As noted on the preceding page, dividing the Martian day, only slightly longer than Earth's day, into hours, minutes, and seconds, produces a second different in length from that used on Earth. Since the Earth second is intimately connected with units of energy, or units used in electronics, it would seem that the Earth second has to be retained.
Also, as noted on the preceding page, from this site, we learn that a Martian day is 88775.260726 seconds long, which is 24.659794646 hours, or 24 hours, 39 minutes, and 35.260726 seconds.
One simple thing to do would be to call the Martian second by another name.
The ratio between the Martian day and the Earth day is very close to 299 to 291, producing a Martian day that is .003 seconds too short.
Could an even closer approximation be obtained by using the product of two ratios of small integers?
For example, if a Martian minute were divided into sixty-one smaller units, rather than sixty, then 94 of those units would be very close to 95 Earth seconds. This wouldn't be quite as accurate, though, as the preceding approximation. However, for this subdivision, another good approximation becomes available. The Martian day would be only .0002 seconds too short if 10,613 of those units were set equal to 10,726 Earth seconds.
There are other possible approximations if the Martian minute were divided into sixty-two units or seventy-seven units, but they would not be quite as close.
Dividing the Martian minute into sixty-five units, equal to 92/97ths of an Earth second leads to an accuracy of one part in thirty million. This makes the day .003 seconds too long. Two or three leap seconds in a Martian year might well be acceptable.
However, using a ratio like this does not really conveniently define Martian time units in terms of the Earth second. An acceptable approximation of this form is also available. The Martian day would be about .00055 seconds too long if the Martian minute were divided into forty-three units, each one defined as 1.433709 Earth seconds. This would still lead to a requirement for a leap-second only once every few years, which is a level of accuracy corresponding to that of terrestrial time-keeping (since not only is the second now defined in terms of an atomic standard instead of the Earth's rotation, but that atomic standard is defined in terms of the length of the solar day in 1900).
So far, this doesn't look encouraging. A simple integer ratio between the Martian time units and the Earth second involving five digit numbers, or a Martian time unit that can be rounded off to the nearest microsecond. Are there other possibilities?
A Martian minute is 61.649486615 seconds long.
Electricity in the U.S. is based on a 60 Hz frequency; if we call 1/60th of a second a "third" (as is done in angular measure), one gets 3698.9691969 thirds in a Martian second. Rounding that off to 3699 or 3698.97 (which leads to a Martian day .02 seconds too long) are still too approximate, so this doesn't seem to lead anywhere.
Some digital watches use quartz crystals tuned to 32.768 kHz, so that seconds can be produced by simple binary division. A Martian minute would consume 2020130.3774 periods of that oscillation. Dropping the part after the decimal point leads to a Martian day about .015 seconds too short, again too great a discrepancy.
The second is defined as 9,192,631,770 periods of an oscillation produced by the cesium atom. (It was the meter, formerly defined in terms of a wavelength of light produced by krypton atoms, that got changed to be defined in terms of the speed of light.) A Martian minute would be 566,721,029,264 of those oscillations, and that involves one digit of precision more than the accuracy to which the length of the Martian day is known actually provides. So does any integer from about 566,721,029,255 to 566,721,029,273 come with a nice small factor?
Searching for a few possibilities turns up:
13,179,558,820 * 43 = 566,721,029,260 33,336,531,133 * 17 = 566,721,029,261 35,420,064,329 * 16 = 566,721,029,264 21,796,962,664 * 26 = 566,721,029,264 37,781,401,951 * 15 = 566,721,029,265 9,605,441,174 * 59 = 566,721,029,266 29,827,422,593 * 19 = 566,721,029,267 15,742,250,813 * 36 = 566,721,029,268
so one possibility would be to divide the Martian minute into 52 units, each defined as 10,898,481,332 of the same cesium atom oscillations as are used to define the Earth second.
