Message in a Bottle Planetary Calculations · 10:34am Jul 8th, 2017
So, for those of you who haven't read the July 8th chapter of Message in a Bottle (G5.05: Satelite), probably go and read that before going on. Either these things aren't going to make sense to you, or they're going to be a major spoiler for that story. So seriously, go read that, then come back here. We good? Good.
So I am not a numbers pony. I can handle some basic calculations, but nothing like what would be required to make sure the megastructure I wanted for Message in a Bottle would work according to known laws of physics (though not necessarily using any known materials). I have a very particular image for what I wanted for the Equestrian homeworld.
Enormous thanks to my friend Star Scraper for his help with all the nitty-gritty math. Hopefully, if you have any questions about what I'm about to show you, he can answer them, because I can't.
It's very important to me to have a story that's as plausible as possible, even given the remarkable things depicted in Message.
So first, a brief word about scale.
This image is more for understanding than actual proportions. The true width of the ring is closer to this:
Because gravity is a product of rotation, it varies as we approach the edges of the ring. If you want some idea of the scale of that variability, it's something like this:
The mathematics used to reach these numbers are entirely beyond me, but for those of you who like to get real deep into the math, I've got the link to all the calculations for you to check them out. That's an album, by the way, not one page.
The narrative isn't going to be about any of these numbers... but I wanted to give those readers who do care a snapshot of the lengths we've gone to in order to make sure the physics of the story make sense.
Wait a minute, let me see if I understood this correctly. The circumference of the Ring is about the same as for our RL Sun? And the width (along the surface) is in the range of RL Earth diameter? Because intuitively it feels like the variability of apparent surface gravity should not be as strong in that case. Also I question the creators' choice of using the star's gravitational pull to induce surface gravity instead of rotation (basically the other way round), but hey, that might be a plot point or just some intruiging background mystery.
In any case, I applaud your and your friend's effort. Well done.
Not gonna go into the mathematics of it all, but assuming a normal red dwarf...
The temperature ought of the star to be less than 4000 K. Let's assume it's at 4000 K.
The radius of a red dwarf is ~ up to 0.6 solar radii, so let's assume that.
The radiant emittance is ~ 15 MW/m2
Thermal radiation power ~ 3.3*1019 MW
The Earth receives ~ 1000 W/m2 energy from the Sun, though some of it is cooled off on the night side of the planet.
Assuming an Earth-like atmosphere and cooling, Equis should receive roughly the same amount.
Assuming Equis is a cylindrical shell (given how narrow it is compared to the radius, not a big loss if it's really a slice of a spherical shell), the receiving area is 2*pi*d*r.
Thermal radiation power at the cylinder's distance is (3.3*1019 MW)*((0.6 solar radii)/r)2.
Thus the total amount received is ((3.3*1019 MW)*(0.6 solar radii)2)/(2*pi*d*r3)
Assuming the width of Equis is one Earth diameter, the cylinder should have a radius of
(((3.3*1019 MW)*(0.6 solar radius)2)/(4*pi*(1 Earth radius)*(1000 W/m2)))1/3 ~ 0.3 AU
Any further, it's too cold. Any closer, it's too hot.
Anyway, if the centripetal force is the main source of "gravity", spinning at 422 km/s at the center circle of the cylindrical shell, the centripetal acceleration is (422 km/s)2/(0.3 AU) ~ 0.4 g
Clearly, the radius needs to be smaller for ~1 g acceleration (or spinning faster), but that's not a problem. Just use a smaller or colder red dwarf.
What about curvature? A 10 km height difference occurs at an angle of
arccos(1 - (10 km)/(0.3 AU)) radians or about 2 arc minutes.
The distance between two such points is the chord length, which is
(0.3 AU)*2*sin(arccos(1 - (10 km)/(0.3 AU))/2) ~ 30,000 km or about 8% of the Moon's orbital radius.
An Earth wide Mount Everest at a distance of 8% to the Earth's Moon, and a height difference at an angular diameter of 1 arc minute.
And the chord distance only gets smaller as the cylindrical radius decreases.
And the radius cannot get larger lest Equis is too cold or the star is not a red dwarf.
In other words...
THE CURVATURE OUGHT TO BE VISIBLE WITH THE NAKED EYE!
And correct me if I'm wrong, but Lucky hasn't noticed that yet?
