This is a report of the process I followed as I tried to put together design specifications for a Spanish airship for the CadizHavana trade. It is provided "for your information," in case anybody might be interested in the process. Other than the hard science, nothing in this article should be considered canon.
I decided on a Spanish airship because someone has to make it, and it might as well be the Spanish. Not the government of course—they'd never be able to put together the will and the money for the project. Instead, I've picked on Don Juan Manuel Pérez de Guzman y Silva (15791636), the eighth duke of Medina Sidonia. Who, as probably the premier duke of the Spains, has the added benefit of being the fatherinlaw of the Duke of Braganza—the man who became John IV, the first king of Portugal of the House of Braganza, in 1640 OTL. Together, these two command the significant resources that such a project will require. Currently the airship is running under the name São Martinho—named for the seventh duke's flagship in the 1588 Armada—but I'm open to alternative suggestions.
The two ports were selected because Cadiz is the main Atlantic port of Spain, while Havana is the formingup port in the New World for the Spanish treasure fleets. The Great Circle (shortest) distance between the ports is 7,326 km (4,552 miles), and the only possible landfall on that route is the Azores, which are Portuguese (technically under Spanish control) and are some 1,934 km (1,202 miles) out from Cadiz.
Getting started
Where to begin?
First we need to set some basic parameters. I started with a desired payload of five thousand kg (5,000 kg)—about a quarter of the Hindenburg (LZ129) payload, which suggests a final airship less than a quarter her size (or, less than 50,000 m^3).
Now we need a structure to carry our payload. Examination of data for historical rigid airships finds that the "weight empty" (deadweight) takes up something like forty to sixty percent of the gross lift (See Appendix 1). The Hindenburg, the last of a long line of Zeppelin designs, and a commercial rather than military airship, is probably the best rigid airship design we can base our calculations on, had a deadweight that was 60% of the gross lift, so we'll use that value for the São Martinho. Thus, with a payload of 5000 kg, we have a deadweight of 7500 kg, and a total gross lift required of 12,500 kg.
Using lift from hydrogen of 1.09 kg/m^3 (at 15 degrees C and 760mmHG: Woodhouse, p.209; Brooks, p.202), we need ~11,468 m^3 of hydrogen to lift the São Martinho off the ground.
If that was all it took, we'd have a pretty small airship. However, we also need to move the São Martinho. To calculate the power we need to move the São Martinho we have two choices. Either we use brute force to follow the shortest route, powering through unfavorable winds approach, or we make use of the prevailing winds just like the sailors of the period do.
Brooks [p.163] indicates that the Graf Zeppelin (LZ127) lacked a high enough cruise speed to fly a scheduled North Atlantic route, but the 32.5 meters per second (mps) (73 mph) cruise speed was considered sufficient to fly the FrankfurtRecife route. Comparing average flight time on the FrankfurtRecife leg (68 hours) with the Great Circle distance (7,688 km), the average speed of 31.4 mps suggests they were flying a route close to the Great Circle route. However, this type of brute force took power, and thus fuel. The Graf Zeppelin (LZ127) carried some 40,000 kg equivalent of fuel (8000 kg of petrol and 30,000 m^3 of Blaugas).
We aren't planning a scheduled route, so we can get away with a lower cruising speed. In 1919 the British airship R.34 completed the first east to west crossing of the North Atlantic. The R.34 had a maximum speed of 26.8mps (60.3mph) and a cruising speed of 20.6mps (46mph). If they could cross the North Atlantic with that cruising speed, then surely the São Martinho can cross the midAtlantic with a cruising speed of 20.6mps. Unfortunately, the R.34 completed that trip consuming nearly all of the 17,500 kg of fuel she carried to cover 5,760km (3,600 miles). This suggests that a São Martinho designed using the brute force approach could be over 70,000m^3 in volume (2,477,000ft^3). Nothing that big has ever been built using timber framing.
This leaves us with the "follow the prevailing winds" approach. After discussions with Iver Cooper (See his companion article in this issue) I have decided that an airship with an endurance of 90 hours at 15mps (33.75mph) should be adequate for the task of delivering 5000 kg of cargo between Cadiz and Havana (See Appendix 2 for the route).
Now we've added speed to the equation, we need to know how much engine power we need to achieve that speed. This is important because engine power defines fuel consumption, and fuel consumption defines how much fuel we need to carry.
A hunt on the internet for how to calculate engine power led me to the Wikipedia and the "drag coefficient", which in turn linked me to "drag equation".
The "force" needed to overcome air resistance is defined by equation one.
Force = (Cd x Rho x V^{2} x A)/2 (Equation 1)
The engine power needed to produce that force is defined by equation two.
Power is = Force x velocity (Equation 2)
Substituting equation one into equation two, we get:
Power = ((Cd x Rho x V^{2} x A)/2) x V (Equation 3)
= ((Cd x Rho x V^{3} x A)/2) (Equation 4)
Where
P is the engine power in watts need to propel an airship at speed v
Cd is the drag coefficient of the airship
Rho is the density of air (we'll use the 15 degrees C, at 760mmHG value of 1.225 kg/m^3)
V is velocity in mps
A is the crosssectional area of the object.
However, because it better reflects the impact of surface area on drag, we are told that "Airships . . . use the volumetric coefficient of drag, in which the reference area is the square of the cube root of the airship's volume" [wiki: drag equation] rather than A. So, substituting the square of the cube root of the airship's volume for A, we get:
P = (Cd x Rho x v^3 x Vol^(2/3))/2 (Equation 5)
Where Vol is the airship's volume in cubic meters.
The essential value, which can only really be obtained by testing in a wind tunnel, is the drag coefficient (Cd). Lacking a wind tunnel, I've used data from Zahm, Smith, and Louden [p.258] to program my spreadsheet to automatically estimate Cd based on the speed and fineness ratio (length / diameter) of the airship. My initial state Cd is 0.05.
