Visual walkthrough — Dynamic stability — pitch - yaw damping derivatives

3.4.11 · D2 · Physics › Rocket Flight Mechanics › Dynamic stability — pitch - yaw damping derivatives

---

Step 1 — "Pitch rate" kya hota hai, aur yeh body ke saath kya karta hai?

KYA. Socho ek rocket right side ki taraf fly kar raha hai. Uski lambi axis horizontal hai. Ab imagine karo ki nose dheere dheere upar tilt ho raha hai jabki tail neeche dip kar raha hai — poori rigid body ek point ke around rotate kar rahi hai jise center of gravity (CG) kehte hain, rocket ka balance point.

Us rotation ki speed pitch rate hai, jise $q$ likhte hain, radians per second (rad/s) mein measure hota hai. Ek radian per second ka matlab hai ki body har second mein roughly $57^{\circ }$ ghoomti hai.

KYUN yahan se shuru karte hain. Damping ek aisi force hai jo rotation se ladte hai. Toh pehle, hume rotation ko hi describe karna hoga aur — sabse important — yeh dekkhna hoga ki yeh body ke har part ko kya velocity deta hai.

PICTURE.

Ek coordinate set karo: origin ko CG par rakho, aur $x$ ko tail ki taraf (aft) distance measure karne do. Toh tail ke paas koi fin kisi positive $x$ par baithti hai; nose par koi point negative $x$ par hota hai.

Rigid-body motion se yeh key fact hai. Jab ek rigid body CG ke around rate $q$ se spin karti hai, toh CG se $x$ distance par ek point sideways (axis ke perpendicular) speed se move karta hai

$w_{\text{local}}=qx.$

Term-by-term padho:

• $q$ — spin rate (zyada spin, zyada tezi se sideways motion),
• $x$ — lever distance (door ke points zyada tezi se move karte hain, jaise merry-go-round ka rim center se zyada tezi se move karta hai),
• $w_{\text{local}}$ — us point ki resulting sideways (yahan, neeche ki taraf) velocity air ke relative.

Tail ($x>0$, bada) tezi se neeche swing karta hai. CG par ($x=0$) koi point sideways bilkul nahi move karta. Nose par points ($x<0$) upar swing karte hain — opposite direction mein. Yeh sign flip baad mein important hoga.

---

Step 2 — Woh sideways motion wind ko tilt kar deti hai: ek local angle of attack

KYA. Rocket speed $V$ se aage fly kar raha hai. Ek tail element ke point of view se jo $w_{\text{local}}=qx$ se neeche bhi slide kar raha hai, oncoming air ab axis ke saath seedhi nahi aati — woh tedi aati hai, thodi neeche se aati hui lagti hai. Yeh tilt ek extra angle of attack hai, local airflow aur body axis ke beech ka angle.

KYUN. Aerodynamic forces us angle se produce hoti hain jis par air kisi surface se milti hai. Rotation ne wind nahi badla — usne yeh badla ki wind kaisi lagti hai har moving element ko. Hum "sideways velocity" ko "extra angle" mein convert karte hain kyunki forces angle se respond karti hain, directly velocity se nahi.

PICTURE.

Chhoti velocity triangle dekho. Axial leg $V$ hai (forward flight). Transverse leg $w_{\text{local}}=qx$ hai (Step 1 ka sweep). Airflow jo element feel karta hai woh diagonal hai — dono ka sum.

Axis se us diagonal ka angle hai

$\Delta \alpha (x)=\arctan \left(\frac{w_{\text{local}}}{V}\right)\approx \frac{w_{\text{local}}}{V}=\frac{qx}{V}.$

$\arctan$ kyun sahi tool hai? $\arctan$ is sawaal ka jawaab deta hai "kaunsa angle hai jiska opposite-over-adjacent ratio yeh hai?" — bilkul wohi jo hume triangle ki do legs ko angle mein convert karne ke liye chahiye. Aur hum ise sirf $\frac{qx}{V}$ mein kyun reduce kar sakte hain? Kyunki real flight mein $qx$ $V$ ke mukable mein bahut chhota hota hai, toh triangle ek patla sliver hai, aur ek patli sliver ke liye angle (radians mein) uske tangent ke barabar hota hai. Yeh small-angle approximation hai.

