Max-Q — maximum dynamic pressure q = ½ρv²; structural limit: Worked examples

3.4.16 · D3 · Physics › Rocket Flight Mechanics › Max-Q — maximum dynamic pressure q = ½ρv²; structural limit

Yeh page Max-Q ke idea ko parent topic se leke har case class ke through hammer karta hai jo tumhare saamne aa sakti hai. Hum pehle poore battlefield ka map banate hain, phir har square ko ek fully worked example se tackle karte hain. Agar koi symbol naya lage, toh hum use yahin build karenge.

Shuru karne se pehle, do main players (parent se):

$ρ$ (Greek letter "rho") = air density, yaani ek cubic metre mein kitne kilograms ki hawa hai. Zameen ke paas dense, upar patli.
$v$ = vehicle ki us hawa mein speed, metres per second mein.
$q = \frac{1}{2} ρ v^{2}$ = dynamic pressure, aero "squeeze," pascals (Pa) mein measure hoti hai. Ek pascal yaani ek square metre pe spread ek newton ki force.

The scenario matrix

Har Max-Q problem in cells mein se ek hai (ya blend). Neeche diya har example us cell ka tag carry karta hai jise woh khatam karta hai.

Cell	Kya cheez ise special banati hai	Killed by
C1 Plain plug-in	Dono $ρ$ , $v$ given hain, $q$ nikalo	Example 1
C2 Degenerate $v = 0$	Liftoff pe speed zero → $q = 0$	Example 2
C3 Degenerate $ρ \to 0$	Zyada altitude pe patli hawa → badi $v$ ke bawajood $q$ tiny	Example 3
C4 Interior peak (calculus)	Kab $q$ maximum hoga yeh nikalo	Example 4
C5 Force / load from $q$	$C A$ se multiply karke newtons nikalo	Example 5
C6 Sensitivity / sign of change	Abhi $q$ badh raha hai ya ghat raha hai? Sign of $d q / d t$	Example 6
C7 Q-alpha (bending) word problem	Gust angle $α$ , real-world side-load	Example 7
C8 Exam twist (throttle-down cap)	Ek $q$ limit di gayi hai, max allowed speed nikalo	Example 8

Worked example Example 1 — Plain plug-in ·

Cell C1 Ek rocket $h = 8 km$ par hai jahan $ρ = 0.53 kg m^{- 3}$ , aur woh $v = 300 m s^{- 1}$ se travel kar raha hai. $q$ nikalo.

Forecast: Guess karo — 41 kPa se bada ya chhota jo parent ne 11 km par nikala tha? (Kam speed, zyada thick air... apna bet lagao.)

Step 1 — Speed ko square karo. $v^{2} = 30 0^{2} = 90000 m^{2} s^{- 2}$ . Yeh step kyun? Formula ko $v^{2}$ chahiye, $v$ nahi; squeeze speed ke square ke saath badhta hai, isliye speed double karne se $q$ chaar guna ho jaata hai.

Step 2 — Density aur one-half se multiply karo. $q = \frac{1}{2} (0.53) (90000) = 0.265 \times 90000 = 23850 Pa \approx 23.9 kPa .$ Yeh step kyun? $\frac{1}{2} ρ$ raw $v^{2}$ ko energy-density (Bernoulli) pressure mein convert karta hai jo parent ne derive ki thi.

Verify: Units: $[kg m^{- 3}] [m^{2} s^{- 2}] = kg m^{- 1} s^{- 2} = Pa$ . ✓ Aur $23.9 < 41$ kPa — kam speed aur peak altitude se neeche hona dono $q$ ko neeche push karte hain, jaise expect tha.

Worked example Example 3 — Degenerate

$ρ \to 0$ · Cell C3 $h = 60 km$ par, $ρ \approx 3.1 \times 1 0^{- 4} kg m^{- 3}$ , lekin rocket ab $v = 2500 m s^{- 1}$ ki speed se daudh raha hai. $q$ nikalo.

Forecast: Abhi tak ki sabse tez speed — surely sabse bada squeeze? (Trap alert.)

