Converting a date to its Julian number

Question

I want to calculate difference between two dates, time and secs, and I know atleast for dates converting it to two Julian Numbers and subtracting it is an efficient way. I am trying to implement this algorithm for that. Below is my implementation for today i.e. 12-22-2015

function g(y,m,d)
m = (m + 9) % 12
y = y - m/10
return 365*y + y/4 - y/100 + y/400 + (m*306 + 5)/10 + ( d - 1 )

function g(2015,12,22)
[9] m = (12 + 9) % 12
[2014.1] y = 2015 - 9/10

735146.5 + 503.525 + 20.141 + 5.03525 + 275.9 + 21
=**735954.10125**

However from this online calculation tool the correct value is: 2457378.877.

What am I doing wrong?

This code doesn't look like C++. Have you implemented it in javascript? Note that your link says "all division is integer division", so y = 2015 - 9/10 should result in 2015, not 2014.1. — Martin Bonner supports Monica, Dec 22 '15 at 09:19
i didn't copied the code but just the calculation steps. why 2015, should be 2014 if we round 2014.1 off to floor? — Maven, Dec 22 '15 at 09:30
No. The calculation needs to go 2015 - 9/10 => 2015 - 0 => 2015. The division 9/10 rounds towards zero. — Martin Bonner supports Monica, Dec 22 '15 at 11:03

Martin Bonner supports Monica · Accepted Answer · 2015-12-22T09:25:49.767

Your fundamental problem is that the algorithm you use has a value of 1 around the start of 1AD. Julian dates start in 4713BC.

For your purposes that may not matter, (but it makes it harder to check online). Excel is good at comparing time points after the start of the 20th Century. If you actually want to convert to Julian date, you will need to search for the algorithm. For example: https://gist.github.com/jiffyclub/1294443 is suitable code in python - you will need to convert to your language of choice

Converting a date to its Julian number

1 Answers1