# Parachute Drag

1. Nov 2, 2008

### Spimon

I'm trying to solve a problem which involves the following scenario. Any help would be great!

A landing aircraft (mass 14 tonnes) is to be slowed from 300km/hr to 180km/hr in 3 seconds, by the deployment of two identical parachutes at its rear. Determine the diameter of each parachute. The landing is done at sea level in standard conditions.

So far:

- Drag coefficient of the parachute is 1.2
- Density of air is 1.23kg/m^3
- 300km/hr = 83.3 m/s
- 180km/hr = 50 m/s

F = M.A
F = 14000 * (83.3-50)/3
F = 15400 N

What I'm unsure of is what to use as V in the drag coefficient equation. I assume this requires me to integrate, but I'm not exactly sure what to integrate.

I'd really appreciate any help offered!

2. Nov 2, 2008

### LowlyPion

Don't you also need the formula for Drag Force?

http://en.wikipedia.org/wiki/Air_resistance#Drag_at_high_velocity

3. Nov 2, 2008

### Spimon

I'm not sure if I need the formula or not. Is that formula applicable to acceleration, or only steady state velocity?

4. Nov 2, 2008

### LowlyPion

I realize your force will vary with speed. But wont the work done result in the desired change in Kinetic energy?

5. Nov 2, 2008

### Spimon

Yes, I have the force required to be 15.4kN. I don't know how to translate this into a projected area. I guess I need to find the area of a parachute that will exert 15.4kN on the aircraft.

Using the Drag Equation, as you kindly posted the link to, I don't have a 'V'. I've successfully used this equation for steady state velocity (for example, finding the area required to maintain a parachute at say, 7m/s.). I'm not sure when the velocity changing.

My initial through was to integrate V^2 from t=o to t=3 and use this as 'V', but this is a huge number (148 929)

6. Nov 2, 2008

### Spimon

Ohhhh, the work approach may be a go'er. I'll see what I can do with that! :D

7. Nov 2, 2008

### LowlyPion

What I'm getting at is that maybe you want to integrate the Force over the distance to arrive at the work done in terms of the area and the other constants - which by the way you already know the work done since you know the change in Kinetic Energy.

8. Nov 2, 2008

### Spimon

I'm still not getting a reasonable answer. Not sure where I'm going wrong. Thanks for the help anyway :)