# Effect of station angle on location errors

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Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 7 221 - 252 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem

If the error in Shoran time measurements is $\pm 0.1\mu s$ , what is the the size of the parallelogram of error in Figure 7.2a when (a) $\theta =30^{\circ }$ and (b) $\theta =150^{\circ }$ ? Take the velocity of radio waves as $3\times 10^{5}$ km/s.

### Background

Shoran is a radio-navigation device which measures the 2-way traveltime between the point of observation and a fixed station. Using two fixed stations, the point of observation can be located by swinging arcs centered at the two stations; for large distances the arcs become nearly straight lines.

The traveltimes are subject to error $\pm \Delta t$ , so the ranges are $V\left(t_{i}\pm \Delta t\right)$ , $i=1$ , 2. Swinging the four arcs corresponding to these time values, we get a parallelogram of error such as that in Figure 7.2a; the location lies somewhere inside this parallelogram.

===Solution ===


In Figure 7.2a, the error in range $=AM=AN=AP=AQ$ {\begin{aligned}=\left(3\times 10^{8}\ \mathrm {m/s} \right)\left(1\times 10^{-7}\ \mathrm {s} \right)=30\ \mathrm {m} .\end{aligned}}  Figure 7.2a  Parallelogram of error for traveltime uncertainty of $\Delta t=\pm 0.1\mu s$ .

The long diagonal $=2AR=2AQ/\sin 15^{\circ }=2\times 30/\sin 15^{\circ }=230$ m.

The short diagonal $=2AS=2\times 30/\cos 15^{\circ }=60$ m.

To get the figure for $\theta =150^{\circ }$ we merely reverse the arrow on ${\textit {AB}}$ or ${\textit {AC}}$ ; therefore the error values are the same as for $30^{\circ }$ .