= Solution
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 $\mu$ gives
$$
\mu=\int_{h_0}^\infty Be^{-bH}\,dH
=\frac{B}{b}e^{-bh_0}.
$$
The mean peak <sea-ice draft> is
$$
h_m=\frac1\mu\int_{h_0}^\infty HBe^{-bH}\,dH
=h_0+\frac1b.
$$
Consequently
$$
\boxed{b=\frac1{h_m-h_0},\qquad
B=\frac{\mu}{h_m-h_0}\exp\left(\frac{h_0}{h_m-h_0}\right),}
$$
with $h_m>h_0$, $b$ 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>:
$$
g(h)=2\cot\delta\int_h^\infty n(H)\,dH
=\frac{2B}{b\tan\delta}e^{-bh},
\qquad h\ge h_0.
$$
Thus
$$
\boxed{A=\frac{2B}{b\tan\delta}.}
$$
This 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, $\tan\delta_{\rm track}=\tan\delta_\perp|\sin\psi|$. The track slope can therefore approach zero at a grazing crossing. https://doi.org/10.1029/95JC00007[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. https://doi.org/10.1016/j.polar.2012.03.002[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. \b[The constant-slope triangle is a useful statistical idealization, not a faithful shape for every old ridge.]
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