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

3.4.1 · D4 · Physics › Rocket Flight Mechanics › Coordinate systems — Earth-Centered Inertial (ECI), Earth-Ce

Ye constants hum baar baar use karte hain:

Neeche do tools baar baar aate hain. Unhe problem ke beech mein quote karne ki bajay, hum unhe yahan ek baar build karte hain, phir reuse karte hain.

Building block A: rotation matrix $R_{z} (θ)$ — ye aata kahan se hai

Formula

$R_{z} (θ)$ ko dot products se derive karna (kuch yaad nahi karna) Maano purane ECI axes $\hat{X}_{I}, \hat{Y}_{I}$ hain. Frame ko CCW $θ$ se ghoomane ke baad, naya $X_{E}$ axis ECI direction $(cos θ, sin θ, 0)$ mein point karta hai, aur naya $Y_{E}$ axis $(- sin θ, cos θ, 0)$ par point karta hai (ye $90°$ aur CCW hai). Ek point $r_{I}$ ka naya coordinate bas har naye axis par uska projection (dot product) hota hai: $x_{E} = r_{I} \cdot \hat{X}_{E} = x_{I} cos θ + y_{I} sin θ,$ $y_{E} = r_{I} \cdot \hat{Y}_{E} = - x_{I} sin θ + y_{I} cos θ, z_{E} = z_{I} .$ In teeno equations ko stack karna hi matrix hai: $R_{z} (θ) = cos θ - sin θ 0 sin θ cos θ 0 001, r_{E} = R_{z} (θ) r_{I} .$ $- sin θ$ / $+ sin θ$ ke do signs arbitrary nahi hain — ye seedha "naya $Y_{E}$ naye $X_{E}$ se $90°$ aage hai" se aate hain. Kyunki rotations length preserve karte hain, $R_{z}^{T} R_{z} = I$ , isliye iska inverse bas iska transpose hai $R_{z}^{T} = R_{z} (- θ)$ .

Building block B: transport theorem — velocity mein extra term kyun chahiye

Formula

$v_{I} = R_{z}^{T} (v_{E} + ω_{e} \times r_{E})$ derive karna Shuru karo is se ki position kaise transform hoti hai (Building block A, inverted): $r_{I} = R_{z}^{T} (θ) r_{E}$ , jahan $θ = ω_{e} t$ waqt ke saath badhta hai. Time ke saath product rule use karke differentiate karo: $v_{I} = \dot{r}_{I} = R_{z}^{T} \dot{r}_{E} + \dot{R}_{z}^{T} r_{E} .$ Pehla term $R_{z}^{T} \dot{r}_{E} = R_{z}^{T} v_{E}$ woh velocity hai jo ECEF mein measure ki gayi, sirf re-expressed ki gayi hai. Doosra term naya hai: $ω_{e}$ rate par $\hat{Z}$ ke around ek rotation differentiate karne se identity milti hai $\dot{R}_{z}^{T} = R_{z}^{T} [ω_{e} \times]$ , jahan $[ω_{e} \times]$ ka matlab hai " $ω_{e} = ω_{e} \hat{Z}$ ke saath cross product lo." To $v_{I} = R_{z}^{T} (v_{E} + ω_{e} \times r_{E}) .$ Term $ω_{e} \times r_{E}$ woh velocity hai jo frame ki spin inject karti hai — merry-go-round effect. Ye exactly eastward launch bonus hai.

Level 1 — Recognition

Ye test karta hai ki kya tum bina heavy arithmetic ke sahi frame / formula ka naam le sakte ho.

Recall Solution 1.1

Answer: ECEF (Earth-Centered Earth-Fixed). ECEF ke axes spinning Earth se chipke hain. Kyunki pad zameen ke saath saath chalta hai, wo kabhi un axes ke relative move nahi karta — isliye uske coordinates fixed rehte hain. ECI mein (jinke axes door ke taaron ki taraf point karte hain) wahi pad ek din mein ek bada circle trace karta hai, isliye uske ECI coordinates lagaataar change hote hain.

Recall Solution 1.2

Answer: ECI. $F = m a$ sirf us frame mein hold karta hai jo na rotate ho na accelerate ho — ek inertial frame. Humare panch frames mein se sirf ECI qualify karta hai (uske axes fixed stars par point karte hain). ECEF ghoomta hai; NED aur body ek curving, turning trajectory pe sawaar hain; launch frame approximately inertial hai lekin sirf ek choti flight ke liye.

