# Does a high staging number have diminishing returns? Is there a way to address that mathematically?

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.

• – Organic Marble Mar 19 at 14:25
• 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. – CPomerantz Mar 19 at 20:41

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

1. The performance of an ideal rocket with zero structural weight does not depend of the number of stages. A real rocket can do no better than that, but a good rocket will get quite close. Therefore, a high specific impulse of the fuel is important for a real rocket, regardless the number of stages.
2. Each stage's payload (which can be another stage) should outweigh the stage's structure, otherwise most of the energy (including the lower stages) is wasted accelerating the structure. Fuel can make up a larger fraction of the weight, as energy used to accelerate fuel is not wasted.
3. Each stage's combined fuel and structure should outweigh its payload. Without this condition, the second and third stage can weight almost as much as the first and each of those needs an engine capable of pushing the full weight. If each stage weights only half or even a fifth of its predecessor, the combined weight of all stages' engines is not much larger than the weight of the first stage's engines alone.

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.

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)
axis([v1(1),v1(end),0,1]);
grid minor
h = legend(arrayfun(@(x)[num2str(x*100),'%'], p, 'UniformOutput', 0));
set(h, 'FontSize', 16);
ylabel('Issp / Isp');
xlabel('∆v / Isp');

• This is a fascinating plot that is going to take me some time to fully comprehend. – Russell Borogove Mar 20 at 0:43
• 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. – TooTea Mar 20 at 9:06
• 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? – BlueCoder Mar 20 at 9:10
• Also: why from 6600 m/s do you infer that this would be a 2-stage vehicle? – BlueCoder Mar 20 at 9:11
• @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. – Steve Linton Mar 20 at 9:41