b=\alpha {\ell }^2$. The moment of inertia is $m{\ell }^2$, soHere $t$ is physical time, ${\omega }_0$ is the undamped small-amplitude angular frequency, and $\gamma$ has units of inverse time. In the normalized equation $\ddot{\theta }+c\dot{\theta }+\sin \theta =0$, dots mean differentiation with respect to $\tau$, not $t$. The parameter $c$ is dimensionless. This separates the chosen damping law from the small-angle approximation: retaining $\sin \theta$ allows finite angles even when the resisting torque is linear in angular velocity.

To motivate the model, release a pendulum without a push and record how its swing shrinks. A published example by Gintautas and Hubler uses a rotary sensor built from a computer mouse to measure the frequency and damping coefficient of a physical pendulum. ^[ref] The following is a proposed experiment with a chosen viscous rotary dashpot; the plotted observations are predictions, not a reproduction of their measurements.

Mount a compact bob on a light rigid rod with a fixed pivot, couple the pivot to an adjustable viscous rotary dashpot, and measure $\ell$ from the pivot to the bob's centre. First record decay with the dashpot disconnected, then repeat with it connected at several settings; this establishes how much decay is already due to other losses. Keep the bob, rod, camera, and release angle unchanged between these trials. The simple model requires rod and damper-rotor moments of inertia to be negligible compared with $m{\ell }^2$. Otherwise measure the total moment of inertia $I$ and the distance $d$ to the assembly's centre of mass: the physical pendulum equation becomes $I{d}^2\theta /d{t}^2+bd\theta /dt+Mgd\sin \theta =0$, where $M$ is the total moving mass. A damper that also exerts an elastic restoring torque would require another term.

Place a camera perpendicular to the swing plane, include the pivot and bob centre in every frame, and determine their coordinates relative to the resting vertical. For horizontal and downward separations $x,y$, use $\theta =atan2(x,y)$; record the frame timestamps rather than assuming a frame rate. Release from rest at about $10^{\circ }$, record through several complete swings, and repeat the release. For every same-side turning point, tabulate its time $t_n$ and angular amplitude $A_n=|\theta (t_n)|$. Measure the oscillation period from differences $t_{n+1}-t_n$; comparing several periods exposes drift. Interpolating around each observed extremum reduces frame discretization error. Check that late peaks still exceed the angular resolution; taking the natural logarithm of tiny, uncertain peaks amplifies their relative errors.

In the small-angle approximation, the damped pendulum becomes a damped harmonic oscillator. For $0<\gamma <2{\omega }_0$ and release from rest at ${\theta }_0$, its exact linear-model solution isThe decaying sinusoidal envelope has coefficient $|{\theta }_0|{\omega }_0/{\omega }_d$; because release is from rest, its successive same-side extrema occur at $t_n=n{T}_d$ and have $A_n=|{\theta }_0|e^{-\gamma n{T}_d/2}$. Thus the envelope and the sequence of actual peak heights are different descriptions of the same trace. The linear damping regimes and this solution are derived in Richard Fitzpatrick, Damped Oscillatory Motion, equations 82–93.

The experiment tests two predictions independently: nearly equal same-side peak spacing and a straight line when plotting $\ln (A_n/A_0)$ against $t_n-t_0$. Its predicted slope is $-\gamma /2$, givingUse several peaks and repeated trials rather than a single ratio, and compare the frequency identity with independently measured $\ell$ and $g$. For the illustrative choices $\ell =1.00\mathrm{m}$, $g=9.81m/{\mathrm{s}}^2$, $c=0.20$, and ${\theta }_0=10^{\circ }$, the linear model gives $\gamma \approx 0.6264{\mathrm{s}}^{-1}$, $T_d\approx 2.0162\mathrm{s}$, and $A_{n+1}/A_n\approx 0.5318$. The first two positive peaks after release are therefore approximately $5.318^{\circ }$ and $2.828^{\circ }$. The undamped small-amplitude oscillation period is about $2.0061\mathrm{s}$: substantial loss per swing need not imply a large change in period.

