Polar view of the planet

[Ixthos]

I did a quick sketch in FreeCAD to get a possible feel for how the planet would look from a polar view and to gauge how close the moons are to the planet and to each other - if anyone could check my maths and my assumptions I would appreciate it. This is what it looks like with only two moons, and I apologise for the roughness of the picture - I have removed the constraints to make it is easier to see how the planet looks:

So each moon orbits above the planet at approximately 1.8928 percent of the planet's diameter, and each is likewise 1.8928 percent of the planet's diameter. The top V shape is 60 degrees from the horizon and likewise 60 degrees between the two arms of the V, and the bottom V that connects to the planet centre is 30 degrees so that twelve of them may fit around the planet for the twelve moons. I used 60 degrees so that each moon would take up one third of the sky when seen from directly below (and this should still apply when considering a circular profile superimposed on a dome), and positioned the moons so that as one begins to dip below the horizon the other begins to rise above it. Again, please bear in mind that this is a rough sketch based on the information so far provided.

This is actually a lot smaller than I thought it would be, and also rather close. If the planet is the same size as Earth then the moons are each - assuming uniform size and distribution - approximately 241 km in diameter and their surfaces are 120.6 km above the surface of the planet, so (again, assuming the planet has the same proportions as Earth) they orbit at a radius of 6491.6 km from the centre of the planet (bearing in mind the planet is 6371 km in radius). This puts it above the Karman line, but only just, and I believe they would experience the effects of atmospheric drag, so this likely is being compensated for in some way. Also, further supporting that conclusion, they would have to be orbiting at a considerable speed to prevent them from colliding with the world. A quick and dirty calculation, again using the Earth as a reference, they would crash into the planet in two and a half minutes, so unless the planet orbits very quickly with a 1.27 hour day they likely are being held in place by some form of magic.

For comparison, our Moon - which is large as moons go - is around 3474.8 km in diameter (over 14 times the diameter of these moons) and orbits at an average distance of approximately 384400 km. Here is a graphic to display those proportions, as well as other objects in the solar system:

They are considerably larger than Phobos (22.2 km) and Demos (12.6 km), the two moons of Mars, which have orbits (if I have my facts straight) of 9377 km and 23460 km, respectively. This puts the distance between the surfaces of Mars and its moons at 5987.5 km and 20700.5 km respectively, just under 50 and 166.5 times the distance respectively between Tress's planet and its moons.

All this is based on the assumption that Hoid's description was mostly accurate, and using Earth as a base.

 

Here is what the planet would look like from a polar orbit with all the moons and with the pole marked:

 

This is a rather interesting configuration, and it almost certainly has to be maintained by magic.

Edited March 3, 2022 by Ixthos
Fixed a sentence and rearranged where some figures were located, and corrected orbit

[iceblade44]

Definitely set up with magic. Though if this was one of the planets Adonalsium set up, unsure how much of the magic still exists to maintain it.

[+Oltux72]

1 hour ago, Ixthos said:

I did a quick sketch in FreeCAD to get a possible feel for how the planet would look from a polar view and to gauge how close the moons are to the planet and to each other - if anyone could check my maths and my assumptions I would appreciate it. This is what it looks like with only two moons, and I apologise for the roughness of the picture - I have removed the constraints to make it is easier to see how the planet looks:

The problem is that the moons are in geosynchronous orbit. We are talking about a habitable planet. That makes geosynchronous orbit not lower than about 20 000km.

[Ixthos]

On 3/3/2022 at 7:10 PM, Oltux72 said:

The problem is that the moons are in geosynchronous orbit. We are talking about a habitable planet. That makes geosynchronous orbit not lower than about 20 000km.

They only have to be at that high due to the period involved, to make sure they orbit ("fall towards") the planet at the same rate the planet is rotating - for a different rotational period they would have a different orbit. I based my calculations on the statement they occupy 1/3 of the sky at any given time, and the assumption that when you stop seeing one you start seeing another. If the occupy 1/3 of the sky at any given time the furthest out they could be would be just over half the planet's diameter away, with orbits just over half the planet's radius - i.e. for Earth they would be touching one another if they orbited at a radius of 13208 km, well below 20000 km, and would collectively out mass the planet by a considerable margin, each being just over half the radius of the planet, and assuming uniform density with one another and the planet (a quick back of the envelop calculation says they would collectively be one and a half times the mass of the planet).