If we plug our values into equation five, we get:
P = (0.05 x 1.225 x 15^3 x 11,468^(2/3))/2
P = 52,562 watts (or 70.5 HP)
So, in theory, we could move our airship at 15 mps using about 70.5 HP. However, we need to carry enough fuel for the voyage—which means we need a larger airship to carry the fuel, which means more air resistance, which means we need more power to overcome the greater air resistance, which in turn means we need more fuel for the same range, which means we need a bigger airship to carry the extra fuel, which needs . . .
Using a Maybach petrol engine (as used on WW1 Zeppelins) that burns 0.250 kg of petrol and oil per HP per hour [Woodhouse, p.198], the São Martinho's engines are burning 17.625 kg per hour to produce 70.5 HP. With an endurance of 90 hours, at 17.625 kg/hr, we would use 1,586.25 kg of fuel.
We add that to our payload, to give a revised disposable load of 6,586.25 kg, but we also need to include water ballast in our disposable load (Brooks, p.186, tells us this was typically assumed to be about four percent of gross lift. We'll use our previous calculation as a basis, so 0.04 x 12,500 = 500 kg).
Normal cruising altitude will be less than 3000 feet, and we will be flying into the tropics. As altitude and air temperature both effect air density, we'll add another ten percent to our disposable load to reflect the reduced lift due to altitude and air temperature (0.10 x 12,500=1,250 kg). With a disposable load of 8,336.25 kg, deadweight is calculated as being 12,504.375 kg. We have a new gross lift requirement of 20,840.625 kg, which requires a gas volume of 19,120 m^3.
Before we start the necessary iterations to calculate a final specification we have to modify our equation. Zahm, Smith, and Louden assume a smooth shape, without gondolas. As soon as we start hanging gondolas off the São Martinho we start increasing drag. I have adjusted the equation to handle this by assuming each gondola is a much smaller version of the hull, and simply add the power needed to propel the gondolas to that needed to propel the hull.
Another problem with Equation 5 is that it assumes that the volume is the total volume of the envelope. If we store cargo internally, then we have to increase the size of the envelope to include it. There is also crew accommodation, and various walkways inside the envelope. These volumes have to be added to the gas volume to obtain the volume value for Equation 5. As it is proposed that the São Martinho could carry passengers, we need more cargo volume than if we were just carrying gold and silver. I've allocated 0.1 m^3 per kg of cargo to account for passengers and services. That means an addition of 500m^3. For walkways and crew accommodation (Hammocks to the side of walkways.), I am allocating 3m^3 per meter of length of the airship.
One hundred iterations later, and we have stabilized on an airship 79.76 m long (261.51 ft), an envelope volume of 23,261 m^3, gas volume of 22,522 m^3, and required engine power for 15mps (33.75 mph) of 64 HP.
But we are missing something . . . a crew. For such a long trip we need at least two shifts. So, a possible crew list is:
1 x Captain
2 x Officers of the watch
2 x helmsmen
1 x navigating officer
1 x engineering officer
6 x riggers (to keep the wires tight, pump the trim ballast or fuel, and repair any damage in flight)
2 x "engineers" per engine. We'll assume 6 engines, so 12 men.
Total crew is 25 men (Compatible with the actual crew aboard the R.34 on her historic west bound crossing), at 75 kg each [Brooks, p.210] this adds 1,875 kg to our disposable load. However, on a multi day voyage crew need an allowance for baggage, food and water, and bedding, so we'll add another 50 kg [Crocco, p.17] for a total crew allowance of 125 kg per man, or 3,125 kg. Fortunately, we are doing the calculations on a spreadsheet, so it's a simple mater of adding the crew and we get . . .
An envelope of 35,913 m^3, a gas volume of 35,041 m^3; engine power for 15 mps of 90 HP; a fuel load of 2,030 kg; a deadweight of 22,346 kg; a gross lift of 37,243 kg; and a final Cd of 0.0293.
That is all it takes to get 5,000 kg of cargo the 7,326km from Havana to Cadiz using prevailing winds.
Of course, that's an ideal, petrol driven world. The first problem is going to be finding engines. There are four possible types of engines that come to mind. Steam engines, with efficiency rates of 610%; hotbulb engines at about 1014%; Petrol engines are 2030% efficient; and diesels are 3040% efficient. Using middle of range values, based on the Maybach petrol engine at 0.25 kg /hp/ hr. Steam would burn 0.78 kg/ hp/ hour, hotbulb would burn about 0.52 kg/ hp/hr, and diesel would burn about 0.18 kg/ hp/ hr.
Table1. The size of the São Martinho using alternative propulsion systems would be:

Length (m) 
Envelope volume (m^3) 
Power (hp) 
Fuel (kg) 
Deadweight (kg) 
Gross lift (kg) 
Steam 
225.50 
63,675 
150 
10,506 
40,874 
68,123 
Hot bulb 
170.36 
48,103 
126 
5,897 
30,798 
51,331 
Petrol 
124.09 
35,041 
90 
2,030 
22,346 
37,243 
Diesel 
116.19 
32,808 
85 
1,369 
20,901 
34,836 
Now to build the São Martinho
We have the basic parameters—mostly "volume"—so now we try building the São Martinho.
We don't have aluminum, let alone duralumin, so we have to use wood framing (More specifically, we'll be using plywood). The biggest known airship design built using wood was the German WW1 period SchütteLanz S.L.20. This class of airship had a gas capacity of 56,000 m^3, a maximum diameter of 22.96 m, and it was 198.3m long. A rough calculation (Gas capacity divided by a cylinder of those dimensions) gives a form factor of 68.2%. This very rough estimate of form factor is used to predict the length of the São Martinho.
