 Solution

For a unilateral finite-source rupture advancing at approximately constant rupture velocity $V_r$, the along-fault propagation length isThere is no supplied $V_r$ or calibrated seismic source time function, so a unique spatial extent cannot be read directly from Figure 4. For illustration, $T\sim 4\mathrm{s}$ and an assumed $V_r\sim 2.5\mathrm{km}{\mathrm{s}}^{-1}$ would give $L\sim 10\mathrm{km}$. Allowing $T\sim 4$–$5\mathrm{s}$ and assumed $V_r\sim 2$–$3\mathrm{km}{\mathrm{s}}^{-1}$ gives lengths of order 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 solution would instead constrain a projected length,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.