Mass budgets — dry mass, wet mass, margin: Worked examples

This page is the drill hall for mass budgets. The parent note gave you the definitions and the key formula. Here we hit every kind of problem that mass budgeting throws at you — small satellites, huge cargo ships, burns that reduce mass, margin that erodes, and the two directions of the rocket equation that trip everyone up.

Before we start, one reminder of the two tools we lean on:

Here $v_{e} = I_{s p} \cdot g_{0}$ with $g_{0} = 9.81 m/s^{2}$ — this links to Specific Impulse. The mass ratio idea is the heart of the Tsiolkovsky Rocket Equation.

The figure below is the mental picture behind every example on this page.

Figure — Mass budgets — dry mass, wet mass, margin

Intuition How to read that figure

The horizontal axis is the ratio $Δ V / v_{e}$ — the exponent that feeds $R = e^{Δ V / v_{e}}$ . Small on the left (easy missions), large on the right (demanding missions). The vertical axis is the propellant's share of the whole vehicle, i.e. $\frac{m _{prop}}{m _{wet}} = \frac{R - 1}{R}$ , a number between 0 and 1.

The coral (red) curve is that fraction. Notice it starts near 0 on the left and bends sharply upward — this is the exponential biting.
The dashed mint line at 1.0 is the impossible ceiling "100 % fuel, no ship left" — the curve approaches but never reaches it.
The two lavender dots mark our worked cases: Example 1 (at $Δ V / v_{e} \approx 0.1$ , only ~10 % fuel) and Example 3 (at $Δ V / v_{e} = 3$ , a crushing ~95 % fuel). Read left-to-right, the picture is the story of why mass discipline matters.

The scenario matrix

Every mass-budget problem is one (or a blend) of these cells. The examples below appear in cell order A → B → C → D → E → F → G → H so the tags read straight down the page.

Cell	What makes it distinct	Covered by
A. Sizing forward	Given dry mass + $Δ V$ , find propellant/wet mass	Ex 1
B. Flight backward	Given wet mass, find mass after a burn	Ex 2
C. Zero / tiny $Δ V$	Degenerate limit: $R \to 1$ , propellant $\to 0$	Ex 3
D. Huge $Δ V$ limit	$R ≫ 1$ , propellant dominates dry mass	Ex 4
E. Margin erosion	Growth eats reserve; recompute $%$ margin	Ex 5
F. Dry-mass-growth penalty	1 kg dry $\to$ many kg propellant	Ex 6
G. Multi-burn / staged	Chain two burns, track mass across each	Ex 7
H. Exam twist	Solve backwards for an unknown ( $I_{s p}$ or $Δ V$ )	Ex 8

The mission $Δ V$ numbers come from a Mission ΔV Budget; the empty-mass fraction ideas connect to Structural Mass Fraction.

Cell A — Sizing forward

Worked example Example 1 — Small comms satellite, size the tank

Given: dry mass with margin $m_{dry} = 420$ kg. Station-keeping over life needs $Δ V = 220$ m/s. Bipropellant thruster, $I_{s p} = 230$ s. Find: propellant mass and launch wet mass.

Forecast: $Δ V$ is small compared with $v_{e}$ , so guess propellant is much smaller than dry mass — maybe tens of kg.

Step 1 — Exhaust speed. $v_{e} = I_{s p} g_{0} = 230 \times 9.81 = 2256.3$ m/s. Why this step? The rocket equation only knows $v_{e}$ , not $I_{s p}$ . We must convert seconds into a real speed first (see Specific Impulse).

Step 2 — Mass ratio. $R = e^{Δ V / v_{e}} = e^{220/2256.3} = e^{0.09751} = 1.1024$ . Why this step? $R$ is the single number that says "loaded is 1.10× the empty ship."

Step 3 — Wet mass (sizing, so multiply). $m_{wet} = m_{dry} \cdot R = 420 \times 1.1024 = 463.0$ kg. Why this step? We know the light end (dry) and want the heavy end (loaded), so we multiply by $R$ .

Step 4 — Propellant. $m_{prop} = m_{wet} - m_{dry} = 463.0 - 420 = 43.0$ kg. Why this step? Propellant is simply the mass we added on top of the dry ship, so we subtract dry from wet to isolate it — that is the number the tank must hold.

