The Midairship is a Project in Service Of Humanity
The project had previously been slated at 5km2, but it will be shifted to 5kmx1.5km, with depth varying on the level of buoyancy. The altitude of this project is expected to be 50,000 feet in accordance with projected FAA and European supersonic flight regulations.
Buoyancy levels of a vacuum are 1.2kg/m3. Helium is 1.0kg/m3. Helium is a volume place -holder in this case. We may also be able to reduce the kPa of the airship in order to make the existing helium work harder, even at 15km altitude.
Consider the Bunker Dryer, details on display at www.timetravelisforsuckers.blogspot.com
. Using this principle, and knowing that saturated air is lighter than dry air because of water vapor density, we could potentially use low-pressure steam, also as described in the science videos on turbines at www.youtube.com/sciencejunkie
. The entire mechanism could be made into a parallel flying drying machine, which would trap the vapor entirely and permanently, to produce buoyancy.
This dryer theory is in works and may replace the original helium strat if it proves useful. For the time being we will economize helium.
Helium has 98%
of the lifting power of hydrogen. Specific sources say 92.64%
'He' priced at $70/100m3 in 1996. The X and Y fields of the craft are expected to come to 7.5 units, with depth Z being estimated at perhaps 400m, depending on functional buoyancy and load. This should require about 3km2 of helium. 30 X 30 X 30 would fill a 3km2 space with 100m3 blocks of helium. That much helium should cost about $1.89 million
. Buying in such a bulk project we could likely acquire the amount for a substantially lower sum from stockpiles.
3km3 of helium: $1.5m-2m
This should produce 27 billion kg
[59.4 billion pounds, or 29.7 million tons] of lift. This is our primary force of lift. We will spend against this in containment and structural materials and planes. What we do not use will be tied down with cable strength. It will be important to not go over this level by more than perhaps a 800,000- 1 million tons, but to remain over it at all times by at least 1000 tons.
The port should be able to support 100-200 commercially sized planes at once. Additional planes often park at airports in their quantity to wait until their flight times. for the airport to serve a metro community meaningfully it should be able to support escalating air traffic. It could be difficult to support a great number of airplanes. This could potentially be limited by charging parking fares for commercial aircraft staying over a certain amount of time.
The fuel in a 747-400 weights 167,000kg. The 747-200's max takeoff weight is 340,000kg/750,000#. I will presume that each plane weighs less than 1 million pounds, accounting for new plane designs in the future. It is actually likely that the planes will require less fuel while taking off from a midairport, and also the advent of scramjet propulsion will further reduce the necessary weight of an aircraft. No plane should ever need to be over 1 million pounds. This is also the heaviest estimate of a plane. It is presumed that the average plane present on the midairport will be either arriving with 20% of its fuel remaining or less, or fuelling or taking off with more than 80% fuelled. A majority of planes will be fuelled. Out of 340,000kg, with fuel being 167,000kg of the total. The average weight per plane should then come to approximately 300,000kg, with an average of 70 planes at the airport at a given time, this would place the average load at around 21 million kg.
It should also be known that fuelling pumps for conventional fuel [someday to be replaced with functionally pure water] will be pumped up to the tower from ground installations. The majority of the airport's fuel can sit on the ground, or in the pipe. The pipe will need to have 2 safety checkpoints to ensure that combustion or contamination either on the ground or at the port will not spread far.
Planes: 21mkg - 45mkg max
This is a very good sum so far. Planes are a substantially heavy portion of an airport.
Control towers and other structures can be built essentially from foams and layered glass and plastic, designed to stop and hold air as an insulator. The airport itself is expected to be rather windy outdoors.
A primarily heavy structure of the airport will be the landing surface. This would likely constitute a landing surface of metal studded/meshed rubber, potentially with traction pads laid down for the planes, along with an aircraft carrier string system arrayed in a mesh to allow planes to land in shorter distances and to aid plane function. The surface would likely be re-lacquered periodically, and would be flexible enough to shift with the ship's buffetings. Below this rubber would be a synchronous system of ferrocement panelings and rubber-encased carbon-fiber rods or metalworks to form a suitable frame upon which planes can land and be supported. How much this will weigh per 10m2 of surface area impacts the size and thickness of the ship dramatically and will not be ironed out entirely until finer engineering is overlaid. Ferrocement reportedly holds up 550kg/cm3 [1210#], but weighs quite a heavy amount. Presuming that the wheels of a plane will be the heaviest load-bearing portions of the airship, how much weight do they need to support? It may be so that an area of about one meter square will need to support 500 tons. From ferrocement alone, this would require 833cm thick FC. This is 8.33meters [27.77'] of cement.* But this is only under 1 single cm2 of cement surface. The real surface of the supporting area will be estimated to 1m2.
