# wall thickness for a very large, low-pressure vessel?

I'm working on a space colony simulation, and I'm at the point where I need to calculate the thickness of the habitat walls, as a function of air pressure and geometry (and assuming some common material).

I've done a fair bit of googling on the topic, and found handy charts like this one: , but these are set up for high pressures (hundreds of psi) and small vessels (a meter or two at most). I'm looking at relatively low pressures -- one atmosphere or less -- but a radius of hundreds, maybe thousands of meters.

To make matters worse, I understand that a cylindrical vessel is most efficiently made using a wound filament design (otherwise, it's twice as strong as it needs to be circumferentially). But I can't find any charts or formulas showing the thickness (or mass) of material needed in this case.

I do understand, though, that the required thickness is directly proportional to both radius and pressure. So if I had an answer for (say) a 1000 m radius and 1 atm, I could simply scale that by whatever radius and pressure I have (right?). Using this calculator, I find a thickness of 44.16 inch (1.12 m). Does that seem right?

• Mylar can withstand several atmospheres. Whatever you're doing in your calculations is way off. Remember dirigibles? – Carl Witthoft Jun 25 '14 at 19:07

If you take a one meter long section of your cylinder and imagine cutting it in half, the force separating the halves is pressure*diameter*1m This is resisted by 2*wall thickness*1m of wall. So the stress in the wall is the ratio of these:$$\text{stress}=\frac {\text{pressure*diameter}}{2* \text{wall}}\\ \text{wall}=\frac {\text{pressure*diameter}}{2* \text{stress}}$$ Note that the wall thickness scales linearly with the diameter.