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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.

A moving ice keel can gouge sediment when its draft exceeds water depth. In an exponential ridge-draft model, a point crossed by ice at relative speed $|V|$ experiences candidate encounters at rate $|V|\mu \exp [-(\max (D,h_0)-h_0)/(h_m-h_0)]$. Actual erosion also depends on grounding, deformation of the keel and sediment strength.

For nonoverlapping triangular sea-ice pressure ridges with a common along-track side slope $\tan \delta$, a ridge whose peak exceeds $h$ contributes $2\cot \delta dh$ of track length in a draft interval $dh$. Summing over peaks proves the identity. Above the cutoff of an exponential ridge-draft model, $g(h)=A{e}^{-bh}$ with $A=2B/(b\tan \delta )$. This draft occupation is not the normalized peak probability density function; level ice and shallower parts supply the remaining probability.