Brooks [p.163] indicates that the Graf Zeppelin (LZ-127) 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 Frankfurt-Recife route. Comparing average flight time on the Frankfurt-Recife 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 (LZ-127) 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 Sao Martinho can cross the mid-Atlantic 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 Sao 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 V2 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 V2 x A)/2) x V (Equation 3)
= ((Cd x Rho x V3 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 cross-sectional 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 Sao 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 Sao 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 Sao 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.