- #1

Andy24

- 6

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I am predicting the altitude of a model rocket based on some testing I did where I found the impulse of a little single stage model rocket motor. I am wondering how to incorporate drag into my calculations to make it more accurate. I know D=Cd*rho*v^2*A and I can definitely find the area with my model rocket and a relatively accurate Cd from the internet, but I cannot find out how to actually incorporate this into my altitude prediction... Just wondering if anyone would be able to give me some help.

My calculations are pretty simple so far, if you're interested:

**Impulse:**1.903 Ns

**Time:**0.67 seconds

**Mass:**0.09228kg

__For the accelerating phase of the model rocket:__

ft=mv

ft=mv

-> 1.903=0.09228*v

-> 1.903/0.09228=v

-> v=20.62m/s

Knowing v=20.62m/s and v=u+at (initial v is 0 as this is the first phase of the rocket)

v=u+at

v=u+at

->20.62=0+0.67*a

->20.62=0.67*a

->20.62/0.67

->a=30.78m/s/s

We know that acc due to gravity is approx. -9.8m/s/s, therefore:

->30.78m/s/s+-9.8m/s/s=20.98m/s/s

Therefore as d=ut+0.5at^2 you just sub in the values for a and v and get the distance traveled in the first section:

d=ut+0.5at^2

d=ut+0.5at^2

->d=0*0.67+0.5*20.98*0.67^2

->d=0.5*20.98*0.67^2

->d=4.7m

__And in the second phase after the motor has stopped burning:__

The initial velocity is now 20.62m/s and the final velocity is now 0m/s, and the acceleration is -9.8m/s/s

d=(v^2-u^2)/2a

d=(v^2-u^2)/2a

->d=(0^2-20.62^2)/2*-9.8

->d=(-20.62^2)/2*-9.8

->d=(-20.62^2)/2*-9.8

>d=approx. 21.7m

Therefore adding the two distances, the total distance becomes 26.4m..

I want to make my predictions more accurate though as when the rocket was tested it only flew about 21m and I'm pretty certain the fact that I haven't taken into consideration the effects of drag have something to do with my inaccuracy...

Thanks heaps, Andrea.