To establish even greater harmony between the Earth second and the Martian time unit, though, we may also note that 9,192,631,770 is 10 times 3 times 49 times 47 times 44,351. Can we use this in any way? The presence of a large factor does make it unlikely that any of the possible values for a subdivision of the Martian minute will stand in a simple integer ratio to the Earth second in addition to being an integer number of cesium oscillations.
As it happens, 12,778,089 times 44,351 is 566,721,025,239. So we are very lucky. Are we quite lucky enough?
That is: an Earth second is 2 * 3 * 3 * 5 * 7 * 7 * 47 times a base unit of 44,351 oscillations. A Martian minute, using the slightly shortened value noted here, is 3 * 19 * 224177 times that same base unit.
So, if we divide the Martian minute into 57 units of 9,942,474,127 cesium oscillations, which would also make the units equal to
224,177 ------- seconds, 207,270
the resulting length of the Martian day is 88775.260096 seconds. This is indeed close enough.
Since 44,351 is a smaller prime integer than 224,177, however, by accepting its presence in the ratio, we might possibly obtain an even simpler ratio than this.
For example, we can start with the fact that
37,781,401,951 * 15 = 566,721,029,265
Now, 803,859,616 times 47 is 37,781,401,952, a very acceptable inaccuracy. We can divide this by 14, to get 57,418,544.
So we have a Martian minute that is 57,418,544 cesium cycles times 2 * 3 * 5 * 7 * 47 at this point.
However, 21 * 43,351 is still four times larger than the factors in our previous ratio. To get around that, we would have to bump the value of the Martian minute by one part in five million.
If we did so, then the Martian minute would be 2,734,216 times 207,270 cesium cycles, which would be 566,720,950,320 cycles. This would not be adequately accurate.
Dividing the Martian minute into 23 units, each 6173/2303 of a second long, and also 24,640,085,070 cesium oscillations in length, leads to a Martian day of 88775.406 seconds, which is not much of an improvement on just using Earth seconds.
Another possibility for improving on the first two attempts is to recall that some watches with sweep-second hands on Earth divide the 60 spaces indicating the minute and second into a further five divisions. This is what made me feel that 200 divisions, but not 2000 or 8000, on a watch for the Burroughs system would be reasonable. Since one purpose of dividing the Martian minute into a number of units other than 60 is to avoid confusion with the second, perhaps looking for smaller divisions of the minute would provide possibilities.
Divide the Martian minute into 113 units, of 5/11ths of an Earth second? Not quite good enough, although accurate to one part in 5,000. Nor is dividing the Martian minute into 137 units of 9/20ths of a second, or 185 units of 1/3rd of a second.
A very accurate Martian day of 88775.26075 seconds is given by dividing the minute into 267 units of 6814/29511ths of an Earth second, but that is not a particularly simple ratio.
Dividing the Martian minute into 92 units of 65/97ths of an Earth second? This produces the same result as 65 units of 92/97ths of an Earth second. However, 92 is 4 times 23, and it is probably of more use to be able to divide a Martian minute into four equal parts than to be able to divide it into five equal parts.
Wait a moment. Why not divide the Martian minute into 20 parts of 299/97ths of an Earth second? That means we're dealing with the basic 299:291 ratio between Earth and Martian time, and could just use a stretched Martian second with the same accuracy.
However, the change from 97 to 291 was done by multiplying by 3, so while dividing the Martian minute into 20 parts corresponds naturally to the ratio between Earth time and Mars time, going to 60 parts isn't really forced by the desire for optimality. Perhaps there is a way to avoid that dangerous stretched second yet.
The Martian minute could be divided into 100 parts, each one 299/485ths of an Earth second in length. Dividing the Martian minute into 100 parts, although not of that precise length (and therefore not for all the same reasons), has been proposed as well by William Woods, and that is listed on this page of the site noted before on different Martian time systems.
Another possibility is to actually use the Earth second directly as a part of Martian timekeeping. With a day that is 88775.260726 seconds long, this would mean, at the very least, accepting a leap second a bit more often than once every four days.
