18-Geom-A1 Surveying · May 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — May 2014 — 04-Geom-A1 Surveying. Closed-book; any non-communicating calculator, ruler and protractor permitted. Format: seven questions are given and any five (20 marks each) constitute a complete paper — all seven are solved below for completeness. Where a datum is implied, elevations are referenced to the Canadian vertical frame (CGVD2013) and azimuths to NAD83(CSRS); US-foot stationing is retained wherever the printed question uses it.
Reference texts: Wolf & Ghilani, Elementary Surveying: An Introduction to Geomatics (15th ed., Pearson); Ghilani, Adjustment Computations: Spatial Data Analysis (6th ed., Wiley); Hofmann-Wellenhof et al., GNSS — Global Navigation Satellite Systems (Springer, 2008).
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
Given. Carrier-phase observations collected simultaneously by two receivers ($A$, $B$) tracking common satellites in a relative (differential) GPS survey.
Find. A sketch and a principle statement for each of single-, double- and triple-differencing.
Carrier-phase positioning is far more precise than code positioning but is corrupted by three dominant nuisance terms: the satellite-clock error, the receiver-clock error, and the integer ambiguity $N$ (the unknown whole number of cycles at lock-on). Differencing forms linear combinations of the raw phase observations $\Phi$ that algebraically remove these terms one layer at a time. The essential requirement is simultaneity: the same satellites are observed by both receivers at the same epochs, so the shared errors are common and cancel.
Subtract the simultaneous phase of the same satellite $k$ observed at the two receivers: $$\Phi^{k}_{AB} = \Phi^{k}_{A} - \Phi^{k}_{B}$$ Because both receivers see the identical satellite clock at that instant, the satellite-clock error cancels. Most of the correlated atmospheric (ionospheric/tropospheric) delay also cancels over short baselines. The receiver-clock difference and the ambiguity difference remain.
Difference two single differences formed to two satellites $k$ and $l$: $$\nabla\!\Delta\Phi^{kl}_{AB} = \left(\Phi^{k}_{A}-\Phi^{k}_{B}\right) - \left(\Phi^{l}_{A}-\Phi^{l}_{B}\right)$$ The second subtraction removes the receiver-clock errors as well (they are common to both single differences). What is left is the geometry plus an integer double-differenced ambiguity $\nabla\!\Delta N^{kl}_{AB}$, which is a true integer and can be "fixed." The double difference is the fundamental observable of precise relative GPS.
Difference two double differences formed at two epochs $t_1$ and $t_2$: $$\delta\nabla\!\Delta\Phi^{kl}_{AB} = \nabla\!\Delta\Phi^{kl}_{AB}(t_2) - \nabla\!\Delta\Phi^{kl}_{AB}(t_1)$$ Provided no loss of lock occurred, the ambiguity $N$ is constant between epochs, so it cancels entirely. The triple difference is therefore ambiguity-free and ideal for an initial approximate solution and for detecting cycle slips (a slip appears as a spike), at the cost of higher noise and correlated errors — hence it seeds, but does not replace, the double-difference fixed solution.
| Method | Combination | Cancels |
|---|---|---|
| Single difference | same satellite, two receivers | satellite-clock error |
| Double difference | two satellites, two receivers | + receiver-clock error (integer $N$ preserved) |
| Triple difference | double difference across two epochs | + integer ambiguity $N$ (flags cycle slips) |