From: "Kareem Lanier" kareemjupiter@.............

Date: Sun, 9 Apr 2000 13:36:10 -0700

Paul, How do you get rid of all the local high-frequency noise? (My delimma) -----Original Message----- From: psn-l-request@.............. [mailto:psn-l-request@............... On Behalf Of S-T Morrissey Sent: Sunday, April 09, 2000 12:56 PM To: psn-l@.............. Subject: seis dynamic range 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) __________________________________________________________ Public Seismic Network Mailing List (PSN-L)

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