Barrowman equations — centre of pressure calculation for finned rockets: Question bank

Definition The vocabulary — nail these before the questions

Angle of attack $α$ : the small angle between the rocket's pointing axis and the actual direction the air streams past it. When the rocket flies dead-straight, $α = 0$ ; a gust tips it to some small $α$ . Measured in radians throughout Barrowman. See Angle of attack and restoring moment.
Normal force $N$ : the sideways (perpendicular-to-axis) air-force that appears once $α \neq = 0$ .
Normal-force coefficient $C_{N} = \frac{N}{\frac{1}{2} ρ v ^{2} A _{r e f}}$ : $N$ made dimensionless ( $ρ$ = air density, $v$ = speed, $A_{r e f}$ = a chosen reference area).
Normal-force slope $C_{N α} \equiv \frac{\partial C _{N}}{\partial α}$ (units: per radian): how fast that coefficient grows per unit of tilt. This — not the force itself — is what every Barrowman term delivers.
CP (centre of pressure): the single point where all the sideways air-force can be imagined to act.
CG (centre of gravity): the single point where all the rocket's weight can be imagined to act. See Centre of gravity determination.
Static margin: the gap $(\overset{ˉ}{X}_{c p} - X_{c g})$ measured in body-diameters ("calibres"). Positive = CP behind CG = stable. See Static stability and static margin.

Figure — Barrowman equations — centre of pressure calculation for finned rockets

True or false — justify

Is the CP always behind the CG on a well-designed rocket?

True. Only then does the air-force act on a lever behind the pivot (CG) — i.e.

\overset{ˉ}{X}_{c p} > X_{c g}

— creating a torque that pushes the tail back into line: a positive static margin.

Barrowman's nose-cone normal-force slope

C_{N α} = 2

per radian depends on the cone's sharpness.

False. Slender-body theory makes the slope depend only on base area (giving 2 per radian with the base as reference); the shape changes only where the CP sits, not how much force appears.

A boat-tail (rear taper narrowing the body) always moves the total CP forward.

False in general. A boat-tail gives a negative

C_{N α}

located near the tail; a negative weight at a large

\overset{ˉ}{X}

can actually pull the weighted-mean CP forward (to smaller

X

), reducing stability — you must compute it, not guess.

At zero angle of attack a symmetric rocket has zero net side force, so Barrowman gives nothing useful.

False. Barrowman deliberately computes the slope

C_{N α} = \partial C_{N} / \partial α

(rate of force per unit tilt), which is exactly what governs the response to a small disturbance even though the force at

α = 0

is zero.

The CP location

\overset{ˉ}{X}_{c p}

changes if you fly faster (still subsonic, still small

α

True in reality but treated as False by Barrowman. The linear equations give an angle- and speed-independent CP; that's an approximation valid only for subsonic, small-

α

flow — near Mach 1 the real CP shifts aft.

Adding a third fin to a two-fin rocket triples the fin normal-force slope.

False. The factor

n

scales it, so going

n = 2 \to 3

multiplies

C_{N α, fins}

3/2

, not 3 — and it also adds tail mass shifting the CG, so stability may not improve as expected.

The body-interference factor

K_{f b} = 1 + \frac{r}{s + r}

can be less than 1.

False. Since radius

r > 0

and semi-span

s > 0

, the fraction is always positive, so

K_{f b} > 1

always — the body always increases the fin's effective force, never decreases it.

Spot the error

"I'll use

C_{N α} = 2

per degree for the nose since the slope is a simple 2."

Error: units. All Barrowman slopes are per radian. Using per-degree secretly divides every force by 57.3, wrecking any comparison between components (see the degrees/radians angle convention).

"My fin term ignores

K_{f b}

— the pure-fin formula is already complete, so my CP is conservative."

Error. Dropping

K_{f b}

shrinks

C_{N α, fins}

, the largest, most-aft weight in

\overset{ˉ}{X}_{c p} = \frac{\sum C _{N α, i} X ˉ _{i}}{\sum C _{N α, i}}

. Reducing that big rear weight pulls the weighted mean forward (smaller

\overset{ˉ}{X}_{c p}

) — so you underestimate stability, which is not automatically "conservative".

"Bigger fins → more restoring force → always more stable."

Error. Bigger fins raise

C_{N α, fins}

and move CP aft (good), but they also add mass at the tail, dragging CG aft (bad). Net static margin can shrink; recompute both CP and CG.

"CP is just the geometric middle of all the aerodynamic surfaces."

Error. CP is the normal-force-weighted mean

\overset{ˉ}{X}_{c p} = \frac{\sum C _{N α, i} X ˉ _{i}}{\sum C _{N α, i}}

; a component with tiny

C_{N α}

barely counts even if it's far away.

"I computed a huge

α = 2 5^{\circ}

manoeuvre with Barrowman and got a clean CP number."

Error. The numbers compute but the theory is invalid: it assumes linear, small-

α

(roughly

α ≲ 1 0^{\circ}

) slender-body flow. Beyond that the real CP moves and the answer is meaningless.

