Visual walkthrough — GNSS — GPS, GLONASS, Galileo, BeiDou

This is the visual companion to the parent topic. The algebra lives there; here the pictures carry the derivation.

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Step 1 — A signal that carries a timestamp

WHAT. A satellite is a small radio station in space. It does exactly two things we care about: it knows where it is, and it broadcasts a message that says "my clock reads $t_{tx}$ right now." The letter $t_{tx}$ just means time of transmit — the instant the signal left.

WHY. To turn a radio message into a distance, we need a stopwatch. Light (and radio waves are light) always travels at the same speed, written $c\approx 3\times 10^8$ metres per second — that is roughly the distance around Earth's equator, seven and a half times, in one second. Because $c$ never changes, time-of-flight is a ruler: if we know how long the signal was in the air, we know how far it flew.

PICTURE. The satellite at the top stamps its clock time onto a wavefront and lets it fly outward. The green ring is the signal expanding at speed $c$.

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Step 2 — The receiver's stopwatch: pseudorange

WHAT. You (the receiver, at unknown position) catch the signal. Your own clock reads $t_{rx}$ at that moment — time of receive. Multiply the elapsed time by $c$ and call the result ${\rho }_i$ (the Greek letter rho, our name for this measured distance to satellite $i$):

$\underset{\text{measured range}}{\underbrace{{\rho }_i}}=\underset{\text{speed of light}}{\underbrace{c}}(\underset{\text{my clock at arrival}}{\underbrace{t_{rx}}}-\underset{\text{sat clock at launch}}{\underbrace{t_{tx}}})$

WHY. Distance = speed × time. Speed is known ($c$), time is the gap between the two stamps, so distance falls out. Simple — except for one lie hiding in $t_{rx}$: your clock is a cheap quartz crystal, not an atomic clock. It is wrong by some unknown amount. So ${\rho }_i$ is not the true distance; we honestly call it the pseudorange — "pseudo" = false-ish.

PICTURE. Two clocks, one in orbit (atomic, correct) and one in your pocket (drifting). The measured travel time is the horizontal gap between the two stamps on the timeline.

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Step 3 — One measurement = one sphere

WHAT. Suppose for a second your clock were perfect. Then ${\rho }_i$ is the true distance $r_i$ to satellite $i$. All points that are exactly distance $r_i$ from a fixed satellite form a sphere of radius $r_i$ centred on that satellite. You are somewhere on that sphere.

The radius comes from the plain 3D distance formula (Pythagoras stretched into three dimensions):

$r_i=\sqrt{\underset{\text{east–west gap}}{\underbrace{(x-x_i{)}^2}}+\underset{\text{north–south gap}}{\underbrace{(y-y_i{)}^2}}+\underset{\text{up–down gap}}{\underbrace{(z-z_i{)}^2}}}$

Here $(x,y,z)$ is you (unknown) and $(x_i,y_i,z_i)$ is the satellite (known, from its broadcast).

WHY. One number ($r_i$) cannot give you three numbers ($x,y,z$). It only shrinks the possibilities from "anywhere in the universe" to "anywhere on this one sphere." That is genuine progress but not an answer. We need more spheres.

PICTURE. One satellite, one shaded sphere. You could be anywhere on the shell.

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Step 4 — Three spheres pin a point (with perfect clocks)

WHAT. Add a second satellite. Two spheres overlap in a circle. Add a third: the third sphere slices that circle at (generally) two points, and one of them is absurd — it is out in space or spinning the wrong way — so we keep the sensible one. Three spheres → one point → your $(x,y,z)$.

WHY. Count the mystery: three unknowns $(x,y,z)$, so three independent facts (three spheres) are needed. This is the perfect-clock answer — and it is the source of the classic mistake "3 satellites are enough." It would be, if your clock were atomic. It isn't.

PICTURE. Sphere 1 ∩ Sphere 2 draws a circle (teal). Sphere 3 stabs that circle at the two orange dots; the real receiver is the filled one.

Recall

With perfect clocks, how many satellites for 3D position? ::: Three — one per unknown $(x,y,z)$.

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Step 5 — The hidden fourth unknown: clock bias $b$

WHAT. Your clock is off. Call the offset $b$ (seconds): "my clock is ahead of true time by $b$." Because you overcounted the travel time by $b$, every pseudorange is too long by $cb$ metres. So the honest relationship is:

$\boxed{{\rho }_i=\underset{\text{true distance}}{\underbrace{r_i}}+\underset{\text{same error on every sphere}}{\underbrace{cb}}}$

Notice $cb$ is identical for all satellites — one clock, one bias. That is the key that lets us solve for it.

WHY. How big is this error? A tiny $1\mu \text{s}$ (one millionth of a second) clock error becomes $cb=(3\times 10^8)(10^{-6})=300$ metres — a whole football stadium. We cannot ignore $b$; we must solve for it as a fourth unknown. Four unknowns $(x,y,z,b)$ need four facts → a fourth satellite.

PICTURE. All three spheres are inflated by the same amount $cb$; they no longer meet at one point but at a fuzzy triangle. The fourth sphere tells us how much to deflate every sphere until they all agree — that shared shrink reveals $b$.