Recall Solution 1.3

(a) → body frame — sensors rocket se bolted hain, isliye wo $x_{b}, y_{b}, z_{b}$ ke along measure karte hain.
(b) → NED — "kitna uuncha" aur "horizon kidhar hai" literally Down aur North/East directions hain.
(c) → ECI — sirf yahi inertial frame hai, jahan $F = m a$ ko koi fictitious forces nahi chahiye.

Level 2 — Application

Ab tum ek chosen formula mein numbers plug karte ho.

Recall Solution 2.1

Ye formula kyun: pad ki position ka sirf wo hissa jo spin axis ke perpendicular ho, wo actually sweep karta hai. Wo perpendicular distance $R_{\oplus} cos ϕ$ hai (latitude ke circle ki radius). Speed = angular rate × radius = $ω_{e} R_{\oplus} cos ϕ$ . Step 1 — cosine: $cos 28.5° = 0.8788$ . Step 2 — multiply: $v = 7.292 \times 1 0^{- 5} \times 6.378 \times 1 0^{6} \times 0.8788 \approx 408.7 m/s .$ To Cape se eastward launch karna rocket ko free mein $\approx 409 m/s$ ki head start deta hai. (Equator par $465 m/s$ se compare karo — boost shrink hota hai jaise north jaate ho kyunki $cos ϕ$ shrink hota hai.)

Recall Solution 2.2

Step 1 — seconds: $t = 4 \times 3600 = 14400 s$ . Step 2 — angle: $θ = 7.292 \times 1 0^{- 5} \times 14400 = 1.0500 rad$ . Step 3 — degrees: $1.0500 \times 57.30 \approx 60.16°$ . Sanity check: $15°/ hr \times 4 hr = 60°$ . ✔️

Recall Solution 2.3

$θ = 90°$ ke saath: $cos θ = 0$ , $sin θ = 1$ .

= \begin{bmatrix}0\cdot R_\oplus + 1\cdot 0\\ -1\cdot R_\oplus + 0\\ 0\end{bmatrix} = \begin{bmatrix}0\\ -R_\oplus\\ 0\end{bmatrix}.$$ **Ye kaisa dikhta hai (neeche figure):** lavender arrows ECI axes hain; mint arrows ECEF axes hain *jo $+90°$ CCW swing kar chuke hain* (Building block A ka sign convention). Coral dot physical point hai — wo space mein kabhi nahi hila, phir bhi naye mint axes ke against padha jaaye to wo $-Y_E$ side par hai. Ye negative sign CCW-axes-⇒-CW-coordinates rule ko concrete karta hai: butter arc wo $+90°$ mark karta hai jinse axes ghume.

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

Level 3 — Analysis

Yahan tum kyun sochte ho — signs, directions, aur structure aisi kyun aati hain.

Recall Solution 3.1

"Down" ye row kyun hai: "Down" seedha Earth ke center ki taraf point karta hai, yaani outward radial direction $\overset{r}{^}$ ka opposite. Outward radial at $(ϕ, λ)$ ke ECEF components $(cos ϕ cos λ, cos ϕ sin λ, sin ϕ)$ hain, isliye Down uska negative hai. Step 1 — plug in $ϕ = 45°, λ = 0$ : $cos 45° = sin 45° = 0.7071$ , $cos 0 = 1, sin 0 = 0$ . $\hat{D} = (- 0.7071, 0, - 0.7071) .$ Step 2 — length check: $0.707 1^{2} + 0 + 0.707 1^{2} = 0.5 + 0.5 = 1$ . ✔️ Ye kaisa dikhta hai (neeche figure): lavender quarter-circle Earth ki surface hai $X_{E}$ – $Z_{E}$ plane mein. Coral dot 45°N point hai; mint arrow outward radial $\overset{r}{^}$ (up) hai; coral arrow $\hat{D} = - \overset{r}{^}$ hai, aadha $- X_{E}$ (equatorial plane ki taraf) aur $- Z_{E}$ (south pole ki taraf) ke beech mein tila hua. Butter arc $ϕ = 45°$ mark karta hai. Dono components ka size equal hona exactly wohi hai jo "45°" produce karna chahiye.

