Optimal staging — equal mass ratios (for same Isp): Worked examples

This page is the "run every case" companion to the parent note on optimal staging. The parent proved why equal mass ratios win. Here we drive that machinery through every kind of input — big and small structural fraction, one stage vs many, the degenerate cases where the maths threatens to break, a real launch problem, and an exam twist. Each example says which cell of the matrix it belongs to.

Before we start, a one-line reminder of the two tools we lean on, so no symbol appears unexplained.

Recall What each symbol means (built in the parent)

$m_{p}$ ::: payload mass — the useful cargo we deliver. $M_{0}$ ::: total mass on the pad at liftoff (payload + all structure + all fuel). $R_{i}$ ::: mass ratio of stage $i$ = (mass when it ignites) ÷ (mass when it burns out). Always $R_{i} \geq 1$ . $ϵ$ ::: structural fraction = structure ÷ (structure + fuel) for a stage. "Dead weight tax." $v_{e}$ ::: exhaust speed. It converts a mass ratio into speed via $Δ v = v_{e} ln R$ . $R^{*}$ ::: the single value all stages share at the optimum.

The scenario matrix

Every problem this topic can throw is one of these cells. The examples below hit each one.

Cell	What varies	Danger it tests	Example
A	$ϵ = 0$ (no dead weight)	Does the product-fixed shortcut hold?	Ex 1
B	$ϵ > 0$ , small ( $n = 2$ )	Solving the nonlinear constraint	Ex 2
C	Equal vs unequal split, $ϵ > 0$	Proving equal is strictly better	Ex 3
D	Growing $n$ (1, 2, 3, 4)	Diminishing returns; the $R_{ma x} = 1/ ϵ$ ceiling	Ex 4
E	Degenerate: $R^{*} \to 1/ ϵ$	What happens as $m_{p} / M_{0} \to 0$	Ex 5
F	Degenerate: $R_{i} = 1$ (empty stage)	Zero-fuel stage adds nothing	Ex 6
G	Real-world word problem	Reach orbit; pick $n$	Ex 7
H	Exam twist: different $v_{e}$ per stage	When the "equal $R$ " rule breaks	Ex 8

Worked example Ex 1 — Cell A: zero structure, product is fixed

$v_{e} = 3000$ m/s, $M_{0} / m_{p} = 16$ , $ϵ = 0$ . Two stages. Show that any split with $R_{1} R_{2} = 16$ gives the same $Δ v$ , and equal ratios $R_{1} = R_{2} = 4$ are one such split.

Forecast: Guess — does the split matter here? (Answer: no, because $ln$ turns a product into a sum.)

Write the constraint at $ϵ = 0$ . Why this step? From the parent, when $ϵ \to 0$ the factor $\frac{1 - ϵ R}{( 1 - ϵ ) R} \to \frac{1}{R}$ , so $(⋆)$ collapses to $\frac{m _{p}}{M _{0}} = \frac{1}{R _{1} R _{2}}$ , i.e. $R_{1} R_{2} = M_{0} / m_{p} = 16$ .
Objective. Why? Total speed adds: $Δ v = v_{e} (ln R_{1} + ln R_{2}) = v_{e} ln (R_{1} R_{2})$ . Only the product enters.
Plug the fixed product. $Δ v = 3000 ln 16 = 3000 \times 2.7726 = 8318$ m/s. Equal ratios $R_{1} = R_{2} = 4$ give $ln 4 + ln 4 = ln 16$ — same number.

Verify: Try the lopsided split $R_{1} = 8, R_{2} = 2$ : $ln 8 + ln 2 = 2.0794 + 0.6931 = 2.7726 = ln 16$ . Identical. ✓ Units: $(m/s) \times (dimensionless ln) = m/s$ . ✓

Worked example Ex 2 — Cell B: real structure, solve the nonlinear constraint (

$n = 2$ ) $m_{p} = 1000$ kg, $M_{0} = 10000$ kg, $ϵ = 0.1$ , $v_{e} = 3000$ m/s. Find $R^{*}$ and $Δ v$ .

