Thread: ERROR 0: Resulting error from shift exceeded tolerance (1e-6)

Started: 2018-08-23 01:17:12
Last activity: 2018-08-23 15:47:22
Topics: SAC Help
Hi,

In order to make the B and KZTIME of the three components consistent, I conducted "synchronize" command. The SAC informed: ERROR    0: Resulting error from shift exceeded tolerance (1e-6): 0.0005. Could you tell me what the error is and how can I fix it, please?

The attached files are the original files that yielded errors.

Thank you!


  • Hi,

    On Mac osx and SACv.101.6a the command synchronise works.
    Here my commands,

    SAC> r 1*

    17.18.15.BMRxxx_RO.BHZ.sac 17.18.17.BMRxxx_RO.BHE.sac
    17.18.21.BMRxxx_RO.BHN.sac

    SAC> synchronize begin on

    SAC> p1

    SAC> qdp off

    SAC> p1

    SAC> lh





    FILE: 17.18.15.BMRxxx_RO.BHZ.sac - 1

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


    NPTS = 22532

    B = 0.000000e+00

    E = 1.126550e+03

    IFTYPE = TIME SERIES FILE

    LEVEN = TRUE

    DELTA = 5.000000e-02

    IDEP = UNKNOWN

    DEPMIN = 8.240000e+02

    DEPMAX = 1.750000e+03

    DEPMEN = 1.237252e+03

    KZDATE = JUL 28 (209), 2018

    KZTIME = 17:18:15.770

    KSTNM = BMR

    CMPAZ = 0.000000e+00

    CMPINC = 0.000000e+00

    STLA = 4.767280e+01

    STLO = 2.349690e+01

    STEL = 2.270000e+02

    STDP = 0.000000e+00

    KEVNM = tmpevent

    EVLA = -2.997600e+01

    EVLO = -1.774360e+02

    EVDP = 3.470000e+01

    KHOLE =

    DIST = 1.735385e+04

    AZ = 3.234513e+02

    BAZ = 4.986711e+01

    GCARC = 1.560796e+02

    LOVROK = TRUE

    NVHDR = 6

    NORID = 0

    NEVID = 0

    LPSPOL = FALSE

    LCALDA = TRUE

    KCMPNM = BHZ

    KNETWK = RO



    FILE: 17.18.17.BMRxxx_RO.BHE.sac - 2

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


    NPTS = 22547

    B = 0.000000e+00

    E = 1.127300e+03

    IFTYPE = TIME SERIES FILE

    LEVEN = TRUE

    DELTA = 5.000000e-02

    IDEP = UNKNOWN

    DEPMIN = -8.740000e+03

    Waiting



    DEPMAX = -6.476000e+03

    DEPMEN = -7.550445e+03

    KZDATE = JUL 28 (209), 2018

    KZTIME = 17:18:15.770

    KSTNM = BMR

    CMPAZ = 9.000000e+01

    CMPINC = 9.000000e+01

    STLA = 4.767280e+01

    STLO = 2.349690e+01

    STEL = 2.270000e+02

    STDP = 0.000000e+00

    KEVNM = tmpevent

    EVLA = -2.997600e+01

    EVLO = -1.774360e+02

    EVDP = 3.470000e+01

    KHOLE =

    DIST = 1.735385e+04

    AZ = 3.234513e+02

    BAZ = 4.986711e+01

    GCARC = 1.560796e+02

    LOVROK = TRUE

    NVHDR = 6

    NORID = 0

    NEVID = 0

    LPSPOL = FALSE

    LCALDA = TRUE

    KCMPNM = BHE

    KNETWK = RO



    FILE: 17.18.21.BMRxxx_RO.BHN.sac - 3

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


    NPTS = 22300

    B = 0.000000e+00

    E = 1.114950e+03

    IFTYPE = TIME SERIES FILE

    LEVEN = TRUE

    DELTA = 5.000000e-02

    IDEP = UNKNOWN

    DEPMIN = -1.258000e+03

    DEPMAX = 4.980000e+02

    DEPMEN = -3.162195e+02

    KZDATE = JUL 28 (209), 2018

    KZTIME = 17:18:15.770

    KSTNM = BMR

    CMPAZ = 0.000000e+00

    CMPINC = 9.000000e+01

    STLA = 4.767280e+01

    STLO = 2.349690e+01

    STEL = 2.270000e+02

    STDP = 0.000000e+00

    KEVNM = tmpevent

    Waiting



    EVLA = -2.997600e+01

    EVLO = -1.774360e+02

    EVDP = 3.470000e+01

    KHOLE =

    DIST = 1.735385e+04

    AZ = 3.234513e+02

    BAZ = 4.986711e+01

    GCARC = 1.560796e+02

    LOVROK = TRUE

    NVHDR = 6

    NORID = 0

    NEVID = 0

    LPSPOL = FALSE

    LCALDA = TRUE

    KCMPNM = BHN

    KNETWK = RO

    HTH,

    Milton


    ************************************
    Milton P. PLASENCIA LINARES

    Centro di Ricerche Sismologiche (CRS)
    OGS - Istituto Nazionale di Oceanografia e di Geofisica Sperimentale