Edited March 5, 2022 by Ixthos
Corrected calculation

[+Oltux72]

4 minutes ago, Ixthos said:

They only have to be at that high due to the period involved, to make sure they orbit ("fall towards") the planet at the same rate the planet is rotating - for a different rotational period they would have a different orbit. I based my calculations on the statement they occupy 1/3 of the sky at any given time, and the assumption that when you stop seeing one you start seeing another. If the occupy 1/3 of the sky at any given time the furthest out they could be would be just over half the planet's diameter away, with orbits just over half the planet's radius - i.e. for Earth they would be touching one another if they orbited at a radius of 13208 km, well below 20000 km, and would collectively out mass the planet by a considerable margin, each being just over half the radius of the planet, and assuming uniform density with one another and the planet (a quick back of the envelop calculation says they would collectively be six times the mass of the planet).

Yes, this is a problem. Did you take each of them filling a third of the sky or all of them taken together filling a third of the sky or as many as you see at once filling a third of the sky?

[Ixthos]

Just now, Oltux72 said:

Yes, this is a problem. Did you take each of them filling a third of the sky or all of them taken together filling a third of the sky or as many as you see at once filling a third of the sky?

I did explain this in the first post  I took it as they always occupy exactly 1/3 of the sky, and so when one begins to drop below the horizon the other begins to rise, so the part that you miss becomes the part that you see on the next one (though  this also means that if you move beyond the equator there are areas you can stand where you can't see them anymore).

In the first image the slightly slanted blue line indicates the horizon, and the red dot on it is where you could stand where the top moon is just touching the horizon - and the further you go to the left the lower it will go - and the left moon is just about to come above the horizon. If you look on the diagram the left moon's top is just touching that blue line, while the top moon's bottom is likewise just touching it.

All this is based on the assumption that Hoid recounted relatively accurate proportions about the moons though, and he likely wasn't prioritising that 

[+Oltux72]

15 minutes ago, Ixthos said:

I did explain this in the first post  I took it as they always occupy exactly 1/3 of the sky, and so when one begins to drop below the horizon the other begins to rise, so the part that you miss becomes the part that you see on the next one (though  this also means that if you move beyond the equator there are areas you can stand where you can't see them anymore).

Yes. And there our problem starts. Let's look at the text:
 

Quote

Big enough to fill a full third of the sky, one of the twelve is always visible, no matter where you travel. 

That is ambiguous. It is also, if you take it extremely literally, impossible.
Either they are low, then we have to conclude that they are kept in orbit by forces beyond unaccelerated orbital mechanics. And even worse, they'd be invisible from polar areas.

Or they are approximately at the height a geostationary orbit would be above the Earth. Then you'll always see more than one and there is no way a single moon could fill a third of the sky, as they would have to touch.

[Zibus]

I take "big enough to fill a full third of the sky" to imply that they fill up to that, but probably there are some places on the planet where they do not appear that large. Just most/a lot of places. Hoid is almost certainly emphasizing their size for dramatic purposes.

I don't know a huge amount about astronomy and how orbiting works, but imagine however we end up slicing it they moons are going to have to be unrealistically close to the surface simply because we get spore fall on relatively localized areas at a pretty constant rate. Course, there very well might be a different delivery mechanism than just falling, but given the description I imagine we're talking dramatically close orbits.

[+Oltux72]

2 hours ago, Zibus said:

I take "big enough to fill a full third of the sky" to imply that they fill up to that, but probably there are some places on the planet where they do not appear that large. Just most/a lot of places. Hoid is almost certainly emphasizing their size for dramatic purposes.

Now that I think about it, there's another issue with an object filling 1/3 of the sky. The climate would be horrible.

In fact we get the indication.

Quote

And so, Tress would sit on her steps in the evenings, watching ships sail toward the horizon. A column of spores would drop from the lunagree, and the sun would move out from behind the moon and creep toward the horizon.