If that is acceptable, 88775 can clearly be divided evenly by 25; so one can divide a Martian day into 25 "hours". The quotient, 3551, is, fortunately, not a prime number. Instead, 3551 equals 53 times 67. So one could have 25 Martian "hours", each divided into 53 Martian "minutes", each divided into 67 seconds. What one would call the Martian hour and minute would be an interesting question. Also, of course, unlike using 24 hours of 60 minutes each on Mars, this changes the units of Right Ascension - but since the Martian North Pole is not pointed in exactly the same direction as Earth's North Pole, one can't use star maps from Earth naïvely upon Mars in any case.
However, somebody else already thought of this one first.
That is one page on this site, which lists a number of proposed Martian time systems.
Dividing the Martian minute into 43 units allowed those units to be an integer number of microseconds, 1.433709 seconds. Using the same technique of restructuring the whole day in order to use the second, since a day that is off by .000726 seconds is acceptable (because a leap-second once a year is comparable to Earthly timekeeping), how can I carve up a day of 88775.26 seconds in order to use a unit that consists of an integer number of hundredths of a second?
That's 18 times 493.07. Unfortunately, 49307 is a prime number.
If I move down a step, though, I can divide the Martian day into 180 units that are an integer number of milliseconds in length. 24 hours of 60 minutes would therefore only require me to divide the millisecond into eight parts. But each minute would then be 6160 7/8 milliseconds long, and would not be susceptible to a further even subdivision.
Even so, a Martian minute that is thus tied to the millisecond might be acceptable. And since one has divided the millisecond into 8 parts, to return to decimal, I suppose the minute should be divided into 25 (or possibly 125) units if this is the basis used. That would be 25 units, each 2.46435 seconds in length. That is slightly better than an integer number of microseconds, but only slightly.
Allowing a limited flexibility in restructuring the Martian time system, can a benefit be derived from dividing the Martian hour (but still having 24 of them in a day) into 48, 72, 84, or 96 parts instead of 60 minutes?
If one has 84 'minutes' in an hour, dividing them into 92 units of 91/97ths of a second, or into 182 units of 46/97ths of a second is simply encountering the 299:291 ratio again, which turns up quite often.
If one has 96 'minutes' in an hour, dividing them into 276 units of 104/291sts of a second again bumps into that ratio.
How about accepting even more inaccuracy, and considering a day of 88775.25 seconds? That is still better than 88775 seconds, and yet it is quite reasonable. Do 355,101 quarter-seconds in a day divide up into any convenient factors?
355,101 is 3 times 41 times 2,887.
Thus, one could divide a day into 41 hours, each divided into 2,887 units consisting of three-quarters of a second. Unfortunately, 2,887 is a prime number.
The next approximation would be to view the Martian day as 88775 and 6/23 seconds.
What factors does 2,041,831 have? Well, it's a multiple of 11. But 185,621 is prime.
Martian clocks composed of unequal units have also been proposed. So one could, I suppose, organize the Martian day as follows: 24 hours of 60 minutes, each composed of 82 units of three-quarters of a second, with 287 left-over units at the end of each day. Or make the minute 80 units long, having a more noticeable 3167 units, presumably composing a short 25th hour. However, I don't really favor such a proposal.
Since any approximation to the day that is off by only a small fraction of a second is acceptable, how about a day of 88775 and 2/7 seconds? Unfortunately, 621,427 is prime. How about a day of 88775 and 5/18 seconds? 1597955 obviously divides by five; but the quotient, 319591, is prime.
If an inaccuracy of about .01 seconds in a day is acceptable, then the inaccuracy acceptable in a Martian minute would be about .000007 seconds. This explains why units based on 1/97th of a second have come up so often: 97 times the number of seconds in a Martian minute is 5980.0002, and 5980 is a nice number with many factors.
Attacking the problem directly, it is seen that fractions involving 97 are about the only really small ones that will serve.
