Altitude compensation methods — nozzle extension, aerospike: Worked examples

This page is the "drill ground" for Altitude compensation methods — nozzle extension, aerospike. The parent note built the ideas; here we grind through every kind of case a problem can throw at you, so no scenario surprises you in an exam or a design meeting.

Before we start, we earn every symbol this page uses.

The single equation we lean on the most:

The scenario matrix

Every problem in this topic lands in one of these cells. The worked examples below are labelled with the cell(s) they cover, and together they hit all of them. (Now that $ϵ$ and its link to $p_{e}$ are defined above, Cell E reads cleanly: pushing $ϵ \to \infty$ means widening the mouth without limit, which drives $p_{e} \to 0$ .)

#	Cell (the case class)	What makes it distinct	Covered by
A	$p_{e} > p_{a}$ — under-expanded	pressure term positive; energy "left on the table"	Ex 1
B	$p_{e} = p_{a}$ — perfectly expanded	pressure term zero; the design sweet-spot	Ex 2
C	$p_{e} < p_{a}$ — over-expanded	pressure term negative; atmosphere pushes back	Ex 3
D	Zero / degenerate: $p_{a} = 0$ (vacuum)	ambient term vanishes; upper-stage regime	Ex 4
E	Limiting value: $p_{e} \to 0$ as $ϵ \to \infty$	diminishing returns, $v_{e}$ saturates	Ex 5
F	Geometry from length ( $ϵ$ from $L$ , $α$ )	conical-nozzle trig, extendable case	Ex 6
G	Real-world word problem: same engine, two altitudes	one nozzle, thrust flips sign of pressure term	Ex 7
H	Exam twist: aerospike vs bell, "which wins where?"	conceptual + break-even reasoning	Ex 8

Constants used throughout: $γ = 1.2$ = heat capacity ratio (a gas property setting how it expands); chamber pressure and areas are given per problem.

Example 1 — Cell A: Under-expanded ( $p_{e} > p_{a}$ )

Worked example A short nozzle at sea level

A nozzle exits at $p_{e} = 150$ kPa. It fires at sea level, $p_{a} = 101$ kPa. Given $\overset{m}{˙} = 200$ kg/s, $v_{e} = 2600$ m/s, $A_{e} = 0.5$ m². Find the thrust and classify the expansion.

Forecast: Guess first — is the pressure term helping or hurting? Since the gas pushes out harder ( $150$ ) than the air pushes back ( $101$ ), it should add thrust. Will it be a big or small correction?

Step 1 — Classify. Compare $p_{e}$ and $p_{a}$ : $150 > 101$ , so $p_{e} > p_{a}$ → under-expanded. Why this step? The sign of $(p_{e} - p_{a})$ decides everything; classifying first tells you the answer's flavour before you compute.

Step 2 — Momentum term. $\overset{m}{˙} v_{e} = 200 \times 2600 = 520, 000$ N. Why this step? This is the forward kick from throwing mass backward — it exists no matter the altitude, so it is the "base" of the thrust we build on.

Step 3 — Pressure term. $(p_{e} - p_{a}) A_{e} = (150, 000 - 101, 000) \times 0.5 = 49, 000 \times 0.5 = 24, 500$ N. Why this step? Convert kPa → Pa (×1000) so units match ( $Pa \cdot m^{2} = N$ ). The term is positive, confirming our forecast.

Step 4 — Total. $F = 520, 000 + 24, 500 = 544, 500$ N. Why this step? Thrust is the sum of the two hands acting on the gas: the momentum push plus the pressure imbalance, exactly as the master equation states.

Verify: Units: $kg/s \cdot m/s = kg\cdotm/s^{2} = N$ ✓, $Pa \cdot m^{2} = N$ ✓. The pressure term is only $\sim 4.5%$ of thrust — under-expansion "wastes" some potential velocity, but at least it never subtracts thrust. Sanity: total > momentum term, as expected for $p_{e} > p_{a}$ .

