Any multi-stage rocket design has to obey three rules to achieve good performance:
- 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.
- 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.
- 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.
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)
h = legend(arrayfun(@(x)[num2str(x*100),'%'], p, 'UniformOutput', 0));
set(h, 'FontSize', 16);
ylabel('Issp / Isp');
xlabel('∆v / Isp');