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Here $n(H)dH$ is the expected number of ridge peaks per track length with sea-ice draft in $[H,H+dH]$. If total line density is $\mu$ and mean peak draft is $h_m>h_0$, integration gives $b=1/(h_m-h_0)$ and $B=\mu b{e}^{b{h}_0}$. The normalized peak draft has a shifted exponential distribution. This is a model over a specified draft range, not a universal description of every ridge.

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.

Sonar uses sound propagation and reflected signals to detect objects or measure distances. Upward-looking sonar can record sea-ice draft along a moving platform's track; beam width, return selection and processing affect the inferred profile.