Petrol and petrol engines are likely to be hard to obtain, so we'll concentrate on the next best thing, hotbulb engines, which are, by canon, being produced downtime. If we assume the São Martinho has the same form factor as the S.L.20 we get a cylinder of 48,103 m^3/.682 = 70,532 m^3. No wood airship was ever built with a maximum diameter greater than 22.96 m, so we'll assume this to be the engineering limit for diameter using wood framing. Divide by the cross sectional area of the S.L.20 (414 m^2), and we get a length of about 170 m—which gives a fineness ratio (length / diameter) of about 7.42.
Are the numbers reasonable?
There has been some discussion between Iver Cooper and myself about the validity of equation 5 (and one or two other assumptions, especially the estimation of Cd values, but we'll ignore those.). If we put what we know about the Hindenburg into Equation 5 (using a Cd of 0.028 based on Zahm, Smith, and Louden), we get a result of 3,285 HP at 34.7 mps (76 mph). Which is close to the 3400 HP we know the engines of the Hindenburg produced at cruising speed [Airships: Flight ops]. Equation 5 is a crude estimator, but it does appear to give something close enough to real world values for the purpose of working out a rough design. The alternative is a lot of very complicated mathematics.
Of the remaining values, "Deadweight" is probably the least reliable. So, how reliable is the value of 30,798 kg? Using the formula from Crocco (see Appendix 3i), I calculated deadweight of 19,357 kg. However, that doesn't take account of the trim ballast, plywood rather than duralumin structure, or the heavier engines. Using very rough estimates, the weight of items contributing to the "Deadweight" are: (For details, see Appendix 3ii)
a) The basic framework: 16,876 kg.
b) The gas bags (including valves): 3,745 kg.
c) The envelope: 4,071 kg.
d) Running rigging for steering and valve controls: 298 kg.
e) Gondola cars: 1,050 kg
f) The engines: 1,560 kg
g) Trim ballast and trim pumps and pipes: 1,260 kg.
h) Fuel tanks, fuel piping, and pumps: 810 kg.
Sub Total: ~29,670 kg
i) Everything else. 30,798 – 29,670 = 1,128 kg
Given how rough the above calculations are, and how close they are to the value I assumed, I think it is safe to assume that the "Deadweight" allocated to the São Martinho is a reasonable compromise value.
The Lifting gas
With no access to helium, this has to be hydrogen—nothing else provides sufficient lift for a reasonable volume. We are using a lift value of 1.09 kg/m^3 (0.068 lb/ft^3) at 15 degrees C and 760mmHg [Brooks, p.202]. Take careful note of the definition. If we travel to Havana, and say the air temperature at sea level is 35 degrees C, then the hydrogen in the gas bags will expand in accordance with the standard gas equations (Equations 6 and 7). Which means the available lift at ground level in Havana will be less than the lift calculated at 15 degrees C.
P1V1/T1 = P2V2/T2 (Equation 6)
With pressure constant: V1 = V2T1/T2 (Equation 7)
V1 = 1m^3 x (273 + 15)/ (273 + 35)
V1 = 288/308 = 0.935% of maximum capacity
For the São Martinho, this means a loss of up to 3,337 kg of gross lift when in Havana—and you were wondering why I put in that 10% penalty for altitude and temperature into the design specification.
The other part of the lift penalty is due to the fact that for every thousand feet of altitude air density drops by about an inch of mercury and air temperature drops about 3.3 degree F. If we set the air density at 100% at 15 degrees C and 760mmHg, at a thousand feet, air density is down to 97.3%. At 2000 ft it's down to 94.5% [Cuneo].
Hydrogen has to be made. There are three probable methods: electrolysis of water, the action of acid on metal, and by forcing steam over red hot iron.
Electrolysis of water has a specific energy per m^3 of hydrogen produced of 4.5 to 5.45 kWh [LookChem], which suggests that electrolysis is not the way to go.
The Union Balloon Corp during the American Civil War [Wiki: Union Balloon Corp] used a pair of wagon mounted sulfuric acid iron filings hydrogen generators per observation balloon. The reaction formula Fe + H2SO4 <=> FeSo4 + H2 indicates that one mole of iron will produce one mole of hydrogen gas. At STP (standard temperature and pressure) one mole of gas occupies 22.4 liters. Therefore, to produce one m^3 of hydrogen, we need (1000/22.4=) 44.64 moles of iron. Iron is 55.85 grams per mole, so one m^3 of hydrogen gas at STP needs the reaction of 2.493 kg (44.64 x 55.85) of iron filings (and 4.375 kg pure acid). Each of the civil war hydrogen generators could produce about 60 m^3 per hour (2166 ft^3) [EB9th: Vol.1 p200]
The steam over redhotiron method has the reaction equation 3Fe + 4H20 <=> Fe3O4 + 4H2. Three moles of iron produces 4 moles of hydrogen, or 2.493 x 0.75 = 1.870 kg of iron per m^3 of hydrogen gas. Normally the iron has to be recharged after use, but there is a hydrogen generator that can reuse the iron. The Lane producers used after 1909 could recycle the iron. They were available in models with production rates of 14 to 284 m^3 (500 to 10,000 ft^3) per hour [Lane].
Using the largest Lane producer, the hydrogen vented after the crossing the Atlantic to allow landing (up to 5,410 m^3), would take about 19 hours to replace. However, that size plant is likely to be exceedingly expensive. Something able to replenish the lost hydrogen in about 48 hours is more likely
The lifting cells
There are several options for gas bag construction, but I'll only consider the two actually used on rigid airships before the advent of modern polymers. There is the tried and proven "gold beater's skin," and then there is elastomer coated fabric—basically rubber, sometimes mixed with gelatin, spread over cotton or silk (Although it wasn't unknown for people to use a combination of goldbeater's skin and rubber.).
