Worldbuilt worlds will, by definition, have their own unique characteristics: rotation and revolution periods; axial obliquity; eccentricity variation; Milankovitch-style cycles; etc., all of which will contribute to making the place exotic and interesting.

One of the challenging but most interesting aspects of a Worldbuilt world is when its "day" and "year" are different from Earth's. Differences in these two periodic cycles mean that civilizations on that planet would have their own unique calendrical systems. When the presence or absence of companion objects such as twin planets or moons, which will have their own periodic cycles, are added into the mix, it can get quite challenging to work out realistic calendrical systems, but also very rewarding and lend a great deal of verisimilitude to the milieu.

One of the challenging but most interesting aspects of a Worldbuilt world is when its "day" and "year" are different from Earth's. Differences in these two periodic cycles mean that civilizations on that planet would have their own unique calendrical systems. When the presence or absence of companion objects such as twin planets or moons, which will have their own periodic cycles, are added into the mix, it can get quite challenging to work out realistic calendrical systems, but also very rewarding and lend a great deal of verisimilitude to the milieu.

## Demeter and Family

Let’s try adding a planet to our own Solar System. We’ll call it Demeter (another name for Ceres, the dwarf planet which occupies a spot in the Sun’s asteroid belt), and place Demeter on Ceres' known orbit of 2.76 AU. Further, let’s make Demeter 1.10 Earth masses and 1.04 Earth radii.

Its surface gravity, then will be:

Its surface gravity, then will be:

... not different enough from Earth to notice in everyday affairs.

Demeter's orbital semi-major axis, being that of Ceres, is 2.76 AU; so, its period is:

Demeter's orbital semi-major axis, being that of Ceres, is 2.76 AU; so, its period is:

This immediately gives us a challenge: this planet’s “year” is over 4½ times that of Earth!

If an inhabitant of this world is 20

As this planet is slightly larger and more massive than Earth, let’s give it a proportionally longer day, say one Earth day multiplied by the mass of the planet:

If an inhabitant of this world is 20

*local*(Demetrian) years old, then she is almost 92 years old in Earth years! In describing characters and historical time periods for this world, it will be*absolutely crucial*to be very clear that local timespans are different—or every timespan will have to be converted to familiar Earth-based ones.As this planet is slightly larger and more massive than Earth, let’s give it a proportionally longer day, say one Earth day multiplied by the mass of the planet:

or 26 hours and 24 minutes.

Remember that axial rotation periods are independent of orbital revolutionary periods (unless the body is tidally locked to a larger object, of course). Also, a planet's rotational period is independent of its mass, radius, density, etc. A good example is Venus, which is quite close to Earth in mass, radius, and density, but rotates on its axismuchmore slowly.

Here, again, a challenge immediately appears. If your characters undertake a journey that comprises 8

Just to get a sense of what we're talking about here, below is an illustration that graphically compares one Earth year with one Demetrian year (measured in Earth days).

*Demetrian*days, they will have travelled for more than 8½ days (8ᵈ 19ʰ 12ᵐ) Earth-time.Just to get a sense of what we're talking about here, below is an illustration that graphically compares one Earth year with one Demetrian year (measured in Earth days).

So, how many

*Demetrian days*are in a Demetrian year?What can we determine, then, about the sorts of months and weeks the Demetrians experience, and how they might organize, mark, and record time?

## Rudiments Of A Demetrian Calendar

Let’s assume the dominant culture of Demetrians—the Naharesa—has a Medieval-level society, without sophisticated astronomical observation and only a limited knowledge of advanced mathematics. Thus, they round their year up to 1523 full days and their day down to a manageable, even 26 Earth-hours. (No, there is no reason why the Demetrians should divide their rotational period into separate units each of which is one Earth-hour long—but we have to start somewhere, don't we?)

Right off the bat, we have yet another challenge: 1523 is a prime number, and will not be evenly divisible by any smaller numbers (including 26), so there are not going to be a convenient number of weeks or months in the Demetrian calendar. (As it happens, 1522, though

Dividing the 1523 days of the Demetrian year into 80 months yields very nearly 19 Demetrian-days per month (19.0375), but 19 is also a prime number, so there’s no convenient way to break a 19-Demetrian-day month into weeks.

(From here on out, when I use "days", "weeks", and/or "months", assume I'm speaking in Demetrian units. I'll explicitly specify, otherwise).

Using a value of 50 months is more promising: for one thing, it permits subdivision into smaller units of 10 by 5, or 2 by 25; doing the division of 1523 by 50 results in 30.46 days per month, which is very close to the same average number of days-per-month we have here, on Earth, so a reader wouldn't have to keep a mental note of a "month" being longer than they're used to in real life.

