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A beta plane retains the leading meridional variation $f=f_0+\beta y$ of the Coriolis parameter.

For a constant wind force $\mathbf{F}$, still water, and linear water drag force $-m\gamma \mathbf{v}$, let $J\mathbf{v}=\hat{\mathbf{z}}\times \mathbf{v}$. The Coriolis parameter $f$ gives equilibrium ${\mathbf{v}}_{\infty }=(\gamma \mathbf{F}-fJ\mathbf{F})/[m({\gamma }^2+f^2)]$. The deviation evolves as $e^{-\gamma t}R(-ft)(\mathbf{v}(0)-{\mathbf{v}}_{\infty })$, a damped inertial oscillation. Actual drag is commonly quadratic, so this is a linear illustrative model rather than a universal iceberg speed law.

An Ekman layer is a viscous boundary layer in a rotating fluid. For constant kinematic viscosity $\nu$ and Coriolis parameter $f$, its characteristic thickness is ${\delta }_E=\sqrt{2\nu /|f|}$, and its velocity turns with distance from the boundary.

An equatorial wave is trapped near the equator by the sign change of the Coriolis parameter. Its meridional structures are Gaussian-weighted Hermite polynomials.

An f-plane approximates the Coriolis parameter by a constant $f_0$ over a limited horizontal region.

In nearly columnar flow between rigid boundaries, a geostrophic contour is a contour of column height divided by the Coriolis parameter. If relative vorticity is negligible, conservation of shallow-water potential vorticity constrains leading geostrophic flow to these contours. Motion across them stretches or squashes columns, generating relative vorticity; for constant $f$, the contours are simply contours of $H$.

Inertial instability occurs when displaced fluid parcels amplify their displacement by conserving absolute momentum in a rotating flow. For a uniformly stratified basic state on an f-plane, a necessary condition is that the Ertel potential vorticity have the opposite sign to the Coriolis parameter.

The quasi-geostrophic approximation isolates slowly evolving, nearly balanced motions in a rotating, stably stratified fluid. Ocean eddies and mid-latitude atmospheric disturbances often have this character: a leading Coriolis force–pressure balance suppresses rapid horizontal acceleration, while hydrostatic balance suppresses rapid vertical acceleration. Small ageostrophic flow nevertheless supplies the vertical motion that changes the leading vorticity. The approximation filters fast inertia-gravity waves while retaining nonlinear transport, vertical shear and Rossby waves.

Use a reference density ${\rho }_0$ and remove the background hydrostatic pressure. Let $\pi =p^{\prime }/{\rho }_0$ and let the upward buoyancy perturbation be $b=-g{\rho }^{\prime }/{\rho }_0$. A vertically varying reference density stratification has $N^2(z)>0$. Under the Boussinesq approximation, the inviscid and adiabatic equations areHere $D/Dt={\partial }_t+{\mathbf{u}}_H\cdot {\nabla }_H+w{\partial }_z$. Density variations are small compared with ${\rho }_0$ but retained in the vertical buoyancy force. This is a useful local, incompressible description; a compressible atmosphere needs the corresponding weighted or pressure-coordinate version.

Let horizontal and vertical lengths be $L,H$, speed be $U$, and take the slow time $L/U$. Take $f_0>0$ for definiteness; use $|f_0|$ in the scale estimates in the opposite hemisphere. The principal small parameter is the Rossby number $\mathrm{Ro}=U/(f_{0}L)\ll 1$ at a latitude where $f_0\ne 0$. For the hydrostatic, strongly stratified regime, $H/L\ll 1$ and $f_0/N\ll 1$. Retaining both horizontal vorticity and vortex stretching gives the Burger numberThe geostrophic balance pressure scale is $f_{0}UL$ and the buoyancy perturbation scale is $B=f_{0}UL/H$. Hence $B/(N^{2}H)=\mathrm{Ro}/\mathrm{Bu}\ll 1$: buoyancy perturbations produce small displacements relative to the background stratification scale. The ageostrophic flow is $O(\mathrm{Ro}U)$ and incompressibility then gives $w=O(\mathrm{Ro}UH/L)$. Vertical acceleration is smaller than hydrostatic terms by order ${\mathrm{Ro}}^2(f_0/N{)}^2$ in this scaling. Weak forcing or diffusion can be added at the slow order, but neither is present here.