Verify: Plug back: $v_{e} ln (463.0/420) = 2256.3 \times ln (1.1024) = 2256.3 \times 0.09751 = 220$ m/s ✓. Propellant (43 kg) is ~10 % of dry mass — matches the small- $Δ V$ forecast. Units: kg throughout, $v_{e}$ in m/s cancels against $Δ V$ inside the log. ✓

Cell B — Flight backward (a burn reduces mass)

Worked example Example 2 — Mass left after a burn

Given: a probe enters a burn with wet mass $m_{i} = 2500$ kg. It executes $Δ V = 1200$ m/s with $I_{s p} = 320$ s. Find: mass remaining after the burn, and propellant used.

Forecast: A burn throws mass overboard, so the answer is less than 2500 kg. We divide by $R$ here, not multiply.

Step 1 — Exhaust speed. $v_{e} = 320 \times 9.81 = 3139.2$ m/s. Why this step? The equation is written in real speed, so we first turn the datasheet's $I_{s p}$ (in seconds) into $v_{e}$ (in m/s) before it can enter any formula.

Step 2 — Mass ratio. $R = e^{1200/3139.2} = e^{0.38226} = 1.4656$ . Why this step? $R$ is the same climb-or-slide factor as always; computing it once lets us either multiply (sizing) or divide (flight) — here we will divide.

Step 3 — Final mass (flight, so divide). $m_{f} = m_{i} / R = 2500/1.4656 = 1705.8$ kg. Why this step? We know the heavy start and want the lighter end, so the lighter mass sits under the division bar — the flight direction of the equation.

Step 4 — Propellant consumed. $2500 - 1705.8 = 794.2$ kg. Why this step? Propellant burned is the mass the ship lost, i.e. start minus finish — this is the number we tick off the tank gauge.

Verify: Run the equation forward from the answer: $v_{e} ln (2500/1705.8) = 3139.2 \times ln (1.4656) = 3139.2 \times 0.38226 = 1200$ m/s ✓. If we had wrongly multiplied, we'd get 3664 kg — heavier after a burn, physically absurd. ✓

Cell C — Zero / tiny ΔV (degenerate limit)

Worked example Example 3 — What happens as ΔV → 0?

Given: a 420 kg satellite performing a pointing tweak of only $Δ V = 2$ m/s, $v_{e} = 2256$ m/s. Find: propellant, and confirm the limiting behaviour.

Forecast: Almost no velocity change means almost no propellant. As $Δ V \to 0$ , propellant should $\to 0$ .

Step 1 — Mass ratio. $R = e^{2/2256} = e^{0.000886} = 1.000887$ . Why this step? We want to see how close to 1 the mass ratio gets — this is the flat, far-left part of the curve in the figure above.

Step 2 — Propellant. $m_{prop} = m_{dry} (R - 1) = 420 \times 0.000887 = 0.372$ kg. Why this step? $R - 1$ is the fractional extra mass; multiply by dry mass to get kilograms.

Step 3 — The limit check. For tiny $x$ , $e^{x} \approx 1 + x$ . So $m_{prop} \approx m_{dry} \cdot \frac{Δ V}{v _{e}} = 420 \times \frac{2}{2256} = 0.372$ kg — the exponential collapses to a straight line. Why this step? It proves the formula is smooth at zero: no division-by-zero, no blow-up. At exactly $Δ V = 0$ , $R = 1$ , $m_{prop} = 0$ — you carry no fuel because you do nothing.

Verify: The linear estimate (0.372 kg) and the exact exponential (0.372 kg) agree to 3 decimals, exactly as expected when $Δ V ≪ v_{e}$ . ✓

Cell D — Huge ΔV limit

Worked example Example 4 — Propellant runs away at high ΔV

Given: an upper stage that must supply $Δ V = 9000$ m/s (near single-stage-to-orbit), dry mass $m_{dry} = 1000$ kg, $v_{e} = 3000$ m/s (chemical). Find: wet mass, and see how badly propellant dominates.

Forecast: $Δ V$ is three times $v_{e}$ . On the figure at the top, this sits at $Δ V / v_{e} = 3$ on the horizontal axis, where the coral curve has already bent up near the mint 100 % ceiling. Guess wet mass is many tonnes.

Step 1 — Mass ratio. $R = e^{9000/3000} = e^{3} = 20.09$ . Why this step? The exponent is $Δ V / v_{e} = 3$ , and $e^{3} \approx 20$ — this single number tells us the loaded ship is twenty times the empty ship, so we compute it before anything else.

Step 2 — Wet mass. $m_{wet} = 1000 \times 20.09 = 20, 090$ kg. Why this step? We know the light (dry) end and want the heavy (loaded) end, so — following "heavy multiplies" — we multiply dry mass by $R$ .

