Since you have an interest in local history AND physics, this may be right up up your alley:
During the late 19th century in Chicago, transplanted French physician Dr. Arthur De Bausset gained considerable attention for proposing a transcontinental airship which he called the “aeroplane”, It consisted of a hollow cylinder with conical ends constructed of 1/44" steel plates - with a total dimension being a whopping 654" long by 258" high. (once described as being as big as the Auditorium). Instead of being filled with lighter-than-air gas like a dirigible, it gained buoyancy by having air removed from the cylinder by a vacuum. The absence of air would make it float, with huge size of the airless chamber negating the weight of a passenger cabin propelled by electric motors, and the weight of the tank itself. Speeds could exceed 120 miles an hour, promising two hour trips between Chicago and New York, and overnight trips to Europe. Promotional materials included testimonials by physicists backing the feasibility of the concept. Investors backed its construction, the US government expressed interest for defense purposes, and a huge tract of land was secured in outlying Worth to build it. Needless to say, nothing ever came of it.
Many questions:
Was the concept feasible from a pure physics standpoint? Although I’d question the ability to successfully construct an air-free chamber of this size using available late 19th century materials and technologies, could it be accomplished today? Would there be an ecological advantage to such a system? The electric power concept seems somewhat unlikely at this early date…but how 'bout now?
“Aeroplane” was used to described early heavier-than-air craft and eventually evolved into the present-day “airplane.” Is this the first use of the term?
Or…was it just an outright fraud? Dr. De Bausset disappears from Chicago, briefly re-surfaces in NYC, then disappears. I’ve never checked his identity as a physician in France.
His treatise accompanied by calculations and testimonials is available on Google Books - here’s the link:
Also, you’ll find some details and discussions via a Chicago Tribune ProQuest search under De Bausset.
I dunno…whatcha think?? I’d try to chase this down further myself, but my physics wattage is remarkable low.
Interesting – the science fiction author Larry Niven mentioned dirigibles filled with vacuum in one or more of his books (although he had the skin made out of a sort of force field). I didn’t know that was proposed in real life!
I have to think you’d have a helluva time keeping the thing from crumpling. In gas and hot-air balloons, the inside pressure is greater than outside, so you can use fabric or other lightweight materials. Here you’d have little or no internal pressure.
I’m pretty sure that small-scale “vacuum balloons” have been made, and in principle, they should scale up just fine. In practice, though, a larger surface area means more places you can get a hole, and anyway, there’s hardly any difference in buoyancy between vacuum on the one hand and hydrogen or helium on the other. Given how easy it is to make a gas-filled balloon, I really don’t see any advantage to a vacuum-filled balloon.
At a guess, he might very well have had a medical degree - and I see no immediate reason to doubt it - but it’s a period where formal professional qualifications meant very little when it came to people claiming to be doctors. Basically any quack could claim to be “Dr.” and usually did.
The classic reverse case is Crippen washing up in London and claiming to be “Dr. Crippen”. By any reasonable 21st century standards he wasn’t, yet that name immediately became an indeliable part of his notoriety.
1/44 inch thick steel plates = 0.02272 inches = 0.0577 cm = 0.577 mm thick.
Plates 1/2 mm thick.
I’m leery of a pressure vessel that size withstanding the compressive forces involved.
Question: how thick is the pressure layer on a commercial airliner? What’s the corresponding pressure differential?
Add to it what Chronos says about surface area and holes, and the effect of a hole in said vacuum balloon being much more catastrophic than a hole in a hot air or helium balloon. I wouldn’t want to get a hole while in flight.
Also consider that the flight itself will induce more stresses in the balloon than just the pressure differential - you have to account for stresses of wind pushing on differently, weight being imbalanced, etc.
I’d need to know what was meant by “late 19th century,” but the OED shows the term used in the UK in 1868:
This is clearly a modern-type airplane and not anything like what De Baussett was proposing. The word was used slightly earlier to mean a curved surface like a wing used to provide lift.
A basic thin-wall pressure vessel calculation is that wall stress = Pr/t, where P=pressure, and r and t are radius and thickness, respectively. Plugging in atmospheric pressure (14.6 psi), r (828 in), and t (1/44 in) gives a stress of 532,000 psi. That’s beyond the yield strength of even the highest-strength modern steel (maybe 350,000 psi), and certainly far beyond anything available in the 1880s. Which means that there’s no way de Bausset correctly tested his 1/44-inch steel plates to mimic the stresses they’d actually see, which means he’s full of crap.
