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An ice-edge band is an elongated concentration of ice floes separated from neighbouring ice by open water. In one mechanism, an off-ice wind opens polynyas; short wind waves then exert wave radiation stress on their downwind floes. Wave forcing compacts floes into bands, opposed by incoming longer-period swell. Collisions and mergers can produce fewer, larger bands.

If a buoyant plume of diameter $d$ supplies oil at volume rate $Q$ to an ice floe moving at speed $v$, the plume sweeps area $dv$ per unit time. Retaining a layer of depth $h$ over that area would require volume rate $dvh$, so conservation of mass gives the painted fraction $\min \{1,Q/(dvh)\}$. Discontinuous painting starts when $v>Q/(dh)$. This estimate assumes all arriving oil is retained and a constant track width equal to the plume diameter; the plume's circular cross-sectional area does not enter the swept-area rate.

The marginal ice zone is the transition between open ocean and compact sea ice, where interacting ice floes and surface gravity waves can dominate the mechanical evolution.

The normalization $\mu =m/A_w$ has units $\mathrm{kg}{\mathrm{m}}^{-2}$. It converts the Coriolis force into a force per water-drag area and allows comparison with wind and water stresses. Icebergs generally have much larger $\mu$ than thin ice floes, but their effective air-to-water drag-area ratio also matters. A geometric depth becomes a mass per area only after multiplication by density.

Pancake ice consists of approximately disk-shaped young ice floes, often with raised edges formed by collisions. Wave-driven collisions and motion in the surrounding frazil ice can dissipate energy.

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.

An oil spill under sea ice first reaches the underside through buoyancy. Oil droplets and gas bubbles rise relative to the surrounding water; their collective buoyancy drives an entraining buoyant plume, while ocean currents bend the plume and carry small droplets away. Droplet size matters: small droplets rise slowly and can be dispersed or partly dissolved before arriving, whereas coalescence produces faster-rising drops. Gas expansion, dissolution and the ambient density stratification change the plume on its ascent. Consequently the arrival footprint can be displaced from the well, and the oil arrival rate at the ice need not equal the discharge rate at the seabed.

The combinations of cover and motion give four distinct collection geometries. Each must then be combined with the four age-and-relief cases below; this accounts for all sixteen combinations without treating age as a substitute for geometry.

For each of these four geometries, the age-and-relief combinations have the following effects.

Thus deformation increases local storage in either age class; motion changes residence time and export in either geometry; and gaps permit direct surfacing in either stationary or moving ice. These distinctions remain important even when two covers have the same mean thickness.

During growth, new sea ice can form around the edges of a lens or at available water-contacting surfaces and enclose oil between successive layers. Oil interrupts direct ice–water contact and changes thermal resistance, so incorporation is not simply freezing a pure oil layer. Oil droplets can be caught in a growing skeletal layer, and connected brine channels allow upward migration when buoyancy overcomes capillary entry resistance. In spring, warming increases the liquid fraction and opens pathways; meltwater and brine circulation can redistribute oil internally. Oil can then reach the surface and melt ponds, darken the ice, or be released to open water when the enclosing ice disappears. Thick ridge pockets may survive longer than flat lenses. The coupling between brine drainage, warming and oil migration is documented by field observations of oil entrainment in first-year ice.

A shorter freezing season gives less time for encapsulation and long-term storage. Less multi-year sea ice, greater open-water exposure and earlier breakup shift more of the spill into surface slicks and mobile fragments. Earlier warming can release stored oil sooner, while reduced cover permits more wave mixing and evaporation. This does not guarantee rapid removal: a plume may still encounter fast ice or deformed reservoirs, and oil-induced darkening promotes ice-albedo feedback. These changes alter exposure, transport and recovery opportunities rather than making every warming effect favourable.

