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Remote External Heat Source & a One-Way Valve ?

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One issue is reducing the pressure of gas inside an airship as altitude increases to allow for higher altitude operations.

A remote energy source (ex : ground based or perhaps orbital Laser) could be used to disproportionately heat any gas remaining within the airship, forcing it to expand and exit via the one way valve. This would decrease moles of gas (& pressure) within the airship (toward a more perfect vacuum). The resulting dynamic decrease in airship density should allow for maximal buoyant lift subject to the strength of materials issues discussed elsewhere in this article. Naturally the process would be inefficient in terms of total energy consumption, but importantly (unlike in a traditional hot air balloon) the energy source does not add to the weight of the airship. (As a practical matter, proper positioning of the one way valve / nossle could also add lift from the discharged gas).

This being said, while such an airship could potentially operate at high altitudes above most of the planet’s atmosphere ( with benefits of high stability, low frictional drag, near vacuum insulation and minimal filtration of high frequency wavelengths of light) it would be only a very small proportion of the planet’s gravity well. Substantial additional stage(s) would be required to achieve orbit. On the plus side, any additional stages could benefit from operating outside most of the atmosphere (potentiality allowing for more exotic approaches like some sort of “space elevator” or an airship based mass thruster). As such it might be a useful platform.

(More practically it could be used for long duration high altitude surveillance and for retransmission of data signals). — Preceding unsigned comment added by 2600:1002:B027:21DA:F091:5C1:6EA8:908F (talk) 04:57, 27 February 2021 (UTC)[reply]


(Again, all of this would be very Energy inefficient, but there may be considerable COST benefits to NOT having to carry all your Energy source with you (even hydrogen has considerable mass), and relying in part on energy sources external to the mass you want to lift).


Calculations

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It would be nice if this article had specific strength/specific modulus calculations. Thank you for showing interest in this stub. — Preceding unsigned comment added by 71.163.181.66 (talkcontribs)

I see a table of calculations has just been added. Because there was no source cited I added {{Citation needed}} tags. I gave this long reason in one: "specific reliable source needed to avoid original research, specifically for the strain figures. Ideally, for the informed layman to make any sense of this we need source that gives specific breaking force of the skin material at different diameters and compares it to the breaking strength of the strongest covalent bond. But no source needed for the trivial square and cube relationship. Note that Pressure vessel#Scaling makes a related, if unclear and poorly sourced, claim that minimum mass scales with volume and not area. This is why cites of reliable sources are needed.". -84user (talk) 20:12, 28 November 2010 (UTC) I have now added an {{Expert}} request, see Wikipedia talk:WikiProject Physics#Vacuum airship. -84user (talk) 21:03, 28 November 2010 (UTC)[reply]
I am not that familiar with airship design. I have some knowledge about pressure vessels, though, and that article is correct about pressure vessels. (I added some explanation to that article as well.)
I imagine that an airship is designed to keep a small positive gauge pressure inside in order to keep the material under tension. Thermal expansion from solar heating and the variation of atmospheric pressure with altitude complicates the design, because the gauge pressure will vary unless that is controlled by pumps or by releasing gas. Further the lesser weight density of air at higher altitudes necessitates that the volume expands to be buoyant at these altitudes.
Even with my ignorance of airship design, the table and the explanation seem gibberish to me. There is no explanation of why the various equations are used for instance. I am tempted to move all of that stuff to the talk page pending discussion. TStein (talk) 20:18, 29 November 2010 (UTC)[reply]

Moved some material including table here pending better explanation that doesn't conflict with established engineering such as pressure vessels

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I removed the following from the article, because it doesn't seem to agree with the well established engineering of pressure vessels, which the vacuum airship must be. The table uses technical terms like stress and strain but the calculation of these in this table don't seem to have anything to do with the technical terms as far as I can tell. For example the stress in the wall is proportional to the pressure and the ratio of the radius to the thickness. In practice the stress is set (to be the maximum allowable for that material) therefore the thickness needs to scale with radius which means that the mass of the airship is proportional to the volume of the air displaced. This is contrary to the assertions of this section. TStein (talk) 20:41, 29 November 2010 (UTC)[reply]

Hi All, I'm the guy that wrote the stuff below. I now understand it was writen in error and that I misinterpreted what Fuller said about the square to cube relationship between surface area and volume. It was an honest mistake. I was only trying to share something I thought would be useful. Sorry for the trouble. Thank you for helping me understand why my thinking was incorrect. John Shearing —Preceding unsigned comment added by Justodian (talkcontribs) 15:23, 1 December 2010 (UTC)[reply]

