Barrowman equations — centre of pressure calculation for finned rockets: Visual walkthrough

Step 1 — What "sideways air push" even means

WHAT. A rocket normally flies with its pointy end straight into the wind. Now imagine it gets nudged, so its axis is tilted by a tiny angle from the airflow. That tilt has a name: the angle of attack, written $α$ (Greek letter "alpha"). See Angle of attack and restoring moment.

WHY. When the rocket is perfectly aligned ( $α = 0$ ), the air slides past symmetrically and pushes equally on all sides — no net sideways force. The instant it tilts, one side of every surface meets the air more head-on than the other, and a sideways force appears. That sideways force is what straightens the rocket (or flips it). So the whole game starts with $α$ .

PICTURE.

The sideways force, drawn as the pink arrow, is called the normal force — "normal" here is the maths word for perpendicular. We write it $N$ : it points at right angles to the rocket's axis.

Step 2 — Turning a force into a pure number

WHAT. The raw force $N$ depends on messy things: how fast you fly, how thick the air is, how big the rocket is. We strip all that out and keep only the shape's own personality by dividing: $C_{N} = \frac{N}{\frac{1}{2} ρ v ^{2} A _{r e f}}$

WHY. Two rockets of identical shape but different sizes, or the same rocket on a windy vs calm day, should be describable by one number. Dividing by $\frac{1}{2} ρ v^{2} A_{r e f}$ — the natural "air pressure scale" — cancels the size, speed and air-density dependence. See Slender-body aerodynamic theory for why this exact denominator is the right one.

PICTURE.

Step 3 — Why we track the slope, not the force itself

WHAT. We do not care about $C_{N}$ at one fixed tilt. We care how fast it grows as the tilt grows. That rate is a slope: $C_{N α} = \frac{\partial C _{N}}{\partial α}$

WHY this tool — the derivative. The symbol $\frac{\partial C _{N}}{\partial α}$ (read "the derivative of $C_{N}$ with respect to $α$ ") answers exactly the question "if I tilt a little more, how much extra sideways force do I get?" A derivative is the perfect tool because stability is a response to a small disturbance — and "response per unit disturbance" is precisely what a slope measures. At $α = 0$ the force is zero, so the number is useless; its rate of change is everything.

PICTURE.

Step 4 — The nose cone gives exactly "2"

WHAT. For any slender nose, slender-body theory gives one clean result: $C_{N α, nose} = 2$

WHY "2" and nothing else. Slender-body theory says the lift slope of a pointed body equals $2 \times \frac{base area}{A _{r e f}}$ . If we choose the nose's own base as the reference area, that fraction is exactly $1$ , leaving $2$ . Remarkably it does not depend on whether the nose is a cone, ogive or parabola — only on how big the back of the nose is.

PICTURE.

Recall Why base area, not tip?

Where does the nose's cross-section stop growing? ::: At the base — beyond that the body is straight and (in this theory) makes no new lift. All the nose's "growing" happens between tip and base, so the base area sets the total.

Step 5 — Where the nose's force acts: the centroid of "growing area"

WHAT. The nose's single balance point is: $\overset{ˉ}{X}_{nose} = k L_{nose}, k = \frac{2}{3} for a cone.$

WHY the centroid, and why the integral. The sideways push is not applied at one spot — it is smeared along the nose. Slender-body theory says the local push per unit length is proportional to $\frac{d S}{d x}$ , the rate at which cross-section area $S$ grows as you move distance $x$ back from the tip. To collapse a smeared load into one equivalent point you take its area-weighted average position — a centroid — which is exactly what an integral computes: $\overset{ˉ}{X} = \frac{\int _{0}^{L} x \frac{d S}{d x} d x}{\int _{0}^{L} \frac{d S}{d x} d x}$

For a cone the radius grows straight-line, $r \propto x$ , so area $S \propto x^{2}$ and $\frac{d S}{d x} \propto x$ . Substituting: $\overset{ˉ}{X} = \frac{\int _{0}^{L} x \cdot x d x}{\int _{0}^{L} x d x} = \frac{L ^{3} /3}{L ^{2} /2} = \frac{2}{3} L .$

PICTURE.

Step 6 — The fins: a bigger slope from thin-airfoil flow

WHAT. Each flat fin behaves like a small wing. Adding up all $n$ fins and correcting for the fat body they sit on: $C_{N α, fins} = K_{f b} \frac{4 n ( s / d ) ^{2}}{1 + 1 + ( \frac{2 ℓ _{f}}{C _{r} + C _{t}} ) ^{2}}$

WHY this shape of formula. A thin flat plate at small $α$ makes lift proportional to its span squared (bigger span = it grabs more airflow) — that is the $(s / d)^{2}$ on top, times $4 n$ for $n$ fins. The ugly square-root denominator is an aspect-ratio correction: short, stubby fins are less efficient than long thin ones, and $ℓ_{f}$ (the slanted mid-chord length) captures the sweep that lowers efficiency. See Slender-body aerodynamic theory.

