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

3.4.1 · D5 · Physics › Rocket Flight Mechanics › Coordinate systems — Earth-Centered Inertial (ECI), Earth-Ce

Shuru karne se pehle, teen pictures mind mein rakho:

ECI = axes door ke taaron se judi hain (ye kabhi rotate nahi karti).
ECEF = axes planet ki crust se chipki hain (ye din mein ek baar planet ke saath spin karti hain).
NED = ek chhota sa tripod tumhare paon ke paas zameen par laga hua: ek pair North, ek East, ek seedha neechey dirt mein Down.

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

Definition Is page par use kiye gaye har symbol — pehle yeh padho

Kyunki yeh ek stand-alone question bank hai, yahan poora glossary aur har sign convention diya gaya hai. Neeche koi bhi aisa symbol use nahi kiya gaya jo yahan listed na ho.

$ω_{e}$ = Earth ka spin rate $\approx 7.292 \times 1 0^{- 5} rad/s$ ( $\approx 15°$ per hour). Hamesha ek positive scalar.
$ω_{e} = ω_{e} \hat{Z}$ = spin vector, jo North spin axis $\hat{Z}$ ki taraf point karta hai. Earth North pole ke upar se dekha jaaye to counter-clockwise ghoomti hai — yahi positive (right-hand-rule) sense hai: right hand ki ungliyon ko uss direction mein curl karo jis direction mein Earth ghoomti hai, thumb $\hat{Z}$ (North) ki taraf point karega.
$R_{\oplus}$ = Earth ka equatorial radius $\approx 6.378 \times 1 0^{6} m$ . Hamesha positive.
$ϕ$ (phi) = geodetic latitude, local vertical ka equatorial plane ke upar ka angle; North positive hai, South negative, range $- 90°$ se $+ 90°$ tak.
$λ$ (lambda) = longitude; Greenwich ke East ki taraf positive hai, West negative, range $- 180°$ se $+ 180°$ tak (yahi sign convention is page ke har formula mein use ki gayi hai).
$θ$ (theta) = Greenwich rotation angle $θ = ω_{e} t$ , ECEF ne $t = 0$ se lekar $\hat{Z}$ ke around ECI se kitna swing kiya hai. Positive $θ$ = $ω_{e}$ ke same right-hand sense mein (North se dekha jaaye to CCW).
$r_{E}$ , $r_{I}$ = kisi point ka position vector jo ECEF aur ECI components mein express kiya gaya ho.
$\overset{r}{^}$ = outward radial unit vector (Earth centre se point ki taraf).
$R$ = ek $3 \times 3$ rotation matrix; yahan specifically woh jo ECI components ko ECEF components mein map karta hai ( $r_{E} = R r_{I}$ ), isliye iska transpose $R^{T}$ ECEF ko wapas ECI mein map karta hai. Yeh satisfy karta hai $R^{- 1} = R^{T}$ aur $det R = + 1$ .

True or false — justify

TF1. ECEF mein pad par baithe rocket ki velocity zero hai, isliye acceleration bhi zero hai.

False — ECEF mein uski coordinate velocity zero hai, lekin ECEF spin karta hai, isliye pad ek circle mein ghoom raha hai; usmein real centripetal acceleration

ω_{e}^{2} R_{\oplus} cos ϕ

hai jo spin axis ki taraf point karti hai.

TF2. Kyunki ek rotation matrix

R^{- 1} = R^{T}

satisfy karta hai, ECEF se ECI mein velocity transform karna bas

v_{I} = R^{T} v_{E}

hai.

False — woh formula sirf position move karta hai. Velocity mein extra transport term aata hai,

v_{I} = R^{T} (v_{E} + ω_{e} \times r_{E})

, kyunki differentiate karte waqt frame khud ghoom raha hota hai.

TF3. NED aur ECEF ka origin ek hi hota hai.

False — ECEF ka origin Earth ka center hai; NED ka origin tumhare current point par surface par hota hai, isliye dono ke beech jaane ke liye rotation ke saath ek translation bhi chahiye.

