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18-Geom-B3 Networks and Precise Engineering Surveys · December 2018

Question 3 of 12: Underground Azimuth Determination with a Gyrotheodolite

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: Closed-book, 3 hours, calculator permitted. TEN questions constitute a complete paper — Part A: all of #1–#8; Part B: one of #9/#10; Part C: one of #11/#12. All twelve questions are solved here for completeness. Most answers are essay-format; Q5, Q6 and Q9 carry short verified numeric illustrations.

Reference texts: Wolf, Ghilani & De Blij, Elementary Surveying: An Introduction to Geomatics (15th ed., Pearson); Mikhail & Gracie, Analysis and Adjustment of Survey Measurements (Van Nostrand, 1981); Kavanagh & Slattery, Surveying with Construction Applications; Hofmann-Wellenhof, Lichtenegger & Wasle, GNSS (Springer, 2008); Kahmen & Faig, Surveying (de Gruyter); Chrzanowski et al. on deformation analysis; USACE Structural Deformation Surveying (EM 1110-2-1009); ISO 17123 field-test procedures. Canadian frame throughout (NAD83(CSRS), CGVD2013).

Question 3: Underground Azimuth Determination with a Gyrotheodolite (8 marks)

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.

Approach. A gyrotheodolite mounts a fast-spinning, gravity-constrained gyroscope on a theodolite; the spin axis seeks the meridian and oscillates about astronomic (true) north, so tracking that oscillation gives the north direction on the instrument’s horizontal circle, from which the azimuth of any marked line follows.

gyrotheodolitespin ωTrue Nu₁ (west reversal)u₂ (east reversal)N̂ = (u₁ + u₂) / 2 → azimuth A = HCR − N̂ + E
Figure 3 — The gyro spin axis oscillates about true north between the west and east reversal points $u_1,u_2$; their mean $\hat N=(u_1+u_2)/2$ is the north reading, and, for a clockwise-graduated circle, the azimuth of the target line is the horizontal-circle reading to the target (HCR) minus $\hat N$, plus the instrument calibration constant $E$.

Field procedure.

  1. Set up and observe the reference line. Centre and level the gyrotheodolite over the underground station, sight the target defining the unknown line, and record the horizontal-circle reading $R$ to that line.
  2. Determine the gyro-north direction by the reversal-point method. Uncage (spin up) the gyro and let the spin axis oscillate. Follow the turning-point (reversal) readings of the oscillation on the horizontal circle. For a symmetric swing the mean of two successive reversal readings gives the gyro-north reading; for the three-turning-point (Schuler) mean: $$N=\tfrac{1}{2}\!\left(u_2+\tfrac{u_1+u_3}{2}\right)$$ where $u_1,u_2,u_3$ are consecutive reversal readings. (Alternatively the transit / time-of-transit method times the swings across a fixed index.)
  3. Apply the instrument calibration (E) constant. The gyro-north reading differs from true north by a fixed instrument constant $E$ (the “$E$-value”), obtained by observing a line of known azimuth $A_0$ (circle reading $R_0$, gyro-north reading $N_0$) on the surface before and after the underground work, so that $E=A_0-(R_0-N_0)$ and any drift of the constant is detected. With a clockwise-graduated circle the azimuth of the target line is: $$A_{\text{gyro}} = (R - N) + E$$ Example: $R=127^\circ14'30''$, $N=359^\circ58'10''$, $E=+1'05''$ give $A=127^\circ14'30''-359^\circ58'10''+360^\circ+1'05''=\boxed{127^\circ17'25''}$.
  4. Reduce to geodetic azimuth. Convert astronomic to geodetic azimuth by applying the Laplace correction (and, if a grid bearing is wanted, the meridian convergence and the arc-to-chord/$t\!-\!T$ correction): $$A_{\text{geodetic}} = A_{\text{astro}} - \eta\tan\phi \quad(\text{Laplace})$$ with $\eta$ the east component of the deflection of the vertical and $\phi$ the latitude.