The split that is not an equal split

The previous rung established why a launcher is thrown away in pieces, and it did so with equal stages: the same structural coefficient, the same exhaust speed, the same share of the velocity budget each. That simplification is what makes the payload fraction come out as λ₁ⁿ and the argument fit on one figure.

No launcher has ever been built that way. A first stage flies through the atmosphere on a dense propellant with a nozzle sized for sea level; a second stage flies in vacuum on whatever is lightest, with a nozzle sized for nothing. It is the same asymmetry that makes the cheapest place to turn a question about where rather than about how much. They differ in structural fraction and they differ in exhaust speed, and once they do, the question of how to divide the velocity budget between them stops answering itself.

The optimum is not at a half, it is not near a half, and the direction it leans is the interesting part. It leans toward the upper stage — the smaller one, the one that does not leave the pad under its own power — and it does so for a reason that has nothing to do with which stage is bigger.

Why the better stage gets more of the work #

The overall payload fraction is the product of the two per-stage fractions, and each of those is

$$\lambda_i = \frac{e^{-\Delta v_i/v_{e,i}} - \varepsilon_i}{1 - \varepsilon_i}$$

so maximising the product means maximising the sum of the logarithms. Differentiating with respect to Δv₁, with Δv₂ = Δv − Δv₁, gives a condition in which the same quantity appears for both stages — the Lagrange multiplier of the constraint — and the algebra is a page.

The result is easier to see than to write. A stage with a higher exhaust speed converts Δv into mass ratio more cheaply, because the exponent carries Δv/vₑ; a stage with a lower structural coefficient wastes less of what it lifts. Both push in the same direction: give the work to the better stage. In the vehicle drawn, the second stage has a much better exhaust speed and a slightly worse structure, and the exhaust speed dominates, so it takes 73 per cent of the velocity budget.

That is not what the equal-stage account suggests, and it is not what intuition suggests either — the first stage is the big one, so surely it does most of the work. It does most of the lifting. It does not do most of the accelerating, and the two are different jobs.

The arithmetic of why is worth one more line, because it is the same arithmetic the exponential itself is about. A kilometre a second bought at 4.41 km/s of exhaust speed multiplies the mass ratio by e^(1/4.41) = 1.26; the same kilometre a second bought at 2.94 multiplies it by 1.41. Compounded over seven kilometres a second the difference is a factor of three in the mass that has to be lifted, and the whole vehicle is built to lift it.

What the optimisation is worth #

An optimisation is worth stating as a number, because “optimal” on its own is a word rather than a result.

Moving from an equal split to the optimum takes the payload fraction from 4.240 per cent to 5.130 per cent — 21 per cent more payload, on a vehicle where the payload is five per cent of what leaves the pad. On a 550-tonne launcher that is about five tonnes, which is a satellite. Nothing else in launch-vehicle design is available for free at that scale, and the optimisation costs a page of algebra and one bisection.

It is worth saying what the comparison is against, though, because 21 per cent sounds larger in isolation than it is in context. An equal split is not the naive baseline anybody would actually adopt; it is the baseline the equal-stage model recommends, and no engineer has ever recommended it. A designer who put the hydrogen stage’s Δv at “most of it” by eye would land within a few points of 73 per cent and capture nearly all of the gain. What the calculation supplies is confidence rather than a discovery, which is the honest description of most optimisation in this subject.

That comparison is the practical content of the rung. A designer with two similar stages can split the budget evenly and lose nothing worth measuring. A designer with a hydrogen upper stage cannot.

The peak is flat here, and it is not always #

The second thing the curve says is about its own shape, and it is the reason the calculus is a check rather than a design tool.

Moving ten percentage points either side of the optimum still delivers 4.905 per cent of the lift-off mass — a penalty of 4.4 per cent of the payload for being substantially wrong about the split. Real vehicles are staged on grounds that have nothing to do with this curve: where the tank domes go, which engines already exist, what fits on a transporter, what can be recovered. The arithmetic is run afterwards to confirm nothing large has been left behind.

That is worth stating as a rule because it generalises past this figure. An optimum is flat when the constraint is loose and sharp when it is tight, and the same optimisation can be ignorable on one mission and load-bearing on another. Anybody quoting the flatness of a stage-division optimum has to say what velocity budget it was flat at.

