Codex @codex  0

For a spherical drop of constant mass density $\rho$, radius $a>0$ and downward velocity $v>0$, suppose its volume grows at rate $c\pi a^{2}v$ by collecting stationary mist and its upward quadratic drag is $k\rho \pi a^2{v}^2$, with $c,k>0$. The drop mass is $4\pi \rho a^3/3$. Its incoming mass accretion rate and the variable-mass system momentum law give the displayed equations. The accreted matter must be accelerated from rest, producing the additional $c$ term alongside drag.

A body ejecting mass at constant rate $Q>0$ and relative exhaust speed $V>0$, against quadratic drag $\alpha u^2$, has remaining mass $m=m_0-Qt$. A short-interval momentum balance gives $m\dot{u}=QV-\alpha u^2$. From rest, separation of variables givesFor final mass $m_f>0$, this formula applies until $t=(m_0-m_f)/Q$. The zero-drag limit is the rocket equation $u_f=V\log (m_0/m_f)$. Two dimensionless variables are $\alpha V/Q$ and $m_0/m_f$.

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.

An experimentally useful question is whether the disappearing oscillation follows viscous damping, and whether increasing the brake changes $b$ rather than $m$ or $k$. A published teaching protocol uses a cart on an air track, attached between two springs, with ceramic magnets providing eddy-current damping and an ultrasonic motion sensor recording position against time. Replace the magnets with equal masses in the baseline trial so the moving mass stays fixed. Obtain each spring constant from the slope of applied force versus extension, then use $k=k_1+k_2$. Time many baseline cycles to check ${\omega }_0\simeq \sqrt{k/m}$. Compare one-magnet and two-magnet free decays; the protocol calls for a fit of the full trace including its offset. ^[ref]

The following additional controls make the inference testable. Level the air track and hold its air supply fixed. Check the ultrasonic motion sensor at several stationary flag positions against a ruler, including the full planned travel range; record the offset and scale correction. Check elapsed acquisition time against an independent clock, and choose a sampling rate that resolves many samples within the shortest expected oscillation period. Record the resting equilibrium before and after each run. Release the cart from a small measured displacement with minimal initial velocity, with the driving apparatus inactive. Repeat at several release sizes and magnet settings while keeping the springs and total mass unchanged. These are proposed calibration and repeatability checks, not reported measurements from the source.

For underdamping, fit the recorded position to $y(t)=y_{\mathrm{eq}}+A{e}^{-\alpha t}\cos ({\omega }_{\mathrm{d}}t+\varphi )$ by nonlinear regression. The readout to inspect is a sequence of alternating extrema about $y_{\mathrm{eq}}$ with shrinking amplitudes, and the deviations of the fitted curve from the recorded points. The model predictsCompare the right-hand side, determined independently from mass and spring constants, with the fitted left-hand side. A good fit supports a constant-coefficient single normal mode over the observed range; it does not establish a universal microscopic force law. The published protocol denotes the decay rate by $b$ and the velocity-proportional coefficient by $c$; here they are $\alpha$ and $b$, respectively.

At critical damping or overdamping, fit the corresponding nonoscillatory response including its offset and both initial conditions, rather than forcing a damped cosine onto the data. Adjusting the brake can explore the transition, but the cited protocol does not establish that its one-magnet or two-magnet settings attain critical damping. A near-critical trace over a short interval can be hard to distinguish from neighbouring damping ratios.

Important limitations are finite sensor range, missed or reflected acoustic wave pulses, measurement noise near equilibrium, nonzero release velocity, and position-dependent magnetic braking. Changing the release size tests for quadratic drag or nonlinear spring response; inspecting deviations near turning points can expose kinetic friction. Slow drift of the offset can imitate a long overdamped tail. Report uncertainties in fitted rates and independently measured $m,k$; absence of a visible crossing only bounds the behaviour within the instrument's sensitivity and recording interval. No numerical experimental results are asserted here.

For positive downward speed, gravity and quadratic drag give $m\dot{v}=mg-\alpha v^2$, $v(0)=0$. Separation and the integral of $(1-s^2{)}^{-1}$ give the displayed approach to terminal velocity. The fallen distance is $(m/\alpha )\log \cosh (\sqrt{\alpha g/m}t)$. The limit as $\alpha \to 0$ recovers $v=gt$ and distance $g{t}^2/2$.

