Warren,
You asked about the dynamic range required of seismometers.

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 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,
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.

Considering the Turkey quake: the record here 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)

	So Ms = log(5.73x10^5/24) + 1.66*log(81deg) -0.18 = 7.37

(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; today they are 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 nanometrers.
(using the 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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