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When you release a payload with the help of a circular parachute, and you want to find out the range in which it would land. And how would you find the descent gradient of the parachute?

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The first thing to figure out is how fast your parachute will fall. From https://www.pcprg.com/rounddes.htm you can compute the falling velocity:

$$ V = \sqrt{\frac{2 W}{\rho C S}} $$

Where

  • $C$ is the parachute drag coefficient which is approx 0.75 for a parachute without holes or slits cut in the fabric; same value in both Metric and English unit systems

  • $S$ is the total surface area of the fabric used to build the parachute, plus the areas of the holes and vents cut in the fabric if present. The units of S are in square feet (English) or square meters (Metric). This definition is such that when vents are cut in the fabric, the value of $S$ remains the same but the value of $C$ becomes smaller.

  • $\rho$ is the air density

    • near sea level its value is given by 0.00237 sl/ft^3 (English units) and 1.225 Kg/m^3 (Metric)
    • near 4000 ft or 1219 m above sea level its value is 0.00211 sl/ft^3 (English units) and approx. 1.07 Kg/m^3 (Metric)
  • $W$ is the weight of the parachute + load, in pounds (English) or Newtons (Metric)

  • $V$ is the vertical descent velocity, here expressed in ft/sec (English) or m/sec (Metric)

You can either assume a constant speed all the way down, or make it a function of $\rho$.

With starting altitude and speed you can find the time it will take to fall.

Then you add the wind contribution. Multiply the wind speed by the descent time and you get the maximum range the parachute can reach.

The calculations will get a little mode tricky if you have a rho profile by altitude and wind speed profile by altitude.

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  • $\begingroup$ Slightly complicated by the inertia of the load - it may be released with a significant horizontal speed and will not immediately adjust to wind shear. At low rates of descent it’s possible for (even round) parachutes to get into a ‘flying’ state where there is some horizontal airspeed and the ‘chute acts an an aerofoil, slowing the rate of descent at the expense of horizontal airspeed. $\endgroup$
    – Frog
    Apr 22, 2021 at 23:43

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