Convert a julian gregorian date code into a regular date

R

Robert

how do I convert a julian gregorian date code into a regular date. I have a
julian gregorian (1582) date code that I want to convert into a regular
month/day/year code. How do I do this? Is there a formula for this? Here
is a sample date code: 2453523.
 
R

Ron Rosenfeld

how do I convert a julian gregorian date code into a regular date. I have a
julian gregorian (1582) date code that I want to convert into a regular
month/day/year code. How do I do this? Is there a formula for this? Here
is a sample date code: 2453523.

It would be helpful for me if you can define exactly what you mean by a "julian
gregorian date code" and give some examples of this code and what you expect
for a "regular date"

A proper Julian Date is an astronomical term. It is defined as a serial number
equal to the number of days elapsed since January 1, 4713 B.C., proposed by
Joseph Scaliger in 1582 and used in astronomical calculations: January 1, 1965,
at noon, Greenwich Civil Time, was Julian Day 2,438,762.0.

Some on this NG use the term to stand for one or more digits which stand for
the current year, followed by the number of days elapsed in that year. But
that doesn't make sense for your example 2453523. That latter would be 1 Jun
2005 12 Noon.

1582, using the astronomical definition, would be 1 May 4709 BCE 12 Noon.

So which do you mean?


--ron
 
R

Ron Rosenfeld

how do I convert a julian gregorian date code into a regular date. I have a
julian gregorian (1582) date code that I want to convert into a regular
month/day/year code. How do I do this? Is there a formula for this? Here
is a sample date code: 2453523.

OK, now I've got it. You are, in fact, interested in the astronomical
calculations. The 1582 finally clued me in :).

For dates on and after March 1, 1900, you can use the formula:

= A1 + 2415018.5 to convert a date/time string in A1 to its Julian equivalent.

To convert the Julian Date to its Gregorian equivalent, you can use the
formula:

=A1-2415018.5

If you require a greater range, you will require complex formulas or VBA.


--ron
 

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