From: Paul Jebb pfjebb@..............

Date: Tue, 11 Apr 2000 10:42:17 -0400

Thanks for that note Sean, Wow! A wavelength the size of the earth. Paul S-T Morrissey wrote: > Paul, > > You correctly estimate of the wavelength of surface waves for a > 20 second wave (0.05hz) with a velocity of 5 km/second as 100 km. > (using v = f * l; or wavelength = velocity * period) > Most people don't realize this. The longest waves seen are the > fundamental free oscillations of the earth at 51 minutes, where the > wavelength is the earths' circumference. Even near field P waves > of 15 hz at 7.5 km/sec have wavelenghts of half a kilometer. > > You also asked about the dynamic range required of seismometers. > Here is a repeat of a previous discussion on the subject. (August 99) > > The range of ground motions expected from earthquakes as recorded > by a seismometer is hugh. Obviously if "you are there" at the fault > scarp, you will need an accelerometer with a range up to 2gs. > > But most instrumental seismology follows general magnitude formulas > of the classic form: > M = log(A/T) + 1.66*log(distance) + constant. > The constant is a regional variable, usually about -0.18. The distance > is degrees (111 km per), A is the ground amplitude in nanometers, > (a nanometer = 10^-9 meter; in older data it was called a millimicron.) > and T is the period of the wave in seconds. For Ms calculations, > A is the sustained peak-peak surface wave amplitude. > > You can plug in numbers for various scenarios. To answer your question > about what will happen in the seismometer, solve for A: > log(A/T) = Ms - 1.66*log(distance) + 0.18 > A = T*log^-1(Ms -1.66*log(d) + 0.18.) > > The first step though is to convert the velocity output of the seismometer > to displacement by dividing by omega, w = 2*pi/period. > > For the Turkey quake: the record here (St. Louis) was 150 microns/second > at 24 seconds. 24 seconds is an w (omega) of 2*pi/24 = 0.262radian/sec. > The velocity is converted to displacement by dividing by w, so > 150/0.262 = 573 microns or 5.73x10^5 nanometers. (0.573 millimeters) > (Every seismometer should have recorded this, and many clipped). > > So Ms = log(5.73x10^5/24) + 1.66*log(81deg) -0.18 = 7.37 > > (solving for A for a maximum magnitude of 8.0 at that distance would > cause 2.46 millimeters of motion at St. Louis) > > But what did I record for the Ms = 5.8 aftershock? > > A = 24[log^-1(5.8 - 3.17 + 0.18) = 15 564 nanometers. (15.5 microns) > > This is a velocity (at 24 seconds) of 4.07 microns/second. > With an output of 5.3mv/micron/sec, the record here was about 22 > millivolts p-p, (or about 10 times the 6-second microseisms; at 0.5 > microns p-p; at the time they were running more than 10 times that due to > the hurricane). > > But what about a nearby quake? Say a Ms 4.0 in the next state? Putting > numbers in the formula with distance = 3 degrees (200 miles) and > T = 0.3 second, the seismometer will record 0.81 microns. For a > magnitude 3, it will only sense 0.081 microns or 81 nanometers. > (using the more accurate local Mblg(3hz) formula gives 43 nanometers) > > If M = 4 and is 1000km away, the motion at 1 second will be 398 nanometers. > If M = 3 and is 1000km away, the motion at 1 second will be 40 nanometers. > > For a threshold event, say a 2.5 at 110 km with T = 0.22 sec (4.5hz), > (this magnitude formula is revised for such local events, but we'll > use it anyhow) the amplitude is 105 nanometers. (using the local > Mblg formula gives 56 nanometers). To record these with useful > signal-to-noise ratio requires a resolution of 1 nanometer from > the seismometer. > > SO.... we need a displacement dynamic range of 1 nanometer to 1 > millimeter, or 10^6. This is one reason seismometers prefer a velocity > output. Looking at such numbers:. > At 10 hz, 1 nanometer is 63nm/sec. At 1 hz, it is 6.3nm/sec. At 20 > seconds (a broadband instrument) 1 millimeter is 314 microns/second. > This is a velocity range of about 50 000 to 1, which is the same as > the voltage output range. > > Until recently this range was difficult to handle with analog > electronics, so multi-level recording was used. A 16-bit digitizer > can realize this if noise is ignored. Providing for noise and > instrumental drift increases the required range by a factor of > 100 to 1000. So modern broadband stations use 24-bit digitizers > with a resolution of 1 part on 16 777 216 with a least count value > of 1.2 microvolts. For a seismometer with a VBB velocity output of > 2000 volts/meter/second, the LSB represents 1.67 nanometers/sec. > The maximum value is 10 millimeters/second p-p. > > Regards, > Sean-Thomas > __________________________________________________________ > > Public Seismic Network Mailing List (PSN-L) > > To leave this list email PSN-L-REQUEST@.............. with > the body of the message (first line only): unsubscribe > See http://www.seismicnet.com/maillist.html for more information. __________________________________________________________ Public Seismic Network Mailing List (PSN-L)

Larry Cochrane <cochrane@..............>