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Best I can figure is the spin is to counteract the pull of gravity as it tries to pull the ring down into the sun - critters live on the outer surface, not the inner surface ala Halo (a widely known example). I can only assume that, as a result, the gravity math checks out different than 0.4g's.
If the outer surface is the livable one then the distances from 4595515 might not be accurate. You would need an artificial star for a satellite though, and that would be glaringly obvious. Also, weird science.
But Star, this is a ring world. You can't treat it as a spherical one for gravitational calculations. Hell, you have to consider that a structure so massive exerts pull from it's other side too. Or freaky space magic.
Actually, it wasn't quite clear enough on the story how this works. The ring spins? Why? Is it considerably smaller than 4595515 suggested? If so, then they are on the outer side. If they are there the diferences in radius from the mass center, if the ring is "flat", are so minute that they become irrelevant for gravitational attraction. They could make it somewhat round, to create an horizon as we know it, and it would have no appreciable effect on the gravity.
Also, gravity is not a product of rotation. If they are on the inner side it could be so, but again as it's a ring world the radius you use is the center of the star. And when you're measuring stuff in AU hundreds of thousands of kilometers are minute diferences. That much diference in G's in your image makes no diference whatsoever. For those to make sense it seems either the calculations took some weird numbers in consideration or this ring world is not nearly a ring world. How big is it, anyway?
Anyway, I'm calling shenanigans. Need more numbers because else it doesn't add up. And yes, I'm considering that WITH the imgur album open.
Edit: Finished current story comments, and Zutcha did make it clear that the inhabited side is the outer one. So with mountain ranges big enough to stop the edges from seeing/interacting with sides, that's one less problem. The ring spinning makes sense if it's considerably smaller, near enough that they need it to counterbalance the stat's attraction. The disparity from 1.2 to 5 G still makes no sense at all. Also, 422.066km/s? Is it 422 + 66 thousandths of a kilometer per second, or 422,066? I imagine that the former, because the latter is faster than light. The former is also about 900 times faster than earth. That's ought to create some tremendous winds. As in perpetual hurricane speeds everywhere.
Of course, there's sufficiently advanced technology to take into consideration. Or magic. One and the same, ain't it? It could explain Equestria's dependency on pegasus-controlled weather. The ring' systems might not be up to snuff on that department and they have to compensate.
All in all, Magic will be the answer, be it in Clarke's meaning or not.
The G's diference still make no sense, though.
4595542
That's arguably worse.
If we assume zero thickness, the "gravity" at the center circle of the cylinder: a = μ/r2 - v2/r
Solving for r gives us (sqrt(v4 + 4aμ) - v2)/(2a).
Based on the stellar classes found on the Wiki page for red dwarf stars, the mass M is roughly proportional to the radius of the star R, thus the standard gravitational parameter μ is also roughly proportional to R.
So if the largest class, M0V, has a μ ~ 8.0*1019 m3/s2, then we can model the parameter of the other classes a fraction of this.
μ = k*μ0, μ0 = 8.0*1019 m3/s2 , 0 < k <= 1
Similarly, we can let R = k*R0, R0 =4.2*108 m.
Thus r = (sqrt(v4 + 4akμ0) - v2)/(2a) > kR0
Solving this inequality requires k < (μ0/R02 - v2/R0)/a ~ 3, for a=1 g.
So we know we won't be inside the sun, at least...
However, assuming a = 1 g, that is still a separation of at most 2*107 m between the cylinder and the surface of the sun and it only decreases for smaller stars.
For reference, 2*107 m is roughly half the circumference of the Earth or 5% of the Moon's orbital radius.
At greatest separation (k=1), the radius of the cylinder would have to be ~ 4.4*108 m or roughly the Earth-Moon distance.
The thermal power received at that distance is in the MW/m2 range.
The smallest known red dwarf is 2MASS J0523-1403, sitting at <0.08 solar masses, ~0.086 solar radii and ~2000 K.
Using this star, r ~ 6*107 m which is basically just on the skin of the sun due to measurement precision. Power emitted is ~4*1016 MW.
Equis would then receive ~ 900 kW/m2. That's 900 times what Earth receives. It's 90 times what Mercury gets! A perfect black body would reach ~2000 K in a steady state, a grey body even more. Tungsten and asbestos would melt!
