Carbon Fiber Vacuum Balloon!
April 7, 2011 7:43 PM Subscribe
Giant carbon fiber ball strong enough to be vacuum pumped. Is that possible?
Here's the theory: create a carbon fiber sphere large enough such that when you vacuum pump the sphere removing all air, the density of the sphere is less than air and it floats!
Is this possible?
I figure that the sphere will have to be gigantic. In theory, according to my napkin calculations, the sphere would have to have radius (40*W)ft. where W is the weight of the carbon fiber for 1 sq. ft.
Would carbon fiber (with the appropriate resin) be strong enough to support a vacuum of that size?
Here's the theory: create a carbon fiber sphere large enough such that when you vacuum pump the sphere removing all air, the density of the sphere is less than air and it floats!
Is this possible?
I figure that the sphere will have to be gigantic. In theory, according to my napkin calculations, the sphere would have to have radius (40*W)ft. where W is the weight of the carbon fiber for 1 sq. ft.
Would carbon fiber (with the appropriate resin) be strong enough to support a vacuum of that size?
@pompomtom: yes, except that, as the Hindenburg showed, hydrogen has this nasty tendency to combust....
posted by brianogilvie at 7:57 PM on April 7, 2011
posted by brianogilvie at 7:57 PM on April 7, 2011
Response by poster: The point is that you do vacuum pump it. That's why it'd be cool. Yeah, helium would work, but that's not the goal.
posted by 47triple2 at 8:00 PM on April 7, 2011
posted by 47triple2 at 8:00 PM on April 7, 2011
Atmospheric pressure is about 15 pounds per square inch, so you can work out what the force will be on it pretty easily.
My guess is it's very hard to do, if not impossible.
posted by pombe at 8:24 PM on April 7, 2011
My guess is it's very hard to do, if not impossible.
posted by pombe at 8:24 PM on April 7, 2011
Carbon-fiber-based materials are prized for their tensile strength; carbon fiber has very poor compressive strength, and compression is the name of the game you want to play here. So unfortunately, no.
posted by mhoye at 8:37 PM on April 7, 2011 [1 favorite]
posted by mhoye at 8:37 PM on April 7, 2011 [1 favorite]
There's a difference between tensile strength and compressive strength. Rope has excellent tensile strength but lousy compressive strength. Concrete is just the opposite.
To create this structure, compressive strength is what's critical. The problem is that carbon fibers are like rope: great under tension, lousy under compression. Just like rope, they fold and bend.
Your sphere would collapse under air pressure. Structural materials capable of supporting that much compression would be too heavy to float on air.
...floating on water is a different thing, of course, and that's why you can build ship hulls out of steel, which has superb compressive strength.
posted by Chocolate Pickle at 8:42 PM on April 7, 2011 [1 favorite]
To create this structure, compressive strength is what's critical. The problem is that carbon fibers are like rope: great under tension, lousy under compression. Just like rope, they fold and bend.
Your sphere would collapse under air pressure. Structural materials capable of supporting that much compression would be too heavy to float on air.
...floating on water is a different thing, of course, and that's why you can build ship hulls out of steel, which has superb compressive strength.
posted by Chocolate Pickle at 8:42 PM on April 7, 2011 [1 favorite]
Response by poster: What about glass? (As per pompomtom's link.) Lightbulbs are made of glass, so let's make a giant sphere of glass! :) Does the pressure on the glass increase with increasing size? (e.g. the pressure "felt" by a glass sphere radius 1" = X lbs/sq.in. the pressure "felt" by a glass sphere of radius 5" = X or Y > X?
posted by 47triple2 at 8:45 PM on April 7, 2011
posted by 47triple2 at 8:45 PM on April 7, 2011
Glass is too heavy.
As to the compression, it's proportional to the surface area. However, as the bulb gets larger, the curvature of the bulb decreases, which means you don't get as much strength from the shape. You have to compensate for that by making the glass thicker. It turns out that the scaling factors work against you: as the bulb gets bigger you lose more than you gain.
Also, note that your light bulbs don't float away.
So far as I know there isn't any material which is both strong enough under compression and simultaneously light enough to make this work, at any size above the level of Brownian motion (which is a different physical effect).
posted by Chocolate Pickle at 8:51 PM on April 7, 2011
As to the compression, it's proportional to the surface area. However, as the bulb gets larger, the curvature of the bulb decreases, which means you don't get as much strength from the shape. You have to compensate for that by making the glass thicker. It turns out that the scaling factors work against you: as the bulb gets bigger you lose more than you gain.