Define the quasi-geostrophic streamfunction by $\psi =\pi /f_0$. Leading geostrophic balance and hydrostatic balance implyDifferentiating horizontally and vertically gives the thermal-wind balanceThus vertical shear is determined by horizontal buoyancy gradients. The homogeneous Taylor–Proudman theorem is recovered only when those gradients vanish. Stable stratification allows sloping density surfaces and different geostrophic flow at different levels, without abandoning leading balance.

The Prandtl ratio of scales makes that modification quantitative. Horizontal and vertical terms in balanced potential-vorticity inversion have sizes $\psi /L^2$ and $(f_0^2/N^2)\psi /H^2$. Equality givesFor constant $N$, $Z=(N/|f_0|)z$ makes the inversion operator an ordinary three-dimensional Laplacian. Equivalently, an interior balanced disturbance of horizontal size $L$ naturally penetrates a depth of order $|f_0|L/N$. With $N\gg |f_0|$ this is a thin, vertically sheared structure, rather than the arbitrarily tall homogeneous columns suggested by strict Taylor–Proudman theorem behaviour. This is an aspect-ratio statement, not the diffusivity Prandtl number.

To allow a weak meridional variation of the Coriolis parameter, use the beta plane $f=f_0+\beta y$, with $\beta L/f_0=O(\mathrm{Ro})$, or equivalently $\beta L^2/U=O(1)$. The small correction to the leading balance is dynamically important on the slow time. Let $\xi ={\nabla }_H^2\psi$ be vertical relative vorticity and let $D_g={\partial }_t+{\mathbf{u}}_g\cdot {\nabla }_H$. At the first nontrivial order, the vertical vorticity equation and buoyancy equation reduce toRelative-vorticity stretching and tilting are higher order here; planetary-vorticity stretching $f_0{w}_z$ must remain. Eliminate $w$ to expose a materially conserved balanced scalar. Because ${\mathbf{u}}_{g,z}=(-b_y/f_0,b_x/f_0)$, the commutator term ${\mathbf{u}}_{g,z}\cdot {\nabla }_{H}b$ is exactly zero. For $N=N(z)$ this givesAdding the two equations proves the quasi-geostrophic potential-vorticity equationAn additive constant such as $f_0$ changes no dynamics. The horizontal streamfunction advection bracket can express this as $Q_t+{\psi }_x{Q}_y-{\psi }_y{Q}_x=0$. Although small departures from balance generated this evolution equation, the leading geostrophic flow itself advects $Q$ nonlinearly. The stretching term couples horizontal motion to vertical displacement of density surfaces.

This gives the practical advection-and-inversion description. Start with interior three-dimensional quasi-geostrophic potential vorticity, advect it with ${\mathbf{u}}_g$, and recover $\psi$ fromThe stratified quasi-geostrophic inversion is elliptic for positive $N^2$. It must be supplied with lateral and vertical boundary conditions and any circulation or mean-flow constraints; interior potential vorticity alone is not sufficient. At rigid horizontal lids, $w=0$ makes boundary buoyancy satisfy $D_{g}b=0$, and $b=f_0{\psi }_z$ supplies the vertical derivative data for inversion. Boundary buoyancy perturbations can therefore support flow even when the interior anomaly is zero. Once $\psi$ is found, its horizontal derivatives give velocity, its vertical derivative gives buoyancy, and the thermodynamic equation diagnoses $w$. This balanced evolution excludes the independent fast-wave initial data present in the full equations.

