Worked examples — Rocket staging — series staging, parallel staging

This page is the practice range for Rocket staging. Every tool it uses — the Tsiolkovsky rocket equation, the structural and propellant fractions, the thrust-weighted exhaust velocity — is used exactly as the parent note built it. If a symbol shows up in an example, we re-state what it means right there before we lean on it.

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The scenario matrix

We use these symbols throughout (all from the parent note):

• $\Delta v$ — the velocity a stage adds, in metres per second (m/s).
• $v_e$ — exhaust velocity: how fast, in m/s, the burnt gas leaves the nozzle. Bigger $v_e$ = more push per kilogram of fuel.
• $m_0$ — mass at the start of a burn (structure + its own fuel + everything it carries).
• $m_f$ — mass at the end of a burn (start mass minus the fuel it burnt).
• $m_{\text{pay}}$ — payload mass: the useful cargo (satellite, capsule) a stage must deliver; from any stage's viewpoint it also includes every stage stacked above it. Measured in kg.
• $\ln$ — the natural logarithm, the function that answers "$e$ to what power gives this number?" It appears because Tsiolkovsky is $\Delta v=v_e\ln (m_0/m_f)$.
• $\lambda$ (propellant fraction) and $ϵ$ (structural fraction): of one stage's hardware mass $S$, the share that is fuel ($\lambda =m_p/S$) versus tanks/engines ($ϵ=m_s/S$), with $\lambda +ϵ=1$. See Structural fraction and propellant fraction.
• $F$ — thrust: the pushing force an engine or engine group produces, in meganewtons (MN). $F_c$ = core thrust, $F_b$ = one booster's thrust.
• $I_{sp}$ — specific impulse, exhaust velocity measured in seconds: $v_e=I_{sp}g$ with $g=9.81{\text{m/s}}^2$. See Specific impulse.

The two figures on this page build the two ideas that are hardest to see: how mass falls in steps (figure s01) and how the mass ratio — not raw mass — controls $\Delta v$ (figure s02). We look at them as we reach them.

Cell	Scenario class	Worked in
A	Clean multi-stage series $\Delta v$ sum	Example 1
B	Parallel boosters + thrust-weighted ${\bar{v}}_e$, two phases	Example 2
C	Degenerate: single stage ($n=1$) — the "no staging" baseline	Example 3
D	Zero / limiting input: $ϵ\to 0$ (perfect tanks) and $ϵ\to 1$ (dead-weight limit)	Example 4
E	Series vs parallel head-to-head on the same hardware	Example 5
F	Word problem: "how many boosters to hit a target $\Delta v$?"	Example 6
G	Exam twist: two students, one wrong mass ratio (spot the error)	Example 7
H	Unit / sanity trap: $I_{sp}$ in seconds vs $v_e$ in m/s	Example 8

Every cell A–H is covered below.

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Example 1 — Cell A: clean two-stage series

Forecast: Guess before computing — will it clear the ~9400 m/s needed for low orbit? Write down yes/no now.

Recall the symbols we need: $\lambda$ is the fuel share of a stage's own hardware $S$, so fuel $=\lambda S$; $ϵ=1-\lambda$ is the structure share, so structure $=ϵS$. $m_0$/$m_f$ are start/end masses, $m_{\text{pay}}$ is the payload (here plus stage 2 when we look at stage 1), and $\Delta v=v_e\ln (m_0/m_f)$.

1. Work top-down: stage 2 first. Keep everything in fractions: structure $={ϵ}_2{S}_2=(1-0.87)\times 8000=0.13\times 8000=1040\text{kg}$; fuel $={\lambda }_2{S}_2=0.87\times 8000=6960\text{kg}$. Why this step? Stage 2 only ever carries the payload, never stage 1's dead mass, so its numbers are self-contained. Using $\lambda ,ϵ$ (not raw masses) lets us re-use this for any $S_2$.

2. Stage 2 masses. Start $m_0=S_2+m_{\text{pay}}=8000+1000=9000\text{kg}$. End (fuel gone) $m_f={ϵ}_2{S}_2+m_{\text{pay}}=1040+1000=2040\text{kg}$. Why this step? Tsiolkovsky needs the before/after mass; $m_f$ keeps only the structure ${ϵ}_2{S}_2$ plus payload once the ${\lambda }_2{S}_2=6960$ kg of fuel is spent.

