Hi,

I'm trying to find the area of a parachute cloth required for a landing. I should specify that it is not homework.

Consider a person of mass m. Let the time required to come down to the surface safely be t. Let the height of the parachute be h. Then,

h=1/2at^2,

a=acceleration during the landing.

As there will be much drag, using Newton's Second Law -

Net force=Weight -Drag

ma=mg-1/2dv^2CA

Here, d=density of air, v=velocity of the parachute, C=drag coefficient and A=area of the parachute.

So, it comes that -

a=g-(dv^2CA)/2m

Substituting the value of a in the first equation -

h=1/2[g-(dv^2CA)/2m]t^2

 

Now to calculate the area required.

The question is: Is my procedure correct?

Well, not quite as I can understand. Using drag formula was also the first thing that popped into my head as I was starting to read your post, but there are different types of parachutes and they will have different maths associated with them. There is a standard round chute, which seems like this way of solving the problem should work for it, but let's not forget that they have a hole in the middle, otherwise oscillations from escaping air would make it very unstable, so in order to be more precise you should consider the effects of this hole since it will effectively reduce the area of chute and hence you'd need the radius of the chute to be slightly larger than without the hole.

 

Ram air type parachutes on the other hand generate both drag and lift (round chutes have no lift) so the calculation for this type should include both effects. It gets pretty complex for this type.

 

Here are some pages you might want to take a look at:

 

http://www.rocketshoppe.com/info/The_Mathematics_of_Parachutes.pdf

https://info.aiaa.org/tac/AASG/ADSTC/RamAir%20Parachute%20Design/Lingard.pdf

http://wwwf.imperial.ac.uk/ssherw/spectralhp/papers/PhDThesis/N_FogellPhDThesis.pdf

Edited October 3, 20169 yr by pavelcherepan

OK. Thnx.