But that’s not the limiting factor. Not anywhere near. The limiting factor is buckling because of the external pressure. I don’t know what the factor is for a cylinder (although I can look it up), but for spherical pressure vessels, the critical buckling pressure is, at best, P = 1.21E(t/r)[sup]2[/sup], where E is the Young’s modulus (stiffness) of the material. For steel, the Young’s modulus is 30 million psi; a vessel of this size should therefore withstand about 0.0226 psi.
Looking up cylindrical shells in my reference, the critical external buckling pressure is P = 0.275E(t/r)[sup]3[/sup]. Yes, that’s (t/r) cubed, which means the cylinder will collapse at 0.00000017 psi differential.
In general, it’s interesting to note that all the important quantities here are scalable, so what will or won’t work at the large scale will hold at smaller scales.
To achieve buoyancy, the mass of steel must be less than the mass of displaced air. For a cylindrical shell, the mass of steel is 2[symbol]p[/symbol]rtL[symbol]r[/symbol][sub]steel[/sub], and the mass of displaced air is [symbol]p[/symbol]r[sup]2[/sup]L[symbol]r[/symbol][sub]air[/sub], so 2[symbol]p[/symbol]rtL[symbol]r[/symbol][sub]steel[/sub] < [symbol]p[/symbol]r[sup]2[/sup]L[symbol]r[/symbol][sub]air[/sub], or t/r < [symbol]r[/symbol][sub]air[/sub]/(2[symbol]r[/symbol][sub]steel[/sub]).
For the classic stress-based pressure vessel calculation, the yield stress of the material has to be larger than the applied stress due to atmospheric pressure. From my post above, that’s [symbol]s[/symbol][sub]yield[/sub] > P[sub]atm[/sub]r/t, or t/r > P[sub]atm[/sub]/[symbol]s[/symbol][sub]yield[/sub]
For buckling, the critical pressure must be less than atmospheric. That’s 0.275E(t/r)[sup]3[/sup] < P[sub]atm[/sub], or (t/r)[sup]3[/sup] > 3.67P[sub]atm[/sub]/E.
So scaling up or down, the ratio of thickness to radius has to be set based on material parameters, and the absolute scale is irrelevent (ignoring manufacturing difficulties, which are moot in any case).
I suspect your browser may not be displaying Greek symbols correctly.
This is a pee->p This is a pi->[symbol]p[/symbol]
This is an ar->r This is a rho->[symbol]r[/symbol]
They all display properly to me.
If they don’t for you, then all the rhos in my previous post have subscripts ([symbol]r[/symbol][sub]air[/sub] and [symbol]r[/symbol][sub]steel[/sub]), as do all the pees (P[sub]atm[/sub]). The ars and pis are non-subscripted.
To answer your next question: no.
A vacuum dirigible requires a high stiffness-to-density ratio. Here’s a chart that lists out engineering materials on a 2D plot of stiffness and density (note the units).
For a cylinder, you can rewrite and combine the two equations in my earlier post, eliminating the (r/t) factor, to get:
E[sup]1/3[/sup]/[symbol]r[/symbol][sub]mat’l[/sub] > 3.08P[sub]atm[/sub]/[symbol]r[/symbol][sub]air[/sub]
or, with P[sub]atm[/sub] = 0.0001GPa and [symbol]r[/symbol][sub]air[/sub] = 0.00123Mg/m[sup]3[/sup]
E[sup]1/3[/sup]/[symbol]r[/symbol][sub]mat’l[/sub] > 116.8
In the chart I linked to, the upper left corner is E = 1000GPa and [symbol]r[/symbol][sub]mat’l[/sub] = 0.1Mg/m[sup]3[/sup], or E[sup]1/3[/sup]/[symbol]r[/symbol][sub]mat’l[/sub] = 100. No material is even close to that, and our actual requirement (116) is literally off the chart.
A spherical shell would be better, with the calculations giving:
E[sup]1/2[/sup]/[symbol]r[/symbol][sub]mat’l[/sub] > 11.1
According to the chart, parallel-grain balsa wood would just barely meet that specification. Of course, you can’t use parallel-grain balsa wood without getting cross-grain in the bargain, but it does show that the specifications are much closer. (And I’m ignoring strength for now in favor of the normally-more-elusive stiffness.)