3. Stage 2 $\Delta v$. $\Delta v_2=v_{e,2}\ln \left(\frac{m_0}{m_f}\right)=3500\ln \left(\frac{9000}{2040}\right)=3500\times 1.484\approx 5190\text{m/s}.$ Why this step? $\ln$ turns the mass ratio into how many "$e$-foldings" of shedding happened; $v_e$ scales that into speed.

4. Stage 1 must lift stage-2 hardware + payload $=8000+1000=9000\text{kg}$. Structure $={ϵ}_1{S}_1=0.15\times 90000=13500\text{kg}$, fuel $={\lambda }_1{S}_1=0.85\times 90000=76500\text{kg}$. Why this step? From stage 1's viewpoint, "payload" is everything above it, so here $m_{\text{pay}}=9000$ kg.

5. Stage 1 masses and $\Delta v$. $m_0=90000+9000=99000$, $m_f=13500+9000=22500$. $\Delta v_1=3000\ln \left(\frac{99000}{22500}\right)=3000\times 1.482\approx 4450\text{m/s}.$ Why this step? We apply Tsiolkovsky to stage 1 alone, exactly as to stage 2: its start mass is the whole stack, its end mass is stage-1 structure plus everything it still carries once its own fuel is gone. The equation is stage-agnostic — same law, new masses.

6. Add them (series $\Delta v$'s add because each stage starts at the speed the last one left it): $\Delta v_{\text{total}}\approx 5190+4450=9640\text{m/s}.$

Recall Verify

Answer $\approx 9640$ m/s > 9400 m/s target, so it reaches orbit. Units: $(\text{m/s})\times (\text{dimensionless }\ln )=\text{m/s}$. ✓

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Example 2 — Cell B: parallel boosters, two phases

Before any algebra, watch the mass fall in steps. Figure s01 plots vehicle mass (vertical axis, tonnes) against time (horizontal axis). The magenta line is phase 1 (core + boosters burning together, mass sliding from $92$ t to $38$ t). At separation the orange dashed cliff drops the empty booster shells ($38\to 32$ t). The violet line is phase 2 (core alone, $32\to 5$ t). Notice the timeline is not one smooth slide — the jettison cliff is the whole point of staging.

Forecast: Will ${\bar{v}}_e$ sit closer to the boosters' $2650$ or the core's $4400$? Guess which engine group "wins."

Symbols here: $F_c$ is core thrust and $F_b$ is one booster's thrust (force, in meganewtons MN), so two boosters give $2F_b$; $v_{e,c},v_{e,b}$ are the core and booster exhaust velocities (m/s); ${\bar{v}}_e$ is the single "effective" exhaust velocity that stands in for both engine groups burning at once.

1. Weight each $v_e$ by its thrust. The boosters push much harder ($2F_b=25$ vs $F_c=5.5$ MN), so they dominate the average. Why this step? Momentum bookkeeping: total thrust $={\dot{m}}_{\text{tot}}{\bar{v}}_e$, and each engine contributes ${\dot{m}}_i{v}_{e,i}=F_i$. Dividing total $F$ by total $\dot{m}$ gives a thrust-weighted mean.

2. Plug in: ${\bar{v}}_e=\frac{F_c{v}_{e,c}+2F_b{v}_{e,b}}{F_c+2F_b}=\frac{5.5\times 4400+25\times 2650}{5.5+25}=\frac{24200+66250}{30.5}=\frac{90450}{30.5}\approx 2966\text{m/s}.$

3. Read the geometry (figure s01): the annotation "uses vbar_e" sits on the magenta phase-1 line — this ${\bar{v}}_e$ governs the whole combined burn. After the orange cliff, the annotation "uses v_e core" sits on the violet line — phase 2 reverts to the core's own $v_{e,c}$.

Recall Verify

${\bar{v}}_e\approx 2966$ m/s lies between $2650$ and $4400$, nearer $2650$ as forecast — boosters carry more thrust. ✓

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Example 3 — Cell C: single stage (degenerate baseline)

Forecast: With less mass to carry, will single-stage beat the $4450$ m/s stage 1 gave in Example 1? Guess higher or lower.