Edge cases (har latitude cover karo):

North pole $ϕ = 90°$ : $cos 90° = 0, sin 90° = 1$ , to $\hat{D} = (0, 0, - 1) = - \hat{Z}_{E}$ . Down seedha south pole ki taraf point karta hai — sahi hai, kyunki north pole par khade ho to "down" axis ke saath Earth ke center ki taraf hota hai.
South pole $ϕ = - 90°$ : $sin (- 90°) = - 1$ , to $\hat{D} = (0, 0, + 1) = + \hat{Z}_{E}$ . Down ab $+ Z$ (north pole) ki taraf point karta hai — phir sahi: south pole se center-ward $+ Z$ direction hai. Note karo ki $z$ -component ka sign $ϕ$ ke sign ke saath flip hua; formula ise automatically $- sin ϕ$ ke through handle karta hai.
Southern hemisphere $ϕ = - 45°, λ = 0$ : $cos (- 45°) = 0.7071, sin (- 45°) = - 0.7071$ , to $\hat{D} = (- 0.7071, 0, + 0.7071)$ . $45°$ N case se compare karo to sirf $z$ -component ka sign badla — Down ab $+ Z_{E}$ (northward-and-inward) ki taraf tilt karta hai, equator ke across mirror image. Length phir bhi $1$ hai. To ek hi expression sab $ϕ \in [- 90°, 90°]$ aur sab $λ$ ke liye valid hai; poles par kuch special nahi hota.

Recall Solution 3.2

Agar $v_{E} = 0$ hai, to naive rule deta hai $v_{I} = 0$ — "pad move nahi kar raha." Lekin transport theorem $ω_{e} \times r_{E}$ rakhta hai. Aao us cross product ko component by component compute karein, sirf uska size quote na karein. Vectors ECEF mein set up karo: spin $Z$ ke along hai, $ω_{e} = (0, 0, ω_{e})$ , aur $X_{E}$ axis par equatorial pad $r_{E} = (R_{\oplus}, 0, 0)$ hai. Cross product ( $\hat{i} \times \hat{j} = \hat{k}$ etc. use karke):

= \hat i(0\cdot 0-\omega_e\cdot 0)-\hat j(0\cdot 0-\omega_e R_\oplus)+\hat k(0-0) = (0,\ \omega_e R_\oplus,\ 0).$$ **Result padhna:** ye $+Y_E$ (eastward) along point karta hai, magnitude $\omega_e R_\oplus$ ke saath. Note karo ki $\hat i$ aur $\hat k$ components vanish ho gaye — cross product automatically **$\vec r_E$ ka wo hissa kill karta hai jo axis ke parallel ho** aur sirf $\vec\omega_e$ ke perpendicular hissa rakhta hai. Equator par poori radius perpendicular hai, isliye kuch nahi kho jaata; equator ke bahar sirf $R_\oplus\cos\phi$ bachega. Isliye sirf perpendicular component matter karta hai. **Number:** $$|\vec v_I| = |\vec\omega_e\times\vec r_E| = \omega_e R_\oplus = 7.292\times10^{-5}\times 6.378\times10^6 = 465.1\ \text{m/s}.$$ Naive student poori $\approx 465\ \text{m/s}$ eastward inertial motion kho deta hai jo rocket ke paas actually hai, sirf isliye kyunki ECEF axes spin kar rahe the ($\dot R\neq 0$).

Level 4 — Synthesis

Ek problem mein multiple transforms ya ideas combine karo.

Recall Solution 4.1

(a) $cos 30° = 0.8660$ , $sin 30° = 0.5$ .

= \begin{bmatrix}0.8660\,R_\oplus\\ -0.5\,R_\oplus\\ 0\end{bmatrix}.$$ **(b)** Spin axis ($Z$) se distance $\sqrt{x_E^2 + y_E^2}$ hai: $$\sqrt{(0.8660R_\oplus)^2 + (0.5R_\oplus)^2} = R_\oplus\sqrt{0.75+0.25} = R_\oplus.$$ **Kyun unchanged:** $Z$ ke around ek rotation kabhi nahi badal sakta ki point $Z$ axis se kitna door hai — wo sirf point ko us axis ke *around* swing karta hai. Rotation matrices lengths preserve karti hain, aur yahan wo length $Z$ se perpendicular distance hai. Ye wohi fact hai jo eastward-boost radius $R_\oplus\cos\phi$ ko time mein constant rakhta hai.

