Calculating the size and distance of Lumar's moons

[alder24]

Since I've read the preview chapters of TotES, I was wondering how big and how far away Lumar's moons have to be to create the world Brandon envisioned. So here I am, finally after reading the book, I can do the math myself and answer that burning question. The book gave us 2 very important pieces of information that makes this math possible. So let's dive into this.

I won't focus on the exact orbit of moons as this was already greatly showed in this topic: 

For me, it is important that only one moon is visible from each sea of spores and that when crossing between two seas, both moons are visible for a short time, before one of them will set behind a horizon. So moons can't be in a geostationary orbit, they have to be in orbit around the entire sphere of the planet. So no matter where you are on the surface of the planet, you can see at least one moon.

The second piece of information, mentioned multiple times in the book, is that moons take 1/3 of the sky. That's a lot. And here we can start doing our math. Bear in mind, I'm not that good in math, so there are mistakes here (few of them I'm aware about), but I think this will give us nice approximations of the size and distance of Lumar's moons. 

 

For now, to have a good comparison, let's assume that each Lumar moon is the size of the Earth Moon. For the Moon to take 1/3 of the sky, he would need to be much closer than it is. And we can calculate how far he needs to be, based on the angular size. Right now, the Moon has an angular size of around 0.5 degree. On Lumar, it would need to have 60 degrees (that's 1/3 of the 180 degrees). And the formula for angular size is:

Angular size in degrees = (size * 57.29) / distance.

We have the size of the moon (diameter = 3,474.8 km), we have angular size (60 degrees), we can reposition the equation to search for the distance.

distance = (size * 57.29) / Angular size in degrees = 3317,85 km.

That's how far away Lumar's moon the size of Earth's Moon would need to be to cover 1/3 of the sky. That's almost 1:1 proportion between size and distance. 

 

But Lumar's moons can't be the size of Earth's Moon. Because only one of them is visible for each sea of spores. The crossing between two seas is very short, and as soon as you cross into a different sea, the moon of the previous sea disappears behind the horizon. Which means we need to look into how much is hidden behind the curve of the Lumar's surface, as this hidden amount would be equal to the distance between Lumar and moon plus the size of a moon (when standing on the edge of a sea of spores). So for simplicity we can assume that moons are like a tower or a mountain stretching from the surface of the planet, obscured by the curvature.

Spoiler

Because there are 12 moons and 12 seas covering the entire planet, 1 sea covers 1/12 of the planet's surface. Assuming that Lumar has the same size as Earth:

Earth's surface = 510100000 km^2
Surface of one sea of spores = 510100000 / 12 =  42508333,33 km^2

So assuming that each sea is circular (which has its own problems, as you can't divide surface of a sphere into circles without them overlapping, so here each sea would have to be bigger than 1/12, and each sea has a pentagonal shape, but let's ignore it for simplicity), each sea of spores would be a surface of a spherical cup. When standing on the edge of this circular sea of spores, the moon of that sea would be fully just under the horizon of the planet.

Spoiler

Here I'm looking for the length of the arch L, so I can use it as a distance between observer and observable object in curve calculator (which, as I see now, gives me another problem, because the distance between observer and observable object behind a curve of the planet is a straight, shortest line between them, not an arch along the surface of the planet, but I don't know how to get that line without having other values like height of observable object etc, so yeah, I did ignore that as well). 

Using the equations presented above for a spherical cup, I have the Area of that cup (1/12 of Earth's surface), the radius of the sphere (Earth's radius R = 6371 km), and pi = 3,1415 so from this I can calculate all values a, h, and angle θ, and from that I can get length of the arch L, L=R*θ.

Doing the math, I got: a =  3521,81 km, h =  1061,91 km, angle θ =  0,585706545, and L =  3731,54 km.

And now having L, I can give it into a curve calculator, which would calculate the hidden height of an object visible behind the horizon (assuming observer height of 10 m, standing on the deck of a ship) - https://www.metabunk.org/curve/

This gave me the value of the hidden height taking into account refraction of 878.62 km. This is the sum of the diameter of Lumar's moon and its distance from the surface of the planet. And here we can go back to the angular size:

Angular size in degrees = (diameter * 57.29) / distance.

Now assuming we're standing right under a moon, we know that the distance from the surface of Lumar to the other side of that moon is 878.62 km, and this is equal to diameter + distance. We need to only have diameter in the equation of angular size, so distance = 878.62 km - diameter, therefore:

Angular size in degrees = (diameter * 57.29) / (878.62 km - diameter).

And the angular size is 60 degrees, so changing the equation to have diameter on one side we get:

60 / 54.29 = diameter / (878.62 km - diameter)
1.05 = diameter / (878.62 km - diameter)
diameter = 1,05 * 878.62 km / 2,05

Diameter of the moon, right below it, is equal to 450,02 km. And the distance between the surface of the planet and a moon is equal to  428,6 km.

