Barrowman equations — centre of pressure calculation for finned rockets: Worked examples

The scenario matrix

Before working anything, let us list all the case-classes this topic can throw. Each worked example below is tagged with the cell it fills.

Cell	Case class	What is special	Example
A	Baseline: nose + fins	The "everyday" rocket	Ex 1
B	Positive transition (flare)	Area grows, term $> 0$	Ex 2
C	Negative transition (boat-tail)	Area shrinks, term $< 0$	Ex 3
D	Degenerate fin: $C_{t} \to 0$	Triangular fin, watch $0/0$	Ex 4
E	Limiting body: $r \to 0$ (skinny)	Interference $K_{f b} \to 1$	Ex 5
F	Limiting body: $r ≫ s$ (fat)	Interference $K_{f b} \to 2$	Ex 5
G	Real-world word problem	Translate prose → numbers	Ex 6
H	Exam twist: solve for CG target	Invert the static-margin formula	Ex 7
I	Sanity: symmetric = zero force at $α = 0$	Why only slopes matter	Ex 8

The symbols used everywhere:

Definition The vocabulary (re-earned)

$C_{N α}$ — normal-force slope: how fast sideways force grows per radian of tilt. Bigger = grabs the air harder.
$\overset{ˉ}{X}$ — the centre-of-pressure position of one component, measured in cm from the nose tip.
$\overset{ˉ}{X}_{c p}$ — the whole rocket's CP: a weighted average of the component $\overset{ˉ}{X}$ 's, weighted by their $C_{N α}$ .
$d$ — reference (body) diameter; $r = d /2$ the body radius.
Transition diameters $d_{1}$ , $d_{2}$ — the fore (front) and aft (back) diameters of a cone-shaped body section (a flare or boat-tail). If $d_{2} > d_{1}$ the body flares out; if $d_{2} < d_{1}$ it tapers to a boat-tail.
$X_{t}$ , $L_{t}$ — the start position and length of a transition section.
Fin numbers: $C_{r}$ root chord (long edge at body), $C_{t}$ tip chord (short edge outboard), $s$ semi-span (how tall one fin stands), $X_{r}$ sweep (how far the tip's front edge sits behind the root's front edge), $n$ number of fins.
$ℓ_{f}$ — the mid-chord length of a fin: the straight distance from the middle of the root chord to the middle of the tip chord. Because a swept fin's lift acts along this slanted line (not along the rocket axis), $ℓ_{f} = s^{2} + (X_{r} + \frac{C _{t} - C _{r}}{2})^{2}$ . We compute it fresh in every fin example.

Formula The tool-box we reuse (from the parent)

Tool 1 — nose: slope fixed at 2, CP is a shape-fraction $k$ of the nose length. $C_{N α, nose} = 2, \overset{ˉ}{X}_{nose} = k L_{nose}$ Tool 2 — transition (flare/boat-tail between fore diameter $d_{1}$ and aft diameter $d_{2}$ , reference $d$ ): lift appears only where the area changes. $C_{N α, trans} = 2 [(\frac{d _{2}}{d})^{2} - (\frac{d _{1}}{d})^{2}]$ Tool 2b — transition CP (the centroid of that linearly-changing load over the truncated cone, starting at $X_{t}$ , length $L_{t}$ ): $\overset{ˉ}{X}_{trans} = X_{t} + \frac{L _{t}}{3} (1 + \frac{1 - d _{1} / d _{2}}{1 - ( d _{1} / d _{2} ) ^{2}})$ Tool 3 — fins (uses $ℓ_{f}$ defined in the vocabulary above): $C_{N α, fins} = K_{f b} \frac{4 n ( s / d ) ^{2}}{1 + 1 + ( \frac{2 ℓ _{f}}{C _{r} + C _{t}} ) ^{2}}, K_{f b} = 1 + \frac{r}{s + r}$ Tool 3b — fin CP (leading-edge root at $X_{f}$ ; first fraction is the taper-weighted centroid, the bracket averages the two chords): $\overset{ˉ}{X}_{fins} = X_{f} + \frac{X _{r}}{3} \cdot \frac{C _{r} + 2 C _{t}}{C _{r} + C _{t}} + \frac{1}{6} [(C_{r} + C_{t}) - \frac{C _{r} C _{t}}{C _{r} + C _{t}}]$ Tool 4 — combine everything by a normal-force-weighted mean: $\overset{ˉ}{X}_{c p} = \frac{\sum _{i} C _{N α, i} X ˉ _{i}}{\sum _{i} C _{N α, i}}$

