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Re: CMIP5 model output [message #83859] Thu, 11 April 2013 15:27
DavidF[1] is currently offline  DavidF[1]
Messages: 94
Registered: April 2012
Member
siumt...@gmail.com wrote:

> how can you convert CMIP5 model output time to the Julian day?

I don't know how CMIP5 does this, but normally time is expressed as "number of days since 1600-01-01", or something like that. So, to create the correct Julian day number you do this:

time = time + Julday(1,1,1600,0,0,0)

Then, when you want months, days, and years, you convert these Julian values with CatDat:

CatDat, time, month, day, year

Cheers,

David
Re: CMIP5 model output [message #83864 is a reply to message #83859] Thu, 11 April 2013 13:09 Go to previous message
Phillip Bitzer is currently offline  Phillip Bitzer
Messages: 223
Registered: June 2006
Senior Member
Not sure what the format of the CMIP5 time is, but I think you'll want to look at the JULDAY function:

http://www.exelisvis.com/docs/JULDAY.html
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