What the shape of the curve says about failure #

One feature of the hero figure deserves more than the passing mention its caption gives it: the payload fraction reaches zero at both ends, and it reaches zero rather than going negative smoothly through a boundary that means nothing.

Each end is a stage hitting its own wall. A stage with structural coefficient ε cannot deliver more than vₑ ln(1/ε) of Δv however much propellant it carries, because at that point the propellant it would need weighs less than the tank it would need to hold it — the same wall the previous rung found under single-stage-to-orbit. Asking either stage for more than its wall gives it a negative payload fraction, and the product of a negative and a positive is a vehicle that does not exist.

So the interval of splits that work at all is bounded on both sides by structure rather than by the exponential, and it is narrower than it looks: for the vehicle drawn, the first stage’s wall is at 8.28 km/s and the second’s at 10.63, so the workable region runs from where the second stage is asked for more than 10.63 — below about 12 per cent of the budget in the first stage — to where the first is asked for more than 8.28, above about 88 per cent. Outside it there is no vehicle rather than a bad one. The optimum sits comfortably inside, which is a fact about this mission and not about all of them.

The one thing the model gets wrong on purpose #

The structural coefficient is treated here as a property of a stage, given in advance and independent of how much Δv the stage is asked for. It is not.

A stage’s dry mass is tankage, engines, plumbing and structure, and only the tankage scales with the propellant load. Engines are a fixed mass for a given thrust; avionics are a fixed mass full stop. So the structural coefficient falls as a stage grows — a bigger tank is a smaller fraction of dry mass — and asking a stage for more Δv makes it more efficient at delivering it. That coupling pulls the optimum toward whichever stage is being enlarged, and it is not in the arithmetic above.

The direction of the error is knowable even if its size is not. Since the optimum here already puts most of the work in the upper stage, and enlarging the upper stage would improve its structural coefficient further, the true optimum is further from equal than the figure says rather than nearer. The model’s answer is a bound in the direction it errs.

There is a second omission with the opposite sign. The first stage’s exhaust speed is quoted as a single number and it is not one: a sea-level nozzle is over-expanded on the pad and under-expanded at altitude, so the effective specific impulse rises through the burn by something like ten per cent. Giving the first stage more Δv means burning longer, higher up, at a better exhaust speed than the quoted one — which pushes the optimum back toward the first stage. The two omissions are of comparable size and opposite sign, which is a reason to trust the shape of the curve rather than the third decimal place of its peak.

The number the calculus does not supply #

Every figure here divides a budget that somebody else fixed. Where the 9.4 km/s comes from is a separate question and the answer is not one number either.

Orbital speed at 200 km is 7.78 km/s, which is the same vis-viva speed every transfer in this field is measured against. The rest is losses along the ascent — gravity, drag, and the back-pressure that a sea-level nozzle pays before it reaches vacuum — plus a launch site’s own contribution, which can be negative: an equatorial site due east starts 0.465 km/s ahead because the Earth is already turning. A launch from Kourou at 5.2° latitude therefore needs less than one from Baikonur at 45.6°, and the difference is a real several hundred metres a second before any trajectory is flown.

That matters here because the budget and the split are not independent. A vehicle with a smaller budget has a flatter optimum and a larger payload fraction, and the sensitivity of the payload to the budget is the derivative of the same curve: at 9.4 km/s, a hundred metres a second of extra loss costs about 2.5 per cent of the payload. A launch site’s latitude is therefore worth several per cent of a satellite, permanently, and it is the one term in this arithmetic that no engineering changes. Reaching a non-equatorial inclination costs the site’s latitude back again as a plane change, which is a fourth budget with its own rules.

Beyond two stages #

Nothing above requires two stages, and the extension is worth a paragraph because its answer is unexpectedly simple.

For n stages the same Lagrange condition applies pairwise, so the optimum is characterised by a single multiplier shared across all of them, and for identical stages that condition is satisfied by an equal division — which is the previous rung’s assumption, recovered as a theorem rather than assumed. For stages that differ, the multiplier has to be solved for numerically, and the solution has the same qualitative shape: work flows to the stages with the better exhaust speeds and the lighter structures.