Quote

Except at the moment it was moonshadow, when the sun hid behind the moon and everything grew a few degrees cooler.

An object that fills a third of the sky by area has to have a radiuseven larger. The areas under the moons would be in darkness half the day.

[AquaRegia]

I'm very happy to see that others noticing the difficulties with the description of this planetary system.  Upon reading it I began trying to visualize this arrangement of moons, and it took about 30 seconds for me to realize "this doesn't make any sense at all."  I mean, I struggled with Roshar's moons - and read some terrific analysis by other 17th Sharders - over a period of weeks before giving up (they also don't work without magic)... but this is a slam dunk.

In addition to all the issues noted by @Ixthos and @Oltux72, there is also the Roche Limit: an orbiting moon can't be too close to its planet without getting torn apart by tidal forces.  The Roche Limit for Earth-Moon system is around 10,000 km.  At 120 km, these would be RINGS, not moons.

In the real universe, there is no way an Earth-sized planet can have 12 objects big enough to block 1/3 of the sky in a synchronous orbit.  Luckily, we have 1) magic, and 2) Hoid to smooth over our objections and allow us to suspend our disbelief.

[+Oltux72]

1 hour ago, AquaRegia said:

In addition to all the issues noted by @Ixthos and @Oltux72, there is also the Roche Limit: an orbiting moon can't be too close to its planet without getting torn apart by tidal forces.  The Roche Limit for Earth-Moon system is around 10,000 km.  At 120 km, these would be RINGS, not moons.

You'd have to assume considerable internal strength. And that means the absolutely have to be geosynchronous or the earthquakes will pulverize the planetary crust, if the moons weigh anything at all.

1 hour ago, AquaRegia said:

In the real universe, there is no way an Earth-sized planet can have 12 objects big enough to block 1/3 of the sky in a synchronous orbit.  Luckily, we have 1) magic, and 2) Hoid to smooth over our objections and allow us to suspend our disbelief.

I'll go as far as saying that nothing that obeys Euclidean geometry can have that.

Though I think we even have the answer in Warbreaker and Rhythm of War. That much Investiture does odd things to light and perception. The picture of the moons you see is not real. Meaning that if we want the true sizes of these objects we need to time the solar eclipses (or in theory stellar eclipses) and calculate the height of the geostationary orbit from the planetary mass and day length.

[Ixthos]

@Oltux72 on your previous point, I just realised one of my assumptions was flawed - because the moons are so close to the planet, and so small with respect to it, as someone moves further away the total angular area they cover in the sky would drop significantly. If someone was moving away from the first moon along the equator, at the point where the first moon begins to disappear under the horizon and the second moon begins to appear above it, the first moon would take up only 1/12 of the sky, so it would be four times as small as when seen from directly below - it still would dominate the sky, but it would be much smaller than 1/3. At the point on the equator where both are seen to be the same size each would take up only slightly more then 1/40 of the sky, and collectively would only cover 1/20. So yes, I agree, the information is very questionable - 1/3 of the sky may mean only when viewed from close to the Lunagree. The only way to make them consistently close to 1/3 would be to make them larger and position them further away, but this wouldn't fully solve the problem either, as the further away and the larger they become the closer together they become as well, and eventually they start to encroach on each other's space.

 

On 3/4/2022 at 8:04 AM, Oltux72 said:

Now that I think about it, there's another issue with an object filling 1/3 of the sky. The climate would be horrible.

In fact we get the indication.

An object that fills a third of the sky by area has to have a radiuseven larger. The areas under the moons would be in darkness half the day.

Agreed, though if they are sufficiently small, and the 1/3 is only near the equator, they probably wouldn't be as impactful as a ring system would be. The equator should be the coldest area - very different to Earth - but if their shadow is sufficiently small it wouldn't be as impactful on the planet as a whole as we might assume if this only applies only to human inhabited areas near the equator.

 

On 3/4/2022 at 8:34 AM, AquaRegia said:

I'm very happy to see that others noticing the difficulties with the description of this planetary system.  Upon reading it I began trying to visualize this arrangement of moons, and it took about 30 seconds for me to realize "this doesn't make any sense at all."  I mean, I struggled with Roshar's moons - and read some terrific analysis by other 17th Sharders - over a period of weeks before giving up (they also don't work without magic)... but this is a slam dunk.