"Buckminster Fuller (Inventor of the Geodesic Dome) noted that even with conventional building materials, the strength required of the materials to stand against atmospheric pressure diminishes as diameter increases.[citation needed] The surface area of a vacuum vessel (which detemines the total crushing force exerted upon the vessel by atmospheric pressure) increases with the square of the diameter.[citation needed] But the volume of the vacuum vessel (which detemines the weight of the structural materials that the vessel can lift) increases with the cube of the diameter. So there is some diameter where the vacuum bottle can support atmospheric pressure at a weight equal to the air it displaces (has neutral buoyancy without collapse). And at any diameter greater than that, the vacuum bottle will rise.
Below is a table, which contains the data used to prove the principle.[citation needed]

This data demonstrates the square to cube relationship between surface area of a sphere and it’s volume as the diameter is increased. The first column lists the diameter in feet. Column 2 shows the surface area of the sphere in square feet. The formula used is as follows: 3.1416 * Diameter^2. Column 3 shows the stress on the surface of the sphere exerted by the atmosphere in pounds per square foot. The formula used is as follows: 2000 * surface area. Column 4 shows the volume of the sphere in cubic feet. The formula is as follows: 5236* Diameter^3. Column 5 shows the lift in pounds generated by the sphere. The formula used is as follows: (.078 * Volume). Column 6 shows the lift relative to the diameter of the sphere. The formula is as follows: Diameter divided by Lift. Please notice that as the diameter is increased a mere 10 times, the lift increases an enormous 100 times. The last column shows the relationship between diameter and strain on structural elements. The formula is as follows: Surface Stress divided by Lift. The understanding of this column is that a small increases in diameter will produce only a small increases in external atmospheric pressure on the exterior of the vacuum vessel but the same small increase in diameter produces large increases in lift. And large increases in lift mean that you can have large increases in the weight of the structural elements. And large increases in the weight of the structural elements mean that you will have large decreases in strain on the structural elements. Please notice that as you increase the diameter 10 times, the strain on the structural elements due to atmospheric pressure is reduced by ten times. The understanding derived from the last two columns is that if you build your vacuum vessel large enough, there is a diameter at which the structure will not collapse under atmospheric pressure and yet will be able to float because it can be built lighter than the air it displaces.

Effect Diameter Has On A Spherical Vacustat In Terms Of Lift And Strain
Diam in Ft Surf Area in Ft^2 Surf Stress in lbs Vol in Ft^3 Lift in lbs RelativeLift RelativeStrain
100 31416 62832000 523600 40841 408 1538
200 125664 251328000 4188800 326726 1634 769
300 282744 565488000 14137200 1102702 3676 513
400 502656 1005312000 33510400 2613811 6535 385
500 785400 1570800000 65450000 5105100 10210 308
600 1130976 2261952000 113097600 8821613 14703 256
700 1539384 3078768000 179594800 14008394 20012 220
800 2010624 4021248000 268083200 20910490 26138 192
900 2544696 5089392000 381704400 29772943 33081 171
1000 3141600 6283200000 523600000 40840800 40841 154

partial vacuum

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Obviously trying to create a "near vacuum" results in incredible forces on the vacuum chamber, but it is silly to dismiss the entire concept as impossible for this reason as a partial vacuum will also create lift. What about trying to evacuate only 50% of the air for 50% of the potential lift? What forces must the vessel then be strong enough to withstand? What if the vessel is huge but only 8% of air is removed? My understanding is that as the size of the vessel increases, the amount of vacuum(air displacement) required for lift is reduced by more than the increased weight of construction materials(surface area). i.e. that larger the vessel, the less one has to worry about strength and weight. 60.241.100.51 (talk) 01:55, 5 January 2016 (UTC)[reply]

My thoughts exactly. Wikiduck81 (talk) 10:24, 17 November 2021 (UTC)[reply]

What is the displacement of a vacuum?

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The answer is nothing.

Congratulations upon the existence of a totally moronic article.

— Preceding unsigned comment added by Mark Lincoln (talkcontribs) 20:43, August 13, 2011

The answer is (per Archimedes of Syracuse) "Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object." Here the "object" is the vacuum. --Kkmurray (talk) 23:09, 15 August 2011 (UTC)[reply]
Does this mean that the inside of Mark Lincoln's head has no displacement? Diss. I'm not sure what you think is moronic, but it's pretty clear that you have no idea what you're talking about. A vacuumed object still displaces whatever its volume is. 69.169.135.10 (talk) 06:30, 23 October 2011 (UTC)[reply]

Proposed for Mars

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These have been proposed for Mars eg [1] and [2] - Could mention ? - Rod57 (talk) 12:15, 28 September 2019 (UTC)[reply]

Designs with rotation?

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The calculations assume a spherical container. How about a "flying saucer" design where coaxial shells and a few internal walls (in the inner shell) rotate in opposing directions, so that the air is pushed out of radial holes/valves in the vessel? The shells will have to be optimized for a different distribution of tensions than a static container and will obviously require some energy to keep rotating. Elias (talk) 14:57, 28 February 2021 (UTC)[reply]

Sounds interesting. Wikiduck81 (talk) 10:24, 17 November 2021 (UTC)[reply]