Where the Clock Offset Hides in the Pseudoranges

Where the Clock Offset Hides in the Pseudoranges

Ada re-derives this chapter’s own numbers step by step, at full precision

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Ada ADA · CALCULATION AUDIT

Where the Clock Offset Hides in the Pseudoranges

GPS turns time into distance, so a receiver-clock error becomes a position error — the chapter notes that one microsecond of clock slip is 300 m off the ground. A cheap receiver clock can never match a satellite’s atomic clock, which the chapter prices near $100,000, yet a fix still needs at least four satellites, one more than the three coordinates alone would suggest. This audit follows that extra pseudorange to ask where the clock offset hides, and why the fourth satellite — not a better clock — is what solves it away.

Companion to the chapter GPS and Outdoor Positioning — every number here comes from that chapter.

See the relationship before changing it

The figure reads from left to right. The blue card is receiver clock slip. The middle card applies the page rule. The green card is common range error. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

Receiver clock slip changes common range error An input card leads through the rule range error = 299,792,458 m/s x clock slip in seconds to the common range error result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. The same clock slip lengthens every pseudorange, so the fourth satellite can solve it.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 50 ns.

  2. 2

    Name the relationship. range error = 299,792,458 m/s x clock slip in seconds

  3. 3

    Substitute with units. 299,792,458 x 50 x 10^-9 = 14.99 m

  4. 4

    Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change receiver clock slip

Try Predict the direction of range error = 299,792,458 m/s x clock slip in seconds. Test another receiver clock slip, then compare common range error.

50 ns
Chapter baseline
Common range error

Observe The same clock slip lengthens every pseudorange, so the fourth satellite can solve it. Reset receiver clock slip to 50 and compare common range error.

Explain The same clock slip lengthens every pseudorange, so the fourth satellite can solve it.

Check yourself

What should you do before trusting a moved-control result?
Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only receiver clock slip moves here. Field effects named in the technical boundary stay fixed.
TryRecompute One microsecond of receiver-clock offset: 299,792,458 x 1e-6 = 299.792458 m, which rounds to the chapter's 300 m.
ObserveTrack The 50-nanosecond slip in Step 2: 299,792,458 x 50e-9 = 14.98962 m, which rounds to the 15 m the chapter says every pseudorange runs long.
ExplainExplain 299,792 x (0.06740 - 0.06733) = 299,792 x 0.00007 = 20.98544 km, so the two ranges genuinely differ by ~21.0 km.

Ready: use the stated baseline inputs, then compare each displayed result.

Ada: The chapter’s headline is that timing is position – a clock slip becomes a distance. Let me push the chapter’s own numbers through and show why the fourth satellite, not a better clock, closes the gap. The speed of light is c = 299,792,458 m/s.

First the sensitivity the chapter quotes:

  • One microsecond of receiver-clock offset: 299,792,458 x 1e-6 = 299.792458 m, which rounds to the chapter’s 300 m.
  • The 50-nanosecond slip in Step 2: 299,792,458 x 50e-9 = 14.98962 m, which rounds to the 15 m the chapter says every pseudorange runs long.

The word every is the whole trick. A receiver clock 50 ns fast does not spoil one range; it adds the same ~15 m to all of them at once. That common shift is separable, because a pure travel-time difference cannot carry it. Taking Sat A at 67.33 ms and Sat D at 67.40 ms with c = 299,792 km/s:

  • 299,792 x (0.06740 - 0.06733) = 299,792 x 0.00007 = 20.98544 km, so the two ranges genuinely differ by ~21.0 km.

So the four pseudoranges are four true geometric distances plus one shared clock offset – four equations for four unknowns (x, y, z, and the offset), which is exactly why four satellites are the minimum. The design meaning is the chapter’s quiet triumph: instead of fitting every device with a ~$100,000 atomic clock, GPS makes the cheap receiver clock’s error the fourth unknown and solves it away with one extra satellite.

Every number above is taken from the chapter’s own material and re-derived step by step.

Technical boundaries. This audit deliberately does not solve a live navigation fix or simulate ephemeris, ionospheric delay, multipath, or satellite geometry. It exposes only how the fixed receiver-clock offset enters the chapter's pseudorange equations.