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On this wiki page (under 'Comparison with heavier-than-air aircraft'), the altitude record of the L-55 Airship states:

The ultimate altitude record for a rigid airship was set in 1917 by the L-55 under the command of Hans-Kurt Flemming when he forced the airship to 7,300 m (24,000 ft) attempting to cross France after the "Silent Raid" on London. The L-55 lost lift during the descent to lower altitudes over Germany and crashed due to loss of lift.

I understand that lift on an airship is created through internal heated air and/or gases, rather than by the movement of air over a wing.

So, how exactly can an airship lose lift? (And how does altitude affect this?)

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    $\begingroup$ L-55 was LZ-101 in the numbering scheme used by Luftschiffbau Zeppelin. The run of airships from LZ-100 to LZ-103 were designed for 7300 m altitude - their gas bags were only ⅓ full at sea level! LZ-104 was even designed for 8000 m … $\endgroup$ Feb 28, 2018 at 20:33
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    $\begingroup$ One solution to your question title: Pop it with a pin... $\endgroup$
    – dalearn
    Mar 7, 2018 at 13:57

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The reason is a temperature difference between the lift gas and the surrounding air, and probably water uptake by the hull when descending through clouds.

A given mass of hydrogen will create a constant lift force, regardless of pressure or altitude, when at equal pressure and temperature with the surrounding air. Therefore, a change in altitude will not change the lift a rigid airship creates. Ideally.

However, air gets colder the higher you climb. The sun heats the atmosphere from below, by way of heating the ground, and space cools it from the top. Therefore, on many days the temperature gradient is bigger than its adiabatic value - that is the way thermals work! L-55 stayed at altitudes where air is -32°C according to the Standard Atmosphere. When descending, the surrounding air got warmer and also warmed up the lift gas, but only slowly. This means that, depending on the rate of descent, the lift gas lagged behind in temperature relative to the air, and this temperature difference reduced its lift capacity.

Note that adiabatic heating will already heat a gas when being compressed. The lapse rate of the atmosphere must be above the dry adiabatic value for this mechanism to work, which it is on many days. Especially behind a cold front. Note that L-55 encountered strong winds - just what you find in and behind a cold front. So it is safe to conclude that L-55 flew in labile air, and when it descended, that motion became unstable, at least close to the ground.

Kapitän Flemming simply descended too quickly. Slowing things down would have warmed the lift gas more, and less lift would have been lost. But delaying the descent has dangers of its own: The gas bags back then were made from goldbeater skin and had a certain amount of seepage. To compensate, Zeppelins started their trips with several tons of ballast water onboard, which was progressively dropped during the many hours of a normal trip. Taking a detour over France delayed the trip, so he was running short on time.

Normally, a loss of lift can be compensated by dynamic lift. With some angle of attack, an airship can create up to 20% of its weight in dynamic lift - as long as all engines are running. L-55 was blown south during the night after attacking Hull and Birmingham, and found itself far more south than assumed when dawn allowed the crew to get a ground fix. When back over Germany, L-55 ran out of fuel and dynamic lift was no longer available to compensate for the lower lift gas temperature. It made a rough landing in the Thuringian countryside near Tiefenort and had to be written off.

Literature: Heinz Urban, Zeppeline der kaiserlichen Marine 1914 bis 1918

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    $\begingroup$ I just thought that the temperature of the gas will decrease on ascent due to approximately isentropic expansion, and vv on descent. If I am right in guessing that this is (almost) the same for H2 and air, the gas temperature/height profile will lie between the dry adiabatic lapse rate and the actual lapse rate. If these assumptions are correct, and the airship starts down with the gas at the same temperature as the surrounding air, will it not reach the ground at least as warm as the air there, on account of compression, given the actual lapse rate is never > adiabatic? On the other hand... $\endgroup$
    – sdenham
    Mar 1, 2018 at 2:55
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    $\begingroup$ @sdenham: Sometimes the air is unstable through most of the troposphere - that is why thunderstorms are possible. If a cold front passes over western Europe, the air is unstable for several days at a time. Only high pressure means stable air with sun-induced instability close to the ground. $\endgroup$ Mar 1, 2018 at 21:39
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    $\begingroup$ @sdenham: I suggest to start gliding and do less theorizing. When you fly gliders you will experience what lapse rate there really is and you will learn quickly that "this sort of weather" is the norm. Also, it's OK if you don't believe my answer, but then please do your own research. $\endgroup$ Mar 2, 2018 at 3:04
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    $\begingroup$ @sdenham: For the accident to happen it would have been enough for the over-adiabatic lapse rate to exist for the last 1000 m. I was mentioning thunderstorms only because you seemed so sceptic that over-adiabatic conditions could range through most of the troposphere. And regarding thermals: For the vertical convection to start it needs a temperature difference of several degrees. An unstable atmosphere will just stay that way if not "kicked" by some bubble of heated air rising from the ground. Thermals becoming stronger with altitude are clear proof of an over-adiabatic lapse rate. $\endgroup$ Mar 2, 2018 at 8:35
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    $\begingroup$ Assuming that this is a significant variance from adiabatic and is the explanation for the L-55 accident, then it is something that could happen to any airship flying in the lower troposphere in such conditions, and is to some extent a matter of insufficient reserve ballast. As the argument is that the L-55's gas was especially cold on account of the altitude it reached, then the question of compression warming exists for the descent from there to the lower troposphere. $\endgroup$
    – sdenham
    Mar 2, 2018 at 12:20
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While Peter Kämpf has addressed the L-55's case, the question asks about loss of lift accidents in general, and the loss of the USS Macon (ZRS-5) provides an additional perspective.