That's small, and very close to the surface. And here perspective would cause a lot of problems, as the further away you are from that moon, the smaller it is in the sky. And when you are 1000 km away from that moon, its angular size is already only around 30 degrees, as that's twice as far, 2000 km - 15 degrees, closer to the edge of the sea of spores 3000 km, and has only 10 degrees of angular size. Nobody in the book ever mentioned a moon shrinking or growing in size in the sky, and this had to be done. Moons can't be further away from the surface of Lumar than 878.62 km, as they would be visible from multiple spore’s seas at once. They can't be on Lumar's geostationary orbit (same as Earth's, around 36,000 km), as again, they would be visible from multiple seas at once, almost from one pole to the other. But only one moon is visible on a one sea, and it hides below the horizon as soon as you leave that sea. This means Lumar's moons have to be very close to the planet, and thus very susceptible to perspective. It's weird that Tress never was amazed by the Emerald moon shrinking, or the Crimson moon growing in size in the sky, as this should be happening, nor did Hoid mention it. Even right below the Crimson lunagree, nobody mentioned that the Crimson moon is huge, like nowhere else on the Crimson sea. And these are very drastic changes, hard to miss. We've been told only that moons take 1/3 of the sky, and that's it.  

Knowing now that the angular size of the moons changes with the distance, we can assume that moons can be bigger than 450,02 km, by decreasing the distance between them and Lumar's surface, but they can't be bigger than 878.62 km (as they would be touching the planet's surface). But this is basically a slide now, the bigger the moons, the closer they get to the planet's surface, the less changeable they're by perspective from a far away, but if you get too close to it, it will cover most of the sky instead, not just 1/3. So there is no good solution to it. 

Ignoring this blazing problem, I will now calculate the Roche limit for the Lumar - moon system. Roche limit is the distance from a celestial body within which a second celestial body, held together only by its own force of gravity, will disintegrate because the first body's tidal forces exceed the second body's self-gravitation.  For the rigid, solid satellites:

Spoiler

Assuming that Lumar's moons have a density of similarly sized moons in our Solar System (Mimas ~400 km, 1,15 g/cm^3), then Roche limit is 15367,61 km.

Assuming that Lumar's moons have a density of Earth's moon (3,344 g/cm^3), the Roche limit for Lumar's moon is the same as the Moon, so ~9482 km.

Now let's go crazy, assuming that Lumar's moons are made out of pure Osmium, the densest element on the periodic table (22,59 g/cm^3), and it's equally dense across its volume, the Roche limit would be equal to 5016,50 km. Which is still far too far to prevent Lumar's moons from disintegrating.

How dense would Lumar's moons have to be, for their surface, closest to Lumar, to be on the Roche limit (making spores fall on the planet), but the rest of the moon to be on a "stable" orbit? So changing the equation and finding the density of the satellite, gave me the value of 36222,46 g/cm3 - which is comparable to a white dwarf star's density (between 10^4 and 10^7 g/cm^3)! That's just crazy!!! A moon with the diameter of 450,02 km and this density, would have a mass of 1,73 * 10^24 kg, which is "only" 3.45 times less than Earth's mass. Therefore all of Lumar's moon would be 3.5 times more massive than Lumar's planet itself. Also, they would be extremely bright and hot. And I don't even want to know what would happen with the orbits of Lumar's system if that was the case. So that's definitely not possible. But that's the only way in which Lumar's moons won't be disintegrated by Lumar's gravity, be that close to its surface, and also make spore fall on the surface of the Lumar "naturally".

 

That's enough math for me. That was a fun Sunday, all of this to prove that Lumar's moons are impossibly close to Lumar's surface, far beyond their Roche limit, on orbits that can't exist in real life. Like we didn't know that already. But still, having some numbers for it is so much more fun (I'm a very weird person). Of course, keep in mind that spore seas aren't spherical, the distance between observer and a moon isn't the same as the distance between observer and the base of a lunagree, going along the surface planet, like I assumed in these calculations, but I think that these calculations provide a close enough approximation of the size and distance of Lumar's moons, to make it "useful". All of my calculations are done in a very chaotic Excel, in which I already got lost, but I'm almost certain there are no basic calculation errors there but feel free to check it out (I hope so, tomorrow I won't recognise that spreadsheet). 

And so here it is, the answer to the question that has been keeping me from sleeping at night for the whole year. How unrealistic is the Lumar planetary system? The answer is - very unrealistic. Thank you for reading this far, if you enjoyed it, you're as crazy as I'm. 

Spoiler

 

Edited March 26, 2023 by alder24

[AquaRegia he/him]

I hope you enjoyed it.  That was an awful lot of work to prove what we already knew: these moons are simply impossible for a variety of reasons, and the only way they can exist is "magic".

