= Solution
Use the northern <S–P time>, approximately $\Delta t_N=6\ \mathrm s$, rather than multiplying an absolute-clock arrival by a <wave speed>. For common approximately straight paths,
$$
\boxed{D_N=\frac{\Delta t_N}{1/V_s-1/V_p}=\frac{V_pV_s}{V_p-V_s}\,\Delta t_N.}
$$
\b[No numerical crustal velocities are specified in Q3.] If the usual illustrative crustal values $V_p=6.0\ \mathrm{km\,s^{-1}}$ and $V_s=3.5\ \mathrm{km\,s^{-1}}$ are adopted, the distance is
$$
\boxed{D_N\simeq8.4\times6\simeq50\ \mathrm{km}.}
$$
These assumed crustal speeds are unrelated to Q1's offshore layer values. They are representative rather than measurements of the Swiss column; the <British Geological Survey> gives broad typical crustal ranges of 5–7 km/s for <P waves> and 3–4 km/s for <S waves>.<How are earthquakes detected by British Geological Survey>[The seismic-wave table provides typical crustal speed ranges, not a velocity model for this particular earthquake.]{ref}
Using the corresponding northern central pulse times gives an effective source time near $13.3-50.4/6\simeq4.9\ \mathrm s$ after midnight, and the S estimate is the same to graph accuracy. The southern central landmarks similarly give a distance near 70 km and an effective source time near 5 s. That consistency supports the example, but pulse landmarks need not mark the rupture's true onset. $D_N$ is a source-to-station path or <hypocentral distance> in the uniform-speed approximation. An <epicentral distance> additionally needs the <hypocenter> depth; it cannot simply be equated to 50 km for an unknown depth.
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