Satellites in view -Used -PDOP -Fix error -fps -

Drag to orbit, wheel or pinch to zoom. Satellite size is exaggerated about 30x. Rings on the ground are the places that are exactly the measured pseudorange away from each satellite.
true receiversolved fixfix if the clock is ignored

GPS Pseudorange Lab

Why a receiver needs four satellites, not three, and why the sky geometry matters.

Presets

Receiver and time

Place

Latitude

Longitude

Time of day (UTC-like, hours)

Error sources

Receiver clock error (ms)

Random range noise sigma (m)

Zenith ionospheric delay (m)

Elevation mask (deg)

Which satellites are used

Selection

Result

Sky plot (azimuth / elevation)

Dilution of precision over 24 h here

Gauss-Newton convergence

Experiments

Press a button. The scatter of many fixes is compared with the prediction sigma x DOP.

Independent checks

The science

Pseudorange. A satellite broadcasts the time it transmitted. The receiver subtracts that from its own clock, giving ρ_i = ‖s_i − r‖ + c·δt + I_i + ε_i: the true distance plus the receiver clock error times the speed of light (1 ms is 299.8 km, 1 ns is 0.2998 m) plus delays and noise. Because the clock error b = cδt is the same for every satellite, it is a fourth unknown next to x, y and z, so four satellites are the minimum.

Why three is not enough. Each pseudorange defines a sphere. With a clock error the spheres do not meet at one point; the rings drawn on the ground miss each other by about b. Removing the solved b makes them meet at the receiver. The red marker shows the best 3-unknown fit that pretends the clock is perfect: its error is of the order of b amplified by the geometry.

Solving. Linearise around a guess x₀ (here the centre of the Earth): Δρ = GΔx, with one row [−u_i^T, 1] per satellite, u_i the unit vector receiver-to-satellite. Gauss-Newton: Δx = (G^TG)⁻¹G^TΔρ, repeated until the step is below 0.1 mm (typically 5 iterations). The convergence plot shows the error falling from 6371 km at the starting guess.

Dilution of precision. With independent noise of sigma on every range, the covariance of the solution is sigma²(G^TG)⁻¹ (G in east-north-up axes). HDOP = √(Q_EE+Q_NN), VDOP = √Q_UU, PDOP = √(Q_EE+Q_NN+Q_UU), TDOP = √Q_bb, GDOP includes all four. Satellites bunched together make G nearly singular and DOP large; spread across the sky they make it small. Vertical is always worse than horizontal because no satellite is below the horizon.

Orbits. Circular orbits of radius 26,560 km (20,200 km altitude), inclination 55 deg, period T = 2π√(a³/μ) = 11 h 58 min, an idealised 24-satellite Walker 24/6/1 pattern. The Earth turns at 7.2921×10⁻⁵ rad/s underneath.

Independent checks. (1) Noise-free pseudoranges must give back the true position to under a millimetre. (2) Bancroft's algebraic closed form (no iteration, a different algorithm) solved on four satellites must agree with the iterative result. (3) A Monte-Carlo run of thousands of noisy fixes must scatter by sigma times the DOP computed from geometry alone. (4) Propagating a satellite for one period must return it to its start.

Simplifications. Spherical Earth; ideal circular, evenly spaced orbits (the real constellation has about 31 active satellites, slightly eccentric and unevenly spaced); no satellite clock or orbit errors, troposphere or multipath; the ionosphere is one thin shell at 350 km with a constant, illustrative zenith delay; ranges are computed at the same instant (the real signal flight time of about 70-90 ms and the Earth's rotation during it, the Sagnac effect, is ignored here); relativistic clock corrections (satellite clocks gain roughly 38 microseconds per day, which would be about 11 km per day if uncorrected: typical published value) are not modelled. Satellite sizes in the picture are exaggerated about 30 times.

Typical values to spot-check before citing: 26,560 km orbit radius, 55 deg inclination, sigma of a few metres, zenith ionospheric delay of a few metres. Independent educational model, not affiliated with any navigation system operator.