I'm about to embark on wroting the most difficult aspect of my novel's background: the technical case for a 19th C global war. The argument revolves around exactly what you wrote, but with one added degree of difficulty (and where the real Steampunk starts): the ships have to reach stratospheric altitude at least close to the Tropopause, in order to use the subtropical jet stream to do a stealthy approach on North America. I have estimated that with the jet stream, one could use an airship to circumnavigate 2/3 of the circumference of the earth in only 6 days, at a latitude of 35 degrees, which is where the weaker subtropical key stream is located.

How much larger would an airship have to be at that altitude to lift the same cargo it can lift closer to the ground?

Bear with me as I'm not feeling too good tonight. Dental (molar) issues as well as sporting inflamed tonsils, either due to allergy or probably fighting an infection due to said molar issue, is making me extremely uncomfortable tonight. So I may be just writing jibberish.

The way a balloon works is that it displaces a volume of air around it. The difference in density between the gas inside the gas bag/envelope and the surrounding air determined the buoyancy of the balloon. As the balloon rises, however, the pressure inside it does not decrease at the same rate as the pressure decreases. Otherwise balloons would continue rising indefinitely. What happens is that the buoyancy force is greater on the ground and then slowly decreases as the balloon rises until it reaches it's "terminal altitude" (I'm going to call it that) when the buoyancy is exactly equal to the load carried by the balloon.

The way to pose the question is to say, if I have just enough volume in this gas bag to just *barely* lift this load mass at sea level (terminal altitude is 0 ft), then what gas bag volume do I need to just *barely* carry the same load at 33000 ft?

The buoyancy of an object in air is defined as the weight *of the air* displaced by the envelope/gas bag. The sum of forces in the y axis is equal to the buoyancy of the balloon minus the weight of the balloon:

ΣF

_{y} = ρ

_{air}(h) g V

_{bag}(h) - m

_{load} g = 0

Where

ρ

_{air}(h) is the density of air as a function of altitude h (i.e. the air density varies with h)

V

_{bag}(h) is the volume of the gas bag as a function of altitude h (i.e. the volume of the bag is also a function of h)

m

_{load} is the mass of the load (all contents of balloon including lifting gas) to be carried by the balloon.

g is the gravitational acceleration (at 33 000 ft there's negligible difference from sea level).

if the load mass is constant at sea level (h=0) as well as altitude h=33000ft = H, the we can say that

ρ

_{air}(0)

~~g~~ V

_{bag}(0) = ρ

_{air}(H)

~~g~~ V

_{bag}(H)

whatever the volume of the balloon V(H) may be at 33000ft (noting that the volume of the gas bag is not a linear function of altitude).

This implies that

V

_{bag}(H) / ( V

_{bag}(0) ) = ρ

_{air}(0) / ρ

_{air}(H)

That is, the ratio in volume depends on the ratio of atmospheric density at sea level vs altitude. We don't even need to mess with the density of the gas inside.

According to the International Standard atmosphere, at 33 000 ft the atmosphere is 0.3325 times as dense as at sea level

http://home.anadolu.edu.tr/~mcavcar/common/ISAweb.pdf (I picked this table for general friendliness to the reader)

so V

_{bag}(H) / ( V

_{bag}(0) ) = 1/0.3325 = 3.007

The volume of the gas bag must be almost 3 times as large at 33 000 ft. Two ways to look at the actual size of the craft:

A) If the envelope is a cylinder of radius R and constant length L, such that V

_{bag} = π R

_{bag}^{2} L then

V

_{bag}(H) / ( V

_{bag}(0) ) =

~~π~~ R

_{bag}^{2}(H)

~~L~~ /

~~π~~ R

_{bag}^{2}(0)

~~L~~ = 3.007

which implies that a constant length diameter requires that

R

_{bag}(H) = √(3.007 R

_{bag}^{2}(0) ) = 1.734 R

_{bag}(0)

The gas bag will be 1.73 times the diameter required at sea level.

B) Another way is just to lengthen the cylinder. At which point L(H) = 3L(0) just a long Zeppelin, 3 times as long as usual.

You can probably balance the diameter and length to make it more reasonable. The challenge will be in dealing with inflating and deflating gas bags during the night/day cycles, and naturally the weight of the structure will always be a limiting factor.

Also note that by the time you get to 60000 ft the airbag grows to an unmanageable size. I'm guessing (need ro run numbers otherwise it's just a wild guess), that 33 kft would be close to the limit for a Duralumin or similar structure.

~ ~ ~

Here's something on stratopsheric unmanned craft today

https://www.nytimes.com/2014/08/26/science/airships-that-carry-science-into-the-stratosphere.html?_r=0