Thread: Re: Insert hedder

ch O GMT 2009 318 13 03 34 300

Dear all,

I have a SAC file converted from SEISAN.

How to insert the origin time(2009-11-14-13:03:34.300) of the event in sac header?

Thanks in advance

Shekar


On Mon, 18 Nov 2013 22:14:49 +0530 wrote

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Today's Topics:



1. Re: sort and plot commands don't play (George Helffrich)

2. Remove Response Instrument (amik Amik)

3. Re: Remove Response Instrument (Arthur Snoke)

4. Re: Remove Response Instrument (Milton P. Plasencia Linares)

5. Calculation of AZ and BAZ in SAC (Fiona Darbyshire)

6. Re: Calculation of AZ and BAZ in SAC (Brian Savage)





----------------------------------------------------------------------



Message: 1

Date: Wed, 13 Nov 2013 09:03:30 +0000

From: George Helffrich

Subject: Re: [SAC-HELP] sort and plot commands don't play

To: Frederik Tilmann

Cc: sac-help<at>iris.washington.edu

Message-ID:

Content-Type: text/plain; charset=us-ascii



Dear All -



Plotting the traces in a record section in SSS displays traces in increasing distance order.



On 13 Nov 2013, at 08:49, Frederik Tilmann wrote:

I have a problem in that the plot commands (p1,ppk) do not show the files in the same order as they are listed in memory after a sort.



What I am trying to do is to display a north- to south record section of noise correlation functions with respect to a target station. The coordinates of both stations are place in evla and stla, but it is not well defined in which variable a given station ends up in (evla, or stla).



So to plot all the functions relative to J10S I use



r sumtraces-nonorm-0.03-0.10/*ZWZW*J10?J[0-9][0-9]*_sum.sac sumtraces-nonorm-0.03-0.10/*ZWZW*J[0-9][0-9]?J10?*_sum.sac

sort evla ascend stla ascend

lh evla stla



Everything looks good, stations are in the right order in the lh list.



But if I then do

p1

or ppk

they come out in the wrong order, plotting the J03-J10 after J10-J11 (correct would have been J09-J10 as in the lh listing). I attach a screenshot demonstrating this effect.

I tried

sort commit evla ascend stla ascend

but this this did not make a difference.



This is on 101.5c

SEISMIC ANALYSIS CODE [02/01/2012 (Version 101.5c)]





Any ideas would be appreciated



Regards

Frederik









Output of lh command



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ23B..HHZHHZ_sum.sac - 1

-------------------------------------------------------------------



evla = 5.253600e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ22B..HHZHHZ_sum.sac - 2

-------------------------------------------------------------------



evla = 5.270155e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ21B..HHZHHZ_sum.sac - 3

-------------------------------------------------------------------



evla = 5.287174e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ20B..HHZHHZ_sum.sac - 4

-------------------------------------------------------------------



evla = 5.305374e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ19B..HHZHHZ_sum.sac - 5

-------------------------------------------------------------------



evla = 5.320426e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ18B..HHZHHZ_sum.sac - 6

-------------------------------------------------------------------



evla = 5.339244e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ17B..HHZHHZ_sum.sac - 7

-------------------------------------------------------------------



evla = 5.355617e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ25B..HHZHHZ_sum.sac - 8

-------------------------------------------------------------------



evla = 5.366659e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ15B..HHZHHZ_sum.sac - 9

-------------------------------------------------------------------



evla = 5.390106e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ14S..HHZHHZ_sum.sac - 10

-------------------------------------------------------------------



evla = 5.411258e+01

Waiting



stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ13B..HHZHHZ_sum.sac - 11

-------------------------------------------------------------------



evla = 5.427260e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ12B..HHZHHZ_sum.sac - 12

-------------------------------------------------------------------



evla = 5.444247e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J10BJ11S..HHZHHZ_sum.sac - 13

-------------------------------------------------------------------



evla = 5.460217e+01

stla = 5.481430e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J09BJ10B..HHZHHZ_sum.sac - 14

-------------------------------------------------------------------



evla = 5.481430e+01

stla = 5.640510e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J08SJ10B..HHZHHZ_sum.sac - 15

-------------------------------------------------------------------



evla = 5.481430e+01

stla = 5.617126e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J07BJ10B..HHZHHZ_sum.sac - 16

-------------------------------------------------------------------



evla = 5.481430e+01

stla = 5.603592e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J04BJ10B..HHZHHZ_sum.sac - 17

-------------------------------------------------------------------



evla = 5.481430e+01

stla = 5.577234e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J03EJ10B..HHZHHZ_sum.sac - 18

-------------------------------------------------------------------



evla = 5.481430e+01

stla = 5.530829e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J02SJ10B..HHZHHZ_sum.sac - 19

-------------------------------------------------------------------



evla = 5.481430e+01

stla = 5.512811e+01



FILE: sumtraces-nonorm-0.03-0.10/mat__ZWZW.J01EJ10B..HHZHHZ_sum.sac - 20

-------------------------------------------------------------------



evla = 5.481430e+01

Waiting



stla = 5.497167e+01

SAC>

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George Helffrich

george.helffrich<at>bris.ac.uk









------------------------------



Message: 2

Date: Sun, 17 Nov 2013 16:07:33 +0800 (SGT)

From: amik Amik

Subject: [SAC-HELP] Remove Response Instrument

To: "sac-help<at>iris.washington.edu"

Message-ID:

<1384675653.16429.YahooMailNeo<at>web190801.mail.sg3.yahoo.com>

Content-Type: text/plain; charset="iso-8859-1"



I am working on a local magnitude research. I want to remove response instrument

from the data and simulate the response of Wood Anderson seismograph. What should I do?