Also, note that your light bulbs don't float away.
So far as I know there isn't any material which is both strong enough under compression and simultaneously light enough to make this work, at any size above the level of Brownian motion (which is a different physical effect).
posted by Chocolate Pickle at 8:51 PM on April 7, 2011
This is one of those questions that engineers and materials scientists tend to dream about. Suffice to say, no one has come up with a practical material which could be pumped to a vacuum, hold its shape and be neutrally buoyant. If there were such a thing, they sky would be full of cheap zeppelins.
Sorry to burst your bubble.
posted by bonehead at 9:12 PM on April 7, 2011 [2 favorites]
Sorry to burst your bubble.
posted by bonehead at 9:12 PM on April 7, 2011 [2 favorites]
Mythbusters made a lead balloon... maybe you should submit this one.
posted by j03 at 10:02 PM on April 7, 2011
posted by j03 at 10:02 PM on April 7, 2011
Response by poster:
Let us suppose that this magical piece of material weighs 100 lbs to support 2160lbs. Then we could say that the sphere would need to be roughly 4,000 feet and it will float.
(This 40*W figure comes from: 4* pi * r^2 * W - 4/3 * pi * r^3 * .075 < 0 where .075 is the density of air and W is the weight of the material for one square foot of surface area. If W = 1, the equation is true for r > 40.)
posted by 47triple2 at 11:01 PM on April 7, 2011
As to the compression, it's proportional to the surface area. However, as the bulb gets larger, the curvature of the bulb decreases, which means you don't get as much strength from the shape. You have to compensate for that by making the glass thicker. It turns out that the scaling factors work against you: as the bulb gets bigger you lose more than you gain.Couldn't you then find a material that is strong enough to support a little more than a ton of weight (15 lbs/in^2 * 12 in. * 12 in = 2160 lbs) over a one foot section? (i.e.: a piece of material is laid flat over a void and compressed at 2160 lbs and the material doesn't flex.) We could then scale the sphere until the density-air is less than 0.
Let us suppose that this magical piece of material weighs 100 lbs to support 2160lbs. Then we could say that the sphere would need to be roughly 4,000 feet and it will float.
(This 40*W figure comes from: 4* pi * r^2 * W - 4/3 * pi * r^3 * .075 < 0 where .075 is the density of air and W is the weight of the material for one square foot of surface area. If W = 1, the equation is true for r > 40.)
posted by 47triple2 at 11:01 PM on April 7, 2011
...Yeah, we could, if such a thing existed. But it doesn't.
Or rather, it exists, but it's too heavy. (It's called "steel plate".)
posted by Chocolate Pickle at 11:04 PM on April 7, 2011
Or rather, it exists, but it's too heavy. (It's called "steel plate".)
posted by Chocolate Pickle at 11:04 PM on April 7, 2011
Couldn't you then find a material that is strong enough to support a little more than a ton of weight (15 lbs/in^2 * 12 in. * 12 in = 2160 lbs) over a one foot section? (i.e.: a piece of material is laid flat over a void and compressed at 2160 lbs and the material doesn't flex.) We could then scale the sphere until the density-air is less than 0.
That isn't how it works. It doesn't scale that way.
It turns out that each section of the surface of your sphere has to support the compressive load of the whole skin, not just its own small piece of it. If you double the diameter of your sphere, the load on every part of the skin quadruples.
posted by Chocolate Pickle at 11:16 PM on April 7, 2011
That isn't how it works. It doesn't scale that way.
It turns out that each section of the surface of your sphere has to support the compressive load of the whole skin, not just its own small piece of it. If you double the diameter of your sphere, the load on every part of the skin quadruples.
posted by Chocolate Pickle at 11:16 PM on April 7, 2011
I think this is the point where I have to hand this off to someone who passed "Statics" when getting a Mechanical Engineering degree....
posted by Chocolate Pickle at 11:17 PM on April 7, 2011
posted by Chocolate Pickle at 11:17 PM on April 7, 2011
Is this even feasible?
http://www.cjfearnley.com/fuller-faq-5.html#ss5.2
posted by roboton666 at 12:54 AM on April 8, 2011 [1 favorite]
http://www.cjfearnley.com/fuller-faq-5.html#ss5.2
posted by roboton666 at 12:54 AM on April 8, 2011 [1 favorite]
I have thought about this sort of thing in the past, not because it is at all practical, but it's a common (small) puzzle.
