# Life on the Broken Ring - an issue of size

One of my ongoing projects is what I think of as the "constructed worlds gallery", a series of Megastructures as settings for stories and games, including things like the "Flying Pie-plate" a world sized dish as suggested by Larry Niven as the starting point for the construction of the habitat in Ringworld, an Alderson Disk galactic lifeboat, and a trefoil mobius knot magicked up by a group of Dragons on the run from Cthulhu. Mainly I like playing with the implied geophysical issues that would otherwise make such structures unhabitable relatively quickly, the strange environments that result from solving them, and the everyday life of their dwellers.

My latest project is a piece, or rather pieces, of a broken Ringworld but I'm having trouble working out how large the habitable zone on such objects will actually be so given the assumptions below...

My understanding is that under the conditions listed below the side walls of the original ring continue to perform their role and the curvature of the ring segment will eventually be sufficient to have a similar effect.

Question: How long do the chunks of a broken Ringworld, per the original design specifications, need to be for the maximum sized pocket of atmosphere to remain in the construct and how big will it be?

Assume that:

• Apart from being separated from their neighbouring ring sections the pieces are otherwise intact.

• The pieces in question have broken across the width not along the length of the ring so both sidewalls are intact.

• The sections are under the standard 0.992 gee acceleration. This is thrust induced, necessary to hold them in an orbit closer to their primary than it should be at its orbital velocity.

• They are the in the same Goldilocks orbit they were built for, or something similar.

I know that the scenario creates appreciable engineering challenges in set up, those are entirely out of scope for this question.

Also please note that while the ultimate result of this question may well be a matter of relatively simple math it is primarily concerned with getting the right math to work from (thus the designation) as I have tried this twice, using different approaches, and gotten consistent results that differ by several orders of magnitude and neither of which looks right when drawn to scale.

It appears that certain assumptions are being made that are not valid let me clarify:

• This is not a ringworld in its original state.

• This is not even necessarily a ringworld in its original setting.

• Think of broken pieces as salvaged objects that have been repurposed as mega-habitats.

• This question is not at all concerned with the fact that this scenario falls outside our current understanding of physics this is a matter of whether the construct as described can hold atmosphere and if so how much.

• Your hard science tag invalidates any answer I could give, but bottom line is, the very best you might get is a tiny "puddle" of atmosphere near the center of a sufficiently long fragment, held by its gravity (due to its mass). No acceleration, because nothing still pulling the fragment around the star. – Zeiss Ikon Jun 26 '19 at 19:03
• The pie plates only hold an atmosphere if they're tied together -- to get your 0.992 G around a G2 star like Sol you need appr. 770 mi/sec rim velocity. That's well above system escape. – Zeiss Ikon Jun 26 '19 at 19:27
• So does this question just devolve to a chord-height calculation to get a blob of atmosphere in the centre of a broken segment? It would be ncie if you could at least mention your two previous attempts that failed, to prevent people going down the same route. – Starfish Prime Jun 26 '19 at 20:12
• So, very long term continuous thrust? Bussard ramscoops burning the solar wind? Your pie plate with a rocket underneath violates your own hard science tag; it's physically impossible with known physics rockets. – Zeiss Ikon Jun 27 '19 at 11:06
• I think the hard-science tag is valid, since "why are the materials unrealistically strong" and "how are the segments accelerated" is not really necessary to answer the question. The design contains all the required mathematical factors, and per this question, the hard-science is valid. – Bilbo Baggins Jun 27 '19 at 19:21

If there is a gap between the sections of the ring, it would allow all the atmosphere to spill through the gap, like so:

The question really isn't one of how long of segments you need (the answer would be "all the way around the circle"), but one of how to prevent the atmosphere from spilling out the ends. Here are three suggestions:

1. Do what Trump wants. "Build a wall! It's gonna be UGE!" Exactly how huge?

Based on the equation on this question and using this calculator, I calculated that the "limit" of your atmosphere is going to be about 99.5 miles above the surface. A 100 mile tall wall would work. Technically this is impossible, but if you have a material that you can build the ring with, you have one you can build the wall with. Here is a diagram (I love diagrams):

1. Manually angle the halves of each segment to trap the atmosphere, like this (exaggerated):

It is possible for life to survive below 5 miles in altitude. Here is the equation for how far from the hinge the segment will be habitable: $$D = \frac{10}{sin(A)}$$. Substituting for $$A$$, $$D = \frac{10}{sin(sin^{-1}(\frac{200}{L}))}$$, which reduces to $$D = \frac{L}{20}$$. One twentieth of the segment will be habitable. "A little puddle" was right.