I have a few questions :

1. Which one the true procedure and command in SAC :

?? a) SAC> trans from evalresp to none

?? ? ?and then

?? ? ?SAC> trans from polezero subtype "file name.pz" to none



?? b) SAC> trans to polezero subtype "wood anderson file name.pz"



?? c) SAC> trans to wa



?? d) SAC> trans from polezero subtype "sation's file name.pz" to wa



?? e) SAC> trans from polezero subtype "sation's file name.pz" to wa to none



?? f) SAC> trans from evalresp fname RESP.SATION to wa



?? g) SAC> SAC> trans from polezero subtype "sation's file name.pz" freq 0.005 0.01 1.0 2.0



2. What the different utilizing of Response File and Polezero File?

?? I think polezero already included in responfile, is it right ?



May you help me? Thank You very much





Sincerely Yours

Ami

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Message: 3

Date: Sun, 17 Nov 2013 07:43:55 -0500 (EST)

From: Arthur Snoke

Subject: Re: [SAC-HELP] Remove Response Instrument

To: amik Amik

Cc: "sac-help<at>iris.washington.edu"

Message-ID:

Content-Type: text/plain; charset="iso-8859-1"; Format="flowed"



I think the best way for you to understand what the various options do is

to try them andsave the results. The TRANSFER help file at

has answers

to your explicit quesitons. If anything is not clear, let me know.



On Sun, 17 Nov 2013, amik Amik wrote:

I am working on a local magnitude research. I want to remove response

instrument

from the data and simulate the response of Wood Anderson seismograph. What

should I do?

I have a few questions :

1. Which one the true procedure and command in SAC :

?? a) SAC> trans from evalresp to none

?? ? ?and then

?? ? ?SAC> trans from polezero subtype "file name.pz" to none



?? b) SAC> trans to polezero subtype "wood anderson file name.pz"



?? c) SAC> trans to wa



?? d) SAC> trans from polezero subtype "sation's file name.pz" to wa



?? e) SAC> trans from polezero subtype "sation's file name.pz" to wa to none



?? f) SAC> trans from evalresp fname RESP.SATION to wa



?? g) SAC> SAC> trans from polezero subtype "sation's file name.pz" freq

0.005 0.01 1.0 2.0



2. What the different utilizing of Response File and Polezero File?

?? I think polezero already included in responfile, is it right ?



May you help me? Thank You very much





Sincerely Yours

Ami

------------------------------



Message: 4

Date: Sun, 17 Nov 2013 14:29:15 +0100

From: "Milton P. Plasencia Linares"

Subject: Re: [SAC-HELP] Remove Response Instrument

To: "amik Amik"

Cc: "sac-help<at>iris.washington.edu"

Message-ID: <20131117142915.910179rx5iy7g6nf<at>webmail.ogs.trieste.it>

Content-Type: text/plain;charset=ISO-8859-1;DelSp="Yes";

format="flowed"



look below,



**********************************

Milton P. PLASENCIA LINARES



Dipartimento Centro di Ricerche Sismologiche

Istituto Nazionale di Oceanografia e di Geofisica Sperimentale - OGS



Borgo Grotta Gigante 42/C

(34010) Sgonico - TRIESTE - ITALIA

Tel: +39-040-2140 256/141

Fax: +39-040-327307



E-mail: mplasencia<at>inogs.it



ASAIN (Antarctic Seismographic Argentinean Italian Network)

*********************************





Quoting "amik Amik" :

I am working on a local magnitude research. I want to remove

response instrument

from the data and simulate the response of Wood Anderson

seismograph. What should I do?

I have a few questions :

1. Which one the true procedure and command in SAC :

?? a) SAC> trans from evalresp to none

?? ? ?and then

?? ? ?SAC> trans from polezero subtype "file name.pz" to none



?? b) SAC> trans to polezero subtype "wood anderson file name.pz"



?? c) SAC> trans to wa



?? d) SAC> trans from polezero subtype "sation's file name.pz" to wa

Option d) This is fine if you use the polezero station file.

?? e) SAC> trans from polezero subtype "sation's file name.pz" to wa to none



?? f) SAC> trans from evalresp fname RESP.SATION to wa



?? g) SAC> SAC> trans from polezero subtype "sation's file name.pz"

freq 0.005 0.01 1.0 2.0



2. What the different utilizing of Response File and Polezero File?

Response file include the filter (FIR) coefficients

?? I think polezero already included in responfile, is it right ?

Yes.

