Calculating Speed Needed to Break a Parachute with Nylon

[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 traveling 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"

Cheers

 

[Nate810]

It looks like you did a good job on your presentation. Your calculations seem to be correct and I can't see any major errors. You may want to double-check the equation you used to make sure it is consistent with the assumptions you made, but other than that, it seems perfect.

 

[ogb p]

Hello,

I appreciate your interest in calculating the speed needed to break a parachute made of nylon. Your assumptions and calculations seem reasonable, but there are a few things to consider. Firstly, the ultimate strength of nylon is typically measured in Newtons per square meter (N/m2) or Pascals (Pa), not in kgm2. This may affect your calculations and the weight at which the parachute would break. Additionally, the thickness of the parachute may affect the strength and breaking point, so it is important to consider the actual thickness of the material used in the parachute. Furthermore, the drag coefficient (Cd) may vary depending on the shape and design of the parachute, so it is important to use an accurate value for this parameter in your calculations. Lastly, it may be helpful to include a safety factor in your calculations to account for any unforeseen variables or errors in your assumptions.

Overall, your approach seems sound and your results are within a reasonable range. However, to ensure accuracy and precision, it may be helpful to consult with a physics or engineering expert or to conduct further research on the specific properties of nylon and parachute design. Good luck with your presentation!