Will converting the difference between 2 datenums to seconds contain leap seconds or not?
5 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
Carl Finney
il 21 Lug 2015
Commentato: Peter Perkins
il 23 Lug 2015
I have a GPS logger that provides a .CSV record of location and speed for every second in the format 'mm/dd/yyyy,HH:MM:SS AM'. Assuming the date and time are in GPS time rather than UTC, when I take the datenum difference between this and the GPS epoch will the leap seconds be included or not?
GPSsecond = 24*60*60*(datenum( strTime, 'mm/dd/yyyy,HH:MM:SS AM' ) - datenum( 1980, 1, 6, 0, 0, 0 ));
Should I add the (current) 17 leap seconds or not?
0 Commenti
Risposta accettata
Brendan Hamm
il 21 Lug 2015
datenum does not handle leap seconds. datetime variables on the other hand can handle this (requires you have 2014b or later) provided you specify the appropriate TimeZone .
2 Commenti
Peter Perkins
il 23 Lug 2015
Just to be clear, datenum does not support leap seconds, but then again GPS time doesn't have leap seconds. If all the arithmetic you're doing is "in" GPS time, and you're never converting to UTC, and not getting the current system time, and not using timestamps from anything other than GPS clocks, then I think datenum actually does what you want. (If you have R2014b or later, so would "unzoned" datetimes, and more elegantly.) It's all a question of what interpretation you put on the values.
If your goal is converting to UTC or some civil time zone, then that's a different story. You'd have to account for the lack of leap seconds in GPS time.
datetime (R2014b or later) provides a version of UTC that completely ignores the existence of leap seconds (lots of people want to act as if leap seconds don't exist), as well as a version that accounts for leap seconds.
Più risposte (2)
the cyclist
il 21 Lug 2015
Modificato: the cyclist
il 21 Lug 2015
An easy test to show that datenum does not handle this is
ds1 = '07/01/1972,00:00:00 AM';
ds2 = '06/29/1972,00:00:00 AM';
GPSsecond = 24*60*60*(datenum( ds1, 'mm/dd/yyyy,HH:MM:SS AM' ) - datenum( ds2, 'mm/dd/yyyy,HH:MM:SS AM' )) - 172800
Note that according to the Wikipedia page on leap seconds, one was inserted on June 30, 1972. So, the above test would have given 1 instead of 0 if the leap second had been in there.
5 Commenti
Steven Lord
il 22 Lug 2015
It does, as described in this Bug Report for release R2014b. Technically you could quibble about whether this is a bug, since the leap second was announced in January 2015 which was after release R2014b shipped, but I think the bug report exists mainly so we could attach a patch.
Vedere anche
Categorie
Scopri di più su Time Series Objects in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!