# Following Meeus's Astronomical Algorithms for the Calculation of JD

For the Calculation of Julian Day (JD), I followed Chapter 7 of Meeus's Astronomical Algorithms. I could get the numbers mentioned in Example 7.a and 7.b.

On the basis of these methodology, proceeded to calculate JD for the Date 1992 October 13 at 0^h TD (Date mentioned in Example 25.a, Meeus). My Code provides JD for this Date as JD = 2448921.50.

However, for this Date (1992 October 13 00:00:00 UT), each of the URL:

provide 2448921.50 as JDN (Julian Day Number).

However, each of the URL:

provide JD = 2448908.5 (Julian Day).

Now, the number 2448908.5 is described as JDE in Example 25.a of Meeus.

Kindly help with the:

1. Calculation of the correct JD corresponding to the Date 1992 October 13 at 0^h TD (Ex.25.a, Meeus).
2. How to obtain JDE from JD.

Thank you.

• Since UT = TD - Delta T; and, Delta T @ 1992 = 58.3s, p.87/488, Meeus, 1992 October 13 at 0^h TD = 1992 October 12 23.984 UT, right. This provides JD = 2448921.50. Still i am not getting it right. May 26 at 11:27
• JDN 2448921 is midday Gregor: 1992-10-25 and Julian: 1992-10-12. I have a converter here: gist.github.com/PM2Ring/… May 26 at 11:55
• FYI: Most programming languages can provide the time in "Unix Time", so a much easier algorithm is $JD= \frac{UnixTime}{86400} + 2440587.5$ and $UnixTime =JD-2440587.5 *86400$. And the language's standard library usually has routines to convert Unix Time to and from Gregorian dates. May 26 at 14:52
• Did you notice the “Gregorian/Julian” radio buttons at quasar.as.utexas.edu/BillInfo/JulianDateCalc.html ? May 26 at 22:33