Coordinate systems — Earth-Centered Inertial (ECI), Earth-Centered Earth-Fixed (ECEF), North-East-Down (NED), launch, bo: Worked examples

3.4.1 · D3 · Physics › Rocket Flight Mechanics › Coordinate systems — Earth-Centered Inertial (ECI), Earth-Ce

Yeh page ek practice arena hai. Parent note ne machinery banayi — paanch frames, rotation matrices, transport theorem. Yahan hum usse har tarah ke input se drive karte hain taaki koi bhi case tumhare liye naya na ho.

Shuru karne se pehle, teen words jo hum baar baar use karenge, seedhi bhaasha mein:

Scenario matrix

Is topic ke har problem ka jawab inhi cells mein se kisi ek mein milta hai. Neeche diye har worked example ko us cell ke saath tag kiya gaya hai jo woh clear karta hai.

#	Cell class	Tricky kyun hai	Example
A	Rotation angle ka Sign / quadrant	$cos, sin$ quadrants mein sign badal lete hain	Ex 2, Ex 4
B	Zero / degenerate input ( $θ = 0$ , $ϕ = 90°$ )	matrix identity ya pole singularity mein collapse ho jaati hai	Ex 5
C	Limiting value ( $ϕ \to 90°$ , $t \to$ ek poora din)	check karo ki answer obvious limit ki taraf jaata hai	Ex 5, Ex 6
D	Latitude scaling ( $cos ϕ$ boost)	boost latitude ke saath constant nahi hoti	Ex 1, Ex 6
E	Velocity vs position transform	$ω_{e} \times r$ term add karna zaroori hai	Ex 7
F	Direction / unit vector ek frame mein	magnitude $= 1$ honi chahiye	Ex 3
G	Real-world word problem	English → frame → formula mein translate karo	Ex 6
H	Exam twist (mid-latitude ECI→ECEF, full chain check)	multiple steps, sign mix up karna aasaan hai	Ex 4, Ex 8

Ab hum poora table clear karte hain.

Warm-up: $R_{z} (θ)$ ko picture se padhna

Numbers se pehle, dekho ki "axes ko $θ$ se rotate karna" asal mein kya karta hai. Point space mein jaisi hai waisi rahi; axes uske neeche ghoom jaate hain.

Figure — Coordinate systems — Earth-Centered Inertial (ECI), Earth-Centered Earth-Fixed (ECEF), North-East-Down (NED), launch, bo

Ex 1 — Equator par Eastward boost · Cell D

Forecast: aage padhne se pehle ek number guess karo. Jet se tez? Ya dheema?

Formula chunno $v = ω_{e} R_{\oplus} cos ϕ$ . Yeh step kyun? Earth ka sirf woh hissa jo spin axis ke perpendicular hai, actually ek circle sweep karta hai. Latitude $ϕ$ par us circle ki radius $R_{\oplus} cos ϕ$ hai (Ex 6 ki figure dekho). Speed = rate $\times$ radius.
$ϕ = 0$ substitute karo toh $cos 0 = 1$ . Yeh step kyun? Equator axis se sabse door hai → sabse bada circle → maximum speed. $v = 7.292 \times 1 0^{- 5} \times 6.378 \times 1 0^{6} \times 1 = 465.1 m/s .$

Recall Verify

Units: $(rad/s) \times (m) = m/s$ ✔ (radian dimensionless hota hai). Magnitude ≈ 465 m/s ≈ Mach 1.4 — yeh woh free speed hai jo ek rocket inherit karta hai. ✔

Ex 2 — 6 ghante baad ECI→ECEF (quadrant sign flip) · Cells A, H

Forecast: quarter-day spin ke baad, point roughly kahan hoga — abhi bhi $+ X$ par, ya kahin aur?

Accumulated angle $θ = ω_{e} t$ , jahan $t = 6 \times 3600 = 21600$ s. Yeh step kyun? ECEF aur ECI mein sirf $Z$ ke around $θ = ω_{e} t$ ka rotation farq hai. $θ = 7.292 \times 1 0^{- 5} \times 21600 = 1.5751 rad \approx 90.2°.$
$R_{z} (θ)$ ko $r_{I} = (R_{\oplus}, 0, 0)$ par apply karo: $x_{E} = cos θ R_{\oplus}, y_{E} = - sin θ R_{\oplus}, z_{E} = 0.$ Yeh step kyun? $R_{z}$ ka sirf pehla column matter karta hai kyunki input mein sirf $x$ component hai.
Numbers. $θ \approx 90.2°$ ke saath: $cos θ \approx - 0.0038$ , $sin θ \approx + 0.99999$ , toh $x_{E} \approx - 2.4 \times 1 0^{4} m (\approx 0), y_{E} \approx - 6.378 \times 1 0^{6} m .$

Recall Verify

$x_{E}^{2} + y_{E}^{2} = R_{\oplus}^{2}$ (rotation length preserve karta hai): $(2.4 \times 1 0^{4})^{2} + (6.378 \times 1 0^{6})^{2} \approx (6.378 \times 1 0^{6})^{2}$ ✔.

