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

Step 1 — Two axes, one spinning under the other

WHAT. Picture the Earth from far above the North Pole, looking straight down the spin axis. We draw two sets of horizontal axes that share the same origin (Earth's center) and the same vertical axis (the spin axis $Z$ , pointing at your eye).

$X_{I}, Y_{I}$ — the ECI axes. They are nailed to the distant stars. They never move.
$X_{E}, Y_{E}$ — the ECEF axes. They are painted on the crust and rotate with Earth.

WHY. Before touching any trigonometry we must be sure the two frames differ by nothing except a turn about $Z$ . Same centre → no shift. Same $Z$ → the whole disagreement lives in the flat $X Y$ plane. That is what lets a single angle describe the relationship — no angle can hide in the $Z$ direction.

PICTURE. The green ECI axes stay put; the blue ECEF axes have swung counter-clockwise by an angle we call $θ$ .

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

Step 2 — Where the rotated axes point (unit vectors)

WHAT. We find the direction of the new axes $X_{E}$ and $Y_{E}$ written in the old ECI numbers. A direction of length $1$ is a unit vector; naming its ECI coordinates is the whole game.

WHY. Any coordinate is just "how far along this axis." So to read a point in the new frame, we must first know which way each new axis points in the language of the old frame. Trig enters here for exactly one reason: a point on a unit circle at angle $θ$ has horizontal coordinate $cos θ$ and vertical coordinate $sin θ$ — that is the definition of cosine and sine. Since $X_{E}$ has been swung to the angle $θ$ on the unit circle, its ECI coordinates are $(cos θ, sin θ)$ .

PICTURE. The blue $X_{E}$ arrow lands at $(cos θ, sin θ)$ ; the blue $Y_{E}$ arrow, a further $90°$ CCW, lands at $(- sin θ, cos θ)$ .

Recall Why

$Y_{E}$ picks up the minus sign Rotate $(cos θ, sin θ)$ by $+ 90°$ ::: A $+ 90°$ turn sends $(a, b) \to (- b, a)$ , so $(cos θ, sin θ) \to (- sin θ, cos θ)$ — the negative lands on the first slot.

Step 3 — Reading a point's new coordinate = projection = dot product

WHAT. Take a physical point $P$ with known ECI coordinates $r_{I} = (x_{I}, y_{I}, 0)$ . We want its ECEF coordinate $x_{E}$ : "how far along the new $X_{E}$ axis does $P$ sit?"

WHY. "How far along an axis" is precisely what the dot product measures. For a unit-length axis $\overset{u}{^}$ , the number $r \cdot \overset{u}{^}$ is the length of the shadow of $r$ on that axis. We use the dot product here and not anything fancier because projection onto a direction is exactly its job — no more, no less.

PICTURE. Drop a perpendicular from $P$ onto the blue $X_{E}$ line; the foot of that perpendicular, measured from the origin, is $x_{E}$ .

Step 4 — Stacking the projections into the matrix

WHAT. We have two scalar equations. A matrix is just a tidy box that applies both dot products at once: each row is one new axis, each column multiplies one old coordinate.

WHY. Writing the two lines as $r_{E} = R_{z} (θ) r_{I}$ is not a new idea — it is bookkeeping. But it exposes the pattern: row 1 = $\hat{X}_{E}$ , row 2 = $\hat{Y}_{E}$ , row 3 = $\hat{Z}$ (unchanged). Once you see rows-are-axes, you never memorize the matrix again — you reconstruct it.

PICTURE. The rows of the box are colour-matched to the axis unit vectors from Step 2.

Step 5 — The orthonormal check (why $R^{- 1} = R^{T}$ )

WHAT. A rotation must preserve lengths and right angles. We check that the rows are orthonormal: each has length $1$ , and any two are perpendicular (dot product $0$ ).

WHY. If lengths were not preserved, we'd be stretching space, not rotating it. The magic payoff: for a matrix whose rows are orthonormal, its inverse equals its transpose — flipping the box across its diagonal undoes it. That is why the parent note says "to go the other way, use the transpose." No matrix inversion algorithm needed.

PICTURE. Row 1 and Row 2 drawn as perpendicular unit arrows; their right angle is the geometric statement of orthonormality.

Step 6 — Edge and degenerate cases (all quadrants of $θ$ )

WHAT. We test the matrix at the four cardinal turns and at the special $t = 0$ , checking that a point starting on $X_{I}$ lands where common sense says.

WHY. A formula you cannot break is a formula you can trust. If any quadrant misbehaved, we'd hit a trajectory bug hours into flight. So we sweep $θ$ all the way around and confirm every landing spot.

PICTURE. A point launched at $(R_{\oplus}, 0, 0)$ in ECI, shown in ECEF for $θ = 0, 90°, 180°, 270°$ — it walks clockwise through all four ECEF quadrants because the frame turns counter-clockwise beneath it.

The one-picture summary

The whole derivation on one canvas: (1) two frames share centre and $Z$ ; (2) the turned axes land on the unit circle at $(cos θ, sin θ)$ and $(- sin θ, cos θ)$ ; (3) a point's new coordinate is its dot with each new axis; (4) stack those dots into rows to get $R_{z} (θ)$ ; (5) orthonormal rows mean the inverse is the transpose; (6) every quadrant checks out and $t = 0$ is the identity.

Recall Feynman retelling — say it like you'd explain to a friend

Imagine a merry-go-round (that's ECEF) spinning inside a still room (that's ECI), both pinned at the same centre pole. A kid stands on the merry-go-round. To the room, the kid hasn't moved — but the merry-go-round's own painted "East arrow" has swung around by an angle $θ = ω_{e} t$ . To figure out where the kid is in merry-go-round language, I ask: how far is the kid along the painted East arrow? That "how far along" is a shadow — a dot product. I do it for both painted arrows, and I stack the two answers into a little grid of numbers — that grid is the rotation matrix. Because the painted arrows are still exactly a right angle apart and still one unit long, undoing the whole thing is as easy as flipping the grid across its diagonal. And if I check the kid at a quarter, half, three-quarter turn, they walk around backwards relative to the spin — exactly the minus sign the grid carries. That's the entire ECI→ECEF story: rows are the turned axes, entries are shadows, the minus sign is the frame moving instead of the point.

One-line version ::: To read a fixed point in the spun frame, dot it with each spun axis; the axes are $(cos θ, sin θ)$ and $(- sin θ, cos θ)$ , so the matrix rows are those, and its transpose spins you back.

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