I suspect that some engineered honeycomb-style panels may offer enough stiffness to work for a spherical vessel.
For a cylindrical dirigble, you’d want something literally off the chart, past the upper left corner. For a spherical balloon, you’d want something to the upper left, just past the line connecting diamond, beryllium, and parallel-grain balsa wood.
zut: I think what’s happening with the letters is that you’re trying to rely on a font I don’t have instead of using Unicode characters. That is, you’re dressing up some 'p’s and 'r’s instead of actually using real 'pi’s and 'rho’s, and that just doesn’t work as well.
Not to worry. I can puzzle out the equations if I want to.
Thanks for the chart and explanations, by the way. It’s kind of surprising how far diamond is from what we’d want.
[noparse][symbol]r[/symbol][/noparse] = [symbol]r[/symbol] and [noparse][symbol]rp[/symbol][/noparse] = [symbol]p[/symbol]
so vBulletin is making whatever choice it makes to display the font. Maybe ask in ATMB?
And, yes, particularly in the context of the original question (a cylindrical dirigible), it’s really non-intuitive just how far away from the target even exotic materials are. But buckling is itself non-intuitive, and the ability of these thin shells to withstand external pressure is purely set by buckling.
I am not aware of ANY materials that would make a “vacuum dirigible” feasible. If you make the pressure vessel thick enough to withstand buckling it will be too heavy to float.
First of all, the “pressure vessel” in a commercial airliner is NOT airtight. What keeps it pressurized is air being pumped into the vessel, not the vessel being sealed. In the event of a power failure the interior air continues out the outlets until the pressure is equalized. If power is restored air is pumped back in faster than it “leaks” out until the desired pressure is restored.
That’s also why a (small) hole in an airliner is not a disaster. It might get windy inside, but that’s about it as long as the mechanism to maintain pressure is intact.
Anyhoo - interior pressure in commercial airliners is usually no less than the equivalent of 5000 feet above mean sea level. Such airliners can fly between 30,000 and 35,000 feet. So… look up the air pressure for both interior and exterior, do a little subtraction, and you’ll have it.
At 5,000 MSL the air pressure is about 12.23 psi. At 30,000 MSl it’s 4.37 for a differential of 7.86 psi.
Air pressure at 35,000 MSL is 3.47, so the differential would be 8.76 psi.
(Air pressure values at various altitudes actually vary somewhat in the real world, air being a gas and pressure affected by various factors, but that should be close enough).
Isn’t everyone here assuming that the interior pressure would be 0 atm? And that’s not true.
The Hindenburg, to use an example, had an envelope volume of about 7 million cubic feet, and the structure weighed 130 tons. [1] A cubic foot of air weighs 0.075 pounds. So at 1 atmosphere, the air in the Hindenburg would weigh 262 tons (had they ever filled her with air).
So correct me if I’m wrong, but if the Hindenburg were rigid and filled with air, then she would float if the envelope pressure were reduced to roughly half an atmosphere, which is still enough pressure to walk around in, possibly with the aid of an oxygen mask. And the bigger you make the envelope, the higher your minimum pressure becomes.
Now I’m not saying that this crazy guy’s idea would work. I’m just saying that it’s not fair to produce equations that demand a material hold up to a complete vacuum.
Everyone is assumong that the pressure is zero absolute because the original text by Arthur De Bausset (linked to in the OP) described it as such.
Actually, it’s more than fair. The failure of an externally-pressurized cylinder is by buckling; to prevent buckling, the critical buckling pressure must be greater than the pressure differential. For a pure vacuum, the pressure differential is simply the atmospheric pressure, and we require:
P[sub]crit[/sub] = 0.275E(t/r)[sup]3[/sup] > P[sub]atm[/sub]
If, as you propose, the zeppelin was filled with air at half atmospheric pressure, then:
P[sub]crit[/sub] = 0.275E(t/r)[sup]3[/sup] > P[sub]atm[/sub]/2
Thus, for the same radius, when the pressure difference is cut in half, the thickness can be reduced only by a factor of the cube root of three = 1.26.
So reducing the pressure difference by two a) decreases the buoyancy force by a factor of two, but b) decreases the weight of the zeppelin only by a factor of 1.26. Decreasing the vacuum actually makes the design worse, not better. Full vacuum is the best case scenario.