May you help me? Thank You very much

My suggest ius use:

transfer from polezero subtype "PZ file" to wa freq 0.005 0.01 1.0 2.0







Sincerely Yours

Ami

Cheers,



Milton







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------------------------------



Message: 5

Date: Mon, 18 Nov 2013 08:01:33 -0500

From: Fiona Darbyshire

Subject: [SAC-HELP] Calculation of AZ and BAZ in SAC

To: sac-help<at>iris.washington.edu

Message-ID:



Content-Type: text/plain; charset=ISO-8859-1



Hello,



Could someone tell me a bit more about how AZ and BAZ are calculated in SAC?



Up to now, e.g. when binning receiver functions, I have used BAZ and

GCARC to group them together, and that's always seemed to work, so

I've always assumed that the BAZ/GCARC (or DIST, I imagine) pair gives

you a unique position from the point of interest (the station).



This time, I have some data output that is expressed in AZ and DIST. I

assumed (perhaps naively?) that this would also allow successful

spatial grouping, but it turns out not to be the case; one can have

the same forward azimuth for several different locations.



Here's the example where I found the issue:

Station lat, lon: 82.5033, -62.35 (ALE)



Event1: lat 80.205, lon -1.091

Event 2: lat 86.876, lon 54.373

The two events are almost 1000km apart.



Event 1: AZ 313, BAZ 72, GCARC 9, DIST 1006

Event 2: AZ 314, BAZ 17, GCARC 9, DIST 1042



I'd like to know more about the calculation, in particular because (i)

I'd like to reassure myself that BAZ/distance does indeed give a

unique location, and (ii) I don't understand how AZ is calculated. One

website I looked at (Matlab) talks about rhumb lines vs great-circles;

is that the issue? Will an AZ/distance pair always be non-unique?

(Also most websites define back-azimuth as simply 180 opposite to

azimuth, but I guess they aren't talking about spherical geometry

then...)



Thanks for any insight.



Fiona Darbyshire.



Centre de recherche GEOTOP, Universit? du Qu?bec ? Montr?al







------------------------------



Message: 6

Date: Mon, 18 Nov 2013 11:38:44 -0500

From: Brian Savage

Subject: Re: [SAC-HELP] Calculation of AZ and BAZ in SAC

To: SAC HELP

Message-ID: <8D3AE955-FCD5-48D4-8643-4E4E25115DEF<at>uri.edu>

Content-Type: text/plain; charset="windows-1252"



Fiona,



The implementation of the computation of Great Circle Distance (degrees), Azimuth, Back Azimuth, and Distance (km) is based off of Rudoe's formula found in Geodesy by R.Bomford (1980). It assumes a Radius of 6378.160 km and a flattening of 1/298.2466081. I can send you the appropriate pages from R. Bomford (1980) if you would like.



The azimuth and back-azimuth will generally not be offset by180 degrees, and some pairs of point will have "odd-looking" (not event close to 180 degrees) azimuth/back azimuth pairs.



Note: There are more robust methods of calculating these quantities and the current implementation in SAC as well as many other implementations, up until recently [1], have trouble with particular pairs of points. For over ~ 99% of point pairs SAC's implementation produces the correct result, but that last 1% can be incorrect (sometimes badly).



I hope this helps.



Brian Savage



[1] Karney, C. F. F. (2013). "Algorithms for geodesics". Journal of Geodesy 87 (1): 43?42. arXiv:1109.4448. Bibcode:2013JGeod..87...43K. doi:10.1007/s00190-012-0578-z (open access). Addenda.





On Nov 18, 2013, at 8:01 AM, Fiona Darbyshire wrote:

Hello,



Could someone tell me a bit more about how AZ and BAZ are calculated in SAC?



Up to now, e.g. when binning receiver functions, I have used BAZ and

GCARC to group them together, and that's always seemed to work, so

I've always assumed that the BAZ/GCARC (or DIST, I imagine) pair gives

you a unique position from the point of interest (the station).



This time, I have some data output that is expressed in AZ and DIST. I

assumed (perhaps naively?) that this would also allow successful

spatial grouping, but it turns out not to be the case; one can have

the same forward azimuth for several different locations.



Here's the example where I found the issue:

Station lat, lon: 82.5033, -62.35 (ALE)



Event1: lat 80.205, lon -1.091

Event 2: lat 86.876, lon 54.373

The two events are almost 1000km apart.



Event 1: AZ 313, BAZ 72, GCARC 9, DIST 1006

Event 2: AZ 314, BAZ 17, GCARC 9, DIST 1042



I'd like to know more about the calculation, in particular because (i)

I'd like to reassure myself that BAZ/distance does indeed give a

unique location, and (ii) I don't understand how AZ is calculated. One

website I looked at (Matlab) talks about rhumb lines vs great-circles;

is that the issue? Will an AZ/distance pair always be non-unique?

(Also most websites define back-azimuth as simply 180 opposite to

azimuth, but I guess they aren't talking about spherical geometry

then...)



Thanks for any insight.



Fiona Darbyshire.



Centre de recherche GEOTOP, Universit? du Qu?bec ? Montr?al



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http://www.iris.washington.edu/mailman/listinfo/sac-help

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