In an Uncle Scrooge (or Scrooge McDuck, not sure of the translation here) comic named "Treasure Under Glass", Scrooge orders a giant glass dome to be manufactured in order to aid his endeavours exploring a 17th shipwreck containing a valuable treasure.

The giant dome, which is Donald's idea is described in the following image by a fine engineering drawing since I couldn't find a pdf of the comic :

enter image description here

The primary objective is to clear the area of sharks so it can be explored on foot.

From the comic, I made the following rough estimates :

  1. The spanish galeon is 60m long (based on the Neptune), which would make the dome bottom diameter approximately equal to 100 m
  2. We will assume the dome is a perfect half-sphere (assuming from the drawings in the comic)
  3. Dome height estimated at 50 m according to the previous statement
  4. The dome is tightened to the bottom of the ocean with several ropes as in the drawing, each rope being fixed into the wet sand at its end. According to the comic, we see that there are 5 ropes extending from one side to another, distributed in an even fashion.
  5. Following 3., the estimated depth at the bottom of the wreck is 100 m

Following these assumptions, would such a thing be feasible from an engineering point of view, assuming close to infinite funds and that Scrooge agrees to pay the bill ?

In particular :

  • What kind of tension would each rope have to withstand ? Would regular ropes be enough or should they have used steel cables ?
  • How would those be tighten into the sand in a reliable way ?
  • What kind of tremendous forces would the dome have to withstand ?
  • $\begingroup$ The load on the ropes depends upon how many ropes, you want to use. How deep? If the dome bottom diameter is 100m and it is a perfect half-sphere, the height from the bottom would be 50m, not 15. If the depth of the sea floor is 30m and the dome is 50m, it will stand above the surface by 20m. $\endgroup$ – user1683793 May 1 '19 at 23:16
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    $\begingroup$ You could research the engineering principles starting with Caisson. Learn how they built the Brooklyn Bridge. But for a quick estimate, determine the volume of the hemisphere, calculate the weight of that amount of displaced water, and that will be the (negative) weight that the ropes would have to hold down. $\endgroup$ – Ray Butterworth May 2 '19 at 1:37
  • $\begingroup$ @user1683793 indeed ! Question edited ! $\endgroup$ – MaximGi May 2 '19 at 8:22
  • $\begingroup$ Nerarly everything is possible with infinite resources. We can even get humans on other starsystems with infinite resources. Note however, not today. $\endgroup$ – joojaa May 2 '19 at 11:36
  • $\begingroup$ @joojaa We are talking Scrooge McDuck rich! $\endgroup$ – StainlessSteelRat May 2 '19 at 16:13

Spheres buckle under external pressure. This is a stability problem. Stability problems are not as clear cut as we might prefer. Everything looks fine and then surprise, it fails, apparently for no reason.

Two sites I found that discuss this (which my Marks' Mechanical Engineering Handbook did not) are: How to calculate external pressure resistance in spheres and cylinders?


They give the same formula and the link from this site says "When tested, however, this formula was found to be completely inaccurate." I don't think we really care about the exact values, right now, however.

But that was not your question.

The buoyancy of the dome minus its weight determines how much stress will be on the ropes. First, let's find how much the dome wants to float. A hemispherical dome 100M in diameter, using the V=4/3 pi R^3 formula for a sphere and knowing that a cubic meter of water has a mass of 1000 kg, the dome wants to float with a force equal to 262,000 metric tons (1000kg). Wow, that's a lot for five ropes to hold. But wait how heavy is the dome?

Water at 100 M exerts a pressure of about 1000 kPa. Waving my hands and saying the pressure will be the same for the whole dome (which it is not), we can use the magic formula from the other web site with the values

  • Modulus of elasticity 50 gPa
  • Poisson's Ratio 0.22
  • Radius 50 M
  • Pressure 1000 kPa

and get a required thickness of glass of 0.2 M. Given that glass weighs around 2500 Kg/M^3, and the surface of the hemisphere would be (A=0.5*(4*PI*R^2) ) about 16,000 M^2, we can calculate the weight of the dome as 7,859,000 kg.

It looks to me as if the weight of the dome exceeds the weight of the displaced water so it will sink.

The answers:

  • Tension? Zero since no ropes are required.
  • Tighten into the sand? Not necessary
  • Tremendous Pressure? 10 bar is roughly ten times atmospheric.

For comparison, my car tires have about 2 bar of (gauge) pressure. My home air compressor can put out about 6 bar. Mind you, 10 bar over an area of 16,000 m^2 is a force of about 1.6 billion kg.

After typing this up, it occurs to me that the density of glass I gave above is only 2.5 times that of water and the volume inside the dome is a LOT greater than the volume of the material that makes up the dome so I expect I lost a decimal somewhere. I encourage readers to comment and correct as appropriate.

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  • $\begingroup$ I visited the Ariane rocket manufacturing facility many years ago, and they explained that the sections of the rocket were automatically kept at an internal pressure of atmospheric + about 5%, otherwise the cylinder would buckle due to the external pressure. $\endgroup$ – Solar Mike May 3 '19 at 10:39
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    $\begingroup$ I think you may've confused yourself by using metric tons and kilograms for your measure of the buoyant force versus the weight of the dome. I've not checked the math for the results you've provided, but unit conversion yields a buoyant force of 262x10^6 kilograms with the weight of the dome being only 7.859x10^3 kg. Therefore, your buoyant force appears to be another massive hurdle. $\endgroup$ – Pyrotechnical May 17 '19 at 16:09

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