Worked examples — Payload fraction as function of Δv and Isp

This page is the "hit every case" workbook for the parent note payload-fraction master equation. If any symbol below feels unfamiliar, that is a signal to re-read the parent first — but we re-state the one formula we lean on so you never have to leave this page.

Everything downstream is just plugging numbers into that box, so the interesting part is knowing which kinds of numbers break it, saturate it, or hide a trap. That is the scenario matrix.

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The scenario matrix

Think of every rocket problem as a point in a grid. Two knobs matter most: how big $R$ is (driven by $\Delta v$ and $I_{sp}$), and how big $ϵ$ is. The corners and edges of that grid each behave differently.

Cell	What makes it special	Example that hits it
A. Easy regime	small $R$ ($R$ close to 1): tiny burn, plenty left over	Ex 1
B. Hard regime	large $R$: $\lambda$ goes negative → mission impossible single-stage	Ex 2
C. Zero-$\Delta v$ degenerate	$\Delta v=0\Rightarrow R=1$	Ex 3
D. Zero-structure ideal	$ϵ=0$: theoretical best case	Ex 4
E. The break-even boundary	find the exact $R$ where $\lambda =0$	Ex 5
F. $I_{sp}\to \infty$ limit	infinite exhaust speed, what $\lambda$ approaches	Ex 6
G. Real-world word problem	pick the engine that delivers a required payload mass	Ex 7
H. Exam twist	given $\lambda$ and $ϵ$, solve backwards for $\Delta v$	Ex 8

The map below plots $\lambda$ against $R$ for a few $ϵ$ values so you can see where each cell lives before we compute anything.

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The worked examples

Forecast: Small $\Delta v$, so $R$ should be close to 1 and $\lambda$ should be large (over 50%). Guess before reading on.

1. Compute the exponent. $\frac{\Delta v}{I_{sp}{g}_0}=\frac{1500}{320\times 9.81}=\frac{1500}{3139.2}=0.4778$. Why this step? The whole formula runs on this dimensionless number; it must come first.
2. Get $R$. $R=e^{0.4778}=1.6125$. Why this step? $R$ is the only thing $\lambda$ depends on once $ϵ$ is fixed.
3. Plug into the box. $\lambda =\frac{1+0.10(1-1.6125)}{1.6125}=\frac{1-0.06125}{1.6125}=\frac{0.93875}{1.6125}$. Why this step? Direct substitution — this is the master equation.
4. Result. $\lambda =0.582=58.2\%$.

Verify: Units — exponent is $\text{(m/s)}/(\text{s}\cdot {\text{m/s}}^2)=$ dimensionless ✓. $R>1$ ✓. $\lambda <1$ and comfortably positive as forecast ✓.

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Forecast: This is the parent note's SSTO case. Huge $\Delta v$ → huge $R$ → we expect $\lambda$ to go negative, meaning impossible.

1. Exponent. $\frac{9400}{350\times 9.81}=\frac{9400}{3433.5}=2.7378$. Why this step? Same first move; large $\Delta v$ makes this big.
2. $R$. $R=e^{2.7378}=15.454$. Why this step? A mass ratio of 15+ means 94%+ of launch mass is propellant.
3. Plug in. $\lambda =\frac{1+0.10(1-15.454)}{15.454}=\frac{1-1.4454}{15.454}=\frac{-0.4454}{15.454}$.
4. Result. $\lambda =-0.0288$.

Verify: Negative $\lambda$ is physically meaningless (you cannot have negative payload). The math is telling us the structure alone exceeds the empty-mass budget. Sanity: this matches the parent note's $-0.029$. Fix in reality = staging. ✓

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Forecast: No burn means no propellant needed, so the whole launch mass could be payload → $\lambda =1$? Let's check whether $ϵ$ spoils that.

1. Exponent = 0. $\frac{0}{I_{sp}{g}_0}=0$. Why this step? Tests the formula's degenerate boundary.
2. $R=e^0=1$. Why this step? No burn ⇒ initial mass equals final mass ⇒ ratio is exactly 1.
3. Plug in. $\lambda =\frac{1+ϵ(1-1)}{1}=\frac{1+0}{1}=1$.
4. Result. $\lambda =1=100\%$.

Verify: With zero propellant, $ϵ\cdot m_{\text{prop}}=0$ structure too, so everything is payload. $\lambda =1$ is exactly right and independent of $ϵ$ — the $ϵ$ term vanishes because $(1-R)=0$. ✓

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Forecast: With no structural tax, $\lambda$ should equal a clean $1/R$ — just the reciprocal mass ratio, the highest a real design could ever approach.

1. Exponent. $\frac{3800}{450\times 9.81}=\frac{3800}{4414.5}=0.8608$.
2. $R$. $R=e^{0.8608}=2.3651$.
3. Plug in with $ϵ=0$. $\lambda =\frac{1+0}{R}=\frac{1}{2.3651}$. Why this step? Setting $ϵ=0$ collapses the box to $\lambda =1/R=e^{-\Delta v/(I_{sp}{g}_0)}$.
4. Result. $\lambda =0.4228=42.3\%$.

Verify: This is the reciprocal of the mass ratio and must exceed the realistic $ϵ=0.08$ answer (37.7% in Ex from parent). $42.3\%>37.7\%$ ✓ — the tax always lowers $\lambda$.

