Two-Stage Weight Allocation — Theory & documentation

Overview

A frequent question in multistage design is how to split the mass between the stages. With cost taken as proportional to mass, two statements of the problem are equivalent:

for a fixed liftoff mass and payload, find the split that gives the greatest ideal velocity;
for a fixed ideal velocity and payload, find the split that gives the least liftoff mass.

The calculator solves the second. The ideal velocity is the velocity the stages would add with no drag or gravity losses. Tsiolkovskii’s rocket equation gives it for each stage, and the stage velocities add:

\Delta V_{ID}=\sum_i c_i\ln\mu_i,\qquad c_i=g_0I_{SPi}

where \mu_i is the mass ratio of stage i (mass at its ignition over mass at its burnout) and c_i its exhaust velocity. To size a real rocket, add an allowance for drag and gravity losses to the burnout velocity you need.

Optimum payload ratios

Stage 1 is at the bottom. For stage i define

the payload ratio \eta_i=M_{0,i+1}/M_{0i}: the mass above the stage over the mass at its ignition (M_{03}=M_{PL});
the structural fraction (structural efficiency) \sigma_i=M_{STi}/(M_{STi}+M_{PRi}): inert mass over loaded stage mass.

Then the mass ratio and the ideal velocity of the stage are

\mu_i=\frac{1}{\eta_i+\sigma_i(1-\eta_i)},\qquad \Delta V_i=-c_i\ln[\eta_i+\sigma_i(1-\eta_i)]

and the total payload ratio is the product \eta_{TOT}=M_{PL}/M_{01}=\eta_1\eta_2. The ideal velocity is maximised subject to \ln\eta_1+\ln\eta_2=\ln\eta_{TOT}, with a Lagrange multiplier \lambda. Setting the derivative of the augmented function with respect to each \eta_i to zero gives

\frac{c_i(1-\sigma_i)}{\eta_i+\sigma_i(1-\eta_i)}=\frac{\lambda}{\eta_i}\quad\Longrightarrow\quad \eta_i=\frac{\lambda\sigma_i}{(c_i-\lambda)(1-\sigma_i)}

Each stage’s optimum payload ratio depends on \lambda alone. The memo then finds \lambda from the total payload ratio. Here the ideal velocity is given instead, so \lambda is the value for which

-c_1\ln[\eta_1+\sigma_1(1-\eta_1)]-c_2\ln[\eta_2+\sigma_2(1-\eta_2)]=\Delta V_{ID}

Both payload ratios rise with \lambda, so the left side falls steadily and there is a single root. It is found by bisection, to machine precision. Each stage may have its own specific impulse.

As \lambda\to0 the liftoff mass grows without bound and the ideal velocity tends to the limit of the two stages:

\Delta V_{lim}=-c_1\ln\sigma_1-c_2\ln\sigma_2

A required velocity at or above this limit is refused.

The multiplier has a useful meaning. At the optimum, \lambda=d\Delta V_{ID}/d\ln M_{01}: adding 1 % to the liftoff mass, with the best split, adds about 0.01\lambda to the ideal velocity.

Stage masses

The payload ratios give the masses directly:

M_{02}=\frac{M_{PL}}{\eta_2},\qquad M_{01}=\frac{M_{02}}{\eta_1}=\frac{M_{PL}}{\eta_{TOT}}

M_1=M_{01}-M_{02},\qquad M_2=M_{02}-M_{PL},\qquad M_{STi}=\sigma_iM_i,\qquad M_{PRi}=(1-\sigma_i)M_i

Every mass is proportional to the payload mass, so the payload only scales the answer.

When a second stage pays

At a fixed total payload ratio the ideal velocity is a concave function of \ln\eta_1, so the stationary point is the maximum. A payload ratio cannot exceed 1, however (a stage cannot have negative mass). At low ideal velocities the stationary point can need \eta_i>1 for the stage with the smaller value of c_i(1-\sigma_i). The best design then leaves that stage out, and the other stage alone gives the velocity:

\eta=\frac{e^{-\Delta V_{ID}/c}-\sigma}{1-\sigma}

The calculator shows this case with a warning, and gives the ideal velocity above which both stages carry mass. That threshold is the ideal velocity at \lambda=\min_i c_i(1-\sigma_i).

