Worked examples — Grain geometry — BATES, star, wagon wheel; neutral - progressive - regressive burn

Before we start, three symbols carried in from the parent, restated in plain words so you never touch an undefined thing:

Definition The five symbols (click to expand)

• $w$ = web burned = how far, in metres, the burning surface has crept perpendicular into the solid. $w=0$ at ignition.
• $A_b$ = burning surface area (m²) = how much propellant is on fire right now.
• $r$ = linear burn rate (m/s) = how fast every burning surface eats inward. Same value everywhere on the surface at a given instant.
• ${\rho }_p$ = propellant density (kg/m³).
• $\dot{m}$ = mass generation rate (kg/s) = ${\rho }_p{A}_{b}r$ (mass = density × area × thickness-per-second).

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The scenario matrix

We must cover every way $A_b$ can behave. Here is the full list, each with the example that hits it:

Cell	Case class	Degenerate / limit?	Example
C1	Progressive — $A_b$ rises (BATES inner-only)	—	Ex 1
C2	Regressive — $A_b$ falls (outer-burning / shrinking annulus)	—	Ex 2
C3	Neutral — $A_b$ ≈ const (end-burner)	zero-slope limit	Ex 3
C4	Competing terms — port grows and ends shrink (full BATES)	sign of $d{A}_b/dw$	Ex 4
C5	Degenerate: web runs out — $w\to w_{max}$, area $\to 0$	limiting/zero input	Ex 5
C6	Perimeter → thrust — star vs wagon wheel, high initial $A_b$	large-input limit	Ex 6
C7	Pressure amplification — the $\frac{1}{1-n}$ exponent bites	non-linear map	Ex 7
C8	Real-world word problem — pick a geometry for a mission	design choice	Ex 8
C9	Exam twist — "which is neutral?" with a hidden regressive tail	trap case	Ex 9

Every numeric answer below is machine-verified.

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C1 — Progressive: BATES inner surface only

1. Inner surface = a cylinder wall. Its area is circumference × length: $A_b(w)=2\pi (R_i+w)L.$ Why this step? When only the inner face burns, the burning surface is the wall of the cylindrical hole. A cylinder wall of radius $\rho$ and length $L$ has area $2\pi \rho L$ (unroll it into a rectangle: width = circumference $2\pi \rho$, height = $L$). The radius grows to $R_i+w$ because the wall recedes outward by $w$.

2. Plug $w=0$: $A_b(0)=2\pi (0.02)(0.30)=0.03770{\text{m}}^2.$ Why this step? We evaluate at ignition first to fix the starting burning area — the baseline every later value is compared against.

3. Plug $w=0.01$: $A_b(0.01)=2\pi (0.03)(0.30)=0.05655{\text{m}}^2.$ Why this step? Same $L$, but radius grew from $0.02$ to $0.03$ — a bigger circle has a longer circumference, so more surface is on fire.

4. Trend: area rose by $50\%$. Since $A_b$ increases with $w$, this is progressive.

Verify: ratio $=\frac{0.03}{0.02}=1.5$ exactly (both share $2\pi L$), so the $50\%$ rise is not a rounding artefact — it's the radius ratio itself. Units: $\text{m}\cdot \text{m}={\text{m}}^2$. ✔

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C2 — Regressive: outer-burning grain

1. Outer surface is again a cylinder wall, but it recedes inward, so its radius decreases: $A_b(w)=2\pi (R_o-w)L.$ Why the minus sign? An outer surface burning inward loses radius as $w$ grows. The geometry is the same cylinder wall; only the sign of the offset flips.

2. $w=0$: $A_b=2\pi (0.05)(0.40)=0.12566{\text{m}}^2.$ Why this step? Fix the ignition baseline, exactly as in Ex 1, so the later value has something to be compared against.

3. $w=0.01$: $A_b=2\pi (0.04)(0.40)=0.10053{\text{m}}^2.$ Why this step? The radius has dropped from $0.05$ to $0.04$; a smaller circle has a shorter circumference, so less surface is on fire.

4. Trend: area fell by $20\%$. Since $A_b$ decreases, this is regressive — thrust droops with time.

Verify: ratio $=0.04/0.05=0.8$, i.e. a $20\%$ drop, purely from the radius ratio. Sign of $d{A}_b/dw=-2\pi L<0$ confirms "decreasing". ✔

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C3 — Neutral limit: the end-burner

1. The burning face is a flat disc, area $A_b=\pi R^2$, and it does not depend on $w$: $A_b(w)=\pi R^2=\pi (0.03{)}^2=2.827\times 10^{-3}{\text{m}}^2.$ Why this step? The disc just translates down the rod. Its radius $R$ never changes (walls inhibited), so the area is frozen. This is the exact zero-slope limit: $d{A}_b/dw=0$ ⇒ perfectly neutral.

