From: sean@...........

Date: Wed, 7 Feb 2001 20:10:01 -0600 (CST)

Ian, We can attempt to calculate the seismic displacement sensitivity of your 5 microradian (urad) resolution tiltmeter. But first we will make some assumptions. We will assume that it is 5 urad p-p. And if this is the threshold of resolution, we will need to assume a minimum signal-to-noise (S/N) ratio of from 20 to 100, requiring data of 100 urad p-p minimum. But note that consistent data requires an S/N of at least 100. Also, the displacement sensitivity of pendulum tiltmeters is inversely proportional to the square of the natural period. We will assume that the sensor is critically damped somehow. Since you don't indicate a large instrument, I will assume that you have something based on or similar to the Fredericks electrolytic bubble, which is viscously damped at one second period. (If you have the Rockwell/Kinemetrics sensor, you would mention two axes with 100x the resolution). Geomechanics uses the Frederics bubbles in their sensors, the best of which they claim to have 0.1 urad resolution. Or you may have a hinged pendulum, ranging from a simple vertical pendulum like the SG design, to a compact garden gate with a displacement detector. In either case, I would expect much greater resolution. A 24.8 cm, 1 second clock pendulum with a 1 nanometer resolution displacement detector (VRDT, etc) has a resolution of 0.004 microradians (neglecting the suspension). For a horizontal tiltmeter: (where z is the displacement in meters, and phi is the tilt) z = (g * Tn^2 / 4 * pi^2) * phi letting Tn = 1 second, and phi = 100 x 10^-6 radians, (g = 9.8) z = 24.8 microns for a 100 urad p-p tilt. So what size quake will cause this: using the standard magnitude formula: Ms = log(A/P) + 1.66*log(distance) - 0.18. , where A is the p-p ground motion in nanometers (10^-9meter), P is the period in seconds, and the distance is in degrees. The value of the corrector term (-0.18) is subject to much debate, and has large regional variations, depending on how urgently one needs funding vs the complaints of the tourism board. For a standard teleseismic surface wave we can assume that P = 20 seconds and that your distance = 90 degrees (~ USA to ~ South Pacific), so Ms = log(24800/20) + 1.66*log(90) -0.18 Ms = 3.09 + 3.24 - 0.18 Ms = 6.16 This is a rather large figure, but is supported by experience. When I operated an array of 10 bi-axial tiltmeters in the Aleutians, we had on-site battery operated strip chart monitors as backup to the digital telemetry systems. The Rustrak charts operated at 1/16" per hour (to get 1 year on a 50 ft roll) at a scale of +,- 2 urads (to accommodate seasonal drift and crustal deformation, which nonetheless required an autonomous digitally controlled auto-zeroing system ). The earth tides are easily evident at 0.1 - 0.2 urad, since I improved the S/N with a 20 second low pass filter to remove the 3 - 6 second microseisms and to reduce storm noise (the Aleutians DO have storms!). The South Pacific teleseisms would show up about monthly (about every 3 - 4 feet of the chart), but they would have to be a M 6+ or greater to show up even at the 0.5 - 1 urad level; we then looked them up in PDE to check for the timing corrections (over a year) of the strip charts for comparisons of the co-sited data. So I would not consider a 5 urad resolution tiltmeter a very interesting instrument if it detects only a few teleseisms per year. __________________________________________________________ Here is a repeat of the calculation of the numbers one can expect for a mechanical pendulum: SO ... Lets consider some formulae of interest for the horizontal pendulum: (assuming that the restoring force by the hinges and/or pivot are minimal): The natural mechanical period: Tn = 2*pi*sqrt(L/(g*sine i)) where L is the boom length in cm, g=980cm/sec^2, i is the angle that the boom makes wrt the horizontal, (if i is measured in radians, (360 degrees = 2*pi radian, or 1 radian = 57.3 degrees), and i is small, sine i = i). For example, a 40 cm boom hanging vertically (mass at the bottom) as a simple tick-tock pendulum ( an SG design at an angle of 90 degrees) has a period of 1.3 seconds. (a one second clock pendulum is 24.8 cm). But the