Step 3 — Propellant. $m_{prop} = 20090 - 1000 = 19, 090$ kg. Why this step? Subtracting dry from wet isolates propellant; here it is 95 % of the whole vehicle. This is exactly why we stage rockets instead of building one giant tank (Launch Vehicle Performance).

Verify: Fraction of propellant $= 19090/20090 = 0.9502$ , i.e. 95 % — the right-hand lavender dot the figure marks at $Δ V / v_{e} = 3$ . Compare Ex 1 (10 %): moving $Δ V / v_{e}$ from 0.1 to 3 pushed the fuel fraction from tiny to overwhelming — the exponential's signature. ✓

Cell E — Margin erosion

Worked example Example 5 — Watching the reserve shrink

Given: lander CBE $= 800$ kg, policy margin 25 %, so MEV $= 800 \times 1.25 = 1000$ kg. After the design review three growths appear: structure +40 kg, avionics +18 kg, harness +12 kg. Find: remaining margin in kg and in percent.

Forecast: Total growth is 70 kg out of a 200 kg reserve — margin should drop from 25 % toward the mid-teens.

Step 1 — New CBE. $800 + 70 = 870$ kg. Why this step? Growth is real hardware on the ship; it raises the CBE (the actual mass), not the fixed MEV ceiling.

Step 2 — Remaining margin (kg). The MEV ceiling stays at 1000 kg. Reserve $= 1000 - 870 = 130$ kg. Why this step? Margin is defined as the gap between the ceiling (MEV) and the current mass (CBE); subtracting the risen CBE from the fixed MEV gives what reserve is left.

Step 3 — Margin percentage. $\frac{130}{870} = 0.1494 = 14.9%$ . Why this step? Margin percent is always reserve ÷ current-best-estimate, so both numerator and denominator move as mass grows.

Verify: Started at $200/800 = 25%$ , now $14.9%$ — dropped as forecast. Sanity: $870 + 130 = 1000$ kg = the unchanged MEV ceiling. ✓

Cell F — Dry-mass-growth penalty

Worked example Example 6 — What one extra kilogram really costs

Given: Mars transfer stage, $Δ V = 6000$ m/s, $v_{e} = 3000$ m/s. Dry mass grows by $Δ m_{dry} = 12$ kg. Find: extra propellant forced by that growth.

Forecast: Big $Δ V$ means a big penalty — 12 kg of growth should blow up into dozens of kg of extra fuel.

Step 1 — Derive the penalty factor on this page. Hold $Δ V$ fixed (it's a mission requirement). The wet mass is always $m_{wet} = m_{dry} R$ with $R = e^{Δ V / v_{e}}$ , so propellant is $m_{prop} = m_{dry} R - m_{dry} = m_{dry} (R - 1)$ . Growing dry mass by $Δ m_{dry}$ changes propellant by $Δ m_{prop} = (m_{dry} + Δ m_{dry}) (R - 1) - m_{dry} (R - 1) = Δ m_{dry} (R - 1) .$ Why this step? Everything cancels except $Δ m_{dry} (R - 1)$ — the extra fuel is directly proportional to the extra dry mass, with the multiplier $R - 1 = e^{Δ V / v_{e}} - 1$ . That is the penalty factor, derived here, not borrowed.

Step 2 — Evaluate the factor. $R - 1 = e^{6000/3000} - 1 = e^{2} - 1 = 7.389 - 1 = 6.389$ . Why this step? Plug the mission numbers into the factor we just built.

Step 3 — Extra propellant. $Δ m_{prop} = 12 \times 6.389 = 76.7$ kg. Why this step? Scale the per-kg penalty by the actual 12 kg of growth.

Verify: The 12 kg problem "grew" into $12 + 76.7 = 88.7$ kg of extra wet mass — an $88.7/12 = 7.39 \times$ amplification, which equals $e^{2}$ , the full mass ratio. Consistent. ✓

Cell G — Multi-burn / staged bookkeeping

Worked example Example 7 — Two burns in a row

Given: cargo ship launches at $m_{0} = 45000$ kg. Burn 1 (trans-Mars injection): $Δ V_{1} = 3800$ m/s. Burn 2 (Mars orbit insertion): $Δ V_{2} = 2100$ m/s. Both use $I_{s p} = 450$ s. Dry mass $= 5000$ kg. Find: mass after each burn; check enough propellant remains.

Forecast: Two burns dividing successively — each division shrinks the ship. After both, we should still be above 5000 kg dry, or the mission fails.

Step 1 — Exhaust speed. $v_{e} = 450 \times 9.81 = 4414.5$ m/s. Why this step? Both burns use the same engine, so we convert its $I_{s p}$ to $v_{e}$ once and reuse it — no need to redo it per burn.

