Rocket staging — series staging, parallel staging: Visual walkthrough

This page is the visual companion to the parent note Rocket staging. It leans on the Tsiolkovsky rocket equation and on Structural fraction and propellant fraction.

Step 1 — What a rocket actually is: three piles of mass

WHY. Every rocket formula is a story about which pile stays and which pile leaves. If we don't separate the piles now, we will confuse "fuel that leaves as exhaust" with "empty tank that we drop later" — and those two do completely different things to our speed. So we draw them apart first.

PICTURE.

Figure — Rocket staging — series staging, parallel staging

Step 2 — The one rule of rockets: throw mass back, get pushed forward

WHY use $v_{e}$ and not "thrust" or "force"? Force changes every second as the rocket lightens, which makes it awkward to add up. But the exhaust speed $v_{e}$ stays roughly constant for a given engine (it's set by Specific impulse). A constant number is far easier to reason with — so we build everything on $v_{e}$ . Typical chemical engines have $v_{e}$ of a few thousand metres per second; we'll use realistic values like $3000$ – $3500$ m/s later.

PICTURE.

Step 3 — Why a logarithm shows up (adding a million tiny kicks)

WHY the logarithm and not simple multiplication? Because each kick is smaller than the last — the rocket is heavier at the start (more mass to move) and lighter at the end. Rearranging Step 2 gives $d v = - v_{e} \frac{d m}{m}$ . The quantity $\frac{d m}{m}$ is a fractional change of mass, and summing fractional changes from $m_{0}$ down to $m_{f}$ produces $ln (m_{0} / m_{f})$ . The natural log $ln$ is precisely "the running total of fractional changes."

PICTURE.

Step 4 — The trap: carrying an empty tank is dead weight

WHY this matters. The mass ratio lives inside the log, and the log flattens fast — doubling the ratio does not double the speed (look again at the shrinking area in Step 3). So a heavy dead $m_{f}$ is punished hard. This is the "vicious cycle" from the parent note, seen as a picture.

PICTURE.

Step 5 — The idea of staging: drop the tank, reset the ratio

WHY it wins. Each stage now sees a fresh, favourable mass ratio, because it never carries a spent tank. The two logs don't multiply — they add, because $Δ v$ is a speed and speeds accumulate. Two decent ratios added beat one crippled ratio.

PICTURE.

After burnout we jettison $m_{s, 1}$ — that gray block falls away and never appears in stage 2's numbers.

Step 6 — Stage 2 sees a cleaner ratio; the speeds add

WHY add and not multiply? Stage 2 begins already moving at $Δ v_{1}$ (velocity is relative; the rocket doesn't "forget" its speed at separation). The new burn adds $Δ v_{2}$ on top. Total speed is the sum.

PICTURE.

Step 7 — Put numbers on it (the worked two-stage rocket)

WHY these particular numbers? The exhaust velocities are realistic chemical-engine values (Step 2 flagged the few-thousand-m/s range): the upper stage uses $v_{e, 2} = 3500$ m/s (upper stages run in vacuum, where nozzles are more efficient → higher $v_{e}$ ), while the lower stage uses $v_{e, 1} = 3000$ m/s (it fights dense atmosphere, so a bit lower). The propellant fractions come from Structural fraction and propellant fraction.

WHY top-down? A lower stage's "payload" is everything above it. You can't know what stage 1 must lift until you know stage 2's total hardware mass $S_{2}$ . So we start from the tip and work down.

PICTURE.

Worked example Two-stage series rocket

Given. Final payload $m_{pl} = 1000$ kg; $v_{e, 1} = 3000$ m/s, $v_{e, 2} = 3500$ m/s; $λ_{2} = 0.87, ϵ_{2} = 0.13$ ; $λ_{1} = 0.85, ϵ_{1} = 0.15$ .