"My rocket has CP exactly at CG, so static margin is zero — that's neutrally fine."

Error. Zero margin means no restoring torque; any gust leaves the rocket tumbling with nothing to correct it. You want roughly 1–2 calibres of positive margin.

"The fin CP

\overset{ˉ}{X}_{fins}

moves forward when I sweep the tips back."

Error. Sweep pushes load aft: the

X_{r}

(sweep distance) term adds to

\overset{ˉ}{X}_{fins}

, moving the fin CP rearward (larger

X

) — which typically helps stability.

Why questions

Why does Barrowman compute force slopes

C_{N α}

rather than absolute forces?

Because a symmetric rocket has zero side force at

α = 0

; stability is about the response to a small tilt, so the meaningful quantity is the rate

\partial C_{N} / \partial α

at which restoring force appears per radian of tilt.

Why is the nose CP a factor

k

times its length, with

k = 2/3

for a cone?

Local lift is proportional to

d S / d x

(how fast cross-section area

S

grows along

x

); for a cone the radius grows linearly so

S \propto x^{2}

and

d S / d x \propto x

, and the CP is the area-weighted centroid

\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

. See the figure below.

Why does the total CP formula divide by

\sum C_{N α, i}

Because CP is defined by "net moment = net force × lever arm":

\sum N_{i} \overset{ˉ}{X}_{i} = \overset{ˉ}{X}_{c p} \sum N_{i}

, and dividing both sides by

\frac{1}{2} ρ v^{2} A_{r e f}

turns each

N_{i}

into

C_{N α, i}

, leaving a weighted average.

Why do fins usually dominate the CP even though the nose also makes force?

Fin slope

C_{N α, fins}

is typically several times larger than the nose's 2, so in the weighted mean the fins pull the CP strongly toward the tail — which is exactly what makes the rocket stable.

Why is a conical transition's slope written as a difference of squared diameter ratios?

Lift is generated only where the area changes; the term

2 [(d_{2} / d)^{2} - (d_{1} / d)^{2}]

measures the net cross-section area added (or removed) between the fore diameter

d_{1}

and aft diameter

d_{2}

Why does the body-interference factor exceed 1?

Air flowing past a fat body is deflected outward around it, so fins mounted on that body "see" extra sideways flow and produce more force than an identical fin in free air — hence

K_{f b} = 1 + \frac{r}{s + r} > 1

. See the deflected streamlines below.

Why is CP location safe to compute even if you accidentally mixed degrees and radians?

Because CP is a ratio of two force-weighted terms; a common per-degree/per-radian scale factor cancels top and bottom. The slope magnitudes (and thus margin comparisons) are what get corrupted, not the CP position.

Edge cases

What is the transition slope if

d_{1} = d_{2}

(a cylinder, no taper)?

Zero.

(d_{2} / d)^{2} - (d_{1} / d)^{2} = 0

: a plain cylinder changes no cross-section area, so slender-body theory predicts no normal force from it.

What happens to the fin term as the semi-span

s \to 0

(fins shrink to nothing)?

The factor

(s / d)^{2} \to 0

, so

C_{N α, fins} \to 0

— vanishing fins give no restoring force, and the CP is set entirely by the nose and body.

For a fin with zero tip chord (

C_{t} = 0

, a triangular fin), does the CP formula still work?

Yes. Substituting

C_{t} = 0

: the taper term becomes

\frac{X _{r}}{3} \cdot \frac{C _{r} + 0}{C _{r} + 0} = \frac{X _{r}}{3}

, and the bracket becomes

\frac{1}{6} [C_{r} - 0] = \frac{1}{6} C_{r}

— every term stays finite and well-defined.

What does a negative total static margin physically mean?

CP sits ahead of CG (

\overset{ˉ}{X}_{c p} < X_{c g}

), so the air-force torque amplifies any tilt — the rocket diverges and tumbles like a badly thrown dart rather than self-correcting.

In the limit of a very fat body where

r ≫ s

, what does

K_{f b}

approach?

K_{f b} = 1 + \frac{r}{s + r} \to 1 + 1 = 2

— the interference nearly doubles the fins' effective force when the body radius dwarfs the fin span.

What if the CG is far forward of the CP, say 10 calibres of margin?

The rocket is over-stable: it weathercocks aggressively into any crosswind, tilting into the wind and flying off-course — too much margin is as unwanted as too little (ideal ≈ 1–2 calibres).

Does a boat-tail's negative

C_{N α}

ever destabilise the rocket?

It can. Its negative force acts near the tail, and in the weighted mean a negative term at large

\overset{ˉ}{X}

can drag the total CP forward (smaller

X

), shrinking the static margin — always check the combined result.

Recall One-line self-test

Cover the answers: can you say why CP is a force-weighted mean, why $K_{f b} > 1$ , and when Barrowman stops being valid? Force-weighted because moment = force × arm ::: divided by total force gives the balance point. $K_{f b} > 1$ because ::: the body deflects flow outward, so fins feel extra deflected air. Barrowman fails when ::: $α$ is large ( $≳ 1 0^{\circ}$ ) or speed is transonic/supersonic.