pendulum supports or hinges can be arranged in a "garden gate" configuration as is the Lehman and most long-period horizontals. When tilted horizontally to about 4 degrees, the period is 5 seconds. At about a 1 degree ((2*pi/360) radian) angle, it is 10 seconds, and at about 0.23 deg. it is 20 seconds. However, if we increase the boom length by times 4 to 160cm, (an impractical 60 inches), we also get a period of 20 seconds. So the period is changing with the square root of the boom length as well as the inverse of the square root of the angle the boom makes with the horizontal. In general, a practical boom length is 10" to 15", with a baseplate of 15" to 24" long and about half as wide at the leveling end (for a horizontal; leveling for a vertical, as shown above, is nowhere as critical, and 4" to 8" widths are workable). It is important to note that the size of the mass determines nothing of the period or sensitivity to tilting. Any reasonable size will work; larger is better for overcoming any torque of the hinges or flexures, to the point where the mass/boom structure begins to distort any part of the suspension. (THe size of the mass IS a factor in a VBB fedback system). The total mass of the boom should be less than 10% of the main mass, which includes the sensor coils. The tilt sensitivity of a seismometer is therefore a function of the square of the operating period Tn. For a HORIZONTAL: (where z is the displacement, and phi is the tilt) z = (g * Tn^2 / 4 * pi^2) * phi For a VERTICAL: z = (g * Tn^2 / 8 * pi^2) * phi^2) (vertical) Note the vertical sensor responds to the SQUARE of the tilt. BUT .. Since the angle is always small and less than 1, the square of a small angle (measured in radians) is smaller than the original number. So conversely the horizontal is MORE sensitive to tilt of the base (at a right angle to the boom) by the square of the tilt angle. SO what does this mean in comparing the tilt noise of a vertical compared with a horizontal of the same period. Suppose the seis is in a corner of the garage or basement. Then suppose that when you walk up to the site you deflect the floor by 1 micron (10^-6 meter) when you are 1 meter away (Or your neighbor parks his Humvee 100 meters away and deflects the neighborhood by 0.1 millimeter.) In both cases the tilt is delta(L)/L or 10^-6 radian. So if your horizontal seis has a period of 10 seconds, the mass will offset 24.8 microns. This is a large number; the 6-second microseisms run about 2 to 4 microns. HOWever: if you have a vertical seis, the displacement from a 10^-6 tilt is 10^-6th of the horizontal. Conversely, it takes a floor deflection if 1mm at 1 meter distance to get the same 24.8 microns movement on a vertical. So when you push the operating period from 10 to 20 seconds, the tilt sensitivity increases by 4. Even a modest period increase demands a good site for the instrument. The WWNSS (worldwide network of standard seismographs) originally tried to operate the long period sensors at 30 seconds, but so many were always at the stops that they backed off to 15 seconds as the standard. You can test the tilt sensitivity using your leveling screw, which I guess is something like 40 threads/inch. In the above formula, "phi" is the angle in radians, so if you turn the screw one turn and if the base support width is 10 inches, the tilt is 1/40" divided by 10", or 0.0025 radians. Using the formula above for a horizontal sensor: The displacement then is 0.062cm times the square of the period. If Tn is 10 seconds, it is 6.2cm; if Tn is 100 seconds, one turn of th 40 tpi screw will try to move the boom 620 cm. Even if 1/100 turn can be used, the displacement is still quite large. THis is why VBB instruments have a feature that allows a shorter period to be switched in for setup at installation, and a motor-driven lever of 100:1 to level the sensor in operation. A typical tilt noise level for the 360-second STS-1 is equivalent to about 6 nanoradians. For practical operation of a home-made VBB, an operating period of 20 to 40 seconds would be preferred. Regards, Sean-Thomas __________________________________________________________ Public Seismic Network Mailing List (PSN-L)

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