2. Duration = web to burn / burn rate. Here the "web" is the whole length $L$: $t=\frac{L}{r}=\frac{0.50}{2\times 10^{-3}}=250\text{s}.$ Why this step? The face travels the length $L$ at speed $r$; time = distance / speed. Constant tiny area ⇒ low thrust but very long burn — the classic sustainer.

Verify: $d{A}_b/dw=0$ (no $w$ in $\pi R^2$) → neutral exactly. $t$ units: $\text{m}/(\text{m/s})=\text{s}$. $250$ s ≈ 4 min, sensible for a sustainer. ✔

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C4 — Competing terms: the full BATES segment

1. Write the two competing surfaces (from the parent note): $A_b(w)=\underset{\text{port}}{\underbrace{2\pi (R_i+w)(L-2w)}}+\underset{\text{two ends}}{\underbrace{2\pi [R_o^2-(R_i+w{)}^2]}}.$ Why this step? Every burning surface must be summed — the inner wall (grows) plus the two ring-shaped ends (shrink as the port eats into them). Missing either term is the classic mistake.

2. Differentiate to get the slope (the tool that answers "rising or falling?"): $\frac{d{A}_b}{dw}=2\pi [(L-2w)-2(R_i+w)]+2\pi [-2(R_i+w)].$ Why a derivative? "Progressive vs regressive" is literally the question the derivative answers: is $A_b$ going up ($>0$) or down ($<0$) right now? No other tool gives the instantaneous trend.

3. Evaluate at $w=0$: $\frac{d{A}_b}{dw}{|}_0=2\pi [L-2R_i-2R_i]=2\pi (L-4R_i).$ Plug numbers: $L-4R_i=0.20-4(0.02)=0.20-0.08=0.12>0.$ So $d{A}_b/dw{|}_0=2\pi (0.12)=0.7540{\text{m}}^2/\text{m}>0.$ Why this step? Setting $w=0$ isolates the ignition trend — the very first instant, where designers most care whether thrust starts rising or falling.

4. Read the sign: positive ⇒ area is rising at ignition ⇒ progressive at the start. The port's growth beats the ends' shrink because $L$ is long relative to $R_i$.

Verify: the neutral crossover is $L=4R_i$, i.e. $L=0.08$ m. Our $L=0.20>0.08$, so progressive — consistent. If we shrank $L$ below $0.08$ m the sign would flip to regressive. ✔

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C5 — Degenerate: the web runs out

1. Radial web is used up when the port reaches the outer wall: $R_i+w_{max}=R_o\Rightarrow w_{max}=R_o-R_i=0.05-0.02=0.03\text{m}.$ Why this step? At that moment there is no propellant left between port and casing — the wall has nowhere further to recede.

2. Evaluate $A_b$ at $w=w_{max}=0.03$:

   • Port term: $2\pi (R_i+w)(L-2w)=2\pi (0.05)(0.20-0.06)=2\pi (0.05)(0.14)=0.04398$ m².
   • End term: $2\pi [R_o^2-(R_i+w{)}^2]=2\pi [0.0025-0.0025]=0$. $A_b(w_{max})=0.04398+0=0.04398{\text{m}}^2.$ Why this step? The ends vanish exactly at burnout because $(R_i+w)=R_o$, so the annulus $R_o^2-(R_i+w{)}^2$ collapses to zero. That's the degenerate/zero case: one whole surface disappears.

3. Thrust behaviour: in reality the burn is not clean here — thin leftover slivers of propellant break up, area collapses toward zero, and thrust falls off sharply — the regressive tail. The model's $A_b$ stays finite because it ignores fragmentation.

Verify: end term $=2\pi [R_o^2-R_o^2]=0$ identically at $w_{max}$ — the annulus is genuinely gone. $w_{max}=0.03$ m is the full radial thickness $R_o-R_i$. ✔

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C6 — Perimeter → thrust: star vs wagon wheel

1. For a case-bonded grain, burning area = perimeter × length (parent's rule, ends inhibited): $A_b=P\cdot L.$ Why this step? With the ends sealed, the only burning surface is the port wall, whose area is (how long the outline is) × (how deep the grain runs) = $P\cdot L$.

2. Star: $A_{b,A}=1.2\times 0.60=0.72$ m². Wheel: $A_{b,B}=3.0\times 0.60=1.80$ m². Why this step? We turn each perimeter into a burning area so we can compare the two grains on the same footing before bringing in the chemistry.

3. Initial mass flow $\dot{m}={\rho }_p{A}_{b}r$:

   • Star: $1800\times 0.72\times 0.006=7.776$ kg/s.
   • Wheel: $1800\times 1.80\times 0.006=19.44$ kg/s. Why this step? Same chemistry (${\rho }_p,r$), so mass flow scales directly with burning area, which scales with perimeter. The wheel's high perimeter buys high initial thrust for a boost phase.