Example 2 — Cell B: Perfectly expanded ( $p_{e} = p_{a}$ )

Worked example The design point

Same engine, now at the altitude where $p_{a}$ equals $p_{e} = 66$ kPa for a redesigned nozzle. Given $\overset{m}{˙} = 200$ kg/s, $v_{e} = 2700$ m/s, $A_{e} = 0.6$ m², $p_{e} = 66$ kPa, $p_{a} = 66$ kPa. Find thrust.

Forecast: With $p_{e} = p_{a}$ , the two hands push equally — the pressure term should vanish. Predict $F = \overset{m}{˙} v_{e}$ exactly.

Step 1 — Classify. $p_{e} = p_{a} = 66$ kPa → perfectly expanded. This is the design sweet-spot for this altitude. Why this step? Perfect expansion means all pressure energy became velocity; nothing is wasted, nothing is fought.

Step 2 — Pressure term. $(66, 000 - 66, 000) \times 0.6 = 0$ N. Why this step? The whole point of "perfect" is this term dies — thrust is pure momentum.

Step 3 — Total. $F = 200 \times 2700 + 0 = 540, 000$ N. Why this step? With the pressure hand exactly balanced, the master equation collapses to just the momentum term — this is the cleanest possible thrust.

Verify: Units ✓. Because the pressure term is zero, $F$ is exactly the momentum term — the cleanest possible case. A fixed nozzle only hits this at ONE altitude; that single point is why altitude compensation exists.

Example 3 — Cell C: Over-expanded ( $p_{e} < p_{a}$ )

Worked example A vacuum nozzle dragged down to sea level

A big nozzle exits at $p_{e} = 40$ kPa but is fired at sea level $p_{a} = 101$ kPa. Given $\overset{m}{˙} = 200$ kg/s, $v_{e} = 2900$ m/s, $A_{e} = 1.2$ m². Find thrust, and comment on danger.

Forecast: Now the air ( $101$ ) pushes back harder than the gas pushes out ( $40$ ). Predict the pressure term is negative — thrust drops below the momentum term.

Step 1 — Classify. $40 < 101$ , so $p_{e} < p_{a}$ → over-expanded. Why this step? This is the dangerous regime the parent note warned about — atmosphere pushes into the nozzle.

Step 2 — Momentum term. $200 \times 2900 = 580, 000$ N. Why this step? Same as always — the forward kick from ejecting mass; it does not care about the outside air, so we compute it independently before folding in the pressure fight.

Step 3 — Pressure term. $(40, 000 - 101, 000) \times 1.2 = (- 61, 000) \times 1.2 = - 73, 200$ N. Why this step? The minus sign is the whole lesson: atmosphere subtracts thrust here.

Step 4 — Total. $F = 580, 000 - 73, 200 = 506, 800$ N. Why this step? We add the two hands as the master equation demands; here one hand pushes forward and one pulls back, so the total falls below the momentum term.

Verify: Units ✓. Thrust is below the momentum term — exactly the over-expansion penalty. Danger note: if $p_{e}$ dropped far enough, the exhaust would separate from the wall (flow separation) causing side loads. Rule of thumb: separation risks appear when $p_{e} ≲ 0.4 p_{a}$ . Here $40/101 = 0.40$ , right at the edge — a real warning flag.

Example 4 — Cell D: Vacuum / degenerate input ( $p_{a} = 0$ )

Worked example The same over-expanded nozzle, now in space

Take the Example-3 nozzle ( $p_{e} = 40$ kPa, $A_{e} = 1.2$ m², $\overset{m}{˙} = 200$ kg/s) but now in vacuum: $p_{a} = 0$ . Its exit velocity in vacuum is slightly higher, $v_{e} = 2950$ m/s (less back-pressure lets gas accelerate a touch more). Find thrust.

Forecast: With $p_{a} = 0$ , the atmosphere's hand disappears entirely. The pressure term becomes purely positive ( $p_{e} \cdot A_{e}$ ). Predict a big jump versus Example 3.

Step 1 — Set $p_{a} = 0$ . This is the degenerate limit; the master equation simplifies to $F = \overset{m}{˙} v_{e} + p_{e} A_{e}$ . Why this step? In vacuum there is nothing pushing back — the "over-expanded" nozzle of Ex 3 is now a good nozzle.

