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Aviation History
1921
1921 - 0013.PDF
JANUARY 6, 1921 has been wasted. This machine is a biplane with two-bay wings and planes of equal span and chord, i.e., 39 ft. 4 ins. and 5 ft. 6 ins. respectively. The gap is 5 ft. 4 ins. and positive stagger 20 ft. The principal areas are as follows :— Total wing area, 404 sq. ft. ; tail plane, 26.; elevator flaps, 22-5.; top fin, 6-9.; bottom fin, 3-8; and rudder 7-2 sq. ft. The total wing area, therefore, Aw = 13-36 C*. or C= -08652 JX. The total wing weight is 362 lbs. The maximum permissibleloaded weight, WT, for this machine is 3,230 lbs., meaning that the maximum permissible wing- stressing load, Ws, is 3,230 — 362 = 2,868 lbs., or 7-1 lbs./ sq. ft. If we alter the scale of the wings throughout, the stress loading will remain the same per sq. ft., whilst the wing weight will vary as the cube of the chord C. But it is imprac- ticable to alter the scale throughout. Thus the wing covering is the [same for all sizes of machines, and amounts to about •14 lb./sq. ft. ofi win* surface. Main spars, internal compression members, metal fittings, internal and external bracing wires will all vary as the cube of the chord. Streamline Struts " The external wing struts will vary as the cube of the chord so long as they are of the same form, but in small sizes they are made solid, while in larger sizes they may be made hollow, and increasingly hollow as the size increases. Assume that, on the cross section of a hollow strut, the internal outline is of the same form as the external outline. Let B represent the maximum breadth of the external outline, and b that of the internal outline. Also, let B, represent the maximum breadth (of outline of section) of a solid strut with cross section of same form, and of same moment of inertia, as that of the hollow strut. Then :— If b = -6Bthen B= 1-0356, ,, ,, = -7B ,, ,, = 1 071 B, „ ,, = -8B ,, „ = 1-141 Bt „ ,, = -9B ,, „ = 1-307 B, and when b = • 6B, area hollow section = • 6S6 A .. = -7B „ = -8B = -58.5 A, = • 469 A, = -324 A, where A, = area of cross section of solid strut of same moment of inertia. " The external, or tween-wing, struts are ' long ' struts, and their strength is directly dependent upon the moment of inertia of cross section, meaning that the hollow struts quoted above should be of the same strength as the equivalent solid ones. We see, therefore, that we can save up to nearly 68 per cent, in weight by hollowing. But as a hollow strut needs a certain amount of external binding, and must have solid ends it is probable that a safe assumption is that 50 per cent, is the maximum that can be saved by using hollow struts. I have assumed, therefore, that solid struts are used up to C = 6-2 ft. ; that 25 per cent, of weight (for solid struts) is saved (by hollowing) from C = 6 • 2 ft. to C = 6 • 7 ft. ; 30 per cent, from C = 6-7 to 7-2 ; 35 per cent, from C = 7-2 to 7-7 ; 40 per cent, from C = 77 to 83 ; 45per cent, from C = 8-3 to 90; 50 per cent, for all sizes above C = 9-0. " The wing ribs, nosing ribs, leading and trailing edges, tip edgings and spacing battens are all parts which cannot remain strictly to scale for all sizes. I have, therefore, made some distinctly empirical assumptions about the weight of these parts. I have taken those of the Bristol ' Fighter ' as being the basic weight for C value of 55 ft., and have assumed that for all values of C below 7-7 ft. the weight varies as C*, while for all values of C above 7-7 ft. the weight then proceeds to vary as C*. Wing Weight Equations " From the foregoing we obtain a series of 7 equations for wing weight expressed as functions of wing chord, each equation referring only to definite limits of chord length. 