1
$\begingroup$

I have calculated the apogee of a 2.5kN avg. thrust sounding rocket once ignoring drag and then assuming maximum constant (magnitude and direction) drag after burnout. Here are the figures I used in my calculations:

Rocket:

  • Carbon body and payload: 6 000 g
  • Propellant: 8 496 g
  • Motor (N2501) at burnout: 4 604 g
  • Average thrust: 2501.8 N
  • Burn time 6.09 s
  • Total drag coefficient: 0.54

Other constants:

  • Nominal launcher elevation angle: 75°
  • Frontal area 𝐴 ≈ 0.0103 m2
  • Constant air density of 𝜌 ≈ 0.631 kg m−3

To my surprise, the results were dramatically different (~47km vs ~4km). Where should I expect the actual apogee to be within this range and why?

$\endgroup$
3
  • $\begingroup$ My first guess was to say it would be in the middle (the average) but the maximal acceleration due to drag I obtained was about 40 times g so I think it will be within the first quarter of the range. $\endgroup$
    – curioso
    Apr 12 at 21:31
  • 1
    $\begingroup$ I'll edit to make it more clear. $\endgroup$
    – curioso
    Apr 12 at 22:20
  • 1
    $\begingroup$ Excellent, thanks for the edit! $\endgroup$
    – uhoh
    Apr 12 at 22:41
3
$\begingroup$

Towards the lower end

  • Your rocket is slamming into a wall of air.
  • Altitude has a quadratic response to reduction in velocity.

Part 1: Getting up to speed with the calculations you have probably already performed.

For instance, it would be nice to know what speed the rocket is moving at when the engine burns out. While gravity and drag also acts upon the rocket during these initial six seconds, the engine should be the most important part here.

The rocket equation comes in handy as usual, provided that we have an initial mass of 19.1kg and a dry mass of 10.6kg. We don't have the exhaust velocity, but that should be the same as thrust divided by mass flow, about 1800m/s.

All in all, I get a propulsive delta-v of 1060 m/s, which if we subtract the ~60m/s of gravitational slowdown over six seconds should be a roughly 1km/s top speed before we start considering drag. Oh, and the vertical component is reduced by some 3.4% by the launch not being straight up.

A parabolic arc should be sufficient to model this. We aren't getting high enough for the weakening of Earth's gravity to start playing any significant role. I too get an altitude of 47.5km from that freefall arc, plus an additional 3km the rocket moved before the engine burned out, for a total of ~50km.


Part 2: a wall of air.

Conveniently, you have provided all the required parameters for the drag equation. This is by far the most imprecise part of it all, as supersonic fluid dynamics is not very "pen and paper" friendly. But sure, a simple drag model can help quantify what forces that are in play.

Contrary to your constant air density approach, I'm going to take a sneak peek at the instantaneous conditions at burnout first:

Firstly, at roughly 3km, the air density is more like 0.91kg/m³, increasing drag by 50% at the point in the flight were it is undoubtedly the most significant.

As we have all the numbers, the drag equation says the initial drag is about 2400N.

Which is nearly the same as the thrust the engine provides!

That means that we have to revisit the initial velocity from part 1, since drag during the initial six seconds is clearly eating a large part of it.

This is the point where differential equations are appropriate, but since this is "estimation", we can instead fiddle with the speeds a bit, and notice that around 700m/s the drag is eating up about half of engine thrust, and given the losses up to that point, that's about as far as its going to make it towards 1km/s.

Ballistically, that's already cutting apogee in half, still assuming the rest of the trajectory is drag free.

"drag free" is far from the case though, currently travelling at over mach 2 somewhere below 3km, drag outmatching gravity by a factor of about thirteen.

Even when the speed is eventually halved, drag is still over 3x stronger than gravity. Meaning that the partition of the flight that should cover 3/4 of the apogee altitude gain lasts for at least 4x less time that it would without drag.

By these observations alone, the upper bound of the apogee is already getting pretty close to your second estimate.

$\endgroup$
1
  • $\begingroup$ Thanks for the thorough answer! That was the kind of analysis I was looking to get help with. $\endgroup$
    – curioso
    Apr 13 at 8:27
3
$\begingroup$

I did a quick'n'dirty numerical simulation with a 0.1-sec ∆t, assuming vertical launch instead of 75°. For altitude the 75° elevation angle will make a little difference (the computed altitude will be a bit high) but it won't make a lot of difference, since sin(75°) is getting pretty close to 1. (But see the caveats below!) The 0.1-sec ∆t will make some difference too, slightly underestimating the drag force during each time interval — I didn't do a second-order drag prediction.

Maximum acceleration came at 3.2 sec into the burn. Beyond that the increase in the drag force was overcoming the decrease in vehicle mass. At burnout the altitude was ~2450 m and the velocity 808 m/s. After the (assumed instantaneous) burnout the acceleration switched from 118 m/s^2 upward to 121 m/s^2 downward, rapidly eating into that 808 m/s. Apogee came at ~38 sec, and just a hair under 10,100 m altitude, so yes, much closer to your smaller estimate than your larger estimate.

CAVEATS

I did not attempt to estimate the variation of $C_D$ in the transonic and supersonic regimes. This would almost certainly reduce the apogee estimate and the peak velocity.

Launching at a non-vertical angle doesn't mean the rocket's path will maintain that angle throughout the burn. The vector sum of acceleration due to the engine and drag plus that due to gravity, is not along the rocket's symmetry axis, and that causes the axis's elevation angle to decrease with time. The longer the burn and the lower the acceleration, the more this occurs.

The engine's actual thrust-time curve will make some difference in the results.

$\endgroup$
3
  • 1
    $\begingroup$ I just did another quick'n'dirty but using ∆t = 0.01 s instead of 0.1 s, and the burnout velocity slowed by 0.33 m/s and the apogee altitude came down by ~132 m. Further decreases in ∆t wouldn't change the result much. $\endgroup$ Apr 16 at 4:17
  • $\begingroup$ Could you please guide me on how you did your "quick'n'dirty' calculation? I am having a hard time incorporating drag into the flight's equation of motion. $\endgroup$ May 14 at 21:46
  • $\begingroup$ @OrangeDurito It might be a few weeks before I can respond. I'm wrapped up in building a large piece of hardware right now. $\endgroup$ May 16 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.