|Sep9-06, 09:52 PM||#1|
Parachute deployment (non-instantaneous)
Here's the problem...
A car of mass 2000 kg, drag coefficient .3, frontal area 1 sq. meter releases parachute of area Pi sq. meters (diameter 2 m). The inital velocity of the car is 100 m/s. Calculate distance and velocity after 10 seconds.
I know how to work this problem for instantaneous deployment. However, if the deployment is non-instantaneous, I need to make a few assumptions. The first is that the area of the parachute increases linearly with time: A(t) = A0 + kt, where A0 is the initial area of the undeployed parachute and k is the time rate of area increase.
***What should I assume for A0???
***Is it reasonable to assume that the parachute can be deployed in 3-5 seconds? Is this too great a time interval? Too small? From this I can determine the approximate value of k (area increase per second).
***I was told I can assume the coefficient of drag remains constant.
Does this sound like a reasonable model for non-instantaneous deployment? If so, I know how to find the desired values for distance and velocity. Am I on the right track???
Thanks for any input!
|Sep10-06, 06:13 AM||#2|
3-5 secs seems ok.
I happen to know that at terminal velocity on a parachute jump i have pulled at 5500ft, on a student canopy (large area), and been under a fully inflated canopy in secs.
They train you to count the inflation before making malfunction checks, and its about 5 sec countdown.
With modern chutes the slider has a big effect on the deployment time. It catches the air and slides up along the risers slowing deployment and reducing Jerk.
|Sep10-06, 06:23 AM||#3|
For area at t0 i would take the vehicles area only ( i.e. the drag that has allways been on the vehicle ).
The drag will be effected as the vehicle slows, but i think thats taken care of in the drag coefficient. Im not familiar with the equation for the coefficient of drag.
Seems like a good start to me, same as i would do, though i would prob try and define a curve for the deployment :P
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