Solution Set for Assignment Exercise : Gutenberg-Richter relationship: log() = a + b. M A) For a time period between January, 90 to December 3, 998 tot in bin [] = 450 6 57 22 7 5 Mid Point M 3.5 3.65 4.5 4.65 5.5 5.65 6.5 6.65 log () = 2.65 2.2.76.34 0.85 0.70 0.00 0.00 log (umber of events) vs Magnitude 000 log() = -0.7923. M + 5.0698 R 2 = 0.982 3 log() = -0.7923. M + 5.0698 R 2 = 0.982 00 2 0 log () 0 0. - Gutenberg-Richter relationship: log() = a + b. M =5.07 0.79. M For the global scale, typical b values are in the range of -2/3 and -. Therefore this value of b (-0.79) fits well the global data. b values describe the proportion of large events to small events. Larger values of b indicate a greater proportion of large events. A2) The data is well fitted with this model, especially for small earthquakes. For higher magnitudes (usually those of greatest interest) the model gives a very good average, but we need to remember that this equation gives the best estimate of what the earthquake will be. It carries uncertainty and therefore, higher and lower magnitudes than the average should be expected. In fact, in this case, even if the average is good, considering only the M<5 events (figure bellow), the model would not have predicted the largest event (M = 6.6). In any case, it is impressive that the rate of occurrence of the earthquakes follow such a simple rule. 2.9<M<3.4 3.4<M<3.9 3.9<M<4.4 4.4<M<4.9 tot in bin [] = 450 6 57 22 Mid Point M 3.5 3.65 4.5 4.65 log () = 2.65 2.2.76.34 Gutenberg-Richter relationship: log() = a + b. M =5.4 0.88. M 4.9<M<5.4 5.4<M<5.9 5.9<M<6.4 6.4<M<6.9 Mid Point M 5.5 5.65 6.5 6.65 predicted 7.83 2.85.04 0.38 observed 7 5
Solution Set for Assignment 000 00 0 Log() = -0.8767. M + 5.4086 R 2 = 0.9996 0000 000 00 0 Log() = -0.8767. M + 5.4086 R 2 = 0.9996 Observed EQ's Gutenberg-Richter model 0. 0. 2 8 89 s a log(89) + b M B) = = 0 89 Using the previously estimated parameters (a = 5.07, b = -0.79, for all 89years) we can estimate the magnitude of the 200-year ( = /200) event: The magnitude of a 200-year event is M = 6.9 Compared to the size and recurrence of the events forecasted to occur on the Bay Area Faults, the prediction seems very accurate. C) First half-period (90-954) tot in bin [] = 65 22 2 9 - - - Mid Point M 3.5 3.65 4.5 4.65 5.5 5.65 6.5 6.65 log () =.8.34.08 0.95 - - - 0.00 00 0 0. Log() = -0.4882. M + 3.20 R 2 = 0.9777 Based only on this 45 years: 45 s a = = 0 45 log(45) + b M And using the previously fitted parameters: a = 3.2 b = -0.49 log() = a + b. M =3.2 0.49. M we can estimate the magnitude of the 200-year ( = /200) event: The magnitude of a 200-year event is M = 7.9 2
Solution Set for Assignment Second half-period (955-998) tot in bin [] = 385 39 45 3 7 5 0 Mid Point M 3.5 3.65 4.5 4.65 5.5 5.65 6.5 6.65 log () = 2.59 2.4.65. 0.85 0.70 0.00-0000 000 log() = -0.88. M + 5.0953 R 2 = 0.9800 Based only on this 44 years: 44 s a = = 0 44 log(44) + b M 00 0 And using the previously fitted parameters: a = 5.0 b = -0.82 log() = a + b. M =5.0 0.82. M 0. 2 8 we can estimate the magnitude of the 200-year ( = /200) event: The magnitude of a 200-year event is M = 7.0 C) In probabilistic analysis, the amount of data often influences the result of the analyses. When increasing the amount of information does not change the results, it is said that the model arrives to convergence. In this case the models have not yet arrived to convergence. It is demonstrated the differences in the parameters that best fit the Gutenberg-Richter equation. The following are the results of the fit in the given time windows: M = 6. 9 90-998: ln ( ) = 0.79 M + 5. 07 M = 7. 9 90-954: ln ( ) = 0.49 M + 3. 2 M = 7. 0 955-998: ln ( ) = 0.82 M + 5. 0 The difference in the estimates is not negligible. The greater prediction given by the first time window (90-954) is due to the fact that a large magnitude event was observed in this time window. This large M was observed in a period of only 45 years, while when considering the complete data (90-989) the same event was seen once in 89 years. 3