You would need some wicked cooling to stave off the heat, at the very least from thermal conductance, if you're on the "dark" side.
Curvature is a bit easier to explain, I guess, as you can't see through the ground. 10 km essentially "disappears" every 2000 km or so.
How you get sunlight, though, is another question. Convex mirrors, perhaps.
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If you read the chapter, there are two satellites outside the ring. Presumably, these are the "Sun" and "moon" that the princesses are moving every day.
Halp! There be evil numbers and letters and symbols attacking me! Me wee brian can't take much more...
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If so... One, I'd expect them to be orbiting opposite each other if they're massive enough to have any serious gravitational pull on the ring due to reason number three.
Secondly, if they're needed to reduce the needed strength of the material to resist the gravitational pull (why? Inertia takes care of it, no?), then what the hell happens in the twilight zones where they barely have any effect (if not even increases net the pull towards the star)?
Three, ringworlds (Niven's ringworlds, at least) are inherently gravitationally unstable. Rotating the ring to orbital speed doesn't help because the ring is rigid and wraps around the star. The smallest perturbation would be disastrous. Thus the satellites, if they're massive enough, need to be of equal mass and orbiting opposite each other, lest the gravitational pull is off-balance, sending the ring into the sun. Stabilizer jets would be needed as well.
But if the jets couldn't counter the satellites...
Well...
Nightmare Moon is insane.
4595799
Poor summer child, such is called engineering. This is but a tiny fraction of what we'd need to build such an structure... IF they are possible. Our current science says nay, therefore space horse magic. Either that or sufficiently advanced space horse technology. One and the same, really.
4595515
Your assumptions are seriously flawed in one respect: Assuming that Equis is being heated directly by the star. Or rather, the section on the exterior surface which has living ponies on it.
This is not the case. The inner layer facing the star could be capturing far more energy, and doing who knows what with it. Considering the size of the ring, there's an enormous land area far away from Equestria (and every other country we've seen or heard about in the show) that could be doing things with the energy, or just radiating it out into space.
Let me just say that it seems highly likely a race capable of building something like this (overcoming the enormous engineering problems one would encounter in the process) instead of just building on planets, probably has more in mind than just a habitat. There are so many awesome things one can do with all that energy, and so many places ponies would never see to vent the heat away when you're done.
Please Specify are the satellites natural or artificial? Are they natural Celestial bodies (Say a rocky moon and a small dwarf star) or are they artificial ones that create some kind of trickery to look natural?
4596123
Artificial. Their surfaces are specifically designed to look like the heavenly bodies ponies observe in their skies (the sun and moon).
4596141
Ah that makes sense.
Questions:
How does the moon work?
and,
What about the edges, has anypony discovered them?
How big is the "sun" compared to the ring? If it is relatively small, the daytimes should be pretty short compared to the night. Unless they have more than one to make up for that.
4595816
I, for one, welcome our new space horse overlords.
Here's a solution to your engineering problems if you like.
Using this method, you can just have the ring be made of iron, be stable, etc. Also, perhaps the heat energy is fed into a magic field that ponies use. That would explain how ponies can do impossibly energy intensive things like teleport or transform things.
4597444
Someone's thinking with portals!
Well'p, I'm confused. I'm sure this is over my head, but I think with it being 3AM right now the "over my head" turns into "a mile over my head". Maybe I can make some sense of this in the day time.....speaking of which, HOW THE FRICK WOULD DAY AND NIGHT WORK?!
So... there are 80 human satellites, several flat satellites, and two heavy satellites?.. Comments seems to imply that those heavy satellites and flat satellites are the same thing, but why "a few" and not "two" then?..
4597444
After thinking on this, I realized that the ringworld would have to be the variety that is rotating fast with stationary magnetic rings inside. They would then act as a compressive force counteracting the tension from the spin. This would reduce the forces acting on the world so much, I think, that it could still be made of iron or steel.
As i understand the science, natural gravity is a product of mass curving space around it, so that objects with mass naturally try to fall towards each other; Centrifugal Gravity and Linear Gravity are both feasible methods of simulating Gravity, but neither of them produces real gravity, and each has its limitations:
- Linear Gravity requires perpetually accelerating in a straight line, which limits its usefulness to things like space ships.
- Centrifugal Gravity works great on a stationary installation, but comes with a fun little tradeoff called the Coriolis Effect.
