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
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