Finally, a gradient in $f$ supplies a restoring mechanism even in an otherwise uniform fluid. Linearizing about rest with constant $N$, the quasi-geostrophic potential-vorticity equation becomesSubstitution of $\psi \propto e^{i(kx+ly+mz-\omega t)}$ gives the Rossby wave dispersionFor $\beta >0$, its zonal phase speed is westward relative to a resting background. A meridionally displaced parcel retains potential vorticity, so it acquires a relative-vorticity anomaly that induces the velocity tending to return the disturbance. Background shear and stratification modify the full potential-vorticity gradient; they can alter propagation or permit instability. On a constant f-plane the planetary contribution vanishes, but gradients of relative vorticity, boundary buoyancy or topography can still support balanced wave motion. Close enough to the equator that $f_0$ vanishes, this mid-latitude quasi-geostrophic approximation and its scaling must be replaced.

Let $h$ be layer thickness and $g$ gravitational acceleration. The unforced shallow water equations on the equatorial beta plane areLinearize about $u=v=0$, $h=h_0$. Put $c_0=\sqrt{g{h}_0}$ and $L=\sqrt{c_0/\beta }$, and scale $(x,y)=L(X,Y)$, $t=(L/c_0)T$, $(u,v)=c_0(U,V)$ and $h=h_0(1+\eta )$. Dropping capitals, the linearized shallow water equations becomeThe choice $\beta L^2/c_0=1$ makes both the wave speed and the coefficient of the dimensionless Coriolis parameter equal to one.

Take Fourier modes $\{u(y),v(y),\eta (y)\}{e}^{i(kx-\omega t)}$ that decay as $y\to \pm \infty$. Their equations areFor $v=0$ and nonzero $k$, the first two require ${\omega }^2=k^2$. The root $\omega =k$ gives $u=\eta$, and meridional balance becomes $u^{\prime }=-yu$. Hence the equatorial Kelvin wave isIt travels eastward at speed $c_0$ in dimensional units. The other sign gives $\eta =-u$ and $u^{\prime }=yu$, so $u\propto e^{y^2/2}$ is not trapped.

For $v\ne 0$, $\omega \ne 0$ and ${\omega }^2\ne k^2$, inversion of the first two equations gives the equatorial wave velocity polarizationSubstitute in the remaining momentum equation. Multiplication by $({\omega }^2-k^2)/i$ leavessoWriting $v=e^{-y^2/2}H(y)$ yields $H^{\prime \prime }-2y{H}^{\prime }+(\lambda -1)H=0$, the Hermite differential equation. The trapped solutions have the Hermite oscillator quantization $\lambda =2n+1$ and $H=H_n$.

For completeness, quantization follows without assuming that every acceptable solution is a polynomial. On decaying smooth functions, define $a=d/dy+y$, $a^{†}=-d/dy+y$ and $\mathcal{H}=-d^2/d{y}^2+y^2=a^{†}a+1$. Integration by parts gives $\lambda \ge 1$ and $\Vert av{\Vert }^2=(\lambda -1)\Vert v{\Vert }^2$. If $av\ne 0$, it is an eigenfunction of $\mathcal{H}$ with eigenvalue $\lambda -2$. Repeated lowering must terminate: otherwise it eventually produces an eigenvalue below one. At termination $av=0$, so the terminal eigenfunction is $e^{-y^2/2}$ and has eigenvalue one. Repeated raising gives $H_n(y)e^{-y^2/2}$ and $\lambda =2n+1$; uniqueness of a decaying solution at one end of this second-order equation fixes each mode up to scale.

Thus, away from the exceptional quotients, all fields are determined byInsert this $v$ and its derivative in the polarization formulae to obtain $u$ and $\eta$. For $n\ge 1$, the cubic ${\omega }^3-(k^2+2n+1)\omega -k=0$ supplies the low-frequency equatorial Rossby wave and two equatorial inertia--gravity wave branches. Their meridional structures are Gaussian-weighted Hermite polynomials.

