Re: [math-fun] An end to leap seconds (was Re: SETI breakthrough: ...)
Dan Asimov <asimov@msri.org> wrote:
1. Suppose Earth is a perfect sphere and that each time zone is a perfect spherical lune of angle 2?/24 that takes 1 hour to traverse an hour of time.
I'm assuming the question mark was meant to be pi. Non-ASCII characters are turned into question marks on this list, at least in its digest form. This is presumably because there are multiple incompatible character sets which senders might be using, and the digest header can specify at most one of them.
Set a stopwatch to begin when it is first mm/dd/yyyy anywhere on Earth, and stop the watch when it is no longer mm/dd/yyyy (where mm/dd/yyyy is an arbitrary day of 24 hours' length).
QUESTION: How long did the stopwatch run for? (Mental math only.)
Twenty-five hours. The default International Date Line is exactly opposite the prime meridian, hence is in the middle of a time zone, not on the border between two of them. So some clocks are 12 hours ahead of Greenwich when others are 12 hours behind.
2. Same question, but with time zones as they actually are. (Any references and writing implements allowed.)
Based on the time zone map in the current World Almanac, 27 hours.
I'm assuming the question mark was meant to be pi. Non-ASCII characters are turned into question marks on this list, at least in its digest form. This is presumably because there are multiple incompatible character sets which senders might be using, and the digest header can specify at most one of them.
I'm not sure its intentional. Emacs turns non-ascii characters to question marks when you write out files, unless you set the variable coding-system-for-write correctly. Whit
Sorry about the delay — I was away from e-mail for a while.
On Apr 3, 2015, at 7:27 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote:
Dan Asimov <asimov@msri.org> wrote:
1. Suppose Earth is a perfect sphere and that each time zone is a perfect spherical lune of angle 2?/24 that takes 1 hour to traverse an hour of time.
I'm assuming the question mark was meant to be pi. Non-ASCII characters are turned into question marks on this list, at least in its digest form. This is presumably because there are multiple incompatible character sets which senders might be using, and the digest header can specify at most one of them.
Yes, I meant 2pi/24.
Set a stopwatch to begin when it is first mm/dd/yyyy anywhere on Earth, and stop the watch when it is no longer mm/dd/yyyy (where mm/dd/yyyy is an arbitrary day of 24 hours' length).***
QUESTION: How long did the stopwatch run for? (Mental math only.)
Twenty-five hours. The default International Date Line is exactly opposite the prime meridian, hence is in the middle of a time zone, not on the border between two of them. So some clocks are 12 hours ahead of Greenwich when others are 12 hours behind.
Nope.
2. Same question, but with time zones as they actually are. (Any references and writing implements allowed.)
Based on the time zone map in the current World Almanac, 27 hours.
I'm not sure of the exact answer, but this can't be right. --Dan ***Just to be clear: "mm/dd/yyyy is an arbitrary day of 24 hours' length" in each time zone.
On 05/04/2015 01:48, Dan Asimov wrote:
On Apr 3, 2015, at 7:27 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote: ... Twenty-five hours. The default International Date Line is exactly opposite the prime meridian, hence is in the middle of a time zone, not on the border between two of them. So some clocks are 12 hours ahead of Greenwich when others are 12 hours behind.
Nope.
Per Keith's observation, at any given instant the nominal times at different points on earth (in the Idealized Dan Asimov Model) are t-24h, t-23h, ..., t for some t. So, at some instant it first becomes midnight-plus-epsilon on day D somewhere. In terms of the previous paragraph, this must be time t. Exactly 24 hours later, the nominal times are t, ..., t+24h, so now in the "slowest" time zone it has just become midnight plus epsilon on day D. All the other time zones are ahead of that. Just under 24 hours later than that, in that time zone it is just about to be the end of day D. All the other time zones are ahead by at least an hour, so in those time zones it is no longer day D. Thus, 48 hours elapse between the first and last instants at which it's day D somewhere on earth. -- g
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
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
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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I agree that if the simplest model I intended is not used, we can get all kinds of different answers. The second part of my original question asked: What is the actual time that it is one actual day D at least somewhere on Earth? For this question, let's assume just the time zones and International Date Line as they currently exist on Earth. (With no mention of Daylight Saving, leap seconds, precession, or nutation, etc.) I do not know the answer to this question, but would like to know. --Dan
On Apr 7, 2015, at 3:57 AM, Seb Perez-D <sbprzd+mathfun@gmail.com> wrote:
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.
