Re: Yes, I'm back

En réponse à Peter Clark <peter-clark@...>:

>         No, nothing interesting happened in the year 0 AD. In fact,
> positively
> nothing happened in that year, since there never was such a year.[*] 1
> BC
> rolls over to 1 AD. And no, to forstall the next obvious question,
> nothing
> particularly note-worthy happened in either of those years, either. (I
> now
> await some historian buff to correct me. :) This is why purists argue
> that
> 2001, not 2000, was the start of the third millenium.
Yep. When the calendar was devised, there was no concept of zero in Europe yet,
and anyway years were counted using *ordinal* numbers, which start at "first".
If you think of "1 BC" and "1 AD" as "first year Before Christ" and "first year
Anno Domini" (with the birth of Christ always taken as starting point), it
becomes obvious that there can be no 0th year in between.

>         We have to thank a monk by the name of Dionysius Exiguus (aka
> Dennis the
> Little), who had the task of figuring out the Easter cycles. In 525
> AD,
> unhappy that the current system of numbering years counted from the
> reign of
> Diocletian, who had persecuted Christians during his reign, decided that
> the
> Church needed a new system, and so (naturally) based it on his
> calculation of
> Christ's birth. Unfortunately, we're a little foggy on why he chose 25
> December 753 AUC[**] (ab urbe condita, i.e. since the founding of
> Rome);
> there are a couple of theories, but I'm not aware of any that have
> been
> decisively proven. In any case, the long and short of it is that Dennis
> the
> Little goofed, and now it's too late to do anything about it.
>
I think he did some calculations by adding the reigns of all the Popes
together, but goofed somewhere (not difficult with all those Roman
numbers ;)))) ).

> [**] The Romans, it should be noted, weren't exactly sure when Rome
> was
> founded, so the margin of error is usually given +-1. Sloppy Roman
> date-keeping.
>
Well, the foundator of Rome had been raised by a wolf. I doubt he learned date-
keeping from her ;))) .

>         Unfortunately, the first half of the problem is unsolvable. In
> theory, almost
> any period is possible for a "day." For instance, a "day" on Mercury is
> 176
> earth days, while a Mercurian year is only 88 earth days, or half the
> day
> length.
Do you mean "day" as "noon to noon", or "complete rotation"? Those are two
different things. Well, just checked, and you meant "noon to noon". Mercury's
actual complete rotation lasts nearly 59 days. But coupled with the rotation of
Mercury around the sun, you get a "noon to noon" days of 176 days :)) (check
out http://www.solarviews.com/eng/mercury.htm). On the other hand, Venus has a
rotational day of 243 Earth days, compared to a year of 225 Earth days. But
since Venus's rotation is retrograde (another funny thing to do if you want
funny rotations. Other possibilities are having tilted planets like Uranus and
Pluto, but I'm not sure if intelligent life could appear on such a planet. If
so, they will have a rather different idea of what day and night and poles
are ;))) ), its solar day lasts "only" 117 days (with a sun rising west and
setting east ;)) ), a little more than half its year. There's a similar effect
on Earth, but since Earth's rotation is direct, its solar day is a bit longer
than its rotation period. Earth's rotation period is 23h56m4s, but its actual
solar day (noon-to-noon period) is 24h00m0.75s, which explains why we don't
have to resynchronise our 24h watches every single day ;))))) .

The calculation of the duration of the solar day when you have the rotational
period and the year length is quite easy, if you're interested (although with
elementary arithmetics you can actually find it yourself I think ;)) ).

>         The second is a "simple" matter of physics. If you assume that
> the star has
> about the same mass as the sun, you can apply Kepler's third law to
> its
> distance from the sun and base your calculation off of Earth's orbit.
> However, if the mass of the star in question is different, then you'll
> need
> some slightly more complicated math.
>
Why? Kepler's third law *includes* the Sun's mass in its formula. Replace it
with the star's mass and you have it.

Christophe.

http://rainbow.conlang.free.fr

Take your life as a movie: do not let anybody else play the leading role.

Reply