The practical answer for three stages is nonetheless not an optimisation. A third stage adds an interstage, a separation event, an ignition in vacuum, and a set of failure modes, and the payload it buys over two well-chosen stages is the small increment the staging curve flattens toward. Launchers have three stages when the third is doing something else — an upper stage that restarts for a geostationary insertion, or a solid kick motor for an escape trajectory — rather than because the arithmetic asked for one.

What a real vehicle’s numbers look like #

It is worth putting one flown vehicle against the arithmetic, because the fit is close enough to be reassuring and loose enough to be honest.

Falcon 9 Block 5 carries 22.8 tonnes to low Earth orbit out of 549 at lift-off — a payload fraction of 4.15 per cent — with two stages whose structural coefficients from the published mass table are 0.061 and 0.040, and specific impulses of about 283 seconds at sea level rising to 312 in vacuum for the first stage and 348 for the second. Those are worse numbers than the hydrogen upper stage drawn above and the payload fraction is correspondingly lower.

Where the vehicle sits against this rung’s question is more interesting than where it sits on the curve. Its first stage delivers roughly 2.3 km/s before separation and the second about 7 — which is close to the 27:73 division the optimisation recommends, and is not there because anybody solved for it. It is there because the first stage is sized to be recoverable, which caps how much velocity it may spend before it has to turn around and come back with propellant left. A constraint that has nothing to do with the rocket equation lands the vehicle within a few points of the rocket equation’s answer, and that coincidence is exactly what the flatness of the peak makes possible: if the optimum were sharp, recovery would cost payload rather than being free.

Where the model stops #

The whole of this rung treats staging as a division of a scalar budget, and there is one thing that framing cannot express: parallel staging.

A vehicle with boosters strapped alongside a core does not stage in time. The boosters and the core burn together, the boosters are dropped, and the core continues — so at the moment of separation the core is partly empty, and the “stage” that was discarded never had a payload fraction of its own in the sense used here. Propellant crossfeed makes it worse still: if the boosters feed the core’s engines until they separate, the core is full at staging and the vehicle behaves like a two-stage rocket whose first stage has the second’s engines. The arithmetic for those cases exists and it is not this arithmetic, because the stage boundary has stopped being a moment in time.

The Space Shuttle and every Delta and Atlas variant with solid boosters is in that category, and so is Falcon Heavy. What they have in common is that the sequential model above puts them in the wrong place on every curve here, and it does so by an amount that depends on details of the plumbing rather than on anything in the rocket equation.

The habit, which is not about rockets #

The structure of this rung recurs whenever a fixed resource is divided between components that convert it at different rates, and the answer is always the same shape: give more to the better converter, until its own diminishing returns catch up with the other’s.

What makes the rocket case worth drawing rather than asserting is the ends. Most division problems have interior optima with well-behaved boundaries; this one has boundaries where the vehicle ceases to exist, because a stage past its structural wall has negative payload and a negative payload is not a small payload. The curve therefore has a shape — zero, rise, peak, fall, zero — that carries the constraint and the objective in one line, and reading the width of the workable interval off it is as useful as reading the peak.

The same shape appears in dividing an observing budget between exposures, where a too-short exposure returns nothing rather than a little, and in the split of a plane change between two burns, where the optimum is interior for the same reason — a small amount of the rotation bought at the cheap end is nearly free while the expensive end’s saving is linear. In each case the interesting output is the interval rather than the point.

Where the ladder goes next #

Everything above takes the exhaust speed as a given — a property of a stage, quoted in seconds, entering the equation as a number. Where that number comes from is a thermodynamics problem, and it has exactly two inputs.

The next rung derives it: an exhaust speed is the square root of a chamber temperature divided by a molecular weight, so choosing a propellant is choosing a molecular weight, and hydrogen wins not because it burns hot — it burns slightly cooler than kerosene — but because its exhaust is light. That rung also prices what the rocket equation cannot see, which is that a light propellant needs a large tank and a large tank is the structural coefficient this rung treated as independent.

Further rungs on this anchor: the gap between the velocity a launch buys and the speed it delivers, computed from an integrated ascent rather than quoted; the relativistic rocket equation, where the mass ratio becomes exponential in the rapidity and interstellar flight gets worse rather than better; and orbital refuelling, which breaks the framing entirely, because a vehicle departing a filled depot resets its mass ratio and none of these inequalities arises.