Visual walkthrough — Grain geometry — BATES, star, wagon wheel; neutral - progressive - regressive burn

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Step 1 — A flame that walks sideways

WHAT. Imagine a solid block of rocket fuel with a hollow tunnel down its middle. We set the inside wall of the tunnel on fire. Fire cannot burn through the whole block instantly — it eats the surface, and that surface retreats, always moving straight into the fuel, perpendicular to itself.

WHY. This "perpendicular retreat" is the single fact everything rests on. A candle flame moves down the candle; a rocket grain's flame moves outward from every burning wall at once. If we know how far the wall has moved, we know exactly how much fuel became gas.

PICTURE. Below, the pale-yellow wall is the burning surface. The blue arrows show it stepping outward by a small distance. That distance — how deep the fire has bitten — gets a name.

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Step 2 — Turning "surface retreats" into "gas is made"

WHAT. In a tiny slice of time $dt$, the whole burning surface — call its area $A_b$ — steps inward by a thin layer of thickness $rdt$. That thin shell of solid becomes gas.

WHY. We want mass of gas per second, because gas is what leaves the nozzle and makes thrust. The cleanest route is pure geometry: volume = area × thickness, then mass = density × volume.

PICTURE. The shaded rind in the figure is the layer that burns away in one instant. Its volume is area times its (tiny) thickness.

$$dm=\underset{\begin{array}{c}\text{fuel density}\\ {\text{kg per m}}^3\end{array}}{\underbrace{{\rho }_p}}\cdot \underset{\begin{array}{c}\text{burning}\\ {\text{area, m}}^2\end{array}}{\underbrace{A_b}}\cdot \underset{\begin{array}{c}\text{shell}\\ \text{thickness, m}\end{array}}{\underbrace{(rdt)}}$$

Divide both sides by $dt$ to get the rate:

$$\dot{m}=\underset{\text{density}}{\underbrace{{\rho }_p}}\underset{\text{area}}{\underbrace{A_b}}\underset{\text{speed}}{\underbrace{r}}$$

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Step 3 — The fire feeds on its own pressure

WHAT. The burn speed $r$ is not a fixed number. Squeeze the chamber to a higher pressure $p_c$ and the surface eats inward faster. The relationship, measured in the lab, is a simple power law:

$$r=\underset{\begin{array}{c}\text{propellant}\\ \text{constant}\end{array}}{\underbrace{a}}\underset{\begin{array}{c}\text{pressure raised}\\ \text{to power }n\end{array}}{\underbrace{p_c^n}}$$

WHY. More pressure means hot gas is jammed harder against the solid, driving heat into it faster, so the solid vaporises faster. This is the Saint-Robert burn rate law. We need this because in Step 2 we secretly used $r$ without knowing what sets it — and $r$ turns out to depend on the very pressure we are trying to find.

PICTURE. The curve below shows $r$ climbing with $p_c$. The exponent $n$ (typically $0.2$–$0.5$) is the steepness: small $n$ = a gentle, well-behaved curve.

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Step 4 — The chamber must not fill up or empty out

WHAT. Gas is being made on the burning surface and gas is leaving through the nozzle's narrowest point, the throat (area $A_t$). In steady running these two must match: whatever the surface makes, the throat lets out.

WHY. If generation beat exhaust, pressure would climb forever; if exhaust beat generation, the fire would starve. Balance is what pins down the pressure $p_c$.

The exhaust side (from nozzle theory, Kn ratio) says gas leaves at a rate proportional to $p_c{A}_t$:

$${\dot{m}}_{\text{out}}=\frac{p_c{A}_t}{c^{\ast }}$$

where $c^{\ast }$ ("c-star") is the characteristic velocity — a fixed property of the burnt gases measuring how efficiently they push out.

PICTURE. A see-saw: gas made (left) versus gas expelled (right). The pressure floats to whatever level balances them.

Set made = expelled ($\dot{m}={\dot{m}}_{\text{out}}$), plugging $r=a{p}_c^n$ from Step 3:

$$\underset{\text{gas generated (Step 2 + 3)}}{\underbrace{{\rho }_p{A}_{b}a{p}_c^n}}=\underset{\text{gas expelled (Step 4)}}{\underbrace{\frac{p_c{A}_t}{c^{\ast }}}}$$

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Step 5 — Solving for pressure, and the hidden amplifier

WHAT. Now isolate $p_c$. It sits on both sides — on the left buried inside $p_c^n$, on the right as a plain $p_c$. Gather the powers of $p_c$:

$${\rho }_p{A}_{b}a{c}^{\ast }=\frac{p_c}{p_c^n}{A}_t=p_c^{1-n}{A}_t$$ $$\boxed{p_c={\left(\frac{{\rho }_{p}a{c}^{\ast }{A}_b}{A_t}\right)}^{\frac{1}{1-n}}}$$

WHY the exponent $\frac{1}{1-n}$? Because $p_c$ appeared as a power on the left, undoing it flips $(1-n)$ up into $\frac{1}{1-n}$. This is not decoration — it is an amplifier. With $n=0.4$, $\frac{1}{1-n}=1.67$, so a $20\%$ rise in area becomes $1.2^{1.67}\approx 1.36$ — a $36\%$ rise in pressure.