Solution Set for Assignment Exercise 2: Mendocino tot in bin [] = 749 44 42 38 0 7 3 2 Mid Point M 3.5 3.65 4.5 4.65 5.5 5.65 6.5 6.65 log () = 2.87 2.64 2.5.58.00 0.85 0.48 0.30 0000 000 log() = -0.794. M + 5.3754 R 2 = 0.9749 Based only on this 44 years: 44 s a = = 0 44 log(44) + b M 00 0 And using the previously fitted parameters: a = 5.38 b = -0.79 log() = a + b. M =5.38 0.79. M 0. 2 8 we can estimate the magnitude of the 200-year ( = /200) event: The magnitude of a 200-year event is M = 7.6 The similar b values in both the Bay Area and Mendocino indicate that the proportion of large events to small events in both areas is similar. Given similar b values, the greater a value in Mendocino indicates that this is a more active region. 0000 000 00 0 0. Bay Area log() = -0.82. M + 5.0 Mendocino log() = -0.79. M + 5.38 2 8 The plot shows the best fit to the seismicity data recorded in Mendocino and Bay Area in the period 955-989. They both have pretty similar b parameter, and therefore almost parallel curves. The main difference is in the a value, where the Mendocino data shows a larger intercept, and therefore a more active zone zone. In fact, the predicted 200 year event based on this time window is larger for the Mendocino area than for the Bay Area: Mendocino: M = 7. 6 Bay Area: M = 7. 0 4
Solution Set for Assignment Exercise 3: tot in bin [] = 2.444E+23 5.4500E+23 9.796E+23 2.2E+24 3.86E+24 2.037E+25.758E+25 8.93E+25 3.5 3.65 4.5 4.65 5.5 5.65 6.5 6.65 log () = 23.39 23.74 23.99 24.34 24.50 25.3 25.25 25.95 I made this calculation using the moment released for each event. You may have found it easier to use the table in exercise and applying the formula: =..5M + 6.05 0 If you used this second (approximation) approach, your numbers will be slightly different from mines. Released Seismic ment vs Magnitude E+26 9E+25 8E+25 7E+25 6E+25 5E+25 4E+25 3E+25 2E+25 E+25 0 3.2 3.7 4.2 4.7 5.2 5.7 6.2 6.7 log() Released Seismic ment vs Magnitude 26.5 26.0 25.5 25.0 24.5 24.0 23.5 23.0 22.5 22.0 3.2 3.7 4.2 4.7 5.2 5.7 6.2 6.7 Fraction of total over 89 years that the M6.6 event released: M 6.6 tot 8.93 0 =.342 0 25 26 = 66.4% Or using the approximation described above: M 6.6 tot.06 0 =.50 0 26 26 = 7% 5
Solution Set for Assignment Exercise 4: A) Magnitude vs time 6 2.0E+22 released in Aftershocks vs time 5.5E+22 Magnitude 4 3 2.0E+22 5.0E+2 0 2 3 4 5 6 Days after mainshock 0.0E+00 2 3 4 5 6 Days after mainshock The left plot shows a high concentration of aftershocks in the close time frame after the mainshock. This can be noticed by the high concentration of events in the first 24 hours after the mainshock. Ignoring the M3.5 event at the end of day 2, a pretty clear trend in the reduction of the maximum magnitude of the aftershocks with time can be noticed. This trend becomes even clearer if we plot the moment magnitude released by the aftershocks in 24hour bins, as shown in the right plot. B) Omori s equation fits reasonably well the data of from the rgan Hill s earthquake. In this case the equation is given by (assuming K = 0): C 06 umber of Aftershocks vs time nt = = P.255 20 ( K + t) t This result is within the usual observed values 00 for the decay rate (.0 < P <.4). umber of Aftershocks 80 60 40 20 0 n t = 05.96. t -.2554 R 2 = 0.9293 2 3 4 5 6 Days after mainshock As shown in the plot, the data is well fitted by the Omori s Law. Observing the graph, a kind of swap between day 3 and day 4 in the frequency of earthquakes becomes apparent. This feature can be explained by observing the data in the Magnitude vs. Time plot. On day 3 a relatively high aftershock was observed, concentrating energy dissipation on it, and at the same time, the events on day 4 could be interpreted as aftershocks of this later event plus aftershocks of the main event. B.) Based on Omori s equation, and knowing that the background rate of M>.0 events before the earthquake was per week, the time it would take before the rate of earthquake occurrence in the area returns to the background rate can be computed as: t = C n t P = 06 7.255 93 days 6.5 months This Law predicts a very long time for the seismicity to go to pre-event rates. 6