For $n=0$, multiplication by $\omega$ factorizes the equatorial shallow-water dispersion relation asThe physically admissible quadratic gives the mixed Rossby-gravity wave, equivalently the Yanai wave:These fields satisfy all three original equations directly and decay at both infinities, including where the quotient formulae are singular.

To test the other factor, set $\omega =-k\ne 0$. The original equations give $u+\eta =-iyv/k$ and $v^{\prime }=-yv$. Put $\delta =u-\eta$; meridional momentum then givesDecay of $u,\eta$ at both infinities requires the integral of the right side to vanish. Hence $C=0$ unless $k^2=1/2$. Ordinarily the extra factor supplies no trapped mode. At the exceptional root of the mixed Rossby-gravity mode, $k^2=1/2$, it coincides with the quadratic root and gives the same physical fields, not an additional branch. This distinction avoids discarding a regular physical mode at the crossing.

These arguments concern nonzero wave angular frequency. At $k=0$, the oscillatory nonzero-$v$ modes have $\omega =\pm \sqrt{2n+1}$, while zero-frequency $v=0$ geostrophic states must be considered in the original equations rather than in expressions divided by $\omega$.

With the free-surface elevation denoted by $\zeta$, the rotating shallow water equations over a fixed bottom areThe momentum gradient is that of the surface elevation, not depth alone; bottom slope enters through the depth and continuity equation. Taking ${\partial }_x$ of meridional momentum minus ${\partial }_y$ of zonal momentum gives the vorticity equationThe terms quadratic in velocity gradients combine into ${\omega }_r{\nabla }_H\cdot {\mathbf{u}}_H$; the constant Coriolis parameter contributes $f{\nabla }_H\cdot {\mathbf{u}}_H$. Continuity gives $D_{H}h/Dt=-h{\nabla }_H\cdot {\mathbf{u}}_H$. Dividing the two equations proves the exact shallow-water potential vorticity invariant

For the quasi-geostrophic approximation, let $U_s$ and $\ell$ be representative velocity and horizontal length scales. Require the Rossby number $\mathrm{Ro}=U_s/(|f|\ell )\ll 1$, a shallow layer, and slow evolution on time $\ell /U_s$. Put $L_R=\sqrt{g{h}_{00}}/|f|$ and assume $\mathrm{Bu}=(L_R/\ell {)}^2$ is order one. Leading momentum then gives geostrophic balance, and a quasi-geostrophic streamfunction can be chosen asIts surface-height scale satisfies $|\zeta |/h_{00}=O(\mathrm{Ro}/\mathrm{Bu})\ll 1$. Take the bottom-height ratio of the same small order, with gentle slopes, so $\eta =\zeta -b$ obeys $|\eta |/h_{00}\ll 1$. Relative vorticity is order $\mathrm{Ro}f$, and expanding the exact potential vorticity givesBecause the leading geostrophic velocity is horizontally nondivergent, ageostrophic advection of the anomaly is one order smaller. At the first evolving order the shallow-water quasi-geostrophic potential vorticity equation isThe constant reference $f/h_{00}$ has zero gradient and drops out. These assumptions also require all topographically induced velocities to retain a small Rossby number; the small bottom-height condition alone is not a full justification of quasi-geostrophic balance.

The uniform eastward current has ${\psi }_0=-Uy$ up to an additive constant, so ${\zeta }_0=-(fU/g)y$. With $b_0=-(fU/g)y$, surface and bottom slopes are equal and $h=h_{00}$. Its relative vorticity vanishes and $Q_0=0$. This sloping background is used locally where its surface and bottom departures remain within the small-height regime.

