Barrowman equations — centre of pressure calculation for finned rockets: Exercises

Intuition One ruler for the whole rocket — read the origin carefully!

All the final CP positions on this page are measured from one fixed origin: the nose tip, with $x$ increasing toward the tail. But the individual component formulas are each written from their own local landmark, and you must add the offset to bring them onto the shared ruler:

Nose: $\overset{ˉ}{X}_{n} = k L_{n}$ is already measured from the tip — no offset needed.
Transition: the formula starts from $X_{t}$ , the point where the transition begins (itself measured from the tip).
Fins: $\overset{ˉ}{X}_{fins}$ starts from $X_{f}$ , the fin root leading edge (itself measured from the tip). The parameters $X_{t}$ and $X_{f}$ are exactly the bridges that carry each local result onto the one shared tip-based ruler. Mixing origins is the single most common CP blunder — always confirm every $\overset{ˉ}{X}_{i}$ is finally quoted from the tip before you average.

Quick reference — the five tools you will reuse, each with its one-line why:

Formula The toolbox (with intuition)

Nose: slope $C_{N α, nose} = 2$ , CP at $\overset{ˉ}{X}_{n} = k L_{n}$ (cone $k = \frac{2}{3}$ , ogive $k = 0.466$ , parabola $k = 0.5$ ). Why 2: slender-body lift slope $= 2 \times$ (base area/reference area) $= 2$ . Why $k L_{n}$ : CP is the centroid of the area-growth loading. Transition: $C_{N α, trans} = 2 [(\frac{d _{2}}{d})^{2} - (\frac{d _{1}}{d})^{2}]$ . Why a difference of squares: lift is generated only where the cross-section area changes, so it counts aft area minus fore area. Here $d$ is the reference diameter (the body diameter used to non-dimensionalise), $d_{1}$ the fore and $d_{2}$ the aft diameter of the cone frustum. Fins (of $n$ fins): $C_{N α, fins} = K_{f b} \frac{4 n ( s / d ) ^{2}}{1 + 1 + ( \frac{2 ℓ _{f}}{C _{r} + C _{t}} ) ^{2}}, K_{f b} = 1 + \frac{r}{s + r}$ Why this shape: start from thin-airfoil lift, correct for finite aspect-ratio (that is the $\dots$ denominator — long slender fins lift more per area, stubby ones less), scale by $(s / d)^{2}$ (bigger span vs. body wins fast) and by the number of fins $n$ ; finally $K_{f b} > 1$ adds the flow the fat body deflects onto the roots. Here $d$ is again the reference diameter. Combine (normal-force-weighted mean): $\overset{ˉ}{X}_{c p} = \frac{\sum _{i} C _{N α, i} X ˉ _{i}}{\sum _{i} C _{N α, i}}$ Why a weighted mean: net moment $=$ net force $\times$ lever arm, i.e. $\sum N_{i} \overset{ˉ}{X}_{i} = \overset{ˉ}{X}_{c p} \sum N_{i}$ ; the piece that pushes hardest (biggest $C_{N α, i}$ ) drags the balance point most toward itself — exactly like a see-saw loaded unevenly. Stability: static margin $= \frac{X ˉ _{c p} - X _{c g}}{d}$ calibres — see Static stability and static margin. Careful: here $d$ plays a different role — it is the body diameter used as the length yard-stick ("one calibre"), converting a distance in cm into a count of body-diameters. Same symbol, but in the slope formulas $d$ normalises area, whereas here it normalises length.

Level 1 — Recognition

Exercise 1.1

State the nose-cone normal-force slope $C_{N α, nose}$ for any slender nose, and explain in one sentence why it does not depend on the nose shape.

Recall Solution

$C_{N α, nose} = 2$ (per radian). Why shape-independent: slender-body theory says the slope equals $2 \times (base area) / (reference area)$ ; taking the nose base as the reference makes that ratio exactly $1$ , so only the base area matters — never the ogive/cone/parabola profile in between.

Exercise 1.2

A cone nose has length $L_{n} = 21$ cm. Where is its centre of pressure, measured from the tip?

Recall Solution

Cone shape constant $k = \frac{2}{3}$ , so $\overset{ˉ}{X}_{n} = \frac{2}{3} L_{n} = \frac{2}{3} (21) = 14 cm from the tip.$ Why $\frac{2}{3}$ : for a cone the cross-section area grows as $S \propto x^{2}$ , the lift-per-length $\propto d S / d x \propto x$ grows linearly, and the centroid of a triangle-shaped load sits at $\frac{2}{3}$ of the length. Because this is already measured from the tip, it needs no origin offset.

