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

WHAT are we describing?

WHY does burning area control thrust? (derive from scratch)

Step 1 — Mass generation rate. The surface recedes at rate $r$ (units m/s). In time $dt$ a layer of thickness $r\,dt$ over area $A_b$ turns to gas. Its mass: $dm = \rho_p \, A_b \, (r\,dt)$

Why this step? Volume burned = area × thickness receded; mass = density × volume. Pure geometry.

So the mass generation rate: $\dot m = \rho_p A_b r$

Step 2 — Burn rate depends on pressure (Vieille's law / Saint-Robert): $r = a\,p_c^{\,n}$

Why? Higher chamber pressure $p_c$ pushes heat into the solid faster, so the surface eats inward faster. $a$, $n$ are empirical constants of the propellant ($n$ usually $0.2$–$0.5$).

Step 3 — Chamber pressure from mass balance. In steady operation, gas generated = gas leaving the nozzle throat ($A_t$). Nozzle mass flow $\dot m_{out}\propto p_c A_t$. Setting generation = exhaust: $\rho_p A_b a p_c^{\,n} = \frac{p_c A_t}{c^*}$ Solve for $p_c$: $\boxed{\,p_c = \left(\frac{\rho_p\, a\, c^*\, A_b}{A_t}\right)^{\frac{1}{1-n}}\,}$

Why the exponent $\tfrac{1}{1-n}$? $p_c$ appears on both sides (left via $r$), so isolating it inverts the $(1-n)$ power. This is why $n<1$ is required for stability — otherwise pressure runs away.

Step 4 — Thrust. $F = C_F\, p_c\, A_t$ Since $A_t$ and $C_F$ are ≈ fixed, thrust tracks $p_c$, which tracks $A_b$. Define the ratio: $K = \frac{A_b}{A_t}$

HOW each geometry shapes $A_b(w)$

1. BATES (cylindrical port, ends inhibited or free)

A hollow cylinder. Burns outward (port radius grows, that surface grows) AND inward from the ends (annulus, shrinks in axial length). For a single unrestricted-end BATES segment of length $L$, inner radius $R_i$, outer $R_o$, burning surfaces are the inner cylinder + two annular ends.

Let web burned $= w$. Inner radius $=R_i+w$, length $=L-2w$. $A_b(w) = \underbrace{2\pi (R_i+w)(L-2w)}_{\text{port grows}} + \underbrace{2\cdot\pi\big[R_o^2-(R_i+w)^2\big]}_{\text{ends shrink}}$

Why does this trend toward neutral? The port term rises while the end term falls; picking $L/D$ correctly makes them cancel → near-neutral. A long, end-inhibited BATES (only inner surface burns) is purely progressive.

2. Star grain

A star-shaped port. The many points and valleys give a large initial perimeter that, as it burns, rounds out — perimeter tends to stay roughly constant.

Why neutral? Early on the pointed "fingers" burn away (area would fall) but the valleys open into the web (area rises); a well-designed star balances these → neutral. Classic choice when you want constant thrust in a case-bonded motor (ends inhibited, so no end-burning to worry about).

3. Wagon wheel

A star with even deeper, thinner spokes — an extreme high-perimeter design.

Why highly progressive/high-thrust-early? Huge initial burning perimeter ⇒ huge initial $A_b$ ⇒ big early thrust. Used when you need a high initial thrust / high mass flow (boost phase). Trade-off: thin spokes leave slivers and give a regressive tail.

Reading the burn class off the geometry

Common mistakes (steel-manned)

Flashcards

Connections

• Saint-Robert burn rate law — the $r=ap_c^n$ that sets the timescale.
• Chamber pressure & throat area (Kn ratio) — $K=A_b/A_t$ drives $p_c$.
• Thrust coefficient and nozzle — $F=C_F p_c A_t$.
• Rocket equation & specific impulse — total impulse = $\int F\,dt$ = area under the curve you shaped.
• Combustion instability — why $n<1$ matters.
• Case-bonded vs free-standing grains — determines whether ends burn.

Concept Map

Hinglish (regional understanding)