The 0.183 m Antenna Height
A six-station static GNSS network is observed for cadastral control in a state capital: dual-frequency receivers, 45-minute sessions, 15° elevation mask, baselines between 8 km and 26 km. Processing and adjustment give horizontal residuals under 18 mm and a loop closure of 1:1,400,000 - comfortably inside specification. However, the height of station DN-04 obtained from GNSS differs from its published levelled value by 0.183 m. The field sheet for the DN-04 session records an antenna height of 1.352 m, while the observation log keyed into the processing software records 1.532 m. A colleague proposes rejecting the entire session and re-observing, costing four days. The client needs coordinates on the national datum and 1:2,000 relative accuracy for a 5 km cadastral scheme.
1. Diagnose the discrepancy and decide whether to re-observe. Show the reasoning.
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Diagnosis The discrepancy is 0.183 m; the antenna-height transcription difference is 1.532 − 1.352 = 0.180 m, with 0.003 m left over - consistent with the expected noise and with the vertical residual pattern of the session. Direction check: if the software was told the antenna was 0.180 m higher than it was, it removes 0.180 m too much, so the computed station height comes out 0.180 m too low. Compare that against the sign of the observed difference before accepting the hypothesis.
Before blaming the antenna height, rule out the two other explanations:
- Geoid model: GNSS gives ellipsoidal heights; levelling gives orthometric heights. If the geoid separation used for DN-04 is wrong or missing, the height will misfit. Check that all six stations were reduced with the same geoid model and that the misfit is confined to DN-04.
- Antenna type: a wrong antenna model in the log applies the wrong phase-centre offset. Check the antenna serial against the log.
If the misfit is confined to DN-04 and the arithmetic is consistent, the cause is the transcription blunder.
Decision Reprocess that one session with the correct 1.352 m and compare. Rejection and re-observation is unnecessary: a 0.180 m height error in relative static positioning is essentially vertical. The horizontal effect of an antenna-height error is zero for a vertical antenna, and even with a 5° antenna tilt it is only 0.180 × sin 5° ≈ 0.016 m - below the 18 mm horizontal residuals already achieved. The horizontal network is unaffected and can be retained.
2. Does the network meet a 1:2,000 relative accuracy requirement for the 5 km scheme? Explain what the requirement actually controls.
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What 1:2,000 means A relative accuracy of 1:2,000 is a ratio of error to distance: 1 m of error per 2,000 m, i.e. 0.5 m per kilometre, or 2.5 m across the whole 5 km scheme.
Measured performance
- Loop closure 1:1,400,000 corresponds to about 0.7 mm per kilometre.
- Residuals under 18 mm on baselines of 8–26 km correspond to roughly 0.7–2 ppm, i.e. between 1:1,400,000 and 1:500,000.
Both are one to three orders of magnitude better than the requirement. The internal geometry is not the problem.
What actually controls the deliverable For a cadastral scheme the controlling factor is not the internal network accuracy but the connection to the national datum. A network can be internally perfect and still be uniformly shifted by metres because of a bad transformation or a bad connection point. So the acceptance test is:
- Internal consistency (passed).
- Agreement at the tie stations to the published national control (must be checked, with residuals reported).
- Agreement at an independent check point not used in the adjustment.
Report all three. Certifying on the strength of the loop closure alone would be certifying the wrong thing.
3. The client's scheme lies 40 km beyond the outermost control station used in the datum transformation. What is your course of action?
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Extrapolating a datum transformation beyond the control that defines it is unsafe, and the risk is not visible in the network residuals.
Reasoning
- A seven-parameter (Helmert) transformation absorbs the systematic difference between two datums. Its parameters are only reliable inside the region spanned by the common points used to derive them.
- Outside that region, the neglected distortion of the local datum (scale and orientation variation across the national network, which is real for older classical networks) grows without any warning in the residuals, because there are no common points there to reveal it.
- A 3-parameter (translation-only) transformation would be even more exposed, since it forces the entire datum difference into a shift.
Action
- Do not simply extend the existing transformation. Obtain or observe at least three to four common stations that surround or lie near the scheme, and derive a local transformation set.
- Hold back one common point as a check and report the transformation residual at that point. If the check point misfits by more than the tolerance your client's accuracy needs allow, the transformation is not fit for the scheme.
- State the transformation used, its parameters, its source, the common points, and the check residual on the plan and in the report.
- If no suitable common control exists within a reasonable distance, tell the client that the connection to the national datum cannot be certified to the required accuracy, and price the additional control survey - do not hide the extrapolation behind a network that looks precise.