1. Extend the orbits of the segments to make the curvature of the segments greater then the curvature of the orbit, holding the atmosphere like a bowl. Here is a diagram.

"What are these things?" The black circle is the orbit of the segment, the gray circle is what the ring would look like fully constructed, and the green arc is an example of a segment of that ring. All the other lines are to help explain the math.

The distance that matters is the distance from the edge of the segment to the orbit circle. I am going to do a lot of math, bear with me.

Here are the variables: $$R_o$$ is the radius of the new orbit (black line), $$R_i$$ is the radius of the old orbit (blue line), and $$A$$ is the angle of the ring formed by the rays from the center of the ring to the ends of a given segment (green arc).

The blue angle is half of the green, so it is $$\frac{A}{2}$$. The length of the green line is $$R_i(\sin {\frac{A}{2}}))$$. The length of the purple line is similar: $$R_i(\cos {\frac{A}{2}})$$ The length of the red line is $$R_o$$ minus the remaining distance from the end of the purple line to the black circle, so it is $$R_o - (R_i - R_i(\cos {\frac{A}{2}}))$$. To find the length of the brown line, we use the Pythagorean Theorem on the green and red lines: $$\sqrt {(R_i(\sin \frac{A}{2}))^2 + (R_o - (R_i - R_i(\cos {\frac{A}{2}})))^2}$$ Now the final step is to find the altitude of the end of the segment by subtracting the brown line from $$R_o$$. So the height of the end of the segment $$H$$ relative to the orbit is: $$H = R_o - \sqrt {(R_i\sin \frac{A}{2})^2 + (R_o - (R_i - R_i\cos {\frac{A}{2}}))^2}$$ $$H$$ must be greater than 100 miles to successfully contain all of the atmosphere. You suggested an orbit a few thousand kilometers wider than original, so with an orbit of $$9.5003x10^8$$ miles, the segments must be longer than 30° of the entire ring to successfully contain the atmosphere. To solve for how much of the ring is habitable, use $$5$$ as $$H$$, since life can survive at less than 5 miles in altitude. The portion of the segment at this orbit ($$9.5003x10^7$$ miles) that would be habitable would be 6° of the ring. So, to maximize the amount of habitable space at this orbit, break the ring into twelve equal chunks and $$\frac{1}{5}$$ of the ring will be habitable.

If we increase the density of the atmosphere, a little more area of the sphere is habitable. According to Wikipedia, humans can survive at 6 atmospheres without any serious or permanent side effects from nitrogen narcosis or the toxicity of oxygen.

Using this calculator, I calculated that the edges of the segment would have to be 150 miles tall to contain 6 atms. Using my equation, the segments would have to be 40° of the ring to reach 150 mi at the edges. Increasing the pressure to 6 atms would make the segment survivable up to 18 miles in altitude, so 12° of the segments would be habitable. At an orbit 3000 miles wider than the original, your ring would ideally be broken up into 9 segments with 6 atmospheres of pressure at the bottom of each, and 30% of the ring would be "habitable."

• Given that you have to have walls on the sides of the ring, walls at the ends of sections shouldn't be too much of a chore – Separatrix Jun 28 '19 at 14:28
• There would be no centrifugal force left. Once the fragments come apart they would fly away from the sun and you'd lose your artificial gravity. – Nosajimiki - Reinstate Monica Jun 28 '19 at 17:55
• The question said "They are in the same Goldilocks orbit they were built for." I assume that means they are somehow going to stay in that orbit. – Bilbo Baggins Jun 28 '19 at 18:44
• Exactly how yuge? – boxcartenant Jun 28 '19 at 19:42
• The circle made by the orbit of the segments is essentially 0 altitude, with moving toward the star increasing altitude, and moving away decreasing. Extending the segments would not increase their altitude since their curvature lines exactly up with their orbit. You need to somehow increase the altitude of the segments on the ends to contain the atmosphere. – Bilbo Baggins Jul 1 '19 at 2:06

There are a number of physics principles at play here.

