Coefficient of Drag on a model rocket

  • #1
6
0

Main Question or Discussion Point

Hello everyone,
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

-> 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

->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 travelled in the first section:

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=(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.
 

Answers and Replies

  • #2
boneh3ad
Science Advisor
Insights Author
Gold Member
3,071
741
How comfortable are you with differential equations?
 
  • #3
6
0
How comfortable are you with differential equations?
I'm in grade 12 math B and C (the two advanced high school math's), so I've done a decent amount but nothing crazy.
 
  • #4
boneh3ad
Science Advisor
Insights Author
Gold Member
3,071
741
Well you are going to have to set it up as a force balance, with the sum of all the force equating to the thrust minus gravity and drag, and drag will depend on the velocity at a given time. The result is a differential equation that, when solved, will give you a time history of the velocity and position. Then you would set up a second equation to solve for after the engine cuts off. In this case, it is probably easiest to just assume constant mass and constant thrust, even though these aren't necessarily true.
 
  • #5
6
0
Well you are going to have to set it up as a force balance, with the sum of all the force equating to the thrust minus gravity and drag, and drag will depend on the velocity at a given time. The result is a differential equation that, when solved, will give you a time history of the velocity and position. Then you would set up a second equation to solve for after the engine cuts off. In this case, it is probably easiest to just assume constant mass and constant thrust, even though these aren't necessarily true.
Thanks heaps boneh3ad, I found this pdf which explores it perfectly I think: http://www.rocketmime.com/rockets/RocketEquations.pdf (it's the first worked equation on the pdf). Is this along the lines of what you were thinking?
 
  • #6
boneh3ad
Science Advisor
Insights Author
Gold Member
3,071
741
I didn't check the solution there but the original equations look correct.
 
  • #7
6
0
I didn't check the solution there but the original equations look correct.
k='drag constants'=(1/2)*rho*A*Cd
Using the formula: Apogee(in acc stage)=(mass/2k)ln((T-mg)/(T-mg-kv^2)) as seen on the site, you actually need the velocity (incl. the effect of drag) to determine the height. So there are actually 2 unknowns as I only have the average velocity without the effects of drag considered... Do you know of any way I can try and get around this?
 
  • #8
3,740
417
If you mean the burnout velocity (vτ), it is calculated on the previous page.

Edit. But I am not sure what they mean by this term. I thought that they mean the maximum speed. But it seems to be a variable.
Edit 2. They are using the same notation vτ for both the integration variable (velocity, v) and the upper limit of the integral. This is confusing.
But the vτ in the formula for altitude should be the maximum velocity, at the end of the burning stage. And can be calculated by the formula in a box.
 
Last edited:
  • #9
russ_watters
Mentor
19,447
5,633
Are you any good with Excel spreadsheet? They are good for this type of thing, to avoid the nasty math.
 

Related Threads on Coefficient of Drag on a model rocket

Replies
5
Views
1K
Replies
6
Views
4K
  • Last Post
Replies
0
Views
4K
Replies
6
Views
7K
  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
988
  • Last Post
2
Replies
33
Views
29K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
12K
Top