The liftoff-mass plot shows the effect. Below the threshold the optimum curve follows the better single stage; above it, staging saves more and more mass as the velocity rises. With equal specific impulses the threshold is low. A weaker upper-stage motor can push it well up: for the memo’s example stages (220 s and 200 s, structural fractions 0.15 and 0.14) a second stage only pays above about 2650 ft/s.

Common specific impulse

If both stages share an exhaust velocity c, the optimum has a closed form. With U=\Delta V_{ID}/c, for two stages

\eta_{TOT}=\frac{\left[e^{-U/2}-\sqrt{\sigma_1\sigma_2}\right]^2}{(1-\sigma_1)(1-\sigma_2)},\qquad \eta_i=\frac{\sigma_i}{1-\sigma_i}\left[\eta_{TOT}\frac{(1-\sigma_1)(1-\sigma_2)}{\sigma_1\sigma_2}\right]^{1/2}

and, inversely, \Delta V_{ID}=-2c\ln\left[\sqrt{\sigma_1\sigma_2}+\sqrt{\eta_{TOT}(1-\sigma_1)(1-\sigma_2)}\right]. The multiplier is \lambda=cR/(1+R), where R=\sqrt{\eta_{TOT}(1-\sigma_1)(1-\sigma_2)/(\sigma_1\sigma_2)}.

The memo applies this closed form with the average (I_{SP1}+I_{SP2})/2 when the impulses differ. The calculator shows that result in the full results table, with the ideal velocity its split really gives when each stage has its own impulse. When the impulses are equal, the general solution and the closed form agree to 10^{-11}.

Verdict and fixes

The optimum split gives the lightest rocket; whether it is light enough depends on what you can build and launch. The calculator asks for your heaviest liftoff mass M_{max} (default 50 lb, a placeholder) and compares: safe up to it, marginal up to 1.25×, unsafe beyond.

Every mass scales with the payload, so the heaviest payload that fits is exact: M_{PL,max}=M_{PL}M_{max}/M_{01}. The fastest ideal velocity within the limit, and the structural fractions (both scaled by the same factor) that fit it, are found by bisection on the full optimum.

An ideal velocity above \Delta V_{lim}=-c_1\ln\sigma_1-c_2\ln\sigma_2 cannot be reached by any two-stage rocket with these stages. Scaling both structural fractions by k moves the limit by -(c_1+c_2)\ln k, so the fractions below which the velocity becomes reachable at all follow exactly; the verdict is unsafe and gives that and the largest reachable velocity.

Verification

tests/twostage.test.js checks the following:

the closed form against the memo’s tabulated example (2000 ft/s, 10 lb payload): total and stage payload ratios, propellant, structure and loaded masses, and liftoff mass, to 10^{-12};
that the general solution reproduces the closed form when the impulses are equal;
that with unequal impulses no other split of the same liftoff mass gives more ideal velocity (brute-force search);
that \lambda=d\Delta V_{ID}/d\ln M_{01};
the single-stage case below the threshold, and continuity at the threshold.

Notes on the method

Each stage’s own specific impulse. The optimum uses the general Lagrange condition with each stage’s exhaust velocity, rather than the average impulse. With the memo’s example stages at 2000 ft/s the true optimum is a single stage of 14.08 lb. The average-impulse split (14.20 lb) gives only about 1977 ft/s with the real impulses.
Two-stage forms. The two-stage expressions above follow from the memo’s N-stage equations with N=2. In the printed two-stage versions the square on \eta_{TOT} and the factor g_0I_{SP} on \Delta V_{ID} are missing.
Multiplier. \lambda comes from the stationary condition. The tabulated example’s value (5446 ft/s) uses \eta_{TOT} in place of its square root; the value consistent with its payload ratios is 5621 ft/s. The multiplier does not affect the masses.
Model. Ideal velocity only: no drag, no gravity loss, no coast. Unburnt or trapped propellant counts as structure.

References

C. P. Hoult, “Optimum Allocation of Weight Between Two Stages (rev. 1),” RS&T memo, 20 March 2015. View in library
M. Vertregt, “Calculation of Step-Rockets,” J. Brit. Interplanetary Soc., Vol. 14, 1955, pp. 20–25.
M. Vertregt, “A Method for Calculating the Mass Ratios of Step-Rockets,” J. Brit. Interplanetary Soc., Vol. 15, 1956, pp. 95–97.
J. J. Coleman, “Optimum Stage-Weight Distribution of Multistage Rockets,” Space Technology Laboratories, STL/TN-60-0000-9036, 1960.

Two-Stage Weight Allocation — theory