Reminder: $S$ is the stage's own hardware mass; $ϵ=0.15$ means $15\%$ of $S$ is structure, so fuel $=\lambda S=(1-ϵ)S$; $m_{\text{pay}}$ is the cargo.

1. Masses. Fuel $=(1-ϵ)S=0.85\times 90000=76500$, structure $=ϵS=0.15\times 90000=13500$. $m_0=S+m_{\text{pay}}=90000+1000=91000$, $m_f=ϵS+m_{\text{pay}}=13500+1000=14500$. Why this step? $n=1$ is Tsiolkovsky with nothing to jettison — the simplest possible case, the baseline everything else must beat.

2. $\Delta v$: $\Delta v=3000\ln \left(\frac{91000}{14500}\right)=3000\times 1.837\approx 5510\text{m/s}.$

3. Interpret. A single stage tops out near $5500$ m/s here — nowhere near orbital $9400$. This is why staging exists: to escape the structural dead weight that caps a single stage.

Recall Verify

$5510$ m/s falls far short of $9640$ m/s from the two-stage rocket, confirming the staging advantage. Units m/s ✓.

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Example 4 — Cell D: zero and limiting inputs

Figure s02 is the visual heart of this example. It plots $\Delta v$ (vertical, m/s) against the structural fraction $ϵ$ (horizontal, from $0$ to $1$) for stage 2. Trace the violet curve left-to-right: at $ϵ=0$ it hits the magenta ceiling ($\approx 7690$ m/s), the real design sits at the orange dot ($ϵ=0.13$, $\approx 5190$ m/s), and at $ϵ=1$ the curve lands on the navy dot at zero. The curve is steep — small increases in dead structure cost a lot of $\Delta v$.

Forecast: In case (a) does $\Delta v$ blow up to infinity or hit a ceiling? In case (b), what should $\Delta v$ be?

Reminder: $m_0=S+m_{\text{pay}}$ never changes here (hardware size fixed at $8000$); only $m_f=ϵS+m_{\text{pay}}$ moves as $ϵ$ changes.

1. Case (a): $ϵ=0$. Then structure $=ϵS=0$, so $m_f=0+1000=1000$ kg and $m_0=8000+1000=9000$. $\Delta v=3500\ln \left(\frac{9000}{1000}\right)=3500\times 2.197\approx 7690\text{m/s}.$ Why this step? $ϵ=0$ is the theoretical best a given tank size can do — the ceiling. It is finite because the $1000$ kg payload stops $m_f$ from reaching $0$ (the log would diverge only if $m_f\to 0$).

2. Case (b): $ϵ=1$. No fuel, so $m_f=S+m_{\text{pay}}=m_0=9000$. $\Delta v=3500\ln (1)=0\text{m/s}.$ Why this step? $\ln (1)=0$ — burning nothing changes nothing. This is the degenerate zero-input check, the navy dot in figure s02.

3. Read the trend (figure s02): as $ϵ$ climbs from $0$ toward $1$, $\Delta v$ slides smoothly from $7690$ down to $0$; the real design ($ϵ=0.13$, orange dot) sits high on that curve at $5190$ m/s. This is the visual argument for chasing low structural fraction in Payload fraction optimization.

Recall Verify

(a) $\ln 9=2.197$, gives $7690$ m/s (finite, above the $5190$ m/s real value). (b) exactly $0$ since $\ln 1=0$. Both limits behave. ✓

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Example 5 — Cell E: series vs parallel, same hardware

Forecast: Series avoids co-lifting dead weight — bet on which wins before computing.

Reminder: for every unit here $ϵ=0.1$, so structure $=ϵS=3000$ kg and fuel $=(1-ϵ)S=27000$ kg per $30000$ kg unit; $m_{\text{pay}}=2000$ kg is the final cargo.

Series layout (unit is a bottom stage that drops before the core burns):

1. Bottom stage carries (core + payload) $=30000+2000=32000$ as its payload. $m_0=S_b+32000=30000+32000=62000$; $m_f=ϵS_b+32000=3000+32000=35000$. $\Delta v_1=3400\ln \left(\frac{62000}{35000}\right)=3400\times 0.5716\approx 1943\text{m/s}.$
2. Core stage: $m_0=S_c+2000=32000$; $m_f=ϵS_c+2000=3000+2000=5000$. $\Delta v_2=3400\ln \left(\frac{32000}{5000}\right)=3400\times 1.856\approx 6311\text{m/s}.$
3. Series total $\approx 1943+6311=8254\text{m/s}$. Why this step? Each dropped stage improves the next mass ratio — the compounding series advantage.

