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Saturn Time |
From: "Shaun Moss" <shaun@earthmultimedia.com>
To: <steve_heaton@ozemail.com.au>
Cc: <marstime@yahoogroups.com>, <time-sig@chapters.marssociety.org>
Subject: [Mars_Time-sig] Titan calendar
Date: Tue, 29 Apr 2003 12:24:36 +1000
[ to Mars Society Time list, from "Shaun Moss" <shaun@earthmultimedia.com> ]
Hey Steve
Well, you've inspired me to start thinking about it!
My initial thoughts are that the major astronomical cycle on Titan will be the solar day, equal to both the period of it's orbit around Saturn and it's rotational period since Titan is tidally locked. This is almost 16 days (15.94542). One way of defining an artificial day (let's call it a T-day) could be to make it one fourteenth of this orbit, which would work well with 7-day weeks - one orbit of Saturn would then equal 2 weeks which is nice and neat. However it's probably more important for the T-day to be as close to 24 hours as possible, so it might make more sense to define a T-day as one sixteenth of an orbit, which results in 23.9 hour days.
Weeks could still be 7 days long, which would match Earth, Mars and maybe Jupiter, in which case they would be out-of-synch with the orbital cycle, which probably isn't a major problem. Alternatively they could be 8 days long so that 1 orbit would equal exactly 2 weeks. I suppose an artificial month could be defined as two orbits (32 days or 4 T-weeks) and an artificial year could be defined as 24 orbits (12 T-months or 384 T-days, about 382.7 Earth days). It would be hard to reconcile with Saturn's orbit around the sun, which is 29.5 years long, but as this cycle would have little effect on Titan (Saturn's orbit is essentially circular) it probably wouldn't be necessary.
Shaun
www.earthmultimedia.com
www.marsengineering.com
Date: Tue, 29 Apr 2003 09:28:07 -0700 (PDT)
From: Thomas Gangale <marcus@martiana.org>
To: <marstime@yahoogroups.com>, <time-sig@chapters.marssociety.org>
Subject: Re: [marstime] Titan calendar
Kaor Shaun!
I concur with your analysis. While there's no obvious reason to depart from the seven-day week on Mars, you make a good case for an eight-day week on Titan. This jibes with what I came up with for the Galileans (see "The Calendars of Jupiter"). Also, Saturn's orbital period is essentially irrelevant in terms of human social timekeeping needs. It's just too long. I came to the same conclusion about the Jovian year. This leads one to the idea that a Titan calendar would be based either on the terrestrial year, or possibly the Martian year if Martian culture turns out to be the predominate influence during the settling of Titan. In terms of Titan days (one-sixteenth of a rev), the year would be:
Terrestrial year: 365.2424 * 16 / 15.94542 = 366.4926
Martian year: 686.9726 *16 / 15.94542 = 689.3241
Since in this case the week has physical meaning (one half of a Titan rev), months and years should contain an integral number of them if at all possible. A 12-month terrestrial Titan calendar could contain ten 32-day months and two 24-day months, totalling 368 days. An eight-day week would need to be subtracted periodically, so the short months could be the 4th, 8th, and sometimes the 12th. A 22-month Martian Titan calendar could contain twenty 32-day months and two 24-day months, totalling 688 days. An eight-day week would need to be added periodically, so the short months could be the 11th and sometimes the 22nd. A Darian Titan calendar (24 months) might not be as convenient: fourteen 32-day months and ten 24-day months, with the short months being the 2nd, 4th, 6th, 8th, 12th, 14th, 16, 18th, 20th, and sometimes the 24th.
Ad Martem!
Tom
From: "Shaun Moss" <shaun@earthmultimedia.com>
To: <time-sig@chapters.marssociety.org>
Cc: <marstime@yahoogroups.com>
Subject: [Mars_Time-sig] RE: [marstime] Re: Titan calendar
Date: Wed, 30 Apr 2003 10:08:01 +1000
There could be a solution there using the number 26. Let's say we had a T-year equal to 26 revs (52 weeks @ 2 weeks/rev) - this would be about 414.58 days - longer than a year but shorter than a mir. Then we find that one orbit of Saturn is approximately 26 T-years.
