The Beacon Geometry

The Beacon Geometry

Ada audits the beacon geometry — one position solve, three residual checks

ADA · CALCULATION AUDIT

Foundations · optional mathematics and physics

The Beacon Geometry

The positioning simulator fixes beacons at (0, 0), (5, 0), and (0, 5) and feeds three RSSI-derived distances of 1.78 m, 4.47 m, and 2.82 m into circle algebra, landing a best-fit point near (0.82, 2.02) with an average residual of just 0.28 m. Yet at a fixed 2 m range the chapter’s own RSSI can swing the estimate from 1.4 m to 5.0 m. This audit reproduces the solve and asks whether that tidy point can be trusted when the RSSI spread dwarfs the 0.28 m residual.

Companion to the chapter Indoor Positioning with BLE Beacons — every number here comes from that chapter.

One position solve, three residual checks, ~4 minutes

The physics problem is not the geometry; it is the radio evidence feeding the geometry. Before trusting the point estimate, reproduce the algebra and then ask how much the RSSI-derived distances can move.

See the relationship before changing it

The chapter figure below shows: Three beacons, three distance estimates, and a best-fit intersection for BLE trilateration

The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only second-beacon distance, so the numeric fixture does not switch without explanation.

Three beacons, three distance estimates, and a best-fit intersection for BLE trilateration

Derive the baseline in four named moves

1. 1

   Name the input. The chapter baseline is 4.47 m.

2. 2

   Name the relationship. x = (1.78^2 - d2^2 + 25) / 10

3. 3

   Substitute with units. (1.78^2 - 4.47^2 + 25) / 10 = 0.819 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 second-beacon distance

Try Predict the direction of x = (1.78^2 - d2^2 + 25) / 10. Test another second-beacon distance, then compare solved x coordinate.

Second-beacon distance

4.47 m

Chapter baseline

Solved x coordinate

Observe The neat coordinate moves sharply with RSSI-derived range, so a small fit residual is not full accuracy proof. Reset second-beacon distance to 4.47 and compare solved x coordinate.

Explain Subtracting the two squared-distance equations leaves the second distance squared with a minus sign, so the calculated horizontal coordinate decreases as that distance increases.

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 second-beacon distance moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only second-beacon distance. It is not a field prediction.

Try

The positioning simulator fixes beacons at (0, 0) , (5, 0) , and (0, 5) and feeds three RSSI-derived distances of 1.78 m , 4.47 m , and 2.82 m into circle algebra, landing a best-fit point near (0.82, 2.02) with an average residual of just 0.28 m . Calculate this case.

Observe

This audit reproduces the solve and asks whether that tidy point can be trusted when the RSSI spread dwarfs the 0.28 m residual. Check shows this.

Explain

The calculation explains why the point looks neat while the claim remains approximate. Clean circle algebra gives a best-fit location near (0.82 m, 2.02 m), but the chapter's own RSSI spread is much larger than the residual for the sample distances, so the release record should keep uncertainty withthe coordinate. Check confirms it.

Technical boundaries

The coordinate calculation fixes the beacon positions and the first range at 1.78 m while changing the second range. It does not account for reflected signals, body shadowing, antenna direction, calibration drift, or the uneven distance errors produced by received signal strength.

Position from the three chapter distances

The simulator starts with beacons at (0, 0), (5, 0), and (0, 5), with distances 1.78 m, 4.47 m, and 2.82 m.

d1^2 = 1.78^2 = 3.1684; d2^2 = 4.47^2 = 19.9809; d3^2 = 2.82^2 = 7.9524

x = (d1^2 - d2^2 + 25) / 10 = (3.1684 - 19.9809 + 25) / 10 = 0.81875 m

y = (d1^2 - d3^2 + 25) / 10 = (3.1684 - 7.9524 + 25) / 10 = 2.0216 m

Residuals and RSSI spread

• Beacon 1 residual: |sqrt(0.81875^2 + 2.0216^2) - 1.78| = 0.40 m.
• Beacon 2 residual: |sqrt((0.81875 - 5)^2 + 2.0216^2) - 4.47| = 0.17 m.
• Beacon 3 residual: |sqrt(0.81875^2 + (2.0216 - 5)^2) - 2.82| = 0.27 m.
• Average residual: (0.40 + 0.17 + 0.27) / 3 = 0.28 m, which is good only for these three supplied distances.
• The chapter's fixed-distance RSSI example spans 5.0 m - 1.4 m = 3.6 m; relative to the real 2 m point, the near estimate is 0.6 m low and the far estimate is 3.0 m high.

The calculation explains why the point looks neat while the claim remains approximate. Clean circle algebra gives a best-fit location near (0.82 m, 2.02 m), but the chapter's own RSSI spread is much larger than the residual for the sample distances, so the release record should keep uncertainty with the position.

Every number above is taken from the chapter's own beacon-geometry example and re-derived step by step.