Example 1 — Cell A: the baseline (nose + fins)

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

Nose. $C_{N α, n} = 2$ , $\overset{ˉ}{X}_{n} = \frac{2}{3} (15) = 10$ cm. Why this step? Slender-body theory fixes the nose slope at 2 regardless of shape; the cone's load grows as $x$ so its centroid sits at $2/3$ of the length (look at the amber load triangle in the figure).
Mid-chord length $ℓ_{f} = s^{2} + (X_{r} + \frac{C _{t} - C _{r}}{2})^{2} = 25 + (4 - 1.5)^{2} = 31.25 = 5.590$ cm. Why? Lift on a swept fin acts along its slanted mid-chord line, not the axis, so we measure that true length (Tool 3's $ℓ_{f}$ ).
Interference $K_{f b} = 1 + \frac{2}{5 + 2} = 1.2857$ . Why? Air deflected around the body hits the fins extra hard; this factor scales up the fin force.
Fin slope $\frac{2 ℓ _{f}}{C _{r} + C _{t}} = \frac{11.18}{9} = 1.2423$ ; $1 + 1.242 3^{2} = 1.5981$ . $C_{N α, f} = 1.2857 \cdot \frac{4 \cdot 3 \cdot ( 5/4 ) ^{2}}{1 + 1.5981} = 1.2857 \cdot \frac{18.75}{2.5981} = 9.279.$ Why? Straight substitution into Tool 3; note $9.28 ≫ 2$ , so fins dominate.
Fin CP (Tool 3b) $\overset{ˉ}{X}_{f} = 50 + \frac{4}{3} \cdot \frac{6 + 2 \cdot 3}{9} + \frac{1}{6} [9 - \frac{18}{9}] = 50 + 1.778 + 1.167 = 52.944$ cm. Why? First term is the taper-weighted centroid; the bracket averages the chords.
Combine (Tool 4, weighted mean): $\overset{ˉ}{X}_{c p} = \frac{2 ( 10 ) + 9.279 ( 52.944 )}{2 + 9.279} = \frac{20 + 491.3}{11.279} = 45.33 cm .$

Recall Verify

Weighted mean must lie between 10 and 52.9 — it does (45.3). Because the fin weight (9.28) is ~4.6× the nose weight (2), CP is pulled strongly toward the fins, matching the forecast. ✓

Example 2 — Cell B: a flare (positive transition)

The figure draws this flare in cyan: the body walls (white) step outward from $d_{1} = 4$ to $d_{2} = 6$ over the shaded cone. The amber dashed line marks where the transition CP lands — read it against the numbers below.

Slope (Tool 2). $C_{N α, t} = 2 [(\frac{6}{4})^{2} - (\frac{4}{4})^{2}] = 2 [2.25 - 1] = 2.5$ . Why? Lift is generated only where cross-section area changes; in the figure the walls splay outward, so the bracket is positive.
CP (Tool 2b) with $d_{1} / d_{2} = 4/6 = 0.6667$ : $\overset{ˉ}{X}_{t} = X_{t} + \frac{L _{t}}{3} (1 + \frac{1 - 0.6667}{1 - 0.666 7 ^{2}}) = 40 + \frac{8}{3} (1 + \frac{0.3333}{0.5556}) .$ $\frac{0.3333}{0.5556} = 0.6$ , so $\overset{ˉ}{X}_{t} = 40 + 2.667 (1.6) = 40 + 4.267 = 44.27$ cm — exactly the amber dashed line in the figure. Why? This is the centroid of the linearly-growing load along the truncated cone.