TF4. Kyunki Earth

\hat{Z}

ke around spin karti hai, ECI→ECEF change ek hi angle

θ = ω_{e} t

se shared

\hat{Z}

axis ke around ek rotation hai.

True — dono frames origin aur spin axis share karte hain, isliye unka fark sirf uss ek Greenwich angle se hai; kuch tilt nahi hota, sirf

X

aur

Y

swing karte hain.

TF5. NED mein Down hamesha Earth ke geometric center ki taraf point karta hai.

Roughly, lekin strictly Down local vertical ke saath point karta hai (reference ellipsoid par geodetic normal ka opposite), jo sirf equator aur poles par exactly center ki taraf aim karta hai — dekho Geodetic vs Geocentric Latitude.

TF6. Launch frame inertial hai, isliye kisi bhi duration ke mission ke liye ise use karna safe hai.

False — yeh sirf short flight ke liye approximately inertial hai; ghanton baad Earth iske neeche se rotate ho jaati hai aur neglect ki gayi rotation errors badhti jaati hain, isliye tumhe true ECI par wapas jaana padega.

TF7. Yaw, pitch aur roll ke liye rotation matrices ko kisi bhi order mein multiply kiya ja sakta hai aur same body attitude milega.

False — finite rotations commute nahi karte, isliye

R_{x} R_{y} R_{z} \neq = R_{z} R_{y} R_{x}

; 3-2-1 (yaw→pitch→roll) order ek convention hai jo fix karta hai ki har angle ka matlab kya hai, dekho Rotation Matrices and Euler Angles.

TF8. Chain ka har frame right-handed hai.

True by construction — har ek right-handed define kiya gaya hai (jaise NED: North×East = Down), jo har transform ke liye

det R = + 1

rakhta hai aur tumhe sign surprises ke bina compose karne deta hai.

TF9. ~465 m/s ka eastward launch bonus thrust se aata hai, geometry se nahi.

False — yeh purely woh surface speed hai jo tumhare paas Earth ke spin se pehle se hai,

ω_{e} R_{\oplus} cos ϕ

; tumhe yeh engines se koi help liye bina free mein milti hai.

TF10. Agar Earth spinning band ho jaaye, to ECI aur ECEF ek hi frame ban jaayenge.

True in effect —

ω_{e} = 0

ke saath Greenwich angle

θ = ω_{e} t

freeze ho jaata hai, isliye unke beech ka rotation ek fixed (ya identity) matrix ban jaata hai aur transport term

ω_{e} \times r_{E}

vanish ho jaati hai.

Spot the error

SE1. "Main

F = m a

ko ECEF mein integrate karunga kyunki pad wahaan fixed hai, phir directly latitude read karunga."

Error yeh hai ki ECEF ko inertial treat kiya ja raha hai; yeh rotate karta hai, isliye

F = m a

mein Coriolis

- 2 ω_{e} \times v

aur centrifugal

- ω_{e} \times (ω_{e} \times r)

terms add karne padte hain — ECI mein integrate karo aur sirf output ke liye transform karo.

SE2. "East bas

\hat{Z}_{E} \times \overset{r}{^}

hai, isliye North pole par East

Y_{E}

ke along point karta hai."

Pole par

\overset{r}{^}

\hat{Z}_{E}

ke parallel hai, isliye unka cross product zero vector hai — East wahaan undefined hai; NED poles par degenerate ho jaata hai, jo ek genuine gimbal-like singularity hai, tumhare algebra mein koi bug nahi.

SE3. "ECEF→NED matrix ke East row mein koi longitude

λ

nahi hai, isliye koi term bhool gayi."

Koi error nahi — upar di gayi matrix ki Row 2 dekho: East unit vector ke ECEF components

(- sin λ, cos λ, 0)

sirf longitude par depend karte hain (jo wahaan hai) aur kabhi latitude

ϕ

par nahi, kyunki north-south move karne se "East" kis direction mein tilt hai yeh nahi badalta.

SE4. "Body→ECI jaane ke liye main same matrices ko same order mein multiply karunga jaise ECI→body ke liye use kiya."