Step 1 — (Badi) speed ko square karo. $v^{2} = 250 0^{2} = 6.25 \times 1 0^{6} m^{2} s^{- 2}$ .

Step 2 — (Tiny) density se multiply karo. $q = \frac{1}{2} (3.1 \times 1 0^{- 4}) (6.25 \times 1 0^{6}) = 0.5 \times 1937.5 \approx 969 Pa \approx 0.97 kPa .$ Yeh step kyun? Ek bahut bada $v^{2}$ bhi aisi density ko rescue nahi kar sakta jo sea level se ~4000 ke factor se collapse ho chuki hai.

Verify: $\approx 1$ kPa, 11 km par 41 kPa se bahut neeche hai — yeh curve ka right end hai. Example 2 ke saath milake (left end = 0) yeh prove karta hai ki peak beech mein hai, exactly jaise interior maximum $d q / d t = 0$ predict karta hai.

Worked example Example 4 — Interior peak, the calculus ·

Cell C4 Isothermal atmosphere $ρ (h) = ρ_{0} e^{- h / H}$ use karo scale height $H = 8.5 km$ ke saath, aur assume karo constant acceleration $a = 25 m s^{- 2}$ rest se, toh $v^{2} = 2 ah$ . Max-Q par altitude aur speed nikalo, phir wahaan $q$ nikalo ( $ρ_{0} = 1.225$ lo).

Forecast: Max-Q altitude compute karne se pehle guess karo — ek scale height, ya zyada?

Step 1 — Max-Q condition likho. Parent se, peak par $\frac{v}{ρ} \frac{d ρ}{d h} + \frac{2}{v} \frac{d v}{d t} = 0.$ Yeh step kyun? Yeh sirf $d q / d t = 0$ ko rewrite kiya gaya hai taaki har term ek fractional rate ho — tug-of-war explicitly dikhaya gaya hai.

Step 2 — Models feed karo. $ρ = ρ_{0} e^{- h / H}$ ke liye, $\frac{1}{ρ} \frac{d ρ}{d h} = - \frac{1}{H}$ . $v^{2} = 2 ah$ ke liye, $\frac{d v}{d t} = a$ . Substitute karne par: $- \frac{v}{H} + \frac{2 a}{v} = 0 \Rightarrow v^{2} = 2 a H .$ Yeh step kyun? Abstract condition ko un numbers mein badalna jo hum solve kar sakein — density-loss term ko speed-gain term ke double ko cancel karna chahiye.

Step 3 — Altitude aur speed solve karo. $v^{2} = 2 ah = 2 a H$ se hum seedha padhte hain $h_{m a x Q} = H = 8.5 km$ . Phir $v = 2 a H = 2 (25) (8500) = 425000 \approx 652 m s^{- 1} .$ Yeh step kyun? Ek scale height classic Max-Q altitude hai; speed constant- $a$ model se aati hai.

Step 4 — Wahaan $q$ compute karo. $ρ (H) = ρ_{0} e^{- 1} = 1.225 \times 0.3679 = 0.4507 kg m^{- 3}$ . $q = \frac{1}{2} (0.4507) (425000) \approx 95770 Pa \approx 95.8 kPa .$ Yeh step kyun? Peak $ρ$ aur $v^{2}$ ko $q = \frac{1}{2} ρ v^{2}$ mein plug karo.

Verify: Figure mein curve $h = H = 8.5$ km par flat-topped hai — red dot exactly peak par baitha hai, aur uske left/right ke neighbours kam $q$ dete hain. Condition $v^{2} = 2 a H$ holds karti hai: $65 2^{2} = 425000 = 2 (25) (8500)$ . ✓

Worked example Example 5 — Pressure se force tak ·

Cell C5 Example 4 ke Max-Q par, $q \approx 95.8 kPa$ . Vehicle ka reference area $A = 12 m^{2}$ hai aur drag coefficient $C_{D} = 0.35$ hai. Drag force nikalo.

Forecast: Kilo-newtons ya mega-newtons?