Repeat at a larger release angle to distinguish two limitations. The relative error in replacing $\sin \theta$ by $\theta$ is about ${\theta }^2/6$ for small angles: about $0.51\%$ at $10^{\circ }$, but about $17.3\%$ at $60^{\circ }$ when compared with $\theta$. Finite-amplitude swing spacing can therefore change as the swing shrinks even if $b$ is constant. Separately, friction at the pivot or quadratic drag can invalidate the assumed linear resisting torque; exponential peak decay should be checked, not imposed on all recorded motion. Peters discusses why real pendulum damping can require nonlinear terms. ^[ref] Repeatability, residual pivot motion, changes in damper temperature, and finite camera resolution are useful checks on this proposed measurement.

The normalized equilibrium points are $(\theta ,\dot{\theta })=(k\pi ,0)$. Even multiples of $\pi$ are attracting: their linearization has eigenvalues $(-c\pm \sqrt{c^2-4})/2$, giving a stable focus for $0<c<2$, repeated negative eigenvalues at $c=2$, and a stable node for $c>2$. These are the underdamped, critically damped, and overdamped small-displacement regimes. Odd multiples of $\pi$ have eigenvalues $(-c\pm \sqrt{c^2+4})/2$ of opposite signs and are saddle equilibria. The energy dissipation of a damped pendulum explains why trajectories in the phase portrait spiral inward near a stable focus instead of closing into the recurring orbit of an undamped simple pendulum.

Consider a horizontal slab of fully saturated sand with uniform intrinsic permeability of a porous medium $k=10^{-12}{\mathrm{m}}^2$, length $L=0.50\mathrm{m}$, total cross-sectional area $A=0.020{\mathrm{m}}^2$, and porosity $\varphi =0.25$. Use one Newtonian fluid with mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$ and dynamic viscosity $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$. Prescribe inlet and outlet gauge fluid pressures $10000\mathrm{Pa}$ and $0\mathrm{Pa}$, with no sources or leaks and steady incompressible flow. The two ends have the same elevation, so gravity has no component along the slab. Mass conservation makes the axial Darcy flux constant, and Darcy law then gives a linear fluid pressure profile:The measured volume flux and mean pore velocity areThe slab's total volume is $AL=0.010{\mathrm{m}}^3$, of which $\varphi AL=0.0025{\mathrm{m}}^3=2.5\mathrm{L}$ is pore volume. In the ideal uniform advective model, a nonreacting tracer has transit time $L/v=6250\mathrm{s}\simeq 1.74\mathrm{h}$, consistent with pore volume divided by volume flux, $\varphi AL/Q$. The incorrect substitution of Darcy velocity for pore velocity would give $25000\mathrm{s}$, four times too long. Real passages have different path lengths and local speeds; these numbers describe a bulk mean rather than every particle's journey.

As an independent check, gravitational acceleration magnitude $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ gives hydraulic conductivity $K=9.81\times 10^{-6}\mathrm{m}{\mathrm{s}}^{-1}$, hydraulic head drop $\Delta h=10000/(1000\times 9.81)\mathrm{m}\simeq 1.01937\mathrm{m}$, and $q=K\Delta h/L=2.0\times 10^{-5}\mathrm{m}{\mathrm{s}}^{-1}$ again. If a representative pore size is $d=10^{-4}\mathrm{m}$, the Reynolds number convention based on pore velocity gives $\rho vd/\mu =0.008$, supporting neglect of pore inertia. This size is an illustrative extra measurement, not something determined by porosity and intrinsic permeability of a porous medium alone.

Regenerate the diagram by running python ../darcy-flux-and-pore-velocity.py from _media; it writes its PNG into the current directory.