Basically you can determine the hypothetical yes/no assuming a perfect hollow sphere with a wall of thickness x, with no dents or anything, then looking at the compressive strength of the material it is made from. Assume a perfect vacuum inside.
Now, imagine the cross section of the sphere that intersects the great circle (a term from astronomy, like the equator, okay there is probably a geometrical term more appropriate here...) thusly dividing the sphere in half.
At the interface between the two halves there is a ring, with the thickness of the wall of the hollow sphere being equal to the thickness of the ring.
You can determine the pressure which is compressing the material here, which is going to be pi*(outermost radius of hollow sphere)^2 multiplied by atmospheric pressure, divided by the total area of the part of the cross section that intersects the material that makes up the wall of the sphere (i.e. if the outer radius of the sphere is Ro, of the inner radius is Ri, pi*Ro^2 - pi*Ri^2).
So that's how you can estimate if a perfect hollow sphere made of a certain material of a certain thickness will get shmooshed by the pressure or not. Obviously if it the pressure exceeds the compressive strength it's not necesarily going to collapse suddenly because the walls would get thicker as it collapsed, but you get the idea.
If the amount of air id displaces weighs more than it does, you're floating, otherwise no.
Turns out BTW that if you make spheres with wall thicknesses is such that the compressive strength of the material is equal to the pressure it has to withstand, whether or not it floats is not affected by the size of the sphere, only the material used.
I think I checked, and there are some materials that would hypothetically work, but I forget what they are. Just lookup a table of compressive strength values.
posted by Nish ton at 1:40 AM on April 8, 2011 [1 favorite]
Basically you can determine the hypothetical yes/no assuming a perfect hollow sphere with a wall of thickness x, with no dents or anything, then looking at the compressive strength of the material it is made from. Assume a perfect vacuum inside.
Now, imagine the cross section of the sphere that intersects the great circle (a term from astronomy, like the equator, okay there is probably a geometrical term more appropriate here...) thusly dividing the sphere in half.
At the interface between the two halves there is a ring, with the thickness of the wall of the hollow sphere being equal to the thickness of the ring.
You can determine the pressure which is compressing the material here, which is going to be pi*(outermost radius of hollow sphere)^2 multiplied by atmospheric pressure, divided by the total area of the part of the cross section that intersects the material that makes up the wall of the sphere (i.e. if the outer radius of the sphere is Ro, of the inner radius is Ri, pi*Ro^2 - pi*Ri^2).
So that's how you can estimate if a perfect hollow sphere made of a certain material of a certain thickness will get shmooshed by the pressure or not. Obviously if it the pressure exceeds the compressive strength it's not necesarily going to collapse suddenly because the walls would get thicker as it collapsed, but you get the idea.
If the amount of air id displaces weighs more than it does, you're floating, otherwise no.
Turns out BTW that if you make spheres with wall thicknesses is such that the compressive strength of the material is equal to the pressure it has to withstand, whether or not it floats is not affected by the size of the sphere, only the material used.
I think I checked, and there are some materials that would hypothetically work, but I forget what they are. Just lookup a table of compressive strength values.
posted by Nish ton at 1:40 AM on April 8, 2011 [1 favorite]
Whoops, cross section that is in the same plane as the great circle.
posted by Nish ton at 1:45 AM on April 8, 2011
posted by Nish ton at 1:45 AM on April 8, 2011
Yeah, I was fascinated by this possibility for a few years, especially after doing the math Nish ton describes and realizing you could make them of arbitrary size. I found a table of compressive strengths and weights and found that fused quartz is theoretically strong enough to work. I then daydreamed mightily of the vast fleets of crystal dirigibles I would one day build, and floating cities which would drift overhead until broken by mischance and falling to the ground in a huge shower of shattered glass.
posted by hattifattener at 1:55 AM on April 8, 2011 [3 favorites]
posted by hattifattener at 1:55 AM on April 8, 2011 [3 favorites]
Oh hey, I even posted in that halfbakery thread! Guess I've liked this idea for longer than I thought.
