# Is this correct?

#### oscarwyatt

Hi,
I have to do a presentation at school and I've chosen to talk about nylon for parachutes. We have to do a calculation and although I've done it, I am a bit worried if its correct or if there's something I've missed, its rather important so I was wondering if anybody could see any problems with it. Briefly explained, the idea is to find out how fast you would have to be falling to make a parachute break, first by calculating how much weight would make a parachute break and then converting this to force and using an equation.

Its based on these assumptions:
- The parachute is a perfect dome
- The straps on the parachute will not break
- The person can be falling at any speed, we will assume his terminal resistance is infinite
- The area of the parachute is 60m2
- We will ignore the weight of the person and that of the parachute itself
- The parachute is formed from a single unbroken mould of a parachute

The ultimate strength of nylon is 75 Mpa – 7,500,000 kgm2
I think a parachute would be about 2 mm thick
Therefore: 15,000,000 = 15,000 kg
1000
The parachute canvas will break when a weight of 15 tonnes is applied

We need to find the velocity at which the drag force will exceed the strength of the parachute.
This can be done by rearranging the following equation:
FD = ½ densityofair Cd A v2
Or
Drag force = ½ x density of air x drag coefficient x area of the parachute x velocity2
To become:
drag force = velocity2
½ x density of air x drag coefficient x area of the parachute

150,000 = velocity2
½ x 1.22 x 1.5 x 60

Therefore:
2,732.22 = velocity2
Therefore, the parachute would only break if you were travelling at 52.27m/s or faster

If anyone wants to see it as it is in the presentation (probably better idea) its http://www.rhphotography.co.uk/nylon.ppt" [Broken]

Cheers

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