Question 7 of 8: Gravity Survey Over a Suspected Tunnel — Horizontal Cylinder Model

Question 7: Gravity Survey Over a Suspected Tunnel — Horizontal Cylinder Model (10 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.

Given. Bouguer anomaly profile (mGal) over position (m), read off the graph at its 13 plotted points; bedrock density $2500\ \text{kg/m}^3$; density contrast for an air-filled tunnel $\rho=2500-0=2500\ \text{kg/m}^3$; depth rule for a horizontal cylinder $z=x_{1/2}$ (the half-width of the residual anomaly).

Find. (a) the regional field and residual anomaly; (b) the tunnel's radius $a$ and depth to top ($z-a$), assuming air-filled; (c) whether the gravity data alone can distinguish an air-filled from a water-filled tunnel.

Approach. Fit a smooth regional trend through the flanks of the profile (the points clearly outside the local low, i.e. away from the dip near $x=30$), subtract it from the observed profile to get the residual anomaly, read the residual's peak value and half-width off that curve, apply the given depth rule $z=x_{1/2}$, then solve the horizontal-cylinder formula (evaluated at the peak, $r=z$) for the radius.

1. Fit the regional field. The profile rises fairly steadily from 0.015 mGal at $x=0$ to 0.198 mGal at $x=60$, except for a local dip between roughly $x=15$ and $x=40$. Fitting a smooth curve through the flanking points ($x=0,5,10$ and $x=45,50,55,60$, all clearly outside the local low) gives the regional estimate plotted as the dashed red curve in the figure below; the two flanks alone do not sit on one straight line (the rise decelerates on the left, then accelerates through the middle before flattening on the right), so the honest hand estimate is a smooth curve rather than a single straight line.
2. Subtract the regional to get the residual. $\text{residual}(x) = g_{obs}(x) - \text{regional}(x)$ at each of the 13 points, plotted as the lower (blue) curve. The residual is essentially zero at both ends of the line (confirming the regional fit is reasonable) and strongly NEGATIVE in the middle — a gravity LOW, as expected for a lower-density (air-filled) void in denser bedrock.
3. Read the peak and half-width off the residual curve. The residual reaches a minimum of $\boxed{\Delta g_{max}\approx-0.122\ \text{mGal}}$ at $x_0\approx30$ m (matching the observed dip). The half-maximum level is $\Delta g_{max}/2\approx-0.061$ mGal, which the residual curve crosses at $x\approx26.0$ m and $x\approx33.9$ m, giving a half-width $x_{1/2}=(33.9-26.0)/2\approx\boxed{3.95\ \text{m}}$.
4. Depth to axis, via the given depth rule. $z=x_{1/2}\approx\boxed{3.95\ \text{m}\approx4.0\ \text{m}}$.
5. Radius, from the horizontal-cylinder formula evaluated at the peak. At the anomaly's peak, $r=z$ (the observation point sits directly above the cylinder axis), so $\Delta g_{max}=2\pi a^2\rho G\,z/z^2=2\pi a^2\rho G/z$. Solving for $a$ (converting $\Delta g_{max}$ from mGal to SI, $\Delta g_{max}=0.122\times10^{-5}=1.22\times10^{-6}\ \text{m/s}^2$): $$a=\sqrt{\frac{\Delta g_{max}\,z}{2\pi\rho G}}=\sqrt{\frac{(1.22\times10^{-6})(3.95)}{2\pi(2500)(6.672\times10^{-11})}}\approx\boxed{2.15\ \text{m}}$$
6. Depth to top of the tunnel. $\text{depth to top}=z-a=3.95-2.15\approx\boxed{1.8\ \text{m}}$.

(c) Air- vs. water-filled — can gravity alone tell? No: the horizontal-cylinder formula shows $\Delta g_{max}\propto\rho\,a^2/z$, so the SAME observed peak anomaly and the SAME half-width (which fixes $z$, independently of $\rho$) can be produced by many different ($\rho,a$) pairs, as long as $\rho a^2$ stays constant. If the tunnel were water-filled instead of air-filled, the density contrast drops from $2500-0=2500\ \text{kg/m}^3$ to $2500-1000=1500\ \text{kg/m}^3$; matching the SAME observed $\Delta g_{max}=-0.122$ mGal at the SAME $z=3.95$ m then requires a LARGER radius, $$a_{water}=\sqrt{\frac{\Delta g_{max}\,z}{2\pi\rho_{water}G}}=a_{air}\sqrt{\frac{2500}{1500}}\approx\boxed{2.77\ \text{m}}$$ i.e. a bigger, water-filled tunnel produces an anomaly indistinguishable from a smaller, air-filled one — the same non-uniqueness ("equivalence") already met for the resistivity of a thin conductive layer in Question 5(b). Gravity data ALONE cannot resolve which is correct; an independent constraint on the tunnel's actual radius (as-built drawings, a borehole intersection, or a second geophysical method such as ground-penetrating radar) would be needed to break the trade-off.