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

This page is a practice arena. The parent note built the machinery — five frames, rotation matrices, the transport theorem. Here we drive it through every kind of input so you never meet a case you haven't already solved.

Before we begin, three words we will lean on constantly, in plain language:

The scenario matrix

Every problem this topic throws at you falls into one of these cells. Each worked example below is tagged with the cell(s) it clears.

We now clear the whole table.

Warm-up: reading $R_z(\theta )$ off a picture

Before numbers, look at what "rotate the axes by $\theta$" does. The point stays nailed in space; the axes swing under it.

Ex 1 — Eastward boost at the equator · Cell D

Forecast: guess a number before reading on. Faster than a jet? Slower?

1. Pick the formula $v={\omega }_e{R}_{\oplus }\cos \varphi$. Why this step? Only the part of Earth perpendicular to the spin axis actually sweeps a circle. At latitude $\varphi$ that circle's radius is $R_{\oplus }\cos \varphi$ (look at Ex 6's figure). Speed = rate $\times$ radius.
2. Substitute $\varphi =0$ so $\cos 0=1$. Why this step? The equator sits farthest from the axis → biggest circle → maximum speed. $v=7.292\times 10^{-5}\times 6.378\times 10^6\times 1=465.1\text{m/s}.$

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

Forecast: after a quarter-day of spin, roughly where does the point end up — still on $+X$, or somewhere else?

1. Accumulated angle $\theta ={\omega }_{e}t$, with $t=6\times 3600=21600$ s. Why this step? ECEF differs from ECI only by rotation about $Z$ through $\theta ={\omega }_{e}t$. $\theta =7.292\times 10^{-5}\times 21600=1.5751\text{rad}\approx 90.2°.$
2. Apply $R_z(\theta )$ to ${\vec{r}}_I=(R_{\oplus },0,0)$: $x_E=\cos \theta R_{\oplus },y_E=-\sin \theta R_{\oplus },z_E=0.$ Why this step? Only the first column of $R_z$ matters because the input has only an $x$ component.
3. Numbers. With $\theta \approx 90.2°$: $\cos \theta \approx -0.0038$, $\sin \theta \approx +0.99999$, so $x_E\approx -2.4\times 10^4\text{m}(\approx 0),y_E\approx -6.378\times 10^6\text{m}.$

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

Forecast: "Down" points to Earth's centre. At 45°N with $\lambda =0$, guess which two ECEF axes it splits between.

1. Grab the 3rd NED row (the Down unit vector): $\hat{D}=(-\cos \varphi \cos \lambda ,-\cos \varphi \sin \lambda ,-\sin \varphi ).$ Why this step? Down $=-\hat{r}$, the inward radial. The outward radial $\hat{r}$ in ECEF is exactly $(\cos \varphi \cos \lambda ,\cos \varphi \sin \lambda ,\sin \varphi )$, so negate it.
2. Plug $\varphi =45°,\lambda =0$: $\cos 45°=\sin 45°=0.7071$, $\sin \lambda =0$, $\cos \lambda =1$: $\hat{D}=(-0.7071,0,-0.7071).$ Why this step? At $\lambda =0$ the point lies in the $X$–$Z$ plane, so no $Y$ component appears; at 45° the direction tilts exactly halfway between $-X_E$ and $-Z_E$ (see red arrow in the figure).

Ex 4 — ECI→ECEF at a mid-latitude with a Y-component · Cells A, H

Forecast: the $z$ coordinate is untouched by a $Z$-spin. Which two coordinates get mixed?

1. Angle in radians: $\theta =30°=\pi /6=0.5236$ rad, $\cos \theta =0.8660$, $\sin \theta =0.5$. Why this step? KaTeX/trig work in radians; degrees are only for our intuition.
2. Apply the rows of $R_z$: $x_E=\cos \theta x_I+\sin \theta y_I=0.8660(4000)+0.5(3000)=4964.1\text{km},$ $y_E=-\sin \theta x_I+\cos \theta y_I=-0.5(4000)+0.8660(3000)=598.1\text{km},$ $z_E=z_I=5000\text{km}.$ Why this step? Rows 1 and 2 mix $x$ and $y$; row 3 copies $z$ because a spin about $Z$ leaves height along $Z$ alone.

