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Comment by hellyeasa | Hacker News Reader

Here's another cool random plane tidbit.

Commercial jetliner fuselage wall thicknesses are typically around 1-2mm. They don't call 'em flying tincans for nothin! Think about that next time you fly!

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To put it another way: Fuselages are 20x thicker than the walls of soda cans.

If you shrank a 777 to the size of a soda can, its skin would be about four times thinner than the average can's.

So it would be unwise to fill a 777 with soda.

This is definitely the important lesson.

This is potentially a pretty stupid question - but are the differences in pressure roughly the same between a full, pressurised soda can and a cruising-altitude pressurised airline cabin?

From a 2-minute google search, it looks like both are ~10 psi pressure difference (psig). The numbers I found were 7-10 psig for a plane and an estimate of 17 psig for the soda can. So they're quite similar and it's not a stupid question.

One of us appear to be wrong :-)

I was just coming back to edit my reply to say that the difference appear to be 0.5:2, or the soda can holds roughly 4x the difference.

Ah, but I'm tired, maybe I read/thought wrong.

Well, the 17 psig number I found was an estimate, so if you found the soda can holds about 4*7 = 28 psig, I'd say we agree.

I'm not sure why you'd think it's a stupid question (what is a stupid question?) -- but according to the Internetz[1,2], I think the answer is no.

The difference of pressure between an soda-can and air (at ground level) would appear to be between to and three atmospheres worth (or like sea-level and a depth of 20-30m/~60-90', if the rule-of-thumb I've learned is about correct; see also: Why is parachuting into water OK, while diving and then flying a bad idea?).

It would appear the interior and exterior and an air-plane typically differ at about half an atmosphere (5m/15' water).

[1] http://hypertextbook.com/facts/2000/SeemaMeraj.shtml

[2] http://en.wikipedia.org/wiki/Cabin_pressurization

I think the internet has just made me a little over-cautious, especially when being around so many smart people as there are on HN :)

So that pretty much answers what I was hesitant to suggest out loud - that even though the fuselage of an airplane is proportionately thinner, the soda can has to withstand a greater pressure difference. Thanks!

More important than pressure differences, is the circumferential stress on the fuselage/can, which is what the walls are actually withstanding. For a pressurized circular container, it is given by p*r/t, where p is the pressure, r the radius and t the wall thickness.

If we consider a 3 m wide airplane, with a 1.6 mm thick fuselage at cruise altitude (around 40k feet), you get around 56.5 kPa of pressure difference, and the above equation gives you approximately 52 MPa. (data from a real aircraft that I cannot mention).

For a coca-cola can, internets say 380 kPa of pressure, for a 6.6 cm diameter can, and 0.15 mm aluminum sheet. That results in approx. 85 MPa.

So as you can see, even though you have roughly 7x more pressure in the soda can, its much smaller diameter severely reduces the stress on the walls, resulting in about 1.5x the circumferential stress.

Hope that answers your question.

edit: several errors in my back-of-the-envelope calculations. sorry.

That's an odd model for me. So as the radius goes to infinity for a fixed thickness and differential the stress goes to infinity too? But take a pressurized cube with reinforced edges. You can say it's faces have infinite radius (or as small as you'd like) -- yet the stress should be finite? How do you reconcile this? Or does that work only for circles? (which would be odd since stress is a local concept to me)

I'm not sure of your background, but it is actually a pretty simple concept. To satisfy static equilibrium on a given transverse "dx" segment of a cylinder, the walls must balance the internal pressure (in this case taken as the pressure difference), so that:

stress * 2 * thickness * dx = p * 2 * r * dx.

[ stress * walls cross-sectional area] = [ internal pressure * projected internal area ]

Solve for stress, you get:

stress = p * r / t

Regarding your question, yes, as the radius goes to infinity, the stress goes to infinity. The area where the pressure is applied grows with r, but the cross-sectional area where the stress is applied is still (w * thickness * dx.) This equations work well for thin-walled cylindrical pressure vessels (r > 5t is general rule of thumb). For a cube, you would have to develop the equations, but keep in mind that you will have a singularity/discontinuity on the walls because of the right angle.

edit: good to have a reference just in case: http://ocw.mit.edu/courses/materials-science-and-engineering... [PDF ALERT]

Ah makes more sense now, thanks for the explanation. I'm an undergrad in EE so mechanics is not my strong point.

So a planar face of a cube cannot satisfy the equilibrium equations? Interesting ... so then a cube will necessarily bulge so the radius is enough to satisfy the equations, right?

> If you shrank a 777 to the size of a soda can, its skin would be about four times thinner than the average can's.

My take.. The difference between inside and outside of soda can is approx 175 (kPa). The difference between inside and outside of aircraft cabin at the cruising altitude is 56.6 (kPa). So Soda can bears differential of approx 3 times than the pressure differential that aircraft cabin structure supports, with material 4 times thinner.

Aircraft is 12 times more safer than a soda can (!)

Interesting.. aircraft safety engineers at work.

Reference:

http://en.wikipedia.org/wiki/Cabin_pressurization

http://www.engineeringtoolbox.com/air-altitude-pressure-d_46...

Well, except that aircraft pressure is applied outside-in while soda can pressure is inside-out. Put some outside-in pressure on a can, and see how well it handles it.

Both are engineered in a way to make pressure not a problem. The can is subject to stress if you refill it, the airplane is a bit overengineered so it's not subject to stress. Increasing the width of any wouldn't lead to an increase on their safety.

Sorry I'm confused by your statement. The air pressure inside an aircraft during flight is much higher than the air pressure outside. So the pressure is also inside -> out, same as a soda can.

I think you got something backwards there. The plane is thinner when you make them the same size.

You are neglecting the fact that a soda can is pressurized once, depressurized when opened for consumption, and then wasted.

An aircraft fuselage needs to withstand THOUSANDS of cycles of pressurization / depressurization during its service life, which is usually much more critical then the static pressure load, due to metal fatigue. This is why aircraft service life is given in flights (which corresponds to one pressurization cycle) and not in flight hours or miles or whatever. This is also why aircraft used on longer routes tend to last for more years (longer flights = fewer pressurization cycles).

edit: If anyone wants a reference, here goes one http://www.airspacemag.com/need-to-know/what-determines-an-a...

edit2: Additionally, consider checking my response to another comment. The diameter of the fuselage is so much bigger than the soda can that the actual stress on the walls are roughly only 1.5x bigger on the soda cans.

Scaling objects doesn't preserve their ability to withstand forces: for example cross-section areas scale with square of scaling factor and masses/weights scale with the cube of scaling factor; so objects are close to collapsing due to their own weight if you scale them up.

That also makes me think about Aloha Airlines Flight 243

http://www.vfrmagazine.net/wp-content/uploads/2012/09/fretti...

http://en.wikipedia.org/wiki/Aloha_Airlines_Flight_243

For comparison, passenger car body panels are somewhat thinner, at 0.6-1mm.

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