For the oil lens capillary balance, distinguish the three interfacial tensions ${\gamma }_{io}$, ${\gamma }_{ow}$ and ${\gamma }_{iw}$, where the subscripts denote ice, oil and water. Replacing an ice–water area by an ice–oil and oil–water pair costs the effective interfacial energyFor a broad, shallow, non-wetting lens of volume $\mathcal{V}=Ah$, neglect the small edge region. The interfacial energy is $\Gamma A$. Because oil is lighter than water, moving it downward through a depth $h$ costs buoyancy potential energy $\frac{1}{2}({\rho }_w-{\rho }_o)gA{h}^2$. At fixed $\mathcal{V}$ their sum isSetting $E^{\prime }(h)=0$ givesThe second derivative is positive when $\Gamma >0$. Equivalently, if $\vartheta$ is the contact angle measured through the oil, the contact-line balance gives ${\gamma }_{iw}-{\gamma }_{io}={\gamma }_{ow}\cos \vartheta$, so $\Gamma ={\gamma }_{ow}(1-\cos \vartheta )$. These are the interfacial quantities required, rather than ${\gamma }_{io}$ alone. If $\Gamma \le 0$, complete wetting replaces this finite-lens balance; if the ice is rough, pocket geometry replaces the flat-lens approximation. oil–ice–water interfacial experiments give the corresponding contact-angle description.

The printed numerical premise is defective. A crude-oil density near $0.85\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ means $850\mathrm{kg}{\mathrm{m}}^{-3}$, rather than the printed $0.85\mathrm{kg}{\mathrm{m}}^{-3}$. Moreover, the specified oil–ice tension does not specify $\Gamma$. Even if $1.5\mathrm{N}{\mathrm{m}}^{-1}$ were assumed to be $\Gamma$, taking ${\rho }_w=1025\mathrm{kg}{\mathrm{m}}^{-3}$ would give $h=1.73\mathrm{cm}$ with the literal density and $h=4.18\mathrm{cm}$ with the plausible density repair. Neither calculation proves the claimed bound. With ${\rho }_o=850\mathrm{kg}{\mathrm{m}}^{-3}$, the exact condition isFor an explicitly defined, illustrative effective tension $\Gamma =0.050\mathrm{N}{\mathrm{m}}^{-1}$, the same formula gives $h=7.63\mathrm{mm}$. This is a consistent sub-centimetre example, not a uniquely recoverable correction of the printed interfacial data.

For ice-floe oil painting fraction, let $Q$ be the oil volume arriving per day and assume complete retention in a layer of specified thickness $h$ on the passing underside. In time $\Delta t$, a plume of width $d$ sweeps a strip of area $dV\Delta t$, apart from the initial end correction. Coating that strip needs $hdV\Delta t$ of oil. The circular plume area cannot replace the swept width: $AVh$ has dimensions of volume times length per time. ThereforeDeposition becomes discontinuous when $V>V_c$. For $d=100\mathrm{m}$ and $V=10^4\mathrm{m}\mathrm{da}{\mathrm{y}}^{-1}$, the four results areThese percentages refer to the swept, $d$-wide corridor. The percentage of a whole floe of unspecified width is not determined. They are volume-equivalent coverage fractions at the stipulated local thickness; real deposition may instead give a thinner continuous smear, and losses reduce the retained fraction. Rough pockets store more oil per painted area, explaining their smaller coverage at fixed $Q$. The oil is exported as a patchy record of the floe's passage, so later melting can release many dispersed patches rather than a single stationary slick. Low areal coverage is compatible with substantial oil mass and locally thick contamination.

Preparedness should follow those transport stages. Before release, install well-control and subsea capping capability, winter-operable access, warmed pumping equipment, oil storage, under-ice survey capability and drifting position beacons; map currents, sea ice motion and underside traps. Once a blowout starts, stopping or collecting oil near the source limits all later exposure. Survey the actual plume landfall with subsea vehicles and suitable acoustic or optical sensors, track both water and ice trajectories, and sample to distinguish an oil layer from ordinary ice roughness.

Under stable fast ice, targeted access holes and suction collectors can recover accessible reservoirs before encapsulation, provided the underside survey identifies their position and connectivity. Under moving ice floes, collectors must follow or intercept the contaminated floes; fixed surface booms cannot recover oil sealed beneath an intact slab. In broken ice and leads, use ice-capable skimmers and containment where relative ice and water motion allow it. Controlled burning is an option only for an accessible surface layer of sufficient thickness; dispersants concern exposed or subsea droplets with adequate mixing, not a sealed lens. During spring release, return to tracked pockets and floes, collect oil reaching leads and melt ponds, and prepare coastal interception along the forecast export route.

