Time is height over descent rate — plus a from-rest correction
If the parachute is already at terminal speed vt, the time to descend a height h is simply t = h/vt, and the horizontal drift in a steady wind is wind × t. A 1,000-foot drop at 1,000 ft/min in a 10-knot wind is therefore 60 seconds and one-sixth of a nautical mile. Terminal speed itself, when you do not type it, is vt = √(2m g / (ρ Cd A)), the same identity used on the terminal velocity page.
Starting from rest under quadratic drag, speed is vt tanh(g t/vt) and the distance fallen is (vt²/g) ln cosh(g t/vt), so the time is t = (vt/g) acosh(exp(g h/vt²)). From a typical opening height that time is already within a second of h/vt, because the canopy reaches terminal speed in a few vt/g time constants. A real ram-air canopy also glides, and a real opening is not a step to Cd A; use the canopy manual for a landing pattern. This page is the vertical-descent and drift estimate those manuals start from.
Sources
- Quadratic-drag terminal velocity; see also this site’s terminal-velocity calculator.
- Integration of m dv/dt = m g − ½ ρ Cd A v² (standard ballistic descent).
- FAA Parachute Rigger Handbook and Airplane Flying Handbook (descent rate and drift — conceptual).
Frequently asked questions
How is this different from the terminal-velocity calculator?
That page finds v_t (and an Allen–Eggers re-entry peak). This one starts from a descent rate or a Cd A, then adds time-to-ground and wind drift, including the from-rest integration.
Does a ram-air canopy really fall straight down?
No. It glides. Treat the “descent rate” here as the vertical component and the “wind” as the net horizontal motion if you already know it; the page will not compute a glide polar.
Why is the from-rest time only a little longer?
The canopy reaches nearly v_t after a few seconds. From a thousand-foot opening height, almost all of the time is already spent at terminal speed.