# Assignment

## Basic data

I was launching a rocket model and I tried to calculate the reached altitude.

• The engine (C6-0) impulse is 10 Ns
• Total weight is 65,7 g (includes the engine)

I calculated speed = 152 m/s

$$\vec F \cdot t = \Delta m \vec v$$

Then I calculated the altitude 1007 m which seems too much to me. I guess something about 200 m (you may see the video)

$$y_{\max} = \frac{v_0^2 \ }{2 g}$$

## Drag

I guess, I have to consider drag

• Diameter of rocket 2,5 cm
• Drag coefficient 0,05 (I guess)

$$F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A$$

But what about now, what is really achieved altitude?

# Engine description

Rocket Altitude Calculation

• You can find the treatment at hyperphysics.phy-astr.gsu.edu/hbase/mechanics/quadrag.html - since this is a "homework-and-exercises" type problem, please try to work through those equations and see how close you get. Likely the drag coefficient is a lot more than 0.05 (for a sphere it's almost 0.5 - getting below 0.1 is really hard). The site has an online calculator for a spherical object as well. If you adjust the density to get the right mass, you should get close. Better would be numerical integration... – Floris Apr 1 '15 at 12:42

I made a simple Excel spreadsheet to calculate this. Some simplifying assumptions:

Mass = 66 gram (during thrust), 33 gram (after burn) Cd = 0.5 (like for sphere) rho = 1.22 (air) Simple numerical (Newton) integration of equation of motion (0.1 second time step)

Resulting curve: Height of about 300 m, total flight time just under 14 seconds. Based on the video (which didn't show the descent) I think that time to peak was about 6 seconds - close to that predicted by this.

Excel file is at http://www.floris.us/physicsSE/rocket.xlsx

It was just a "rough" calculation... I know much better calculators exist "out there".

• CD of 0.5 is really high. Sleek missiles have around 0.15. – Kenira Apr 1 '15 at 13:20
• @Kenira - based on the video, the time to peak was about 6 seconds. The thrust is given, as is the mass. The only variable I have to play with is the Cd. I tried to make it match the only observable data point (time to peak). I agree it's high. But it seems to be what the data says... – Floris Apr 1 '15 at 13:21
• With that assumption i would agree, but at least i can pretty much only see the smoke from the engine and not the rocket itself. Even if you could see more it'd still be hard to tell when it has reached it's highest point. I think it is more reasonable to assume a CD between 0.15 and 0.2. – Kenira Apr 1 '15 at 13:25
• @Kenira - I agree with you that there is insufficient (and potentially conflicting) data to reach a conclusion, and further that the only reasonable approach is numerical integration using the best data available. Total time of flight can be a good way to calibrate for a single unknown parameter - but that data point is not given. – Floris Apr 1 '15 at 13:28
• It seems more realistic. Could you, please, share the excel? Weight after burn is cca 53 g – banterCZ Apr 1 '15 at 13:46

You have to integrate all the forces over time. As soon as the rocket is flying, drag acts as a downward force that reduces altitude and since drag depends on velocity there is no simple equation you can just plug it all into.

I recently started programming a rocket simulation, if i plug in the numbers into that:

• delta mass = 10.22g
• Isp = 99.74s
• assuming constant 5N thrust for 2s (should not cause much of an error since it's such a short burst)
• CD = 0.15 (0.05 would be really low, at least for military missiles and real rockets it's more like 0.15 to 0.25 so i went with 0.15 for now)

i get a maximum altitude of 730m, speed of 137 m/s at burnout, flight time 24s. Simulation interval is 0.001s.

CD of 0.1 gives 830m, 139 m/s and 25s.

CD of 0.2 gives 650m, 134 m/s and 23s.

Without knowing the CD better, this guess is as good as it gets i am afraid.

• "This guess is as good as it gets". Correct. Need more data to calibrate. When I put your numbers into my model I get a very similar result. – Floris Apr 1 '15 at 13:34
• According to the engine chart, the flight time is 1.8s only. – banterCZ Apr 1 '15 at 13:36
• @banterCZ The engine burns for 1.8s, but what counts is the total impulse of 10Ns - 5N for 2s are 10Ns. And since it only takes such a short amount of time, this simplification will not cause much of an error (which would be caused by either gravity or aerodynamic drag, both are tiny on these timescales) – Kenira Apr 1 '15 at 13:38