*F0r some reason, this figure seems unlikely. Traditional theory would have theorized that the area could have been supported under perhaps 10' thick ferrocement undercarriage. Maybe it is so that not the entire weight is supported by a single set of tires. That would be a difficult load for the plane's structure to support anyway. I am going to presume that the weight of the plane be cut at least in half, and likely the heaviest load on a single 1m2 section of wheel is only 2/5 of the plane's entire weight. From this figure, we will presume that an area of 1m2 of the landing surface will need to support 500000 pounds, on 90cm2. 90 x 90 x 1200# = 9,720,000#, 4860 tons, evenly supported at ~1cm. This figure seems unusually small. I would not think that 1cm thick ferrocement would support 9.7 million pounds, even if distributed evenly over an entire square meter. Thankfully, FC is remarkably strong.
I am going to place the surface structure thickness of the FC, underneath 20-30cm of meshed rubber, at 90cm thick, barring weight requirements. This means that this 90cm3 block of FC should support 437,400 tons of weight, and thereby provide the majority of the structural strength of the airport even in high winds. I would still expect to sectionalize the cement to allow the airport to shift shape a small amount.
The airport's shape should be slightly shiftable by an exterior facade to lessen winds. The ship may also benefit from enormous hanging clear plastic wind shields, which may double as projectile shields or sensors. I would expect these hanging at 50' outside the facade down the length of the ship, and meeting around the center cables. Beyond these shields lies very, very good wind farm real estate, with nearly unlimited finspan and midatmospheric wind levels. Surrounding a structure with a 2D perimeter of 13km, this is a lot of outstanding territory, with lots of space to hang down more self-foiling 10m+ mills from. According to this kid's exercise
, wind can blow Westerly at over 300mph in the atmosphere, especially near the Hadley and Ferrel cell's conversion points at 30* and 60*, but also anywhere flowing north-south. Winds in Antarctia blow at 200mph. The wattage from these mills will be estimated later. Their cost will be added above the top of the project. Solar panels can also be effectively oriented to capture the sun's rays very well from any latitude at this altitude.
Concrete weighs about 3000# per 90cm3. This sum could be even slightly lighter considering the amount of mesh that would be utilized in the FC, but 3000#/90cm3 will be the estimate. Also, I will estimate out the flex-zones from the plan, in favor of heavier concrete. 5km times 1.5km = 7.5 million blocks. That is 112.5 million tons. Which is 225 billion pounds. 102 billion kg of weight. This is too heavy.
The structure can only support 27 billion kg. Fortunately the 90cm thick figure supporting 437,000 tons of weight was a gross overstatement. The foundation of a house does not need to be that thick, and can in fact support a large house with only one foot. I will just have to trust the 500kg/cm3 figure and go with supporting a weight of 1 million pounds in one square meter, which would be /90cm /90cm /1200 kg/cm = the last figure in cm. 0.1cm thick x 90cm x 90cm will allegedly support 1 million pounds, by volume. This figure is dubious. If 90cm thick is 102 billion kg, and I would like a figure some 1/10 that much, I suppose going with 10cm would be an easier sum on the eyes. 4 inches of FC times the structure would weigh 11.34 billi0n kg
, totalling about 38% of the structure's buoyancy. This should support the heaviest planes, though, with an average square meter of surface supporting 97.2 million pounds, or 48,600 tons, and providing a large amount of support for the structure. It could potentially be reduced to a 3 inch structure. Certain area of the structure could have a firm support network of only about 2", such as indoors, which would cover a substantial portion of the structure.
Buildings should not come to greater than several hundred tons, perhaps 50 million kg considering comfortably appointed but foamlike building materials not requiring substantial strength but to wind, and using air to insulate for heat. The airport itself may be able to provide substantial heat if the He is pressurized. Each building should have plant life inside it to help provide a more oxygenated atmosphere. Doors to the outside of the airport should not be public access and should be in airlock. The planes will also taxi to suitable points to link up to tunnels to release and pick up passengers. While the outdoors is not dangerous to stand in, it would probable be unpleasantly cold, windy, and the air would be rather thin. People working in this environment should be properly equipped.
Several hundred tons is a meaningless figure coming to fractions of a % of buoyancy of the airport. The entire complex would float on top of a balloon of helium 400 meters thick all around the structure. Cement framework would mostly provide a frame to divide the weight of an object in the area among the collective buoyancy.
I would expect the interior chambers to be sectionalized, with column sections made of strong materials running the height of the structure, with airtight dividers inbetween, and log cabin style horizontal wires.
The next major weight the structure will be 15km of cables able to hold the structure down. This too will be an unknown figures and will work variably with the total weight and overbuoyancy of the structure. This will also include fuel and water pumps, netting and modest wind shielding around the channel, an electrical system, and a system of electric tramlines capable of ferrying passengers, cargo, and weight at high speeds up and down the channel from an original highrise ground structure rising perhaps 100 stories into the air. The groundworks will be a seperate figure, but since it is mandatory and contributes materially to the airport's functionality, it will be included in the fiscal estimation.