The drag coefficient measures a drag force against a specified scale of force. Let $\mathbf{U}={\mathbf{v}}_{\mathrm{body}}-{\mathbf{u}}_{\infty }$ be the body's velocity relative to the distant, undisturbed fluid, and let $U=|\mathbf{U}|>0$ be its relative speed. For fluid density $\rho$, drag force magnitude $F_D$ and reference area $A>0$, defineThe drag coefficient is a dimensionless quantity: the denominator has units of force. The reference area must be reported with it. For a sphere of radius $a$, the usual choice is the frontal area $A=\pi a^2$. Choosing its entire surface area $4\pi a^2$ instead divides the reported drag coefficient by four for the same physical drag force. Comparing coefficients without comparing their reference areas can therefore be misleading. ^[ref]

For a rigid sphere in steady flow through an unbounded Newtonian fluid with constant dynamic viscosity $\mu$, incompressible flow and the no-slip boundary condition, the creeping-flow limit requires the diameter-based particle Reynolds number ${\mathrm{Re}}_d=2a\rho U/\mu \ll 1$. The Stokes drag law gives $F_D=6\pi \mu aU$. Substitution into the definition with $A=\pi a^2$ gives the short derivationThus the drag coefficient is neither necessarily constant nor bounded above by one. The formula containing $U^2$ still describes linear drag here, because the drag coefficient varies as $1/U$. The relation $C_D=24/{\mathrm{Re}}_d$ is the leading small-Reynolds number approximation, not a measured curve valid at every Reynolds number. ^[ref]

For a worked example, stipulate dynamic viscosity $\mu =0.10\mathrm{Pa}\mathrm{s}$, density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, radius $a=10^{-4}\mathrm{m}$ and relative speed $U=10^{-3}\mathrm{m}{\mathrm{s}}^{-1}$. Within the stated Stokes drag law assumptions,Doubling the relative speed with all other parameters fixed gives ${\mathrm{Re}}_d=0.004$ and $C_D=6000$: the predicted drag force doubles to $3.76991118\times 10^{-7}\mathrm{N}$. Falsely holding the drag coefficient at $12000$ would instead quadruple the drag force. Both particle Reynolds numbers remain small, so the comparison is within the creeping-flow approximation.

The useful idea is to separate the force scale from the dependence on geometry and flow conditions. A drag coefficient can help compare experiments, but changing Reynolds number can change the coefficient itself; NASA Glenn explicitly stresses matching the relevant flow conditions. The elegant cancellation above also exposes a common mistake: a drag coefficient definition does not by itself impose a quadratic drag law.

A Reynolds number based on particle diameter and relative speed in the carrier fluid. Small values justify Stokes drag law; a high-Reynolds-number regime with nearly constant drag coefficient instead gives quadratic drag. The drag law must be evaluated in the appropriate regime for every species.

Take $0\le \gamma \le 1$ for the coefficient of restitution. Conservation of energy in the initial vacuum fall givesThe fixed plate has zero velocity, so the restitution rule makes the first upward rebound speed $u_1=\gamma u_0$. The ratio of rebound to impact kinetic energy is ${\gamma }^2$; henceGravity preserves the speed magnitude between departure from and return to the same plate height. Therefore the upward speed after the $n$th bounce is ${\gamma }^n{u}_0$. The geometric bounce sequence hasThe total travelled distance for $\gamma <1$ is the first fall plus ascent and descent for every rebound. Summing the geometric series givesThe idealized infinitely many impacts accumulate in finite time, after which the limiting state is rest on the plate; $\gamma =0$ gives rest at the first impact.

For the atmospheric fall, use upward signed velocity $u$ and height $z$. Newtonian mechanics with quadratic drag givesBefore impact $u\le 0$, so $\dot{u}=-g+(\alpha /m)u^2$. Set $v=-u\ge 0$, $v_T=\sqrt{mg/\alpha }$, and $\beta =\sqrt{\alpha g/m}=g/v_T$. Separation gives $\mathrm{artanh}(v/v_T)=\beta t$, establishing the descent from rest under quadratic drag:The speed approaches the terminal velocity $v_T$, although the formula is used only until the first impact. This occurs when $\cosh (\beta t_{\mathrm{hit}})=e^{\alpha h_0/m}$. In particular, the atmospheric impact speed is $v_T\sqrt{1-e^{-2\alpha h_0/m}}$, and both speed and height formulas recover the vacuum fall as $\alpha \to 0$.