1 - The ring world is not "full" of air. The Barometric formula tells us how to model the distribution of 1 atmosphere of air in 1 G worth of gravity. Your ring world does not have exactly 1 G so this is only going to be an approximation but a very close one. If you graph out the Earth's atmospheric density by elevation, you'll see that a significant majority of the air is going to be in the bottom 20km of your ring's walls and reach an approximately space like vacuum by about 100km, but your ring has ~1609km tall walls. If you were to evenly distribute your atmosphere inside of that space, you'd be at 0.512% of Earth's atmospheric density. Not truly a space like vacuum, but close enough for most practical purposes that most people would consider the ring decompressed before you actually lose any meaningful amount of air.

2 -There is no such things as a mathematical point of complete depressurization, when you use a decompression algorithm we measure how long it takes to go from one air density to another. You lose pressure slower as you approach zero without ever hitting zero; so, hitting an uninhabitable atmosphere, and hitting the density of space are two VERY different time scales.

3 - Air can never decompress faster than the speed of sound. Most decompression algorithms don't account for this because they measure for a small vessel losing air through a hole small enough for this to not be an issue. Mathematically, this station should be able to lose 1/2 of its air in just a few seconds, but it's air can't move fast enough to cover the hundreds of miles it takes to even get to any holes in your ring in that amount of time.

4 - When your ring breaks apart, the pieces will be going about 40 times the rotational speed of the Earth sending the fragments off into deep space and taking away your artificial gravity. This lose of gravity means that your air is not going to be pouring towards your side holes nearly so much as it will be dissipating upward and out the top.

5 - The ring is under a LOT of tension. While it's easy to say scrith makes hand waving away that much stress possible, any breaking apart that it might do would be violent. Like a giant snapping guitar string, you would expect there to be massive waves of oscillations throughout your structure flinging off most of the atmosphere in an instant, large sections of the ring would curl or crumble and everything would be so chaotic on such an incompressible scale that without a very detailed explanation of scrith's properties, it would be very hard to say what would happen. For purposes of this question, I will assume scrith is also infinitely rigid otherwise the answer to this question becomes very open-ended.

Given all these factors, we know we can not use a typical decompression algorithm because we first need to find out just how long it's going to take for the air to expand to fill the ring before it can even start to escape.

Since the upper atmosphere will expand slower than the higher density lower atmosphere, we can get away with simplifying this equation by averaging out the starting atmosphere and still get a very close answer to if we tried to model out the exact expansion of a non-linear gradient pressure since it will all diffuse pretty evenly by the time it expands enough to reach the top of your wall.

• Density of air is @ sea level = 1.225 kg/m3
• The mass of Earth's atmosphere = 10,092.139 kg/m3

So, we can estimate you have a starting body of gas that is 8238.481 meters high at a density of 1.225 kg/m3 that will expand as fast as it can upward to get out of the ring.

Next we need to come up with a decompression formula that works for the speed of sound as you loose density. The air at 1 atmosphere of pressure can expand at a rate of ~344 m/s, but as air loses density, it will expand slower. So by the time your atmosphere's averaged height doubles to about 16,476m, the rate of expansion will be halved to about 172m/s, so on and so forth.

Below is a simple JavaScript program that calculates this:

<div id="output"></div>
<script>
speed = 344;
height = 8238.481;
startheight = 8238.481;
endheight = 1609340;
time = 0;

while (height < endheight){
if (height*2 < endheight){
period = 2;
heightC = height;
} else {
period = endheight / height;
heightC = endheight - height;
}
time += heightC/-((1-Math.log(2)*speed)-speed);
height *= period;
speed *= 1/period;
}
document.getElementById("output").innerHTML = 'FILLS RING AT<br>Time: ' + Math.round(time) + ' sec<br> Height: ' + Math.round(height/1000) + ' km<br> End Speed: ' + speed.toFixed(5) + ' m/s<br>Pressure: ' + (startheight/endheight).toFixed(5) + 'Atm';

• FILLS RING AT
• Time: 241626 sec
• Height: 1609 km
• End Speed: 1.76099 m/s
• Pressure: 0.00512Atm

This means, if you could find a way to restore gravity within 2.8 days, you'll be able to keep most of the air just fine. However, by this point your fragment has already drifted out of the Goldilocks zone and pressure has dropped so much everyone is dead.