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Forecast: From Ex 2 we saw $R=15.45$ already went negative; break-even should be a bit below that. Guess ~11.

1. Set numerator to zero. $\lambda =0\Rightarrow 1+ϵ(1-R)=0$. Why this step? $\lambda$ is a fraction with $R>0$ in the denominator, so it is zero exactly when the top is zero.
2. Solve for $R$. $1+0.10-0.10R=0\Rightarrow R=\frac{1+ϵ}{ϵ}=\frac{1.10}{0.10}=11$. Why this step? Clean algebra gives the break-even mass ratio $R^{\star }=(1+ϵ)/ϵ$.
3. Back out $\Delta v$. $\Delta v=I_{sp}{g}_0\ln R=350\times 9.81\times \ln 11=3433.5\times 2.3979=8233\text{m/s}$. Why this step? Invert Tsiolkovsky to translate the critical $R$ into a mission speed.

Verify: Plug $R=11$ back: $\lambda =\frac{1+0.10(1-11)}{11}=\frac{1-1}{11}=0$ ✓. And $8233<9400$ (Ex 2 went past it into negatives) ✓. Forecast of "~11" nailed it.

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Forecast: Infinite $I_{sp}$ means you need almost no propellant, so intuitively $\lambda \to 1$. But does the $ϵ$ term leave a scar? Guess.

1. Exponent goes to 0. As $I_{sp}\to \infty$, $\frac{\Delta v}{I_{sp}{g}_0}\to 0$. Why this step? The denominator explodes, killing the exponent.
2. $R\to e^0=1$. Why this step? Infinitely efficient engine ⇒ negligible propellant ⇒ mass ratio ≈ 1.
3. Take the limit. $\lambda \to \frac{1+ϵ(1-1)}{1}=1$. Why this step? Same collapse as the zero-$\Delta v$ case — $(1-R)\to 0$ erases $ϵ$.
4. Result. $\lambda \to 1=100\%$.

Verify: Numerically at $I_{sp}=10^6\text{s}$, $\Delta v=3800$: exponent $=3800/(10^6\cdot 9.81)=3.87\times 10^{-4}$, $R=1.000387$, $\lambda =\frac{1+0.10(1-1.000387)}{1.000387}=0.99958\approx 1$ ✓. This is why chemistry-beating drives (ion, nuclear) are so prized — see Isp.

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Forecast: From the parent's Ex 2, this configuration gives $\lambda =37.7\%$. So $m_0$ should be roughly $2000/0.377\approx 5300$ kg. Guess before computing.

1. Find $\lambda$. Exponent $=\frac{3800}{450\times 9.81}=0.8608$, $R=e^{0.8608}=2.3651$. Why this step? We need the fraction before we can invert it to a mass.
2. Plug in. $\lambda =\frac{1+0.08(1-2.3651)}{2.3651}=\frac{1-0.10921}{2.3651}=\frac{0.89079}{2.3651}=0.3767$. Why this step? This is the same 37.7% the parent found — good cross-check.
3. Invert the definition. $\lambda =m_{\text{pay}}/m_0\Rightarrow m_0=m_{\text{pay}}/\lambda =2000/0.3767$. Why this step? $\lambda$ is a ratio; solving for $m_0$ just divides.
4. Result. $m_0=5309\text{kg}$.

Verify: Payload check $0.3767\times 5309=2000$ kg ✓. Matches forecast (~5300). Units: kg ÷ (dimensionless) = kg ✓.

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Forecast: $\lambda =25\%$ is middling, so $R$ is moderate (bigger than the easy cases, well short of break-even $R^{\star }=(1.08)/0.08=13.5$). Expect $\Delta v$ a few km/s.

1. Rearrange the box for $R$. $\lambda R=1+ϵ-ϵR\Rightarrow R(\lambda +ϵ)=1+ϵ\Rightarrow R=\frac{1+ϵ}{\lambda +ϵ}$. Why this step? We want $R$ alone so we can take a log; group the $R$ terms.
2. Insert numbers. $R=\frac{1.08}{0.25+0.08}=\frac{1.08}{0.33}=3.2727$. Why this step? Turns the given $\lambda$ into the mass ratio.
3. Invert Tsiolkovsky. $\Delta v=I_{sp}{g}_0\ln R=340\times 9.81\times \ln (3.2727)=3335.4\times 1.1857=3955\text{m/s}$. Why this step? $\Delta v$ lives inside the exponent; the natural log undoes it.
4. Result. $\Delta v\approx 3955\text{m/s}$.

Verify: Forward-check: $R=e^{3955/3335.4}=e^{1.1857}=3.273$ ✓, then $\lambda =\frac{1+0.08(1-3.273)}{3.273}=\frac{0.8182}{3.273}=0.250$ ✓. And $R=3.27<R^{\star }=13.5$, so $\lambda >0$ as it should be ✓.

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Recall Quick self-test

Break-even mass ratio formula ::: $R^{\star }=(1+ϵ)/ϵ$ What does negative $\lambda$ physically mean? ::: No valid single-stage rocket exists at those parameters — you need staging. As $I_{sp}\to \infty$, $\lambda \to$ ::: $1$ (100%), because $R\to 1$. With $ϵ=0$, the formula collapses to ::: $\lambda =1/R=e^{-\Delta v/(I_{sp}{g}_0)}$.

Related: Tsiolkovsky Rocket Equation · Mass Ratio · Optimal Staging · Oberth Effect.