Step 2 — Burn 1 (divide). $R_{1} = e^{3800/4414.5} = e^{0.86081} = 2.3650$ . $m_{1} = 45000/2.3650 = 19, 027$ kg. Why this step? Flight direction: after the burn the ship is lighter, so we divide.

Step 3 — Burn 2 (divide again from $m_{1}$ ). $R_{2} = e^{2100/4414.5} = e^{0.47569} = 1.6091$ . $m_{2} = 19027/1.6091 = 11, 824$ kg. Why this step? The second burn starts from whatever mass survived burn 1 — you always chain from the previous result.

Step 4 — Feasibility. $m_{2} = 11, 824$ kg $> 5000$ kg dry, so $11824 - 5000 = 6824$ kg propellant remains for later maneuvers. Mission viable. Why this step? The whole point of a mass budget is a go/no-go check: comparing surviving mass against dry mass tells us whether any propellant is left, and hence whether the mission can still be flown.

Verify: Total $Δ V$ from a single combined ratio: $R_{1} R_{2} = 2.3650 \times 1.6091 = 3.806$ , and $v_{e} ln (3.806) = 4414.5 \times 1.3365 = 5900$ m/s $= 3800 + 2100$ ✓. Chaining the ratios reproduces the sum of $Δ V$ s — the log turns products into sums. ✓

Cell H — Exam twist (solve backwards)

Worked example Example 8 — Given the fuel, find the required

$I_{s p}$ Given: you must achieve $Δ V = 4000$ m/s. Your dry mass is fixed at 600 kg and you can only load 900 kg of propellant, so the start (wet) mass is $m_{i} = 600 + 900 = 1500$ kg and the finish (dry) mass is $m_{f} = 600$ kg. The exam asks: what minimum $I_{s p}$ makes this possible? Find: $I_{s p}$ .

Forecast: This inverts every previous example — we know the masses and $Δ V$ , we want the engine. A demanding $Δ V$ with only a modest fuel-to-dry ratio means we need a high-performance engine, so expect $I_{s p}$ well above a cold-gas value.

Step 1 — Mass ratio from the masses. $R = m_{i} / m_{f} = 1500/600 = 2.5$ . Why this step? The masses are fixed, so the achievable ratio is fixed — no engine choice changes it. This is the number the engine must "cash in" as velocity.

Step 2 — Required exhaust speed. Start from $Δ V = v_{e} ln R$ and isolate $v_{e}$ by dividing both sides by $ln R$ : $v_{e} = \frac{Δ V}{l n R} = \frac{4000}{l n 2.5} = \frac{4000}{0.91629} = 4365.5 m/s .$ Why this step? $ln$ is exactly the operation that undoes the exponential in $R = e^{Δ V / v_{e}}$ ; dividing $Δ V$ by $ln R$ recovers the exhaust speed the engine must deliver.

Step 3 — Convert to $I_{s p}$ . $I_{s p} = v_{e} / g_{0} = 4365.5/9.81 = 445.0$ s. Why this step? Engine datasheets quote $I_{s p}$ in seconds, not $v_{e}$ ; dividing by $g_{0}$ turns our exhaust speed into the shopping-list number.

Verify: Forward check with $I_{s p} = 445$ s: $v_{e} = 445 \times 9.81 = 4365.5$ m/s, $Δ V = 4365.5 \times ln 2.5 = 4365.5 \times 0.91629 = 4000$ m/s ✓. An $I_{s p}$ of ~445 s means a high-efficiency bipropellant engine — a cold-gas thruster ( $I_{s p} \approx 65$ s) would fall hopelessly short. ✓

Recall Quick self-test

In these reveal lines, everything before the triple colon is the question; everything after it is the hidden answer — cover the right-hand side and try to answer first.

Sizing vs flight: which operation for each? ::: Sizing multiplies dry mass by $R$ ; flight divides wet mass by $R$ . As $Δ V \to 0$ , what does propellant approach? ::: Zero — $R \to 1$ , so $m_{prop} = m_{dry} (R - 1) \to 0$ . Two burns of $Δ V_{1}$ then $Δ V_{2}$ : how do their mass ratios combine? ::: They multiply, $R = R_{1} R_{2}$ , because the logs of $Δ V$ add. Penalty factor per kg of dry mass growth? ::: $e^{Δ V / v_{e}} - 1$ .

Related deep pages: Propellant Management, Center of Mass Control, Mass Properties Measurement, Systems Engineering V-Model.