    Borgo Grotta Gigante 42/C
    (34010) Sgonico - Trieste - Italia
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    E-mail: mplasencia<at>inogs.it

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    ASAIN (Antarctic Seismographic Argentinean Italian Network)
    **************************************


    On Wed, 22 Aug 2018 at 18:07, Wenkai Song <wsvwf<at>mst.edu> wrote:

    Hi,



    In order to make the B and KZTIME of the three components consistent, I
    conducted "synchronize" command. The SAC informed: ERROR 0: Resulting
    error from shift exceeded tolerance (1e-6): 0.0005. Could you tell me what
    the error is and how can I fix it, please?



    The attached files are the original files that yielded errors.



    Thank you!



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  • Dear All -

    These files have a BEGIN time offset (B header value) that is 5x10**-4 sec, or 0.5 milliseconds.

    The ZERO time in a SAC file is only precise to 1 millisecond. The header stores the zero hour, minute, second, and millisecond values as integers.

    When you SYNCHRONIZE the files, you’re asking SAC to align the times to a 0.5 millisecond precision, which, for it, is impossible - hence the error message.

    If you want to retain the full precision of the BEGIN values in your files (meaning you *really* believe that 0.5 ms time offset), you must do the alignment manually. Here’s how, using the ALLT option of the CHNHDR command. Notice that the 0.5 ms offset is still preserved in the B values for each file:

    SAC> r /tmp/*sac
    /tmp/17.18.15.BMRxxx_RO.BHZ.sac ...17.18.17.BMRxxx_RO.BHE.sac ...17.18.21.BMRxxx_RO.BHN.sac
    SAC> lh files all nzmin nzsec nzmsec


    FILE: /tmp/17.18.15.BMRxxx_RO.BHZ.sac
    -------------------------------------
    nzmin = 18 nzsec = 15
    nzmsec = 769

    FILE: /tmp/17.18.17.BMRxxx_RO.BHE.sac
    -------------------------------------
    nzmin = 18 nzsec = 17
    nzmsec = 769

    FILE: /tmp/17.18.21.BMRxxx_RO.BHN.sac
    -------------------------------------
    nzmin = 18 nzsec = 21
    nzmsec = 269
    SAC> ch file 1 allt -5.5
    SAC> ch file 2 allt -3.5
    SAC> lh files all nzmin nzsec nzmsec


    FILE: /tmp/17.18.15.BMRxxx_RO.BHZ.sac
    -------------------------------------
    nzmin = 18 nzsec = 21
    nzmsec = 269

    FILE: /tmp/17.18.17.BMRxxx_RO.BHE.sac
    -------------------------------------
    nzmin = 18 nzsec = 21
    nzmsec = 269

    FILE: /tmp/17.18.21.BMRxxx_RO.BHN.sac
    -------------------------------------
    nzmin = 18 nzsec = 21
    nzmsec = 269
    SAC> lh default


    FILE: /tmp/17.18.15.BMRxxx_RO.BHZ.sac
    -------------------------------------
    NPTS = 22532 B = -5.49950
    E = 1121.05 IFTYPE = TIME SERIES FILE
    LEVEN = TRUE DELTA = 0.50E-01
    IDEP = UNKNOWN DEPMIN = 824.0
    DEPMAX = 1750.0 DEPMEN = 1237.25
    KZDATE = JUL 28 (209), 2018 KZTIME = 17:18:21.269
    ...

    FILE: /tmp/17.18.17.BMRxxx_RO.BHE.sac
    -------------------------------------
    NPTS = 22547 B = -3.49950
    E = 1123.80 IFTYPE = TIME SERIES FILE
    LEVEN = TRUE DELTA = 0.50E-01
    IDEP = UNKNOWN DEPMIN = -8740.0
    DEPMAX = -6476.0 DEPMEN = -7550.46
    KZDATE = JUL 28 (209), 2018 KZTIME = 17:18:21.269
    ...

    FILE: /tmp/17.18.21.BMRxxx_RO.BHN.sac
    -------------------------------------
    NPTS = 22300 B = 0.50E-03
    E = 1114.95 IFTYPE = TIME SERIES FILE
    ...


    On 23 Aug 2018, at 01:06, Wenkai Song <wsvwf<at>mst.edu> wrote:

    Hi,

    In order to make the B and KZTIME of the three components consistent, I conducted "synchronize" command. The SAC informed: ERROR 0: Resulting error from shift exceeded tolerance (1e-6): 0.0005. Could you tell me what the error is and how can I fix it, please?

    The attached files are the original files that yielded errors.

    Thank you!

    <17.18.15.BMRxxx_RO.BHZ.sac><17.18.17.BMRxxx_RO.BHE.sac><17.18.21.BMRxxx_RO.BHN.sac>
    ----------------------
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    George Helffrich
    george<at>elsi.jp


12:11:35 v.01697673