PICTURE.

Step 7 — Where the fins' force acts (sweep drags it aft)

WHAT. The fin centre of pressure: $\overset{ˉ}{X}_{fins} = X_{f} + \frac{X _{r}}{3} \cdot \frac{C _{r} + 2 C _{t}}{C _{r} + C _{t}} + \frac{1}{6} [(C_{r} + C_{t}) - \frac{C _{r} C _{t}}{C _{r} + C _{t}}]$

WHY these three pieces.

$X_{f}$ — where the fin's leading edge starts on the body. Everything is measured from here.
The middle piece with $X_{r}$ (the sweep: how far the tip is set back from the root) — a taper-weighted centroid. If the tip is swept back, its load acts further aft, so CP moves back.
The bracket piece — a chord-averaging correction blending root and tip chords.

PICTURE.

Recall Degenerate check: a rectangular, unswept fin

Set $C_{t} = C_{r} = C$ and $X_{r} = 0$ . The middle term vanishes; the bracket becomes $\frac{1}{6} [2 C - \frac{C}{2}] = \frac{C}{4}$ . ::: So $\overset{ˉ}{X}_{fins} = X_{f} + \frac{C}{4}$ — the CP sits at the quarter-chord, exactly the classic thin-airfoil result. Good, the formula reduces correctly.

Step 8 — Combining everything: the weighted balance point

WHAT. We have, for each part $i$ , a slope $C_{N α, i}$ (how hard it pushes) and a location $\overset{ˉ}{X}_{i}$ (where it pushes). The whole rocket's CP is: $\overset{ˉ}{X}_{c p} = \frac{\sum _{i} C _{N α, i} X ˉ _{i}}{\sum _{i} C _{N α, i}}$

WHY a weighted average — from balancing torques. Torque = force × lever arm. The total torque of all parts must equal one imagined force at one point $\overset{ˉ}{X}_{c p}$ : $\sum_{i} N_{i} \overset{ˉ}{X}_{i} = \overset{ˉ}{X}_{c p} \sum_{i} N_{i} .$ Divide both sides by $\frac{1}{2} ρ v^{2} A_{r e f}$ and every $N_{i}$ becomes $C_{N α, i}$ . Solve for $\overset{ˉ}{X}_{c p}$ — and out drops the boxed formula. A part with a big slope (fins!) drags the balance point strongly toward itself.

PICTURE.

Then compare with the CG to get the static margin: with CG at $38$ cm and $d = 4$ cm, margin $= \frac{45.3 - 38}{4} = 1.8$ calibres — comfortably stable.

Step 9 — Edge cases you must never trip over

WHAT / WHY / PICTURE together. The formula has quiet corners:

Boat-tail (body narrows), $d_{2} < d_{1}$ . The transition slope $2 [(d_{2} / d)^{2} - (d_{1} / d)^{2}]$ goes negative — a destabilising forward pull. It's a real effect; keep the sign.
Zero angle of attack, $α = 0$ . Every $N_{i} = 0$ , yet the slopes $C_{N α, i}$ and the locations $\overset{ˉ}{X}_{i}$ are unchanged. CP location does not depend on $α$ — that's why we can quote it as a fixed number.
All slopes zero (a finless, noseless tube). The denominator $\sum C_{N α, i} \to 0$ and $\overset{ˉ}{X}_{c p}$ is undefined — physically there's no restoring force and no CP. The maths honestly refuses to answer.
Beyond small $α$ or near Mach 1. The straight-line assumption of Step 3 breaks; the true $C_{N}$ curve bends. Barrowman is only trustworthy for $α ≲ 1 0^{\circ}$ and subsonic.

The one-picture summary

Everything on one board: each part contributes a push strength (its slope $C_{N α, i}$ , drawn as arrow thickness) at a position ( $\overset{ˉ}{X}_{i}$ ). The rocket's CP is their strength-weighted balance point. Place it behind the CG → the restoring moment straightens the flight, feeding a stable trajectory into any 6-DOF simulation.

Recall Feynman retelling of the whole walkthrough

Tilt the rocket a hair — that tilt is $α$ . Because it's now slightly sideways to the wind, each surface gets a little sideways shove, $N$ . We scrub away the speed and air-density to keep a clean number, $C_{N}$ , and then ask the only question stability cares about: how fast does that shove grow as I tilt more? — that's the slope $C_{N α}$ . The nose, by a neat theorem, always scores exactly 2 and pushes at two-thirds of its length (the centre of its growing width). Each fin acts like a tiny wing; big spans push hard, sweep drags their push spot backward, and sitting on a fat body makes them push even harder ( $K_{f b}$ ). Finally, to find the single spot where all these shoves balance, we do the same thing you'd do to find where several kids balance a seesaw: a strength-weighted average of their positions. The fins push hardest, so they win the tug-of-war and pull the balance point near the tail — which is exactly where you want it, safely behind the weight balance, so the rocket flies like a well-thrown dart.