That said, the edge of the Earth's atmosphere @ 100km (The point we start to call space) has a pressure of .00001 atmospheres. So a to get to this point, you make this change:

endheight = 823848100;


Ooops, the time output for this blows WAY past the max float size allowed in JavaScript; so, need to run these calcs in something that allows for larger numbers to give you an exact answer, but let's just say it is a very very long time. Someone else with access to MATLAB or something similar can probably get you an exact answer, but this gets you close enough to what you need for your story.

One last factor is gravity. In a smaller fragment, there will not be enough gravity to matter, but let's you have a nice big chunk of the ring. Like in the picture below. The ring's gravity will actually make you lose your atmosphere ever so slightly faster because your center of gravity will be above the surface of the ring. The good news is though, that your atmosphere will eventually coalesce into icy planet. For a 1/3rd segment like the image below, this will result in frozen gas planet about the mass of Earth.

To answer your question about salvageability, the shape of the ring fragment would make it a useless habitat, but a future civilization may perhaps scrap parts of it to make a much smaller halo superstructure around the Earth sized gas planet. They could mine the ice ball for a nearly inexhaustible source of water, air, and hydrogen to run their fusion reactors for power, and the ring fragment could give them all the soil, metals, and minerals they would need.

As for you new requirement: "This is thrust induced, necessary to hold them in an orbit closer to their primary than it should be at its orbital velocity." While this new requirement pretty much kills my answer, I will leave this up as a point of reference for future inquiries that may not rely on propulsion to maintain the orbit.

• Have you actually read Niven's novel (series)? With the size of a full-blown Ringworld, you could punch an Earth-sized hole in the floor and it would take millennia for the atmosphere loss to become ring-wide hazardous. – Zeiss Ikon Jun 28 '19 at 10:53
• If the fragments are scattering, they lose the spin "gravity" and depend only on the mass-related gravity of each fragment to hold its atmosphere. Air breaks or no, most of the atmosphere will just float away into space where there isn't enough gravity (near the ends of the fragments) to hold it, or it'll pool so deeply near the "downhill" center of each fragement that much of it escapes over the edge walls. – Zeiss Ikon Jun 28 '19 at 14:48
• In fact, engineers capable of building a full Ringworld wouldn't need to depend on airlocks; they depend on scale and impossibly strong materials to ensure the ring can't break up. – Zeiss Ikon Jun 28 '19 at 14:50
• Not sure if "can't break" is ever a good assumption for engineers (see the Titanic), but you are right about the "top" being the biggest vulnerability here. Revised my answer accordingly. – Nosajimiki - Reinstate Monica Jun 28 '19 at 17:50
• @ZeissIkon The fragments, in the scenario stated by the OP, also have their added thrust-generated gravity, since they've got rockets thrusting them in towards the star at 1G to counterbalance the forces trying to throw them out away from the star at 1G. – nick012000 Jul 7 '19 at 16:57

Frame Challenge

One of the criteria you gave is:

The sections are under the standard 0.992 gee acceleration. This is thrust induced, necessary to hold them in an orbit closer to their primary than it should be at its orbital velocity.

I propose that this doesn't make sense and should be discarded.

First off... the energy output of such a drive is ludicrous. If your "chunk" is merely "squarish" (about as long as it is wide), we're talking about 5e25 Newtons of thrust. You didn't supply enough information to convert this to energy output, but without some extreme hand-waving, there's a good chance we're talking about stellar energy levels. (In fact, IIUC, this drive produces about 1 Solar output if the chunk is moving at a piddling — by the astronomical standards we are talking about — 10 m/s. By comparison, Earth's orbital velocity is 3e4 m/s.) No matter how much hand-waving you want to employ, that drive is probably going to produce some heat. Time for even more hand-waving to explain how you can dissipate all that without cooking your habitat. (Maybe forget the star and just have it accelerating through empty space?)

