Re: time conversion

• To: araligin@xxxxxxxxxxxxx
• Subject: Re: time conversion
• From: t.hume@xxxxxxxxxx
• Date: Fri, 25 Feb 2005 09:46:46 +1100

Hi,

You want to be really careful with converting hours since 1-1-1 into
year-month-day.

I think you will be aware that years are not exactly 365.25 days long.
To get around this you need to drop the leap years at the century,
unless the century is divisible by 400 (1700, 1800 and 1900 were not
leap years, but 2000 was). This is called (I think) Gregorian time, and
I guess it was introduced by Pope Gregory. Before that, Julian
time was used, where every year divisible by four was a leap
year. However, not all countries implemented the newer system at the
same time (I don't think Russia introduced it until approximately the
time of the Revolution, from memory), so that adds another complication.

You should be OK negelecting leap seconds ... but I really don't know.
I'm not sure if leap seconds were "invented" before the 1970s.
However if they were, after 2000 years, the number of leap seconds will
be close to one hour.

Really, it might be better to keep your original units. You probably
don't want to open up the can of worms which the time and calender
system is.

Tim.

On Thu, 24 Feb 2005 12:17:02 -0800
araligin@xxxxxxxxxxxxx wrote:

> 
> Hi,
> 
>   I have my time variable reading below for the netcdf file
>     double time(time) ;
>                 time:units = "hours since 1-1-1 00:00:0.0" ;
>                 time:long_name = "Time" ;
>                 time:actual_range = 17461560., 17549208. ;
>                 time:delta_t = "0000-00-01 00:00:00" ;
> 
> 
>     my first data is from 01 Jan 1993. I read the time as double
>     precision and
> convert it to integer and give input to the subroutine which converts
> into UTC. But I get the first date as 27 Nov 1993.  following is the
> subroutine.
> 
>  subroutine  ymd(jn)
>     integer W,X,A,B,C,D,E,F
>     integer mnt,yr,day
>       integer  jn,Z
>        write(*,*) jn
>       jn=int(jn/24.0)
>       z=jn
>         w=int((z - 1867216.25)/36524.25)
> c       w=z/36524.25
> c       w=z
> C       x=int(w/4)
>         a=z+1+w-int(0.250*w)
>         write(*,*) w,x,a
>         b=a+1524
>         c=int( 6680.0+((b-2439870)-122.1)/365.25)
>         d = 365*c+int(0.25*c)
>         e= int((b-d)/30.6001)
>         f= int(30.6001*e)
>         write(*,*) b,c,d,e,f
>         day= b - d - f
>         if(e.lt.13.5) then
>                mnth=e-1
>          else
>                mnth=e-13
>         endif
>         if(mnth.le.2) yr = c -1
>         if(mnth.gt.2) yr = c-2
>         write(12,*) yr,mnth,day
>         return
>         END
> 
> 
> Thanking you,
> Nilesh
> 

• Follow-Ups:
  • Re: time conversion
    • From: John Caron

• References:
  • time conversion
    • From: araligin