Term-by-term:

• $qx$ — sideways velocity (numerator, "opposite"),
• $V$ — forward speed (denominator, "adjacent"),
• $\Delta \alpha (x)$ — station $x$ par wind ka extra tilt, radians mein.

Dhyaan do: $\Delta \alpha$ $x$ ke saath linearly badhta hai. Peechle elements sabse zyada wind-tilt dekhte hain.

---

Step 3 — Tedi wind push karti hai: extra normal force

KYA. Ek surface jo angle $\Delta \alpha$ par air se milti hai woh ek force feel karti hai body axis ke perpendicular — ek normal force. Zyada tilt → zyada force, proportion mein (chhote angles ke liye).

KYUN. Yeh aerodynamics ka basic linear model hai: force = (air kitni takat se push karti hai) × (kitna area) × (woh surface angle ke liye kitni sensitive hai) × (angle). Force milne se pehle hume torque nahi mil sakta.

PICTURE.

Station $x$ par area $dA$ ki ek chhoti surface patch ke liye, incremental normal force hai

$dN(x)=\underset{\text{dynamic pressure}}{\underbrace{\frac{1}{2}\rho V^2}}\underset{\text{angle-sensitivity}}{\underbrace{c_{n_{\alpha }}(x)}}\underset{\text{tilt}}{\underbrace{\Delta \alpha (x)}}\underset{\text{patch area}}{\underbrace{dA}}.$

Term-by-term:

• $\frac{1}{2}\rho V^2$ — dynamic pressure, air ka "punch". $\rho$ air density hai (kg/m³), $V$ flight speed. Tez ya dense air zyada takat se push karti hai.
• $c_{n_{\alpha }}(x)$ — local lift-curve slope: us shape ke liye angle ke har radian par kitne units of normal-force-coefficient milte hain. Fins ka bada $c_{n_{\alpha }}$ hota hai; smooth body tube ka chhota $c_{n_{\alpha }}$ hota hai.
• $\Delta \alpha (x)=\frac{qx}{V}$ — seedha Step 2 se laya gaya.
• $dA$ — tiny patch area.

$\Delta \alpha =qx/V$ substitute karte hain:

$dN(x)=\frac{1}{2}\rho V^2{c}_{n_{\alpha }}(x)\frac{qx}{V}dA=\frac{1}{2}\rho Vq{c}_{n_{\alpha }}(x)xdA.$

Ek $V$ cancel ho gaya. Har patch par force $x$ ke proportional hai ($\Delta \alpha$ ke through). Yeh single power of $x$ yaad rakho — ek aur abhi aane wala hai.

---

Step 4 — Har force ek lever par act karti hai: torque, aur $x^2$ ka janam

KYA. CG se $x$ distance par act karne wali force $dN$ CG ke around ek torque (twisting moment) produce karti hai: torque = force × lever arm = $xdN$. Hum saari patches ko add up karke total damping moment $M$ nikalte hain.

KYUN. Hume rotation ki parwaah hai, aur rotation torque se drive hoti hai, force se nahi. CG ke paas wali force barely twist karti hai; wahi force door se zyada twist karti hai. Lever arm $x$ woh hai jo force ko twist mein convert karta hai.

PICTURE.

Ek patch ka moment $xdN$ hai. Poore rocket par sum (integrate) karte hain:

$M=-\int xdN=-\frac{1}{2}\rho Vq\int c_{n_{\alpha }}(x)\underset{x^2}{\underbrace{x\cdot x}}dA=-\frac{1}{2}\rho Vq\int c_{n_{\alpha }}(x)x^{2}dA.$

Yeh poore page ka dil hai. Do alag powers of $x$ ek saath multiply ho rahe hain:

• $x$ ki ek power Step 1–2 se aayi (door elements air ko tezi se sweep karte hain, toh bada $\Delta \alpha$, badi force),
• $x$ ki doosri power Step 4 se aayi (door elements ek lamba lever par act karte hain).

