Lumar size and shape

[king of nowhere]

Some things don't add up about it.

 

First of all, I assumed that the 12 moons, being in geostationary orbit, would all be on the equator. yes, they cannot be in geostationary orbit that close to the surface unless the planet rotates really, really fast, or it has extremely low gravity, there's clearly magic at work here. Still, a geostationary orbit away from the equator would be even harder to achieve - it would basically require the moon to be standing still in freefall all the time.

Still, 12 pentagons cover a sphere, so while there is no specific mention on how the moons are disposed, it seems more likely the idea that they cover all the planet evenly and not only the equator. And in that case, one would have to go through 6 pentagons to circle the whole planet.

Anyway, when tress becomes captain, salay says it would take "one week" to reach the midnight sea. They were currently in the middle of the crimson sea, so half a pentagon away from the border. We also don't know how long a "week" lasts on Lumar; we could go by our 7-day week, but given how important moons are on Lumar, a 12-day week seems appropriate. I couldn't find any reference in the text as to which is correct.

I also looked at how long a sail boat can travel in a day, and I found 100 nautic miles as a good average with favorable wind. So, 185 km.

Going with the most likely explanations - 12 days for a week, 6 seas to circle the planet - one would get a circumference of 185*12*6*2=26640 km, for a radius of 4240 km.

Earth has a radius of 6370 km. Lumar would be much closer to the size of Mars (3390 km of radius, one third of Earth gravity) than to that of Earth. That would be small enough that the planet would not be able to retain its atmosphere over astronomical timespans. And that's assuming Tress could sail with favorable winds all the time, or the planet would be even smaller.

On the plus side, a small planet with low gravity would help tress resist 15 minutes hanging from the ship.

 

The only way to get an estimate consistent with Earth-like size would be to have the moons around the equator, in which case it would take 12 seas to circle the planet, and the previous estimate would double - but it could be brought down to Earth level by assuming the wind wasn't always favorable.

Of course, assuming a 7-days week would be even worse, resulting in a "planet" slightly bigger than our Moon.

Another assumption that could bridge the gap would be that Lumar has a longer day. With a 40-hours day, we'd get the estimate just right.

Most likely, Brandon and his assistants didn't make an exact calculation here. But assuming the numbers are good, what do you think it implies for Lumar's size and/or calendar?

 

[AquaRegia he/him]

I've also been bothered by the astronomical "realities" of the Lumar system.  Back in the "SP1" days, there were several discussions about it;  I'll pull over some quotes.

 @cometaryorbit

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If they are large moons, and of conventional matter, there would need to be some active effect keeping their mutual gravitational attraction from messing up their orbits.

But if Tress's planet has an Earthlike day length and is roughly Earth sized, to fill "a third of the sky" from geosynchronous orbit would mean moons the size of a large planet (our Moon is a bit more than 10x geosynchronous orbit distance, and has an apparent size of half a degree, so would be about 5 degrees at geosynchronous distance).

So I would think either the visual appearance of the moons doesn't correlate to their real size (or Hoid is exaggerating "third of the sky"), or the moons are not conventional matter and have very little mass.

@platnumkid

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In our solar system the largest moon relative to it's planet is our moon (by a large margin) at roughly 27% the diameter of earth.  If we stretch the definition of moon and planet we would come to charon and pluto where the ratio is about 50%, now we're talking about something that can take up 1/3 the night sky.

However even at that size if all 12 moons were located around the equator to make them naturally occurring geostationary orbits at the low orbit required to fill up the sky you wouldn't be able to see any of them from the poles.

So I found an equation to find the coordinates of roughly equally spaced n-points.  I plotted the results for 12 points.  It looks much more like you would imagine from the statement "you can always see one of the 12 moons no matter where you travel.

Therefore I am concluding that we have 12 geostationary moons that scattered roughly equidistance from each other around the planet, and they are using investiture to maintain their relative positions.  I haven't decided if spiritual mumbo-jumbo is required for them to maintain their spacing, of this is achieved through the use of a simple constant adjustment to their natural orbits to stay geostationary.  Maybe they used spiritual mumbo-jumbo to get in position but now don't need it.