The first one not based on subdividing 1/97th of a second involves dividing the minute into 72993 units of 1/1184th of a second. 72993 isn't a prime number, it's equal to 3 * 29 * 839. So the minute could be divided into 87 units of 839/1184ths of a second.
The next possibility would be 151 units of 523/1281sts of a second.
Then, 51 units of 1783/1475ths of a second. 199 units of 487/1572nds of a second. 43 units of 2671/1863rds of a second. 123 units of 1031/2057 of a second, and 123 is at least 3 times 41.
While 2134 is a multiple of 97, 2154 isn't, and it leads to 97 popping up on the other side of the fraction. Dividing the second into 2154 parts leads to the minute containing 132793 of them, and that number is 37 * 37 * 97. Thus, one has the choice between dividing the minute into 97 units of 1369/2154 of a second or 37 units of 3589/2157 of a second. Thus, if one used 97 units, each of those units could be divided into a further 37 parts, each one 37/2157 of a second, so Martian electrical power could be provided at a frequency of 2157/37 Hz which would also be 37 cycles per Martian time unit.
Still, since nothing very useful seems to turn up using any other units, dividing the Martian minute into 100 units of 299/485ths of a second seems best. 299 is not prime, it is 13 times 23, so Martian electrical power could be provided at a frequency of 485/13 Hz, which is about 37.3 Hz, at 23 cycles per Martian time unit, or at twice that frequency. 74.6 Hz would be fairly close to the standards used on Earth. That frequency would assume the same phase with respect to Earth seconds once every thirteen seconds as well as dividing the Martian time unit of one one-hundredth of a Martian minute evenly.
Incidentally, at this site, a suggestion was made that was far too strange for me to think about.
Simply use Earth hours on Mars. Have a six day week, where four days are 25 hours long, and the two days on the weekend are 24 hours long.
While my suggestions may be weird and idiosyncratic, one thing they are not is particularly bold like that one. Another proposal simply uses 20 hours a day of 74 minutes, and has people set their clocks back every 180 days; after all, we set our clocks this often for daylight savings time. I have ventured to suggest nothing nearly so bold.
Nachum Dershowitz and Edward Reingold have designed a calendar that goes beyond my calendar with intercalary months to tie Martian dates closely to Earth dates both in terms of the month and day on the one hand, and the day of the year on the other; only an abstract of their presentation, however, and not a complete description of their system, is on the web site by Thomas Gangale with information about Martian time systems that I have pointed to in several other links here.
If it is possible to have a Martian time system with a six day week in which two days have twenty-four hours while the rest have twenty-five, I suppose that I can hazard the following scheme, but merely for purposes of illustration. I have no wish to inflict such a system on the Martian colonists.
Given a Martian day of 88775.260726 earth seconds, one can have a Martian week of seven or eight days in which two of the days are 88776 seconds long, while the others are 88775 seconds long.
The 88775 second-long days would be divided by means of the scheme mentioned above into 25 short hours of 53 long minutes of 67 seconds each.
The 88776 second-long days, on the other hand, instead of just sneaking in a leap-second, would be divided into 24 long hours of 137 short half-minutes of 27 seconds each.
The 88775 second-long days would be divided by means of reversing a scheme mentioned above to obtain 25 short hours of 67 short minutes of 53 seconds each.
The 88776 second-long days, on the other hand, instead of just sneaking in a leap-second, would be divided into 24 long hours of 27 long double-minutes of 137 seconds each.
In this way, at least the double-minutes would be evenly divisible into three parts. Reversing the scheme used for the 88775 second days makes one minute long, and the other minute short, to avoid confusion. Although confusion would be plentiful with a chronometric system (the word system is used advisedly, or perhaps that should be ingenuously, of course) such as the one here described.
An eight-day week would be .08581 seconds too short.
A seven-day week would be .17492 seconds too long.