Level 2 — Application

Exercise 2.1

A body transition (shoulder) grows from fore diameter $d_{1} = 3$ cm to aft diameter $d_{2} = 5$ cm. The reference (body) diameter is $d = 5$ cm. Find the transition's normal-force slope. Is it stabilising (positive) or destabilising (negative)?

Recall Solution

$C_{N α, trans} = 2 [(\frac{5}{5})^{2} - (\frac{3}{5})^{2}] = 2 [1 - 0.36] = 2 (0.64) = 1.28.$ Positive ⇒ it adds normal force. Why: a flare ( $d_{2} > d_{1}$ ) turns flow outward and catches air like a small extra nose; only where area increases is lift generated.

Exercise 2.2

A boat-tail shrinks from $d_{1} = 5$ cm to $d_{2} = 3$ cm, with reference $d = 5$ cm. Compute $C_{N α, trans}$ and interpret the sign.

Recall Solution

$C_{N α, trans} = 2 [(\frac{3}{5})^{2} - (\frac{5}{5})^{2}] = 2 [0.36 - 1] = 2 (- 0.64) = - 1.28.$ Negative — a boat-tail removes normal force at the tail, which nudges CP forward and is mildly destabilising. Why: the area now shrinks, so the flow closes inward and the local pressure loading points the opposite way compared with a flare.

Exercise 2.3

Three fins ( $n = 3$ ) sit on a body of radius $r = 2$ cm; each fin has semi-span $s = 6$ cm. Compute the body-interference factor $K_{f b}$ .

Recall Solution

$K_{f b} = 1 + \frac{r}{s + r} = 1 + \frac{2}{6 + 2} = 1 + 0.25 = 1.25.$ Why $> 1$ : flow bends around the fat body and spills onto the fin roots, so each fin feels more deflected air than it would in isolation — a $25%$ boost here.

Level 3 — Analysis

The figure below is the map for this whole level. It shows one trapezoidal fin flattened out: the horizontal axis runs along the chord (nose-to-tail direction), the vertical axis runs out along the span (away from the body). Look for the violet outline (the fin), the magenta double-arrows marking the root chord $C_{r}$ at the bottom and the tip chord $C_{t}$ at the top, the orange vertical arrow marking the semi-span $s$ , and the navy arrow at the top marking the sweep $X_{r}$ (leading-edge to leading-edge). The dashed orange line joining the two orange dots is the mid-chord line whose length is $ℓ_{f}$ — the single quantity Exercise 3.1 asks you to measure. Notice how the mid-chord line does not run straight up: it leans back by more than $X_{r}$ because the shorter tip chord shifts its own mid-point forward.

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

Exercise 3.1

A trapezoidal fin (see figure) has root chord $C_{r} = 8$ cm, tip chord $C_{t} = 4$ cm, semi-span $s = 6$ cm, and sweep distance $X_{r} = 5$ cm (how far the tip leading edge lies behind the root leading edge). Compute the mid-chord sweep length $ℓ_{f}$ .

Recall Solution

The mid-chord line runs from the middle of the root chord to the middle of the tip chord (the two orange dots in the figure). Horizontally it moves by $X_{r} + \frac{C _{t} - C _{r}}{2}$ (sweep plus the shift because the tip is shorter), vertically it climbs the semi-span $s$ : $ℓ_{f} = s^{2} + (X_{r} + \frac{C _{t} - C _{r}}{2})^{2} = 6^{2} + (5 + \frac{4 - 8}{2})^{2} = 36 + (5 - 2)^{2} = 36 + 9 = 45 = 6.708 cm .$ Why the $\frac{C _{t} - C _{r}}{2}$ term: the mid-chord point of a shorter tip chord sits further forward relative to its leading edge, so the horizontal run of the mid-chord line is not just $X_{r}$ — exactly the lean you see in the dashed orange line.

Exercise 3.2

Using the fin above with $n = 3$ , body radius $r = 2$ cm, reference diameter $d = 4$ cm, compute $C_{N α, fins}$ .

