I was reading this question:

Help me understand what Farside, a ten "stage" rockoon looked like? How was it configured?

Comments link to Highest stage counts in actual launchers? but here I'm interested in understanding if there are diminishing returns in going to a high stage number, and how this could be addressed mathematically.

  • What are the limiting factors of staging aside from the added cost/weight of additional engines, coupling components?

  • Is there a way to formulate an equation to express the relationship of staging benefits versus number of stages? Obviously it will vary between designs, but I'm wondering if there's something to generalize like the Tsiolkovsky rocket equation, but for staging.

  • $\begingroup$ Related: space.stackexchange.com/questions/8719/… $\endgroup$ Commented Mar 19, 2019 at 14:25
  • $\begingroup$ I would imagine one of the more relevant costs would be up-front development. Even if it's not the biggest long-term cost, in some cases it might stop space agencies and companies from considering rockets with more stages. $\endgroup$
    – CPomerantz
    Commented Mar 19, 2019 at 20:41
  • $\begingroup$ google lagrange mulipliers in connection with rocket staging. As I understand it there isn't a closed form solution (not my topic) though it may shed some light on it. $\endgroup$
    – Puffin
    Commented Nov 15, 2019 at 1:06

1 Answer 1


Any multi-stage rocket design has to obey three rules to achieve good performance:

  1. Fuel type and engine design must allow for a high specific impulse. This is equally valid for single and multi staged rockets.
  2. Each stage's payload (which can be another stage) should outweigh the stage's structure, otherwise most of the energy is wasted accelerating the structure.
  3. Total weight should decrease rapidly with each stage separation, so that the lower stage's heavy engines and interstage don't need to be duplicated.

A rocket with too few stages pushes a lot of empty tankage around and eventually violates rule #2. A rocket with too many stages and too little ∆v per stage has lots of duplicated engines and interstages and violates rule #3. Somewhere in between there's an optimum.

To design such a rocket, start with a given specific impulse and structural mass fraction. Distribute the remaining mass between fuel and payload and calculate ∆v per stage. Finally, calculate the number of stages to achieve the desired total ∆v. The tricky part is to chose the optimal fuel and payload fractions and to measure efficiency of the design.

To help that, I created the below plot. It shows mass efficiency (system specific impulse) over ∆v per stage for different structural mass fractions. Everything is normalized with respect to specific impulse of the fuel.

enter image description here

Example usage:

Your fuel's specific impulse is 3000m/s, you want a system specific impulse no less than 2500m/s to keep launch mass low and you can achieve a structural mass ratio of 4%. You look up 2500m/s (83.3% of 3000, y-axis) on the 4%-line, the result of which is a ∆v of 6600m/s (220% of 3000, x-axis) per stage.

6600m/s per stage allows a two-stage-to-orbit vehicle. More stages yield diminishing returns, as maximum efficiency of the 4%-line never goes above 90%. If you use too many stages with little ∆v per stage, efficiency will even drop.

I created the plot using Matlab/Octave as follows:

p = [0.02, 0.04, 0.07, 0.10, 0.15, 0.20, 0.30, 0.40];    # structural mass ratio
v1 = (0.0:0.01:4)';                                      # v_end / I_sp
ma = 1.0 ./ exp(v1);                                     # non-fuel mass ratio
na = ma - p;                                             # payload mass ratio
na(na < 0) = 0;
v2 = v1 ./ log(1.0 ./ na);                               # I_ssp / I_sp

h = plot(v1, v2, 'LineWidth', 3);
set(gca, 'fontsize', 16)
grid minor
h = legend(arrayfun(@(x)[num2str(x*100),'%'], p, 'UniformOutput', 0));
set(h, 'FontSize', 16);
ylabel('Issp / Isp');
xlabel('∆v / Isp');
  • 3
    $\begingroup$ This is a fascinating plot that is going to take me some time to fully comprehend. $\endgroup$ Commented Mar 20, 2019 at 0:43
  • 1
    $\begingroup$ I'm not sure I understand Rule #3. Why is it fuel > structure + payload and not fuel + structure > payload? After a stage burns out, its structure is discarded and thus the next stage doesn't need to carry it. $\endgroup$
    – TooTea
    Commented Mar 20, 2019 at 9:06
  • $\begingroup$ I have some questions: When finding the intersection between the 4% line and 0.83%, why do you look at 220%, instead of (roughly) 20%? 220% is clearly better, but why the 4% line and the imaginary horizontal line at 0.83% intersect twice? What does the first intersection represent? $\endgroup$
    – BlueCoder
    Commented Mar 20, 2019 at 9:10
  • $\begingroup$ Also: why from 6600 m/s do you infer that this would be a 2-stage vehicle? $\endgroup$
    – BlueCoder
    Commented Mar 20, 2019 at 9:11
  • 1
    $\begingroup$ @BlueCoder essentially the y-axis of the curve measures the loss in efficiency (distance from the top) due to the need to carry tanks, engines, etc. compared to the perfect efficiency given by the fuel. The x-axis corresponds to how much fuel (and so tankage and everything else) you carry. If you are too far to the left, the mass of engines is too great and you don't push enough fuel through them to use the effectively. If you are too far to the right, you spend too long pushing empty fuel tanks. $\endgroup$ Commented Mar 20, 2019 at 9:41

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