I'm not a numbers guy either, but i know enough of the basic theory, to sense that you've got a serious potential problem lurking in this wings; the numbers may match up fine for balancing natural against artificial gravity to arrive at an Earth equivalent gravity; but counter balancing such extreme forces is certain too cause all manner of strange looking Coriolis Effects, objects dropping through air may bend their flight path in tune with the rotation, and other similar phenomena.
But Rotating Reference Frame shenanigans are a mere side show attraction, compared to the real problem:
Once any object or being leaves the ground of this Ring World, it is no longer being centrifuged away from the star's gravity, and only air resistance will slow its descent as it comes back down under the full might of natural gravity...
sciencebasedlife.files.wordpress.com/2012/06/its-not-the-fall-that-kills-you_o_22020.jpg
Ow.
My brain hertz.
This level of math and science is WAAAYYYYYY over my head.
You and your co-conspirator have created a strange inverse Niven Ring.
As for the thermal loading... Maybe your dwarf star is either an off-the-charts small/cold red dwarf or an off-the-charts hot brown dwarf. Neither is ENTIRELY beyond the realm of possible. We haven't seen EVERYTHING there is to see in the universe.
Hello, everyone. I'm the one who ran the numbers if you're not already aware. I'm on a trip right now, and even when I'm not, I don't have all the time in the world. (Curse my procrastination in writing my own fics!)
But I do love talking about this stuff, so I'll reply to a little bit at a time, in and out.
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The reason the math is so complicated is because it takes into account the full 3d coriolis effect by examining straight lines in regular 3d (Cartesian space) in a spherical polar coordinate system.
Or did I use cylindrical? I think I used spherical, off the top of my head.
The g values on the graphic are derived from that. Smaller radius due to holding a constant r from the sun's center means less centrifugal acceleration, thus less to counteract the star's immense gravity. Edit: Also an angle in-between the centrifugal acceleration and gravitational acceleration develops. See my later comments for more explanation.
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All I can tell you is that math beats intuition, and your intuition is wrong about the gravity. I've done the math, it is there for you to see.
Remember that centrifugal force has a direction along the radial coordinate in a cylindrical coordinate system, while gravity is along the radial coordinate in a spherical coordinate system.
If the latitude is zero in the spherical coordinate system, the two are parallel. If you go off that equator, though, an angle develops in-between the two. The surface gravity of the smallest Red Dwarves is around 700g's. Even a small angle in-between the two means a lot of gravity.
That, qualitatively, is why the gravity is so strong at such a small latitude.
Secondly, it is actually a slice of a sphere, not a ring. I discussed having the edges "flare up" into sections of a cone, I think Starscribe went with that, iirc. Since "uphill" is an increasing r coordinate in spherical coordinates, in order for the surface to seem "flat" (ie, I'm not going uphill when I go North or south from the equator) under the star's gravity, it must be a section of a spherical surface.
Switching to cylindrical, or even a sphere with less curvature, would make that section go "uphill" as you move away from the equator. This would be a nice, more natural looking way of retaining the atmosphere at the edges.
Furthermore, on the topic of heating - extracting energy from a system cools it by definition, if you are taking thermal energy out to store it. I don't know how much spoiler Starscribe wants me saying, so I'll err on the side of caution, but this is one seriously cool structure, guys. Whoever built this knew how to prepare for the eternal cold.
Anyways, consider a rocket engine. If you go read up on them, in ideal rocket engine actually has exhaust at absolute zero - all the thermal kinetic energy is transferred to the vehicle.
Of course real rockets don't do that, but I did run the situation by the heat pump equation for max theoretical efficiency and yes, with a cold temp of the CMB and hot temp of a small Red Dwaf's surface, you can get it very, very cold, using power from the temperature gradient.
An example of this type of system is a closed-loop nuclear reactor. Using power from nuclear decay, even on a closed loop you can have reactor core temperatures at very hot temperatures, but have the power plant be cool, and cool the reactor core with the power it produces.
The reactor core is the star, and the powerplant is the ring. Same thermodynamic system. It's possible. Just saying that preemptively because I've had people tell me it's not.
4595505
No, it is a very small Red Dwarf. The strength in the gravity was carefully calculated with the appropriate math and physics. See my previous comment.