Step 2 — Momentum term. $200 \times 2950 = 590, 000$ N. Why this step? We recompute it because $v_{e}$ rose slightly (no back-pressure to fight), so the forward kick is a touch larger than at sea level.

Step 3 — Pressure term. $(40, 000 - 0) \times 1.2 = 48, 000$ N. Why this step? Same nozzle, but with $p_{a} = 0$ the sign flipped from $- 73, 200$ N (Ex 3) to $+ 48, 000$ N — that's the altitude story in one number.

Step 4 — Total. $F = 590, 000 + 48, 000 = 638, 000$ N. Why this step? Both hands now push forward, so the master equation's sum is the momentum kick plus the full exit-pressure push — the largest thrust this nozzle can give.

Verify: Units ✓. Compare: Ex 3 gave $506, 800$ N at sea level, Ex 4 gives $638, 000$ N in vacuum — the identical nozzle produces $\sim 26%$ more thrust in space. This is exactly why big- $ϵ$ nozzles live on upper stages.

Example 5 — Cell E: The limiting value ( $p_{e} \to 0$ , diminishing returns)

Before the numbers, we earn three new symbols this example needs:

Now we earn the formula itself before using it — where it comes from and why the strange exponent appears.

Intuition Where the exhaust-velocity formula comes from (and its two assumptions)

The parent note's boxed formula is not magic — it is energy accounting for the gas, resting on exactly two assumptions:

Ideal gas — the exhaust obeys $p = ρR T$ , so its energy per kilogram is a clean $c_{p} T$ . This lets us talk about "thermal energy" as just a temperature.
Isentropic (loss-free, adiabatic) expansion — no heat leaks through the walls and no friction wastes energy; every joule of thermal energy the gas gives up becomes ordered kinetic energy of the jet.

Under those two, energy conservation from chamber (subscript $0$ , nearly still) to exit (subscript $e$ , fast) reads $thermal energy in chamber c_{p} T_{0} = thermal left at exit c_{p} T_{e} + kinetic gained \frac{1}{2} v_{e}^{2} \Rightarrow \frac{1}{2} v_{e}^{2} = c_{p} (T_{0} - T_{e}) .$ Why energy conservation and not force balance? We want a speed, and the cleanest bookkeeping of "how much heat turned into motion" is energy in = energy out. Force balance would need the full pressure profile down the nozzle; energy only needs the two endpoints.

Intuition Why the exponent is

$(γ - 1) / γ$ — the isentropic relation We still need $T_{e}$ in terms of the pressures. For an isentropic (constant-entropy) ideal-gas process, temperature and pressure are locked together by $\frac{T _{e}}{T _{0}} = (\frac{p _{e}}{p _{0}})^{(γ - 1) / γ} .$ Why this exact power? Entropy staying constant forces $p V^{γ} = const$ (gas that expands cools by a fixed rule set by $γ$ ). Combining that with the ideal-gas law $p V = R T$ to eliminate volume $V$ leaves temperature tied to pressure by the leftover exponent $(γ - 1) / γ$ . So the odd-looking power is simply "what's left of $γ$ after you trade volume for temperature." A bigger $γ$ (stiffer gas) means a smaller exponent, so the gas cools less for the same pressure drop.

Substituting this $T_{e}$ back into $\frac{1}{2} v_{e}^{2} = c_{p} (T_{0} - T_{e})$ with $c_{p} = γ R / (γ - 1)$ gives the boxed result below — every symbol now earned.

Worked example Doubling the expansion twice

Using the parent note's boxed formula $v_{e} = \frac{2 γ}{γ - 1} R T_{0} [1 - (\frac{p _{e}}{p _{0}})^{(γ - 1) / γ}],$ with $T_{0} = 3500$ K, $R = 400$ J/(kg·K), $p_{0} = 3$ MPa, $γ = 1.2$ , compute $v_{e}$ for three exit pressures: $p_{e} = 30$ kPa, then $15$ kPa, then $7.5$ kPa (each halving). Show that $v_{e}$ gains shrink each time.

Forecast: Halving $p_{e}$ each time — do you get equal velocity jumps? Guess: no, the gains get smaller because $v_{e}$ saturates.