Saturn's orbit (S-year) is 29.458 years long. I've assumed (perhaps wrongly) that this figure refers to tropical years of 365.242199 days each, which means an S-year equals about 10759.3 days. This is equal to 674.7583 revolutions of Titan. 26 T-years x 26 revs = 676 revs is very close to this. Making it line up neatly would require an intercalation scheme which dropped 1.2417 revs on average from the 26x26 cycle.
One such scheme could be:
1. A regular calendar S-year with 25 T-years of 26 revs each, plus one short T-year of 25 revs. This would give a calendar S-year of 675 revs.
2. Every fourth S-year an extra rev is dropped at the end of the 13th T-year, i.e. 12 T-years of 26 revs, followed by one of 25 revs, followed by another 12 T-years of 26 revs, finishing with one of 25 revs. This would mean every fourth S-year would have 674 revs.
3. Each 100th S-year, the extra rev is not dropped from the 13th T-year, meaning a regular S-year of 675 revs.
With these 3 rules the average S-year would be equal to 674.76 revs, which is very close. The calendar would only be off by 0.17 revs (1.36 T-days) after 100 S-years (2946 Earth years). (That's if I got the length of the S-year right!)
Interesting that this intercalation scheme is very similar to Earth's except that revs are being dropped rather than leap days being added!
The problem with this method of synching revs with S-years is the old leap-baby problem. It means that everyone born in the final week of a normal 26-rev T-year does not have a birthday in short T-years. (Admittedly this would probably only happen every 26 T-years.) 16 dates would be missing from the calendar in those years. It would be like Earth years being sometimes 52 and sometimes 53 weeks - everyone born in the 53rd week would not have a birthday in 52-week years. The way to minimize the problem is to do what we've done in the Gregorian calendar, which is to not synch weeks and years and to have an intercalation scheme that inserts days rather than weeks (or revs).
So, if a T-day is 1/16 of a rev, this means there are 674.7583 * 16 = 10796.1328 T-days in an S-year. If we keep the idea of 26 T-years per S-year, this gives an average of about 415.236 T-days per T-year (which is 25.95 revs). An intercalation scheme could be:
1. Regular T-years of 415 T-days each.
2. Every fourth T-year is a leap T-year of 416 days, giving an average of 415.25.
3. Every 100th T-year is not a leap T-year, giving an average of 415.24
This scheme is not based on the S-year, so S-years would have an irregular pattern varying numbers of T-days depending how many leap T-years fell within the 26 T-year cycle. The main problem here is dates. With weeks not synched to the T-years, how do we specify a certain date? Without months, the only way would be by using a single number from 1..416. This actually lends weight to the first system - with revs and T-years synched, we could write dates as rev-day where rev would be from 1..26 (or perhaps A-Z) and day would be from 1..16. In light of this issue the problem of the leap-babies (or Z-babies) is not so severe - in the infrequent short T-years they could celebrate their birthdays on New Year's Eve.
Using the first system, dates could be written as follows: SY-AA-DD where:
SY is the current orbit of Saturn (S-year). This would only change every 29.5 years.
AA specifies the rev within the 26x26 scheme. The first 'A' refers to the T-year as A..Z, and the second 'A' refers to the rev within the T-year as A..Z.
DD would be the T-day within the rev, from 1..16.
So a typical date might be written as: 2-KF-12
Shaun
Date: Wed, 30 Apr 2003 09:54:22 -0700 (PDT)
From: Thomas Gangale <marcus@martiana.org>
To: <marstime@yahoogroups.com>, <time-sig@chapters.marssociety.org>, <shaun@earthmultimedia.com>
Subject: Titan calendar
Kaor Shaun!
A very elegant solution! Good on you, mate! However, we need to get the constants right and tweak the system a bit. According to the following sources:
http://solarsystem.nasa.gov/features/planets/saturn/saturn.html
http://solarsystem.nasa.gov/features/planets/saturn/titan.html
Saturn's sidereal orbital period is 29.447498 terrestrial sidereal years.
Titan's (presumably sidereal) orbital period is 15.94542068 days.
