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The science
What this lab shows. How a leatherback turtle (Dermochelys coriacea), the deepest-diving reptile, spends a limited oxygen budget on a dive. Four ingredients are coupled: hydrostatic pressure, the lung gas that is squeezed by it (which sets buoyancy), the drag of swimming through water, and the oxygen stores that pay for the metabolism.
1. Pressure. Seawater is slightly compressible, so density grows with pressure, ρ = ρ0 exp(β(p - p0)), and dp/dz = ρ g. Solving gives a closed form:
p(z) = p0 - ln(1 - β ρ0 g z) / β, ρ0 = 1025 kg/m³, β = 4.4×10-10 Pa-1
At 1280 m (the deepest recorded leatherback dive) this is about 13.0 MPa, roughly 128 atmospheres. The page checks the formula by integrating the ODE with Runge-Kutta.
2. Lung gas and buoyancy. The air in the lungs obeys Boyle's law, p V = constant, down to the depth where the alveoli collapse (a few tens of metres; the exact value is uncertain, so it is a slider). Below it the lung holds only the airway volume. The body displaces tissue plus gas:
Vgas(z) = V0 p0 / p(z) (z < zc), FB = g [ ρw(z) (M/ρT + Vgas) - M ]
With the default tissue density of 1.030 g/cm³ and 80 % inflated lungs the turtle is about 230 N buoyant at the surface (it must swim down), becomes neutral near 120 m, and is about neutral after the lungs collapse. Where FB is negative the turtle can glide. The work spent against buoyancy on the way down appears in the readout ("buoyancy work").
3. Drag and thrust. Along the path at pitch θ: meff dv/dt = T - ½ρ CD A v² — FB sinθ (descending; the sign of the last term flips on the way up). The controller holds the cruise speed with the least thrust ≥ 0, so with enough negative weight the turtle simply glides and speeds up to its terminal speed vt = sqrt(2 |FB| sinθ / (ρ CD A)). A = 0.16 m² x (M/450)2/3, CD = 0.12 (a ridged carapace is streamlined). At depth the turtle swims at 0.5 m/s and the flippers must also supply lift |FB| at a lift-to-drag ratio of 5.
4. Oxygen budget. Stores: lung gas (21 % O2, three quarters usable) + blood + myoglobin, together about 20 to 25 ml/kg here. Metabolic power:
W = M w0 Q10(Tb - 28)/10 (1 - depression) + Pmech/η, O2 rate = W / 20.1 J per ml
Body temperature relaxes towards the water temperature with a time constant of about 40 min (a big animal has thermal inertia), which is why a deep dive into 4 to 10 C water slows the metabolism only gradually. The aerobic dive limit is the dive length at which the O2 stores reach zero: ADL = stores / mean O2 rate. The lung O2 can only be used above the collapse depth; at greater depth the remaining lung O2 is stranded until the turtle climbs again, so the real limit can be shorter than the simple ratio. At the surface uptake is a multiple of the resting rate, which sets the recovery time.
Checks in the page. (1) RK4 integral of the pressure ODE vs the closed form; (2) incompressible vs compressible pressure at 1280 m differ by the expected 0.3 %; (3) the neutral-buoyancy depth found by bisection equals a fixed-point Boyle solution; (4) integrating the equation of motion reproduces the terminal glide speed; (5) the oxygen drawdown equals the integral of the metabolic rate; (6) p V stays constant above the collapse depth.
Simplifications. One body compartment; fixed CD; depression, collapse depth, Q10 and efficiency are plausible values, not measurements of a particular animal. Leatherbacks also use gliding, heat exchange, bradycardia and tolerate some anaerobic metabolism; none of the numbers here are a forecast for a real turtle. The record depth (about 1280 m) and the longest dives (about an hour and a half) come from different animals and tags. Educational model only.