Exercises — Static stability — weather-cocking tendency

Compute $\ell =X_{cp}-X_{cg}$ for each and read the rule: $\ell >0$ stable, $\ell <0$ unstable, $\ell =0$ neutral.

• A: $\ell =0.55-0.40=+0.15\text{m}>0$ → CP behind CG → stable ✅
• B: $\ell =0.60-0.60=0$ → CP exactly on CG → neutrally stable (no restoring moment either way).
• C: $\ell =0.62-0.70=-0.08\text{m}<0$ → CP ahead of CG → unstable ❌

What we did: turned three position pairs into one signed lever arm each. Why: the ordering of CP and CG, captured by the sign of $\ell$, is the whole story for static stability.

(a) Weather-cocking (weather-vaning). (b) It proves the rocket is statically stable — the aerodynamic force produced a restoring moment.

Step 1: $\ell =0.92-0.80=0.12\text{m}$. Why: positive ⇒ CP behind CG ⇒ stable in principle. Step 2: $\text{SM}=\ell /d=0.12/0.04=3.0$ calibers. Why divide by $d$: one caliber = one diameter, giving a size-independent number. Result: $\text{SM}=3.0$ — stable but over-stable (well above 2). Expect aggressive weather-cocking into crosswinds.

First convert the tilt: $\alpha =4^{\circ }=4\times \pi /180=0.06981\text{rad}$. Why: $C_{N_{\alpha }}$ is "per radian", so $\alpha$ must be in radians. (a) $q=\frac{1}{2}\rho V^2=\frac{1}{2}(1.2)(80^2)=\frac{1}{2}(1.2)(6400)=3840\text{Pa}$. Why: dynamic pressure sets the force scale. (b) $N=qA{C}_{N_{\alpha }}\alpha =3840(1.5\times 10^{-3})(2.5)(0.06981)=1.005\text{N}$. (c) $M=-N\ell =-1.005(0.10)=-0.1005\text{N⋅m}$. Restores? Yes — the minus sign means the moment opposes the positive tilt, pushing $\alpha$ back to zero.

Reasoning: $M=-\frac{1}{2}\rho V^{2}A{C}_{N_{\alpha }}\ell \alpha$. Only $V$ changed, and $M\propto V^2$. Doubling $V$ multiplies $M$ by $2^2=4$. Predict: $M_{\text{new}}=4\times (-0.1005)=-0.402\text{N⋅m}$. Verify: $q_{\text{new}}=\frac{1}{2}(1.2)(160^2)=15360\text{Pa}$; $N=15360(1.5\times 10^{-3})(2.5)(0.06981)=4.021\text{N}$; $M=-4.021(0.10)=-0.402\text{N⋅m}$. ✔ Insight: stability authority grows with the square of speed — a slow rocket just off the pad has the weakest restoring moment, which is why low launch speed is the dangerous moment.

Step 1: $\ell =\text{SM}\times d=1.5\times 0.05=0.075\text{m}$. Why: $\text{SM}=\ell /d\Rightarrow \ell =\text{SM}\cdot d$. Step 2: $\ell =X_{cp}-X_{cg}\Rightarrow X_{cg}=X_{cp}-\ell =1.00-0.075=0.925\text{m}$. Why: CG sits ahead of CP by the lever arm. Result: $X_{cg}=0.925\text{m}$ from the nose.

Step 1 — target lever arm: need $\ell =\text{SM}\times d=1.5\times 0.04=0.06\text{m}$. Step 2 — required CG: $\ell =X_{cp}-X_{cg}\Rightarrow X_{cg}=X_{cp}-\ell =0.58-0.06=0.52\text{m}$. Step 3 — sanity: original $X_{cg}=0.60$, target $0.52$ ⇒ CG must move forward by $0.08\text{m}$. Nose ballast moves CG forward, so this is achievable. ✔ Result: move the CG to $X_{cg}=0.52\text{m}$. Now $\ell =+0.06\text{m}>0$ and $\text{SM}=1.5$. See Fin Design and Sizing for the alternative of moving CP rearward instead.

Step 1: need $\ell =1.5\times 0.04=0.06\text{m}$. Step 2: $\ell =X_{cp}-X_{cg}\Rightarrow X_{cp}=X_{cg}+\ell =0.60+0.06=0.66\text{m}$. Result: move CP to $X_{cp}=0.66\text{m}$ (bigger/further-aft fins per Barrowman Equations for CP location). Same margin, opposite lever — either forward CG or aft CP works because both increase $\ell$.

Convert margins to lever arms: $\ell =\text{SM}\cdot d$, so $\ell \in [1.0\times 0.05,2.0\times 0.05]=[0.05,0.10]\text{m}$. Turn into CG positions: $X_{cg}=X_{cp}-\ell$.

• At $\ell =0.05$: $X_{cg}=1.10-0.05=1.05\text{m}$ (that is the most rearward allowed CG, $\text{SM}=1$).
• At $\ell =0.10$: $X_{cg}=1.10-0.10=1.00\text{m}$ (the most forward allowed CG, $\text{SM}=2$). Result: allowed band $X_{cg}\in [1.00,1.05]\text{m}$. Larger $\ell$ (more forward CG) ⇒ more margin ⇒ closer to over-stable; smaller $\ell$ ⇒ closer to neutral.

On the pad: ${\ell }_0=1.00-0.90=0.10\text{m}$, ${\text{SM}}_0=0.10/0.05=2.0$ calibers — comfortably stable. In flight: ${\ell }_1=0.93-0.85=0.08\text{m}$, ${\text{SM}}_1=0.08/0.05=1.6$ calibers — still positive, still stable. Comment: the forward CP drift ate margin, but the simultaneous forward CG shift from fuel burn gave some back. Net result stays inside $1$–$2$ calibers, so stability survives the whole flight. The lesson (see Thrust Vectoring vs Passive Stability): you must check the margin at the worst instant, not just on the pad, because both CP and CG move.

Lever arm: $\ell =0.70-0.70=0$. Static margin: $\text{SM}=0/d=0$ calibers. Moment: $M=-\frac{1}{2}\rho V^{2}A{C}_{N_{\alpha }}\ell \alpha =-(\cdots )\times 0\times \alpha =0$ for every $\alpha$. Physically: with zero lever arm the air force passes through the pivot, so it produces no torque. The rocket is neutrally stable: a gust tilts it and it simply stays at the new angle — it neither corrects nor diverges. It will drift off course. This is why designers demand a strictly positive margin, not merely non-negative.