[Treamayne]

Thanks for the math

On 3/26/2023 at 5:08 PM, alder24 said:

That's enough math for me. That was a fun Sunday, all of this to prove that Lumar's moons are impossibly close to Lumar's surface, far beyond their Roche limit, on orbits that can't exist in real life. Like we didn't know that already. But still, having some numbers for it is so much more fun (I'm a very weird person). Of course, keep in mind that spore seas aren't spherical, the distance between observer and a moon isn't the same as the distance between observer and the base of a lunagree, going along the surface planet, like I assumed in these calculations, but I think that these calculations provide a close enough approximation of the size and distance of Lumar's moons, to make it "useful". All of my calculations are done in a very chaotic Excel, in which I already got lost, but I'm almost certain there are no basic calculation errors there but feel free to check it out (I hope so, tomorrow I won't recognise that spreadsheet). 

Some things that may have changed your calculations:

In Ch 54 Tress does note that the moons are smaller on the horizon and larger at the Apex:

Spoiler

As it emerged from the horizon Tress feared, irrationally, that it would keep growing—that the Midnight Moon wouldn’t be the size of the others, but would turn out to be a vast darkness that consumed the entirety of the sky.

Also, keep in mind the Horizon Visual Illusion (such as Earth's moon appears larger to the eye at the horizon than it does in the open sky - but if you measure it the visible diameter never changes) - this implies that the moons appear larger at the horizon because of visual illusion making up some of the difference lost to distance.

In Ch 58 Tress notes that after only 1 hour of rowing the Crow's Song is far enough away to have dropped below the horizon and no longer be visible from within the Midnight sea:

Spoiler

And now the sun was going down. Tress turned and looked backward longingly—but she had rowed herself out here for a good hour or so. Her arms were burning as proof.

The Crow’s Song wasn’t even visible, nor was the Crimson Sea. She was alone.

So, if the planet is small enough that a 15-20(ish) meter vessel can fall behind the horizon from a viewing height of "rowboat + seated occupant" (call it 2 meters' of viewing height). Then the angles and radii might all be fairly different than your example.

Also note that it is approximately 9 days' sailing from the spore crossing to the Lunagree*, compared to the 5 week voyage for multi-masted vessels crossing the Atlantic circa Columbus (granted that was not a straight distance - Europe to the Carribean)

• Spoiler

  I arrived at the number based on the following (not that Hoid is the most reliable narrator):

  • 1 Day after the Crossing, Tress Cooks for Fort
  • About 3 days later, Doug Pakson is killed in the storm
  • About 3 days later, Tress completes her Flare gun, saves the Crow's Song, and is taken to Xisis
    • Crow said they were 2 days away from the Lunagree when the go to see the dragon

  Assuming a similar base average sailing speed, that would put the Lunagree for each sea at about 900-1000 nm from the flat side of a sea's pentagon-shape to its center - which would yield a planetary circumference somewhat less than half of Earth (somewhere between 9500-11500 nm - based on a 100-125nm per day average sailing distance)

Edited March 28, 2023 by Treamayne
SPAG

[alder24]

9 hours ago, Treamayne said:

Thanks for the math

Some things that may have changed your calculations:

In Ch 54 Tress does note that the moons are smaller on the horizon and larger at the Apex:

  Reveal hidden contents

As it emerged from the horizon Tress feared, irrationally, that it would keep growing—that the Midnight Moon wouldn’t be the size of the others, but would turn out to be a vast darkness that consumed the entirety of the sky.

Also, keep in mind the Horizon Visual Illusion (such as Earth's moon appears larger to the eye at the horizon than it does in the open sky - but if you measure it the visible diameter never changes) - this implies that the moons appear larger at the horizon because of visual illusion making up some of the difference lost to distance.

In Ch 58 Tress notes that after only 1 hour of rowing the Crow's Song is far enough away to have dropped below the horizon and no longer be visible from within the Midnight sea:

  Reveal hidden contents

And now the sun was going down. Tress turned and looked backward longingly—but she had rowed herself out here for a good hour or so. Her arms were burning as proof.

The Crow’s Song wasn’t even visible, nor was the Crimson Sea. She was alone.

So, if the planet is small enough that a 15-20(ish) meter vessel can fall behind the horizon from a viewing height of "rowboat + seated occupant" (call it 2 meters' of viewing height). Then the angles and radii might all be fairly different than your example.