Recall Verify

$C_{N α, t} = 2.5 > 0$ ✓ (flare adds force). CP $44.27$ lies inside the flare $[40, 48]$ ✓ and matches the amber marker. The flare, sitting behind the nose, would pull $\overset{ˉ}{X}_{c p}$ aft — good for stability.

Example 3 — Cell C: a boat-tail (NEGATIVE transition)

Slope (Tool 2). $C_{N α, t} = 2 [(\frac{2.5}{4})^{2} - (\frac{4}{4})^{2}] = 2 [0.3906 - 1] = - 1.2188$ . Why? Aft diameter smaller than fore ⇒ bracket negative ⇒ the boat-tail removes normal force where it sits (the amber walls in the figure pinch inward).
CP (Tool 2b) with $d_{1} / d_{2} = 4/2.5 = 1.6$ : $\overset{ˉ}{X}_{t} = 44 + \frac{6}{3} (1 + \frac{1 - 1.6}{1 - 1. 6 ^{2}}) = 44 + 2 (1 + \frac{- 0.6}{- 1.56}) .$ $\frac{- 0.6}{- 1.56} = 0.3846$ , so $\overset{ˉ}{X}_{t} = 44 + 2 (1.3846) = 44 + 2.769 = 46.77$ cm. Why? Same centroid formula; note the ratios stay valid even when $d_{1} > d_{2}$ (both numerator and denominator flip sign, giving a legitimate positive fraction).

Recall Verify

$C_{N α, t} = - 1.219 < 0$ ✓. CP $46.77$ inside $[44, 50]$ ✓. A negative weight in the average is legal; if $∣ \sum C_{N α} ∣$ stays positive overall the mean is still meaningful.

Example 4 — Cell D: degenerate fin ( $C_{t} \to 0$ , triangular)

Mid-chord $ℓ_{f} = 6^{2} + (8 + \frac{0 - 8}{2})^{2} = 36 + 16 = 52 = 7.211$ cm. Why? $\frac{C _{t} - C _{r}}{2} = - 4$ ; add to $X_{r} = 8$ gives $4$ ; the slant length is honest even for a delta.
Interference $K_{f b} = 1 + \frac{2.5}{6 + 2.5} = 1.2941$ .
Slope (Tool 3) $\frac{2 ℓ _{f}}{C _{r} + C _{t}} = \frac{14.42}{8} = 1.8028$ ; $1 + 1.802 8^{2} = 2.0592$ . $C_{N α, f} = 1.2941 \cdot \frac{4 \cdot 4 \cdot ( 6/5 ) ^{2}}{1 + 2.0592} = 1.2941 \cdot \frac{23.04}{3.0592} = 9.746.$ Why? $C_{r} + C_{t} = 8$ is finite even with $C_{t} = 0$ ; no division problem here.
Fin CP (Tool 3b) — check the danger term. $\overset{ˉ}{X}_{f} = X_{f} + \frac{X _{r}}{3} \cdot \frac{C _{r} + 2 C _{t}}{C _{r} + C _{t}} + \frac{1}{6} [(C_{r} + C_{t}) - \frac{C _{r} C _{t}}{C _{r} + C _{t}}] .$ With $C_{t} = 0$ : $\frac{C _{r} + 0}{C _{r}} = 1$ and $\frac{C _{r} C _{t}}{C _{r} + C _{t}} = \frac{0}{8} = 0$ — not $0/0$ , because the denominator $C_{r} + C_{t} = 8 \neq = 0$ . $\overset{ˉ}{X}_{f} = 60 + \frac{8}{3} (1) + \frac{1}{6} [8 - 0] = 60 + 2.667 + 1.333 = 64.00 cm .$ Why this step? The apparent singularity dissolves: the product $C_{r} C_{t}$ hits zero while its denominator stays positive.

Recall Verify

$\frac{C _{r} C _{t}}{C _{r} + C _{t}} = 0$ (finite), so no blow-up ✓. CP $64.0$ cm sits inside the fin's chord $[60, 68]$ ✓. A delta fin's CP lands at exactly $C_{r} /2$ back plus sweep-share here.

