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The science
A falling body is acted on by weight W = m g and aerodynamic drag F = 1/2 rho Cd A v^2. Newton's second law gives m dv/dt = m g - 1/2 rho Cd A v^2. Speed saturates when the two forces balance, at the terminal velocity vt = sqrt(2 m g / (rho Cd A)). In constant air density the exact solution is v(t) = vt tanh(g t / vt).
Air density falls with height; this page uses the isothermal barometric law rho(h) = 1.225 exp(-h / 8500 m), so a diver at 4000 m has less drag and a higher terminal speed than near the ground. Opening the canopy raises the product Cd A by a factor of roughly 100 over a deployment time of a second or two (modelled here as a smooth ramp), so the diver decelerates at several g (the opening shock) and then settles to the canopy's terminal velocity.
Simplifications: one dimension (vertical only), constant drag coefficients, no canopy inflation overshoot, no wind and no added-mass effect. The 3D page animates the same sequence; this page integrates the equations. Educational model only, not for planning real jumps.
What to try
- Press Tandem skydive and watch the opening shock peak in the g-load graph.
- Choose a small canopy (4 m) and a high landing speed appears; a 12 m canopy lands softly.
- Set the deploy time to 0.5 s: the opening shock grows because the same speed is removed in less time.