1
$\begingroup$

If we assume that a small scale rocket is launched perpendicularly to the ground and ignore the drag, how can I calculate the average acceleration given that I know the burn time, the average thrust, and the average mass?

update: There are some calculations in this model rocketry tutorial but I wonder how well this applies to real launch vehicles.

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ Rockets are usually between 85% to 95% propellant by mass, so "average mass" as a concept is somewhat indirect. Acceleration can change from 0.2g to 5 or 10g over several minutes, so average acceleration is not the best way to look at it. Nonetheless once a propellant mass fraction is given or assumed, the Tsiolkovsky rocket equation can be used to generate a velocity vs time curve and can be differentiated to produce acceleration (or it can be numerically integrated directly) and then averaged, but fun is in the shape of the curve! $\endgroup$
    – uhoh
    Commented Dec 30, 2020 at 22:59
  • $\begingroup$ "but I wonder how well this applies to real launch vehicles." - the equations listed, (limited by the assumptions), are the same basic equations that go into real launch planning. (though simplified). The biggest difference here is they will only be applied to a short phase of flight from lift off to a low altitude burn out. The primary difference with real flight planning is in the details... such as atmospheric conditions, use of actual vehicle data (such as an actual Cd) and how each, thrust, attitude, etc.. vary with time, altitude, etc.. They are actual basic equations. $\endgroup$
    – David C. Rankin
    Commented Jan 2, 2021 at 10:45

2 Answers 2

Reset to default
1
$\begingroup$

I would like to give a simple example. Lets assume a rocket with 1 kN thrust and 1000 kg weight. I calculate the acceleration for some weights.

  1000 kg    1.000 m/s^2
   875 kg    1.142 m/s^2
   750 kg    1.333 m/s^2
   625 kg    1.600 m/s^2
   500 kg    2.000 m/s^2
   375 kg    2.667 m/s^2
   250 kg    4.000 m/s^2
   125 kg    8.000 m/s^2

Do you really think an average acceleration is useful for such a non linear relation?

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Well, I have seen some resources on the internet using it for rough apogee estimates. This pdf nakka-rocketry.net/articles/altcalc.pdf kind of answered my question. $\endgroup$
    – curioso
    Commented Dec 30, 2020 at 22:39
  • $\begingroup$ @funeselmemorioso oh that's a great link, I'm going to add it to your question because it represents prior research (which is good to include when asking questions) and it makes answering easier. The rocket in the tutorial has a huge dry weight ("dead weight") and tiny propellant mass, and as mentioned here real launch vehicles are totally the opposite! $\endgroup$
    – uhoh
    Commented Dec 30, 2020 at 23:10
  • $\begingroup$ Oh, thank you for the edit! $\endgroup$
    – curioso
    Commented Dec 31, 2020 at 7:36
1
$\begingroup$

Too long for a comment:

I wonder how meaningful an average would be considering a rocket goes from 0 to 6 gs during launch.

It doesn't help that acceleration depends on the wildly variable mass of the rocket. Fuel accounts for the bulk of the weight, and most of it is gone two minutes into flight.

Nor does it help that the rocket gradually pitches down from vertical to horizontal. Close to vertical you lose close to 1g just counteracting gravity, and close to horizontal you lose close to nothing.

So even when you simplify by ignoring drag and other details of launch, the basic physics you're left with still make it very challenging to estimate things like acceleration---and therefore averages of those things, to the extent that they're meaningful :)

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ I meant this as a comment, sorry. I'll delete it in a bit so please don't add more comments below. And yes, calculating averages is straightforward---if you have data to calculate them from. But acceleration during launch is nonlinear and very dependent on variables that are themselves nontrivial to estimate. But even if you get your average, remember you're not dealing with some statistical distribution. How meaninful is it to say that your average is 3gs when the instantaneous values vary rapidly from 0 to 6gs? But you do you. $\endgroup$
    – user36480
    Commented Jan 2, 2021 at 2:45
  • $\begingroup$ I think your comment is quite reasonable, no reason for "sorry" or deletion! In Stack Exchange we can discuss stuff openly and others may disagree with me completely! From time to time answer posts are used for longish, helpful comments. Let's try this edit and see how it goes? $\endgroup$
    – uhoh
    Commented Jan 2, 2021 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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