Coefficient of Drag on a model rocket

In summary, Andrea was trying to predict the altitude of a model rocket based on some testing she did. She found the impulse of a little single stage model rocket motor and determined that v=20.62m/s. She also calculated that d=ut+0.5at^2. However, she is not comfortable with differential equations and wants to try and figure out a way to incorporate drag into her calculations. She found a pdf that explored this concept, and Boneh3ad told her that she would need to set it up as a force balance with the sum of all the force equating to the thrust minus gravity and drag.
  • #1
Andy24
6
0
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 traveled 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.
 
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  • #2
How comfortable are you with differential equations?
 
  • #3
boneh3ad said:
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
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
boneh3ad said:
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
I didn't check the solution there but the original equations look correct.
 
  • #7
boneh3ad said:
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
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
Are you any good with Excel spreadsheet? They are good for this type of thing, to avoid the nasty math.
 

Related to Coefficient of Drag on a model rocket

1. What is the coefficient of drag on a model rocket?

The coefficient of drag on a model rocket is a measure of the resistance the rocket experiences as it moves through the air. It is a dimensionless number that is dependent on the shape and size of the rocket, as well as the properties of the surrounding air.

2. How is the coefficient of drag calculated?

The coefficient of drag is typically determined through wind tunnel testing, where the rocket is subjected to different air speeds and the resulting forces are measured. It can also be calculated using mathematical equations that take into account the rocket's shape and air properties.

3. Why is the coefficient of drag important for model rockets?

The coefficient of drag is an important factor in determining the stability and flight characteristics of a model rocket. A higher coefficient of drag means the rocket will experience more resistance and may not fly as far or as straight as a rocket with a lower coefficient of drag.

4. How does the coefficient of drag affect the flight of a model rocket?

The coefficient of drag affects the flight of a model rocket in several ways. It can determine the rocket's maximum altitude, as well as its stability and trajectory. A higher coefficient of drag can also result in a slower descent due to increased air resistance.

5. Can the coefficient of drag be reduced on a model rocket?

Yes, the coefficient of drag can be reduced on a model rocket through various methods such as streamlining the shape of the rocket, using smoother materials, and minimizing protrusions and edges. However, it is also important to maintain a certain level of drag for stability and control during flight.

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