Stage 2 — choose hardware $S_{2} = 8000$ kg:

$m_{p, 2} = λ_{2} S_{2} = 0.87 \times 8000 = 6960$ kg, $m_{s, 2} = ϵ_{2} S_{2} = 0.13 \times 8000 = 1040$ kg
$m_{0} = S_{2} + m_{pl} = 8000 + 1000 = 9000$ kg, $m_{f} = m_{s, 2} + m_{pl} = 1040 + 1000 = 2040$ kg $Δ v_{2} = 3500 ln (\frac{9000}{2040}) = 3500 \times 1.4839 \approx 5194 m/s$

Stage 1 — choose hardware $S_{1} = 90000$ kg; it must lift stage 2 + payload $= S_{2} + m_{pl} = 8000 + 1000 = 9000$ kg:

$m_{p, 1} = λ_{1} S_{1} = 0.85 \times 90000 = 76500$ kg, $m_{s, 1} = ϵ_{1} S_{1} = 0.15 \times 90000 = 13500$ kg
$m_{0} = S_{1} + 9000 = 99000$ kg, $m_{f} = m_{s, 1} + 9000 = 22500$ kg $Δ v_{1} = 3000 ln (\frac{99000}{22500}) = 3000 \times 1.4816 \approx 4445 m/s$

Total: $Δ v \approx 5194 + 4445 = 9639 m/s$ — enough (with Gravity losses budgeted) for low Earth orbit.

Step 8 — Edge cases: what if we don't stage, or stage badly?

Case A — one stage, never drop the tank. Then $m_{f}$ always contains the whole structure. As you add propellant, $m_{0} / m_{f}$ rises but saturates: the log flattens, $Δ v$ hits a ceiling. This is the Step-4 trap in numbers.

Case B — zero structure ( $ϵ \to 0$ ). A magical weightless tank. Then $m_{f} \to m_{pl}$ and the ratio can be huge — the ideal every engineer chases via Payload fraction optimization. Unreachable, but it shows structure is the enemy.

Case C — all structure, no fuel ( $λ \to 0$ ). Then $m_{0} = m_{f}$ , so $m_{0} / m_{f} = 1$ and $ln 1 = 0$ : zero speed. A stage with no propellant does nothing — it's pure dead weight. This is the degenerate floor.

Case D — infinitely many tiny stages. Dropping structure continuously would give the theoretical maximum, but each separation is a failure point, so real rockets use 2–3 stages (Falcon Heavy, Saturn V).

PICTURE.

The one-picture summary

Recall Feynman retelling — the whole walkthrough in plain words

A rocket is three piles: fuel, empty structure, and payload. It goes faster by throwing fuel out the back at a fixed speed $v_{e}$ . Each little scoop of fuel gives a little kick; when you add up all the kicks (that's the integral from $m_{0}$ down to $m_{f}$ ) you get a logarithm of the mass ratio — Tsiolkovsky, $Δ v = v_{e} ln (m_{0} / m_{f})$ . The problem: the log flattens fast, and an empty tank sitting in $m_{f}$ chokes your ratio — with $10%$ structure you can't beat a ratio of $10$ , capping you below orbital speed. So we cheat: build the rocket in stacked stages, burn one, then drop its dead tank so the next stage starts light. Each clean stage gets its own healthy log, and because velocities just add, the two logs stack up into a big total. Working top-down (stage 2 carries only the payload $m_{pl}$ ; stage 1 carries stage 2's hardware $S_{2}$ too), our example with $v_{e, 1} = 3000$ and $v_{e, 2} = 3500$ gives about 5.2 km/s + 4.4 km/s ≈ 9.6 km/s — orbit. And the edge cases keep us honest: no fuel means $ln 1 = 0$ (nothing), no structure means an enormous ratio (dream), and never dropping the tank means you hit a wall no amount of fuel can climb.

Recall Quick self-test

Why do stage $Δ v$ 's add instead of multiply? ::: Because $Δ v$ is a velocity; the rocket keeps its speed through separation, so each burn adds its gain on top of the previous. What makes staging beat a single stage? ::: Dropping the empty structure keeps $m_{f}$ small for every stage, so each mass ratio (and its log) stays large. What is $Δ v$ for a stage with no propellant? ::: Zero — $m_{0} = m_{f}$ gives $ln (1) = 0$ . What does $S_{i}$ mean? ::: The hardware mass of stage $i$ , $S_{i} = m_{p, i} + m_{s, i}$ (its own propellant plus structure, not counting the payload it carries).