Exercises — Pitch program — open-loop pitch-over

Open-loop: commanded pitch angle ek fixed function of time ${\theta }_c(t)$ hota hai, flight se pehle decide kiya gaya, wahan se recompute nahi hota jahan rocket actually hai. (Ek attitude loop us command ko track zaroor karta hai — open-loop command ke baare mein hai, na ki loop ke baare mein.) Rocket ko kya turn karta hai: gravity. Ek chhoti si initial kick ke baad, gravity ka sideways component $g\cos \gamma$ velocity vector ko bend karta hai; engine body axis ke along pointed rehta hai near-zero angle of attack ke saath.

$\cos 90^{\circ }=0$, isliye $\dot{\gamma }=-\frac{g\cdot 0}{v}=0.$ Ek bilkul vertical rocket ka zero turn rate hota hai — gravity exactly velocity ke opposite hai, uska koi bhi hissa sideways nahi. Yahi wajah hai ki tumhe ek kick chahiye: thoda vertical se tipe bagair, $\cos \gamma$ $0$ hi rehta hai aur rocket kabhi turn shuru nahi karta.

$\cos 89^{\circ }=0.017452$. $\dot{\gamma }=-\frac{9.8\times 0.017452}{120}=-0.001425\text{rad/s}.$ Convert karo: $1\text{rad}=57.2958^{\circ }$, isliye $-0.001425\times 57.2958=-0.0817^{\circ }/\text{s}.$ Rocket sirf kareeb $0.082^{\circ }$ har second mein turn karta hai — na ke barabar. Yahi wajah hai ki initial kick ko actively turn shuru karna padta hai.

$\dot{\gamma }=-\frac{9.8\times 0.017452}{60}=-0.002850\text{rad/s}=-0.1633^{\circ }/\text{s}.$ Exactly double L2.1 rate, kyunki $\dot{\gamma }\propto 1/v$ aur humne $v$ half kar diya. Ek slow rocket gravity-turn faster karta hai, isliye usse ek chhoti initial kick chahiye.

$\cos 45^{\circ }=0.707107$. $\dot{\gamma }=-\frac{9.8\times 0.707107}{800}=-0.008662\text{rad/s}=-0.4962^{\circ }/\text{s}.$ Bhaale hi $v$ bahut bada hai, $\cos 45^{\circ }$ $\cos 89^{\circ }$ se kaafi bada hai, isliye turn rate kick-over se zyada hai. Turn accelerate hota hai jaise $\gamma$ girta hai — jab tak $v$ itna na badh jaaye ki usse slow kar sake.

Rate ko $v$ ke function mein likho: $\dot{\gamma }(v)=-\frac{g\cos \gamma }{v}$. $v$ ko $v/2$ se replace karo: $\dot{\gamma }(v/2)=-\frac{g\cos \gamma }{v/2}=2\left(-\frac{g\cos \gamma }{v}\right)=2\dot{\gamma }(v).$ $v$ half karne se magnitude double hoti hai — confirmed. $4v$ ke liye: $\dot{\gamma }(4v)=\frac{1}{4}\dot{\gamma }(v)=\frac{1}{4}(-0.001425)=-0.000356\text{rad/s}$.

Sideways component $g\cos \gamma$ hai. Hum chahte hain $g\cos \gamma =\frac{1}{2}g$, yaani $\cos \gamma =\frac{1}{2}\Rightarrow \gamma =\arccos (0.5)=60^{\circ }.$ ($\arccos$ poochh raha hai "kaunse angle ka yeh cosine hai?" — yeh $\cos$ ko undo karta hai.) $\gamma =60^{\circ }$ par, gravity ka aadha hissa velocity turn karne mein busy hai aur aadha path ke saath climb ke against lad raha hai. Jaise $\gamma$ $0^{\circ }$ ki taraf girta jaata hai, $\cos \gamma \to 1$ aur gravity ka poora hissa arrow turn karta hai — lekin tab tak $v$ bohut bada hota hai, isliye actual turn rate phir bhi modest hoti hai.

