If you start in the first time-zone to be yyyy/mm/dd, at 0:00, and travel west at the same speed as the sun, you'll get to the last time-zone in 23 hours, with a local time of still 0:00. You can then spend 24 hours in that last time zone and will stop the chronometer with 47 hours, which agrees with Adam's answer. However, if you choose your day carefully and time it with the move off daylight savings time, you can hope to get one hour more. If DST is not allowed, then on a specific day we might win one UTC leap second (in line with the subject of this email), bringing the total to 47 hours and 1 second. Cheers, Sébastien On 7 April 2015 at 12:46, Adam P. Goucher <apgoucher@gmx.com> wrote:
I presumed that the answer would be 47 hours, since there are 24 time zones (naively -- I'm disregarding those annoying half-hour time zones) so we want to take the sum-set:
{1, 2, 3, ..., 24} + [0, 24) = [1, 48)
which has measure 47.
Sincerely,
Adam P. Goucher
Sent: Sunday, April 05, 2015 at 3:07 AM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] An end to leap seconds (was Re: SETI breakthrough: ...)
On Apr 4, 2015, at 6:42 PM, Gareth McCaughan < gareth.mccaughan@pobox.com> wrote:
(in the Idealized ... Model) . . . . . . Thus, 48 hours elapse between the first and last instants at which it's day D somewhere on earth.
Nope.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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