Apologies to not having all the exact numbers off the top of my head, I just saw a link to this and my laptop isn't set up so I'm replying on my phone while trying to get ready for stuff tonight. I can fill in details later if y'all want.
4597990
Also, coriolis effects are inversely proportional (don't know the power) to radial distance and proportional (also don't know the power) to the rate of spin. (Though all the equations actually should be in the math Starscribe linked)
Earth has coriolis effects, of course, hence hadley cells, hurricanes, and, well, weather.
This thing is spinning a lot faster but is also bigger.
And then how much of this happens will depend on viscosity and heating differences, as well.
In a nutshell, you might be right, but the exact calculations to show one way or another are beyond what either of us is willing or capable of doing (with current knowledge, at least, and I'm not sure if it's serious enough to warrant some weeks studying or days intensely studying. If I'm going to study, it's going to be relativity. And thermal gradients would be largely guesswork, anyways. Fortunately, as it happens, relativity study gave me the mathematics to study the coriolis effects. So awesome! :rainbowsoawesome: ). What I worked out is a starting point, but then you basically need to incorporate that into a numerical solution to compute fluid dynamic effects to see how those tidal forces would effect the weather - and by how much.
But my hunch is that it won't be so extreme, largely due to the fact that the ring is so narrow, greatly restricting the extremes, unlike, say, Jupiter's bands that go over the entire sphere. I think the smooth gradients at the edges (curving up instead of a solid wall) will also cause a lot of friction with the air and reduce the effect as well, where it would otherwise be most prominent.
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Basically what Neece said. If the structure were dead, it would be thousands of degrees and uninhabitable, so some systems must still be running. Among those, you are right - would have to be station-keeping systems.
But it's by no means a show stopper. We've done station keeping since Gemini in the early 60's. This is on a bigger scale, but building big obviously isn't a problem for whoever built this.
Also, like the station-keeping we humans have done, It's a positive feedback system. Which means a lot of how well you can fight it depends on how well you can sense and predict it. I'd guess these guys are pretty good at the required computer modeling to do a fantastic job predicting and sensing it, so as to do it with great efficiency.
4599086
<<<Edit: the problem with how long it takes me to type out these complicated responses, is i cant see if if a second response rolled in while i was typing; the Coriolis effects may indeed remain small to the point of irrelevant, but my primary concern still stands.>>>
The design of this ring world is counterbalancing two absurdly extreme forces against each other, its a tug of war, and the poor ponies are the rope.
The sheer size of the ring means that if only Centrifugal gravity was was being used, the Coriolis effects would be barely noticeable too most observers, and spin induced nausea effects are unlikely to ever be a concern, even at the insane rate this thing is spinning.
But then you've got the Centrifugal forces being counterbalanced against an intense gravitational field from proximity to the star; and now things start to get... interesting...
Math was never my strong suit, and i do not doubt that the numbers you crunched balance out neatly enough for making the effective gravitational pull at any point on the rings surface equal the expected amount; but can they account for what happens when an object or living being leaves contact with the spinning ground for longer than say, a few seconds, while they are still being dragged down by the gravity field?
One of the inherent limitations of Spin Gravity, is that the effect decreases and then disappears altogether if you slow down relative to the spin surface; while standing on the spinning surface this is no concern, but as soon as you leave that surface you become vulnerable to slowing down and losing the gravity simulation, as the effect is not a constant field.
In a Zero-G environment, losing contact with the "ground" in a spinning habitat is merely inconvenient, as you just end up floating; but this "habitat" also has a natural gravity field present, one powerful enough to cause terminal velocity in an uncontrolled fall from high up to be much greater than is encountered here on Earth.
4599185
Interesting point. Looking into it...
About 30 mph will make like a 15% difference. Fascinating. This wouldn't be too noticeable in a pre-industrial world (barring high speed ponies at times, like Rainbow), but get jet airlines and it'll effect flight times.
4599227
The human characters have such an aircraft. What would they observe while using it?
So wait... Halo? Can't wait to see chief and Cortana make an appearance.
4599281
Good news! I was off by a factor of 10 because I was thinking I was using g's when I was actually using m/s, hah. Corrections are as follows:
They'd observe gravity getting stronger or weaker depending on which direction they fly. About every 1,454 mph (660 m/s about) would change the gravity by 1 g. I can get a graph for you once my laptop is set up, but it *should* be *roughly* linear since we're taking a small differential to our overall speed, 422 km/s.