Forecast: With $ϵ = 0.1$ , will $R^{*}$ be bigger or smaller than the zero-structure value $10 \approx 3.16$ ? (Dead weight costs you, so $R^{*}$ must be larger to squeeze out the same payload fraction.)

Compute $k$ . Why? The parent's closed form uses $k = (m_{p} / M_{0})^{1/ n} = (0.1)^{1/2} = 0.31623$ .
Plug into $R^{*} = \frac{1}{ϵ + ( 1 - ϵ ) k}$ . Why this step? It solves the $n$ -th-root form of $(⋆)$ directly. $R^{*} = \frac{1}{0.1 + 0.9 \times 0.31623} = \frac{1}{0.38461} = 2.600$ .
Speed. $Δ v = n v_{e} ln R^{*} = 2 \times 3000 \times ln 2.600 = 6000 \times 0.9555 = 5733$ m/s.

Verify: Rebuild masses. Stage-1 fuel $= M_{0} (1 - 1/ R^{*}) = 10000 (1 - 0.38461) = 6154$ kg; burnout $3846$ ; structure $\frac{0.1}{0.9} \times 6154 = 684$ ; hand-off $3846 - 684 = 3162$ kg. Then $M_{1} / M_{0} = 3162/10000 = 0.3162 = k$ ✓, and repeating gives exactly $1000$ kg payload ✓ (matches the parent's worked check).

Worked example Ex 3 — Cell C: equal

strictly beats unequal when $ϵ > 0$ Same rocket as Ex 2 ( $M_{0} = 10000$ , $m_{p} = 1000$ , $ϵ = 0.1$ , $v_{e} = 3000$ ). Compare optimal equal split ( $R_{1} = R_{2} = 2.600$ ) against a forced unequal split. To keep it fair, both must satisfy the true constraint $(⋆)$ and deliver exactly $1000$ kg.

Forecast: Ex 1 said splits tie at $ϵ = 0$ . Now $ϵ > 0$ — guess whether the tie is broken.

Pick an unequal pair on the constraint. Why? We cannot pick $R_{1}, R_{2}$ freely; they must satisfy $\frac{1 - ϵ R _{1}}{0.9 R _{1}} \cdot \frac{1 - ϵ R _{2}}{0.9 R _{2}} = 0.1$ . Set $R_{1} = 3.2$ and solve for $R_{2}$ : the first factor is $\frac{1 - 0.32}{0.9 \times 3.2} = \frac{0.68}{2.88} = 0.23611$ ; we need the second $= 0.1/0.23611 = 0.42353$ . So $\frac{1 - 0.1 R _{2}}{0.9 R _{2}} = 0.42353 \Rightarrow 1 - 0.1 R_{2} = 0.38118 R_{2} \Rightarrow R_{2} = 1/0.48118 = 2.0782$ .
Compute both $Δ v$ 's. Why? Direct comparison. Equal: $Δ v_{e q} = 6000 ln 2.600 = 5733$ m/s. Unequal: $Δ v_{u n} = 3000 (ln 3.2 + ln 2.0782) = 3000 (1.16315 + 0.73139) = 3000 \times 1.89454 = 5684$ m/s.
Read the gap. $5733 - 5684 = 49$ m/s lost by going unequal.

Verify: $5733 > 5684$ , so equal ratios win when $ϵ > 0$ ✓. The advantage vanished at $ϵ = 0$ (Ex 1) and appears at $ϵ = 0.1$ — exactly what the Lagrange argument predicted.

Worked example Ex 4 — Cell D: how

$R^{*}$ and $Δ v$ change with $n$ Fixed $M_{0} / m_{p} = 30$ (i.e. $m_{p} / M_{0} = 1/30$ ), $ϵ = 0.1$ , $v_{e} = 3000$ . Tabulate $n = 1, 2, 3, 4$ .

Forecast: Does one giant stage ( $n = 1$ ) even work? (Recall $R_{ma x} = 1/ ϵ = 10$ ; we need effective 30. Watch what breaks.)

