Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 72 / 3

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.

Let $H$ denote a ridge's peak sea-ice draft, reserving $h$ for the draft at a randomly sampled position. In the exponential ridge-draft model, normalization by the line density $\mu$ givesThe mean peak sea-ice draft isConsequentlywith $h_m>h_0$, $b$ having dimensions inverse length and $B$ inverse length squared. The normalized peak probability density function is a shifted exponential distribution.

For the triangular argument, interpret the common ridge shape as geometrically similar triangles with common along-track slope $\tan \delta$ and variable peak height. Literal congruence would require identical sizes and could not coexist with an exponential peak-draft distribution. Each side of a triangle has $dx=|dh|/\tan \delta$. A ridge reaching draft $H\ge h$ therefore contributes $2\cot \delta dh$ of horizontal track in the interval $[h,h+dh]$. Summing this occupation length over all qualifying peaks proves the triangular ridge occupation identity:ThusThis is a tail relation for sampled draft occupation, not an instruction to normalize $n$ and $g$ identically. Below $h_0$, the ideal triangles contribute $2\mu \cot \delta$ rather than the same exponential; level ice and gaps contribute their own draft distributions. If triangular keels are referenced to a level-ice base, the vertical coordinate must be shifted consistently. We also require nonoverlapping occupation: arbitrary choices of $\mu$, mean draft and slope can otherwise demand more than the available track length.

Observed mean keel slopes are typically of order $20^{\circ }$–$30^{\circ }$, with broad individual variation rather than a single universal angle. Orientation matters: if a track crosses a straight ridge at angle $\psi$ to the crest, $\tan {\delta }_{\mathrm{track}}=\tan {\delta }_{\perp }|\sin \psi |$. The track slope can therefore approach zero at a grazing crossing. A sonar morphology study found location-dependent mean slopes about $22^{\circ }$–$27^{\circ }$ after correcting for ridge orientation.

Young sea-ice pressure ridges often have recognizably triangular sections with angular, porous rubble and comparatively continuous crests. Melting, refreezing and repeated cracking modify older ridges: their blocks can become rounded and consolidated, and their keel or crest can fragment into separated hummocks rather than retain one triangular shape. A pre-exam multibeam study found first-year sea ice ridge slopes averaging roughly $27^{\circ }$, while multi-year sea ice ridges often consisted of irregular separated smooth blocks. Multi-year sections can be broader or locally shallower, but age alone does not determine one slope angle. The constant-slope triangle is a useful statistical idealization, not a faithful shape for every old ridge.

Use the exponential ridge-draft model along the direction of motion relative to the seabed. During time $dt$, the stationary point samples track length $|V|dt$. Peaks deeper than the seabed have line densityThe candidate seabed gouging by ice encounter rate is thereforeFor the usual case $D\ge h_0$, replace $\max (D,h_0)$ by $D$. For $D<h_0$, every counted ridge is deep enough and ${\nu }_D=|V|\mu$.

The units are inverse time. This is the number of potentially scouring keel passages per unit time, not sediment volume removed per unit time. The latter also requires keel width, sediment properties and an erosion law. The expression assumes the ridge statistics remain applicable during drift and grounding does not arrest or reshape the deep keels before they reach the point. If speed fluctuates, use mean relative speed under independence from ridge occurrence, not the magnitude of a vector-mean drift that might cancel reversing motion.

The independent, spatially homogeneous random-placement idealization gives a Poisson process of intersections along the sampling line. With line intensity $\mu$, an interval of length $s$ contains no ridge with probability $e^{-\mu s}$. The spacing $X$ thus has exponential distributionRandom orientation changes the intersection intensity, which has already been represented by the measured $\mu$. Merely saying that positions are random does not prove a Poisson process: independence and homogeneity are additional assumptions. Finite segments, clustering or excluded widths can invalidate them.

For a shifted lognormal distribution, write $Z=\log (X-\theta )\sim N(m,{\sigma }^2)$ with $\sigma >0$. Applying the change of variables formula, $dZ/dX=1/(X-\theta )$, givesThe threshold $\theta$ is a minimum observable separation. Finite keel widths impose geometric exclusion, and the sonar's footprint and ridge-identification criterion can suppress a shallow peak near a deeper peak. This sonar ridge shadowing or resolution effect means that $\theta$ need not be a universal physical distance between every pair of ridges; it can partly reflect how the profiles were processed.

A lognormal distribution is compatible with products of many positive factors: taking logarithms turns multiplicative changes into sums, which can approach a normal distribution. Repeated deformation, breakup, convergence and merging across scales are plausible contributors. The observed form therefore suggests correlated or multistage ridging rather than the simplest independent-intersection model. It does not identify a unique ridging mechanism. For example, directly generating $X=\theta +\exp Z$ with a normal distribution for $Z$ produces the same spacing law without specifying any particular mechanics. Distributional agreement must be supplemented by dynamical and spatial evidence.