Darcy law is a linear relation between the bulk Darcy flux and the net driving force per unit volume. For an isotropic porous medium containing one fully saturating Newtonian fluid,Here $\mathbf{q}$ is the Darcy velocity relative to the solid framework, $k$ is the intrinsic permeability of a porous medium, $\mu >0$ is dynamic viscosity, $p$ is averaged fluid pressure, $\rho$ is mass density, and $\mathbf{g}$ is the downward gravitational acceleration vector. The pressure gradient exerts force density $-\nabla p$, while gravity contributes $\rho \mathbf{g}$; the minus sign makes the discharge follow their sum. The symbols $\mathbf{u}$ and $\mathbf{q}$ are both commonly used for Darcy velocity; neither denotes pore velocity here.

With upward coordinate $z$ and positive gravitational acceleration magnitude $g$, $\mathbf{g}=-g{\mathbf{e}}_z$. Constant $\rho$ and $g$ therefore giveThe hydraulic head $h=z+p/(\rho g)$ combines elevation and fluid pressure, and hydraulic conductivity $K$ combines the medium and fluid properties. Hydrostatic pressure satisfies $\nabla p=\rho \mathbf{g}$, so it produces zero Darcy flux even though fluid pressure increases downward. Conversely, equal pressures at the two ends of a vertical column do not imply zero flow: their elevations give different hydraulic heads.

The appeal of Darcy law is that a complicated connected pore geometry enters a macroscopic calculation through a measurable coefficient. It predicts filter discharge and aquifer flow; combined with mass conservation, it gives an equation for the evolving fluid pressure. Its simple linear form requires a small pore-scale Reynolds number, slow adjustment compared with pore-scale inertial relaxation, and a valid continuum approximation. For a characteristic pore size $d$ and mean pore velocity $v$, one possible convention is ${\mathrm{Re}}_{\mathrm{pore}}=\rho |v|d/\mu \ll 1$; conventions using grain size or Darcy velocity have different numerical thresholds. Merely calling flow laminar does not guarantee negligible inertia. Unsaturated pores, several fluid phases, appreciable inertia, or a fluid whose stress is not Newtonian require further modeling. Woessner and Poeter discuss these limits in their authored hydrology text.^[ref]

Henry Darcy's 1856 book Les fontaines publiques de la ville de Dijon was published by Victor Dalmont; the University Grenoble Alpes digital collection verifies this bibliographic record and supplies a digitized copy.^[ref] Woessner and Poeter's account connects Darcy law to experiments on water flowing through sand filters, relating discharge to cross-sectional area, hydraulic head difference and sample length.^[ref] The modern intrinsic permeability of a porous medium notation above should not be read as a quotation of Darcy's original notation.

Dimensional analysis checks the physical dimensions of quantities and identifies combinations unchanged by the choice of units. It catches impossible equations and reveals scaling before a full dynamical calculation. Adding two quantities requires the same physical dimension; replacing metres by centimetres must change numerical values consistently while preserving physical predictions.

For the ideal simple pendulum, suppose its oscillation period $T$ depends only on mass $m$, length $\ell$, gravitational acceleration $g$, and dimensionless angular amplitude ${\theta }_0$. Write $T=m^a{\ell }^b{g}^{c}F({\theta }_0)$. Since $[m]=M$, $[\ell ]=L$, and $[g]=L{T}^{-2}$, matching physical dimensions gives $a=0$, $b+c=0$, and $-2c=1$. ConsequentlyThe angular amplitude is dimensionless when expressed in radians. Dimensional analysis establishes the length/gravity scaling and the absence of a mass factor under these assumptions, but cannot determine $F$. The harmonic oscillator equation gives $F\to 2\pi$ at small amplitude; the full simple pendulum supplies its finite-amplitude correction. A model with drag or a finite massive support has additional parameters, so the same dimensional argument does not automatically apply.

The result is useful for comparison: at fixed angular amplitude and $g$, lengths in the ratio $1:4$ give periods in the ratio $1:2$. The ratio $T\sqrt{g/\ell }$ should be the same for both in this ideal model. A failure to collapse onto the same value points to measurement error or an omitted parameter. Correct physical dimensions are a necessary consistency check, while the coefficient and the adequacy of the model require more information.