In addition to all the issues noted by @Ixthos and @Oltux72, there is also the Roche Limit: an orbiting moon can't be too close to its planet without getting torn apart by tidal forces.  The Roche Limit for Earth-Moon system is around 10,000 km.  At 120 km, these would be RINGS, not moons.

In the real universe, there is no way an Earth-sized planet can have 12 objects big enough to block 1/3 of the sky in a synchronous orbit.  Luckily, we have 1) magic, and 2) Hoid to smooth over our objections and allow us to suspend our disbelief.

Ahhh! I can't believe I forgot about the Roche limit! I'd been doing research on it a few months back, thank you for reminding me about it! I agree - they definitely have to be held together by molecular forces or magic - gravity alone would be overcome at those distances. That and Hoid, as you noted. He certainly doesn't strike me as someone who would prioritise accurate facts over embellishments for poetic effect.

 

On 3/4/2022 at 10:27 AM, Oltux72 said:

I'll go as far as saying that nothing that obeys Euclidean geometry can have that.

Though I think we even have the answer in Warbreaker and Rhythm of War. That much Investiture does odd things to light and perception. The picture of the moons you see is not real. Meaning that if we want the true sizes of these objects we need to time the solar eclipses (or in theory stellar eclipses) and calculate the height of the geostationary orbit from the planetary mass and day length.

I'm reminded of a short story I read a few years back, [found the name: The Pacific Mystery by Stephen Baxter], but basically, while the Earth was a sphere it was a sphere with non-Euclidean geometry, where you could keep going west and never wrap around again, a combination of a sphere and a helix in a sense. I like your suggestion about how magic could warp their perception as well, that objects can be distorted in apparent size due to magic.

Edited March 5, 2022 by Ixthos

[+Oltux72]

2 hours ago, Ixthos said:

@Oltux72 on your previous point, I just realised one of my assumptions was flawed - because the moons are so close to the planet, and so small with respect to it, as someone moves further away the total angular area they cover in the sky would drop significantly. If someone was moving away from the first moon along the equator, at the point where the first moon begins to disappear under the horizon and the second moon begins to appear above it, the first moon would take up only 1/12 of the sky, so it would be four times as small as when seen from directly below - it still would dominate the sky, but it would be much smaller than 1/3.

Are you giving those fractions as fractions of the area or fractions of the arc?

[Ixthos]

1 hour ago, Oltux72 said:

Are you giving those fractions as fractions of the area or fractions of the arc?

Fractions of the arc, but under the assumption that the area, once rotated, is identical in proportion. Here is my reasoning, though note its been a while since I've done calculations of this nature, so if anyone can see any errors in this conclusion please let me know.

If each moon always form a circle when looked at then they occupy a circular area on the "dome" of the sky, as that dome is the typical view one was of the sky from the surface - a dome ringed by the horizon. If you then shift that circle to be directly overhead then haven't changed the area, only shifted where it is. Now, if it is overhead and you take an infinitesimal slice through it from one point of the horizon, passing overhead through the very top of the dome (where the circular view of the object has been re-positioned), and to the point directly opposite that starting point, you can work out how much of that slice is taken up by the arc corresponding to the circle in that slice. If you then take another slice from any other point on the horizon through the centre and back to the horizon you will have the same result, the same proportion taken up - thus the sum of each infinitesimal slice will preserve the relationship between how much is empty sky and how much is taken up by the circle, as with sufficiently small slices the area at which they overlap within the circle tends towards zero.

[Edit] Disregard the above - I hadn't thought this through as well as I'd thought; for example the area of a circle half the size of another circle is a quarter of the size of that previous circle. I will have to go over this more thoroughly later, as our Moon's area in the sky is apparently 0.2 degrees^2 but it as an arc of 0.5 degrees on average.

Edited March 5, 2022 by Ixthos

[Fritochip]

With your assumptions about the gravity of the moons, would we be able to figure out the approximate height of the tide under the moon. I suspect The Rock must be a very tall mountain as it would sit at near the peak of the tidal area under the Verdant Moon. The oceans must then be very full as the low area between moons would be much, much lower. 