Step 1 — Recall the formula (just derived above). $v_{e} = \frac{2 γ}{γ - 1} R T_{0} [1 - (\frac{p _{e}}{p _{0}})^{(γ - 1) / γ}]$ Why this step? This is the correct scaling — the naïve $v_{e} \propto ϵ$ is wrong. Everything that varies lives inside that bracket; the rest is a fixed constant. The bracket is literally the fraction of chamber thermal energy converted to motion (from the energy balance above).

Step 2 — The prefactor. $\frac{2 γ}{γ - 1} R T_{0} = \frac{2.4}{0.2} \times 400 \times 3500 = 12 \times 1, 400, 000 = 16, 800, 000$ . Why this step? Compute the constant once; only the bracket changes with $p_{e}$ , so this saves repeating work three times.

Step 3 — The bracket at each $p_{e}$ (exponent $(γ - 1) / γ = 0.2/1.2 = 1/6$ , the isentropic power from above):

$p_{e} = 30$ kPa: $p_{e} / p_{0} = 0.01$ , bracket $= 1 - 0.0 1^{1/6} = 1 - 0.4642 = 0.5358$
$p_{e} = 15$ kPa: $p_{e} / p_{0} = 0.005$ , bracket $= 1 - 0.00 5^{1/6} = 1 - 0.4136 = 0.5864$
$p_{e} = 7.5$ kPa: $p_{e} / p_{0} = 0.0025$ , bracket $= 1 - 0.002 5^{1/6} = 1 - 0.3686 = 0.6314$ Why this step? The bracket is the fraction of thermal energy that got converted to motion; it grows toward $1$ as $p_{e}$ falls, so its growth is what we track.

Step 4 — Velocities $v_{e} = 16, 800, 000 \times bracket$ :

$v_{e} (30) = 16.8 M \times 0.5358 \approx 3000$ m/s
$v_{e} (15) = 16.8 M \times 0.5864 \approx 3139$ m/s → gain +139 m/s
$v_{e} (7.5) = 16.8 M \times 0.6314 \approx 3257$ m/s → gain +118 m/s Why this step? Taking the square root converts the energy bracket into an actual speed, and comparing successive speeds exposes the shrinking gains.

Verify: The two gains ( $139$ then $118$ m/s) are for equal-ratio halvings, yet the second gain is smaller — diminishing returns, confirming $v_{e}$ saturates as $p_{e} \to 0$ . Units: $J/kg = m^{2} / s^{2} = m/s$ ✓. See the flattening curve below.

Figure walkthrough (below): The horizontal axis is exit pressure $p_{e}$ in kPa; the vertical axis is exhaust velocity $v_{e}$ in m/s. The blue curve rises steeply on the right (high $p_{e}$ ) but flattens toward the left as $p_{e} \to 0$ . The three coloured dots (green $30$ , orange $15$ , red $7.5$ kPa) sit ever closer together in height even though each halves the pressure — that visual bunching is the diminishing return. Follow the blue curve leftward: it approaches a ceiling and never shoots up, which is why doubling $ϵ$ forever buys less and less speed.

Figure — Altitude compensation methods — nozzle extension, aerospike

Example 6 — Cell F: Geometry — expansion ratio from cone length

Worked example Extendable nozzle: stowed vs deployed

A conical nozzle has throat radius $R_{t} = 0.10$ m and half-angle $α = 1 5^{\circ}$ . Stowed length $L_{1} = 1.0$ m; deployed length $L_{2} = 2.5$ m. Find $ϵ_{1}$ and $ϵ_{2}$ .

Forecast: Longer cone → wider mouth → bigger $ϵ$ . Guess whether deploying (2.5× the length) gives roughly $2.5 \times$ , more, or less than $2.5 \times$ the expansion ratio.

Step 1 — Exit radius grows linearly with length. $R_{e} = R_{t} + L tan α$ . Why this step? In a straight-walled cone, walk a distance $L$ down the axis and the wall has climbed $L tan α$ outward. Here $tan α$ = opposite/adjacent = (how far the wall rises) / (how far you walked) — that ratio is the wall's steepness. Look at the red wall in the figure.

Step 2 — Compute $tan 1 5^{\circ} = 0.2679$ .