The research questions are specific: detection accuracy against natural underside roughness; droplet-size and plume-landfall prediction; capillary entry into saline pores; trapping capacities and remobilization thresholds of ridges; oil-dependent growth, viscosity and pumping rates at low temperature; timing of spring channel opening; and recovery, burning or dispersion efficiency and environmental effects under realistic ice and mixing conditions. A plan must quantify recoverable mass and operating windows rather than assume any one technique works in all ice states. Arctic response-gap research explicitly treats weather, ice, access and technique limitations.

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.

It is important to distinguish loss of forward-going surface-gravity-wave energy from conversion of mechanical energy into heat. Scattering attenuation by ice floes redirects energy; it can attenuate a coherent transmitted wave without dissipating the total energy.

For fixed floe geometry, increasing frequency usually increases attenuation over the relevant swell range: shorter wavelengths respond more strongly to the contrast between water and the elastic plate, and to repeated floe edges. Long surface gravity waves have weak curvature and often penetrate much farther. This is a trend over a specified frequency range, not a theorem excluding resonances.

The diameter dependence is governed by $d/\lambda$. An ice floe much smaller than the wavelength moves nearly with the water and scatters weakly. Scattering becomes appreciable when floe size is comparable with the wavelength, and interference between its two edges can give maxima and minima. At fixed ice concentration, larger ice floes also mean fewer edges per unit propagation distance, roughly proportional to $p/d$. Consequently the attenuation coefficient need not increase monotonically with diameter: the single-floe reflection and the number of encounters must both be considered. Thickness increases areal inertia as ${\rho }_{i}h$ and bending stiffness as $h^3$, generally increasing wave mismatch and reflection, although detailed frequency-dependent resonances again prevent a universal monotonic law.

When $d\ll \lambda$, particularly for frazil ice and pancake ice, weak individual scattering leaves other processes dominant. Relative crystal and water motion causes viscous dissipation; an aggregate layer can behave as a viscous or viscoelastic material, and pancake ice collisions, rubbing and overwash remove energy. Their importance depends on concentration and wave amplitude.

For a uniform continuous sheet with horizontal dimensions much greater than the wavelength, there are no repeated floe edges in its interior. A perfectly elastic sheet over inviscid water supports undamped flexural-gravity waves, so internal scattering is not an explanation of decay there. Real attenuation can instead arise from internal ice anelasticity or viscoelasticity, a dissipative sub-ice viscous boundary layer, turbulence, cracks and brine-related processes. Small-floe mixtures and continuous sheets require dissipation models beyond the isolated-floe scattering picture.

Take $x$ positive in the off-ice wind direction, from the compact pack towards the open sea. The wind initially separates the outer ice floes, creating irregular polynyas. A larger opening gives more open-water fetch, so stronger short wind waves develop before reaching its downwind edge. Their reflection supplies a positive force on the downwind ice floes; these catch their neighbours and compact into an ice-edge band. Incoming longer swell exerts force in the opposite direction. The short waves can exert substantial force despite their smaller amplitude because a small ice floe reflects them much more effectively than it reflects the long swell.

The surface-gravity-wave energy of the short waves grows with fetch in the windward polynya, then decays rapidly across the band. Swell enters from the seaward side and usually attenuates more slowly. This gives inward forcing from the two sides: short-wave forcing is largest at the windward face and swell forcing is largest at the seaward face. In the next open polynya, short wind waves regrow from the weak transmitted component. Reflected waves also enhance energy locally on the incident side, with interference that is omitted from a smooth, phase-averaged sketch.

Each plotted component is normalized by its own incident energy; the curves do not assert equal absolute wind-wave and swell energies. The band is partially transmitting. For a perfectly opaque reflector, the transmitted component would instead vanish.