So going just under the speed of sound (340 m/s) could make you 50% heavier (going against the spin of the ring) or 50% lighter (gong with the spin of the ring). That of course massively effects takeoff weight, cargo capacity and cruising altitude. Lower altitudes mean slower, too, though. Lift is proportional to speed. Denser air means more drag but also more lift so you can fly while heavier, but it's easier to go faster higher because less drag. Jet engines are actually more efficient at higher speeds, but cannot combust supersonic air (supersonic jets reduce the mach number in the engine through slowing the air with drag, and a greater speed of sound in compressed air, thus lower mach number at the same speed - though that's not why they compress air (see: Brayton Cycle), it is an advantage of it. They also have little sections so combustion is in a "backwater" sort of region where air is slower. The end effect is turbojet aircraft can go supersonic but the air in the combustion portion of the engine isn't supersonic. I think the SR-71s, though, were kind of special in that their engine functioned partially as a ramjet, allowing even faster flight while retaining subsonic combustion. Scramjets are unique from where their name derives, "Supersonic Combustion RAMJET", thus "SC RAMJET")
(also, making assumptions off underlying physics, I think that may be off - and the air must be subsonic long before combustion and supersonic combustion itself isn't an issue, because scramjets supersonicly combust, but a classic turbofan or turbojet engine compresses air with fan (compressor) blades, and I don't think that would even compress air at supersonic speeds, though the underlying shaft housing geometry might, the drag might be too much. Whatever the case, I just happen to know that supersonic air in the combustion portion of a classic turbojet engine will make it not work, and a turbofan can't even have it supersonic on the fan ie, you can't get a vehicle supersonic with a turbofan or prop, you must use a turbojet or something else, though turbofans are way more efficient, hence turbofans on airliners and turbojets on fighters).
So, higher is faster and more efficient (that's why airliners go so high they need a pressurized cabin) but you have to be lighter to do it.
If you're heavier, you can still fly lower in denser air but you're less fuel efficient and slower. And of course it reaches a point where surface air isn't dense enough and you can't take off at all.
I'm not sure how all that applies to prop engines with regards to fuel efficiency. The rest is aerodynamics irrelevant of engine type, but the fuel efficiency part I'm not sure of.
As for helicopters, they really struggle to fly very high at all. It's basically unheard of for one to have or even need a pressurized cabin I think. Rotor wings of course generate lift by spinning, not from the vehicle's airspeed, so that other stuff doesn't apply (?).
But on the upshot, it means shorter runways if you take off with the direction of the spin of the ring, and greater lift/weight. Perhaps you could even gain speed and altitude "Eastward" (spinning like Earth to define "East") before turning to a more westerly heading, where you'd lose altitude, gain speed, and reach a new faster and/or lower equilibrium (because you got heavier when you turned west).
Not sure what exactly it would be... Lower means more drag, but dropping gives you speed. So initially faster, but I'm not sure where you'd come to equilibrium. Suddenly gaining weight is a funny thing, lol. Basically you'd gain speed as you drop, but the gained speed would make more lift, and the dropping would make more drag which would take the speed off.
Gliders do that, but of course, being unpowered, they don't come to a constant-altitude equilibrium like we would here.
If you reduce power in a plane, you'll lose speed and drop, then come to equilibrium lower and slower, but that's reducing thrust, not increasing weight...
Anyways, it might be savvy to fly in a great right triangle if you need to go west (with the hypotenuse going directly west), so that much of your velocity vector is north or south so you don't weigh as much. It's no longer than a sawtooth path but keeps you from gaining too much weight, better for comfort and cargo capacity, though it does make the flight longer.
Do you need to know what happens to speed and altitude when you gain weight? Should be easy enough to throw in a simulator.
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Important differences: around a star, and live on the outer surface with gravity generated by the star, and mostly counter-acted by the spin, not generated entirely by the spin.