$n = 1$ . Why start here? It exposes the ceiling. $k = (1/30)^{1} = 0.03333$ , $R^{*} = \frac{1}{0.1 + 0.9 \times 0.03333} = \frac{1}{0.13} = 7.6923$ . One stage of ratio $7.6923$ gives $Δ v = 3000 ln 7.6923 = 6121$ m/s — not enough for orbit and the required payload fraction is barely reachable ( $R < R_{ma x} = 10$ , so it's just physically possible but weak).
$n = 2$ . $k = (1/30)^{1/2} = 0.18257$ , $R^{*} = \frac{1}{0.1 + 0.9 \times 0.18257} = \frac{1}{0.26432} = 3.7833$ , $Δ v = 2 \times 3000 ln 3.7833 = 7982$ m/s.
$n = 3$ . $k = (1/30)^{1/3} = 0.32183$ , $R^{*} = \frac{1}{0.1 + 0.9 \times 0.32183} = \frac{1}{0.38965} = 2.5664$ , $Δ v = 3 \times 3000 ln 2.5664 = 8481$ m/s.
$n = 4$ . $k = (1/30)^{1/4} = 0.42729$ , $R^{*} = \frac{1}{0.1 + 0.9 \times 0.42729} = \frac{1}{0.48456} = 2.0637$ , $Δ v = 4 \times 3000 ln 2.0637 = 8703$ m/s.

The figure below plots these four points. Read it left to right: the blue curve is total $Δ v$ , each point tagged with its $R^{*}$ ; the orange numbers between points are the gains $+ 1861, + 499, + 222$ — visibly shrinking. The red dashed line is the single-stage ceiling $v_{e} ln (1/ ϵ) = 6908$ m/s, which even $n = 1$ here sits below (because our payload is not negligible).

Verify: $Δ v$ climbs $6121 \to 7982 \to 8481 \to 8703$ — increasing but with shrinking gaps ( $1861$ , then $499$ , then $222$ ). That's the "diminishing returns beyond 3–4 stages" the parent claimed. ✓ See Staging basics.

Worked example Ex 5 — Cell E: the ceiling

$R^{*} \to 1/ ϵ$ as payload shrinks $ϵ = 0.1$ , one stage ( $n = 1$ ). Push $m_{p} / M_{0}$ toward $0$ and watch $R^{*}$ .

Forecast: As you demand a tinier payload fraction, does one stage's $R^{*}$ grow without bound, or hit a wall?

Take the limit. Why? $R^{*} = \frac{1}{ϵ + ( 1 - ϵ ) k}$ with $k = (m_{p} / M_{0})^{1/1} = m_{p} / M_{0} \to 0$ . Then $R^{*} \to \frac{1}{ϵ} = 10$ .
Interpret. Why? Even with zero payload, a single stage cannot exceed ratio $10$ : it must lift its own $10%$ structure. Max single-stage speed $= v_{e} ln (1/ ϵ) = 3000 ln 10 = 6908$ m/s.
Consequence. Any mission needing $M_{0} / m_{p} > 10$ is impossible with one stage — you must stage. This is the core lesson.

Verify: Plug $m_{p} / M_{0} = 0.001$ : $R^{*} = \frac{1}{0.1 + 0.9 \times 0.001} = \frac{1}{0.1009} = 9.911$ — creeping toward $10$ but never reaching it ✓. Ties to the parent's Structural fraction ceiling.

Worked example Ex 6 — Cell F: a degenerate "empty" stage does nothing

Two stages, but suppose we accidentally load stage 2 with no fuel ( $m_{f 2} = 0$ ), so $R_{2} = 1$ . $R_{1} = 4$ , $v_{e} = 3000$ .

Forecast: Does the dead stage help, hurt, or do nothing to $Δ v$ ?