Galat — rotations ki ek chain ko reverse karne ke liye transpose karna aur order reverse karna dono zaroori hain:

(R_{3} R_{2} R_{1})^{T} = R_{1}^{T} R_{2}^{T} R_{3}^{T}

; order same rakhne se bilkul alag, invalid rotation milta hai.

SE5. "Maine

ω_{e} \times r_{E}

term drop kar di kyunki yeh chhoti hai."

Surface par yeh

ω_{e} R_{\oplus} cos ϕ \approx 465

m/s equator par hai — bilkul bhi negligible nahi; ise silently drop karne se poora rotational launch advantage chhoot jaata hai aur orbit insertion corrupt ho jaati hai. Dekho Rotating Reference Frames and Coriolis Force.

SE6. "NED matrix mein latitude

ϕ

wahi angle hai jo Earth's center se meri position tak hai."

Generally nahi — map geodetic latitude use karta hai (geodetic normal / local vertical ka equatorial plane se angle), jo geocentric angle se ~0.19° tak differ karta hai kyunki Earth ek ellipsoid hai, per Geodetic vs Geocentric Latitude.

SE7. "

X_{I}

Sun ki taraf point karta hai."

Nahi —

X_{I}

vernal equinox ki taraf point karta hai, jo taaron ke beech ek fixed direction hai; Sun saal bhar ecliptic par move karta hai, isliye woh ek inertial axis anchor nahi kar sakta.

SE8. "Maine ek launch site 120° West ko

λ = + 120°

NED matrix mein daala aur East vector galat direction mein point kar raha tha."

Sign-convention error — is page par East longitude positive hai, isliye 120° West ko

λ = - 120°

daalna chahiye; galat sign daalne se

sin λ

flip ho jaata hai aur East row mirror ho jaati hai, yeh ek classic quadrant trap hai.

Why questions

WHY1. Ek single coordinate frame rocket flight ka har kaam kyun nahi kar sakta?

Alag-alag tasks alag "still" points chahte hain — Newton's law ko non-rotating frame chahiye (ECI), fixed pad ko Earth-fixed frame chahiye (ECEF), horizon ko local frame chahiye (NED), aur thrust ko rocket ke apne axes chahiye (body).

WHY2. Equations of motion ko integrate karne ke liye ECI use kyun karte hain, ECEF kyun nahi?

Sirf ECI genuinely non-rotating aur non-accelerating hai, isliye

F = m a

sirf real forces ke saath hold karta hai — koi fictitious Coriolis ya centrifugal terms add nahi karne padte, dekho Rocket Equations of Motion.

WHY3. ECI aur ECEF ko relate karne ke liye sirf ek hi angle kyun kafi hai?

Dono same origin aur same spin axis

\hat{Z}

share karte hain, isliye jo freedom bachi hai woh sirf yeh hai ki

X

aur

Y

us axis ke around kitna swing kar chuke hain — poori tarah Greenwich angle

θ = ω_{e} t

se capture ho jaata hai.

WHY4. Hum eastward aur equator ke kareeb se launch kyun karte hain?

Eastward launch rocket ko Earth ki existing surface velocity ke saath align karta hai taaki woh free speed rakh sako; equator ke kareeb woh speed

ω_{e} R_{\oplus} cos ϕ

maximum hoti hai kyunki tum spin axis se sabse door hote ho, dekho Launch Azimuth and Orbital Inclination.

WHY5. Velocity transform karte waqt transport term kyun aata hai, position mein kyun nahi?

Position ek single snapshot hai, isliye sirf axes ki orientation (matrix

R

jo ECI→ECEF map karta hai) matter karti hai; velocity ek rate hai, aur derivative lete waqt woh axes khud ghoom rahi hoti hain (

\dot{R} \neq = 0

), jo

ω_{e} \times r_{E}

correction inject karta hai.

WHY6. Thrust aur IMU readings express karne ke liye aerospace body frame kyun prefer karta hai?

Woh quantities physically vehicle se tied hain — thrust nose se baahir point karta hai, gyros/accelerometers structure mein bolted hain — isliye unke components body frame mein constant aur simple hote hain, phir attitude ke through rotate kar diye jaate hain.

