= Solution
For a unilateral <finite-source rupture> advancing at approximately constant <rupture velocity> $V_r$, the along-fault propagation length is
$$
\boxed{L=V_rT.}
$$
There is no supplied $V_r$ or calibrated <seismic source time function>, so \b[a unique spatial extent cannot be read directly from Figure 4]. For illustration, $T\sim4\ \mathrm s$ and an assumed $V_r\sim2.5\ \mathrm{km\,s^{-1}}$ would give $L\sim10\ \mathrm{km}$. Allowing $T\sim4$–$5\ \mathrm s$ and assumed $V_r\sim2$–$3\ \mathrm{km\,s^{-1}}$ gives lengths of order \b[8–15 km]. These are explicitly conditional scale estimates, not measured fault dimensions.
If equal-and-opposite ray geometry and phase pulse durations were known, the directivity model in <3/e/iii/solution> would instead constrain a projected length,
$$
L_\parallel=\frac c2\left(D_S^{(\mathrm{phase})}-D_N^{(\mathrm{phase})}\right),\qquad L_\parallel=L\cos\psi,
$$
where $\psi$ is the angle between the rupture direction and the northern ray. Distinct P and S <wave speeds>, unknown takeoff angles, smooth pulse boundaries and the clipped southern S tail prevent a reliable independent numerical inversion here. This observation is consistent with a few-kilometre projected extent while the actual along-fault length can be larger. A rupture length is not the accumulated <fault slip>, the width of the whole <fault plane>, or the approximately 50 km station distance.
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