A terrestrial sidereal year is 1.0000174 Julian years:
1.0000174 * 365.25 = 365.25646 days
Saturn's sidereal year, expressed in days:
29.447498 * 365.25646 = 10755.886 days
One question we need to ask before we proceed is whether we want to base a Titan calendar on the Saturn sidereal year. On Earth, (and presumably on Mars) the purpose of most calendars (the Islamic calendar being a notable exception) is to keep civic and cultural life in synch with the seasons. Thus the accuracy of terrestrial calendars is usually measured against the tropical year (which averages the precession of all equinoxes and solstices) or one particular equinox or solstice year. In "Marking Time," Duncan Steel asserts that the purpose of the Gregorian reform was to reference the calendar to the vernal equinox, the basis for calculating Easter, therefore the proper benchmark is the vernal equinox year of 365.2424 days, not the tropical year of 365.2422 days, thus the Gregorian calendar year of 365.2425 is more accurate than most commentators give it credit (theoretically, it would only be off by one day in 10,000 years, but the length of the vernal equinox year will have changed by then!). Extending this logic, a Martian calendar beginning its year on the vernal equinox would properly be referenced to the Martian vernal equinox year of 668.5906 sols. So, on what type of Saturn year should we base out Titan calendar? Sidereal? Tropical? Vernal equinox?
According to:
Saturn's spin axis precession of 1.8*10e6 years. Thus the tropical year would be:
10755.886 * (1 - (1 / 1.8*10e6)) = 10755.880 days
I believe the deviation of the vernal equinox year from the tropical year is influenced by the precession of the apsides and the change in the ellipticity of Saturn's orbit. This deviation is probably small, so using the tropical year may be good enough.
Titan revs in a Saturn tropical year:
10755.880 / 15.94542068 = 674.54350
Note that this is the number of Titan sidereal revolutions in a Saturn tropical year. The number of Titan solar days in a Saturn tropical year is exactly one less than this, or 673.54350 days. The length of the Titan solar day is:
10755.880 / 673.54350 = 15.969095
From these numbers you can rework the intercalation scheme.
Dropping a Titan solar day from the calendar year periodically is not an insurmountable cultural problem. Consider the tens of millions of Asians who are born during the 13th month of an embolismic year on the Chinese lunisolar calendar, but who have no birthdays in 12-month years, which are roughly 12 out of every 19 years. They get by.
Ad Martem!
Tom
From: "Shaun Moss" <shaun@earthmultimedia.com>
To: <marstime@yahoogroups.com>, <time-sig@chapters.marssociety.org>
Cc: "Tom Gangale" <marcus@martiana.org>
Subject: RE: [marstime] Titan calendar
Date: Thu, 1 May 2003 10:11:41 +1000
Thanks for the new numbers, Tom! I still have a lot to learn about astronomy. The only thing I don't get is how the "The number of Titan solar days in a Saturn tropical year is exactly one less than this, or 673.54350 days." Is it just a coincidence that it's exactly one less? Or is there an astronomical reasons for this?
Ok now we have the new numbers, I want to try them on a slightly different pattern, with 24 revs per T-year and 28 T-years per S-year. I think this is a more useful system. For one thing, the more significant cycle will be the shorter T-year, and by having it comprised of 24 revs, we have a very familiar number of divisions. We can then easily divide the T-year by 2, 3, 4, or 6, just as we frequently do on Earth and will probably also do on Mars. The number 26 is not as useful in this way. 24 revs at 15.969095 days per rev comes to 383.25828 days (384 T-days), not much longer than a terrestrial year. (Fortunate that Titan's solar day is so close to a useful integer number of days. With 16 T-days per rev, a T-day would thus be 23.9536425 hours long - or 23:57:13.113. With 24 T-hours per T-day, a T-hour would therefore be only 7 seconds shorter than a regular hour.)
The number 28 is also more useful than 26 in terms of the S-year, because it then allows us to divide the S-year by 4 for the purpose of marking Saturn's seasons (imagine 7-year-long seasons!!). Since Titan is affected by Saturn's seasons this will probably be of interest to the colonists. Every 7th T-year will be a seasonal marker and perhaps an excuse for a bigger New Year's Eve party than usual.
28 T-years is then 672 revs, which is fairly close to the desired average of 673.54350. An intercalation scheme is required to add an average of 1.5435 revs per S-year. Let's say:
1. A regular S-year is made up of 27 regular T-years of 24 revs each, followed by a leap T-year of 25 revs. This comes to 673 revs.