A dispersion relation specifies which angular frequencies and wavevectors a linear wave model admits. Fix the plane wave conventionHere $a$ is a constant complex amplitude, $\mathbf{k}$ is the wavevector, and $\omega$ is the angular frequency. Substitution into a constant-coefficient linear partial differential equation replaces ${\partial }_t$ by $-i\omega$ and $\nabla$ by $i\mathbf{k}$, turning the wave problem into an algebraic condition. A system can have several roots; $\omega =\omega (\mathbf{k})$ means one chosen branch, rather than a promise that the full dispersion relation is single-valued. For real $\mathbf{k}$, $\mathrm{Im}\omega >0$ means temporal growth in this convention, while $\mathrm{Im}\omega <0$ means damping. The real-branch discussion of phase velocity and group velocity below assumes negligible loss.

For the one-dimensional wave equation $u_{tt}=c^2{u}_{xx}$, the dispersion relation is ${\omega }^2=c^2{k}^2$. The branch $\omega =ck$ carries right-moving travelling waves for $c>0$; both its phase velocity and group velocity are $c$. One may instead use the positive-angular frequency branch $\omega =c|k|$, with right-moving components at $k>0$ and left-moving components at $k<0$. A real wave pairs each $(k,\omega )$ with $(-k,-\omega )$; that conjugate pair represents the same travelling pattern. Branch and sign choices therefore matter when interpreting a dispersion diagram.

A particularly tangible dispersive example is a small-amplitude surface gravity wave on still water of uniform depth $h$. Neglect surface tension and viscosity, assume incompressible flow and irrotational flow, and impose a flat impermeable bottom. With $k>0$ and positive angular frequency,Here $g$ is gravitational acceleration and $\tanh$ is the hyperbolic tangent. Deep water means $kh\gg 1$: $\omega \simeq \sqrt{gk}$ and longer wavelengths have larger phase speed. Shallow water means $kh\ll 1$: $\omega \simeq \sqrt{gh}k$, an approximately nondispersive wave. Thus depth relative to wavelength, rather than depth alone, determines whether crests and wave packets travel together.^[ref]

The dispersion diagram below also includes the ideal deep-water capillary wave limit $\omega =\sqrt{\gamma /\rho }{k}^{3/2}$, where $\gamma$ is surface tension and $\rho$ is mass density. This limit requires the surface tension restoring term to dominate gravity; it reverses the trend of the surface gravity wave, giving larger phase speed at shorter wavelength. The combined gravity-capillary wave has ${\omega }^2=gk+(\gamma /\rho )k^3$.^[ref]

The curves use dimensionless normalizations chosen separately for each model; they are analytical diagrams, not experimental data. Their beauty is that a single dispersion relation encodes several observable behaviours: its value gives the angular frequency, its origin-to-point slope gives the one-dimensional phase velocity, its tangent slope gives the group velocity, and variation of that tangent slope predicts distortion of a wave packet. A straight line through the origin describes a nondispersive wave; an affine branch $\omega =ck+b$ with $b\ne 0$ has constant group velocity but phase velocity $c+b/k$, so those properties must be distinguished.

A double pendulum consists of two pendula coupled because the second pivot moves with the first mass. It has two degrees of freedom: even when the upper pivot is fixed, moving the upper arm carries the lower pivot along. This makes the double pendulum a compact example of how mechanical constraints couple motion, and how Lagrangian mechanics converts geometry into dynamics.

Use two equal point masses $m>0$ and two massless rigid rods of equal length $\ell >0$, with joints that allow free rotation in one vertical plane. The upper pivot is fixed; gravity is uniform, with gravitational acceleration $g>0$. Neglect joint friction and drag force. Both angles ${\theta }_1,{\theta }_2$ are measured from the downward vertical, positive toward the right, in radians. In particular, ${\theta }_2$ is an absolute angle, not the bend between the rods: that bend is ${\theta }_2-{\theta }_1$.

The double pendulum Lagrangian below gives the exact dynamics of this ideal model. Near the downward stable equilibrium, normal modes of a double pendulum replace the coupled motion with two independent harmonic oscillators. Their fixed amplitude ratios explain which releases make the rods swing together and which make them swing in opposite directions.