Example 5 — Cells E & F: skinny vs fat body ( $K_{f b}$ limits)

Skinny (E): $K_{f b} = 1 + \frac{0.1}{5.1} = 1.0196 \approx 1$ . Why? With almost no body, there is nothing to deflect the flow around, so fins act as if in free air ( $K_{f b} \to 1$ ).
Fat (F): $K_{f b} = 1 + \frac{100}{105} = 1.9524 \approx 2$ . Why? A huge body doubles the effective flow the fins see; the ratio $\frac{r}{s + r} \to 1$ as $r \to \infty$ , so $K_{f b} \to 2$ .
Consequence. The fin slope scales linearly with $K_{f b}$ , so a fat body nearly doubles the fin's contribution to $C_{N α}$ versus a skinny one — CP shifts noticeably aft.

Recall Verify

$lim_{r \to 0} K_{f b} = 1$ ✓, $lim_{r \to \infty} K_{f b} = 2$ ✓. Both computed values bracket exactly these limits: $1.0196$ and $1.9524$ .

Example 6 — Cell G: real-world word problem

Translate prose → symbols. $d = 5.4$ , $r = 2.7$ ; $n = 4$ , $C_{r} = 7$ , $C_{t} = 2$ , $s = 5.5$ , $X_{r} = 5$ , $X_{f} = 68$ ; $L_{n} = 16$ , $k = 0.466$ . All cm. Why? Half the battle is naming: "across" = diameter, "standing out" = semi-span, "behind the root's front edge" = sweep.
Nose (Tool 1): $C_{N α, n} = 2$ , $\overset{ˉ}{X}_{n} = 0.466 \cdot 16 = 7.456$ cm. Why? Ogive uses its own $k$ ; slope is still 2 (shape-independent).
Mid-chord $ℓ_{f} = 5. 5^{2} + (5 + \frac{2 - 7}{2})^{2} = 30.25 + (2.5)^{2} = 36.5 = 6.042$ cm.
Interference $K_{f b} = 1 + \frac{2.7}{5.5 + 2.7} = 1.3293$ .
Fin slope (Tool 3) $\frac{2 ℓ _{f}}{C _{r} + C _{t}} = \frac{12.083}{9} = 1.3426$ ; $1 + 1.342 6^{2} = 1.6739$ . $C_{N α, f} = 1.3293 \cdot \frac{4 \cdot 4 \cdot ( 5.5/5.4 ) ^{2}}{1 + 1.6739} = 1.3293 \cdot \frac{16.596}{2.6739} = 8.251.$
Fin CP (Tool 3b) $\overset{ˉ}{X}_{f} = 68 + \frac{5}{3} \cdot \frac{7 + 4}{9} + \frac{1}{6} [9 - \frac{14}{9}] = 68 + 2.037 + 1.241 = 71.278$ cm.
Combine (Tool 4): $\overset{ˉ}{X}_{c p} = \frac{2 ( 7.456 ) + 8.251 ( 71.278 )}{2 + 8.251} = \frac{14.912 + 588.12}{10.251} = 58.83$ cm.

Recall Verify

CP between 7.46 and 71.3 ✓. As forecast, the longer body pushes CP aft (58.8 vs 45.3 in Ex 1) ✓. Ogive's smaller $k$ barely matters since the fins hold most of the weight.

Example 7 — Cell H: exam twist (solve for the CG target)

Invert the formula. $SM = \frac{X ˉ _{c p} - X _{c g}}{d} \Rightarrow X_{c g} = \overset{ˉ}{X}_{c p} - SM \cdot d$ . Why? Static margin is a definition we can algebraically rearrange — CP and $d$ are known, solve for $X_{c g}$ .
1.5 calibres: $X_{c g} = 58.83 - 1.5 (5.4) = 58.83 - 8.1 = 50.73$ cm.
1.0 calibre (minimum): $X_{c g} = 58.83 - 1.0 (5.4) = 58.83 - 5.4 = 53.43$ cm. Why? Smaller margin means CG allowed closer to CP, i.e. further aft (53.4 > 50.7) — matching the forecast.
Reading it: to keep margin $\geq 1.0$ , CG must be at or forward of 53.43 cm; for the target 1.5, aim for 50.73 cm.