I think I was probably inspired by the Buckminster Fuller thing.
posted by hattifattener at 1:58 AM on April 8, 2011
I think I was probably inspired by the Buckminster Fuller thing.
posted by hattifattener at 1:58 AM on April 8, 2011
calculating the compressive stress in your spherical shell isn't too hard (σ = pr/2t), but the problem is the failure mode won't be compressive failure it'll be buckling. all the online references I can find with some trivial googling cover buckling failure in columns not spherical shells, and a quick look through the structures references I have handy also only cover columns, so I can't give you one of the empirical "here's your overpressure limit" calculations. I *can* tell you that buckling failure will occur at a significantly lower (by orders of magnitude) compressive stress than traditional compressive failure, especially at the large volumes and hence low curvatures that this would involve.
short form: we don't have the materials to do this, by a long long shot.
posted by russm at 3:12 AM on April 8, 2011
short form: we don't have the materials to do this, by a long long shot.
posted by russm at 3:12 AM on April 8, 2011
ah, OK.
∆p = 2E/3 · (t/D)³
so the overpressure at which your material will buckle is 2/3 of Young's modulus times the cube of the thickness to diameter ratio. your thickness to diameter ratio will be tiny, and when you cube that it'll be miniscule. buckling failure will kill this dead.
posted by russm at 3:27 AM on April 8, 2011 [1 favorite]
∆p = 2E/3 · (t/D)³
so the overpressure at which your material will buckle is 2/3 of Young's modulus times the cube of the thickness to diameter ratio. your thickness to diameter ratio will be tiny, and when you cube that it'll be miniscule. buckling failure will kill this dead.
posted by russm at 3:27 AM on April 8, 2011 [1 favorite]
Best answer: damn, now I'm curious.
the highest Y value in this table is diamond, so lets go with that. substituting in, assuming standard 1atm overpressure, we have
101325Pa = 2/3 1220000000000Pa · (t/D)³
t/D = ∛(3/2 · 101325 / 1220000000000)
≈ 0.005 ≈ 1/200
so a perfect diamond sphere can withstand a 1atm overpressure if it has a minimum thickness to diameter ratio of 1/200
diamond density is 3.5g/cm³, so the mass of our sphere is 3.5g/cm³ · 4πr² · t
mass of air displaced is 1.22521 kg/m³ · 4/3 πr³
to be neutrally bouyant, 3.5g/cm³ · 4πr² · t = 1.22521 kg/m³ · 4/3 πr³
which eventually gives us a required thickness to diameter ratio of t/D = 1/17140000 independant of diameter (cool!)
so we need a material that has a stiffness approximately 85700 times that of flawless diamond in order to be able to do this.
bummer. :(
posted by russm at 3:59 AM on April 8, 2011 [2 favorites]
the highest Y value in this table is diamond, so lets go with that. substituting in, assuming standard 1atm overpressure, we have
101325Pa = 2/3 1220000000000Pa · (t/D)³
t/D = ∛(3/2 · 101325 / 1220000000000)
≈ 0.005 ≈ 1/200
so a perfect diamond sphere can withstand a 1atm overpressure if it has a minimum thickness to diameter ratio of 1/200
diamond density is 3.5g/cm³, so the mass of our sphere is 3.5g/cm³ · 4πr² · t
mass of air displaced is 1.22521 kg/m³ · 4/3 πr³
to be neutrally bouyant, 3.5g/cm³ · 4πr² · t = 1.22521 kg/m³ · 4/3 πr³
which eventually gives us a required thickness to diameter ratio of t/D = 1/17140000 independant of diameter (cool!)
so we need a material that has a stiffness approximately 85700 times that of flawless diamond in order to be able to do this.
bummer. :(
posted by russm at 3:59 AM on April 8, 2011 [2 favorites]
or, I guess, something as strong as diamond but ∛85700 the density (79kg/m³, roughly enough the density of styrofoam)
posted by russm at 4:10 AM on April 8, 2011
posted by russm at 4:10 AM on April 8, 2011
It has to be air-tight, it doesn't have to be perfectly spherical. Can fractal geometry work here, a'la a Buckminster dome where each panel is supported by a lightweight suspension system, where carbon-fiber could do it's thing?