I am glad you are talking about the Roche limit. This whole system feels impossible without magic to me. You could solve some problem by changing the gravitation of the moons, like if Kaladan gave them a 1/3 lashing up. This could explain a situation in which the moon pulls on the planet at a different gravitation than the planet pulls on the moon.

Edited March 5, 2022 by Fritochip

[Ixthos]

4 minutes ago, Fritochip said:

With your assumptions about the gravity of the moons, would we be able to figure out the approximate height of the tide under the moon. I suspect The Rock must be a very tall mountain as it would sit at near the peak of the tidal area under the Verdant Moon. The oceans must then be very full as the low area between moons would be much, much lower. 

I am glad you are talking about the Roche limit. This whole system feels impossible without magic to me. You could solve some problem by changing the gravitation of the moons, like if Kaladan gave them a 1/3 lashing up.

Yup, though tide may be a little misleading in this sense, as the moons remaining fixed means they may not notice their are different levels, each area remaining at a fixed height.

Indeed - this world really is only possible with magic. I've been looking over my figures and while I think I've made several mistakes, those mistakes are ones that imply I've made the moons too small and too far away, so the actual results have the moons as both larger and closer than I've determined them to be - if anything, their orbits are even more extreme than my model shows.

Edited March 5, 2022 by Ixthos

[AquaRegia]

I, for one, have reached the point where I don't feel like it makes much sense to continue trying to explain this system using actual physical laws, at least, not until we get more information.  Those moons (and the seas of spores, for that matter) are just flat-out magic.

And I thought the Rosharan moons were a problem! pffff  (protip: they are)

[Ixthos]

10 minutes ago, AquaRegia said:

I, for one, have reached the point where I don't feel like it makes much sense to continue trying to explain this system using actual physical laws, at least, not until we get more information.  Those moons (and the seas of spores, for that matter) are just flat-out magic.

And I thought the Rosharan moons were a problem! pffff  (protip: they are)

Yup! Even dropping the spores in a column doesn't make much sense. I also am starting to wonder if the moons are supposed to be located all around the planet, not just on the equator, but I need to reread the descriptions again. If that is the case though then they also can't be orbiting, they would have to be floating above the surface of the planet somehow.

[+Oltux72]

21 minutes ago, Ixthos said:

Yup! Even dropping the spores in a column doesn't make much sense. I also am starting to wonder if the moons are supposed to be located all around the planet, not just on the equator, but I need to reread the descriptions again. If that is the case though then they also can't be orbiting, they would have to be floating above the surface of the planet somehow.

The spores "filter down". We must assume that the moons eject them on purpose for an unknown reson by an unknown mechanism.

But I am afraid the eclipses dictated that they are on the equator and Diggen's Point is a tropical island.

Quote

And so, Tress would sit on her steps in the evenings, watching ships sail toward the horizon. A column of spores would drop from the lunagree, and the sun would move out from behind the moon and creep toward the horizon.

Regular occultations require a satellite over the tropics.

[Ixthos]

1 hour ago, Oltux72 said:

The spores "filter down". We must assume that the moons eject them on purpose for an unknown reson by an unknown mechanism.

But I am afraid the eclipses dictated that they are on the equator and Diggen's Point is a tropical island.

Regular occultations require a satellite over the tropics.

I agree on that, I'm just noting that for the spores to land below the moon they would have to be specifically targeted at that area - compare how the moons drop the spores vs how Threadfall happened in the Dragonriders of Pern series.

We know that the Emerald Sea had such eclipses, but it isn't clear if all inhabited areas have those moonshadows. I don't doubt Tress's home is on the equator, I'm just wondering if ALL the inhabited areas are by the equator, and if the moons are only over the equator.

[+Oltux72]

19 minutes ago, Ixthos said:

I agree on that, I'm just noting that for the spores to land below the moon they would have to be specifically targeted at that area - compare how the moons drop the spores vs how Threadfall happened in the Dragonriders of Pern series.

It may just be economics. It is the shortest distance.