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Step 6 — The full observation equation (four satellites)

WHAT. Put Steps 2–5 together, once per satellite $i=1,2,3,4$:

${\rho }_i=\sqrt{(x-x_i{)}^2+(y-y_i{)}^2+(z-z_i{)}^2}+cb$

Four such equations, four unknowns $(x,y,z,b)$. Solve → you get position and the true time for free.

WHY. Each satellite contributes one equation and there are four unknowns, so four satellites is the honest minimum. This is the parent note's central result — now every piece of it has been built from a stopwatch and Pythagoras.

PICTURE. Four satellites in the sky, four line-of-sight arrows converging on the receiver, one of them labelled with the full equation.

Recall

Why four satellites, not three? ::: The receiver's clock adds a fourth unknown $b$; four unknowns need four equations.

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Step 7 — Solving it: aim, measure the miss, correct, repeat

WHAT. The square roots make the equations nonlinear — you can't just do high-school elimination. So we do what a dart player does: guess, see how far off, nudge, repeat.

1. Guess a starting position ${\mathbf{x}}_0$ (even the centre of the Earth works).
2. For each satellite, the direction from your guess to that satellite is a unit vector ${\hat{\mathbf{u}}}_i$ — an arrow of length 1 pointing at the satellite. Its ingredients come from the distance formula's slope: $\frac{\partial r_i}{\partial x}=\frac{x-x_i}{r_i}=-{\hat{u}}_{ix}$ This says: "if I step one metre east, how much does my distance to satellite $i$ change?" — and the answer is the east-component of the pointing arrow. That is why a derivative appears: it converts a curved sphere into a flat local slope we can solve.
3. Stack these arrows (plus a column of 1's for the shared clock term) into a table $\mathbf{H}$, the geometry matrix, and solve the now-linear correction $\Delta \mathbf{x}=({\mathbf{H}}^{\mathsf{T}}\mathbf{H}{)}^{-1}{\mathbf{H}}^{\mathsf{T}}\Delta \mathbf{\rho }$.
4. Apply the correction, re-guess, repeat two or three times until it stops moving.

WHY. Linearizing turns an unsolvable curved problem into a solvable straight-line one near the guess; iterating walks that guess onto the true answer. With 5+ satellites the system is overdetermined and we use Least Squares Estimation to find the point that best fits all of them — and a Kalman Filter does this continuously through time, fused with an Inertial Navigation System (INS).

PICTURE. A guess dot, an arrow of correction $\Delta \mathbf{x}$ pointing toward the true dot, and the residual "misses" $\Delta \rho$ shrinking after each iteration.

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Step 8 — The degenerate case: bad geometry (DOP)

WHAT. Four satellites are enough only if they are spread across the sky. If they all huddle in one patch, their pointing arrows are nearly parallel, and the geometry matrix ${\mathbf{H}}^{\mathsf{T}}\mathbf{H}$ becomes nearly impossible to invert. A small range error then explodes into a huge position error. That amplification is the Dilution of Precision.

${\sigma }_{\text{position}}=\text{DOP}\times {\sigma }_{\text{measurement}}$

Here $\sigma$ (sigma) means "typical size of the error." DOP is a pure multiplier: DOP = 1 is ideal, DOP = 10 means your metre of ranging error becomes ten metres of position error.

WHY. Same reason a table with legs bunched in the centre wobbles while a table with legs at the corners is rock-steady. Widely separated lines-of-sight brace your position estimate from many independent directions. This is the real payoff of using GPS + GLONASS + Galileo + BeiDou together — more satellites, better spread, lower DOP, especially in urban canyons.

PICTURE. Left: four spread-out satellites, arrows fanning out, tight small error box (good). Right: four clustered satellites, near-parallel arrows, huge stretched error box (bad).

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The one-picture summary

Everything above compressed into a single frame: a satellite stamps a time (Step 1–2), that becomes a sphere (Step 3), four inflated spheres plus one shared clock error meet at your point (Steps 4–6), and the spread of the satellites sets how sharp that point is (Step 8).

Recall Feynman retelling — the whole walk in plain words (click to open)

You're lost in a huge dark field. Four friends stand on tall, far-apart towers. Each friend knows exactly where their tower is, and each shouts "NOW!" at the same true instant, using perfect wristwatches.

Your own watch is a little fast or slow — you don't know by how much. When you hear a "NOW!", you note the time on your watch. The friend farther away sounds like they shouted later, because their voice took longer to reach you. Turn each delay into a distance (delay × speed of sound), and each friend gives you a ring of "you're this far from me."

Three rings would cross at one spot if your watch were perfect. It isn't — so every ring is drawn a little too big by the same amount (your watch error). A fourth friend gives you the fourth clue you need to figure out both where you are and exactly how wrong your watch is — all at once. And if all four friends are bunched together on one side of you, your fix is mushy; spread them around and it snaps into a sharp point.

Swap "sound" for "radio at the speed of light" and "friends on towers" for "satellites in orbit," and that is GNSS.

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Back to the parent topic · related: Trilateration and Multilateration, Special Relativity — Time Dilation, General Relativity — Gravitational Time Dilation, Orbital Mechanics — MEO/GEO/IGSO, Signal Modulation — CDMA & FDMA.