posted by Slap*Happy at 4:57 AM on April 8, 2011
posted by Slap*Happy at 4:57 AM on April 8, 2011
It also doesn't have to be hollow, as you point out, an internal framework can support the outer shell, as long as your density (structure + shell) is still less than air...
posted by defcom1 at 5:03 AM on April 8, 2011
posted by defcom1 at 5:03 AM on April 8, 2011
Carbon-fiber-based materials are prized for their tensile strength
Not necessarily. Carbon fibre when cured in resin (and with paper/kevlar/aluminium honeycomb reinforcements ) is often used in torsional and compressive installations. Carbon fibre race car chassis, aerodynamic elements and bodywork and aircraft fuselages have carbon fibre composites in compressive loadings with no other supports (ie steel or other reinforcements). The more severe loaded areas (is roll protection structures and major suspension mounting points) are usually made with a bonded in aluminium or otherwise structure but it is not at all purely a tensile stuctural material.
Noseboxes on race cars and rear intrusion protection piece are purely carbon fibre and honeycomb. Doesn't get much more compressive than that when it comes to crash protection.
In this askme instance, of course, the carbon structure would be able to support a sizeable negative pressure, but there's no way it'd support a vacuum with sufficiently low mass of carbon that it'd achieve negative bouyancy in air.
posted by Brockles at 5:05 AM on April 8, 2011 [1 favorite]
Not necessarily. Carbon fibre when cured in resin (and with paper/kevlar/aluminium honeycomb reinforcements ) is often used in torsional and compressive installations. Carbon fibre race car chassis, aerodynamic elements and bodywork and aircraft fuselages have carbon fibre composites in compressive loadings with no other supports (ie steel or other reinforcements). The more severe loaded areas (is roll protection structures and major suspension mounting points) are usually made with a bonded in aluminium or otherwise structure but it is not at all purely a tensile stuctural material.
Noseboxes on race cars and rear intrusion protection piece are purely carbon fibre and honeycomb. Doesn't get much more compressive than that when it comes to crash protection.
In this askme instance, of course, the carbon structure would be able to support a sizeable negative pressure, but there's no way it'd support a vacuum with sufficiently low mass of carbon that it'd achieve negative bouyancy in air.
posted by Brockles at 5:05 AM on April 8, 2011 [1 favorite]
Slap*Happy - not with any materials we know of. according to New Scientist (according to some guy at the halfbakery), vacuum is 14% more buoyant than helium. so imagine a normal everyday helium balloon, then weigh it down with paperclips or something until it's just hovering there. then add 14% more stuff. that's how much stuff you have to make a vessel the size of that balloon that will withstand a 1atm overpressure.
posted by russm at 5:22 AM on April 8, 2011
posted by russm at 5:22 AM on April 8, 2011
I like defcom's idea - what if you made a very strong (insert fancy materials science stuff here) wire armature, and covered it with a rubber balloon?
posted by aimedwander at 8:20 AM on April 8, 2011
posted by aimedwander at 8:20 AM on April 8, 2011
aimedwander - think of it this way: the helium inside a balloon is what stops the balloon from collapsing. if you want to replace the helium with a vacuum, you need to make a rigid truss structure to stop the skin from collapsing, and it has to have the same mass as the helium you're removing.
any solid will be much much denser than helium gas. Dupont E130 carbon fiber is 2.15 g/cm³ (dunno if that includes the matrix), and helium is 0.0001786 g/cm³. so this hypothetical truss structure can fill no more than 0.000083 of the volume of the balloon.
of course, "insert fancy materials science stuff here" can solve anything... just make out of unobtainium! :)
posted by russm at 6:25 PM on April 8, 2011
any solid will be much much denser than helium gas. Dupont E130 carbon fiber is 2.15 g/cm³ (dunno if that includes the matrix), and helium is 0.0001786 g/cm³. so this hypothetical truss structure can fill no more than 0.000083 of the volume of the balloon.
of course, "insert fancy materials science stuff here" can solve anything... just make out of unobtainium! :)
posted by russm at 6:25 PM on April 8, 2011
What about an exo skeleton frame using new materials in a graduated decompression....two or three layers dropping pressure gradually say 5psi per layer?
posted by kitbuoy at 11:00 AM on November 28, 2011
posted by kitbuoy at 11:00 AM on November 28, 2011
This thread is closed to new comments.
posted by pompomtom at 7:55 PM on April 7, 2011