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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.

Sea-ice free drift neglects internal stress transmitted from neighboring ice. An isolated floe is governed by air and water drag forces, the Coriolis force and sea-surface tilt; steady drift also neglects acceleration.

// Target: geophysics.bigb