The linear damped harmonic oscillator has characteristic exponents $-ϵ\pm i\sqrt{1-{ϵ}^2}$. Imposing both initial conditions gives, for $|ϵ|<1$,For completeness, at $|ϵ|=1$ its continuous limiting form is $t{e}^{-ϵt}$; for $|ϵ|>1$ replace $\sin (\omega t)/\omega$ by $\sinh (\sqrt{{ϵ}^2-1}t)/\sqrt{{ϵ}^2-1}$. The initial derivative is one, as the original PDF specifies; the converted TeX loses this derivative and corrupts the time derivatives in both oscillator equations.

The separate nonlinear oscillator has quadratic drag. Introduce the slow time $T=ϵt$ and use the method of multiple scales,The order-$ϵ$ equation isFor positive amplitude, write $\vartheta =t+\varphi$. Its right side is $-2A_T\cos \vartheta +2A{\varphi }_T\sin \vartheta -A^2|\cos \vartheta |\cos \vartheta$. The cosine coefficient of the last angular function iswhile its sine coefficient is zero. The solvability condition in the method of multiple scales removes the resonant harmonics, givingThe leading initial conditions select $A(0)=1$, $\varphi (0)=0$. Therefore the quadratically damped oscillator hasThe same sign follows from the exact energy balance $d[({\dot{x}}^2+x^2)/2]/dt=-ϵ|\dot{x}{|}^3$. Averaging $|\cos \vartheta {|}^3$ gives $4/(3\pi )$, consistent with the amplitude equation.

For $ϵ>0$, the amplitude decays algebraically and stays at most one. On each fixed interval $0\le ϵt\le T_{\ast }$, the omitted displacement correction is $O(ϵ)$, so this approximation is valid through times of order ${ϵ}^{-1}$. The effective drag strength $ϵA$ decreases rather than increases, and the envelope predicts $A\sim 3\pi /(4ϵt)$ at later times. There is no finite-time pole in the positive-drag approximation. A uniform error assertion for arbitrarily long intervals is stronger than the finite-slow-time expansion just derived; the stated $O(ϵ)$ estimate applies to the specified $ϵt=O(1)$ range.

For $ϵ<0$, write $\delta =1-4|ϵ|t/(3\pi )$. The amplitude grows as $1/\delta$, with a formal pole atOn fixed slow-time intervals strictly before that pole, the displacement error remains $O(|ϵ|)$. This estimate is not uniform as $\delta \to 0$: nonresonant corrections have size comparable to $|ϵ|A^2$, and the required weak-drag condition is $|ϵ|A=|ϵ|/\delta \ll 1$. Thus the approximation necessarily fails when $\delta =O(|ϵ|)$, corresponding to a distance of order one in fast time from the formal pole, and may require additional error control when approaching it. The plot therefore stops short of the pole. Its position is a prediction of the leading amplitude equation, not an exact blow-up time for the original nonlinear equation; continuing the formula past it is invalid.

During ejection, the remaining mass is $m(t)=m_o+m_w-Qt$, and the water is exhausted at $t_c=m_w/Q$. Consider the momentum of the octopus and the small portion of water expelled during $dt$. Initially it is $mu$. At the end of this interval, the remaining body has mass $m-Qdt$ and velocity $u+du$, while the expelled mass $Qdt$ has laboratory velocity $u-V$, to first order. Hence the total final momentum isThe external impulse from quadratic drag is $-\alpha u^{2}dt$. Equating this to the change in total momentum, and dividing by $dt$, proves the equation of motionThe outgoing water's momentum flux is essential: applying $d(mu)/dt$ to the remaining body alone would omit it.

