I want to convert the Julian date 273.791667 to calendar day and time

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I want to convert the Julian date 273.791667 to calandar day and time
  1 Comment
James Tursa
James Tursa on 20 Jul 2017
Edited: James Tursa on 20 Jul 2017
As a nit, that number should be referred to as "day of year" and not "Julian date". I know there are segments of the scientific community that call it a "Julian date", but that usage is confusing at best. E.g., the definition of the term from the USNO:
You will find variations of this term in the literature depending on the underlying time scale (e.g., UT or TT) and 0 point, but even these are not "day of the year" types of measures.

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Accepted Answer

Chad Greene
Chad Greene on 20 Jul 2017
You'll have to specify a year, but for 2017 you can get a date vector like this:
datevec(datenum(2017,0, 273.791667))
or a date string like this:
datestr(datenum(2017,0, 273.791667))
  8 Comments
James Tursa
James Tursa on 20 Jul 2017
datevec returns a vector. You cannot stuff a vector into a single element.

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More Answers (1)

Steven Lord
Steven Lord on 20 Jul 2017
Use datetime. You can 'ConvertFrom' 'juliandate' (Number of days since noon UTC 24-Nov-4714 BCE (proleptic Gregorian calendar)) or 'modifiedjuliandate' (Number of days since midnight UTC 17-Nov-1858).
If you want to determine what the 273.791667th day of the year is, try something like this. Check the day of the year using the day function with the 'dayofyear' option:
x = datetime(2017, 1, 0) + days(273.791667)
day(x, 'dayofyear')
  2 Comments
Thishan Dharshana Karandana Gamalathge
Now when I import the local time data (converted from the Julian date), it is restricted to only two decimals. How can I have it for 6 decimals?

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