Rounding a month down to 30 days per month gives 1500 days, 23 days short of a year; rounding up to 31 days per month gives 1550 days, which is 27 days over.

Perhaps they, like Earthlings, would develop a complex but regularized system of alternating month lengths? However, simply making half of the months 30 days and the other half 31 days doesn’t quite work:

Right off the bat, we have yet another challenge: 1523 is a prime number, and will not be evenly divisible by any smaller numbers (including 26), so there are not going to be a convenient number of weeks or months in the Demetrian calendar. (As it happens, 1522, though

*not*prime, isn’t much better—do the math.)Dividing the 1523 days of the Demetrian year into 80 months yields very nearly 19 Demetrian-days per month (19.0375), but 19 is also a prime number, so there’s no convenient way to break a 19-Demetrian-day month into weeks.

(From here on out, when I use "days", "weeks", and/or "months", assume I'm speaking in Demetrian units. I'll explicitly specify, otherwise).

Using a value of 50 months is more promising: for one thing, it permits subdivision into smaller units of 10 by 5, or 2 by 25; doing the division of 1523 by 50 results in 30.46 days per month, which is very close to the same average number of days-per-month we have here, on Earth, so a reader wouldn't have to keep a mental note of a "month" being longer than they're used to in real life.

Rounding a month down to 30 days per month gives 1500 days, 23 days short of a year; rounding up to 31 days per month gives 1550 days, which is 27 days over.

Perhaps they, like Earthlings, would develop a complex but regularized system of alternating month lengths? However, simply making half of the months 30 days and the other half 31 days doesn’t quite work:

... still a full 2 days too long.

Okay ... what if they divide the year into two halves of 25 months each? There are 2 extra days in the alternating calendar just postulated, so what if they subtract one day from the 1st month and one day from the 25th month [1]? They now have 27 months of 30 days each and 23 months of 31 days each, or:

Okay ... what if they divide the year into two halves of 25 months each? There are 2 extra days in the alternating calendar just postulated, so what if they subtract one day from the 1st month and one day from the 25th month [1]? They now have 27 months of 30 days each and 23 months of 31 days each, or:

... which works out perfectly!

In fact, a better solution is 76 months, comprising 4 weeks of 5 days each, which amounts to 304 weeks per year. This would make each quarter of Demeter's year approximately 76 weeks, or 380.75 days long. However, the discussion above is completely in keeping with a way to work out near-comensurability for non-Terran time systems.

Have you noticed what’s potentially missing from this scenario? What if Demeter has moons? How might a moon or moons affect the Demetrian reckoning of time?

## The Moons Of Demeter

Let’s take the two moons postulated in the discussion of synodic periods in a previous blog and assign them to Demeter. We’ll make the outer one the larger of the two and call it Persephone and the inner, smaller one will be Kora (another name for Persephone) [2]. Looking back, we specified that these moons had orbital periods of 11.72 and 24.36 Earth days. These translate to 10.6545 Demetrian days for Kora and 22.1455 Demetrian days for Persephone.

Assuming the Demetrians round these up and down respectively to 11 and 22 Demetrian days, we immediately see that neither of these periods fits neatly into a 30- or 31-day Demetrian month, or a 1523-day Demetrian year.

Such rounding

There would be approximately 76 such periods per Demetrian year (another good reason for using the alternative calendar system mentioned in the sidebar above). This synodic period of 76 Demetrian days would divide into two halves of 38 days each, or four quarters of 19 days. The 19.5413 Earth-years it takes these two moons to return to exactly the same point in the sky translates to 4.262 Demetrian-years, four cycles of which equal just a fraction over 17 Demetrian years (17.04706).

It might be that some other Demetrian culture would base their calendar on these periods, rather than the year, in much the same way that some Earth cultures use months as the basis for their calendar.

And what about Demetrian clocks? A Demetrian hour is 1.1 Earth-hours long (26.4 ÷ 24 = 1.1), or 66 minutes, so a statement like, “We waited five-and-a-half hours”—if those are

We also have said nothing whatsoever about Demeter’s obliquity, so let's talk about that next.

Assuming the Demetrians round these up and down respectively to 11 and 22 Demetrian days, we immediately see that neither of these periods fits neatly into a 30- or 31-day Demetrian month, or a 1523-day Demetrian year.

