6
$\begingroup$

For simple rockets, using the rocket equation is simple enough.

$$ \Delta v = v_e ln\left(\frac{m_i}{m_f}\right) = I_{sp} \ g \ ln\left(\frac{m_i}{m_f}\right) $$

plug in full and empty weights and Isp, and you're done.

But what to do when a rocket uses boosters? You get 2 Isp values. I expect you can't just add the delta-V of the main stages to the delta-V of the individual boosters.

Is there a way to use the Rocket equation on rockets that use boosters?

$\endgroup$
  • 1
    $\begingroup$ I'd imagine you'd do a weighted average based on the fuel mass. Just a guess on my part, though. $\endgroup$ – UIDAlexD Feb 9 '18 at 15:52
  • $\begingroup$ Not at all an answer, but there's some deriving going on here. $\endgroup$ – uhoh Feb 9 '18 at 16:04
11
$\begingroup$

For each phase of flight (stage or throttle variation or what have you) you weight the specific impulse (= exhaust velocity) by the propellant mass flow rates (i.e. consumption rates) of the different engines operating.

So if, for example the boosters are consuming 5000kg/s at 240s specific impulse and the core is consuming 1000kg/s at 310s, it works out to

$$ \frac {5000 \frac {kg} {s} * 240 {s} + 1000 \frac {kg} {s} * 310{s} } {5000 \frac {kg} {s} + 1000 \frac {kg} {s} } = 251.66 {s} $$

For 251.66s specific impulse.

$\endgroup$
  • 3
    $\begingroup$ What does that equation mean? What are the units? $\endgroup$ – kingledion Feb 9 '18 at 20:32
  • 3
    $\begingroup$ I reckon the units are (kg/s * s) / (kg/s) = s, yielding 251.66s combined Isp. Just a guess. $\endgroup$ – Brian Feb 9 '18 at 21:28
  • $\begingroup$ @Brian Good guess. $\endgroup$ – Russell Borogove Feb 10 '18 at 1:05
5
$\begingroup$

You have to consider each "phase" of flight.

For each phase of flight you need to work out the initial mass, the final mass and the overall specific impulse (total thrust/total propellant flow rate). You can then apply the rocket equation to each phase of the flight and add up the results.

$\endgroup$
  • $\begingroup$ This is also true, but the problem is that the boosters can have different Isp and dm/dt also. $\endgroup$ – uhoh Feb 9 '18 at 16:34
  • 1
    $\begingroup$ If you install the Flight Engineer mod for Kerbal Space Program, it will tell you the ISP, the dV, the TWR, and burn time for every stage. That's how I get it to work ;) $\endgroup$ – Mazura Feb 10 '18 at 0:03
5
$\begingroup$

The derivation is based on the rate of change of total momentum

$$\frac{dp_{tot}}{dt}$$.

For each engine,

$$\frac{dp}{dt}=v_{ex}\frac{dm}{dt},$$

the exhaust velocity times the rate that the propellants' mass is ejected (kg/sec). You calculate that for each engine, then add them up.

So where you would normally have

$$m \frac{dV}{dt} = -v_{ex}\frac{dm}{dt},$$

you would have instead

$$m \frac{dV}{dt} = -\Sigma \left(v_{ex, i}\frac{dm_i}{dt}\right),$$

where $v_{ex,i}$ and $\frac{dm_i}{dt}$ are the exhaust velocity and mass rate of the ith engine. You can then introduce the weighted exhaust velocity:

$$ v_{ex,m} = \frac{\sum v_{ex,i}\frac{dm_i}{dt}}{\sum \frac{dm_i}{dt}} = \frac{\sum v_{ex,i}\frac{dm_i}{dt}}{\frac{dm}{dt}}$$

Hence you have

$$m\frac{dV}{dt} = -v_{ex,m}\frac{dm}{dt}$$

which is the same equation as in the single engine case. Integrating this equation yields the following rocket equation:

$$\Delta V = v_{ex,m} ln\big(\frac{m_i}{m_f}\big)$$

$\endgroup$
  • $\begingroup$ Oh, that's much nicer, as well as pretty obvious in retrospect. Thank you Community♦ $\endgroup$ – uhoh Feb 10 '18 at 0:19
  • 1
    $\begingroup$ "exhaust" - such a depressing term. $\endgroup$ – Organic Marble Feb 10 '18 at 2:14
  • $\begingroup$ @OrganicMarble lol, indeed. $\endgroup$ – uhoh Feb 10 '18 at 3:34

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.