Recall Solution

Step 1 — interference: $K_{f b} = 1 + \frac{2}{6 + 2} = 1.25$ . Step 2 — the aspect-ratio bracket: $\frac{2 ℓ _{f}}{C _{r} + C _{t}} = \frac{2 ( 6.708 )}{8 + 4} = \frac{13.416}{12} = 1.118$ , so $1 + 1.11 8^{2} = 2.25 = 1.5$ . Step 3 — geometric core: $\frac{4 n ( s / d ) ^{2}}{1 + 1.5} = \frac{4 ( 3 ) ( 6/4 ) ^{2}}{2.5} = \frac{12 ( 2.25 )}{2.5} = \frac{27}{2.5} = 10.8$ . Step 4 — apply interference: $C_{N α, fins} = 1.25 \times 10.8 = 13.5.$ Why fins dominate: the $(s / d)^{2} = 2.25$ factor and the $\times n$ multiply up quickly, so fins routinely swamp the nose's slope of $2$ .

Exercise 3.3

For the same fin, find the fin CP location $\overset{ˉ}{X}_{fins}$ if the root leading edge sits at $X_{f} = 55$ cm.

Recall Solution

$\overset{ˉ}{X}_{fins} = X_{f} + \frac{X _{r}}{3} \cdot \frac{C _{r} + 2 C _{t}}{C _{r} + C _{t}} + \frac{1}{6} [(C_{r} + C_{t}) - \frac{C _{r} C _{t}}{C _{r} + C _{t}}] .$ Here $X_{f} = 55$ cm (the fin root leading edge, measured from the tip) is the offset that lifts the local fin result onto the shared tip-based ruler. Term 2 (taper centroid): $\frac{5}{3} \cdot \frac{8 + 2 ( 4 )}{8 + 4} = \frac{5}{3} \cdot \frac{16}{12} = \frac{5}{3} (1.3333) = 2.2222$ cm. Term 3 (chord-averaging): $\frac{1}{6} [12 - \frac{8 \cdot 4}{12}] = \frac{1}{6} [12 - 2.6667] = \frac{1}{6} (9.3333) = 1.5556$ cm. $\overset{ˉ}{X}_{fins} = 55 + 2.2222 + 1.5556 = 58.78 cm from the tip.$ Why aft of $X_{f}$ : sweep and taper push the load rearward and outward, so the CP lands about $3.8$ cm behind the root leading edge.

Level 4 — Synthesis

The figure below is the master picture for Level 4: a side view of the whole rocket, tip on the left, tail on the right, drawn on the one shared tip-based ruler (the horizontal axis is "distance from nose tip, cm"). Spot the magenta dot (nose CP at $12$ cm), the violet dot (fin CP at $58.8$ cm), the orange triangle (the combined CP the weighted mean produces, at $52.7$ cm), and the navy diamond (the CG at $47$ cm). The double-arrow between CG and CP is the static margin. The key thing to glean: the combined CP sits far to the right, close to the fins — because the fins carry most of the total slope, they drag the balance point strongly toward themselves.

Exercise 4.1

A full rocket has:

Cone nose: $L_{n} = 18$ cm ( $k = \frac{2}{3}$ ), so $\overset{ˉ}{X}_{n}$ from tip.
Fins: from Exercises 3.2–3.3, $C_{N α, fins} = 13.5$ at $\overset{ˉ}{X}_{fins} = 58.78$ cm.

Find the total CP location $\overset{ˉ}{X}_{c p}$ and the total slope $C_{N α, total}$ .

Recall Solution

Nose CP: $\overset{ˉ}{X}_{n} = \frac{2}{3} (18) = 12$ cm from the tip; nose slope $= 2$ . Both CPs are already on the shared tip-based ruler, so we may average directly. Total slope: $C_{N α, total} = 2 + 13.5 = 15.5$ . Weighted mean: $\overset{ˉ}{X}_{c p} = \frac{2 ( 12 ) + 13.5 ( 58.78 )}{2 + 13.5} = \frac{24 + 793.5}{15.5} = \frac{817.5}{15.5} = 52.74 cm from the tip.$ Why so close to the fins: the fins carry $13.5/15.5 \approx 87%$ of the total slope, so the balance point is dragged almost all the way back to them.

Exercise 4.2

The rocket's CG is at $X_{c g} = 47$ cm and $d = 4$ cm. Compute the static margin and state whether it is stable, and whether it lies in the "healthy" $1$ – $2$ calibre band.

Recall Solution

$static margin = \frac{X ˉ _{c p} - X _{c g}}{d} = \frac{52.74 - 47}{4} = \frac{5.74}{4} = 1.435 calibres .$ Here $d = 4$ cm is the length yard-stick (one calibre $=$ one body diameter), turning the $5.74$ cm gap into a count of body-diameters. Positive ⇒ stable (CP behind CG). And $1.435$ sits comfortably inside the $1$ – $2$ calibre "stable but not over-stiff" range — see Angle of attack and restoring moment for why too large a margin makes a rocket weathercock hard into gusts.

