Amount of material required for a pressure tank I read the answer for the question Why is a hot air balloon “stiff”? and thought something sounded ridiculous.  My engineering requirement is that the walls be strong enough.  Here $T$ will be the tension (for a surface, not sure about those units) and $R$ is radius of curvature of the wall.  Requirement is:
$$\Delta p < 2 \frac{T}{R}$$
Let $d$ be the thickness of the wall and $\sigma$ the material tensile strength.
$$T \propto d \sigma$$
This would indicate that thickness increases linearly with scale.  That sounds ridiculous.
Why it sounds so silly


*

*Volume scales as $R^3$ and surface area as $R^2$.  SA x (thickness) = material volume = constant, so that implies there are no economies of scale for pressure tanks in terms of pressurized volume divided by structural materials.  That sounds nonsensical.  That means a chemical plant wouldn't save any materials by buying a large tank as opposed to 1,000 tiny tanks.

*Say that I have a tank shape in mind.  If I build a small tank and a large tank, they will be geometrically congruent.  That is, if thickness if 5% of the diameter of the small one, it will be 5% of the diameter of the large one.


Please prove me wrong.  And if you can't prove me wrong, please establish a physical intuition as to why this should be the case.
 A: The formulas look correct. As for economies of scale, they also depend on the weight of the pressure vessel and on heat exchange (for example, cryogenic vessels do provide the economies of scale, as far as I know).
A: A similarity to a prior Physics SE question struck me:
Physics of scaling up an animal: the neck
I think there was another question that I commented on like this, but the gist of both of them is that weight scales with volume, $L^3$ while the yield strength of the supporting structures scales with cross sectional area $L^2$.  This is why elephants and dinosaurs have proportionally thick legs.  Similar logic applies to buildings.  In other words, the diameter of a leg must grow compared to a measurement of the body as $L$.  In the case of pressure vessels, the factor is $1$.  Note, my problem with this conclusion was that I wanted it to be less than $1$.  For both of these we should ask why.
I think the key is to take a cross section.  For a building, you will take the cross section across the supporting structures.  For a pressurized sphere, you will take it across the equator of the sphere.  The sphere's cross section has another dimension to grow in.
I tried to think of an example that demonstrates actual physical examples of economies of scale, but nothing so far.
