Re: "fast" earthquakes more destructive?

For our hypothetical 100-km super-shear rupture those factors would all
remain equal. But, how could a super-shear, then be MORE destructive? As you
point out, duration is an important factor in destructiveness. This is
because the force, or acceleration and amplitude, of the shaking tends to
max out at a fairly low magnitude, leaving only duration to vary. The
article seems to suggest that in super-shear quakes the maximum acceleration
exceeds that value found in normal quakes, and does so sufficiently to more
than offset the reduced duration.

So, all other factors being equal (rock modulus, rupture area, total
displacement, site conditions), the super-shear rupture will last about half
as long, and be more destructive in spite of its short duration due to the
higher "intensity" of the resulting motions.

Mike Williams
Arroyo Grande, CA USA

This is exactly right - it is the magnitude (technically, the moment
magnitude) that captures the size of the overall motion in an
earthquake. So halving the duration of a given magnitude event does
not effect the amount of energy released much. Actually, a hidden
property, the absolute stress against which the fault is moving,
matters, but I'll ignore that.

Changing the rupture velocity doesn't change the amount of energy
much, but it does change how it is radiated. My impression is that
rupture velocity near the shear wave speed is actually the worst case
- a very strong S or Love wave pulse is radiated directly in the
direction that the rupture is propagating. Super-shear rupture
velocities send more energy out to the sides, but do not generate the
forward-directed "directivity pulse". Rupture velocity much slower
than the shear wave velocity generates less shaking, and perhaps more
(harmless to people) heating on the fault surface.

So which is more damaging partly depends where the cities lie relative
to the fault, rupture direction, and style of faulting (my comments
apply only to strike-slip faults).

.