Parallel layout (both boosters + core burn in phase 1; boosters drop; core finishes): 4. Phase 1: all engines fire. Same $v_e$ everywhere so ${\bar{v}}_e=3400$. We stipulate a cross-feed design in which the core's tanks stay full until separation — the boosters feed the core so only booster propellant is drawn in phase 1. Start mass $=$ core $30000$ + two boosters $60000$ + payload $2000=92000$ kg. Boosters burn all their fuel $=2\times 27000=54000$ kg; core burns none yet: $m_1=92000-54000=38000$. $\Delta v_1=3400\ln \left(\frac{92000}{38000}\right)=3400\times 0.8842\approx 3006\text{m/s}.$ Why this step? The "core burns nothing in phase 1" assumption is not arbitrary — it is the explicit cross-feed rule we chose so the two layouts start the core burn with identical full tanks, making the comparison fair. (Without cross-feed the core would also drain in phase 1, giving a different but similarly computable split.) 5. Separation. The boosters are now empty shells; jettison them. Their dropped mass is $2\times ϵS_b=2\times 3000=6000$ kg, so $m_2=m_1-6000=38000-6000=32000$ kg. Why this step? Separation removes booster structure only — the fuel already left as exhaust in phase 1. This cliff is what parallel staging buys you. 6. Phase 2: core burns its full $27000$ kg. Start $m_2=32000$; final $m_{2,f}=ϵS_c+m_{\text{pay}}=3000+2000=5000$. $\Delta v_2=3400\ln \left(\frac{32000}{5000}\right)=3400\times 1.856\approx 6311\text{m/s}.$ Why this step? After jettison the vehicle is just the core carrying the payload, so Tsiolkovsky applies to the core alone with its own untouched fuel. 7. Parallel total $\approx 3006+6311=9317\text{m/s}$.

Compare: Series $8254$ vs parallel $9317$ m/s — parallel wins here.

Recall Verify

Series $8254$ vs parallel $9317$ m/s. Here parallel wins because in the series case the core had to lift the second unit's full fuel through the first burn as payload, hurting its ratio; the cross-feed parallel burn drains that fuel low and early. The "series is always more efficient" slogan holds only when stages are separately optimised — this shows why you must actually compute. ✓

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Example 6 — Cell F: word problem, "how many boosters?"

Forecast: Guess $n$: is $1$ enough, or do you need several — or is it even possible?

Reminder: $m_0$ = liftoff mass (fixed core $60000$ + $25000$ per booster); $m_1$ = mass at booster burnout ($m_0$ minus the $20000n$ kg of booster fuel).

1. We need $\Delta v_1\ge 9400-7000=2400\text{m/s}$. Why this step? The core already supplies $7000$; boosters must cover the gap.

2. Try $n=3$: $m_0=60000+75000=135000$; fuel burnt $=3\times 20000=60000$; $m_1=75000$. $\Delta v_1=1000\ln \left(\frac{135000}{75000}\right)=1000\times 0.5878=588\text{m/s}.$ Too small.

3. Try $n=8$: $m_0=60000+200000=260000$; fuel $=160000$; $m_1=100000$. $\Delta v_1=1000\ln \left(\frac{260000}{100000}\right)=1000\times 0.9555=956\text{m/s}.$ Still short — and note it grew slowly. Diminishing returns are showing.

4. Take the limit $n\to \infty$. Write the ratio explicitly and divide top and bottom by $n$: $\frac{m_0}{m_1}=\frac{60000+25000n}{60000+5000n}=\frac{60000/n+25000}{60000/n+5000}\underset{n\to \infty }{\to }\frac{25000}{5000}=5.$ Why this step? As $n$ grows, the fixed core mass $60000$ becomes negligible next to the booster terms that scale with $n$, so the ratio is governed purely by booster fuel-to-total. The $60000/n$ terms vanish, leaving the pure booster ratio $25000:5000=5$. This is the ceiling. $\Delta v_1^{\max }=1000\ln 5=1000\times 1.609=1609\text{m/s}.$

5. Conclusion: since even $n\to \infty$ gives only $1609<2400$ m/s, no number of these boosters reaches the target. You must raise the booster $v_e$ or lower their structural mass, not add more.