Ex 3 — "Down" direction as a unit vector · Cell F

Forecast: "Down" Earth ke centre ki taraf point karta hai. 45°N aur $λ = 0$ par, guess karo ki yeh kaun se do ECEF axes ke beech split hoga.

NED ki 3rd row lo (Down unit vector): $\hat{D} = (- cos ϕ cos λ, - cos ϕ sin λ, - sin ϕ) .$ Yeh step kyun? Down $= - \overset{r}{^}$ hai, yani inward radial. ECEF mein outward radial $\overset{r}{^}$ exactly $(cos ϕ cos λ, cos ϕ sin λ, sin ϕ)$ hota hai, toh usse negate karo.
$ϕ = 45°, λ = 0$ daalo: $cos 45° = sin 45° = 0.7071$ , $sin λ = 0$ , $cos λ = 1$ : $\hat{D} = (- 0.7071, 0, - 0.7071) .$ Yeh step kyun? $λ = 0$ par point $X$ – $Z$ plane mein hai, isliye koi $Y$ component nahi; $45°$ par direction exactly $- X_{E}$ aur $- Z_{E}$ ke beech halfway tilt hoti hai (figure mein red arrow dekho).

Recall Verify

$∣ \hat{D} ∣ = 0.707 1^{2} + 0^{2} + 0.707 1^{2} = 0.5 + 0.5 = 1$ ✔ — ekdum genuine unit vector.

Ex 4 — Y-component ke saath mid-latitude par ECI→ECEF · Cells A, H

Forecast: $z$ coordinate $Z$ -spin se untouched rehta hai. Kaun se do coordinates mix honge?

Angle radians mein: $θ = 30° = π /6 = 0.5236$ rad, $cos θ = 0.8660$ , $sin θ = 0.5$ . Yeh step kyun? KaTeX/trig radians mein kaam karta hai; degrees sirf humari intuition ke liye hain.
$R_{z}$ ki rows apply karo: $x_{E} = cos θ x_{I} + sin θ y_{I} = 0.8660 (4000) + 0.5 (3000) = 4964.1 km,$ $y_{E} = - sin θ x_{I} + cos θ y_{I} = - 0.5 (4000) + 0.8660 (3000) = 598.1 km,$ $z_{E} = z_{I} = 5000 km .$ Yeh step kyun? Rows 1 aur 2 $x$ aur $y$ mix karte hain; row 3 $z$ copy karta hai kyunki $Z$ ke around spin $Z$ ke saath height ko akela chodh deta hai.

Recall Verify

Length preserved: $x_{E}^{2} + y_{E}^{2} = x_{I}^{2} + y_{I}^{2}$ ⇒ $4964. 1^{2} + 598. 1^{2} \approx 400 0^{2} + 300 0^{2} = 25, 000, 000 km^{2}$ ✔.

Ex 5 — Degenerate aur pole cases · Cells B, C

Forecast: zero ka rotation kuch nahi karna chahiye. Aur pole par, "North kaun si taraf hai" ajeeb ho jaata hai — guess karo kyun.

(a) $θ = 0$ set karo: $cos 0 = 1, sin 0 = 0$ , toh $R_{z} (0) = 100010001 = I .$ Yeh step kyun? Zero turn = identity = coordinates unchanged. Yeh woh sanity floor hai jo har rotation ko pass karna chahiye.
(b) $ϕ = 90°$ set karo NED rows mein: $cos 90° = 0, sin 90° = 1$ . North row $(- cos λ, - sin λ, 0)$ ban jaati hai aur Down row $(0, 0, - 1)$ ban jaati hai — Down seedha $- Z_{E}$ ke saath (centre ki taraf) point karta hai, jaisa hona chahiye. Yeh step kyun? Pole par "Down" unambiguous hai, lekin "North" ab $λ$ par depend karta hai — sab longitudes pole par milte hain, isliye North/East split arbitrary ho jaata hai. Yeh classic pole singularity hai.

Recall Verify

$R_{z} (0) = I$ : check karo $R_{z} (0)^{T} R_{z} (0) = I$ trivially ✔. $ϕ = 90°$ par, Down $= (0, 0, - 1)$ ki magnitude $1$ hai ✔.