Also not a superweapon to kill all life in the galaxy. At least not that I know of
4599116
(won't argue anything about heating because honestly, space horse magic. Any such structure would be chockfull of it just to exist)
On the gravitational part, how damn close is this thing to the star? Because it's a force balance. The result of both forces acting on the ring outer surface would result in that sweet 1G at the center (so around 699G of centrifugal force). But at the edges? You lose 4G for going some 5,000km north/south (considering the image Star posted)? The ring would've to be tiny as hell for that. As in... that's a 6.28° diference on the centrifugal force to the gravity, to cause a change form 669 to 665. For 5,000km to cause that the radius from the star center to the middle of the ring (curved or not) would be of only... 45.700km. That's barely over seven times earth's radius. So it's a really tiny ring world? Spoiler'ing this because it can be a major one for the story.
(I'm aware that "around 700g's" is not nearly precise enough for using two significant digits, but I believe it makes my math easier to understand in this situation).
Of course, that considers the habitable surface to keep a mostly equal distance to the center of the star (therefore curved), but that wouldn't change the order of magnitude of the ring' size. I was imagining something bigger, not something that's pretty much hugging it. Was this the intention? Was my force balance consideration wrong? What am I not considering?
4599442
Three questions:
1. Suppose pegasai are flying west at the highest typical flight speeds. (let's assume 60mph, since that's what a duck can do). Ignoring ponies like Ranbow Dash flying at absurd speeds, would ponies notice anything at these lower speeds?
2. I'm not quite bridging the gap in mind. The ring is spinning around 420 km/h, but you can double your weight by flying less than 2 km/s? Why does it take so little speed? Is it just that there's such an enormously large amount of gravity that even a little fraction of that amount can be crippling?
3. Suppose the crew of a 747 accelerates to typical 747 speeds and crusing altitude, then abruptly turns west and is determined to keep flying in that direction no matter what. Suppose the aircraft was flying at its typical maximum load to begin with. What happens to it? It falls out of the sky like a rock as it can't generate enough lift? Does friction get involved with the increased mass to eventually reach an equilibrium?
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4599481
Gosh dangit, I'm freakin' derping a lot recently. I blame the stress of this trip and using my phone to answer questions :p
Okay, for reference, this is an M9V-class Dwarf Star, using the typical values given on the table here on wiki. Its radius is 8% of the Sun's, and its mass is 7.5%.
Surface gravity is 327.6 g's (using 9.8 m/s^2 for g), not 700. I was remembering wrong, but digging up where I did all my math, I computed everything with the correct value. It was surprising at first, but considering there's less fusion, it makes sense that gravity can compact the atmosphere further, making the star more dense.
Surface gravity can actually be expressed as directly proportional to density multiplied by radius (both to the first power - within Newtonian gravity, which is perfectly okay here, even. 422 km/s and 330 g's are intense, but peanuts compared to relativistic domains around white dwarfs, neutron stars and black holes, and actually relativity shows that g's aren't the most insightful way to define gravity). A red dwarf star is very small, but more than makes up for it with that density, I suppose.
And yes, the ring hugs the stellar surface. Tbh, I'm not sure how that table is defining the "radius" of a star - is that the depth where 50% of light is absorbed (iirc, I think that's how it's defined?) - or where the ambient atmosphere density or pressure at equivalent to Earth's, or what? So I simply ran with the ring at that radius for a close-enough estimate. Truth be told, a few km won't make a huge difference on orbital velocity, and it will still be around 422 km/s. At 1,000 km above that height, it's still 419 km/s.
But anyway, the greater temperature gradient means you can extract more power from the star. Also there are other reasons why it's so close, I'm asking Starscribe if he minds me revealing this aspect of the ring.
As for the 6.28 degrees... One thing, I was off (with my memory, but not with the calculations I used to send Starscribe numbers) as mentioned. Secondly, hugging the star, as mentioned. And thirdly, I commend the effort, but it's little more than a rough estimate. Keep in mind the rotation rate of the ring is such that it cancels to 1g at the "equator" of the star (not necessarily the actual equator (but being on the equator may have some advantages), but for visualization purposes, it may help to think of it as the equator). As you go along the curve, the angle is one factor, but also the decreased radius at the same angular speed means that the centrifugal force is weaker.
So remember the angle is changing and the force is getting weaker, too.
And no, the structure wouldn't need magic to work. The only real stretch here is the material strength - and even then, it's not as bad as you might think. Both the gravitational and centrifugal forces are inertial forces, thus they are not inherently felt. Astronauts in orbit do not feel stretched, squashed or anything by gravity fighting centrifugal force that keeps them in orbit. Same here. The only force the structure would "feel" (ie, all it needs the material strength to fight) is the 1 to 5g's from its center to edges. Which is still remarkably strong, but not nearly as intense as, say, Larry Niven's ring world.