Add the compact ridge and set $\psi =-Uy+\varphi (x)$. Then $u=U$, $v={\varphi }^{\prime }(x)$, and $Q$ depends only on $x$. The steady equation becomes $U{Q}_x=0$. Uniform incoming potential vorticity fixes that constant to zero, giving the cosine-ridge shallow-water geostrophic responseLet $a=\pi L/2$, $\lambda =L_R^{-1}$ andThe forcing is even, so the decaying response is even. A cosine particular solution inside the ridge leaves an even homogeneous term; outside, retain only the decaying exponential. Since the forcing is bounded and contains no delta function, both $\varphi$ and ${\varphi }^{\prime }$ are continuous at the edges. Write $\varphi =A\cos (x/L)+D\cosh (x/L_R)$ inside and $\varphi =B{e}^{-(|x|-a)/L_R}$ outside. Matching at $x=a$ givesSolving yields $D=A(L_R/L)e^{-a/L_R}$, henceA globally decaying homogeneous solution that is continuously differentiable is zero, so this matched solution is unique.

A steady streamline is a constant-$\psi$ curve. If its upstream ordinate is $y_{\infty }$, it therefore hasFor $f>0$, $\varphi$ is positive: the flow first turns toward increasing $y$, passes the crest at its largest deflection, then returns to its upstream ordinate. Reversing the sign of $f$ reverses the deflection. For $L=3L_R$, the center deflection is $A[1+e^{-3\pi /2}/3]/U$ and each tail decays on scale $L_R$. The figure shows several streamlines and the ridge support; the deflection extends beyond the ridge.

One compatible realization of the initial condition has $h(x)=h_{00}-\hat{b}(x)$ and $\zeta =b_0(y)$. Its stated zonal velocity then satisfies $u(x)h(x)=U{h}_{00}$, so the initial mass flux is constant and $v=0$. The data given only far upstream do not uniquely determine the initial height everywhere; this is the natural mass-flux-compatible choice. It is not fully balanced over the ridge: $fu(x)$ differs from $-g{\zeta }_y=fU$. The resulting transverse acceleration initiates geostrophic adjustment and launches fast inertia-gravity waves.

Initially the relative vorticity is zero, but exact shallow-water potential vorticity is $f/[h_{00}-\hat{b}(x)]$, not the uniform upstream value. That localized anomaly is materially transported downstream, approximately at $U$ in the leading quasi-geostrophic approximation; it is not annihilated by adjustment. Fresh incoming parcels of uniform PV subsequently supply the ridge region, whose balanced potential-vorticity inversion approaches the steady solution above. The fast waves can carry unbalanced energy away on the inertia-gravity time scale, short compared with the advective time at small Rossby number. In an open domain this can leave the local balanced response; an ideal closed or reflecting domain need not settle monotonically, because undamped waves can return. Radiation, advection of the initial PV anomaly, and any weak damping must be distinguished from irreversible loss of PV.

Let $\zeta$ denote free-surface height and $h=\zeta -b$ the water depth. The inviscid rotating shallow water equations with constant Coriolis parameter $f$ areIt is $\zeta$, not $h$, that supplies the horizontal pressure gradient. For a fixed bottom, mass conservation equivalently gives $D_{H}h=-h(u_x+v_y)$.

Write $\xi =v_x-u_y$ for relative vorticity. Differentiate the second momentum equation in $x$ and the first in $y$, then subtract. The pressure terms cancel. Expanding the differentiated advection terms givesConsequently the absolute vorticity $q_a=\xi +f$ satisfies $D_H{q}_a=-q_a{\nabla }_H\cdot {\mathbf{u}}_H$. Combining this with mass conservation yields exact shallow-water potential vorticity conservation:A sloping bottom changes the depth and hence the stretching, but contributes no curl to the gravitational pressure force.