$R_{e, 1} = 0.10 + 1.0 \times 0.2679 = 0.3679$ m
$R_{e, 2} = 0.10 + 2.5 \times 0.2679 = 0.7698$ m Why this step? Plugging the two lengths into the linear rule gives the mouth radius in each configuration — the raw geometry we need before squaring.

Step 3 — Expansion ratio is area ratio, and area $\propto R^{2}$ : $ϵ = (\frac{R _{e}}{R _{t}})^{2}$ Why this step? $ϵ = A_{e} / A_{t} = π R_{e}^{2} / π R_{t}^{2}$ ; the $π$ cancels, leaving a squared radius ratio.

$ϵ_{1} = (0.3679/0.10)^{2} = 3.67 9^{2} = 13.5$
$ϵ_{2} = (0.7698/0.10)^{2} = 7.69 8^{2} = 59.3$

Verify: Deploying multiplied length by $2.5$ but $ϵ$ by $59.3/13.5 = 4.4 \times$ — more than linear, because $ϵ$ scales with radius squared. Units: ratio is dimensionless ✓. Geometry below.

Figure walkthrough (below): The horizontal axis is distance $x$ along the nozzle axis (m); the vertical axis is radius (m), drawn above and below the dotted grey centre-line so you see the full funnel. The blue double-arrow at $x = 0$ marks the throat width $2 R_{t}$ . The red wall is the stowed cone (throat to $L_{1}$ ); the dashed orange wall is the deployed extension ( $L_{1}$ to $L_{2}$ ). The green arrow measures the stowed mouth radius $R_{e, 1}$ and the black arrow the deployed mouth radius $R_{e, 2}$ — notice the mouth roughly doubles in radius but, because area goes as radius squared, $ϵ$ more than quadruples.

Example 7 — Cell G: Real-world word problem (one nozzle, two altitudes)

Worked example Sea level vs 30 km — same fixed nozzle

A first-stage engine has a fixed nozzle: $p_{e} = 70$ kPa, $A_{e} = 0.9$ m², $\overset{m}{˙} = 300$ kg/s. Exhaust velocity $\approx 2800$ m/s (treat as roughly constant here). Compare thrust at sea level ( $p_{a} = 101$ kPa) and at 30 km altitude ( $p_{a} \approx 1.2$ kPa).

Forecast: At sea level $p_{e} < p_{a}$ (over-expanded, penalty). High up, $p_{e} > p_{a}$ (under-expanded, bonus). Predict thrust rises with altitude for this fixed nozzle.

Step 1 — Momentum term (same both places). $300 \times 2800 = 840, 000$ N. Why this step? $\overset{m}{˙}$ and $v_{e}$ barely change with altitude, so this forward kick is a fixed baseline; only the pressure term moves as we climb.

Step 2 — Sea level pressure term. $(70, 000 - 101, 000) \times 0.9 = - 31, 000 \times 0.9 = - 27, 900$ N. → over-expanded, negative. Why this step? At sea level the dense air out-pushes the exit gas, so this hand pulls thrust down.

Step 3 — 30 km pressure term. $(70, 000 - 1, 200) \times 0.9 = 68, 800 \times 0.9 = 61, 920$ N. → under-expanded, positive. Why this step? High up the air is thin, so the same exit gas now out-pushes it and this hand adds thrust — the sign has flipped.

Step 4 — Totals.

Sea level: $840, 000 - 27, 900 = 812, 100$ N.
30 km: $840, 000 + 61, 920 = 901, 920$ N. Why this step? Adding the fixed momentum kick to each altitude's pressure term (via the master equation) turns the sign-flip into the real thrust each place delivers.

Verify: Thrust climbs from $812, 100$ N to $901, 920$ N as the rocket ascends — a real effect every launch shows. Gain $= 89, 820$ N ( $\sim 11%$ ). This is exactly the inefficiency altitude compensation attacks: the fixed nozzle is only "right" at one crossover altitude in between (where $p_{a} = 70$ kPa). Units ✓.

Example 8 — Cell H: Exam twist — aerospike vs bell break-even

Worked example "Which nozzle wins, and where?"