Second... if your thrust is really continuous, and not varying in something like a sinusoidal cycle, then, to be useful, it has to be at a constant angle relative to your star, which means your chunk of Ringworld is (effectively) tidally locked. However, once this is true, I can see no benefit to mucking about with your orbital velocity in the first place, other than "because we can". When you're tidally locked, you don't have seasons, and "year" doesn't mean much unless you're practicing astrology.

If you still want to play with your orbital velocity... then I think the question is unanswerable without additional information. Namely, the answer will depend on your actual velocity and how your actual orbital mechanics modifies the effects of your star's gravity.

Let's assume, instead, that your chunk is tidally locked facing the star, with your thrust directly toward the star such that the perceived gravity in the center of your chunk is 0.992G. Let's also assume that you've chosen an orbit such that the combination of stellar illumination and waste heat from your drive makes your chunk "comfortable". This feels like a much more plausible scenario, and fortunately, it has an easy answer:

Since we're talking about near-Earth "effective" gravity, we can assume that the atmospheric "depth" will be comparable. Atmosphere, as elsewhere noted, doesn't "stop" at any particular point, but Earth's atmosphere is anyway regarded as being about 500km deep, so Ringworld's 1000km walls should retain this pretty well. The answer to your question, then, is that the curvature should be such that the ends of the arc are about 1000km "higher" than the center (i.e. the distance between the midpoint of the arc and the midpoint of the line between the ends is about 1000km). Calculating the required angle for this is left as an exercise for the reader. In fact, because your thrust accounts for only part of the perceived "gravity" towards the ends, actual gravity from the mass of the chunk itself will have the effect of "flattening" the arc somewhat in terms of its apparent gravity. (The actual calculations for this effect requires moderately complicated calculus or approximation via FEA.) On the one hand, this will increase the length of the arc needed to match the side walls. On the other hand, 1000km may be more than necessary for atmospheric retention.

• The only reason orbital velocity is increased is to have centripetal force that allows for enough continuous counter thrust to simulate gravity, if you're saying I don't need that then cool but I'm not sure the physics is on your side there. Heat dissipation is an engineering issue. The only thing this question is asking for is a determination, mathematically verified, of the angle (thus the distance) you're saying should be "left as an exercise for the reader". – Ash Jul 9 '19 at 16:53
• If your thrust is purely for effective gravity, then cool, you're in my "let's assume, instead..." section. (Working out the actual thrust needed, in combination with the star's gravity and the chunk's own gravity, probably merits a seprrate question.) Just what orbit you're in (if you care), probably likewise. (I take it, then, that you are always facing the star, with either no night, or some other mechanism to block the sun a la original Ringworld?) – Matthew Jul 9 '19 at 17:08
• As to "left as an exercise for the reader", what I was saying is that I believe, under the conditions I assume at the end, the answer is "such that the distance between the midpoint of the line connecting the ends, and the midpoint of the arc, is 1000km". From there, it is just simple trigonometry. Although I don't think you can really, correctly answer this question without a correct model of the chunk's gravity, because I believe you are dealing with enough mass that that will matter. (Apologies; was typing this when you replied... ran out of room in prior comment.) – Matthew Jul 9 '19 at 17:16
• Yeah okay definite point, it does weigh several orders of magnitude more than Earth. – Ash Jul 9 '19 at 17:20

I'm going to take a bit different approach to this based on the revisions... unfortunately I don't have math to back this up, but it should at least get you started thinking in a useful direction. (So far, I haven't seen anyone else accounting for the chunk's own gravity, and my own calculus is too rusty, so I asked P.SE to help us.)

Assuming you are talking about a relatively small chunk of ring (say, 45° or less; bigger than that, I expect the orbital mechanics just get... interesting¹), you are essentially dealing with a really oddly-shaped planetoid. In particular, this chunk has to be in orbit around the star, because otherwise it is almost by definition not going to stick around very long.