Exercise 4.3

Now add a boat-tail at the rear: $d_{1} = 4$ cm → $d_{2} = 3$ cm, reference $d = 4$ cm, starting at $X_{t} = 62$ cm with length $L_{t} = 4$ cm. Recompute the total CP.

Recall Solution

Slope first. $C_{N α, trans} = 2 [(3/4)^{2} - (4/4)^{2}] = 2 [0.5625 - 1] = - 0.875$ .

Now the transition CP — let us build the formula, not just quote it. A conical transition is a truncated cone (frustum). Just like the nose, its lift-per-length follows the area growth $d S / d x$ , which for a cone-shaped wall varies linearly with position. The CP is the centroid of that linear loading over the frustum from $X_{t}$ to $X_{t} + L_{t}$ . Doing the same centroid integral we did for the cone nose — but now over a slice that starts at a non-zero radius (because the fore end already has diameter $d_{1}$ ) — gives the standard result $\overset{ˉ}{X}_{trans} = X_{t} + \frac{L _{t}}{3} (1 + \frac{1 - d _{1} / d _{2}}{1 - ( d _{1} / d _{2} ) ^{2}}) .$ How to read it: if $d_{1} = 0$ (a full cone, no truncation) the fraction collapses and the bracket becomes $1$ , giving $\overset{ˉ}{X} = X_{t} + \frac{L _{t}}{3}$ — but measured from the wide base backwards this is the same $\frac{2}{3}$ -centroid we found for the nose. The extra term $\frac{1 - d _{1} / d _{2}}{1 - ( d _{1} / d _{2} ) ^{2}}$ is the correction for chopping off the cone's tip: a truncated frustum has its area-growth spread differently, shifting the centroid. And $X_{t}$ is simply the offset that puts the answer on the shared tip-based ruler.

Ratio $d_{1} / d_{2} = 4/3 = 1.3333$ (here the fore diameter is larger, since it is a boat-tail). $\overset{ˉ}{X}_{trans} = 62 + \frac{4}{3} (1 + \frac{1 - 1.3333}{1 - 1.333 3 ^{2}}) = 62 + 1.3333 (1 + \frac{- 0.3333}{1 - 1.7778}) .$ Inner fraction: $\frac{- 0.3333}{- 0.7778} = 0.4286$ , so bracket $= 1.4286$ , and $\overset{ˉ}{X}_{trans} = 62 + 1.3333 (1.4286) = 62 + 1.905 = 63.905$ cm from the tip.

New total slope: $2 + 13.5 + (- 0.875) = 14.625$ . New CP: $\overset{ˉ}{X}_{c p} = \frac{2 ( 12 ) + 13.5 ( 58.78 ) + ( - 0.875 ) ( 63.905 )}{14.625} = \frac{24 + 793.5 - 55.917}{14.625} = \frac{761.58}{14.625} = 52.07 cm.$ Why CP moved forward: the boat-tail contributes a negative force way at the back — subtracting a rear load pulls the balance point slightly forward (from $52.74$ to $52.07$ cm). It shaved the static margin by about $0.17$ calibres.

Level 5 — Mastery

Exercise 5.1 (design)

You have the nose ( $C_{N α} = 2$ at $12$ cm) and fins ( $C_{N α, fins} = 13.5$ at $58.78$ cm) of Exercise 4.1, giving $\overset{ˉ}{X}_{c p} = 52.74$ cm. Your CG is fixed at $X_{c g} = 47$ cm, $d = 4$ cm. Your customer demands a static margin of exactly $2.0$ calibres. By how much must you move the fin set (change $X_{f}$ ), assuming the fin CP moves rigidly with $X_{f}$ and the fin slope is unchanged?

Recall Solution

Required CP: $\overset{ˉ}{X}_{c p} = X_{c g} + 2.0 d = 47 + 2 (4) = 55$ cm from the tip. Let the fin CP shift by $Δ$ . The new weighted mean must equal $55$ : $\frac{2 ( 12 ) + 13.5 ( 58.78 + Δ )}{15.5} = 55.$ Numerator must equal $55 \times 15.5 = 852.5$ . Currently it is $817.5$ , so we need extra $852.5 - 817.5 = 35$ in the numerator. That extra comes only from $13.5 Δ$ : $13.5 Δ = 35 \Rightarrow Δ = \frac{35}{13.5} = 2.593 cm .$ Move the fins $2.59$ cm rearward. Why it works cleanly: since only the fin term carries $Δ$ and the denominator is unchanged, the shift in $\overset{ˉ}{X}_{c p}$ is just $\frac{13.5}{15.5} Δ$ — the fins transmit $87%$ of their own movement to the CP.