$\cos 88.5^{\circ }=0.0261769$, $\cos 89.5^{\circ }=0.0087265$. $\dot{\gamma }(88.5^{\circ })=-\frac{9.8\times 0.0261769}{100}=-0.0025653\text{rad/s},$ $\dot{\gamma }(89.5^{\circ })=-\frac{9.8\times 0.0087265}{100}=-0.0008552\text{rad/s}.$ Ratio hai $0.0261769/0.0087265=3.0$ — $88.5^{\circ }$ kick teen guna tez turn karta hai shuru se hi. Kyunki $\cos \gamma$ $90^{\circ }$ ke paas bahut chhota aur steep hai, ek tiny angle change initial rate ko bahut zyada swing kar deta hai; yeh ek self-amplifying bend ko seed karta hai (zyada turn → bada $\cos \gamma$ → zyada turn), isliye burnout angle wildly different ho jaata hai. Yahi sensitivity hai jis wajah se pitch-over ko carefully tune karna padta hai.

Setup (WHAT): point mass, speed $v$, flight-path angle $\gamma$, thrust $T$ $\vec{v}$ ke along (zero angle of attack), gravity $mg$ neeche. Forces ko "along $\vec{v}$" (tangential) aur "across $\vec{v}$" (normal) mein split karo. Normal balance (WHY): sirf woh gravity ka hissa velocity vector ko rotate kar sakta hai jo uske perpendicular ho. Figure s01 ke triangle se, woh hissa $mg\cos \gamma$ hai, $\vec{v}$ ko downward bend karne ki direction mein point karta hua. Path ke across Newton's law (centripetal form) hai: $mv\dot{\gamma }=-mg\cos \gamma .$ Left side $mv\dot{\gamma }$ hai mass $\times$ (arrow ke turning ka rate $\times$ speed) — yaani sideways acceleration. Minus sign: gravity $\gamma$ ko neeche turn karti hai. $m$ cancel karo (WHY): mass dono sides par aata hai, isliye drop out ho jaata hai — turn rate is baat par depend nahi karta ki rocket kitna bhaari hai, sirf $g$, $v$, $\gamma$ par: $\dot{\gamma }=-\frac{g\cos \gamma }{v}.■$

Target convert karo: $-0.1^{\circ }/\text{s}=-0.1/57.2958=-0.00174533\text{rad/s}$. $v$ ke liye formula solve karo: $v=-\frac{g\cos {\gamma }_0}{\dot{\gamma }}=-\frac{9.8\times \cos 89^{\circ }}{-0.00174533}.$ $\cos 89^{\circ }=0.0174524$ ke saath: $v=\frac{9.8\times 0.0174524}{0.00174533}=97.98\text{m/s}\approx 98\text{m/s}.$ Toh kick tab command ki jaani chahiye jab vehicle $98\text{m/s}$ ke paas ho taaki woh turn rate mile.

Time eliminate karo (WHY): hum jaanna chahte hain ki speed angle par kaise depend karti hai, clock par nahi, isliye dono rate equations divide karo — $dt$ cancel ho jaata hai: $\frac{dv}{d\gamma }=\frac{\dot{v}}{\dot{\gamma }}=\frac{(T-mg\sin \gamma )/m}{-g\cos \gamma /v}.$ $T/m=ng$ likho (kyunki $n=T/(mg)$):

$dv/d\gamma =0$ ke liye numerator $v(n-\sin \gamma )=0$ chahiye. Kyunki $v\ne 0$: $n-\sin \gamma =0\Rightarrow \sin \gamma =n=\mathrm{1.5.}$ Lekin $\sin \gamma \le 1$ hamesha hota hai, aur yahan $n=1.5>1$ hai, isliye aisa koi $\gamma$ nahi — numerator kabhi zero nahi hota. Physically: kyunki thrust path ke har jagah weight se zyada hai ($n>1$), along-path acceleration $g(n-\sin \gamma )$ har angle par positive rehta hai ($\sin \gamma \le 1<n$), isliye rocket hamesha speed up ho raha hai. Sirf agar $n<1$ ho (thrust weight se kam) toh $\sin \gamma =n$ ka koi solution ho sakta hai, ek angle mark karta hua jahan climbing losses thrust ko exactly kha jaate hain aur $v$ momentarily badhna band kar deta hai.

$\dot{\gamma }$ at $\gamma =90^{\circ }$ ::: Zero — koi sideways gravity nahi, kick ke bagair koi turn nahi. Effect of halving $v$ on turn rate ::: Use double kar deta hai ($\dot{\gamma }\propto 1/v$). Does mass appear in $\dot{\gamma }$? ::: Nahi — yeh normal force balance mein cancel ho jaata hai. Sign of $\dot{\gamma }$ during ascent ::: Negative — $\gamma$ $90^{\circ }$ se $0^{\circ }$ ki taraf girta hai.