4. Ratio: $19.44/7.776=2.5$ — exactly the perimeter ratio. The wagon wheel delivers $2.5\times$ the launch thrust — at the cost of thin spokes that leave a regressive sliver tail.

Verify: ${\dot{m}}_B/{\dot{m}}_A=P_B/P_A=3.0/1.2=2.5$ (all other factors cancel). ✔

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C7 — The $\frac{1}{1-n}$ amplification bites

Before the numbers, two symbols enter that we must define on this page:

1. Recall the pressure law (Chamber pressure & throat area (Kn ratio)): $p_c\propto A_b^{\frac{1}{1-n}}.$ Why this exponent, not just $A_b^1$? Because $p_c$ sits on both sides of the mass balance — it feeds back through the burn rate $r=a{p}_c^n$ (see Saint-Robert burn rate law). Isolating $p_c$ inverts the factor $(1-n)$, so the geometric change is amplified by the exponent $\frac{1}{1-n}$.

2. Compute the exponent: $\frac{1}{1-n}=\frac{1}{1-0.4}=\frac{1}{0.6}=1.6\overline{6}.$ Why this step? We need the exact power that maps an area ratio to a pressure ratio before we can raise anything to it.

3. Map the $20\%$ area rise (a factor $1.20$) through the exponent: $\frac{p_c^{\text{new}}}{p_c^{\text{old}}}=1.20^{1.6667}=\mathrm{1.3551.}$ So chamber pressure rises about $35.5\%$, not $20\%$. Why this step? Raising the ratio to $1.667$ magnifies it — that is the whole point of the non-linear feedback.

4. Thrust $F=C_F{p}_c{A}_t$ with $C_F,A_t$ fixed, so thrust tracks $p_c$: also up $\approx 35.5\%$. A "small" area drift becomes a big thrust drift — this is why designers chase true neutrality.

Verify: $1.2^{1/0.6}\approx 1.3551$; the amplification factor $1.3551/1.20\approx 1.13>1$ confirms pressure moves more than area. Required $n<1$ so the exponent stays finite and positive (self-regulating). ✔

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C8 — Real-world design word problem

1. Boost phase → wagon wheel. Why? Its enormous initial port perimeter gives the largest initial $A_b$, hence the largest initial $\dot{m}$ and thrust (Ex 6). Its regressive sliver tail is acceptable because boost is meant to be brief and front-loaded.

2. Cruise phase → end-burner (or star). Why? The end-burner's disc area is frozen (Ex 3): neutral and long-burning, giving a low, flat, steady thrust for the whole cruise. A well-designed star is the alternative when higher (but still flat) thrust is wanted, since it holds $A_b$ ≈ constant.

3. Why not swap them? A wagon wheel as a cruiser would burn out too fast and droop badly (regressive). An end-burner as a booster gives far too little thrust. Class dictates role: progressive/high-perimeter for boost, neutral for cruise.

Verify: consistent with the matrix — wagon wheel = high initial $A_b$ (C6), end-burner = neutral (C3). No numeric to check; logic maps each phase to its burn class. ✔

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C9 — Exam twist: "which is neutral?" (the trap)

1. State what neutral really means: $A_b$ ≈ constant over the main burn — an approximation, not an identity. Why this step? The label "neutral" describes the bulk behaviour, not every instant; pinning down its true meaning is what disarms the trap.

2. Find the hidden non-neutral parts. As the star's points and valleys round out, small ripples appear; and near burnout, thin slivers of propellant between spokes leave the port and $A_b$ collapses — a regressive tail. Why this step? No real geometry keeps $A_b$ literally frozen from ignition to burnout; the tail is unavoidable when web is exhausted (echoing C5).

3. Correct answer: "No — the thrust is ≈ constant over the main burn, with a mild ripple and a regressive sliver tail at the end. Neutral is an approximation." The student's "exactly flat" is the trap.

Verify: consistent with the parent's mistake callout ("neutral is approximate; real stars have a regressive sliver tail"). ✔

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Recall Quick self-test (click to reveal)

Progressive means $A_b$ does what over time? ::: Increases → rising thrust. Why is an end-burner exactly neutral? ::: Its flat disc face has area $\pi R^2$ with no $w$-dependence, so $d{A}_b/dw=0$. Full BATES: neutral crossover length in terms of $R_i$? ::: $L=4R_i$ (from $d{A}_b/dw{|}_0=2\pi (L-4R_i)=0$). Area up $20\%$ with $n=0.4$ → pressure up by? ::: About $35.5\%$, because $p_c\propto A_b^{1/(1-n)}=A_b^{1.667}$. At $w_{max}$ in a BATES, what happens to the end-annulus term? ::: It goes to zero — the annulus vanishes ($R_i+w=R_o$).

See also: Saint-Robert burn rate law · Chamber pressure & throat area (Kn ratio) · Thrust coefficient and nozzle · Rocket equation & specific impulse · Combustion instability · Case-bonded vs free-standing grains