For the quasi-geostrophic approximation, choose constant reference depth $H_0$, speed $U$, and horizontal length $L$. Require Rossby number $\mathrm{Ro}=U/(|f|L)\ll 1$, slow evolution on the advective time $L/U$, smooth horizontal scales, and $L/L_D=O(1)$ or smaller, where $L_D=\sqrt{g{H}_0}/|f|$ is the Rossby deformation radius. Write $\zeta ={\zeta }_0+\eta$, $b=b_0+\delta b$, and $h=H_0+\eta -\delta b$. The leading momentum balance is geostrophic balance:It gives $\eta /H_0=O(\mathrm{Ro}(L/L_D{)}^2)$, while we require $\delta b/H_0=O(\mathrm{Ro})$ and slopes on the chosen horizontal scale. Relative vorticity divided by $f$ is also $O(\mathrm{Ro})$. ThusThe derivative of the constant first term vanishes. In advecting the first-order anomaly, the ageostrophic flow contributes only at the next order. Therefore the shallow-water quasi-geostrophic potential vorticity equation isThis retains the balanced vortical branch, including Rossby waves driven here by the topographic potential-vorticity gradient. It filters the freely oscillating inertia–gravity branches of the full shallow water equations, and does not retain independent boundary Kelvin wave dynamics. Small Rossby number alone is insufficient for the reduction: fast waves must be absent at leading order, as in well-prepared balanced initial data.

In the channel take $H_0=h_{00}$ and measure the bottom anomaly from mid-channel: $\delta b=\epsilon (y-L/2)$. Define $\beta =\epsilon f/H_0$. The resting potential vorticity is $\beta (y-L/2)$, and the linear Rossby wave equation isFor nonzero zonal wavenumber $k$, wall conditions $\psi =0$ selectThe $k=0$ modes are stationary. Write $K^2=k^2+(n\pi /L{)}^2+L_D^{-2}$ and take real $A$ for the following sketch. The total potential vorticity is $q=\beta (y-L/2)-K^2\psi$. For a contour whose undisturbed position is $y_0$, its small meridional displacement isThus potential-vorticity conservation says a northward displacement in a positive background gradient carries a negative anomaly, $q^{\prime }=-\beta \delta y$. Potential-vorticity inversion turns this into positive $\psi$ and anticyclonic motion around the bulge. At a positive cosine crest, $v={\psi }_x$ is northward on its western flank and southward on its eastern flank. Advection raises the contour on the west and lowers it on the east, so the crest moves west. Algebraically, $\delta y_t=v$ and $\delta y=(K^2/\beta )\psi$ give $\omega /k=-\beta /K^2$, the same phase velocity as above. For $\beta <0$ all propagation directions reverse.

For a square basin use a nonzero-frequency normal mode $\psi =\mathrm{Re}\{\Psi (x,y)e^{-i\omega t}\}$, with $\Psi =0$ on all walls. Its equation isSet $\alpha =\beta /(2\omega )$ and $\hat{\Psi }=e^{i\alpha x}\Psi$. Substitution cancels the first derivative and givesThe square's Dirichlet Laplacian eigenfunctions are $\hat{\Psi }=A\sin (m\pi x/L)\sin (n\pi y/L)$, with $m,n$ positive integers. Consequently the square-basin topographic Rossby mode hasFor $\beta >0$, the positive-frequency branch therefore tends to $\beta L/[2\pi \sqrt{m^2+n^2}]$ as $L/L_D\to 0$, as required. The negative-frequency solution is its complex-conjugate representation for a real disturbance; the physical frequency magnitude is $|\beta |/[2\sqrt{{\pi }^2(m^2+n^2)/L^2+L_D^{-2}}]$.

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 $\mathbf{v}$ be horizontal iceberg velocity, ${\mathbf{u}}_a$ the wind and ${\mathbf{u}}_w$ the current. Air and water drag forces act on the relative velocities, typically with magnitudes proportional to their squares. Their exposed and immersed areas and drag coefficients differ. For iceberg mass $m$, the horizontal balance can be written schematically aswhere $f$ is the Coriolis parameter and $\zeta$ the sea-surface elevation. The last term represents the large-scale pressure gradient; contact with pack ice and grounding require additional forces. Iceberg free drift neglects such contact constraints, not the water drag or Coriolis force.