Exercise 5.2 (limiting behaviour)

Take the fin geometry of Exercise 3.2 but imagine the body becomes infinitely thin ( $r \to 0$ ) and separately infinitely fat ( $r ≫ s$ ). What are the limiting values of $K_{f b}$ , and what do they mean physically?

Recall Solution

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

$r \to 0$ : $K_{f b} \to 1 + 0 = 1$ . Meaning: with no body, fins act in isolation — no extra deflected flow, no interference boost.
$r ≫ s$ : $\frac{r}{s + r} \to 1$ , so $K_{f b} \to 2$ . Meaning: on a very fat body the fins see the maximum possible deflected flow; interference doubles their effectiveness. The factor can never exceed $2$ in this model — a built-in ceiling that keeps the estimate physical. Compare with the assumptions in Rocket flight simulation (6-DOF) where such closed-form limits are used as sanity checks.

Exercise 5.3 (synthesis + judgement)

A student doubles every fin dimension ( $C_{r}, C_{t}, s, X_{r}$ all $\times 2$ ) to "make the rocket more stable," keeping $n = 3$ , $d = 4$ cm, $r = 2$ cm, $X_{f} = 55$ cm. Show what happens to $C_{N α, fins}$ , feed it back into the CP and static margin, and explain why bigger fins do not guarantee more static margin.

Recall Solution

Doubling all fin lengths scales $s \to 12$ , $C_{r} \to 16$ , $C_{t} \to 8$ , $X_{r} \to 10$ , so $ℓ_{f} \to 2 (6.708) = 13.416$ cm.

Bracket: $\frac{2 ℓ _{f}}{C _{r} + C _{t}} = \frac{2 ( 13.416 )}{24} = 1.118$ — unchanged (both parts scaled together). So $1 + 1.11 8^{2} = 1.5$ as before.
Core: $\frac{4 n ( s / d ) ^{2}}{2.5} = \frac{12 ( 12/4 ) ^{2}}{2.5} = \frac{12 ( 9 )}{2.5} = 43.2$ .
$K_{f b}$ unchanged at $1.25$ (body untouched). $C_{N α, fins} = 1.25 (43.2) = 54.0 - four times the original 13.5.$ Feed it back into the CP. The fin CP also moves aft because chords and sweep doubled; recomputing Exercise 3.3 with the doubled numbers: $\overset{ˉ}{X}_{fins} = 55 + \frac{10}{3} \cdot \frac{16 + 2 ( 8 )}{16 + 8} + \frac{1}{6} [24 - \frac{16 \cdot 8}{24}] = 55 + 4.444 + 3.111 = 62.56 cm .$ New total slope $= 2 + 54.0 = 56.0$ ; new CP: $\overset{ˉ}{X}_{c p} = \frac{2 ( 12 ) + 54.0 ( 62.56 )}{56.0} = \frac{24 + 3378.2}{56.0} = \frac{3402.2}{56.0} = 60.75 cm .$ Static margin if CG stayed at $47$ cm: $\frac{60.75 - 47}{4} = 3.44$ calibres — huge on paper. But the parent note's Centre of gravity determination warning bites: doubling fin dimensions roughly quadruples fin mass placed at the tail, dragging $X_{c g}$ rearward. Suppose the heavier fins pull CG back to $X_{c g} = 58$ cm; then static margin collapses to $\frac{60.75 - 58}{4} = 0.69$ calibres — worse than the original $1.4$ , and dangerously close to neutral. Conclusion: the fin slope quadrupled and CP moved aft, yet the static margin can still shrink, because margin is the difference $\overset{ˉ}{X}_{c p} - X_{c g}$ and the added tail mass moves CG the same direction as CP. Always recompute CP and CG together.

Recall Self-test cloze — the essentials

Nose slope is always ::: $C_{N α, nose} = 2$ (per radian) Cone CP sits at ::: $\frac{2}{3}$ of the nose length from the tip A boat-tail's transition slope is ::: negative (destabilising) Body interference factor ::: $K_{f b} = 1 + \frac{r}{s + r}$ , bounded between $1$ and $2$ All final CP positions are measured from ::: the nose tip (one shared ruler) CP is computed as the ::: normal-force-slope-weighted mean of component CPs Healthy static margin band ::: $1$ – $2$ calibres