Hmmm, am I wrong in understanding that, by pretending leap seconds don't exist, the unix time actually represents how many seconds passed rather than what time time is now? (Where a second is as long as every other second, not smearing out the leap second across the whole year. I.e. by the cesium standard definition.)

    user@debian:~$ qalc "$(date -d 2022-01-01T00:00:00+0000 +%s)"/3600/24/365.25
    ((1640995200 / 3600) / 24) / 365.25 = 52

Precisely 52 years' worth of seconds, which I guess also tells me that there have been an integer number of leap years since then. Either way, at 1666666666, 1666666666 seconds will have passed rather than ...693 no? The extra 27 seconds are either for keeping the sun at its zenith when the clock strikes noon (not that that is even close to true where I live, it's off by more than an hour) or for keeping winter in December. I can never remember which, although we already have leap days for keeping December in winter (in this hemisphere) so I think it must be the 24-hour variable rotation speed thing.

If you want to define 1666666666 as having +27 seconds passed since 1970 started, you would have to redefine the length of the second every time a leap second is announced, I think.

I find this all tough to wrap my head around, though, so it can very well be that I'm wrong here.