Get total number of leap seconds relative to gps time

I have a gps time stamp from a gps receiver, I'd like to convert that to GMT/UTC date and time.

GpsTime = 1667089162.11771

DayToSecs = 24*60*60;

TimeZero = datetime(1970,1,1, 'Format', 'd-MMM-y HH:mm:ss.SSS Z');

Tvec = GpsTime/DayToSecs + TdmsTimeZero;

Do I need to also adjust for leap seconds? If so, How do I return the total number of leap seconds from 'leapseconds' instead of the timetable?

Answer 1

[David Hill]

As long as the gps collection date is after 1 January 2017, then just subtract 18 seconds (37-19) see

https://hpiers.obspm.fr/eop-pc/index.php?index=TAI-UTC_tab&lang=en

Although your gps time seems to be in the future (not sure why).

GpsTime = 1667089162.11771;
utc=datetime(1980, 1, 6)+seconds(GpsTime-18);
utc=datetime(utc,'Format','yyyy-MM-dd HH:mm:ss.SSS')

utc = datetime
   2032-11-03 00:19:04.117

Answer 2

[James Tursa]

To convert a GPS timestamp to a UTC time, use the UTCLeapSeconds time zone which automatically accounts for leap seconds in the background calculations. E.g.,

GpsTime = 1667089162.11771;
GPS_epoch = datetime(1980,1,6,'TimeZone','UTCLeapSeconds')

GPS_epoch = datetime
   1980-01-06T00:00:00.000Z

GPS_epoch + seconds(GpsTime)

ans = datetime
   2032-11-03T00:19:04.117Z

My advise is to stick with the UTCLeapSeconds time zone exclusively when working with continuous time scales such as GPS, TT, TAI, etc.

(Same comment as David ... why is this in the future?)

If the timestamp in question is actually seconds from Modern UTC epoch and not GPS time as your post seems to allude to, then the calculations would be:

UtcTime = 1667089162.11771;
UTC_epoch = datetime(1970,1,1,'TimeZone','UTCLeapSeconds')

UTC_epoch = datetime
   1970-01-01T00:00:00.000Z

UTC_epoch + seconds(UtcTime)

ans = datetime
   2022-10-30T00:18:55.117Z

Now we get something reasonable for the year 2022, but still in the future compared to your post date, so this is still confusing.