PSN-L Email List Message
Subject: Re: Large Response Coming in Now at 18:00 UTC
From: Brett Nordgren brett3nt@.............
Date: Mon, 03 Aug 2009 21:00:27 -0400
No WinQuake magic here, just used the calculator.
Given that the period, P, of the largest trace was 13 seconds, which I got
by measuring the surface wave in WinQuake, then Peak Displacement = Peak
Velocity * Period / 2 Pi. Or in my case = 3mm/s * 13s / 2Pi = 6.2mm.
That works in general, though if you had a smaller, much longer wave, it
might compute to have a larger displacement than waves with the highest
velocity peaks, but I wouldn't worry too much about that happening except
when you are looking at a local event.
And in your case, if the period was also 13 sec like mine, Peak
displacement = 2.6mm/s * 13s / 2Pi = 5.37mm peak or nearly 1.1 cm p-p.
At 06:28 PM 8/3/2009 -0600, you wrote:
>Hi Brett, How do I set up WinSDR or Winquake to view as movement? ( I
>recorded 2.6 mm of movement on the vertical channel)
>----- Original Message ----- From: "Brett Nordgren"
>Sent: Monday, August 03, 2009 5:56 PM
>Subject: Re: Large Response Coming in Now at 18:00 UTC
>>Now that I corrected the y-scale the plot from our vertical
>>http://bnordgren.org/seismo/090803.175000.bhz-pl.psn.gif shows roughly
>>3mm/s peak, which agrees pretty well with the values (assuming you meant
>>mm/s) that you were seeing. This is from near LA which is at about 6.5
>>deg. It's interesting that if you assume the predominant frequency was
>>1/13 seconds, that should imply that the vertical ground motion at Dave's
>>vault was about 6.2mm peak or 1.24 cm p-p. That's a lot of motion!
>>For laughs, I uploaded the output from the high gain channel
>>http://bnordgren.org/seismo/090803.175000.bhz-ph.psn.gif which clips at
>>300um/s and indeed it clipped. However, the low gain channel should be
>>able to handle up to 1.5cm/sec peak.
Public Seismic Network Mailing List (PSN-L)
[ Top ]
[ Back ]
[ Home Page ]