2. Every second (odd) S-year is a leap S-year with 674 revs. The extra leap T-year can be the 14th T-year in the 28. A leap S-year would therefore be 2 x (13 regular T-years of 24 revs + 1 leap T-year of 25 revs). This rule brings the average to 673.5 revs/S-year.
3. If an S-year is divisible by 20 but not 100 then it's also a leap S-year. This adds an extra 4 leap S-years per 100 which brings the average S-year length to 673.54 revs.
4. A very similar rule - if an S-year is divisible by 200 but not 1000 then it too is a leap S-year. This adds an extra 4 leap S-years per 1000 which brings the average to 673.544 revs, very close to what we want.
To summarize - an S-year is a leap S-year of 674 revs if it ends in 1, 3, 5, 7, 9, 20, 40, 60, 80, 200, 400, 600, or 800. This pattern gives 544 leap S-years per 1000, equal to 1544 leap T-years per 28000.
I think these are enough intercalation rules for the moment. With our current numbers this would mean an error of just 0.05 revs (19 hours) per 100 S-years (2945 years).
It may seem that in leap T-years of 25 revs we lose the benefit of having 24 revs per T-year, as 25 does not divide evenly by anything but 5. As leap T-years occur relatively infrequently (1544/28000 = about 11/200), perhaps we could just treat the extra rev as a bonus or holiday period outside the regular yearly pattern and divide up the other 24 revs as per usual.
Shaun
Date: Thu, 1 May 2003 08:47:37 -0700 (PDT)
From: Thomas Gangale <marcus@martiana.org>
To: <marstime@yahoogroups.com>, <time-sig@chapters.marssociety.org>, <shaun@earthmultimedia.com>
Subject: [Mars_Time-sig] Sidereal Day vs. Solar Day
[ to Mars Society Time list, from Thomas Gangale <marcus@martiana.org> ]
Shaun wrote:
The only thing I don't get is how the "The number of Titan solar days in a Saturn tropical year is exactly one less than this, or 673.54350 days." Is it just a coincidence that it's exactly one less? Or is there an astronomical reasons for this?
-----
The easiest way to explain this is in terrestrial terms. A sidereal day is 23 hours 56 minutes. A solar day is 24 hours. What is the difference? A sidereal day is the time it takes Earth to complete one rotation referenced to a fixed star. The fixed star is at a distance that approaches infinity in relation to the size of Earth's orbit around the sun, so the direction from Earth to the fixed star can be assumed to be constant. As Earth completes one rotation, it also travels a little less than one degree in its orbit around the sun, and the direction from Earth to the sun changes by the same angle. Thus, "from sun to sun" takes Earth a bit more time to rotate through this additional one degree of arc, 1/365 of a day, or about four minutes. This is the solar day of 24 hours. In the course of a year, the diverging angle between the direction from Earth to the fixed star and the direction from Earth to the sun accumulates to 360 degrees, or one entire rotation, and those daily four-minute increments sum to one full day. So there is exactly one less solar day than there are sidereal days in a year. There are 366.2564 sidereal days in a sidereal year. The number of solar days in a sidereal year is exactly one less than this, or 365.2564.
Since the Martian year is nearly twice as long as Earth's, but the Martian day is only a trifle longer that a terrestrial day, the difference between the sidereal day and the solar day is only about half: a bit more than half a degree of arc, or about two minutes. Again, there is one less solar day in the year than there are sidereal days, because the daily half-degree increments sum to 360 degrees in the course of a Martian year.
A number of unwary would-be Martian calendar designers have been tripped up by the difference between the Martian sidereal day and the Martian solar day (sol). They looked in an astronomy book, saw that the rotational period of Mars is 24 hours 37 minutes, and came up with calendars based on 669.6 Martian sidereal days instead of 668.6 sols. Absolutely wrong! There are some really bad Martian calendars out there, some authored by respected scientists who should have known better but were careless. This isn't rocket science (and as an aerospace engineer I can say that with some authority). The math is easy, but people do need to understand the concepts behind the numbers and think through all of the necessary steps in logic.
Ad Martem!
Tom
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Saturn Time |