Gold beater's skin is well described in the article by Chollet, and after reading that article, you'll understand why it isn't still used. The process is drawn out, and labor intensive. You get two skins per animal, and an airship such as the ZR1 Shenandoah (gas capacity: 60,915m^3) needs about 750,000 skins [Wiki: Goldbeater's skin; Steadman]—Remember, the São Martinho has a gas capacity of 47,092 m^3, so it would need something of the order of five hundred and eighty thousand skins (And only you get two skins per fully grown cow).
An acceptable alternative is to dissolve the raw rubber and mix it with gelatin, and spread it on cotton fabric. This was the material used for the gas bags on the Hindenburg and many other airships. The weight of this material was about 180 gsm, compared with a standard goldbeater's skin gas cell, which weighed about 145 gsm. Simply spreading rubber over fabric produced gas tight cells that weighed 240 gsm [Cooper].
Gold beater's skin is considered "tight". That means the diffusion of hydrogen through the layers is slow—only a few liters per m^2 per 24 hours. On the São Martinho, with gas bag surface area of 19,116 m^2, that's a loss of 38134 kg of lift per day. The latex coated cotton is good for less than 9 liters per m^2 per 24 hours [Woodhouse, p.211], or no more than 172 kg per day. Generally, lift freed by burning fuel will be enough to counter this loss on a voyage, but hydrogen will need to be replenished on a regular basis.
Through his relationship with the Duke of Braganza, it is hoped the Duke of Medina Sidonia can obtain natural rubber from the Amazon for his airship.
Fuel
What fuel is to be used? Well, the final design for the São Martinho has assumed we'd use hotbulb engines that can burn nearly any flammable liquid. However, no matter what propulsion system is used, the fuel will be in liquid form. This is a simple matter of energy density and ease of handling.
The steam engine variant is the only propulsion system that might consider solid fuel, but that involves using a much less efficient boiler system—one that needs constant stoking, and clearing away of the ashes. There are also problems with the fuel. Solid fuel has to be manually moved, or you use heavy automated systems. It is also heavier for its energy content. Coal is about half the energy per pound of petrol, and as for wood, that's about a third the energy per pound. A cord of wood (3.6 m^3) has about as much energy as a 450 kg of petrol, and it weighs about 1360kg. Using solid fuel doubles or triples the mass of fuel required for a steam propulsion system. Worse yet, moving the mass of fuel around the airship will cause significant trim management problems.
One thing to remember with liquid fuels aboard an airship is that, as the airship climbs, the ambient air temperature drops. That means fuel will thicken. Air temperature hits freezing at about 8,000ft. Heaters, or something may have to be added to allow the fuel to flow if the airship is to regularly fly at higher altitudes.
Engines:
We have stated that the São Martinho will use hotbulb engines. These are heavy and not very economical compared with petrol or diesel engines. However, they are being made downtime in the desired power range (at least 40 HP) as early as 1634 ("The Boat" By Kerryn Offord, GG#30).
We could use petrol sparkignition or diesel engines, but the Spanish are unlikely to be able to purchase uptime built engines. There are currently no new diesel engines being built, which leaves new build petrol fueled sparkignition engines. The best bet would be newbuild variations on the radial engines in "The Spark of Inspiration" by Gorg Huff and Paula Goodlett (GG#13), or "The Boat" by Kerryn Offord (GG#30). These are nominally 125 hp engines, and they will tend to be less economical than the water cooled inline Maybach engines we've been basing our petrol engine calculations on. However, they have significantly better power to weight ratios than the Hotbulb engines. Two such engines could easily provide all the propulsion the São Martinho needs, releasing the weight of 4 gondolas and engines (1,510 kg) and removing the drag of four gondolas—something to look forward to when the Hotbulb engines are upgraded to petrol sparkignition engines sometime in the future.
Operating Ceiling
When you research airships, you might see a value called "static ceiling". This is the altitude at which an airship's gas capacity is at 100%, and it is only lifting the deadweight.
For the São Martinho, that happens where air density is 60% of sea level. From tables we can find that this happens at about 14,000 ft. Note that this does not mean that the São Martinho can actually climb to 14,000 ft in normal operations (because there should always be some disposable load on board).
Something else to consider is the reduction in engine efficiency when the air density reduces with altitude. For example, when air density is at 50% of sea level density, engine performance is also down 50%, so to maintain the same delivered HP you had at sea level will take something like twice the fuel.
The normal operating altitude of the São Martinho will be about in the range 100200 m, as this offers significant fuel economies over higher altitudes. The Hindenburg was usually operated at about 650 ft (198.25 m), so "we are not alone". However, the São Martinho has an "altitude and air temperature" allowance of 10% of gross lift. That means the São Martinho can fly, fully laden, in conditions at sea level of 15 degrees C, 76mmHg, to an altitude with an air density of 0.90—about 915 m (3,000 ft)—without having to vent hydrogen. For every hour of flight at cruise power the São Martinho will gain about 40 m^3 of buoyancy due to fuel being consumed, and if nothing is done to prevent it, she will naturally gain altitude. At about 915 m the gas bags will reach 100% inflation, and as more fuel is burned, the São Martinho will want to climb higher. To prevent the gas bags rupturing, safety valves will automatically vent hydrogen. At the static ceiling, gas volume will be 100%, but we will have vented almost 40% of the hydrogen we started with.
Ground operations
You don't absolutely HAVE to have a hangar to store an airship. However, it is nice to have somewhere safe to put your airship, especially in bad weather. This is especially so for timberframed rigid airships. Wood is naturally hydroscopic (will absorb water). Irrespective of what water might do to the glue holding the airship together, there is the added weight of absorbed water. That is one reason why all wood surfaces have to be waterproofed with paint or varnish. However, paint scratches—enough said.