Speed from the empty stage. Why? $Δ v_{2} = v_{e} ln R_{2} = 3000 ln 1 = 3000 \times 0 = 0$ . A ratio of $1$ contributes nothing — $ln 1 = 0$ .
Worse than nothing. Why? Its structure $m_{s 2} > 0$ is still dead mass that stage 1 had to lift. So the fueled stage now has a smaller effective $R_{1}$ than if that structure weren't there. The empty stage strictly hurts.
Lesson. Every stage in an optimal design has $R_{i} = R^{*} > 1$ ; a stage with $R = 1$ should be deleted, not flown.

Verify: $ln 1 = 0$ exactly ✓, so $Δ v_{s t a g e 2} = 0$ ; total speed comes entirely from stage 1: $3000 ln 4 = 4159$ m/s, and any structure on the dead stage only reduces that. ✓

Worked example Ex 7 — Cell G: real-world — get a satellite to LEO

A satellite $m_{p} = 1500$ kg must reach low Earth orbit needing $Δ v_{r e q} = 9400$ m/s (incl. gravity + drag losses). Engines give $v_{e} = 3200$ m/s, $ϵ = 0.08$ . How many equal stages, and what launch mass $M_{0}$ ?

Forecast: Guess $n$ before computing. (Orbit needs $M_{0} / m_{p}$ big; single stage caps at $R = 1/0.08 = 12.5$ , i.e. $Δ v \leq 3200 ln 12.5 = 8083$ m/s — short of 9400. So $n \geq 2$ . But is $n = 2$ best, or should we go higher?)

Rule out $n = 1$ . Why first? A single stage cannot exceed $Δ v = v_{e} ln (1/ ϵ) = 3200 ln 12.5 = 8083$ m/s $< 9400$ . Infeasible. So $n \geq 2$ .
For each $n$ , find the launch mass it demands. Why this step? "How many stages" is an optimization over $n$ : each feasible $n$ needs a different $M_{0}$ , and we want the smallest $M_{0}$ (cheapest rocket). For given $n$ , the per-stage ratio is fixed by the mission: $R^{*} = exp (\frac{Δ v _{r e q}}{n v _{e}})$ , and then $(⋆)$ gives $\frac{M _{0}}{m _{p}} = (\frac{( 1 - ϵ ) R ^{*}}{1 - ϵ R ^{*}})^{n}$ .
Tabulate. Why? Compare launch masses directly.
- $n = 2$ : $R^{*} = exp (9400/6400) = 4.344$ ; feasible since $4.344 < 12.5$ . Factor $\frac{0.92 \times 4.344}{1 - 0.08 \times 4.344} = \frac{3.99648}{0.65248} = 6.1251$ ; $M_{0} / m_{p} = 6.125 1^{2} = 37.52$ ; $M_{0} = 1500 \times 37.52 = 56276$ kg.
- $n = 3$ : $R^{*} = exp (9400/9600) = 2.6635$ ; feasible. Factor $\frac{0.92 \times 2.6635}{1 - 0.08 \times 2.6635} = \frac{2.45042}{0.78692} = 3.1140$ ; $M_{0} / m_{p} = 3.114 0^{3} = 30.20$ ; $M_{0} = 45305$ kg.
- $n = 4$ : $R^{*} = exp (9400/12800) = 2.0844$ ; feasible. Factor $\frac{0.92 \times 2.0844}{1 - 0.08 \times 2.0844} = \frac{1.91765}{0.83325} = 2.30140$ ; $M_{0} / m_{p} = 2.3014 0^{4} = 28.05$ ; $M_{0} = 42069$ kg.
Read the trade-off. Launch mass falls as $n$ rises ( $56.3 \to 45.3 \to 42.1$ t) but with shrinking savings, exactly the diminishing returns of Ex 4. Since each added stage brings real complexity and interstage mass (parent's caveat), $n = 2$ or $n = 3$ is the practical sweet spot; $n = 3$ trims $\approx 11$ t of launch mass over $n = 2$ and is the recommended choice.

Verify: Check the $n = 2$ design satisfies the mission: $Δ v = 2 \times 3200 \times ln 4.344 = 6400 \times 1.46875 = 9400$ m/s ✓, payload fraction $1500/56276 = 0.02666$ ✓. And $n = 3$ : $3 \times 3200 ln 2.6635 = 9600 \times 0.97917 = 9400$ m/s ✓ with $M_{0} = 45305$ kg. Launch masses strictly decrease with $n$ ✓. Uses Payload fraction.