Recall Solution 4.2

Step 1 — Kourou: $cos 5.2° = 0.99589$ , to $v_{K} = 7.292 \times 1 0^{- 5} \times 6.378 \times 1 0^{6} \times 0.99589 = 463.2 m/s$ . Step 2 — Cape: Ex 2.1 se, $v_{C} = 408.7 m/s$ . Step 3 — difference: $Δ v = 463.2 - 408.7 = 54.5 m/s$ . Interpretation: equator ke $23°$ kareeb hone se Kourou ko lagbhag $55 m/s$ free velocity milti hai — ek concrete reason ki equatorial launch sites itni precious hain.

Level 5 — Mastery

Poore multi-step problems jo exam mein aa sakte hain.

Recall Solution 5.1

$v_{E} = 0$ ke saath, $v_{I} = R_{z}^{T} (ω_{e} \times r_{E})$ . Rotation ( $R_{z}^{T}$ ) magnitude preserve karta hai, isliye $∣ v_{I} ∣ = ∣ ω_{e} \times r_{E} ∣ = ω_{e} (R_{\oplus} cos ϕ),$ kyunki (jaise Ex 3.2 mein dikha) cross product sirf axis-perpendicular part $R_{\oplus} cos ϕ$ rakhta hai. $∣ v_{I} ∣ = 7.292 \times 1 0^{- 5} \times 6.378 \times 1 0^{6} \times cos 28.5° = 408.7 m/s .$ Direction: $ω_{e} \times r_{E}$ , $ω_{e}$ spin axis ke upar ke saath, eastward point karta hai (latitude ke circle ki tangent). To "motionless" pad actually inertial space mein $\approx 409 m/s$ east ki taraf ja raha hai — Ex 2.1 ke same, jo confirm karta hai ki boost ye hi inertial velocity hai.

Recall Solution 5.2

Step 1 — period: $T = 2 π / ω_{e} = 6.2832/7.292 \times 1 0^{- 5} = 86166 s$ . Step 2 — hours: $86166/3600 = 23.935 h \approx 23 h 56 min$ . 24 h se chota kyun: ek solar day (24 h) wo time hai jab Sun wapas upar aata hai. Lekin Earth Sun ke around bhi orbit karta hai, isliye use Sun ke saath catch up karne ke liye har din thoda extra spin karna padta hai. Isliye pure star-relative (sidereal) spin lagbhag 4 minute chota hota hai. ECI stars se tied hai, isliye $ω_{e}$ sidereal rate hai.

Recall Solution 5.3

3rd row $= (- cos ϕ cos λ, - cos ϕ sin λ, - sin ϕ)$ . Step 1 — trig evaluate karo. $ϕ = 0$ ke saath: $cos ϕ = 1, sin ϕ = 0$ . $λ = 90°$ ke saath: $cos 90° = 0, sin 90° = 1$ . Step 2 — substitute: $\hat{D} = (- 1 \cdot 0, - 1 \cdot 1, - 0) = (0, - 1, 0) .$ Step 3 — length check: $0^{2} + (- 1)^{2} + 0^{2} = 1$ . ✔️ Step 4 — direction interpret karo. Ye exactly $- \hat{Y}_{E}$ hai. $90°$ E par equator par ek point $+ Y_{E}$ ke along bahar nikla hua hai; "seedha neeche" wahan se matlab Earth ke center ki taraf jaana, yaani $- Y_{E}$ direction mein. Koi $Z$ component nahi hai kyunki equator par "down" poori tarah equatorial plane mein hai — Ex 3.1 ke pole cases se consistent, jahan saara kuch $Z$ component mein tha. To wohi 3rd-row formula smoothly poles par all- $Z$ se equator par all-horizontal tak interpolate karta hai.

Wrap-up recall

Recall Ladder ka ek-line summary

L1: frame ka naam lo. L2: boost/rotation formula plug karo. L3: signs aur extra transport term explain karo. L4: transforms sahi se chain aur invert karo. L5: sidereal timing aur transport theorem use karke true inertial velocity nikalo.

Related deep dives jinhe aage explore kar sakte ho: Rotating Reference Frames and Coriolis Force · Rotation Matrices and Euler Angles · Launch Azimuth and Orbital Inclination · Geodetic vs Geocentric Latitude · Rocket Equations of Motion