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The pillars are spread out about a mile apart each, and they all come straight from the ground, slowly gets just a smidge smaller in the middle, and then get bigger near the top, reaching a height of two miles tall. And also would an extremely thick liquid (that isn't lava) underneath hell help?

essentially the world is a massive pillar in general, with a layer of floating heaven being held up by pure white magic that goes up near infinitely, which underneath heaven is a flat earth about 200 miles in diameter, and about 1 mile deep. under that is hell, that seems to go on for thousands of miles, so the space of hell definitely accounts for all of earth. then under hell is an extremely thick black liquid going down near infinitely

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    $\begingroup$ To the downvoters: downvotes without comments are distinctly unhelpful. $\endgroup$ – Starfish Prime Nov 15 '19 at 12:27
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    $\begingroup$ (nitpick: they can't be stalagmites unless something else is holding up the surface, because of the way stalagmites form. They're just rock pillars) $\endgroup$ – Starfish Prime Nov 15 '19 at 12:28
  • $\begingroup$ I guess I kinda meant "like stalagmites" because they are just pillars of rock, but I found pillars to be boring a while back and have started calling them stalagmites, so I guess I should call them pillars for now until I find a better less confusing name $\endgroup$ – michael griffin Nov 15 '19 at 12:30
  • $\begingroup$ Not "tough", use "strong". It's an important technical difference. $\endgroup$ – Separatrix Nov 15 '19 at 12:38
  • $\begingroup$ I actually thought about using strong in the first place but thought that it might be wrong, so let me go fix that $\endgroup$ – michael griffin Nov 15 '19 at 12:41
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TL;DR: For the pillars: nephritic jade. Sounds good, looks good, is super tough. Better hope you don't have earthquakes though.

The ceiling is where you're gonna have real problems. A plain flat ceiling will collapse unless you use a roofing material with a really high tensile strength. You probably want to vault it, too.


Lets look at a single pillar first. Though you've defined pillar spacing, you haven't mentioned pillar layout so it isn't entirely clear quite what proportion of the surface world each pillar supports. Lets imagine that it is a cylinder a mile across, and a mile high. I'm going to move to using metric units now, because imperial ones are weird. With a volume of about 3.27 billion cubic metres, and assuming it has the same density of continental crust, it'll have a weight of about 9.26 billion tonnes. Assuming the pillars are perfectly cylindrical (you said they weren't, but didn't give a minimum diameter) their cross-sectional area will be 2.03 million square metres and therefore the pressure they'll be under will be about 178.65MPa (for reference, the compressive strength of concrete can be as high as 25.57MPa, so your pillars need to have stronger than that).

There's a frequently cited, but never actually referenced figure that suggests that the rock with the strongest compressive strength is nephritic jade at 400MPa. I'll take that figure as true, though it could be a load of cobblers for all I know (a figure cited here suggests more like 370MPa, but it is still just about high enough for our needs). It has a density around 3.15g/cm3. The weight of the pillar and the earth above it combined give a pressure at the foot of the column of 278MPa. Using nephritic jade would give you a safety margin of about 1.43. Note that this is a static safety margin... if your hollow world experiences earthquakes, you might have more serious problems, but that's a question for another day.

What I'm less sure about is the bulging of your pillars as they're squished from the top. I think they might be OK, but I'm quite far from certain on that. Maybe some nice metal reinforcing bands might help?


The next problem is the unsupported span of earth between two pillars... there's at least a half-mile gap between the edge of one pillar and the start of the next (I say at least, because depending on how your pillars are laid out the gaps might be larger... in a square grid, the diagonal spacing is more like .91 miles) and that's a big span for rock to cover. Even nephrite with its impressive compressive strength isn't nearly as strong in tension and has a tendency to fracture instead of yielding.

I threw some figures into the simple bending equation, and they didn't come out well. I'm not a structural engineer, so I won't repeat my amateur scribblings here because they're so far from an accurate assessment of the situation but: that roof ain't gonna hold, even if it were pure jade. The unsupported span is just too great. It'll crack in the middle, and everything will crash down into hell, crushing everything. Sure, the surface world will be ruined but hey! no hell anymore!

You might be able to fix this with a carefully vaulted ceiling, but I wouldn't want to say for sure. Really, you need to be making that ceiling out of some kind of reinforced composite if you don't want it to be crashing down on you.

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    $\begingroup$ oh.. ok. btw I updated the question with a bunch of requested information from commenters, I assume you didn't reload the page, but the information added certainly doesn't seem to help the situation any. $\endgroup$ – michael griffin Nov 15 '19 at 14:51
  • $\begingroup$ also, having it vault sounds like a REALLY cool idea actually. it would help with releasing some of the pressure. $\endgroup$ – michael griffin Nov 15 '19 at 14:54
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Let's proceed with some approximation.

Per your statement, the columns are equally spaced, 1 mile apart.

If they are arranged in a square lattice, it means each column hold a cube of 1 cubic mile.

The weight of a cubic mile of Earth, assuming it is basalt, will be given by its volumes times its density.

Basalt density ranges around 2700 and 3100 $kg/m^3$, let's make it 3000, and 1 mile is 1609 m, which gives for our cube a mass of $M=1609^3 \cdot 3000 = 12.5 \cdot 10^{12} kg$, or $12.5 \cdot 10^{13} N$.

Let's also simplify the shape of the column to be a cylinder 1 mile in diameter, its cross section will be $2 \cdot 10^6 m^2$, therefore the load will be $P=$$12.5 \cdot 10^{13} \over 2 \cdot 10^6$$=62.5 MPa$.

That is $62.5 \cdot 10^3 kN/m^2$, out of my mind that's about thousand times what concrete can withstand.

I think no real material can withstand that compressive load.

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    $\begingroup$ Compressive strength of granite is 200MPa and quartz can reach 1GPa. Cast iron exceeds 700MPa. Your comment "no real material can withstand that compressive load" is incorrect. $\endgroup$ – Starfish Prime Nov 15 '19 at 14:49

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