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

Forecast: a rotation of zero should do nothing. And at the pole, "which way is North" gets weird — guess why.

1. (a) Set $\theta =0$: $\cos 0=1,\sin 0=0$, so $R_z(0)=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I.$ Why this step? Zero turn = identity = coordinates unchanged. This is the sanity floor every rotation must pass.
2. (b) Set $\varphi =90°$ in the NED rows: $\cos 90°=0,\sin 90°=1$. The North row becomes $(-\cos \lambda ,-\sin \lambda ,0)$ and the Down row becomes $(0,0,-1)$ — Down points straight along $-Z_E$ (to the centre) as it should. Why this step? At the pole "Down" is unambiguous, but "North" now depends on $\lambda$ — all longitudes meet at the pole, so the North/East split becomes arbitrary. This is the classic pole singularity.

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

Forecast: 28.5° is a third of the way to the pole. Do you lose a third of the boost, or less?

1. Same formula, general latitude: $v(\varphi )={\omega }_e{R}_{\oplus }\cos \varphi$. Why this step? The sweeping radius shrinks by $\cos \varphi$ as you move off the equator (red radius in figure).
2. Plug $\varphi =28.5°$, $\cos 28.5°=0.8788$: $v=7.292\times 10^{-5}\times 6.378\times 10^6\times 0.8788=408.7\text{m/s}.$
3. Fraction of equatorial boost = $\cos 28.5°=0.8788$, i.e. 87.9%. Why this step? The ${\omega }_e{R}_{\oplus }$ factor cancels; only $\cos \varphi$ survives in the ratio.

Ex 7 — Velocity transform: don't forget ${\vec{\omega }}_e\times \vec{r}$ · Cell E

Forecast: it's "not moving," so is its ECI velocity zero? Careful.

1. Write the transport theorem: ${\vec{v}}_I=R_z^T(\theta )({\vec{v}}_E+{\vec{\omega }}_e\times {\vec{r}}_E)$, ${\vec{\omega }}_e={\omega }_e\hat{Z}=(0,0,{\omega }_e)$. Why this step? A point still in a rotating frame is genuinely moving in the inertial frame — the cross-product term carries that motion.
2. Cross product ${\vec{\omega }}_e\times {\vec{r}}_E=(0,0,{\omega }_e)\times (R_{\oplus },0,0)=(0,{\omega }_e{R}_{\oplus },0)$. Why this step? $\hat{Z}\times \hat{X}=\hat{Y}$: the spin pushes the point in the $+Y$ (eastward) direction.
3. With $\theta =0$, $R_z^T=I$: ${\vec{v}}_I=(0,{\omega }_e{R}_{\oplus },0)=(0,465.1,0)\text{m/s}.$

Ex 8 — Exam twist: round-trip must return the start · Cell H

Forecast: if rotation matrices are honest, a round trip should return you home exactly.

1. Forward gave ${\vec{r}}_E=(4964.1,598.1,5000)$ km (Ex 4). Why this step? Reuse the checked result rather than redo arithmetic.
2. Inverse is the transpose: ${\vec{r}}_I=R_z^T(\theta ){\vec{r}}_E$, because $R_z^T{R}_z=I$. Why this step? Orthonormal matrices invert by transposing — no separate inverse needed. $x_I=\cos \theta x_E-\sin \theta y_E=0.8660(4964.1)-0.5(598.1)=4000\text{km},$ $y_I=\sin \theta x_E+\cos \theta y_E=0.5(4964.1)+0.8660(598.1)=3000\text{km},$ $z_I=5000\text{km}.$

Recall

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