For short periods (days), there is no real problem in leaving an airship outdoors attached to a mooring mast. Just as long as it is a low one, as the airship virtually needs to be flown (trim etc maintained) at all times while moored to a high mast [Brooks, p.146]. Certainly, on the South America run, the Zeppelins didn't have a hangar in South America until the Brazilians built one at Rio de Janeiro in late 1935, and they never built one at the Recife stopover, where they just used a low mooring mast.
However, you do HAVE to have a hangar to build your airship. These are enormous, and thus expensive structures, as witnessed by the willingness of the Germans to limit the size of the Graf Zeppelin (LZ127) to the dimensions of the available hangar, rather than build a larger hangar. Even though they knew the resulting airship would be a suboptimal design.
The São Martinho is based on the maximum diameter of the SchütteLanz S.L.20 class, and they were built in hangars between 26 m – 38 m wide, and 25 m – 35 m high, and up to 240 m long. The larger hangar was in Berlin (was 38 m wide, 35 m high, and 240 m long), and it took 6 months during wartime to build.
It'd be a tight fit putting a 22.96 m diameter airship into something 26 m wide and 25 m high, but a 30 m by 30m opening might be a more comfortable fit. It also has to be over 170 m long. To give an example of how big this is, St. Paul's Cathedral in London has a nave 37m wide and 30m high, and the cathedral is about 175 m long. The airship hangar for the São Martinho is about the volume of the main building (excluding the dome and the transepts).
The United States built two, mostly timber construction, airship hangars at Tustin, in California [RDF Consulting]. Each hangar was about twice as long and three times as wide as what is predicted necessary for the São Martinho and took about six months to build under wartime urgency. Each hangar required about three million board feet of wood, of which 750,000 board feet was needed for the 51 roof arches, 79 tons of bolts and washers, and 30 tons of ring connectors. It sounds a lot, but each hangar only used about 33 tons of structural steel, whereas a regular design would have required 4,000 tons of structural steel. The airship base, with two hangars and other buildings, cost about US$10 million Y1943 dollars, that's about US$100 million in Y2000, suggesting a price for the São Martinho, purely on a per square meter basis, of around 80,000 thaler, which is probably an under estimate of what the hangar might cost downtime. And yes, it could be possible to build a hangar for the São Martinho downtime in six months, although a year to eighteen months would be a more reasonable schedule.
Ground handling is where a lot of accidents happen, and trying to thread 170 m of airship through a small opening—into or out of the hangar—is "difficult". The Germans "tied" their airships to ground vehicles on rails to run them in and out of their hangars. This is what I'd like to see for the São Martinho.
Outside the hangar, ground handling is still mostly done by humans. The purpose of a ground party is partly to move the airship, and partly to provide ballast to hold down a buoyant airship. With all disposable load removed the São Martinho has excess buoyancy of forty percent of gross lift, or 20,532 kg. At 75 kg per man, that is a minimum of 274 men.
Costs:
There are two basic airship designs. Either you have a cylinder with curves at each end, or you have a modified "teardrop" shape. The cylinder design has the advantage that most of your ring frames are exactly the same size and shape. This provides savings in time and money, as you can use jigs. The downside is that the aerodynamics of the cylinder design is less than optimal.
The teardrop design has a better drag coefficient (being a more aerodynamic design). However, this is at the cost of ease of manufacture. In the teardrop design, nearly every ring frame is a different size, so you can't just use a standard jig to build every ring frame. That adds costs in construction, but the lower drag coefficient means that for a given size airship (gas capacity) you need less power for a given cruising speed, which means you use less fuel. Meaning you can carry more payload. For these reasons the São Martinho will be a modified teardrop design.
Brooks [p.199] suggests that the late WW1 airships needed something of the order of 800,000 – 1,000,000 hours of "direct shoplabor" to build the first airship of a new type, while the post WW1 airships took two million, or three, or even more direct shoplabor hours to build. The São Martinho is probably big enough to require at least one million hours of shoplabor. That equates to something like 100,000 man days. At the standard rate for skilled tradesmen (carpenters and blacksmiths) of 0.3 thaler per day (USE$30), that is 30,000 thaler. We know the Vasa (lost 1628) cost about 40,000 thaler [wiki: Vasa], but that was for a completed warship. The São Martinho's labor cost alone is threequarters what the completed Vasa cost. As you can see, airships are going to be very expensive downtime.
As we can't find period prices for most of the components, any pricing beyond the base labor content would be just be guessing. However, we know labor is usually a fraction of any construction cost, so the materials aren't going to be cheap.
There is some good news. From the German experience, it has been shown that airship construction follows the "learning effect" [Brooks, p.199]. For every doubling of production of a design, man hours required drops by twentypercent.
Cost in operations. These include fuel to propel the São Martinho, and to produce the hydrogen gas, the labor cost of flight and ground crews, depreciation to reflect repairs and maintenance, and then there are "opportunity costs". For now, we'll just look at the cost of fuel and hydrogen generation.
Fuel for each inbound or outbound flight is budgeted at 5,897 kg (~6,700 liters). That's about 1,770 US gallons, and if we cost it as Greenland Whale oil (0.35 thalers/gal) we get close enough to 620 thaler (USE$62,000) each way.
