Anthony replied to me: > Well, you have the problem of the fan rules being a gross approximation. > A ducted fan produces 18N thrust per kW, implying that air is being > accelerated from about 0 to 100 m/s. Air density ranges from about 0.6 > to 1.2 kg/m^3 over the appropriate altitude range for a helicarrier, > we'll assume we can operate at high altitude with no excess lift, so > air flow rate at top altitude is 60 kg/m^2/s and thrust is 6,000N/m^2 or > .06 bar. > > A 20,000 ton ship then needs 180,000,000N lift, meaning 30,000m^2 > (260,000 sf) fan area. With 6 fans, each one has a diameter of 240 feet; > a row of 3 on each side means you have something 720' long and 480' wide, > plus the actual width of the flight deck. That's shorter than a Nimitz, > but much much wider. Also, the flight deck is going to be suffering from > crosswinds of up to 200 mph, which is going to make takeoffs rather > challenging.
Hello Anthony, there could be more and smaller fans, which can be arranged in a hexgrid pattern. But that saves only a couple of percent. The next question would be if the cross-section and the shape of the fans, the intakes, and the nozzles would have to be the same. That won't help with the crosswinds, of course. A trimaran hull like the Battlestar Galactica? Either the fans are in the outriggers, away from the flight deck, or the flight decks are in the outriggers, away from the engine rooms. Regards, Onno _______________________________________________ GurpsNet-L mailing list <[email protected]> http://mail.sjgames.com/mailman/listinfo/gurpsnet-l
