Covert binary 64-bit timestamp offset from the GPS epoch to python datetime object

Question

I'm trying to figure out what I think should be an 8byte/64-bit timestamp.

import datetime
GPS_EPOCH = datetime.datetime(1980, 1, 6)
t1 = "\x00\x00\xBF\x13\xDB\x79\xC0\x00" # expected: 2012-10-04 01:00:51.759
t2 = "\x00\x00\xC0\x13\xDB\x79\xC0\x00" # expected: 2012-10-04 01:00:51.760
t3 = "\x00\x00\xC2\x13\xDB\x79\xC0\x00" # expected: 2012-10-04 01:00:51.763
t4 = "\x00\x00\x80\xE7\xFB\x79\xC0\x00" # expected: 2012-10-04 01:45:40.960

I believe the value(s?) resulting from t1 and t2 should be offset from the GPS_EPOCH. However, I can't seem to get the result to match the expected result datetime.

I've been reading up and it seems logical that this would be split into 2 parts, with one perhaps being fractional and the other seconds (4 bytes each?). However, I haven't found any reference for timestamp formats that are based off the GPS epoch.

Any ideas how this could be transformed into the expected result?

Answer 1

I have it. You provided just enough examples.

>>> t1 = b"\x00\x00\xBF\x13\xDB\x79\xC0\x00" # expected: 2012-10-04 01:00:51.759
>>> import struct
>>> import datetime
>>> GPS_EPOCH = datetime.datetime(1980, 1, 6)
>>> t1_unpacked = struct.unpack('<q', t1)[0]
>>> t1_seconds = t1_unpacked // 52428800
>>> t1_us = int(round((t1_unpacked % 52428800) / 52.428800, 0))
>>> GPS_EPOCH + datetime.timedelta(seconds=t1_seconds, microseconds=t1_us)
datetime.datetime(2012, 10, 4, 1, 0, 51, 758750)

Putting it all together:

def gps_time(timestamp):
    unpacked = struct.unpack('<q', timestamp)[0]
    seconds = unpacked // 52428800
    microseconds = int(round((unpacked % 52428800) / 52.428800, 0))
    return GPS_EPOCH + datetime.timedelta(seconds=seconds, microseconds=microseconds)

>>> gps_time(t2)
datetime.datetime(2012, 10, 4, 1, 0, 51, 760000)
>>> gps_time(t3)
datetime.datetime(2012, 10, 4, 1, 0, 51, 762500)
>>> gps_time(t4)
datetime.datetime(2012, 10, 4, 1, 45, 40, 960000)