Hydrogen, made by blowing steam on to red hot iron filings (See Appendix 5 for calculations) to replace the vented hydrogen (5,897kg fuel at 1.09 kg/m^3 = 5,410 m^3), will need about two thaler worth of firewood to generate the steam. Keeping the iron filings red hot will use a bit more fuel, but probably not enough to bring the cost of hydrogen generation much above two thaler (about five cents per cubic meter of hydrogen). However, at 1.870 kg/ m^3 of iron filings, that's 10,117 kg of iron filings. Maybe they can be reused, but that's about 602 thaler (USE$50,500) worth of iron, based on period wholesale prices (5.95 thaler/ 100 kg of Swedish bar iron). There are also manpower costs to consider, even if we have a good Lane Hydrogen producer, and it only takes 48 hours, that's likely to add another 20 thaler (USE$2,000) per fill. For a rough estimate of 625 thaler (USE$52,700) per fill.
Then we have to pay for crew and maintenance personnel. But how much do you pay an airship crewman? We have nothing to compare the skill sets with, so any number would just be guessing. However, twentyfive men at an average of, say, 0.4 thaler (USE$40) a day (About 150 thaler pa) is a total of about 3,750 thaler pa. If they make one return flight per month, that's 156.25 thaler per crossing. On any trip carrying firstclass passengers, there will be a need for passenger service crew. They'll add a further 25 thaler per trip. Then there are airship and hangar maintenance personnel. We don't really know how many are needed, nor how much they should be paid.
We can make a reasonable guess at the cost of the massive ground crew needed for launching and landing the airship. The job is mostly low skill. A handful of men who know what to do can control several hundred unskilled laborers. Still, ground parties will cost around 50 thaler any time you want to "ground handle" the airship (Assuming standard European pay rates), or 100 thaler for each trip (launch and landing parties).
A rough guess would suggest, ignoring wages of the maintenance staff, the direct labor and consumables cost of a oneway crossing of the Atlantic would be about 1,526.25 thaler (USE$152,625) per trip.
Conclusion
With an allowance per passenger of 150 kg [Crocco], the São could carry thirtythree passengers. Make eight of those passenger service crew, and we have space for maybe twentyfive paying customers. At 100 thaler per head (about a year's wages for a tradesman, or a month's pay for a military captain/ lieutenantcolonel) there is a gross income per flight of 2,500 thaler. With one return trip per month, that is a revenue stream of 60,000 thaler, which we hope would cover expenses.
Alternatively, the São Martinho can carry high value freight (like the gold, silver, pearls, and jewels of the fabled treasure fleets). The 1628 treasure fleet captured by Piet Heyn carried about 80,300 kg of silver and 30 kg of gold. The São Martinho might not be able to carry that all at once, but she could carry it all if she could manage 17 return trips a year (About one return trip every 21 days.). The savings in ship days, provisions, and wages for ships to carry the treasure across the Atlantic should easily justify using an airship for the task. It'll also provide a more regular revenue stream, which will make balancing the books easier, and keep the bankers happier.
This analysis suggests that a rigid airship large enough to service the CadizHavana trade could plausibly, and profitably, be built by the Spanish using downtime resources. It won't be easy, and it will take time. A suitable hangar needs to be built. The plywood framing has to be developed by trial and error (before it goes into the first airship). Gas bags need to be made, engines need to be produced, and there is a need for suitably skilled workers to build the airship. If we assume the Spanish started working with airships in 1632, and start building a hangar in 1634, it is possible that work on the São Martinho could start as early as 163637.
Appendix 1.
The "weight empty" or "deadweight" is the weight of the airship after removing everything that can and is normally removed between voyages (Things like water, crew, food, payload, fuel and oil for the engines, and ballast—usually water or sandbags.)
Table 2. Percentage of Gross lift given to Weight Empty for the last of the rigid airships (using numbers from Brooks). Gross lift is calculated at 1.09 kg/m^3 of gas capacity The ratio is calculated by dividing Weight empty by Gross lift
Yes, some of the WW1 Zeppelins had ratios below fifty percent, but that was achieved by seriously weakening the structure—essentially, removing a third of the ring frames—and halving the number of layers of goldbeater's skin used on the gas bags. That might be acceptable for war time expedients, but we don't want what happened to LZ114 to happen to our airship. (LZ114, a war prize, being operated by the French as Dixmude, suffered a structural failure in flight and fell, burning, into the Mediterranean, with the loss of all fifty aboard. [Brooks, p.108].).
The Hindenburg was the last rigid airship design, and we can assume she was designed using all of the available knowledge built up over the years, making her suitable for commercial operations. Therefore, for this exercise, we will allocate the same sixty percent of gross lift to "weight Empty."
Appendix 2: The route, based on Iver's Route 2.
Iver's article assumes that the power delivered by the propeller (i.e., the propulsive power implicit in airship motion) is 71% of the power output from the engine [Cooper]. I used this setting in his spreadsheet when calculating this route.
Depart Cadiz,
1) Head south through Variables for 5 hours at an engine setting of 211.26 HP, at an altitude of 100m
2) Continue south to Trades for 12 hours at 101.81 HP, at 200m.
3) Follow Trades westward for 157 hours at 20.64 HP, at 200m.
Arrive in Havana after 174 hours, with reserves of 2,620 kg fuel.
Depart Havana,
4) Head north to Variables for 24 hours at 207.74 HP, at 100m.
5) Head north to Westerlies for 12 hours at 101.81 HP, at 200 m.
6) Follow Westerlies for 66 hours at 9.86 HP, at 915 m.
Arrive in Cadiz after 102 hours, with reserves of 2,664 kg fuel.
You will have noticed that the reserves of fuel are about 45%. However, those flight times and power requirements are based on average winds. It is hoped that this reserve will be sufficient to ensure safe transit under most conditions.
Appendix 3: Calculating the Deadweight
Appendix 3i: Formula from Crocco [p.10].