Step 1 — Universal aero equation apply karo. $F_{D} = C_{D} q A .$ Yeh step kyun? $q$ ek pressure hai; force ke liye area chahiye, aur dimensionless $C_{D}$ (drag coefficient, "yeh shape kitna draggy hai") real, non-flat geometry ko account karta hai. Dekho Drag Force and Drag Coefficient.

Step 2 — Plug in karo. $F_{D} = 0.35 \times 95770 \times 12 \approx 402234 N \approx 402 kN .$ Yeh step kyun? Teeno factors ki seedhi multiplication.

Verify: Units: $[] \cdot [Pa] \cdot [m^{2}] = [N m^{- 2}] [m^{2}] = N$ . ✓ Lagbhag 402 kN — roughly 41-tonne mass ke weight ke barabar; isliye airframe skin precisely yahin moti hoti hai.

Worked example Example 6 — Kya

$q$ abhi badh raha hai ya ghat raha hai? · Cell C6 $h = 5 km$ par rocket ki $v = 400 m s^{- 1}$ , acceleration $a = 25 m s^{- 2}$ , aur (isothermal, $H = 8.5$ km) $\frac{1}{ρ} \frac{d ρ}{d h} = - \frac{1}{H}$ hai. Kya dynamic pressure abhi bhi badh rahi hai? $d q / d t$ ka sign determine karo.

Forecast: Peak se neeche (jo 8.5 km par hai) — toh badh raha hai ya ghat raha hai?

Step 1 — $d q / d t$ ki fractional form use karo. $d q / d t$ ko $q$ se divide karne par $\frac{1}{q} \frac{d q}{d t} = \frac{v}{ρ} \frac{d ρ}{d h} + \frac{2}{v} \frac{d v}{d t} .$ Yeh step kyun? Hume sirf sign chahiye; $q > 0$ hamesha, isliye $d q / d t$ ka sign is bracket ke sign ke barabar hai.

Step 2 — Har term evaluate karo.

Density term: $\frac{v}{ρ} \frac{d ρ}{d h} = - \frac{v}{H} = - \frac{400}{8500} \approx - 0.0471 s^{- 1} .$
Speed term: $\frac{2}{v} \frac{d v}{d t} = \frac{2 a}{v} = \frac{2 ( 25 )}{400} = 0.125 s^{- 1} .$

Yeh step kyun? Ek term negative hai (hawa patli ho rahi hai), ek positive (abhi bhi accelerate ho raha hai) — unka sum sab kuch decide karta hai.

Step 3 — Add karo. $\frac{1}{q} \frac{d q}{d t} \approx - 0.0471 + 0.125 = + 0.0779 s^{- 1} > 0.$ Yeh step kyun? Positive sum ⇒ $q$ abhi badh raha hai.

Verify: Positive hai, toh hum abhi Max-Q se pehle hain — $h = 5 < 8.5$ km ke consistent. Speed gain (0.125) density loss (0.047) ka almost teen guna hai; peak tab aata hai jab dono balance ho jaate hain, aur upar jaake. ✓

Worked example Example 7 — Q-alpha side-load (real-world gust) ·

Cell C7 Max-Q ke paas, $q = 41 kPa$ hai. Ek wind gust vehicle ko angle of attack $α = 4^{\circ}$ par le jaata hai (nose direction aur actual airflow ke beech ka tilt). Reference area $A = 12 m^{2}$ par side-force coefficient per radian $C_{N α} = 2.0$ hai. Lateral (bending) force nikalo. Dekho Angle of Attack and Q-alpha Loads.

Forecast: Kya 4° harmless lagta hai? Pehle newtons guess karo.

Step 1 — Angle ko radians mein convert karo. Aerodynamic derivatives radians use karte hain: $α = 4^{\circ} \times \frac{π}{180} \approx 0.0698 rad .$ Yeh step kyun? $C_{N α}$ "per radian" hai; degrees mix karne se answer ~57× badh jaata.

