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 LZ-114 to happen to our airship. (LZ-114, 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 3ii-a) The basic frame. 16,876 kg
The British R.31 class airships (Which are of a similar gross lift and gas capacity to the Sao 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 radius 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 (170-117 =) 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 3ii-b) 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 latex-Gelatin 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 3ii-c) 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 3ii-d) 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 3ii-e) 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 3ii-f) Engines 1,560 kg
Woodhouse [p.209] allocates 260 kg per 100 hp engine, for engine, mufflers, radiator, water, propeller, and hub. Hot-bulb 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 Sao Martinho can provide the required cruise power of 126 HP even with two of the six engines out for repairs and maintenance.
Appendix 3ii-g) 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 Sao 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 3ii-h) 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 red-hot 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 (100-10) = 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 double-boiler 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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****
The Future of The Field
Written by Kristine Kathryn Rusch
Maybe it’s my age. Maybe that’s why I’ve recently taken refuge in the history of science fiction. Or maybe it’s just the realization that I’m now one of the old-timers. I can actually remember meeting or knowing many of the legends of the field, now gone. And I have been in this field for a long time, even though it seems like a nanosecond to me.
Some of this is the rapidity of change. Readers of The Grantville Gazette appreciate history, and know that sometimes lives, institutions, countries, and everything else can change in the blink of an eye. If you look at a world map from five years ago, it’s different than a world map from fifty years ago, and different from a world map from five hundred years ago. And let’s not even talk about how those maps were made.
Some of the reasons I’m reading about the past, though, is the breadth of time. When you’re a kid, adults say, “Wow, you’ve grown up so fast,” and you think, “Fast? Are you kidding? This day is already a year long.”
But as an adult you realize that time really does speed up, and all of the other time periods live in your head as real, not imagined places. (Which probably explains why time travel books are so popular.)
I got hit with the breadth of time-again-in May. I stood on a stage in Hollywood, California, at the annual Writers of the Future award ceremony, and read my lines off the TelePrompTer: Twenty-five years ago, I . . .
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