Apparently (http://en.wikipedia.org/wiki/Time_zone) there is a time zone that is UTC -12 (Baker Island and Howland Island, both uninhabited) and another that is UTC +14 (Line Islands, part of Kiribati). Wikipedia says that since the UTC-12 islands are uninhabited the time zone is unspecified but we can go with this time zone anyway. Kiribati does not use daylight savings time, and I would venture that neither do Baker and Howland Islands. Since the international date line goes through the middle of a timezone ( http://en.wikipedia.org/wiki/International_Date_Line#/media/File:Internation...), it would seem that the "real world" maximum length of day is thus 50 hours. Cheers, Seb On 7 April 2015 at 21:35, Dan Asimov <asimov@msri.org> wrote:
I agree that if the simplest model I intended is not used, we can get all kinds of different answers.
The second part of my original question asked: What is the actual time that it is one actual day D at least somewhere on Earth?
For this question, let's assume just the time zones and International Date Line as they currently exist on Earth. (With no mention of Daylight Saving, leap seconds, precession, or nutation, etc.)
I do not know the answer to this question, but would like to know.
--Dan
On Apr 7, 2015, at 3:57 AM, Seb Perez-D <sbprzd+mathfun@gmail.com> wrote:
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.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Yes. Start at the earliest time zone. It takes 24 hours to finish the day in question. It takes 1 more hour, respectively, for each of the succeeding 23 time zones to finish the same day. --Dan
On Apr 7, 2015, at 3:46 AM, 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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 07/04/2015 17:44, Dan Asimov wrote:
Yes. Start at the earliest time zone. It takes 24 hours to finish the day in question.
It takes 1 more hour, respectively, for each of the succeeding 23 time zones to finish the same day.
So you intended a model in which the date line is not opposite the prime meridian, but on some border between two of your 24 time-zones? Fair enough, but I do think that in that case the answer to my proposed solution should have been not "Nope." but something more like "That would be right if we put the date line in the middle of a time zone, but that isn't the model I intended." I wasted some time trying to figure out what was wrong with my reasoning, and it turns out nothing was wrong with it. -- g
Well, it assumed facts not in evidence. But the post-defensive me knows you are right. Sorry. --Dan
On Apr 7, 2015, at 7:31 PM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On 07/04/2015 17:44, Dan Asimov wrote:
Yes. Start at the earliest time zone. It takes 24 hours to finish the day in question.
It takes 1 more hour, respectively, for each of the succeeding 23 time zones to finish the same day.
So you intended a model in which the date line is not opposite the prime meridian, but on some border between two of your 24 time-zones?
Fair enough, but I do think that in that case the answer to my proposed solution should have been not "Nope." but something more like "That would be right if we put the date line in the middle of a time zone, but that isn't the model I intended." I wasted some time trying to figure out what was wrong with my reasoning, and it turns out nothing was wrong with it.
On 08/04/2015 04:06, Whitfield Diffie wrote:
So you intended a model in which the date line is not opposite the prime meridian, but on some border between two of your 24 time-zones?
I don't think the date line is even straight.
It's very far from straight, but Dan's model with 24 time zones perfectly spaced at pi/12 intervals seems to demand a straight date line. He didn't say explicitly where to put it; having read Keith Lynch's comments I thought putting it down the middle of a time zone (which is kinda-sorta where the real one is on average) would be (1) close to maximally simple, (2) morally consistent with reality, and (3) a wrinkle to make the puzzle tricky :-). (*Before* reading Keith Lynch's comments I had simply not thought about the date line, and therefore thought the answer was 47 hours. Which of course was the answer Dan had in mind.) I can totally see Dan's point, though; a date line that doesn't run along the boundary between time zones can be thought of as turning one time zone into two, so arguably saying "exactly 24 time zones spaced at pi/12 intervals" implicitly says that the date line is on a time zone boundary. -- g
One of the Ringworld stories uses the "long day" idea: The story opens with the main character celebrating his 200th birthday for about two days by skipping from time zone to time zone. The fine points are ignored. --Rich
participants (8)
-
Adam P. Goucher -
Dan Asimov -
Dan Asimov -
Gareth McCaughan -
Keith F. Lynch -
rcs＠xmission.com -
Seb Perez-D -
Whitfield Diffie