Confirmed in the live stream (they are working on the scientific justification for the non-equatorial geostationary orbit).
Also from the live stream, someone from the planet has tried to go to the moons, but tress doesn't know about that.
Each moon has different coloured spores

This entire topic, kudos to @Ixthos, @Oltux72, and others:

FANTASTIC work by @Oltux72 here:

Flat-topped: this configuration has 3 fold symmetry: 4 offset rings of 3 moons each.

I'm fairly confident that this is closest to what Brandon has in mind.  Each moon has 5 closest neighbors, consistent with each sea being a pentagon.  Among the many problems, of course, are that they are impossibly close to the planet (Roche limit), and moving impossibly slowly.  If they are big enough to "cover 1/3 of the sky", they must be practically touching the planet.

Not only do the various "facts" we have about Lumar and its moons not make sense, they literally cannot be made to make sense.  Even with "magic", we're really stretching things.

Edited January 7, 2023 by AquaRegia
found that 3rd topic

[king of nowhere]

4 hours ago, AquaRegia said:

I've also been bothered by the astronomical "realities" of the Lumar system.  Back in the "SP1" days, there were several discussions about it;  I'll pull over some quotes.

 

yes, but all that correlates to the impossible orbits. ok, the orbits are impossible and require magic, we all agree with it. given that the moons are entirely taken over by heavily invested entities, it is possible. it is certainly intended; brandon knows it's physically impossible and decided to make it that way and make it work with magic.

the planet looking way too small bothers me more at this point

 

[AquaRegia he/him]

39 minutes ago, king of nowhere said:

the planet looking way too small bothers me more at this point

Sorry, I neglected to actually make the point I intended to.

10 hours ago, king of nowhere said:

Lumar would be much closer to the size of Mars (3390 km of radius, one third of Earth gravity) than to that of Earth. That would be small enough that the planet would not be able to retain its atmosphere over astronomical timespans.

Very true, and I agree with your estimates.  A sailing ship that small traversing ~30% of the planet should take several months, not a few weeks.  But given the number of other physical impossibilities we see in the cosmere, this one seems like small potatoes.  Fine, it's a Mars-sized planet whose atmosphere is magically retained and/or renewed.  Maybe it's more dense than average, for higher surface gravity.  I don't recall any mentions in the text of how long the day/night cycle is, so that could be significantly different than cosmere standard as well.

The size of Lumar and the stability of its atmosphere over astronomical time are open questions, agreed.  They bother me less than the twelve streams of spores being endlessly dumped on the planet... and don't forget that said spores must be flying UPWARD from the surfaces of the moons somehow. 

I guess my real point is that the number of things for which the answer has to simply be "magic" is VERY large for Lumar.

[Captain Coffee]

I have wondered if Lumar is a planet that has been, for some reason, entirely moved into the cognitive realm. That could account for weird physics.

 

[+Oltux72 he/him]

19 hours ago, king of nowhere said:

First of all, I assumed that the 12 moons, being in geostationary orbit, would all be on the equator. yes, they cannot be in geostationary orbit that close to the surface unless the planet rotates really, really fast, or it has extremely low gravity, there's clearly magic at work here. Still, a geostationary orbit away from the equator would be even harder to achieve - it would basically require the moon to be standing still in freefall all the time.

Yes, the moons are not in orbit. They are hovering. How they manage that feat is unclear.

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Anyway, when tress becomes captain, salay says it would take "one week" to reach the midnight sea. They were currently in the middle of the crimson sea, so half a pentagon away from the border.

And here I am afraid we are reaching the first hard roadblock. The oceans are pentagons. That means the surface of the planet is kind of a deformed dodecahedron (projected onto a sphere). That means that the moons form an icosahedron, with a moon right in the center of the pentagon. In other words, if you are in the exact middle of an ocean, you are at the lunargree and thereby dead.

Hence the "middle" here is taken too literally.

 

15 hours ago, AquaRegia said:

FANTASTIC work by @Oltux72 here:

Ixthos did it. Honor to those who deserve it.