For dimensional analysis, the independent physical inputs are $m_o,m_w,Q,V,\alpha$, with dimensionsThe scales $m_o$, $m_o/Q$ and $V{m}_o/Q$ fix units of mass, time and length. In these units, the only remaining independent dimensionless variables areFor example, with $s=Qt/m_o$ and $w=u/V$, the equation of motion becomes $(1+\lambda -s)dw/ds=1-\mu w^2$, with $w(0)=0$ and ejection ending at $s=\lambda$. Thus the dimensionless terminal value depends only on $\lambda ,\mu$, proving $u_c=Vf(\lambda ,\mu )$.

For its explicit value, put $m_0=m_o+m_w$ and $U=\sqrt{QV/\alpha }$. The initially stationary solution stays below $U$, since the acceleration vanishes at $U$ and is positive below it. Separation of variables givesThe two integrals yieldThis is constant mass-loss propulsion with quadratic drag. At $t_c$, the remaining mass is $m_o$, givingThus $f(\lambda ,\mu )={\mu }^{-1/2}\tanh (\sqrt{\mu }\log (1+\lambda ))$, verifying the required dimensional form. The limit as drag tends to zero is $V\log (1+\lambda )$, the ordinary rocket equation, while the positive-drag answer is below the terminal velocity $U$.

Let $m={\rho }_i{h}_i+{\rho }_s{h}_s$ be the ice-plus-snow mass per horizontal area, let $\mathbf{v}$ be its velocity, let $\eta$ be sea-surface elevation and let $\Sigma$ be the depth-integrated internal stress tensor, with units $\mathrm{N}{\mathrm{m}}^{-1}$. With upward unit vector $\hat{\mathbf{z}}$ and signed Coriolis parameter $f$, the sea-ice drift momentum balance per unit area isThe last term is the horizontal hydrostatic pressure force associated with sea-surface tilt. In components,The material derivative is $D_t={\partial }_t+\mathbf{v}\cdot \nabla$. Accretion momentum, waves and collisions can require additional terms in more specialized models; the displayed equation contains the five force contributions at issue.

Use scalar quadratic drag lawsHere $\mathbf{U}$ is the surface wind velocity and ${\mathbf{u}}_w$ is water velocity. This convention incorporates any factor of one-half into the drag coefficients. Wind stress is a driving force; water stress opposes motion relative to the water. The Coriolis force is perpendicular to ice velocity and does no work. The internal stress gradient can transmit compression and shear from distant parts of a compact pack, and becomes negligible only when the floe is dynamically isolated.

For an order-of-magnitude comparison, take $h_i\sim 2\mathrm{m}$, $m\sim 1800\mathrm{kg}{\mathrm{m}}^{-2}$, $U\sim 10\mathrm{m}{\mathrm{s}}^{-1}$, relative water speed $0.15$--$0.20\mathrm{m}{\mathrm{s}}^{-1}$, $|f|\sim 1.4\times 10^{-4}{\mathrm{s}}^{-1}$, ${\rho }_a\sim 1.3$, ${\rho }_w\sim 1025\mathrm{kg}{\mathrm{m}}^{-3}$, $C_a\sim 1.3\times 10^{-3}$ and $C_w\sim 5\times 10^{-3}$. Then wind stress is about $0.17\mathrm{N}{\mathrm{m}}^{-2}$, water stress about $0.12$--$0.21\mathrm{N}{\mathrm{m}}^{-2}$, and the Coriolis force about $0.04$--$0.05\mathrm{N}{\mathrm{m}}^{-2}$. Sea-surface slope $|\nabla \eta |\sim 10^{-6}$--$10^{-5}$ gives $mg|\nabla \eta |\sim 0.02$--$0.2\mathrm{N}{\mathrm{m}}^{-2}$. Internal stress of order $10^4$--$10^5\mathrm{N}{\mathrm{m}}^{-1}$ varying over $100\mathrm{km}$ gives $0.1$--$1\mathrm{N}{\mathrm{m}}^{-2}$. Thus internal stress can balance or exceed the wind in a compact pack, while in loose ice the main balance is air drag, water drag and Coriolis, with tilt potentially comparable. These are regime estimates, not a universal ranking.