Also note that it is approximately 9 days' sailing from the spore crossing to the Lunagree*, compared to the 5 week voyage for multi-masted vessels crossing the Atlantic circa Columbus (granted that was not a straight distance - Europe to the Carribean)

•   Reveal hidden contents

  I arrived at the number based on the following (not that Hoid is the most reliable narrator):

  • 1 Day after the Crossing, Tress Cooks for Fort
  • About 3 days later, Doug Pakson is killed in the storm
  • About 3 days later, Tress completes her Flare gun, saves the Crow's Song, and is taken to Xisis
    • Crow said they were 2 days away from the Lunagree when the go to see the dragon

  Assuming a similar base average sailing speed, that would put the Lunagree for each sea at about 900-1000 nm from the flat side of a sea's pentagon-shape to its center - which would yield a planetary circumference somewhat less than half of Earth (somewhere between 9500-11500 nm - based on a 100-125nm per day average sailing distance)

Ok, ok, I knew about these estimates before, but ignored them to make math easier, but I'm up to the challenge. So based on what you've written, still assuming sea of spores is spherical for simplicity etc:

For an average sailing speed of 125 nm/day, arch L = 2083,5 km, and the radius of the planet is R = 3557,36 km.
For an average sailing speed of 100 nm/day, arch L = 1666,8 km, and the radius of the planet is R = 2845,89 km.
For an average sailing speed of a ship from the 14th-16th century, around 4 knots, which is 7.4 km/h, daily 177,6 km/day, arch L = 1598,4 km, and the radius of the planet is R = 2729,1 km.
 

It keeps getting smaller. Now, Tress rowed for 1 hour and the ship disappeared behind the horizon. Assuming viewer height of 1 m (she is sitting) and the ship's height of 20 m, taking into account refraction, and a faster than average rowing speed of 13 km/h (average is around 5.6–7.4 km/h, for 7 km/h planet's radius would have to be ~700 km to make that ship disappear, that's a no-no, Tress knows how to row fast), Lumar have to be ~2400 km in radius, for the ship to fully drop beyond the horizon. That is almost the size of Mercury, which has a radius of 2439,7 km. That's small and Mercury doesn't have an atmosphere but it's possible for Mercury to have it - Saturn's moon Titan is ~5x less dense than Mercury and holds mostly a nitrogen atmosphere with surface pressure of around 60% higher than Earth's, so Mercury could easily hold Earth's atmosphere. That's at least checking out.

Now doing the same calculation as before, I will assume Lumar is a Mercury-like planet, the same radius, density etc, arch L = 1428,9 km, which gives Crow's Song average sailing speed of around 158,77 km/day (85,77 nm - why do people have to keep using so many different, weird units for every, single thing???), very close to an average sailing speed of 4 knots (precisely 3.57 knots), and that gives us a hidden height of 335,16 km, which, using angular size of 60 degrees to determinate the true diameter of the moon, means that Lumar's moons has around 171,67 km in diameter, and are 163,49 km away from Lumar's surface. And perspective works almost similar to Earth's radius size, as close to the edge of the sea of spores, its moon would only be around 7 degrees in angular size, and that's the same value for Earth's calculations. Lumar's moons would still have to drastically visually change in angular size to make up for that very close distance to the planet.

And because of that, Lumar's moons can be theoretically almost as big as 330 km, which would take 60 degrees of the sky around 300 km away from it, but right at the Lunagree, the moon would cover literally the entire sky, and that's way too big, and way too dark. For "some" fragment of the sky to be something else than a moon, it would have to be at most ~250 km in diameter, which would left like 3 degrees of normal sky visible from Lunagree. As you can see, there isn't much room to change moons diameter to make them look sensible. 

But there is a bigger (or rather smaller) elephant in the room. You see, that diameter is so small, that it means Lumar's moons can't be gravitationally spherical. The smallest spherical body in a Solar System is Mimas (I think), the moon of Saturn, with a diameter of almost 400 km. And 400 km is a lower limit for icy bodies to become spherical, for rocky it's around 600 km. So those moons are now too small to be round (which is still the smallest of all problems).

Roche limit for those moons - for density of Earth's moon - 3612,91 km, for density of Mimas - 5156,8 km, for density of Jupiter's moon Amalthea of 0,857 g/cm3 (which has similar diameter of 170 km but isn't spherical) - 5687,91 km, or have almost the same density of 36086 g/cm3 to make it work with new numbers. So yeah. This is still a bigger problem.

But Mercury-sized Lumar with moons with a diameter of 170 km is as impossible as Earth-sized Lumar with moons having 450 km in diameter. Moons are already impossible, so them being spherical changes nothing. Lumar is a truly magical world.

 

17 hours ago, AquaRegia said:

I hope you enjoyed it.  That was an awful lot of work to prove what we already knew: these moons are simply impossible for a variety of reasons, and the only way they can exist is "magic".

Oh, I'm still having fun  Knowing moons are impossible wasn't enough for me, I wanted numbers to see how much are they impossible! Now I have so many of them, and I'm satisfied.