Deadweight (kg) = (0.1759 + 0.00002275 x velocity^2) x Envelope volume +
(0.09994 x Number gas cells + 3.075) x Envelope volume^(2/3) +
(0.0019725 x Envelope volume^(4/3) + (Gross HP x 2.150)
Where Envelope is the volume of the envelope, and one gas cell per 9 m of length.
Inserting numbers = (0.1759 + 0.00002275 x 15^2) x 49,114 +
(0.09994 x 17 + 3.075) x 49,114^(2/3) +
(0.0019725 x 49,114^(4/3) + (240 x 2.150)
= 8890.644 + 6402.561 + 3547.834 + 516
= 19,357
Note that this value doesn't include trim ballast, and it uses lighter engines, and Duralumin rather than plywood for structure
Appendix 3ii
Appendix 3iia) The basic frame. 16,876 kg
The British R.31 class airships (Which are of a similar gross lift and gas capacity to the São Martinho.) were built using spruce plywood using three panels 10' long and 10" wide formed into equilateral triangles[ Airship Heritage Trust: R31]. Assuming 28 lbs per cubic foot for air dried spruce, plywood half an inch thick, and that the holes cut into the panels leave 30% of the panels, each 10' girder is 0.03125 ft^3 or 8.75 lbs.
However, wood is hydroscopic (absorbs water) so we need to seal the plywood. We'll want to apply at least two coats of varnish, at 9 lbs per 450 ft^2. Each girder has a basic surface area of about 15ft^2, so that’s an extra 0.60 pounds for 9.35 pounds per girder (4.25kg). Note that I'm ignoring nails, glue, and any internal framing.
As everything else is metric, we'll convert these to 3m girders weighing 4.18 kg. How many do we need? Let's approximate. The envelope volume is 49,114 m^3. The diameter of our structure is 22.96m. We could call our airship a cylinder with a cross section area of 414m^2 and ~117 m long (rounding down the actual value of 118.63 to give a value divisible by 3 and 9).
If we put a major ring frame every 9m, and minor rings every 3m, and we have a lateral every 3 meters, then:
a) each ring is ~72 m circumference, needing about 24 girders
b) each lateral is 117 m, or 39 girders
c) there are 39 ring frames (13 being major frames)
d) there are 24 laterals.
e) double frames for major rings 13 frames at 24 girders
a x c + b x d + e = (24 x 39) + (39 x 24) + (13 x 24) = 2,184 girders = 9,129 kg (1).
Admittedly there is another (170117 =) 53 meters of hull to frame. A wild guess would be ad an extra 25% (53 m is ~31% of the cylinder, but it is made up of two curved shapes)—2,282 kg (2).
Next we add the fin surfaces. Based on the Hindenburg fins [Brooks, p.180], these are aerofoil shapes, having two layers of girders. Laying out the girders I estimate there are 80 x 3m girders per fin. With four fins, that is 320 girders, or 1,338 kg (3).
There are Keel and central walkways on the Hindenburg. Each running the full length of the airship, if we just call that one girder wide for ~170 m or 57 girders per walkway (114 girders), 477 kg (4)
Then we have to add the lattice that actually stops the gas bags pushing against the outer envelope. This is made of cable strung between the girders, and in a photograph of R29 under construction (Brooks, p.117) they look quite thick. If we call it 3/8" manila at 23 ft to the pound (0.065kg/m), and make a 0.3m lattice on our cylinder (72/.3 = 240 circumference runs (240 x 117 = 28,080), plus 117/0.3 laterals (390 x 72m = 20,080), we need about 56,160 m of "rope", or about 3,650 kg. (5)
Total structure (Sum(1:5)) = 16,876 kg
Appendix 3iib) The gas bags 3,745 kg.
The surface area of the gas bags can be estimated based on a cylinder based on the gas volume of 48,103 m^3. It is ~116 m long, with a circumference of 72m. Each end (two per bag) is 414m^2.
Each gas bag is 9m, we have 13 gas bags, so area of gas bags is:
116 x 72 + 13 x 2 x 414 = 19,116 m^2. Using the Hindenburg standard latexGelatin formulation to produce gas tight fabric at 180 gsm gives a total gas bag weight of 3,441 kg (1). To which we have to add the actual (170/9=) 19 bleed valves etc at 16 kg each =304 kg (2).
(1) + (2) = 3,745 kg.
Appendix 3iic) The Envelope 4,071 kg.
Woodhouse [p.211] talks about an envelope being made of rubberized fabric with protective coatings of heavy spar varnish, "Valspar" or its equivalent. Actually, he says the fabric should have at least five applications of nitrocellulose, and final coat of "Valspar". The protective coating should be at least 70 grams per m^2, and no piece of fabric should weigh more than 440 gsm.
Just using the surface area of our cylinder (From Appendix 3a), we have 117 m x 72 m + 2 x 414 (ends) = 9,252 m^2 at 440 gsm = 4,071 kg.
Appendix 3iid) The Running Rigging 298 kg.
The Command gondola must be able to transmit signals to the tail surfaces, and it must also be able to vent gas from each gas bag at will. Assuming steel wire at 10m /kg. Assume two cables per fin. Four fins (upper and lower rudder, left and right elevator) for 8 wires. Assume the controls have to travel 170 m, to give 1,360 m.
Gas vents. Assume 19 vents (there should be one per gas bag, and one gas bag per 9 m), there is one cable per valve, and an average distance from the control gondola of 85 m, to give 1,615 m.
Running rigging is 2,975 m or 298 kg.
Appendix 3iie) The Gondolas 1,050kg
I have no idea. Gondolas need to be able to carry the engines and people servicing them, plus the command gondola. Woodhouse [p.209] allocates about 150 kg each. We have decided on 6 engines and a command gondola, so, 7 x 150 = 1050kg.
