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Help With A Julian Date Conversion Script (calendar->julian)

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roololing
 PostPosted: Mon Jun 20, 2011 3:23 pm   

Joined: Mon Jun 20, 2011 3:00 pm
Posts: 1
I'm trying to write a program in which the user enters a calendar date such as 2011 06 20 11 04 45 (YYYY MM DD hh mm ss), and gets back a Julian date such as 2455732.961632 (I test results with the conversion tool at http://aa.usno.navy.mil/data/docs/JulianDate.php). The day occasionally strays, from 1 day to around maybe 50 days off, with no discernible pattern, at least not chronologically. The time, the numbers after the decimal, is always correct. I was wondering if anyone knew of some vague equation I've forgotten. I've already written the code for leap years, the different amounts of days in each month, the absence of a year 0, etc. Is there anything I may be missing, or something that I'm likely to have miscalculated?

Any help would be appreciated.
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jsz
 PostPosted: Mon Jun 20, 2011 5:46 pm   
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Joined: Wed Jun 08, 2011 8:27 am
Posts: 159
Location: outer Shpongolia
I did it with awk(1), using the formulas from this website.

Code:
#!/usr/bin/awk -f

BEGIN \
{
    for (i = 1; i < ARGC; i++) {
        if (ARGV[i] ~ /[^[:digit:]]/)
            j++
    }   
    if (j || ARGC-1 != 6) {
        printf("usage: YYYY MM DD hh mm ss\n") > "/dev/stderr"
        exit 1
    }   
    else if (ARGV[1] == 0) {
        printf("There is no year 0 in the Julian system!\n") > "/dev/stderr"
        exit 1
    }   
   
    Y  = ARGV[1]; M  = ARGV[2]; D  = ARGV[3]
    HH = ARGV[4]; MM = ARGV[5]; SS = ARGV[6]
   
    printf("%f\n", 367*Y - int(7*(Y + int((M+9)/12))/4) \
                   - int(3*(int((Y+(M-9)/7)/100)+1)/4) \
                   + int(275*M/9) + D + 1721028.5 \
                   + (HH+(MM+SS/60)/60)/24)
}
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