Recall Verify

$n=3\to 588$, $n=8\to 956$, limit $\to 1609$ m/s, all below the required $2400$. Correct conclusion: unachievable with these boosters. ✓

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Example 7 — Cell G: exam twist, spot the wrong mass ratio

Forecast: Which student double-counts the fuel already burnt in phase 1?

Reminder: $m_{p,c}^{(1)}$ = propellant burnt in phase 1; $m_{p,c}^{(2)}$ = propellant left for phase 2; $m_2$ = mass at the start of phase 2 (after boosters have gone).

1. Split the propellant. $m_{p,c}^{(1)}=\frac{1}{3}\times 36000=12000$ kg. Remaining $m_{p,c}^{(2)}=36000-12000=24000$ kg. Why this step? Phase 2 can only burn what's left — Student A's formula pretends the $12000$ kg is still onboard.

2. Correct phase-2 masses (Student B): $m_2=m_{s,c}+m_{p,c}^{(2)}+m_{\text{upper}}=4000+24000+6000=34000$; final $m_{2,f}=m_{s,c}+m_{\text{upper}}=4000+6000=10000$. $\Delta v_2=v_{e,c}\ln \left(\frac{m_2}{m_{2,f}}\right)=3200\ln \left(\frac{34000}{10000}\right)=3200\times 1.2238\approx 3916\text{m/s}.$

3. Student A's wrong value. Using all $36000$ kg: $m_0=4000+36000+6000=46000$, giving $3200\ln \left(\frac{46000}{10000}\right)=3200\times 1.526=4884\text{m/s}$ — inflated by $\approx 968$ m/s of phantom fuel. Why this step? Comparing the two numbers pins the error: the surplus is exactly the $\Delta v$ the already-spent $12000$ kg would falsely provide.

Recall Verify

Correct $\Delta v_2\approx 3916$ m/s (Student B). Student A's $4884$ m/s overcounts the $12000$ kg already spent — exactly the "whole-propellant" error the parent note flags. ✓

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Example 8 — Cell H: unit trap ($I_{sp}$ seconds vs $v_e$ m/s)

Forecast: What did the student forget to multiply by?

Reminder: $I_{sp}$ is exhaust velocity expressed in seconds; to feed Tsiolkovsky you need $v_e$ in m/s, related by $v_e=I_{sp}g$ with $g=9.81{\text{m/s}}^2$. See Specific impulse.

1. Convert $I_{sp}$ to exhaust velocity. $v_e=I_{sp}g=452\times 9.81=4434\text{m/s}$. Why this step? Tsiolkovsky wants $v_e$ in m/s, not $I_{sp}$ in seconds; using $452$ directly gives an answer in seconds$\times$(dimensionless) — the wrong units, off by the factor $g$.

2. Now apply Tsiolkovsky. The mass ratio $m_0/m_f=200000/60000=10/3\approx 3.333$. $\Delta v=v_e\ln \left(\frac{m_0}{m_f}\right)=4434\ln \left(\frac{200000}{60000}\right)=4434\times 1.2040\approx 5339\text{m/s}.$

3. Student's raw error. $452\times 1.204\approx 544$ — this is exactly $5339/9.81$, i.e. the right answer shrunk by the missing $g$. Why this step? Recognising that the wrong number equals the right one divided by $9.81$ confirms the single cause was the dropped $g$ factor.

Recall Verify

$v_e=452\times 9.81=4434.12$ m/s; $\Delta v\approx 5339$ m/s. The bad answer $544\approx 5339/9.81$ confirms the missing $g$. Units check: $(\text{s})\times ({\text{m/s}}^2)=\text{m/s}$ for $v_e$. ✓

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Recall Self-test

Series total $\Delta v$ from Example 1 ::: about 9640 m/s Thrust-weighted ${\bar{v}}_e$ in Example 2 ::: about 2966 m/s Why does $ϵ\to 1$ give $\Delta v=0$? ::: no fuel, so $m_0=m_f$ and $\ln 1=0$ The unit slip in Example 8 ::: forgot $v_e=I_{sp}\times g$

Related: Tsiolkovsky rocket equation · Gravity losses · Falcon Heavy · Saturn V · Payload fraction optimization.