(¹ The whole ring is easy to reason about because it is balanced and many forces will cancel out. This is also why a whole ring can mostly ignore gravity and spin as fast or as slow as you like, subject only to torsional stresses.)

This being the case, unless your chunk is really close, relatively speaking, to the star, or really big in relation to its orbital distance, you mostly only have to worry about the chunk's own gravity, because the other forces are going to be comparatively weak. (Just as the Earth doesn't lose its atmosphere due to centrifugal force.)

What you have, essentially, is a tidally-locked planet (i.e. in a fairly "normal" orbit, with zero axial tilt and a rotation period of exactly one revolution per orbit). Note that this has to be the case, because, unlike a complete ring which is gravitationally stable at any rotational speed, a fragment on its own is either in a regular orbit, or, by definition of not being in an orbit, isn't going to stick around your star for very long.

At a large enough size (and we're almost certainly talking about such a size), we are mostly looking at atmosphere retention being primarily a function of the segment's gravity. A 1/300 chunk has roughly Earth-mass, and that's slim; Ringworld's width is 1/625 it's circumference, so we're talking about a chunk only twice as "long" as wide (and you said we aren't cutting it along the width).

This is where things get weird and hard. Because your segment is curved, your point of maximum gravity is going to be above the surface... but because of the distances involved, they aren't going to exert much effect. The strongest gravity is going to tend to 'hug' the surface to a fair degree.

Now... I haven't developed my model far enough to actually prove this, but I think you ought to be able to simply park an atmosphere on the thing, and it will mostly stay put just via gravity. Doing things like bending the walls in may actually make things worse, since, as others have noted, that shifts the point of maximum gravity up away from the surface. Moreover, it's going to exacerbate a problem you're going to have anyway, which is that dirt (and rock) near the edges is going to want to slide toward the center... which is going to tend to turn your system even more into a "regular" planet. In any case, you're going to have thinner air toward the edges, but I think you'll have plenty of air towards the center and even spread out a bunch.

For bonus points, this means you can also spin your chunk to give it a day/night cycle. In fact, you can probably give it an atmosphere on both sides. (This might be a very good idea, because it will eliminate atmosphere loss in case you get a hole through it.)

Your real problem, of course, is keeping this monstrosity from collapsing under its own weight. Although if the intact ring managed this, you're probably okay. (Also, you mentioned that the structural problems are out of scope...) On the other hand, if you let it collapse into a ball (this would end up looking like two hemispheres in the middle of a big honking plate), well then, you've got yourself a nice little planetoid that will have no problem retaining atmosphere. (The weather might be interesting!)

• You have not accounted for the fact that the object is under nearly one gee of thrust. – Ash Jul 7 '19 at 11:05
• Ah... Right. I'm going to challenge the point of that. First off, if you have an engine that can produce that sort of thrust (note: at least 1e24 N, up to 2e28 N), why are you mucking about with bits of someone else's Ringworld rather than building your own? Second, why does the orbital velocity matter? If the chunk isn't tilted, it isn't clear why you'd care how long a "year" is. Moreover, this assumes that either your star is either much larger or much smaller than Sol, or you want years that are nowhere near the length of Terran years. – Matthew Jul 8 '19 at 14:22
• Moreover, unless your thrust is always pointed exactly along the Ringworld radius, good luck keeping any atmosphere. Which means that a) your orbit is much faster than normal orbital velocity, and b) your "habitable" section is always facing directly toward the sun. Not to mention, c) if your engine fails, you will have a Very Bad Day. IMHO not mucking about with weird orbits (and the ludicrous engine needed to provide it — something like 1e20 space shuttles!) is much more sensible. – Matthew Jul 8 '19 at 14:30
• Having a reactionless drive, or similar, does not automatically beget the matter conversion technology needed to manufacture the materials necessary to build a Ringworld. You want gravity, the year length reduction is a consequence of getting it nothing else. Once again the question is not concerned with the issues arising in setting up the scenario but one effect. – Ash Jul 8 '19 at 15:04
• (I'm going to post a second answer to discuss this; it needs more space than a comment...) – Matthew Jul 9 '19 at 15:07