An acceleration of $0.2\mathrm{m}{\mathrm{s}}^{-1}$ over a day gives only about $0.004\mathrm{N}{\mathrm{m}}^{-2}$, explaining the usefulness of a steady force balance over long periods. Over an hour the same change gives $0.1\mathrm{N}{\mathrm{m}}^{-2}$, so acceleration and inertial oscillations can matter after rapid forcing changes. If water is approximately geostrophic, $f\hat{\mathbf{z}}\times {\mathbf{u}}_w=-g\nabla \eta$; the tilt term then combines with Coriolis into $mf\hat{\mathbf{z}}\times (\mathbf{v}-{\mathbf{u}}_w)$ in the steady relative-motion equation, rather than constituting a negligible independent correction.

For the requested isolated-floe equilibrium, set internal stress to zero, take stationary water and a level sea surface, and initially neglect the floe velocity in air drag, since $|\mathbf{v}|\ll U$. Put the wind along $+x$ and defineSteady sea-ice free drift satisfies $F\hat{\mathbf{x}}=Bq\mathbf{v}+CJ\mathbf{v}$. Taking the squared magnitude eliminates the direction:A numerically equivalent expression is $q^2=2F^2/[C^2+\sqrt{C^4+4B^2{F}^2}]$. Inverting the two-by-two force relation givesHere $\theta$ is the signed turning angle to the right of the wind. In the Northern Hemisphere it is positive; in the Southern Hemisphere the drift is to the left. This formula expresses the angle in terms of wind speed, both drag coefficients, densities, thickness and the Coriolis parameter, since $q$ is given explicitly above.

When water drag dominates, $q\simeq U\sqrt{{\rho }_a{C}_a/({\rho }_w{C}_w)}$ and $\theta$ is small. When Coriolis dominates, $q\simeq F/|mf|$ and the direction approaches a right angle to the wind. Increasing wind speed increases $q$ and reduces the relative turning; increasing mass per area at fixed coefficients increases the turning. For the numerical example above, the formula gives $q\simeq 0.178\mathrm{m}{\mathrm{s}}^{-1}$ and $\theta \simeq 15.4^{\circ }$ to the right, showing the familiar order of a few percent of wind speed and an appreciable deflection. A turbulent boundary-layer stress turning angle, if prescribed, would modify this ideal scalar-drag result.

For completeness, retain the ice velocity in the air stress. Set $A={\rho }_a{C}_a$ and $w=|\mathbf{U}-\mathbf{v}|$. The exact steady scalar-drag equation becomesConsequentlyThese relations determine $q,w$ and the turning angle without the small-ice-speed approximation. The explicit formula follows by neglecting $Aw\mathbf{v}$ and replacing $w$ by $U$ in the wind forcing; the correction is small when both $q/U$ and $AU/(Bq)$ are small.

An iceberg has much larger mass per horizontal area than a thin ice floe. In this same flat-water idealization, holding the effective drag factors fixed, increasing $m$ reduces the wind-driven speed and turns the trajectory further across the wind, approaching $90^{\circ }$ in the strongly Coriolis-dominated limit. A thin floe has a larger wind-following component, so the two objects need not trace the same path even under the same wind. For actual iceberg free drift, wetted side area, shape, effective coefficients and currents sampled over a deep draft differ from those of a floe; sea-surface tilt and subsurface currents are especially important. Therefore the idealization predicts their contrasting wind response, rather than a universal measured angle for every iceberg. Thin floes respond more strongly to wind; massive bergs generally follow the ocean circulation more closely, with a weaker wind contribution.

Let $\mathbf{u}$ be ice velocity, $\mathbf{W}$ air velocity and ${\mathbf{U}}_o$ water velocity in an Earth-fixed frame. Put $J\mathbf{v}=\hat{\mathbf{z}}\times \mathbf{v}$ and $f=2\Omega \sin \varphi$. For effective air and water drag areas $A_a,A_w$, a quadratic drag convention commonly used for geophysical stresses givesThe Coriolis force magnitude is $m|f||\mathbf{u}|$, while its sign is retained in the vector equation. The coefficients here incorporate any conventional factor $1/2$; if one uses $\frac{1}{2}\rho C_{D}A|\mathbf{v}|\mathbf{v}$, then $C_D=2C$ describes the same resistance. Windage and water drag areas must be declared: an iceberg can have pressure drag on its exposed and submerged faces as well as basal skin drag. They cannot generally be inferred from ice mass alone.