Appendix 3iif) Engines 1,560 kg
Woodhouse [p.209] allocates 260 kg per 100 hp engine, for engine, mufflers, radiator, water, propeller, and hub. Hotbulb engines aren't known for their power to weight ratios, we'll assume 260 kg for every complete 40 HP engine. Our airship needs 104 HP for 15 mps as a cruise speed. We want some spare power for emergencies, so we'll assume 6x 40 hp = 240 hp. 1,560 kg (6.5 kg/ hp). Giving a top speed in still air of 16 – 19 mps (37.9 – 42.7 mph)—the lower value represents performance with only 71% power being delivered as thrust. The six engines means the São Martinho can provide the required cruise power of 126 HP even with two of the six engines out for repairs and maintenance.
Appendix 3iig) Trim ballast and trim pumps and pipes 1,260 kg.
An airship flying level is an airship flying economically. If there is an angle of attack—using the hull to gain lift—more drag is experienced, and more power (thus more fuel) is needed for a given forward velocity. To maintain level trim an airship has trim ballast. This is usually water that can be pumped to various holding tanks to balance the ship (The crew can double as emergency trim ballast.). Normal ballast, which is dumped to reduce weight when trying to gain height (averaging 4% according to Brooks [p.186]), is not trim ballast.
Woodhouse says a navy patrol blimp has 90 lbs of trim ballast on a 5,275 lb gross lift airship (1.7%). Using this value, the São Martinho would have trim ballast of ~873 kg (1).
We now have to be able to move it to tanks forward and aft. We'll assume 5/8" inch copper tube at about 0.5kg/m, and we’ll want about 340 m—170 kg (2).
Hand operated pumps at, call it 25 kg (3)
Copper tanks to hold 1,500 liters. 6 x 275 liter drums at 16kg = 96 kg (4).
(We need some empty space so we can move water from place to place, hence available volume exceeds actual volume of trim ballast.)
We also need water tanks for the ~2,097 kg of ballast water in say, 6 x 300 liter tanks at about 16kg each = 96 kg (5).
Totaling 1,260 kg.
Appendix 3iih) Fuel tanks, fuel piping, and pumps 810 kg.
We have 5,897 kg of fuel. Assume it is petrol, and we have ~6,700 liters. We want to spread them along the length of the keel, so 12 x 600 liter tanks at 20 kg each = 240 kg (1)..
Plumbing for the fuel, 85 m x 10, plus two runs of 170 m to connect all tanks = 1,020 m at 0.5 kg per m 510 kg (2).
Pumps per tank 12 x 5 kg = 60 kg (3)
Total 810 kg.
Appendix 4: Conversion rates
All currency conversions are based on an exchange rate of one guilder = USE$40.
One Guilder = 36 kruezer)
One Guilder = USE$40
One Thaler = 90 kruezer
One Thaler = USE$100
One Spanish Ducat = 110 kruezer
One Spanish Ducat = USE$122
One Spanish Reale (a "piece of eight") = 10 kruezer
One Spanish Reale = USE$11
One Venetian Ducat = 88 kruezer
One Venetian Ducat = USE$98
One English Pound = 400 kruezer
One English pound = USE$444
One French Livre = 36 kruezer
One French Livre = USE$40
One Florin = 60 kruezer
One Florin = USE$67
One Gulden = 60 kruezer
One Gulden = USE$67
One Rixdaler (Swedish) = 90 kruezer
One Rixdaler = USE$100
One Rigsdaler (Danish) = 90 kruezer
One Rigsdaler = USE$100
(Note: there are 3 kruezer to the groschen, and 20 groschen to the thaler.)
One cubic meter (m^3) = 35.315 cubic feet (ft^3)
One kilogram (kg) = 2.204 pounds (lbs)
One meter per second (mps) = 2.237 miles per hour (mph)
One KW (1000 watts) = 1.34 horse power (HP)
One Horse Power (HP) = 746 watts
Appendix 5: Calculating energy to produce one m^3 of hydrogen by blowing steam over redhot iron filings:
The chemical equilibrium equation is:
3Fe + 4H2O <=> Fe3O4 + 4H2
This tells us that for every water molecule input, we get a H2 molecule.
At STP one mol of a gas occupies 22.41 lt. One m^3 of H2 at STP needs 1,000/22.41 lt/mol = 44.623 moles.
Atomic weight H2O = 18.01528 g
44.623 mol x 18.01528 g = 803.92g (0.80392 kg)
To heat 1kg water from say, 10 degrees C to 100 degrees C
Specific heat = 1 cal/ gram/ degree C. 1000 g x (10010) = 90,000 cal
Latent heat of vaporization (convert water to steam at 100 degrees C) = 540 cal/g @100 degrees C. 1000 g x 540 = 540,000 cal
Total energy to get 1 kg of water from 10 degrees C to steam is: 90k + 540 k = 630,000 cal/ kg
@ 4.1813 j/cal = 2,634,219 j/kg
Energy to make steam to produce 1 m^3 of H2 = 2,634,219 x 0.80392 = 2,117,701 j.
At 2,117,701 j per m^3, we are replacing all the hydrogen vented to cover burned fuel (5,897 kg, / 1.09 = 5,410 m^3) = 11,456,762 kj.
Charcoal at 29,600 kj/ kg = 387 kg at 4.2 thaler / 1000 kg = 1.63 thaler
Coal at 27,000kj/kg = 424 kg at 2.28 thaler / 1000kg = 0.97 thaler
Dry wood at 15,000 kj/kg = 764 kg at 2.64 thaler/ 1000kg = 2.02 thaler
A very big, permanent doubleboiler system might produce 284 m^3 (10,000 ft^3) per hour. Portable systems, similar in size to the wagon mounted systems used by the Union in the ACW might have a capacity of up to 60 m^3 per hour. Refilling airships is a slow process.
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