In still water with a fixed wind force $\mathbf{F}$, a transparent transient model linearizes water drag as $-m\gamma \mathbf{v}$. Put $J\mathbf{v}=\hat{\mathbf{z}}\times \mathbf{v}$. ThenThe damped free drift of an iceberg isThus an initially resting berg accelerates under the wind, turns under the Coriolis force, and approaches a balance of wind, water drag and rotation through damped inertial oscillations. Its equilibrium drift is to the right of the wind in the Northern Hemisphere and to the left in the Southern Hemisphere. With quadratic drag, the drag coefficient in the equilibrium balance depends on speed; the linear formula illustrates the turning and relaxation rather than prescribing an exact percentage of wind speed.

For a moving ocean, the background pressure gradient associated with geostrophic balance must also be included. Large tabular icebergs have substantial submerged area and respond strongly to currents; smaller bergs can have a larger wind-driven velocity relative to the water. Near Antarctica, the mainly westward Antarctic Coastal Current can carry bergs along the continental margin. Export into offshore gyres and then the eastward Antarctic Circumpolar Current changes their trajectories; bathymetry, winds, fronts and evolving size all matter. Gladstone and Bigg's Weddell Sea tracking illustrates the contrast between coastal and offshore drift. The Antarctic Convergence separates colder Antarctic surface water from warmer water farther north, where an exported iceberg can encounter enhanced melting. No single wind direction determines the complete route.

Let $f(y)$ denote the signed vertical component of the Coriolis parameter; it is negative in the southern hemisphere. In the steady, linear, inviscid interior, depth integration of horizontal momentum giveswhere $\mathbf{U}={\int }_{-H}^0(u,v)dz$ and $\Pi ={\int }_{-H}^{0}pdz/{\rho }_0$. Horizontal curl eliminates the pressure. Its left side is $f{\nabla }_h\cdot \mathbf{U}+\beta V$. Incompressibility and impermeable flat top and bottom imply ${\nabla }_h\cdot \mathbf{U}=w(-H)-w(0)=0$, givingThis is Sverdrup balance. It assumes steady small-amplitude motion, a Boussinesq approximation, negligible nonlinear advection, negligible Rayleigh and bottom/lateral drag in the interior, a flat fixed-depth domain and no normal top/bottom flux. The applied surface wind stress remains a forcing, even though interior friction is neglected. With cooling absent, the steady buoyancy equation additionally requires $B_z=0$, if that equation is imposed; the curl balance itself does not depend on the cooling coefficient. In the closed, $x$-independent channel below, depth-integrated $V=0$, so a nonzero wind-stress curl cannot be balanced by an inviscid interior alone.

Steady forced channel. Now take $r>0,\alpha >0$. Put $a=\pi /L$, $s=(z+h{)}_{+}=\max (z+h,0)$ and write ${\tau }_0$ for the positive wind-stress amplitude. The prescribed profiles giveThe buoyancy jumps across the sharp forcing interface in this simplified model; pressure and the circulation below are continuous. Let $P=p_y/{\rho }_0$, $D=f^2+r^2$, and $b_y^{\mathrm{top}}=-a{B}_0\sin ay/(\alpha h)$. Hydrostatic balance gives $P=C(y)+b_y^{\mathrm{top}}s$. The horizontal momentum equations and their solution areThe no-normal-flow conditions imply ${\int }_{-H}^{0}vdz=0$: this integral is independent of $y$ by incompressibility and is zero at the meridional walls. It determines the otherwise free depth-uniform pressure gradient,DefineA meridional overturning streamfunction satisfying $v=-{\Psi }_z$, $w={\Psi }_y$ isIt vanishes on all four boundaries. Together with the expressions for $b,P,u$, this solves all the steady equations. The formula permits variable $f(y)$; freezing it at a negative $f_0$ gives the simple schematic $w=aA\cos ayG(z)$. No unspecified latitude dependence is silently needed for the sketch.