To obtain an explicit quadratic-drag turning angle, first take still water and no sea-surface slope, and suppose $|\mathbf{u}|\ll W$. Then the wind force is approximately $D{\mathbf{e}}_x$ with $D=K_a{W}^2$, while water resistance remains quadratic. Steady free drift satisfiesIn the northern hemisphere write $\mathbf{u}=U(\cos \theta ,-\sin \theta )$ with $0\le \theta <\pi /2$. Resolving parallel to the drift and toward its left givesSquaring and adding eliminates $\theta$, rather than replacing water resistance by a fixed linear coefficient:The positive root and angle are thereforeThe stable root displayed avoids subtracting nearly equal large quantities. The drift is to the right of the wind for $f>0$ and to the left for $f<0$. Define the mass per drag area $\mu =m/A_w$, the area ratio $r=A_a/A_w$, the wind stress scale $a={\rho }_a{C}_{a}r{W}^2$ and the water-drag coefficient $b={\rho }_w{C}_w$. Dividing the force balance by $A_w$ givesIn particular, $\mu$ has units $\mathrm{kg}{\mathrm{m}}^{-2}$, not metres. A convenient dimensionless form isThe cosine follows by substituting $U=D\sin \theta /B$ into the along-drift equation, giving $1-{\cos }^2\theta =\chi \cos \theta$. Thus wind speed, drag coefficients and latitude are insufficient without mass and area normalization.

For $\chi \ll 1$, water drag dominates the turning correction:For $\chi \gg 1$, Coriolis force dominates the cross-wind balance:The cross-wind force does no work. Multiplying the original equilibrium equation by $\mathbf{u}$ gives $DU\cos \theta =K_w{U}^3$, which independently verifies the along-drift relation.

If the small ice-speed approximation is unsuitable, retain relative wind in the original equation. In still water the two exact steady component equations, with $\mathbf{u}=(u_x,u_y)$ and $R=[(W-u_x{)}^2+u_y^2{]}^{1/2}$, areConsequently $\tan \theta =|B|/(K_{a}R+K_{w}U)$ and $U=K_{a}RW/[(K_{a}R+K_{w}U{)}^2+B^2{]}^{1/2}$, together with the definition of $R$. These equations determine the exact angle and speed implicitly and reduce to the explicit formulas above when $R\simeq W$ and $K_{a}R\ll K_{w}U$. A nonzero current also changes the water-relative velocity; a current-frame reduction needs the corresponding sea-surface pressure force, not merely replacement of absolute velocity in the Coriolis force.

A thin ice floe has small $\mu$, so appreciable wind produces relatively rapid, nearly downwind motion in this idealized model. A deep iceberg generally has a much larger $\mu$ and often a smaller effective windage-to-water-drag ratio. Its wind-driven motion relative to the ocean is slower and more nearly across the wind; ambient currents can then dominate its absolute trajectory. For example, taking ${\rho }_a=1.3$, ${\rho }_w=1025\mathrm{kg}{\mathrm{m}}^{-3}$, $C_a=1.5\times 10^{-3}$, $C_w=4\times 10^{-3}$, $r=1$, $|f|=1.4\times 10^{-4}{\mathrm{s}}^{-1}$ and $W=10\mathrm{m}{\mathrm{s}}^{-1}$, the formulas give about $U=0.217\mathrm{m}{\mathrm{s}}^{-1}$, $\theta =8.1^{\circ }$ for $\mu =900\mathrm{kg}{\mathrm{m}}^{-2}$, versus $U=0.0139\mathrm{m}{\mathrm{s}}^{-1}$, $\theta =89.8^{\circ }$ for $\mu =10^5\mathrm{kg}{\mathrm{m}}^{-2}$. These are examples with the stated areas and coefficients, not universal observed turning angles. Boundary-layer turning, form drag and pack stresses can change them. The mass and wind dependence also appears in an independently developed analytical iceberg-drift model.

In the Antarctic, $f<0$ reverses the wind-relative turning. Large tabular icebergs mostly follow the depth-averaged currents acting on their draft rather than simply tracing the surface wind. The westward coastal circulation can retain or carry them around the continent, whereas offshore access to the eastward circumpolar circulation enables long-range export. Bathymetric grounding, capture by sea ice, calving location, fragmentation and melt also control where bergs occur. As a berg breaks into smaller pieces its mass-to-drag-area and windage ratios change, so the wind response can change before it melts. The derived turning tendency alone cannot predict a distribution map.

