^
Too many, that’s one reason why foundation drilling and structural seismics is done.
Well, if there are lots of them, and a few a year collapse then the chance in any given year of any one collapsing is small then isn’t it?
That was the point of discussing large cave voids in the first place. There are large underground structures that naturally occur that last a long time. If those exist, besides any real economic reason to do so, it certainly seems possible to build some very large and long lasting manmade underground spaces.
And any engineer will tell you he/she would rather excavate on solid granite or metasediment than use a natural limestone cave.
Really?
That STILL misses the point. And your claim that caves are unstable over a human lifespan is laughably wrong.
Technically, all of the Grand Central complex is cut and cover. For several blocks to the north, the buildings are built on stilts that fit between the tracks of the rail yard. The streets and sidewalks are similarly built on platforms. Here is a drawing of the railyards without any buildings:
https://flavorwire.files.wordpress.com/2013/02/grand-central-3.jpg
Here is a neat view of the actual excavation:
Grand Central was actually built on the site of an older station, was built in stages, as the original was gradually demolished. The original station (the building with domed towers to the right in the photo) had a street level rail yard which still partially exists next to the excavations. In the middle is the partially built structure holding the two levels of track and platforms. About a quarter of the current station has been built, seen in the middle of the photo at the head of the track structure. (building to the left directly behind the pole is either the Graybar building or the Post Office, but not Grand Central itself).
with the current stations two underground track structure exposed to the in the middle, with a quarter of Grand Central build at the end.
One could build an indefinitely large underground structure if you are wiling to include support columns that divide the space. However, the deeper you go, the wider the columns would become (until the columns all literally melt together in the mantle).
As for a single bay chamber, a dome would be your best bet. The dome would be limited by the compression and sheer strength of the stone it is carved out of, or material used to line and support the dome.
To get a ball park figure, let us make a few simplifying assumptions. We know, for instance, that a dome will transfer all the weight it supports to its base.
Thus we only need to know how much weight the base can support. This is a factor of the material’s compression strength “C” in pounds per square foot, and its area “A” in square feet. The area at the base of a dome is based on its thickness “T” between the outer radii (“R”) inner radii (R-T)). A circle’s area is pi*r^2 (pi=3.14…). A=pi(r^2-(r-T)^2)
Thus the maximum weight the dome can support is C*A or:
Max=C*A=pi(r^2-(r-T)^2)
We also need to estimate how much weight the dome will support. This includes all the material directly above the dome to the surface. To simplify, we will ignore the void under dome, and estimate the weight of material from the floor of the dome to the surface. Using the outer radii, “R”, and a depth “D” to the dome floor, we get: volume of material = piR^2D; multiply the density of the material “M” by the volume, we get
Max weight = MpiR^2*D
We now must solve for R, which will tell us the size of the dome we could build.
Since we can hold no more than “Max”, our “Weight” must be less than or equal to “Max”; thus:
Max = CA=pi(r^2-(r-T)^2) = Mpi*R^2 = Weight
Doing the algebra and simplifying, we get a neat little equation! That is:
R^2-2TR-T^2=C/(D*M)
Assuming we build the dome of granite, and that the earth about is also granite, we can look up the values for C,and M.
For granite:
C=2.7x10^6 psf (pounds per square foot)
M=168 pcf (pounds per cubic foot)
We will propose a sample problem, a dome of depth
D=1000 feet
and thickness
T=20 feet.
**
Plugging into our equation we get a maximum radius of 48 feet, or 96 feet across!!!**
Using different values for the depth and thickness or material will vary the size.
This of course is maximum theoretical size. Engineering standards would only permit a fraction of this size, a “factor of safety”, to account for various other loads that are difficult to account for. Due to the geometry of the dome, there will also be “sheering” loads that must be accounted for using the shearing strength of granite.
**Do not try this under your home! 
Congrats on doing some serious math to get an answer. And I am not being snarky here.
But something is horribly wrong with your numbers. Or assumptions or something.
There are cave rooms in Florida bigger than that. And these aren’t precision engineered voids. These are random rooms formed in shitty limestone that is only one step above chalk as a structural material.
Move to places where the limestone is decent and you have rooms many times that size.
Cecil have mercy, I did find a large error in my algebra. I reported myself to the Mod to have this offending post removed while I correct my math.
That’s more embarrassing than when Dr Evil asked for $20 MILLION dollars to ransom the world 
No worries. Hope you can work this out. I’d really like to see a ballpark estimate of how large a manmade void could be made.
Corrected Version!!!*
Please disregard my original post above
One could build an indefinitely large underground structure if you are wiling to include support columns that divide the space. However, the deeper you go, the wider the columns would become (until the columns all literally melt together in the mantle).
As for a single bay chamber, a dome would be your best bet. The dome would be limited by the compression and sheer strength of the stone it is carved out of, or material used to line and support the dome.
To get a ball park figure, let us make a few simplifying assumptions. We know, for instance, that a dome will transfer all the weight it supports to its base.
Thus we only need to know how much weight the base can support. This is a factor of the material’s compression strength “C” in pounds per square foot, and its area “A” in square feet. The area at the base of a dome is based on its thickness “T” between the outer radii (“R”) inner radii (R-T)). A circle’s area is pi*r^2 (pi=3.14…). A=pi(r^2-(r-T)^2)
Thus the maximum weight the dome can support is C*A or:
Max=C*A=pi(r^2-(r-T)^2)
We also need to estimate how much weight the dome will support. This includes all the material directly above the dome to the surface. To simplify, we will ignore the void under dome, and estimate the weight of material from the floor of the dome to the surface. Using the outer radii, “R”, and a depth “D” to the dome floor, we get: volume of material = piR^2D; multiply the density of the material “M” by the volume, we get
Max weight = MpiR^2*D
We now must solve for R, which will tell us the size of the dome we could build.