Saath mein: torque $\propto x^2$. Iseelie fin ki CG se distance double karne par uski damping chaar guna ho jaati hai — woh famous $x^2$ jo parent note mein baar baar aata tha.

Minus sign. Leading $-$ kyun? Physics trace karo: nose upar pitch kar raha hai ($q>0$), tail neeche swing kar raha hai, toh tail neeche se aati air se milta hai, jo tail ko upar push karti hai. Tail ko upar push karna nose ko neeche rotate karta hai — original pitch-up ke opposite. Moment motion se ladate hai. Koi bhi moment jo hamesha rotation ko oppose kare uska sign $q$ se opposite hona chahiye, isliye $-$.

Kyunki $\int c_{n_{\alpha }}{x}^{2}dA$ squares ka sum hai (hamesha positive), $M$ ka sign $q$ se opposite hai har possible rocket ke liye. Yeh kabhi bhi accidentally rotation ko drive nahi kar sakta. Yahi ise damping banata hai, reverse nahi.

---

Step 5 — Saaf karna: non-dimensional derivative

KYA. Engineers koi moment newton-metres mein quote nahi karte jo speed aur altitude ke saath badlata rahe. Woh ek coefficient quote karte hain — ek pure number jo sirf shape capture karta hai. Hum messy physical factors ko divide out karte hain.

KYUN. Hume ek single geometry number chahiye jo kisi table mein rahe aur kisi bhi speed par apply ho. Non-dimensionalising $\rho$, $V$ aur reference size ko strip away karta hai taaki shape akela rahe.

PICTURE.

Do standard definitions:

• Moment coefficient $C_m=\frac{M}{\frac{1}{2}\rho V^{2}Sd}$, jahan $S$ reference area hai aur $d$ reference length (diameter) hai. Yeh moment ko ek pure number mein badalta hai.
• Non-dimensional pitch rate $\hat{q}=\frac{qd}{2V}$ — pitch rate ko ek natural time se scale kiya gaya, "body kitna rotate hoti hai us time mein jitne mein air ek radius travel karti hai".

Damping derivative $C_m$ ka $\hat{q}$ ke against slope hai:

$C_{m_q}\equiv \frac{\partial C_m}{\partial \hat{q}}.$

$M=-\frac{1}{2}\rho Vq\int c_{n_{\alpha }}{x}^{2}dA$ ko $C_m$ mein daalo, phir $\hat{q}$ ke respect mein differentiate karo (matlab $q$ ko $2V\hat{q}/d$ se replace karo aur $\hat{q}$ ka coefficient padho). Saare $\rho$, $V$ cancel ho jaate hain, aur bahar aata hai:

$\boxed{C_{m_q}=-\frac{2}{S{d}^2}\int c_{n_{\alpha }}(x)x^{2}dA<0}$

Term-by-term:

• $-$ — hamesha rotation ko oppose karta hai (Step 4 se).
• $\frac{2}{S{d}^2}$ — upar ke do definitions se normalising factor.
• $\int c_{n_{\alpha }}{x}^{2}dA$ — CG ke around second moment of aerodynamic area: geometry, aur sirf geometry.

Isi argument ko $90^{\circ }$ ghuma do (nose left/right swing kar raha hai upar/neeche ki bajaye), yaw deta hai $C_{n_r}=C_{m_q}$ ek axisymmetric rocket ke liye — do planes geometrically identical hain.

---

Step 6 — Edge cases (koi scenario kabhi na chhodein)

KYA. Formula ko uski extremes par test karte hain taaki kabhi surprise na ho.

PICTURE.

