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Aviation History
1942
1942 - 1256.PDF
596 FLIGHT JUNE IITH, 1942 FLYING BOAT PROJECT the hull tail. Decking should be curved to assist shedding of spray. 13. Hall Cross-section.—At the main step we will fix the deadrise angle of 15 deg. In this matter it should be noted that a deep vee-sectioned planing bottom gives a lower air drag than flatter vee-sections, but the water performance can be seriously impaired if the angle of dead- rise exceeds 25 deg. at the main step. Hollow-vee sections, which until quite recently were widely used, and which definitely make for cleaner running, have now been dis carded on account of their high air drag. Assuming the use of 7ft. diameter airscrews, we can estimate the hull depth, bearing in mind that a deep hull—within limits— is a good feature. The airscrews should have at least 3ft. clearance between their tips and the water. To be on the safe side this will be made 4ft., and this distance plus 3ft. 6in. (airscrew radius) is 7ft. 6in.—say 8ft. Assuming the thrust line to be in line with the top of the root rib (which we know to be 2.2ft. deep), and also deciding to sink the wing into the top of the hull, we have a total hull depth above the load water line of 8ft. — 2.2ft., say 6ft. The height from keel to chine, with 6ft. beam and 15 deg. deadrise, will be 0.81ft., so that the depth of hull at the main step, from keel to lower surface of the wing, will be 6.8ift. 14. Control Surface Areas.—In a trial layout such as ours, it is general to use what are termed " control volumes " to determine the areas of fin, rudder, elevators, and ailerons. These control volumes are arrived at as follows :— For stabiliser and elevator ST + SE L = 0.40. ==. 0.45. For elevator SE x /L« Sw " V^E, For fin and rudder For rudder For ailerons SF +SR a X ,Sw Sw \sj SA flV Sw- X \s) — 0.06. 0.04. = 0.06= Sw ST SE SF SR SA L I The symbols have the following meaning : CE = mean chord (feet). = wing area. = area of stabiliser. = "area of elevators. = area of fin. = area of rudder. = area of ailerons. = moment arm (for fin, rudder and elevator. The distance back from C.G. to c.l. of control surface hinge). = moment arm (for ailerons: from C.P. of ailerons to c.l. of aircraft), s = semi-span. The volumes given above have been obtained from an analysis of existing successful flying boats of a similar type. By the use of the above, and without going into detail, our areas will be : Stabiliser 58 sq. ft. ) _ ., , ,A Elevator 46 sq. ft. f Tal1 Plane IO4 s1- ft- Rudder 26 sq. ft. \. _.. ; .. .. Fm 24 sq. ft. J Fm and rudder. 50 sq- ft. Ailerons 24 sq. ft. In proportioning these areas on the layout we bear in mind that long narrow surfaces are more efficient than short wide ones—aspect ratio again—but a compromise must be struck to gain economy in structure weight. This is particularly necessary with regard to the fin and rudder. A high fin means a high C.P. for that surface, causing large torsion loads in the thinned-down sections of the tail portion of the hull. The tail plane will probably be 20ft. span with a chord of 5ft. giving an A.R. of 4. Elevators will have a span of 20ft. or so, and a chord of 2.3ft. with an A.R. of 8.70. For control effectiveness and efficiency, large aspect ratios should be used wherever possible. 15. Tail Unit. General.—The fin and rudder, together with the tail plane, should be mounted so as to be well clear of solid water during take-off and alighting and also so that they are reasonably safe from spray buffeting. 16. Engines.—It is assumed that two engines will be used, mounted in the wing leading edge and driving tractor airscrews. These engines will be of 400 h.p. each ajji turn 7ft. diameter 3-bladed airscrews through gearing The engines will be air-cooled radials, to minimise piping runs, installation troubles, weight, and also to reduce servicing time. Their fuel consumption, at a.rating of 300 h,p., wil1 be taken as 0.50 lb./h.p./hour. If, as stated in our specification, we are to provide for a duration, at cruising speed, of six hours, then fuel to be carried will be 0.50 x 300 x 6 = 900 lb. Taking fuel to weigh 7.5 lb. per gallon (87 octane), the tankage required will be 900 =120 galls. 7-5 • With allowances for starting, taxying, and climbing, this can usefully be increased to 150 galls. The tanks will be in the wing centre section. One oil tank will be installed in each nacelle and will have a capacity of 0.04 lb. x 100 X 2 gallons = 8 galls. It is necessary to boost this up and we should allow for 10 galls. + 2 galls, air space. As far as possible, bearing in mind para. 13, effort should be made to keep the thrust line low, to avoid troubles in engine-on and engine-off flight and to avoid aggravating any porpoising tendencies. It is better, from considerations of air drag, to keep most of the engine-cowling below the wing and have as little as possible on the upper surface. The engine mountings will project from the front spar of the wing torsion-box in such a way that the vertical c.l. of the airscrews is at least one-quarter of the chord ahead of the wing leading edge. 17. Lateral Stability—To prevent the boat rolling over sideways on the water, because of its " built-in " instability we have to provide means to aid lateral stability. We will choose, from the five or so possible methods, wing tip floats. They will be placed at 70 per cent, of the semi- 30 X 70 span outboard from the hull c.l. Thus 100 = 2Jft. or gft. from each wing-tip. Their volume will be estimator from R.VV (k + *-\/W) sin 0 lb. ft. where W = all-up weight of boat in lb. (10,000 lb.). h = negative metacentric height of hull in upright condition in feet (1.6 for our boat). 6 — angle of heel or roll to submerge completely a wing float (7 deg. generally). R =1.0 for our example. Minimum righting moment then for one float = 1.0 X 10,000 (1.6 , = 10,000 4- 11.6 X Minimum displacement of one float = lb. per cu. X 3\/10,000) x 0.12187. .12187 = 14,137 lb. ft. 14.137 21 672.5 lb. ft., the volume of Since sea water weighs 64 our wing-tip float will be : 672 5 = 10.5 cu ft. 64 This ro.5 cu. It. represents the length, breadth, and depth of the float multiplied together. We can alsoteay ii we assume a block coefficient of 0.50 that 0.50 x (L x B X D) = volume = 10.5ft.11 and 0.50 x (4B x B x B) = 10.5ft.3 2B3 10.5
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