Thus, one would usually have two eight-day weeks followed by one seven-day week, and that would be .0033 seconds too long, so one would have an extra eight day week every 26 cycles of three weeks (or sometimes after 25 cycles).
Obviously, a Martian timekeeping system where a completely different system of keeping time is used on weekends, while suitable for comic relief, is one that cannot be seriously entertained as a practical proposition.
However, the fact that 27 is divisible by three suggests yet another type of Martian clock, one based on a day of 88775.25 Earth seconds, and a unit of 3/4 second.
Let the Martian day be divided into 25 long periods, each consisting of 53 short divisions of 67 ticks, each 3/4 of a second long, and 24 short periods, each consisting of 9 long divisions of 137 ticks.
Midnight, then, would be when one long period would directly follow another without an intervening short period.
How would one possibly design a geared clock for such a scale? One way would be this: have an hour hand that goes around the face of the clock at a uniform speed, once each day. Have a minute hand that goes around a face showing one long period and one short period. During the two consecutive long periods, marked in red on the hour scale, have the minute hand move at half-speed, and divide the long period area on the minute hand face into red half-divisions.
As an expedient to allow the use of a unit of 3/4 seconds on Mars, instead of 299/485ths of a second, of course, this remains too extreme.
Subtracting 25 units from each of the 24 short periods, and adding 24 units to each of the 25 long periods could be done repeatedly. Would that make either of them into nicer round numbers?
Obviously, that couldn't make the number of units in a short period divisible by 5, or the number of units in a long period divisible by 2 or 3, so it seems like it would only be a frustrating process.
Well, then, starting from the fact that 88775.25 seconds is 118367 units of 3/4 second, perhaps if I put 23 odd units in between 24 periods, I can get something that works out nicely in a dual-interval scale of this type.
118367 = 4931 * 24 + 1 * 23 4908 * 24 + 25 * 23 4885 * 24 + 49 * 23 ... 4816 * 24 + 121 * 23 4816 = 16 * 301 4793 * 24 + 145 * 23 4770 * 24 + 169 * 23 4770 = 90 * 53 169 = 13 * 13 ... 4632 * 24 + 313 * 23 4632 = 24 * 193 ... 4540 * 24 + 409 * 23 4540 = 20 * 227 ... 4425 * 24 + 529 * 23 4425 = 75 * 59 529 = 23 * 23 ... 4356 * 24 + 601 * 23 4356 = 36 * 121 ... 4080 * 24 + 889 * 23 4080 = 60 * 68 889 = 7 * 127
One can divide the day, therefore, into 24 segments, each consisting of 60 periods of 68 units of 3/4 second each, with 23 small segments between them of 7 long periods of 127 units of 3/4 second each.
For purposes of civil timekeeping, the system suggested by Dr. Robert Zubrin of simply dividing a Martian solar day into 24 Martian hours of 60 Martian minutes, each of which is 60 Martian seconds long, does have much to commend it.
If we do want to retain the Earth second, how simple can we keep a system of Martian timekeeping so that it approximates this ideal?
A Martian day is 88775.260726 seconds long. Divide that by 24. A Martian hour could be 3698 or 3699 seconds long, with one hour in the day slightly long or short to make up odd seconds and the odd fraction of a second. Having one odd hour in each day means that accurate Martian clocks would be too complicated to easily implement in clockwork, but nearly all clocks have been electronic for quite some time now.
3699 is 27 times 137. 3698 is 2 times 43 times 43.
So one simple approach would be to divide the Martian hour into 86 Martian minutes of 43 standard SI seconds. As this yields a Martian day of 88752 seconds, one hour a day - perhaps the one between 2 AM and 3 AM, the one used for changing in and out of Daylight Saving Time on Earth - would simply have an extra 23.260726 seconds added to it.
Other modifications of this approach are possible, such as adding almost, but not quite, one second to every hour in the day. To the use of clockwork, just with a dial with markings that are not fully uniform, one would instead have to add a tiny fraction of a second to the end of every minute.