Question

An average movement of about 5 m (16 ft) along the San Andreas Fault was associated with the devastating 1906 San Francisco earthquake that killed many people and destroyed properties. Assuming that all displacement along the SAF was produced by Earth motion of this magnitude (5 m offset per earthquake), how many earthquakes must have occurred in order to account for the total displacement?What would the recurrence interval be, or in other words, every how many years would such an earthquake, on average, occur?

Answer 1

Answer: The overall lateral lenght of the San Andreas Fault is estimated by measuring the distance on similar terranes on both sides of the fault (continental x pacific). The lenght is increased by eathquake occurrence (here given as 5m per event) and a regular annual displacement rate (estimated in 0,10m per year).

Also, it is acceptable to consider a time lapse of 45 years between each earthquake (considering the major events of 1857, 1906, 1957 and 1989).

Those values of lenght on the fault are considered to vary from 150 to 350 miles. If we apply simple median arithmethical value of 250 miles between these numbers, against the obtained displacement rate of 5,1 meters (let´s set the annual displacement rate aside to balance) we have: 250 miles divided for 0,003 miles (5,1 meters converted to miles) is equal to: 83.333.

So, more than 80.000 thousand eathquakes (considering the movement was mostly made by these earth intercurrences) would be an acceptable number to account for the total actual displacement on the SAF.