• CG par exactly ek surface ($x=0$). Toh $\Delta \alpha =q\cdot 0/V=0$: koi sweep nahi, koi extra force nahi, koi torque nahi. Contribution $\propto 0^2=0$. Balance point ke paas wali surfaces damping ke liye kuch nahi karti.
• CG ke aage ek surface ($x<0$, nose). Uska sweep sign flip karta hai (upar move karta hai jab tail neeche jaata hai), aur uska lever arm sign flip karta hai — do sign flips. Lekin $x^2=(-x{)}^2>0$: yeh integral mein positive contribute karta hai, yaani abhi bhi damping add karta hai. Distance-squared ko koi farq nahi padta kis side hai.
• Air density $\rho \to 0$ (high altitude). Coefficient $C_{m_q}$ unchanged rehta hai — yeh pure geometry hai. Lekin physical moment $M=-\frac{1}{2}\rho V{d}^{2}S{C}_{m_q}\hat{q}$ mein $\rho$ aage hai. Jaise $\rho \downarrow$, real damping fade hoti hai. Rocket ka shock absorber kamzor hota jaata hai jaise woh chadhta hai, bhale hi table number kabhi nahi badlta.
• Zero spin ($q=0$). Koi rotation nahi, koi sweep nahi, koi damping moment nahi — damping sirf tab exist karti hai jab body rotate kar rahi ho. Yeh motion ka response hai, static force nahi.

---

Ek-picture summary

Chain ko left se right padho: spin $q$ → sideways sweep $qx$ (ek $x$) → wind-tilt $\Delta \alpha =qx/V$ → force $dN$ → lever $x$ par torque (doosra $x$) → total $\propto x^2$, spin ko oppose karta hai → negative $C_{m_q}$.

Recall Feynman retelling — poori walkthrough seedhe shabdon mein

Rocket ko thoda spin karo: nose upar jaata hai, tail neeche jaata hai. Body ka koi bhi part jo balance point se door hai woh air mein sideways drag hota hai — aur jitna door hai, utni tezi se drag hota hai (Step 1). Us part ko ab wind thodi tedi lagne lagti hai (Step 2). Tedi wind surface par push karti hai (Step 3), aur lambe arm par push bada twist banata hai (Step 4). Twist hamesha nose ko wapas neeche push karta hai — yeh spin se ladate hai, wohi minus sign hai. Aur kyunki "door" ka matlab hai "tezi se sweep" bhi aur "lamba arm" bhi, effect distance squared ke saath badhta hai — fins ko do guna peechhe le jaao, chaar guna shanti milti hai. Air density aur speed hata do aur sirf shape bachti hai, toh tumhe $C_{m_q}$ milta hai, rocket ka built-in shock absorber (Step 5). Balance point par zero hai, kisi bhi side se kaam karta hai, aur upar jaane par air patli hone ke saath quietly kamzor hota jaata hai (Step 6).

Recall Quick self-check

$x$ ki do powers kahan se aati hain? ::: Ek sweep velocity $qx$ se (door points tezi se sweep karte hain → bada angle → badi force); ek lever arm $x$ se (torque = force × distance). $C_{m_q}$ hamesha negative kyun hota hai? ::: Integral $\int c_{n_{\alpha }}{x}^{2}dA$ squares ka sum hai (positive), leading minus ke saath — toh moment hamesha $q$ ko oppose karta hai. CG par exactly baitha surface kya contribute karta hai? ::: Kuch nahi — $x=0$ zero sweep aur zero lever deta hai, toh $x^2=0$. Kya geometry number $C_{m_q}$ altitude ke saath badlta hai? ::: Nahi, lekin physical damping moment badlta hai, kyunki woh $\rho V$ ke saath scale karta hai.

---

Prerequisites & neighbours: Parent topic · Static Stability — Center of Pressure & Margin · Damped Harmonic Oscillator · Fin Design & Sizing · Atmospheric Density vs Altitude · Barrowman Equations · Moments of Inertia of a Rocket