Grantville Gazette 36 gg-36

by Paula Goodlett

  Unfortunately, it will be some years before there's a good network of weather stations across the shipping routes of interest, and it will be even longer before we have reliable weather forecasting.

  The next stage in the evolution of pressure pattern flying was made possible by the invention of the radar altimeter. Previously, the aircraft altimeter inferred the altitude on the basis of the barometric pressure. Since the radar altimeter measured altitude directly, that meant that a barometer could be used to measure the air pressure at the known altitude. Winds, especially at high altitude, tend to follow the curves of constant pressure (isobars) so you could use the barometer (pressure altimeter) , adjusted for your altitude, to follow the wind. If you found that your pressure was dropping, it would warn you that you were getting closer to the center of the storm.

  Predicting when the 1632 universe will develop radar altimeters is well beyond my area of competence, but I would not expect them until the 1640s at the earliest.

  The Graf Zeppelin determined height "by firing a shotgun and calculating altitude from the time delay of the returning echo." (Miller; Dick 61). The Hindenburg had a fancier sonic altimeter, using "a compressed air whistle at station 228, whose sound bounced off the ground and was picked up by a receiver at station 188. (Dick 88; airships.net). Yet another trick, used on both the Graf Zeppelin and the Hindenburg, was to drop a water bottle overboard, and time its fall with a stopwatch (Dick 61-62).

  The Graf Zeppelin also is said to have dropped smudge pots and observed the smoke to determine wind direction and speed (Miller) , but that would only speak to surface wind velocity.

  ****

  It's important not to expect too much of pressure pattern navigation. Even if you know exactly where the lows and highs, and the associated winds, are, it doesn't guarantee that it's worth detouring to exploit them. It all depends on the size and intensity of the pressure system, its location relative to that of the airship at a given time, and its own speed and direction of movement relative to the airship's intended course.

  During World War II, pressure pattern navigation typically reduced transatlantic flying time "an average of 10% compared to a great-circle track, with occasional savings exceeding 25%. . . ." (Kayton 12).

  Conclusion

  "The Bozo people, who live on the southern fringes of the Sahara, believe that Wind wrestled with Water, and Water lost. . . ." (DeVilliers 12). If the new airships of the 1632 Universe, fight the wind, instead of exploiting it, they too will lose.

  Author's Note: Appendices, bibliography and the spreadsheet will be posted to www.1632.org in the Gazette Extras section.

  ****

  A Trans-Atlantic Airship, Hurrah

  Written by Kerryn Offord

  This is a report of the process I followed as I tried to put together design specifications for a Spanish airship for the Cadiz-Havana 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 Perez de Guzman y Silva (1579-1636), the eighth duke of Medina Sidonia. Who, as probably the premier duke of the Spains, has the added benefit of being the father-in-law 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 Sao 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 forming-up 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 (LZ-129) 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 Sao 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 Sao Martinho off the ground.

  If that was all it took, we'd have a pretty small airship. However, we also need to move the Sao Martinho. To calculate the power we need to move the Sao 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 (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.

  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 6-10%; hot-bulb engines at about 10-14%; Petrol engines are 20-30% efficient; and diesels are 30-40% 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, hot-bulb would burn about 0.52 kg/ hp/hr, and diesel would burn about 0.18 kg/ hp/ hr.

  Table1. The size of the Sao Martinho using alternative propulsion systems would be:

  Now to build the Sao Martinho

  We have the basic parameters-mostly "volume"-so now we try building the Sao 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 Schutte-Lanz 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 Sao Martinho.

  Petrol and petrol engines are likely to be hard to obtain, so we'll concentrate on the next best thing, hot-bulb engines, which are, by canon, being produced down-time. If we assume the Sao 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 Sao 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 Sao 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].