Hellbound
Merchant of Doom
http://en.wikipedia.org/wiki/Close-packing_of_equal_spheres
You're wrong, quarky. It's approximately 26% void, not 27%.
Ah, JREF. Where you can never get so high that someone can't knock you down a peg or two
Engineering a vacuum zepelin?
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Ah, JREF. Where you can never get so high that someone can't knock you down a peg or two
Ah, the relentless humiliation that is jref.
In my defense, (Christ, quarky, must you?) I must reiterate that the, ahem, 26% is with optimal packing, which is unlikely in the extreme.
I come at this from cement and concrete. In normal gravel and sand, the voids can be as much as 40%. This is why a variety of aggregate sizes are indicated.
Fine sand, for instance, counter-intuitively, has more vacancies (40% plus) than coarse sand.
Why?
Because coarse sand contains large and small particles...the small, filling spaces within the large. In making mortars and concrete, the cement's purpose is to fill these voids.
For economic (and other) reasons, one want's to minimize the voids.
For those less familiar with such arcane and boring engineering, I use this anology to drive home the boredom:
Picture a large room filled with bowling balls:
Packed in their tightest configuration (ahem, seriously unlikely) there is 26% vacancies.
Now, imagine a crap-load of golf balls added. They fit nicely in the vacancies, thereby reducing the overall emptiness...quite dramatically...especially if you golf and bowl, god forbid.
In my defense, (Christ, quarky, must you?) I must reiterate that the, ahem, 26% is with optimal packing, which is unlikely in the extreme.
I come at this from cement and concrete. In normal gravel and sand, the voids can be as much as 40%. This is why a variety of aggregate sizes are indicated.
Fine sand, for instance, counter-intuitively, has more vacancies (40% plus) than coarse sand.
Why?
Because coarse sand contains large and small particles...the small, filling spaces within the large. In making mortars and concrete, the cement's purpose is to fill these voids.
For economic (and other) reasons, one want's to minimize the voids.
For those less familiar with such arcane and boring engineering, I use this anology to drive home the boredom:
Picture a large room filled with bowling balls:
Packed in their tightest configuration (ahem, seriously unlikely) there is 26% vacancies.
Now, imagine a crap-load of golf balls added. They fit nicely in the vacancies, thereby reducing the overall emptiness...quite dramatically...especially if you golf and bowl, god forbid.
rwguinn
Penultimate Amazing
Back to the cclaims/defense of such structures ...One of the problems is that you have to maintain the internal volume while increasing the external pressure. This calls for a very rigid outer shell. Fabric won't do the trick...
Be nice if a balloon could be held rigid by tensional rather than compressive members- ie attached to an external frame.
How about a vacuum bubble inside a traditional gas filled dirigible, with the inner balloon anchored to the dirigible frame by tensioned wires? Lighter than comprssive columns, I'd think. (Though the compressive load is only 1 atmosphere. Can't kevlar spokes support that?)
Also are support columns made of plastic tubes full of pressurised helium complete SF?
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rwguinn
Penultimate Amazing
1atmospere times the area each wire is connected to. A mere 160 square inches (<15X15)gives you a ton of load in the tension member...
Ugh.
I'm embarrassed by how much I've thought about this, when I could have been making money.
For the poster that "beat me to it', I've pondered this for 30 years, at least.
However, there is evidence of the notion being tossed about in the 1600's.
Nothing new under the sun.
Except for the new stuff.
Soapy Sam, I may return to address your point.
(I apologize, pre-emptively. You seem like a decent bloke.)
rwguinn said:
Yes, but that's why high tensile steel is so called. The thickness of a drilling rig derrick leg, compared to the diameter of the drilling line illustrates the difference between the amount of steel needed to support the same load in compression and in tension.
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Don't think so-- just enlarge those helium support columns until you're left with a small token vacuum chamber in the center, and you're done!
phunk
Illuminator
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A pure tension structure is impossible.
Sure, but in the chase, you might get a mostly tension structure.
We have suspension bridges and bike wheels. And it's just a beginning.
macdoc
Philosopher
toss an evacuated vacuum tube in water - will it float? - there is your answer.
BenBurch
Gatekeeper of The Left
Ziggurat
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I went through the scaling on this once. It doesn't help: the force that the outer structure needs to exert on the inner structure will need to be supported by one of two things: either higher gas pressure or stiff structural support. If it's structural support, then you face exactly the same strength to weight problem you have without putting your vacuum balloon inside another balloon. If you're trying to support the outer balloon with increased gas pressure inside the surrounding balloon, then the density of your gas-filled balloon increases, meaning that you're adding weight. And the amount of weight you add by doing so is going to exactly match the weight you displaced with your vacuum balloon, ignoring the weight of the vacuum balloon itself (under the assumption that tensile strengths are potentially so high that it's negligible). So you gain nothing by putting a vacuum balloon inside a gas balloon, you might as well just stick with the gas-only balloon.
It's not SF, it's just not any better (and significantly harder) than using a simple helium balloon.
rwguinn
Penultimate Amazing
???
Ziggurat
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It's no use. The higher pressure needed in the outer balloon in order to support the structure means higher gas density and more gas weight. You gain nothing from it over a simple gas balloon.
We can even figure out the scaling involved. Suppose we have an outer balloon of radius R and an inner vacuum balloon of radius r. Strings from the outer surface to the inner surface will keep the inner vacuum balloon expanded, so long as the outer balloon stays expanded as well.
Let's call the pressure in the outer balloon P, and the pressure of the outside atmosphere as A. This pressure pushes inwards on the vacuum balloon with a net force of P*4pi*r2. It also presses on the outer shell with a net force of P*4pi*R2. But the net force on the outer shell must resist both the external pressure (total force of A*4pi*R2) and the pull from the vacuum balloon. The lowest pressure which will accommodate this (ie, no extra force held by the outer balloon membrane) happens when
P*4pi*R2 = A*4pi*R2 + P*4pi*r2
or, dropping our factors of 4pi,
P*R2 = A*R2 + P*r2
Now we solve for P:
P*(R2-r2) = A*R2
P = A*R2/(R2-r2)
Obviously for r > 0, we must have P >= A. This increases the density of our gas by a factor of P/A over what would have been required to inflate the balloon with no inner vacuum. If d and m are the density and mass of gas inside a barely inflated gas balloon with no inner vacuum balloon and D and M are the density and mass of gas inside a barely inflated balloon with an inner vacuum balloon, then we have
m = d*(4/3)pi*R3M = D*[(4/3)pi*R3-(4/3)pi*r3]=D*(4/3)pi*[R3-r3]
But as above, we know D=d*P/A
M = d*(P/A)*(4/3)pi*[R3-r3] = d*(R2/(R2-r2))*(4/3)pi*[R3-r3]
A little bit of juggling gives us
M = d*(4/3)pi*R3*(R2-r3/R)/(R2-r2)
M = m*(1-r3/R3)/(1-r2/R2)
For any 0 < r < R, r3/R3 < r2/R2, so 1-r3/R3 > 1-r2/R2 and M > m. So actually I misremembered: this will always shoot you in the foot. The higher density needed to support this structure will always cost more in mass than you gain by reducing the gas volume.
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Ziggurat
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If the inside balloon pressure is nonzero, then the weight penalty is reduced, but you're still going to have a penalty. But reducing the penalty doesn't ever turn it into a benefit, so this still won't help.