Recall Verify

Plug back: $(58.83 - 50.73) /5.4 = 8.1/5.4 = 1.5$ ✓; $(58.83 - 53.43) /5.4 = 5.4/5.4 = 1.0$ ✓. Higher margin ⇒ CG forward, confirmed. See Static stability and static margin and Centre of gravity determination for the CG side.

Example 8 — Cell I: the sanity case ( $α = 0$ gives zero force)

Force at zero tilt. Normal force on each component is $N_{i} = \frac{1}{2} ρ v^{2} A_{r e f} C_{N α, i} α$ . At $α = 0$ , every $N_{i} = 0$ . Why? A symmetric body pointing straight into the wind feels equal push on both sides — they cancel, so the net side force is exactly zero.
So absolute forces are useless for finding CP: the moment balance would read $0 = \overset{ˉ}{X}_{c p} \cdot 0$ , i.e. $0/0$ — undefined. Why? You cannot locate a balance point of forces that are all zero.
Slopes survive the limit. Divide the moment balance $\sum N_{i} \overset{ˉ}{X}_{i} = \overset{ˉ}{X}_{c p} \sum N_{i}$ by $\frac{1}{2} ρ v^{2} A_{r e f} α$ — every factor of $α$ cancels, leaving $\overset{ˉ}{X}_{c p} = \frac{\sum C _{N α, i} X ˉ _{i}}{\sum C _{N α, i}},$ which is finite and angle-independent. Why this step? This is exactly why the parent said "every Barrowman term is a $_{α}$ quantity" — it is the derivative that stays alive as $α \to 0$ .
Numeric demonstration. Take Ex 1's numbers ( $C_{N α, n} = 2$ at $\overset{ˉ}{X}_{n} = 10$ ; $C_{N α, f} = 9.279$ at $\overset{ˉ}{X}_{f} = 52.944$ ). Pick any tilt, say $α = 2^{\circ} = 0.0349$ rad, and let $q = \frac{1}{2} ρ v^{2} A_{r e f}$ . Then $N_{n} = q (2) (0.0349), N_{f} = q (9.279) (0.0349) .$ Weighted mean: $\overset{ˉ}{X}_{c p} = \frac{N _{n} ( 10 ) + N _{f} ( 52.944 )}{N _{n} + N _{f}}$ . Both $q$ and $0.0349$ appear in every term and cancel, giving $\frac{2 ( 10 ) + 9.279 ( 52.944 )}{2 + 9.279} = 45.33$ cm — identical to Ex 1, for any $α$ you had chosen. Why? It proves CP is a purely geometric property; the tilt only sets the size of the force, never where it acts.

Recall Verify

Set $α = 0$ : numerator and denominator forces both vanish ( $0/0$ ), but their slopes do not — the ratio is well-defined and equals 45.33 cm. Choosing $α = 2^{\circ}$ gives the same 45.33 cm because $q α$ cancels. ✓ (This links to Angle of attack and restoring moment.)

Recall One-line recap of the matrix

Baseline weighted mean :: Ex 1 — CP 45.3 cm Flare = positive term :: Ex 2 — $C_{N α} = + 2.5$ Boat-tail = negative term :: Ex 3 — $C_{N α} = - 1.22$ Delta fin $C_{t} \to 0$ is safe :: Ex 4 — CP 64.0 cm, no $0/0$ $K_{f b}$ runs $1 \to 2$ :: Ex 5 — skinny 1.02, fat 1.95 Word problem :: Ex 6 — CP 58.8 cm Solve for CG :: Ex 7 — 50.73 cm at 1.5 cal Only slopes matter at $α = 0$ :: Ex 8 — CP 45.3 cm for any $α$

See also: Slender-body aerodynamic theory · Nose cone shapes and drag · Rocket flight simulation (6-DOF).