WHY $n<1$ matters. If $n\ge 1$ the exponent $\frac{1}{1-n}$ becomes infinite or negative — pressure runs away or misbehaves. $n<1$ keeps the motor self-regulating: a pressure blip makes gas leave a touch faster than it is made, damping the blip. (More on failures of this in Combustion instability.)

PICTURE. Two dials geared together: turn the small $A_b$ dial a little, the $p_c$ dial swings further. The gear ratio is $\frac{1}{1-n}$.

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Step 6 — From pressure to thrust

WHAT. Thrust is what the throat-plus-nozzle turns pressure into:

$$F=\underset{\begin{array}{c}\text{thrust}\\ \text{coefficient}\end{array}}{\underbrace{C_F}}\underset{\text{pressure}}{\underbrace{p_c}}\underset{\text{throat area}}{\underbrace{A_t}}$$

WHY. $A_t$ and $C_F$ (set by nozzle shape — see Thrust coefficient and nozzle) are essentially fixed hardware. So the only mover is $p_c$, and from Step 5 the only mover behind $p_c$ is $A_b$. Chain it all:

$$A_b(w)\Rightarrow p_c\propto A_b^{\frac{1}{1-n}}\Rightarrow F\propto p_c$$

Therefore: the shape of the thrust-vs-time curve is the shape of $A_b(w)$, amplified. This is why grain geometry is design, not chemistry.

PICTURE. The full pipeline, left to right: a port shape ➜ an area-vs-web curve ➜ a pressure curve ➜ a thrust curve, each an amplified echo of the one before.

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Step 7 — Edge and degenerate cases (never skip these)

WHAT & WHY. A derivation you trust must survive its extremes. Three matter here.

PICTURE. Three mini-panels, each a limiting shape.

• End-burner ($A_b=\pi R^2$, constant). Only one flat circular face burns. Area never changes ⇒ exactly neutral, low thrust, very long burn (the whole length must be eaten). The perfect sustainer.
• Burnout / sliver ($A_b\to$ small, then $0$). When receding walls finally meet, leftover thin "slivers" of fuel keep burning with shrinking area ⇒ a regressive tail. So "neutral" is only ≈ constant over the main burn, never dead flat to the very end.
• $n\to 1$ (forbidden). As $n$ climbs toward $1$, $\frac{1}{1-n}\to \infty$: the tiniest area change explodes the pressure. The motor cannot self-regulate. This is why real propellants live at $n\approx 0.2$–$0.5$.

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The one-picture summary

The whole page in one frame: a receding wall makes gas ($\dot{m}={\rho }_p{A}_{b}r$), gas made must equal gas expelled, that balance pins pressure ($p_c\propto A_b^{1/(1-n)}$), and thrust rides on pressure ($F=C_F{p}_c{A}_t$). The port shape enters at the far left and its every wiggle is echoed — amplified — all the way to the thrust curve.

Recall Feynman retelling — say it in plain words

Picture fire eating a tunnel through solid fuel. The fire never jumps; it just crawls straight into every wall at a steady crawl-speed. In one heartbeat it turns a paper-thin shell of solid into gas — how much gas depends only on how big the burning surface is, because the crawl-speed is the same everywhere. That gas has to get out somewhere: the nozzle throat. If the fire makes gas faster than the throat can spit it out, the chamber pressure rises — and higher pressure makes the fire crawl a little faster still, but the throat also spits harder. The two settle into a balance, and that balance fixes the pressure. Here's the sneaky bit: because pressure feeds back into the burn speed, a small change in burning-surface size doesn't nudge the pressure — it shoves it, by a power of $1/(1-n)$. Finally, thrust is just pressure times a fixed nozzle, so thrust is a magnified copy of the burning-surface-size story. And the burning-surface-size story is set by one thing you get to design: the shape of the hole. Circle → hole grows → thrust climbs. Clever star → area holds steady → flat thrust. That is the entire art of grain geometry.

Recall Self-test

Why is thrust a magnified copy of area, not a straight copy? ::: Because pressure appears on both sides of the mass balance; solving gives $p_c\propto A_b^{1/(1-n)}$, and thrust $\propto p_c$. The exponent $1/(1-n)>1$ amplifies. What does the web burned $w$ measure? ::: The perpendicular distance the burning surface has receded into the solid. Why must $n<1$? ::: So $1/(1-n)$ stays finite and positive — the motor self-regulates instead of running away. Why is an end-burner exactly neutral? ::: Its one flat face has constant area $\pi R^2$ as it recedes, so $A_b$ never changes.