Drag makes swimming cost grow as speed cubed, while the body burns energy even at rest, so there is a most economical speed. Oxygen stores then set how far a narwhal can swim under ice between two breathing gaps.
Chance that a swimmer crosses the whole ice zone, against ice concentration, from the closed form and from a Monte Carlo of random floe lengths (1,000 crossings per point).
The science
Swimming cost. Drag power on a body of wetted area S is Pdrag = ½ ρ Cd S v³. Muscles supply it at overall efficiency η, and the animal also pays a maintenance rate Pb, taken as a multiple of Kleiber's basal rate 70 M0.75 kcal/day. Total power is P(v) = Pb + k v³ with k = ½ ρ Cd S/η.
Cost of transport COT = P/v (joules per metre) is smallest at v* = (Pb/2k)1/3, found by setting d(COT)/dv = 0. Slower swimming wastes maintenance energy, faster swimming wastes drag.
Dive limit. Oxygen stores VO2 (mL per kg of body mass) hold energy E = VO2·M·20.1 kJ/L. The aerobic dive limit is ADL = E/P and the longest ice run a single breath allows is Dmax = v·ADL = E/COT, so the speed that minimises COT also maximises Dmax.
Ice. Along the track, floe lengths are drawn from an exponential law with mean D̄; concentration c = D̄/(D̄+g) fixes the mean gap g. In a zone of width W there are n ≈ W/(D̄+g) floes, each longer than Dmax with probability e−Dmax/D̄, so an animal crosses with probability (1 − e−Dmax/D̄)n.
Simplifications, stated honestly. One-dimensional track, independent exponential floe lengths, constant speed, no recovery time at the surface, no sideways routes through leads, no wave drag, identical animals apart from a small oxygen spread in the animation. Mass 1500 kg, maintenance of 2x Kleiber, Cd = 0.004, η = 0.25 and 40 mL/kg oxygen stores are typical or illustrative values, not measurements for narwhals; real dives are often much longer and animals use leads sideways. Fat energy is taken as 37 MJ/kg at 90 % assimilation.