As mentioned elsewhere, rigid airships flew with the lifting gas at the ambient pressure: their lifting cells within the envelope were only partially inflated on the ground, and as the craft gained altitude, they expanded. If they reached their full extent, any additional gain of altitude beyond this ‘pressure height’ would lead to gas being released through safety valves, in order to avoid an overpressure that could rupture a cell.

The demise of the Macon started with the failure of the incompletely-repaired upper fin, which punctured one or more of the rear gas cells. The officer-in-charge’s response was to drop large quantities of ballast and fuel, causing the ship to zoom well above its pressure height, which was less than 3000 ft., leading to an additional loss of gas. This climb was exacerbated by the change in pitch resulting from the loss of gas from the rear cells, which was not fully compensated for by the elevator man, and which produced additional dynamic lift.

As pointed out by Peter Kämpf, venting gas from exceeding the pressure height is not generally sufficient to leave the ship deficient in buoyancy: given that the amount of gas remaining is sufficient to support it at this altitude, it is sufficient to support it at any lower altitude, so long as the gas is no colder than the surrounding air. This is a consequence of the ideal gas law (and the fact that hydrogen, helium and air are all very nearly ideal gases at atmospheric pressures and temperatures): a mole of one gas will displace a mole of another if they are at the same pressure and temperature, regardless of what that temperature and pressure is, and so, by Archimedes’ principle, will create a similarly-independent buoyancy equal to the weight of one mole of the displaced gas.

In the Macon’s case, however, losing additional gas would not have helped in dealing with the leakage from the punctured cells, and some forty minutes later, it settled on the water. It is the opinion of historian Richard K. Smith that the excursion above the pressure height was decisive, and without the additional loss of lifting capacity that it caused, the Macon may well have remained airborne. He believes that mishandling of the ship led to dynamic lift contributing to the zoom, in which case the above analysis is not necessarily sufficient, as in the presence of dynamic lift (or upwards momentum), we cannot assume the ship was buoyant above the pressure height.

This is essentially the mirror image of Peter Kämpf's argument: if the airship did not contain sufficient gas to be statically buoyant at the apex of its trajectory, then it did not contain sufficient gas to be so at any lower altitude, a situation that could only be remedied through dynamic lift or by jettisoning weight - something the crew was working on until almost the last minute. Once it became apparent that a crash was likely, the commander had to confront the choice between slowing down or endangering everyone on board, with the former robbing the craft of dynamic lift.

In the case of the L-55, launching with its cells only one-third filled, the pressure height would have been about or perhaps a little above its record-breaking altitude, where the density is about one third of that at sea level. The pressure height of an airship is not fixed in construction, but by the degree to which it is filled before launch.

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    $\begingroup$ Flying close to the maximum height was avoided in order to not have the pressure-limiting valves vent the hydrogen gas. Bursting cells were much less of a concern - that needed failed overpressure valves first. And the temperature lag effect works both ways: In case of the Macon, the lift gas was warmer than the surrounding air and accelerated the climb, once that got started. This effect is even sometimes called the aerostatic phygoid. $\endgroup$ Mar 2, 2018 at 16:25
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    $\begingroup$ @PeterKämpf I do not think anyone is suggesting there was a danger of cell bursting, given the presence of automatic relief valves. It subsequently occurred to me, however, that in the absence of such valves, perhaps the biggest danger of over-expanding cells might be to the structure of the airship rather than to the cells themselves - especially in a lightly-built height-climber (though the Macon was more strongly built, following the loss of the Shenandoah.) $\endgroup$
    – sdenham
    Mar 6, 2018 at 15:02

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