Worked example Ex 8 — Cell H: exam twist — stages with

different $v_{e}$ The "equal $R$ " rule assumed the same $v_{e}$ . Now stage 1 has $v_{e 1} = 2800$ m/s, stage 2 has $v_{e 2} = 3400$ m/s, $ϵ \to 0$ , product fixed $R_{1} R_{2} = 16$ . Should the ratios still be equal?

Forecast: Which stage deserves the bigger mass ratio — the efficient one or the weak one?

Write the objective and constraint. Why? Now $Δ v = v_{e 1} ln R_{1} + v_{e 2} ln R_{2}$ (unequal weights on the two $ln$ terms), and at $ϵ \to 0$ the constraint is $ln R_{1} + ln R_{2} = ln 16$ . Introduce variables $x = ln R_{1}, y = ln R_{2}$ so the objective is linear: $Δ v = v_{e 1} x + v_{e 2} y$ , constraint $x + y = ln 16$ , with the physical bounds $x \geq 0, y \geq 0$ (ratios $\geq 1$ ).
Build the Lagrangian. Why? To locate stationary points. With multiplier $λ$ for the equality: $L (x, y, λ) = v_{e 1} x + v_{e 2} y - λ (x + y - ln 16) .$ $\partial L / \partial x = v_{e 1} - λ = 0$ and $\partial L / \partial y = v_{e 2} - λ = 0$ demand $λ = v_{e 1} = v_{e 2}$ simultaneously — impossible when $v_{e 1} \neq = v_{e 2}$ . So there is no interior stationary point; the maximum of a linear objective on a line segment must sit at an endpoint (a KKT corner where one bound $x = 0$ or $y = 0$ activates).
Check the corners. Why? The feasible segment $x + y = ln 16, x, y \geq 0$ has two endpoints. Since $v_{e 2} > v_{e 1}$ , put all the "budget" into the higher- $v_{e}$ stage: $x = 0 (R_{1} = 1), y = ln 16 (R_{2} = 16)$ .
Compare. Equal ( $R_{1} = R_{2} = 4$ ): $Δ v = 2800 ln 4 + 3400 ln 4 = (2800 + 3400) \times 1.3863 = 8595$ m/s. Corner ( $R_{1} = 1, R_{2} = 16$ ): $Δ v = 2800 \times 0 + 3400 ln 16 = 3400 \times 2.7726 = 9427$ m/s.
Lesson. With different $I_{s p}$ , equal ratios are no longer optimal — the maximum is pushed to the boundary, and the high- $v_{e}$ stage should carry the larger mass ratio (here it takes the entire fuel budget). The clean "equal $R^{*}$ " theorem is a special privilege of the same- $I_{s p}$ assumption; drop that assumption and the whole answer changes. This is precisely the regime handled by Variable specific impulse staging.

Verify: $9427 > 8595$ ✓ — the corner beats the equal split, confirming the "same $I_{s p}$ " hypothesis is essential to the parent theorem. (Method: Lagrange multipliers with boundary/KKT check.)

Recall Rapid self-test

If $ϵ = 0$ and the product $R_{1} R_{2}$ is fixed, does the split change $Δ v$ ? ::: No — $ln$ makes only the product matter (Ex 1). Why does equal beat unequal once $ϵ > 0$ ? ::: The constraint $(⋆)$ becomes nonlinear; unequal ratios waste structural mass (Ex 3). What is the maximum mass ratio of a single stage with structural fraction $ϵ$ ? ::: $R_{ma x} = 1/ ϵ$ (Ex 5). A stage with $R_{i} = 1$ contributes how much $Δ v$ ? ::: Zero, and its structure even hurts (Ex 6). When does the "equal $R$ " rule fail? ::: When stages have different $v_{e}$ — then favour the high- $v_{e}$ stage (Ex 8).