Hi Ted,

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.

Brett


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)
>
>Thanks, Ted
>
>
>----- Original Message ----- From: "Brett Nordgren" 
>To: 
>Sent: Monday, August 03, 2009 5:56 PM
>Subject: Re: Large Response Coming in Now at 18:00 UTC
>
>
>>Bob
>>
>>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.
>>
>>Brett


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