19 minutes ago, Ixthos said:

We know that the Emerald Sea had such eclipses, but it isn't clear if all inhabited areas have those moonshadows. I don't doubt Tress's home is on the equator, I'm just wondering if ALL the inhabited areas are by the equator, and if the moons are only over the equator.

Technically the sample chapters never say that the moons are over the equator or even that they are regularly distributed over the planetary surface. That follows only from orbital properties. If you are willing to take the Verdant moon to be on the equator by chance and/or the moons to hover by a force other than gravity, all bets are off.

[cometaryorbit]

Yeah, I think it would make the most sense if the "third of the sky" is some kind of visual distortion (or Hoid exaggerating). On Earth, a geosynchronous orbit is about 23,000 miles up - about a tenth of the distance to our Moon - so if the Moon was in geosynchronous orbit, it would look about ten times its current apparent diameter (a hundred times its current apparent area) - about 5 degrees across, nowhere near a third of the sky. The Earth is about four times the Moon's diameter, so even another Earth wouldn't take up a third of the sky.

So unless this planet rotates way faster than Earth so geosynchronous orbit is much closer - which would mean a really weird day-night cycle - the moons can't literally be that large and still be moons.

[LewsTherinTelescope]

On 3/5/2022 at 9:58 AM, Ixthos said:

Yup! Even dropping the spores in a column doesn't make much sense. I also am starting to wonder if the moons are supposed to be located all around the planet, not just on the equator, but I need to reread the descriptions again. If that is the case though then they also can't be orbiting, they would have to be floating above the surface of the planet somehow.

He mentions that "one of the twelve is always visible, no matter where you travel", which I took to indicate that they're not just on the equator, but it's not really certain. But combined with the fact that only one moon is visible where Tress is, it feels to me like they'd have to be spread out, because I don't think someone closer to the poles would see any, in that case? But this is entirely gut feeling and I know exactly nada of the physics involved, so I could be wrong.

I feel like it's extremely likely that future drafts of the book will give up on trying to make it at all a true orbit, going off all the issues raised here, tbh.

[Ixthos]

3 hours ago, LewsTherinTelescope said:

He mentions that "one of the twelve is always visible, no matter where you travel", which I took to indicate that they're not just on the equator, but it's not really certain. But combined with the fact that only one moon is visible where Tress is, it feels to me like they'd have to be spread out, because I don't think someone closer to the poles would see any, in that case? But this is entirely gut feeling and I know exactly nada of the physics involved, so I could be wrong.

I feel like it's extremely likely that future drafts of the book will give up on trying to make it at all a true orbit, going off all the issues raised here, tbh.

That does seem likely. We'll just have to see what Brandon ends up saying. If they aren't in a true orbit and instead are maintaining their position through magic then they may be making a duodecahedron configuration, with a moon possibly floating above each pole. It would space them out some more and ensure each part of the planet has one visible, though if Brandon wants them to orbit and be seen from all parts of the planet they would have to be orbiting much further out, and would have to occupy much less of the sky - for now I'll assume Hoid was exaggerating the percentage they take up (he is the sort to do that - consider the Duke's speech when introducing his new heir )

[+Oltux72]

10 minutes ago, Ixthos said:

That does seem likely. We'll just have to see what Brandon ends up saying. If they aren't in a true orbit and instead are maintaining their position through magic then they may be making a duodecahedron configuration, with a moon possibly floating above each pole.

If you do that either the regular solar eclipses or the extreme apparent size due to a position very close to the planet go away. In that configuration, objects will be at the poles and at 30° north and south of the equator. But for a regular eclipses by objects so close, the sun needs to be close to the zenith. You cannot combine that with a terrestial axial tilt.

10 minutes ago, Ixthos said:

It would space them out some more and ensure each part of the planet has one visible, though if Brandon wants them to orbit and be seen from all parts of the planet they would have to be orbiting much further out, and would have to occupy much less of the sky - for now I'll assume Hoid was exaggerating the percentage they take up (he is the sort to do that - consider the Duke's speech when introducing his new heir )

Indeed. Every single feature of the moons Brandon has described is possible. But you cannot have them all and preserve Euclidean geometry.

I also like the exaggeration explanation best, but that is personal preference.