Ex 6 — Word problem: Cape Canaveral vs equator par boost · Cells C, D, G

Forecast: 28.5° pole ki taraf ek tehaai ka rasta hai. Kya boost ka ek tehaai hi khooge, ya kam?

Same formula, general latitude: $v (ϕ) = ω_{e} R_{\oplus} cos ϕ$ . Yeh step kyun? Sweeping radius equator se door jaate waqt $cos ϕ$ se shrink karta hai (figure mein red radius).
$ϕ = 28.5°$ daalo, $cos 28.5° = 0.8788$ : $v = 7.292 \times 1 0^{- 5} \times 6.378 \times 1 0^{6} \times 0.8788 = 408.7 m/s .$
Equatorial boost ka fraction = $cos 28.5° = 0.8788$ , matlab 87.9%. Yeh step kyun? $ω_{e} R_{\oplus}$ factor cancel ho jaata hai; ratio mein sirf $cos ϕ$ bachta hai.

Recall Verify

$408.7/465.1 = 0.879 = cos 28.5°$ ✔. Units m/s ✔.

Ex 7 — Velocity transform: $ω_{e} \times r$ mat bhoolna · Cell E

Forecast: woh "move nahi kar raha," toh kya uski ECI velocity zero hai? Careful.

Transport theorem likho: $v_{I} = R_{z}^{T} (θ) (v_{E} + ω_{e} \times r_{E})$ , $ω_{e} = ω_{e} \hat{Z} = (0, 0, ω_{e})$ . Yeh step kyun? Ek rotating frame mein still point, inertial frame mein genuinely move kar raha hota hai — cross-product term woh motion carry karta hai.
Cross product $ω_{e} \times r_{E} = (0, 0, ω_{e}) \times (R_{\oplus}, 0, 0) = (0, ω_{e} R_{\oplus}, 0)$ . Yeh step kyun? $\hat{Z} \times \hat{X} = \hat{Y}$ : spin point ko $+ Y$ (eastward) direction mein push karta hai.
$θ = 0$ ke saath, $R_{z}^{T} = I$ : $v_{I} = (0, ω_{e} R_{\oplus}, 0) = (0, 465.1, 0) m/s .$

Recall Verify

$∣ v_{I} ∣ = ω_{e} R_{\oplus} = 465.1$ m/s, Ex 1 se exactly match karta hai (ek stationary ground point hai hi Earth ki surface) ✔.

Ex 8 — Exam twist: round-trip start wapas lana chahiye · Cell H

Forecast: agar rotation matrices honest hain, toh round trip tumhe bilkul wahi wapas lana chahiye.

Forward ne diya $r_{E} = (4964.1, 598.1, 5000)$ km (Ex 4). Yeh step kyun? Checked result reuse karo, arithmetic dobara mat karo.
Inverse transpose hai: $r_{I} = R_{z}^{T} (θ) r_{E}$ , kyunki $R_{z}^{T} R_{z} = I$ . Yeh step kyun? Orthonormal matrices transpose karke invert hoti hain — alag se inverse ki zaroorat nahi. $x_{I} = cos θ x_{E} - sin θ y_{E} = 0.8660 (4964.1) - 0.5 (598.1) = 4000 km,$ $y_{I} = sin θ x_{E} + cos θ y_{E} = 0.5 (4964.1) + 0.8660 (598.1) = 3000 km,$ $z_{I} = 5000 km .$

Recall Verify

Recover hua $(4000, 3000, 5000)$ = original input ✔.

Recall

Velocity ECEF→ECI transform karte waqt kaun sa term add karte hain?

ω_{e} \times r_{E}

(transport-theorem term) — position ko aisa koi term nahi chahiye, velocity ko chahiye.

28.5°

latitude par equatorial boost sirf

\approx 12%

kyun khooti hai?

Kyunki loss factor

cos ϕ

hai, aur cosine

0

ke paas flat hota hai, toh chhote latitudes bahut kam cost karte hain.

Orthonormal rotation matrix ko invert kaise karte hain?

Uska transpose lo, kyunki

R^{T} R = I

NED frame mein

ϕ = 90°

par kya singular ho jaata hai, aur kya theek rehta hai?

North/East split arbitrary ho jaati hai (sab longitudes milte hain); Down direction

(0, 0, - 1)

par well-defined rehti hai.

See also: Rotating Reference Frames and Coriolis Force, Rotation Matrices and Euler Angles, Launch Azimuth and Orbital Inclination, Geodetic vs Geocentric Latitude.