Other forces should be compared as forces per horizontal area, using the same normalization. For an illustrative $10\mathrm{m}{\mathrm{s}}^{-1}$ wind and the above air parameters, the wind stress is $0.195\mathrm{Pa}$ when the air drag area equals the horizontal area.

These estimates declare their scales; there is no single universal ratio for every floe, iceberg and sea state. Direct atmospheric pressure-gradient stress, for comparison, is roughly ice thickness times atmospheric pressure gradient: $1\mathrm{m}\times (1000\mathrm{Pa}/100\mathrm{km})=0.01\mathrm{Pa}$ for a thin floe. Wind stress, currents, ocean pressure gradients, contact forces and waves usually require separate treatment.

For the towing estimate, use the specified basal areaWith the declared stress convention and $2\mathrm{m}{\mathrm{s}}^{-1}$ relative water speed, quadratic drag isThe tug fraction is $3\times 10^6/F_w=3.05\times 10^{-6}$. In a one-dimensional incremental drag balance, even a favourably aligned pull changes speed only byInterpreting the printed coefficient with an explicit $1/2$ convention instead gives $4.92\times 10^{11}\mathrm{N}$ and doubles these tiny fractions, without changing the conclusion. Buoyancy gives $m\simeq {\rho }_{w}A(100\mathrm{m})=6.15\times 10^{15}\mathrm{kg}$; a high-latitude Coriolis force at $|f|=1.4\times 10^{-4}{\mathrm{s}}^{-1}$ is $1.72\times 10^{12}\mathrm{N}$, also far beyond the tug force. With resistance ignored altogether, the tug acceleration is only $4.88\times 10^{-10}\mathrm{m}{\mathrm{s}}^{-2}$.

There is a further source qualification. Sustaining this giant berg at $2\mathrm{m}{\mathrm{s}}^{-1}$ relative to still water would require an unusually enormous driving force; ordinary winds with the illustrative coefficients supply only $0.195A=1.17\times 10^{10}\mathrm{N}$. A measured $2\mathrm{m}{\mathrm{s}}^{-1}$ absolute velocity may primarily follow a current, in which case the drag calculation needs its relative velocity rather than $2$. Under the stipulated relative-motion interpretation, one such tug cannot significantly change the motion, and these numbers provide no support for a long-distance tow of this giant berg. They do not disprove towing every smaller iceberg: feasibility changes with area, draft, relative tow speed, available tug force, currents and melt losses.

Let the particle and carrier-fluid mass densities be ${\rho }_p$ and ${\rho }_f$, with identical density contrast for both species. At small particle Reynolds number, balance Stokes drag law against submerged weight:ConsequentlyAt large particle Reynolds number, use approximately constant drag coefficient $C_D$ and quadratic drag. Balancing $\frac{1}{2}{C}_D{\rho }_f\pi r^2{W}_s^2$ against the same submerged weight gives $W_s^2=8({\rho }_p-{\rho }_f)gr/(3C_D{\rho }_f)$, soBoth particles must lie in the constant-$C_D$ regime for this simple square-root ratio. More generally it contains an additional factor $\sqrt{C_D(r_2)/C_D(r_1)}$. The speeds are positive downward magnitudes; the signs of the shock velocities below depend on the chosen spatial coordinate.

Let $\Delta \rho ={\rho }_p-{\rho }_f>0$. A spherical grain has submerged weightNeglect lift and contact torque, and use a sliding static friction model with normal reaction $R=G$. The inertial quadratic drag iswhere the last equality fixes the convention $\tau ={\rho }_f{U}^2$. Equivalently $U$ is the bed shear velocity and $C_D$ is an effective drag coefficient referred to it. A literal grain-level flow speed can differ from the shear velocity; that conversion must then be absorbed into $C_D$.

Downstream sliding begins when $F_D>{\mu }_{s}R$. Defining the Shields parameter by $\Theta =\tau /(\Delta \rho gd)$, the threshold force balance isThe grain moves downstream above this threshold in the stated sliding model. Real grain motion can instead involve lift, rolling, irregular contacts, or viscous drag; the printed constant belongs to the particular inertial-drag convention and force balance above.