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Flat-topped: this configuration has 3 fold symmetry: 4 offset rings of 3 moons each.

Note that while the distribution of the moons has to be as depicted here, the actual position is arbitrary. We can only know that Diggen's Point must be located polewards of the Emerald Moon for the sun to be in line with its center.
 

 

Edited January 8, 2023 by Oltux72

[king of nowhere]

56 minutes ago, Oltux72 said:

 

And here I am afraid we are reaching the first hard roadblock. The oceans are pentagons. That means the surface of the planet is kind of a deformed dodecahedron (projected onto a sphere). That means that the moons form an icosahedron, with a moon right in the center of the pentagon. In other words, if you are in the exact middle of an ocean, you are at the lunargree and thereby dead.

"in the middle" in this case means "close enough that we can approximate"

[lacrossedeamon]

Regarding the atmosphere not being retained due to weak gravity, would the air vents and zephyr aether be enough to offset losses?

[king of nowhere]

34 minutes ago, lacrossedeamon said:

Regarding the atmosphere not being retained due to weak gravity, would the air vents and zephyr aether be enough to offset losses?

oh, right. more than enough, actually.

we only have to figure out where that air comes from...

[lacrossedeamon]

What if the pancakelike beings Hoid mentions live in the center of Lumar and the air is from their flatulence?

[+Oltux72 he/him]

21 hours ago, king of nowhere said:

"in the middle" in this case means "close enough that we can approximate"

It may also mean equidistant between edge and lunagree. And that is a factor of 2. And we need that kind of accuracy for such a claculation.

[king of nowhere]

5 hours ago, Oltux72 said:

It may also mean equidistant between edge and lunagree. And that is a factor of 2. And we need that kind of accuracy for such a claculation.

I may be wrong, but I recall that xisis lived very close to the lunagree.

in any case, to get to the midnight sea tress had to cross all of the crimson

[+Bzhydack he/him]

On 8.01.2023 at 2:43 PM, king of nowhere said:

we only have to figure out where that air comes from...

I think Spores need to be part of some Cycle. Otherwise, such a small planet will be completly drown under them. Maybe Spors have limited lifespan, and they spontanously transform into Investiture, and release some Air in the process? Considering fact Air is definitly made IN THE OCEANS (during Seethe) I see thin as plausible.

[Leuthie]

They aren't actually moons? They're mostly Investiture with very little if any actual mass they hold themselves in place using Investiture and have little to no effect on the gravitation and atmosphere of Lumar. They're "supernatural" beings that spit out spores and look search for water.

[AquaRegia he/him]

On 1/7/2023 at 9:03 AM, king of nowhere said:

I also looked at how long a sail boat can travel in a day, and I found 100 nautic miles as a good average with favorable wind. So, 185 km.

I saw this video today, in which we see real world experiments with fluidized sand.

I'm struck by how nearly frictionless the motion of a boat sliding on the surface is; it's more like air hockey than sailing.  It seems reasonable that a ship moving on something like fluidized spores might go MUCH faster than a ship in water.  If so, "one day's sail" might be a considerably longer distance, allowing a significant increase in our estimate of Lumar's size.

On 1/7/2023 at 9:03 AM, king of nowhere said:

The only way to get an estimate consistent with Earth-like size would be to have the moons around the equator

This would be ruled out by Hoid's assertion that the moons take up "fully 1/3 of the sky", among other logical and geometric considerations.  In the case of equatorial moons, at least 3 or 4 would be visible simultaneously from anywhere near the equator... and they'd be touching or overlapping each other LOL

Edited January 25, 2023 by AquaRegia
grammar

[king of nowhere]

1 hour ago, AquaRegia said:

I saw this video today, in which we see real world experiments with fluidized sand.

I'm struck by how nearly frictionless the motion of an boat sliding on the surface is; it's more like air hockey than sailing.  It seems reasonable that a ship moving on something like fluidized spores might go MUCH faster than a ship in water.  If so, "one day's sail" might be a considerably longer distance, allowing a considerable increase in our estimate of Lumar's size.

Point taken. this could bring lumar to a more normal size