Fullmetal Alchemist (2003) features a giant (and I mean stupidly big) cave under a large and busy metropolis. Nobody is aware of its existence, which means that at least since the city's founding (let's say 150 years) there must have been no subsidence of any kind.

What little I know of cave-ins tells me this is rather unlikely, but for story-reasons I need to know how shallow I can make it before shit starts falling down.

  • The cave is unsupported by any columns, and there is a very large space from the floor to the roof (perhaps 80-100m). The floor diameter is about 2-3km. [pic]

  • The cave was formed by wild magic, so the erosion/seepage found in karst caves need not be an issue.

  • The time period is equivalent to early 20th century, and there is some light automobile traffic. My admittedly superficial research suggests that foundations in early 20th-century London were pretty shallow, so foundation depth is unlikely to have any impact.

  • A river runs around the outskirts but not through it. Earthquakes are unlikely.

  • The cave should run deeper than 4m (depth of sewers)

In this admittedly improbable scenario, my question is: What would be the minimum thickness for the cave ceiling so that, assuming ideal geological conditions, there is no risk of collapse under the weight of the city above?


Sarawak chamber (Malaysia) is the largest known cave room (700x400m) with an unsupported roof span of 300m. No word on roof thickness.

Son Doong Cave (Vietnam) has the largest known passage. According to Wikipedia, it is 4.6 km long, 80m high and wide over most of its length, but over 140m high and wide for part of its length.

A very interesting paper on the small (<11m) man-made caves in the sandstone under Nottingham recommends a depth of >50% of the cave width, which is pretty damn deep. This is for flat roofs in karstic limestone (stronger than sandstone).

  • $\begingroup$ I am not familiar with "Fullmetal Alchemist", but is the roof flat or domed? Flat 3km wide roof would be scientifically impossible. $\endgroup$
    – Alexander
    Jul 18, 2017 at 22:23
  • 1
    $\begingroup$ Dome-shaped cave will be stable and theoretically big enough for your scale. I don't know however what the maximum span-to-height ratio can be. 2:1 is definitely good. 6:1 (similar to arched bridges) maybe too aggressive. $\endgroup$
    – Alexander
    Jul 18, 2017 at 23:44
  • 1
    $\begingroup$ a dome transfers pressure. the arch of the dome cannot stand by itself, it needs to transfer the vertical and the horizontal forces into the next possible planet. the thickness of the top of the arch, i.e. the thinnest point, can be next to zero, so given you need 4 meters anyway, you should be safe. The overall stability of your system is limited by how much compression the material near the sides and base of the cave can stand. $\endgroup$
    – Burki
    Jul 20, 2017 at 8:33
  • 2
    $\begingroup$ You might be able to use a snowload calculator (while substituting the weight of snow with that of stone/dirt) to at least give you a lower bound on the weight above a geodesic dome, but that will put you into a catch 22 of playing with the weight/thickness. We don't stack stuff on domes, try arches. $\endgroup$
    – Mazura
    Jul 21, 2017 at 0:18
  • 1
    $\begingroup$ @Notiophilus obviously a material thickness of 0 supports 0 weight. the rest depends on the material. a good first estimate is to sketch the setup, and look how the forces are led into the ground. if you can get vectors at 45° you are probably safe. for materials with high tensile strengts (very much UNLIKE rock) shallower angles can work. For more details, a civil engineer studies for years, for a reason :-) $\endgroup$
    – Burki
    Jul 21, 2017 at 6:54

3 Answers 3


The weight of a city is insignificant next to the forces that the cave has continually exerted on itself. Any safety factor that would definitely keep the cave from randomly collapsing in on itself, would make that little bit of extra weight negligible.

I'm having a pretty hard time trying to interpret stone lintel load charts, but given the type of stone and the length, you should be able to extrapolate a rough estimate. You'd better put a pretty large safety factor into it though, as any good load chart will be based off of independently tested samples.

Meaning, there's no way to be sure, and that's why things with a sufficient safety factor "are purposefully built much stronger than needed for normal usage to allow for emergency situations, unexpected loads, misuse, or degradation (reliability)."

If this cave were under Chicago, it'd need to be about 3000 feet down, or we'd have found it when we drilled wells. I'm guessing that if it weren't for modern seismologic surveys, there could be one down there right now...

I've stuck to your title and ignored most of your criteria, because built on bedrock and there's no earthquakes sounds like Chicago to me. Build your city basically right on top of the bedrock, and next to a nearly inexhaustible source of fresh water. Nobody likes drilling through bedrock, and if you don't need to for water, you won't.

The SWL values in our load span tables are often subject to load ratios. These ratios represent the ratio of load that the lintel can bear as inner leaf to outer leaf. The ratios are different for the different applications.

  • 1:1 – Lintels supporting masonry only
  • 3:1 – Lintels supporting masonry and timber floors
  • 5:1 – Lintels supporting concrete floors
  • 19:1 – Lintels for eaves applications


I can't make heads or tails of that (I think you'd need to convert the weight to either kNm or kN/m anyway, to make use of those charts). I'd say somewhere over 5:1, which is NASA's ratio. But I believe the construction industry rather's a factor of seven, especially for lifting equipment.

For story-reasons I need to know how shallow I can make it.

I dunno; do the math. The question is, what safety factor can you get away with. Again, I dunno. A SF of 1 wouldn't, if you put anything on it.

Do more than 1 and less than 20; it's a cave not a cove (supported at each end; not an eave).

  • 1
    $\begingroup$ I would also mention that thickness only helps if it is maintained down to the sides of the cavern. I.e. a thick roof and no walls is worse than a thin roof and sturdy walls. Throw in all that junk about domes and buttresses and you can fool most readers into accepting this answer! ;-) $\endgroup$ Jul 21, 2017 at 3:23

How about forgetting architecture and putting your city on a lava dome? You could pick a dome like this one in Andes, and excavate by magic (or just very advanced mining) below it to form your cave. It should hold, because you have hundreds of meters of volcanic rock above your cave.

There are huge cave chambers such as Sarawak Chamber which is 600x435 meters. If the cave ceiling is arched, there is a good chance it could be bigger.

  • $\begingroup$ As I said in my question, I'm working from an existing story so I can't change canonical facts, however silly they may be. I am aware of Sarawak Chamber. $\endgroup$ Jul 20, 2017 at 21:36

I'm not an architect, but I think it is not the thickness, thickness adds weight but at the same time adds stability.

Consider brick ceilings, I don't see why this cave is not possible.

  • $\begingroup$ What worries me isn't so much the weight of the rock ceiling (although I hadn't really thought of that, thanks for bringing it up) as the weight of the buildings bearing down on it. Basically - how thick do I need my brick ceiling so I can put some furniture up there? $\endgroup$ Jul 18, 2017 at 23:26
  • $\begingroup$ @Notiophilus you cannot know, you don't know the material in it or the structure of the stone pressing agains eachother. It is not an engineered structure. But it should be at least 100 meters, just a guess. Sorry. $\endgroup$
    – Aus
    Jul 19, 2017 at 0:01
  • $\begingroup$ Downvoting for lack of numbers/research/any supporting evidence on a science-based question. $\endgroup$
    – Frostfyre
    Jul 19, 2017 at 12:53
  • $\begingroup$ The weight thing is correct enough, but this answer does lack details about what would be helpful to the question. A piece of this answer is probably better as part of another answer, methinks. $\endgroup$ Jul 21, 2017 at 3:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .