Hi Beth, good points, and if you know the answers, you should post them

But, OK: to others: don't worry about meteor impacts: that was just an analogy to show how unlikely it is that there could be a M9 earthquake on the San Andreas fault.

Magnitude is related to two things: the area of the fault, and the average slip across the fault through that area. The M8.9 quake is on a fault that is almost flat. The San Andreas fault is very steep. In California continental crust the crust is only brittle enough to support an earthquake down to 15 or 20 km depth. So, that dimension of the fault is only 15 to 20 km. If 20 km, a 400 km-long earthquake would have a fault rupture area of 8000 sq km. The slip on a San Andreas fault earthquake is maximum 10 m. 1 to 4 m is more typical.

The Japan quake may have been 500 km-long and 100 km-wide. So, the fault area is sq 50,0000 compared to 8000 sq km max for San Andreas. A more typical near great earthquake on San Andreas might be 300 km-long, 15 km deep, which is only 4500 sq km.

I'll see if I can get HW to post on the Mendocino triple junction...OK, she said no but commented:

The Mendocino triple junction is the intersection of 3 plates (kind of like 4 corners is the intersection of 4 states). Each plate is moving a different direction. North America is moving slowly to the northwest(?) relative to the deeper earth, Pacific Plate is moving faster to the northwest, and the Juan de Fuca is moving northeast. There is therefore convergence between the oceanic Juan de Fuca plate and the continental North American plate. The oceanic plate is denser and drags itself beneath North America.

The Pacific plate is moving mainly parallel to North America and does not subduct beneath it.

It would really take an animation to make this triple junction clear. Beth: want to find an animation and link to it?
Chris