Two engines share the same momentum term $\overset{m}{˙} v_{e} = 800, 000$ N and same $A_{e} = 1.0$ m². The bell is fixed with $p_{e} = 20$ kPa (vacuum-optimised). The aerospike self-adjusts so that $p_{e} \approx p_{a}$ at every altitude (perfectly expanded everywhere, ideally). At what ambient pressure does the bell catch up to the aerospike, and who wins at sea level ( $p_{a} = 101$ kPa)?

Forecast: The aerospike's pressure term is always $\approx 0$ (perfectly expanded), so its thrust $\approx 800, 000$ N always. The bell's pressure term is $(20, 000 - p_{a}) \times 1.0$ . Guess: the bell only ties/beats the aerospike when $p_{a}$ is low enough.

Step 1 — Aerospike thrust. $F_{s p ik e} = 800, 000 + (p_{a} - p_{a}) \times 1.0 = 800, 000$ N at all altitudes. Why this step? Its defining property — $p_{e} = p_{a}$ — kills the pressure term at every altitude. That's the whole selling point of an aerospike.

Step 2 — Bell thrust. $F_{b e l l} = 800, 000 + (20, 000 - p_{a}) \times 1.0$ . Why this step? The bell has a fixed $p_{e} = 20$ kPa, so its pressure term swings with $p_{a}$ instead of tracking it — that's why it can only be right at one altitude.

Step 3 — Break-even. Set $F_{b e l l} = F_{s p ik e}$ : $20, 000 - p_{a} = 0 \Rightarrow p_{a} = 20$ kPa (roughly $12$ km altitude). Why this step? Equating the two thrusts isolates the crossover ambient pressure — above $20$ kPa the bell loses, below it the bell ties/wins, because that is exactly where its fixed $p_{e}$ finally matches ambient.

Step 4 — Sea level. At $p_{a} = 101$ kPa: $F_{b e l l} = 800, 000 + (20, 000 - 101, 000) \times 1.0 = 800, 000 - 81, 000 = 719, 000$ N, versus aerospike $800, 000$ N. Why this step? Plugging the sea-level ambient into the bell's swinging pressure term shows the concrete penalty the fixed nozzle pays low down, where the aerospike is designed to shine.

Verify: At sea level the aerospike beats the bell by $800, 000 - 719, 000 = 81, 000$ N ( $\sim 11%$ ) — its headline advantage. Break-even at $p_{a} = 20$ kPa matches $p_{e} = 20$ kPa (the bell is perfectly expanded exactly there, Cell B). Above $12$ km the fixed bell is under-expanded (Cell A) and effectively matches the ideal aerospike. Units ✓.

Conclusion / interpretation: This is the standard exam takeaway. The aerospike wins the entire low-altitude climb because it self-compensates and never pays a pressure penalty; the fixed bell only rivals it high up, once thin air finally matches its single design pressure. That is why aerospikes are attractive for single-stage-to-orbit vehicles that must be efficient from the pad to space, while bells with extendable sections (Examples 4–6) are the practical choice for upper stages that only fire near vacuum.

Recall Self-test before you leave

The sign of the pressure term when $p_{e} < p_{a}$ ::: Negative (over-expanded — atmosphere subtracts thrust) What happens to $(p_{e} - p_{a}) A_{e}$ in vacuum ::: $p_{a} = 0$ , so it becomes $+ p_{e} A_{e}$ (always positive) Why doubling $ϵ$ gives shrinking $v_{e}$ gains ::: The bracket $[1 - (p_{e} / p_{0})^{(γ - 1) / γ}] \to 1$ , so $v_{e}$ saturates $ϵ$ from cone geometry ::: $ϵ = (R_{t} + L tan α)^{2} / R_{t}^{2}$ Why fixed-nozzle thrust rises with altitude ::: $p_{a}$ falls, flipping the pressure term from negative to positive The aerospike's pressure term at every altitude ::: Approximately zero, since it keeps $p_{e} \approx p_{a}$ The two assumptions behind the $v_{e}$ formula ::: Ideal gas and isentropic (loss-free, adiabatic) expansion Origin of the exponent $(γ - 1) / γ$ ::: The isentropic relation $T_{e} / T_{0} = (p_{e} / p_{0})^{(γ - 1) / γ}$