The initial bands have unequal floe inventories, widths and forcing. Differential drift and collisions merge some into composite bands. Larger bands tend to shield smaller downstream accumulations from the short-wave forcing needed to keep them separate, while sufficiently wide intervening polynyas can generate fresh wind waves and maintain separation. Finite available ice, available wave fetch, the opposing swell, wind strength and duration, floe size and thickness, and subsequent mergers limit the number of persistent bands. The wave-force formula alone does not select a universal count. This mechanism and its merger interpretation are supported by the original ice-band study.

There is a normalization issue in the requested stress calculation. Let $\mathcal{E}={\rho }_{w}g{a}^2/2$ be standard linear surface-gravity-wave energy for crest amplitude $a$. The deep-water wave radiation stress is $S_{xx}=\mathcal{E}/2$. Hence standard momentum balance givesFor lossless reflection with amplitude reflection coefficient $R$, $r^2=R^2{a}^2$ and $t^2=(1-R^2)a^2$, so $F_{\mathrm{phys}}={\rho }_{w}g{R}^2{a}^2/2$.

In the usual independent-floe, weak-reflection approximation, there are about $p/d$ effective layers per unit distance. Each removes a fraction $R^2$ of the forward energy, givingThis yields the intended wave-driven ice-band compaction:The attenuation approximation retains the leading term in $R^2$; an independent discrete-layer model instead gives an energy coefficient $-(p/d)\log (1-R^2)$. Neither coefficient follows from the single-object force formula without this additional scattering closure.

For ordinary crest amplitudes, the force printed in the paper is $4F_{\mathrm{phys}}$. With that printed normalization and the same attenuation law, its derivative is four times the requested stress. More generally, if $a^2=a_0^2{e}^{-\kappa x}$, the printed lossless force gives $-F^{\prime }=2{\rho }_{w}g{R}^2\kappa a^2$. Recovering the stated stress from it requires $\kappa =p{R}^2/(4d)$ instead, a different unspecified attenuation convention. The supplied force and stress cannot both be derived from standard amplitudes and the usual floe-layer attenuation law. Restoring the factor $1/4$ in the force gives a consistent intended model.

Using the printed force for the numerical question, perfect reflection gives $r=a$, $t=0$, and thereforeUsing the requested stress formula for partial reflection givesFor comparison, the consistently normalized perfect-reflection force is $50.3\mathrm{N}{\mathrm{m}}^{-1}$, and the printed force with standard attenuation would give $0.628\mathrm{N}{\mathrm{m}}^{-2}$ for the partial-reflection derivative. The approximate stress is a force per horizontal area, not a direct measure of the three-dimensional ice-skeleton stress.

The inward force gradient helps maintain a coherent, close-packed band, especially near the incident-wave faces. It is modest enough that wind, current, swell changes or mergers can disrupt the arrangement; it does not guarantee permanent mechanical stability. Perfect reflection estimates a bounding force, whereas the much smaller partial-reflection stress varies as $R^4$. No mechanical-strength law is supplied, so stability can be assessed only qualitatively.

A sea-ice pressure ridge forms where horizontal convergence compresses sea ice. The initially thinner sheet or colliding ice floes fracture and ride over or under one another. Continued convergence piles broken blocks into an emergent sail and a submerged keel. The submerged volume is normally larger because buoyancy supports the pile. Pores and brine-filled gaps initially make the rubble unlike a solid intact sheet; refreezing can consolidate it.

A sea-ice shear ridge develops along a fracture where neighbouring ice moves tangentially in opposite directions or at different speeds. Rough edges interlock, crush and locally converge, producing chains of piled blocks along the shear boundary. Thus the large-scale strain is mainly shear, but the actual production of ridge rubble involves local compression. Pure sliding of perfectly smooth parallel surfaces need not create a ridge. Pressure ridging is driven by convergence; shear ridging is driven by relative tangential motion with local crushing and convergence.

Interfaces between ice floes and open water reflect and redirect surface gravity waves. The effective attenuation depends on floe thickness, diameter, spatial arrangement and wave period. For very small diameter relative to wavelength, a floe responds nearly with the water and scatters weakly. Individual-floe interference and the number of interfaces per unit distance prevent a universal monotonic diameter law.