The thermodynamics aren't violating any laws to my knowledge, and I know thermo somewhat well.
For what it's worth, that crazy math is a coordinate transform from regular cartesian coordinates to spherical polar coordinates and solved the geodesic equations in the spherical polar coordinate system. So what's funny is I'm using general relativistic principles to solve for the centrifugal gravity in regular Newtonian mechanics
Basically, I'm asking what straight lines look like to spherical polar coordinates. The "inertial forces" in a rotating frame come about due to the fact that things want to move along straight lines, but your coordinates aren't straight.
General Relativity does the exact same but it works with spaces that are actually intrinsically curved and using time as a fourth dimension. So it's actually the exact same mathematics and procedure as finding the geodesic equations in General Relativity.
But just out of curiosity, are you studying physics or engineering or something?
4599481
1. Not unless they're carrying precise, purpose-made instruments to detect it :P
2. Pretty much, yes - it's because even a tiny fraction of such a speed, is a tiny fraction of such strong gravity, and even a tiny fraction of such strong gravity is quite a bit. On the upshot, you could pretty much reach orbit with jet engines, and just a tiny bit of rocket propulsion to do a little more once in vacuum.
3. It would slowly descend like a powered glider unless/until it can be throttled up enough to go faster so it has more lift to balance the extra weight.
It would weigh 1.39 times as much (going About 490 knots). Since lift is proportional to velocity squared, it would need to go about 18% faster, so 578 knots, to keep altitude, or go to air that's 1.39 times denser while keeping speed constant. Either one would require 1.39 times as much thrust from the engines. Looking it up - max mach number of 0.92, speed of sound at 30,000 feet being 90% sea level, that puts the maximum speed at 30,000 feet to 552 knots.
So, it would slowly lose altitude until it veers north or south to lose weight.
4599811
Electrical Engineering actually. One of the reasons I found it weird as hell that you started by converting the regular coordinates to spherical ones, instead of going spherical from the beginning. But yeah, if it DOES hug the star then it does make sense.
I'd say the initial confusion was because when one imagines a ring world it's usually at least in the tenths of AU in radius, not almost glued to the star itself.
Kudos for that idea, by the way. Decidedly novel and inventive
That does raise a bunch of questions though. But in the story, not the physics of it. Like how did the probe manage to land on over 320g's (before the spin)? How did it decide to land on what's almost the surface of a star? Did we program it to consider such mega structures? How will the ponies face the reveal, be it in the story or in the future?
4600228
Oh, okay, and thanks! I mean, you get power from temperature gradients, so why make it unnecessarily huge and less efficient at the same time? It makes sense if life is on the inside portion, but it works on the outside, too. Less area, but more efficient power usage.
As for the probe, if it's coming down on the ringworld then just like the ponies on it, it would only experience 1 to a few g's.
Hold on, I get why the gravity would vary in a sphere like planet at the edges, but in a ring world like the one described in the story gravity should be about that same all throughout the surface unless the surface starts to bend in an angle. In which case it's would be just like any slope, hill or mountain just at a larger scale.
Now the source for magic in equestria can still work in the way described by the story in which is weaker the further you get from the source, but gravity should stay the same.
Another thing, while the source of gravity in a ring facing inwards like halo would be the rotation, I'm under the impression the ring in the story faces outwards.
That would mean the source of gravity would have to be the star in the middle more than the rotation. If anything the rotation counters part of the stars enormous gravity just leaving some to have a surface to walk on.
So which is it? Does the ring face outwards and has a satellite sun and moon in addition to the star in the middle to cycle from day and night, or does it face inwards and the star in the middle is replaced by a moon at night?
Since this is science fiction, either one could work. But each one differently so whichever one you choose stick with it. Otherwise it doesn't add up.
In short,
If the ring faces inwards the star has to be small otherwise it would cover the entire sky, and if the ring faces outwards then the source of gravity would be the star more than the rotation.
~Leonzilla
Your image links are broken, fyi.
4643850
Yes, gravity is a combo of the star pulling down, and rotation countering it, to live on the "outside" of the ring.