Another significant figure will be the weight of the support structure of the aircraft itself, rather than the runway. Substantial strength can be derived from the runway, which appears mathematically to be grossly overstrength for the weight of any plane by a factor of approximately 100X. The structural measures should not come to a weight of greater than 3000# per m2 of surface area, and should hold helium indefinately. The total filling cost for helium is so ridiculously low on the figure of commerce that the airport will generate. It's electricity alone could produce enough helium to refill it entirely every month. Ideally it should never be refilled, but an upkeep cost of $~2m every 18 months is a worst-case sustained loss rate.
The ship will also be equipped with a pair of 360* adjustable wings, and be equipped with powerful electric turbines. I speculate that the windmills can double as propellers if current is inlaid through them instead of drawn out through the alternator. A few large jet turbines as well positioned strateigically midway up the side surface all around should contribute to resisting the force of ambient winds. Cables will contribute to and provide stability, as well as a series of few-hundred-ton winged ballast weights like those used on ancient ships such as Noah's Ark for stability, but the ship should ideally be independently navigable. It may be possible to use the wings and the profile of the ship as a wing to steer through wind. It may be possible to use alternating low/high pressure structures in the wings through wind to pull and push the ship in the direction of the wind with a kind of <><><><><> or >>>>> wing arrangement with flaps.
These wings and the electric motors to operate them, as well as the airport electrical system for collecting and routing electrical energy, the wind turbines, and optional non-silicon solar arrays, may weigh on the order of 2 billion kg.
Since total weight of planes is not a factor in this estimation, I will increase the maximum number of parked aircraft from 100 to 220, including their hangars.
Buoyancy: 27 billion kg
[59.4 billion pounds, or 29.7 million tons]
11.34 billion kg -concrete landing area, 5km x 1.5km x 10cm [max figure]
2.00 billion kg
-cables and linkage between groundworks and airport
2.00 billion kg -wings, windmills, solarworks, turbines, and electrical systems
2.00 billion kg - miscellaneous weights, passengers, cargoes, airport vehicles
1.00 billion kg -structural frame and partitions of the aircraft [at ~ 3000# per m2 surface]
00.1 billion kg [100 million kg] - maximum planes [220 fully fuelled]
00.1 billion kg [100 million kg] - weight of structures, control tower, their furniture, facilities
---------------------------------------18.36 billion kg
[40,392 million pounds, 20.196 million tons] estimated total weight. The structure itself has been estimated to support 27 billion kg. It would be untenable to cable down almost 9 billion kg of force. The airport would tear away from very strong supports and supports required would be difficult. The airport should be made to hold in this situation, perhaps 20-22 billion kg in rough engineering. Finer engineering should place the overbuoyancy at a figure high enough to keep the cables taut and floating surely even at maximum load, which would probable be only a few million kg over rather than billions. One million kg is equal to 1100 tons. This much force is supercilious.
If the structure was estimated at 400m thick with a total of 3km3 [$1.5-2m] of helium, which produces 27 billion kg of buoyancy, and only it seems 20 is needed, we can reduce the thickness of the airport by 25%, to 300m thick, which will produce 20.25 billion kg of buoyancy, and cost only $1.41m. Finer engineering will probably bring these figures down, as the greatest quantity has been sought where uncertainty lies.
The groundworks is a simpler structure. The electrical system of the midairport should be fitted with a ribbing of metal poles to attract lightning in the area and channel it down around lines on the exterior of the channel to the groundworks, in which there would be a coiled superconducting system capable of handling and storing a lightning-bolt-sized charge of electricity, which can be enough to power 50,000 homes for about a year. The cost of this supercapacitance could be very high input, but it should serve the state's interests and will also be charged by wind power from the airport.
Groundworks merely needs to have deep enough foundations to ground the cables to hold down the few thousand spare tons of weight from the airport plus tug from any gusts. It might be possible to engineer a ground-level ballast system that would automatically apply weight to the ship near the ground when it was necessary, and detach ballast when it needed more lift. This could be far superior to and eliminate the need for continual dramatic overbuoyancy, and account for the landing of individual planes on the weight of the whole structure. Somewhat less He would be required and somewhat reduced cable strength required as well.
Beyond this it will be the aforementioned customs region, luggage and loading docks, and a subterranean parking garage should adjoin it in honor of or as part of the building's foundation.
It should reduce planes' fuel consumption and pollution by a substantial portion by reducing the need for them to climb, and should dramatically shorten plane trips for the same reason, along with the higher flying altitude enabling groundquiet supersonic flights. These airports should be safer accounting for their netting and even-grade landing capability.
They will also enjoy the benefit of providing electricity for their region, and freeing square miles of urban ground real estate by moving it into the clouds. This will raise the real estate value density of the region considerably, and provide better service for the urban environment and take a large bite out of urban pollution.