The hydrogen inside an airship has a mass that is approximately 14 times lower than the mass of air and this difference results in lift.

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However the solid material inside an airship including the frame, skin, walkways, gasbag material etc. would also displace air. Would the air that these materials displace also be counted towards lift? I'm not sure.

If the extra displacement is counted towards lift is there any reasonable estimate as to how much space the solid material might fill? 1%, 2%, 5%?

A U Class Zeppelin has an outside volume of approximately 59,000 cubic meters. However if 2% of this volume was solid material that's 1,180 cubic meters of atmosphere displacement or an extra 1,445 Kg of lift. Although not decisive I'm sure that any little bit helps.

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    $\begingroup$ Air has mass, so, an empty airship is lighter than an airship full of air. $\endgroup$ Oct 16 '20 at 9:27
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    $\begingroup$ "and this [density] difference results in lift" That's inaccurate, at best. The lift comes from the hydrogen (or the frame, or anything) having volume, regardless of the mass. However, the difference in density results in positive buoyancy, meaning that lift is greater than the weight. Once you make that split clear (volume + density of surrounding fluid -> lift, mass + gravity -> weight) the answer should be quite intuitive. $\endgroup$
    – Frax
    Oct 16 '20 at 10:14
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    $\begingroup$ Everything that displaces air is affected by buoyancy. Fun fact: when calibrating equipment that can measure weight/forces in the hundreds of tonnes you have to take into account the air displaced because the displaced air "lifts" the calibration weight: youtu.be/_k9egfWvb7Y?t=231 $\endgroup$
    – slebetman
    Oct 17 '20 at 3:25

It would, but it is so little it is not noticeable

Duraluminium has a density of 2780 kg/m3, plain air has a density 1.3 kg/m3. So the buoyancy is in the order of 0.05%. The dry weight of the Hindenburg for example was 100 tons, meaning that the buoyancy of the solid material is about 50 kg.

This is so little that other effects on lift — i.e. changes in temperature, internal gas pressure, external atmospheric pressure, dynamic lift from relative wind speed, ballasting, spending fuel, and more — will completely drown out this effect.

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    $\begingroup$ It's also worth noting that the buoyancy is already accounted for -- when you stick the uninflated Hindenburg on a scale to weigh it, the number it produces is 50 kg lower than it would be in a vacuum. $\endgroup$
    – Mark
    Oct 16 '20 at 20:07
  • $\begingroup$ @Mark -- absolutely-- of course you could argue that what you read on the scale is not really "weight", because it is not exactly the same as mass * gravity $\endgroup$ Oct 17 '20 at 11:44
  • $\begingroup$ @quietflyer What you read on a scale is called Apparent Weight. $\endgroup$
    – bogl
    Oct 17 '20 at 17:23

Short answer yes. Clearly any solid material must displace an equal volume of air. However, the mass of a structure is typically determined by measuring its weight, and that is typically measured in air and so the buoyancy provided by the atmosphere is already factored in. For dense materials such as metals this is typically negligible, but clearly for low-density materials such as hydrogen it is important, where a litre of hydrogen has a true mass of around 80mg but a weight in free air of approximately -10mN.


Buoyancy is best defined as weight of displaced air minus weight of the stuff that is displacing the air. If the "stuff" doing the displacing is heavier than air, then the buoyancy contributed by that displacement is negative rather than positive. Still, if we want to calculate total buoyancy and we know total mass or total weight, strictly speaking we should not neglect the weight of the volume of air that is displaced by the heavy parts of the airship. As other answers have pointed out, the resulting contribution to buoyancy is negligible, but not zero.

This answer is using the strict physics definition of "weight", i.e. mass times gravity. If by "weight" we mean what we read when we put an object on a scale, which is called "Apparent Weight", then the buoyancy created by the displacement of air is already accounted for, at least for the atmospheric conditions prevailing at the location of the scale. The fact that in reality we almost never consider the difference between these two kinds of "weight" in the context of heavy metal objects, shows how slight the difference actually is.

  • $\begingroup$ What are you people suggesting, here? Metal displaces the same volume of air and that creates lift? A hollow tank with thin sides might but a sizeable chunk of metal? $\endgroup$ Oct 16 '20 at 23:42
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    $\begingroup$ "...displace more air per unit volume than..." WHAT? A think of X volume displaces that volume of air, irrespective of the thing's density. All reasoning that follows a misstatement like this is highly suspect. $\endgroup$
    – Ralph J
    Oct 17 '20 at 13:58
  • $\begingroup$ @RalphJ -- I edited to this "Note that airplanes are much denser than airships, so they displace MORE air (per unit volume enclosed by the outer surface of the aircraft) than an airship with empty gas cells does. Of course that "advantage" is completely offset by the weight of the stuff that is doing the displacing.", then decided to just delete the first two sentences entirely, not really essential to the main point, and leads to useless arguments as to whether or not one is intending to include the volume occupied by the air inside the aircraft. $\endgroup$ Oct 18 '20 at 14:51

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