Dynamic pressure is the bulk kinetic energy per unit volume associated with fluid motion at speed $U$ in a specified reference frame:Here $\rho =m/V$ is mass density. The word "pressure" identifies its units and its use as a pressure scale; $q$ is not an extra isotropic stress to add to the local static pressure in every surface traction. A stationary fluid can have large static pressure with $q=0$. Conversely, knowing $U$ and $\rho$ fixes $q$ without fixing static pressure. The definition remains useful in viscous fluid flow and compressible flow, even when the simplest Bernoulli equation is inapplicable. ^[ref]

For stipulated mass density ${\rho }_{\infty }=1.20\mathrm{kg}{\mathrm{m}}^{-3}$ and reference speed $U_{\infty }=20\mathrm{m}{\mathrm{s}}^{-1}$, $q_{\infty }=240\mathrm{Pa}$. At $40\mathrm{m}{\mathrm{s}}^{-1}$ with the same mass density, $q_{\infty }=960\mathrm{Pa}$. This quadratic scaling makes dynamic pressure a useful normalization for the lift coefficient, drag coefficient and pressure coefficient; it predicts four times a particular force only if its corresponding coefficient remains unchanged.

The connection with stagnation pressure requires hypotheses. For steady flow, constant mass density, inviscid flow, and gravity as the only body force, the Bernoulli equation gives, along one streamline,Here $p$ is static pressure, $g$ is gravitational acceleration, and $z$ is height. Bring that flow to rest without losses at the same height and the limiting static pressure is the stagnation pressure $p_0$:There must be no intervening pump supplying work or other energy input. Constancy along one streamline does not assert equal $B$ on all streamlines; irrotational flow in a connected region supplies that additional conclusion. Changing height requires retaining $\rho gz$. In compressible flow the stopping pressure difference generally differs from $\rho U^2/2$, and viscous dissipation can reduce recovered stagnation pressure. ^[ref]

A Pitot-static tube turns this relation into a measurement: its upstream-facing opening senses stagnation pressure, while separate side openings sample static pressure. Connect these to the positive and negative ports of a differential pressure transducer or manometer. Under the hypotheses above, the measured difference $\Delta p$ givesThus the stipulated $240\mathrm{Pa}$ and $1.20\mathrm{kg}{\mathrm{m}}^{-3}$ imply $20\mathrm{m}{\mathrm{s}}^{-1}$, not an independently measured speed. Align the probe with the incoming flow and keep its passages clear: blockage and pinched tubing invalidate the comparison. At very low speed, the difference can be smaller than instrument error; at high Mach number, compressibility and possibly a shock wave require a different reduction. ^[ref]

A useful control is differential-pressure calibration. For a digital manometer, zeroing establishes the reading at equal port pressures; it does not establish the gain away from zero. The manufacturer's Series 475 Handheld Digital Manometer: Specifications - Installation and Operating Instructions by Dwyer Instruments, Inc. directs the user to vent both ports to the atmosphere before zeroing. Its stated accuracy is a fraction of full scale, not of the reading, and depends on instrument temperature. Select a suitable range and record the instrument variant and units; the manual covers several ranges.^[ref]

An independent physical check can use a liquid-column manometer: apply a stable pressure difference to both instruments and compare the digital reading with the hydrostatic pressure inferred from the vertical level difference. For equal-height connections carrying a gas of mass density ${\rho }_g$ above a manometric liquid of mass density ${\rho }_m$,This follows by balancing hydrostatic pressure down and up the two connected columns. Use vertical height, even if a tube is inclined. Known mass densities, measured height and gravitational acceleration set the comparison scale. Check several differences across the working range and reverse the port connections to check the sign. This is a proposed independent check, not a claim that the cited teaching experiment performed a full instrument calibration. Recheck zero after reconnecting the sealed tubing; a zero adjustment alone cannot detect a blocked passage or an incorrect scale.