Such rounding

*does*present a problem: it*implies*a perfect 2:1 mean motion resonance between the moons’ periods, which our earlier discussion showed is definitely not the case. It might happen then, that rather than basing time reckoning on either of the individual moons’ orbits, they might base a time unit on their synodic period, in Demetrian units, of course, which would be 20.53355 Demetrian days.There would be approximately 76 such periods per Demetrian year (another good reason for using the alternative calendar system mentioned in the sidebar above). This synodic period of 76 Demetrian days would divide into two halves of 38 days each, or four quarters of 19 days. The 19.5413 Earth-years it takes these two moons to return to exactly the same point in the sky translates to 4.262 Demetrian-years, four cycles of which equal just a fraction over 17 Demetrian years (17.04706).

It might be that some other Demetrian culture would base their calendar on these periods, rather than the year, in much the same way that some Earth cultures use months as the basis for their calendar.

And what about Demetrian clocks? A Demetrian hour is 1.1 Earth-hours long (26.4 ÷ 24 = 1.1), or 66 minutes, so a statement like, “We waited five-and-a-half hours”—if those are

*Demetrian*hours—means 6.05 Earth-hours (6ʰ 3ᵐ). And, anyway—as we said earlier—why would the Demetrians divide their day into 26.4 units each one Earth-hour long, and not into some other number of units that would make more sense to them?We also have said nothing whatsoever about Demeter’s obliquity, so let's talk about that next.

## Obliquity Of The Demeter's Axis

What if we set the obliquity of Demeter's axis to 30°? In this previous blog we saw that this would cause Demeter’s seasons (which on average will be 1.146 Earth

It is also worth noting that at 2.76 AU Demeter is 1.39 AU outside the Sun’s habitable zone, so we’d really have to put it into a stellar system of its own with a proportionately more luminous star in order for it to be habitable at all.

*years*long) to have somewhat greater temperature variation than Earth’s seasons.It is also worth noting that at 2.76 AU Demeter is 1.39 AU outside the Sun’s habitable zone, so we’d really have to put it into a stellar system of its own with a proportionately more luminous star in order for it to be habitable at all.

## Building a Star for Demeter

We can work backward from Demeter's semi-major axis to determine the needed luminosity of the star (let's call it Zeus), by declaring 2.76 AU to be the

*nucleal habitable zone orbit*, and, restructuring the equation for the nucleal habitable zone, thus:Again, working backward, we can determine the mass and radius of Zeus. Because Zeus' luminosity puts it in Luminosity Regime 2, we use the following equation to determine its mass:

... which is just slightly over the maximum limit of 1.65 solar masses, but let's ignore that for the moment. This mass gives Zeus a radius of:

... which actually falls

However, if we set Demeter’s orbit as the

*well*beyond the 1.2 Solar-radii upper limit, by 0.3791 solar radii. We might have been able to overlook a slight overage in mass, but this overage in "waistline" is too much.However, if we set Demeter’s orbit as the

*optimistic outer habitable zone orbit*instead of the nucleal orbit:... then Zeus’s mass and radius come out to 1.24873 and 1.22130 solar radii, both reasonable: its mass is well within the 1.65 maximum, and its radius is close enough to 1.2 solar radii that the extra size is tolerable, remembering that mass is the more critical characteristic.

Note that we have not changed Demeter’s

Note that we have not changed Demeter’s

*orbital period*, so the length of its year is still the same, and our earlier calculations for its days, months, etc., remain valid. We have simply*defined the***optimistic outer habitable zone**limit for this star system to be equal to Demeter’s orbit. We have built the star, Zeus, to accommodate the planet, Demeter.## The Spectral Class of The Star

Zeus' luminosity value gives a surface temperature of:

... which is, of course, in solar units, so multiplying by the Sun's surface temperature of 5839.872 K:

... we get an effective temperature for Zeus of 6751.5 K, which in the spectral type F.

Doing the math:

Doing the math:

... we find that Zeus is an F4.849 spectral class.

## Conclusion

It can be great fun working all of this out, but, as you can see, it can be very challenging and time‑consuming, and communicating it all to one’s readers or the players of one’s game can be complicated and confusing.

So, while it is good for you, the Worldbuilder to know these details, it is best that they are shared with the viewer in the form of the interesting and curious details they provide to the narrative.

So, while it is good for you, the Worldbuilder to know these details, it is best that they are shared with the viewer in the form of the interesting and curious details they provide to the narrative.

1. Subtracting from the 25ᵗʰ and 50ᵗʰ months might at first seem more attractive, but in this scheme, the 50ᵗʰ month already has only 30 days, and subtracting a day leaves it with 29, which breaks an otherwise largely symmetrical solution—call me pedantic.

2. Yes, I know there is already an asteroid named Persephone and a Jovian moon named Kore, but this is hypothetical. In any case, these aren't the names the Demetrians would give their moons.

2. Yes, I know there is already an asteroid named Persephone and a Jovian moon named Kore, but this is hypothetical. In any case, these aren't the names the Demetrians would give their moons.