Since we can hold no more than “Max”, our “Weight” must be less than or equal to “Max”; thus:
Max = CA=pi(r^2-(r-T)^2) = Mpi*R^2 = Weight
Doing the algebra and simplifying, we get a neat little equation! That is:
MD(R^2) + 2CT*(R) + C*T^2 = 0
Assuming we build the dome of granite, and that the earth above is also granite, we can look up the values for C,and M.
For granite:
C=2.7x10^6 psf (pounds per square foot)
M=168 pcf (pounds per cubic foot)
We will propose a sample problem, a dome of depth
D=1000 feet
and thickness
T=20 feet.
**
Plugging into our equation we get a maximum radius of 2500 feet, or 5000 feet across!!!**
That is just about a mile! Using different values for the depth and thickness or material will vary the size.
Now, this is a thin, flimsy granite dome and this is of course a maximum theoretical size. **Do not try this under your home! 
Engineering standards would only permit a fraction of this size, a “factor of safety”, to account for various other loads that are difficult to account for (such as minute variations in dome geometry). Due to the geometry of the dome, there will also be “sheering” loads that must be accounted for using the shearing strength of granite. (Indeed, sheering stresses in a dome this size would be the “governing” (largest) load)
Voids in solid rock could be potentially even larger!
Much better!
And WOW 
IMO there’s a fundamental error in using standard CE dome calculations for this: What is the outer diameter of a “dome” which is really just (more or less) solid rock extending for miles away in every direction from our (more or less) dome-shaped hollow?
Which also explains billfish678’s point about much larger natural hollows.
I feel obligated to point out that a mile wide dome at 1000 feet below would likely crest well above the surface…
One suspects you math still needs some work
Corrected Version!!!*
Please disregard my original post above
One could build an indefinitely large underground structure if you are wiling to include support columns that divide the space. However, the deeper you go, the wider the columns would become (until the columns all literally melt together in the mantle).
As for a single bay chamber, a dome would be your best bet. The dome would be limited by the compression and sheer strength of the stone it is carved out of, or material used to line and support the dome.
To get a ball park figure, let us make a few simplifying assumptions. We know, for instance, that a dome will transfer all the weight it supports to its base.
Thus we only need to know how much weight the base can support. This is a factor of the material’s compression strength “C” in pounds per square foot, and its area “A” in square feet. The area at the base of a dome is based on its thickness “T” between the outer radii (“R”) inner radii (R-T)). A circle’s area is pi*r^2 (pi=3.14…). A=pi(r^2-(r-T)^2)
Thus the maximum weight the dome can support is C*A or:
Max=C*A=pi(r^2-(r-T)^2)
We also need to estimate how much weight the dome will support. This includes all the material directly above the dome to the surface. To simplify, we will ignore the void under dome, and estimate the weight of material from the floor of the dome to the surface. Using the outer radii, “R”, and a depth “D” to the dome floor, we get: volume of material = piR^2D; multiply the density of the material “M” by the volume, we get
Max weight = MpiR^2*D
We now must solve for R, which will tell us the size of the dome we could build.
Since we can hold no more than “Max”, our “Weight” must be less than or equal to “Max”; thus:
Max = CA=pi(r^2-(r-T)^2) = Mpi*R^2 = Weight
Doing the algebra and simplifying, we get a neat little equation! That is:
MD(R^2) + 2CT*(R) + C*T^2 = 0
Assuming we build the dome of granite, and that the earth above is also granite, we can look up the values for C,and M.
For granite:
C=2.7x10^6 psf (pounds per square foot)
M=168 pcf (pounds per cubic foot)
We will propose a sample problem, a dome of depth
D=1000 feet
and thickness
T=20 feet.
Plugging into our equation we get a maximum radius of 80 feet, or 160 feet across!!!
Using different values for the depth and thickness or material will vary the size.
Now, is of course a maximum theoretical size. Do not try this under your home!
Engineering standards would only permit a fraction of this size, a “factor of safety”, to account for various other loads that are difficult to account for (such as minute variations in dome geometry). Due to the geometry of the dome, there will also be “sheering” loads that must be accounted for using the shearing strength of granite. (Indeed, sheering stresses in a dome this size would be the “governing” (largest) load)
Voids in solid rock could be potentially even larger!
:smack: I am having a bad math day…
Maybe we need to call Randy from Southpark ?
Proceed with caution…
Really great stuff !!
My very first day of work, and then first gig, in the film industry was on a documentary on Grand Central.
Wound up shooting jobs in there as well. The Great Room will forever be evocative for me.
Neat fact: One of the 7 sub-levels was for many years used as a secondary gold bullion reserve, additional to Ft. Knox.
Recrunching my numbers, I actually got a dome that would be 1000 feet across. Such a stone dome would be impossible; due to the sheering force of the weight, a 300 foot diameter disk would come crashing down from the top center, and without the key stone, the rest would come crumbling down as well.
In perfectly solid rock, such a void might be possible, although I think it rare a void that size would exist without interruption.