Step 2 — Side-force formula. Lateral load product $q α$ ("Q-alpha") ke saath scale karta hai: $F_{side} = C_{N α} α q A .$ Yeh step kyun? Bilkul drag $= C_{D} q A$ jaisa, lekin coefficient khud $α$ ke proportional hai, isliye dangerous quantity $q α$ hai.

Step 3 — Plug in karo. $F_{side} = 2.0 \times 0.0698 \times 41000 \times 12 \approx 68700 N \approx 68.7 kN .$ Yeh step kyun? Seedhi multiplication.

Verify: Units: $[] \cdot [rad] \cdot [Pa] \cdot [m^{2}] = N$ (radian dimensionless hai). ✓ Sirf 4° ka tilt airframe ko ~69 kN se sideways bend karta hai — exactly isliye guidance $α$ ko tiny rakhta hai jab $q$ large ho.

Worked example Example 8 — Exam twist: formula ko ulta chalao ·

Cell C8 Mission rules dynamic pressure ko $q_{m a x} = 35 kPa$ par cap karte hain structure ki protection ke liye. $h = 11 km$ par ( $ρ = 0.36 kg m^{- 3}$ ), maximum allowed speed kya hai — aur kitna throttle karna padega agar unthrottled speed $480 m s^{- 1}$ hoti?

Forecast: Kya cap badi speed cut force karega ya choti?

Step 1 — $q = \frac{1}{2} ρ v^{2}$ ko $v$ ke liye invert karo. $v_{m a x} = \frac{2 q _{m a x}}{ρ} .$ Yeh step kyun? Hume output $q$ diya gaya hai aur input $v$ chahiye; algebra formula ko ulta solve karta hai. Square-root $v^{2}$ ko undo karta hai.

Step 2 — Plug in karo. $v_{m a x} = \frac{2 ( 35000 )}{0.36} = 194444 \approx 441 m s^{- 1} .$ Yeh step kyun? Inverted formula ka numeric evaluation.

Step 3 — Unthrottled speed se compare karo. Vehicle 480 m/s chahta tha; ise 441 m/s tak rokna hoga, yaani slower by $\frac{480 - 441}{480} \approx 0.081 = 8.1%.$ Yeh step kyun? Fractional speed cut guidance ko batata hai ki "thrust bucket" kitna deep hona chahiye.

Verify: 441 m/s wapas feed karo: $q = \frac{1}{2} (0.36) (44 1^{2}) = 0.18 \times 194481 \approx 35007 Pa \approx 35 kPa$ . ✓ Exactly cap. Aur 441 < 480, toh throttle-down genuinely required hai.

Recall Har example kaun sa cell hit karta hai?

Example 1 kaun sa cell hit karta hai? ::: C1 — plain plug-in. Example 2 aur 3 milke kya prove karte hain? ::: Max-Q peak interior hai: $v = 0$ par $q = 0$ aur $ρ \to 0$ par $q$ tiny hai, toh maximum beech mein hai. Example 4 kaun se do models use karta hai? ::: Isothermal density $ρ = ρ_{0} e^{- h / H}$ aur constant-acceleration $v^{2} = 2 ah$ . Example 6 kya poochta hai, aur ise sasta kaise karo? ::: $d q / d t$ ka sign; fractional form use karo kyunki $q > 0$ hai, toh sirf bracket ka sign matter karta hai. Example 8 kaun sa trick hai? ::: $q = \frac{1}{2} ρ v^{2}$ ko ulta chalana, structural cap ke neeche max allowed speed nikalne ke liye.

Active recall

Liftoff par

q

kya hota hai, aur kyun?

0

Pa, kyunki

v = 0

(koi momentum flux nahi) chahe hawa dense hi kyun na ho.

q

cap di gayi ho toh max allowed speed kaise nikaalte ho?

v_{m a x} = 2 q_{m a x} / ρ

Max-Q altitude se neeche

d q / d t

ka sign?

Positive — speed-gain term, density-loss term se zyada hai.

Gust se dangerous bending quantity?

Product

q α

("Q-alpha").