WHY7. Attitude ke liye Euler angles se quaternions par switch kyun kiya jaa sakta hai?

Euler angles singularities hit karte hain (gimbal lock) jahan do axes align ho jaati hain aur ek angle undefined ho jaata hai; quaternions mein aisa koi gap nahi aur woh smoothly interpolate karte hain, dekho Quaternions for Attitude.

WHY8. "Down" ko

- \overset{r}{^}

define kyun kiya gaya hai, "Up" choose kyun nahi kiya?

Convention ek right-handed North-East-Down set choose karta hai taaki

\hat{N} \times \hat{E} = \hat{D}

ho; Up choose karne se handedness flip ho jaati aur transform chain mein jaabjaah sign changes force ho jaate.

Edge cases

EC1. Exactly North ya South pole par NED frame ka kya hota hai?

North aur East directions ambiguous ho jaati hain kyunki pole par "North kaun sa direction hai" undefined hai — local tangent plane apna reference kho deta hai, isliye NED wahaan singular hai aur usse specially handle karna padta hai.

EC2. Pole (

ϕ = 90°

) se launch ke liye eastward rotation bonus kitna hai?

Zero — factor

cos ϕ = 0

hai, kyunki spin axis par ka point move nahi karta; polar launches ko koi rotational head-start nahi milta, yahi ek wajah hai ki equatorial sites itne prized hain.

EC3.

t = 0

par, ECI→ECEF rotation matrix kya reduce ho jaata hai?

Identity mein —

θ = ω_{e} \cdot 0 = 0

ke saath,

cos θ = 1

aur

sin θ = 0

, isliye ECI aur ECEF momentarily coincide karte hain usse pehle ki Earth unhe alag kar de.

EC4. Agar ek spacecraft Earth ke exact center par ho (ek thought experiment), to NED kyun break down karta hai?

Radial direction

\overset{r}{^}

r = 0

par undefined hai, isliye Down, aur isliye poora North-East-Down tripod, construct nahi kiya ja sakta — is frame ko exist karne ke liye ek nonzero position vector chahiye.

EC5. Ek poore sidereal day mein ECI→ECEF angle ka behaviour kaisa hota hai, aur iska reuse ke liye kya implication hai?

θ = ω_{e} t

poora

2 π

sweep karta hai aur phir frames realign ho jaate hain, isliye transform periodic hai; tum kisi bhi ek frozen matrix ko us span mein valid nahi treat kar sakte — tumhe

θ

continuously recompute karna hoga.

EC6. Bahut chhoti flight (seconds se couple minutes) ke liye launch frame ko inertial treat karna acceptable kyun hai?

Us window mein Earth thoda sa hi ghoomti hai (

ω_{e} t

bahut chhota hota hai), isliye neglected rotation aur transport corrections trajectory tolerances se neeche rehti hain — approximation sirf isliye safe hai kyunki elapsed time chhota hai.

EC7. Spin axis par exactly baithe kisi object ke liye

ω_{e} \times r_{E}

kya evaluate karta hai?

Zero vector, kyunki

r_{E}

ω_{e}

ke parallel hai wahaan aur parallel vectors ka cross product vanish hota hai — consistent with ek point on the axis having no rotational speed.

Recall wrap-up

Recall Quick self-test

Kaun sa frame truly inertial hai, aur integration ke liye yeh kyun matter karta hai? ::: ECI — iske axes na rotate karte hain na accelerate, isliye $F = m a$ sirf real forces ke saath hold karta hai, jo ise equations of motion integrate karne ka correct frame banata hai. Ek jagah batao jahaan NED undefined hai. ::: Poles par (North/East ambiguous) aur Earth's center par (radial direction undefined). Ek sentence mein: ECI aur ECEF ke beech velocity, position se alag kyun transform hoti hai. ::: Kyunki derivative lete waqt frame rotate ho raha hota hai, jo plain rotation ke upar transport term $ω_{e} \times r_{E}$ add karta hai. Is page par $ϕ$ aur $λ$ ka sign convention kya hai? ::: $ϕ$ positive North, $λ$ positive East of Greenwich — yahi right-handed choice har formula mein use ki gayi hai.