Re: If a person weighs 220lb how big would the parachute have to be?

The question: The reason why i'm asking is because i was wondering how big I would have to make my parachute to get an egg to safety. If the egg weighed 1lb for example how big would the surface area have to be in comparison to weight. If possible note formula for future reference.

Hello Jasmine,

Thanks for an interesting question!

The following formula may be used to calculate the decent rate of a (round) parachute:

                   _____________
                  /
                 /   2 W
1)       V = \  /  --------
              \/    rho CD A


Where V = vertical decent velocity in ft/sec
      W = weight of the load + parachute in pounds
      rho = air density.  Near sea level use the value .00237 slug/ft^3
      CD = parachute drag coefficient.  Use a value of .75 for a round parachute
      A =  surface area profile of the parachute.

The formula is a bit over simplified, but should give you some reasonable results. The density units "slug/ft^3" may be a little confusing! A "slug" is the English unit of mass, comparable to 'kilogram' used in the metric system. You can get more familiar with this unit if you check out the following link:
http://en.wikipedia.org/wiki/Slug_(mass)

Let’s try out the formula for the 220 lb person:

For “A”, assume the parachute produces a profile area of a 24 foot diameter circle (R = 12’) as it falls through the air. ‘A’ would then be:
A = pi R^2 = 3.141 x 12^2 = 452 ft^2

Using formula 1): V = ( 2 x 220 / .00237 x .75 x 452) (SQRT) = 23.4 ft/sec
(I use “(SQRT)” to indicate ‘square root’ of the values in parentheses since it is so difficult to try to type out the square root symbol, and “X^2” to mean “X squared”)

This says that our skydiver will hit the ground with a speed of 23.4 ft/sec.

To give this number some meaning (which will also help with your diving egg), lets compare it to speeds generated by jumping off of different height objects:

There is a classic physics formula that gives a final velocity with only acceleration and distance given:

2)  V = (2 a d) (sqrt)
    a = acceleration of gravity = 32.2 ft/sec^2 
    and “d” = height in ft.

Lets compare the parachute speed with the speed hitting the ground after jumping off of a 9 foot high wall:
V = (2 x 32.2 x 9) (sqrt) = 24.1 ft/ sec

This result says that jumping off of a 9 foot wall will give about the same impact speed hitting the ground as in our 24-foot-diameter parachute!

NOW, lets get back to your egg:

The first thing you should find is how great an impact speed your egg can endure without breaking: Experiment to find the greatest height that you can drop the egg and keep it intact. (This may take a few suicidal test pilot eggs!) Of course, the surface that it lands on will be very important - is it grass, or concrete? And, I hope that you are allowed to give your egg a little “space suit” made of bubble wrap, or foam rubber, etc. Either way, if you don’t change the landing spot or the space suit, the Drop Test will let you calculate the maximum speed that the parachute can allow to keep the egg intact.

Lets say, that your maximum drop height is 2.0 feet. From formula 2) you calculate the impact speed:
V = (2x 32.2 x 2ft) (sqrt) = 11.3 ft/ sec.
If you solve formula 1) for “A”, it becomes:
A = 2 W / (rho x CD x V^2 )
Using your value of ‘1 pound’ for the egg:
A = 2 x 1 / (.00237 x .75 x 11.3^2) = 8.81 ft^2
Since a = pi x r^2 for a circle, the radius = (a/ pi)(sqrt)
Radius = (8.81/ 3.141)(sqrt) = 1.67 ft.
That is, a 1.67 foot radius parachute should slow your egg to the same speed as the 2 foot free drop.

I hope that you can use this information to make your project successful! Please write back to Mad Science and let us know how it all works out!

Best Regards,

Jay Shapiro