1 • 466 Cs 4-3 • 888 C« = Ww for C<6-2ft. i-4OoC» + 3S88C2 = Ww 6-2 <C<6-7 „ • - 1386C* + 3-888C* = Ww 6-7<C<7-2 „ . •• ' - 1-373 C»+ 3-888 C» = Ww 7-2<C<7-7 .. 'v c -.., i-589C»+ 2119C = Ww 7-7<C<8-3 .. ••••-'• '•-•• I-576C3 + 2-1190*=- Ww 8-3<C<9'O ,, 1563C*+. 2119C* = Ww C>9ir! .. where Ww = total wing weight, in lbs. ; C = wing chord in feet, and Aw = 13-36 C», in sq ft.* " These equations refer strictly to wing weights for a two- bay biplane of the proportions of the ' B.F.' They will, however, not be very far from accurate for biplanes of other proportions, provided that the basic numerical value of C be taken as . / __i_. , not as the real chord length. Fuselage Weight " We come now to consider the weight of the fuselage. By 'fuselage' I mean only the girder structure complete with its covering. "The fuselage alse is a structure, which, if perfectly designed throughout, should remain exactly to scale in all its parts. For manufacturing and other reasons, it cann6t do this. I have taken as basic figures the weights of the ' Bristol Fighter.' In this machine the fitselageis 18-1 ft. long, 2 • 73 ft. wide, and 2 -95 ft. deep. It weighs 143 lbs., composed as follows •— Wood fairing, battens, for- mers, etc., 11 -3 lbs. Sheet aluminium covering, 13-2 lbs. Fabric covering complete, 12 -o lbs. 4 longitudinals, 23-1 lbs. All struts, 31-4 lbs. Plywood, decking and floor- ing, etc., 20-5 lbs. All metal fittings, 20-3 lbs. All bracing tie-rods, II-O lbs. " I have retained as a basic figure, C, the wing chord—C being of course of numerical value 5 • 5 for this basic fuselage of 143 lbs. weight. I have then assumed that :—Fabric covering, 3-ply decking and flooring and one-half of the weight of wood fairing and sheet aluminium covering, vary as C2. All metal fittings and tie rods and the other half of the weight of wood fairing and sheet aluminium covering vary as Ca. Longitudinals and struts vary as C8, but that beyond certain value of C weight can be saved by hollowing these. " Now the longitudinals and struts of a fuselage must be regarded as ' short' struts, and their strength determined by /cxA the Rankine-Gordon formula P = 1 + C (L)* W) Taking fc = 5,000 lbs. per sq. in., constant C as , and3000 a value for _ of 90, we get P = 1,350 A for a square section solid strut of length equal to 26 times its thickness ; K for a solid square section being equal to -289/. If we substitute a square section hollow strut, with a square section bore, and wall thickness equal to one-tenth of the strut siding -.—Cross sectional area A = • 36**. Radius of gyration, " We find that we must make the siding of this hollow strut about 1-25 times that of the equivalent solid strut to get the same ultimate strength, and the weight of the hollow strut will be about 56 per cent, of that of the solid. " I have made the following approximations for possible lightening due to hollowing of longitudinals and struts ; no hollowing is practicable up to C = 6-2 ; 20 per cent, of weight can be saved for values of C between 6-2 and 6-7 ; 25 per cent, for values between 6-7 and 7-2 ; 30 per cent, between 7-2 and 7-7 ; 35 per cent, between 7-7 and 8-3 ; 40 per cent, for all values of C greater than 8-3. " From the foregoing are obtained five equations for total weight of tutelage — WF = •5893~C« + i -479 C* C<6-2 „ =-52380*+ „ 6-2<C<6-5 „ =-5<>75C»+ „ 6 5<C<7-2 „ =-49o7C»+ „ 7"2<C<7-7 ,, =•• -4744 C»+ „ C>7-7 Weight of Tail " The weight of the tail unit WT is much the same problem. " I have again taken the basic weight for this element from the ' Bristol Fighter,' and obtained two equations in terms of wing chord C. WT = -2446 C» + -5587 c* for C<7-.7 „ = -2764 C* + -314 C2 „ C>7-7 Landing Gear Weight " The last structural element is the landing gear, consisting of main undercarriage and tail skid. In the ' Bristol Fighter ' the complete undercarriage weighs 97 lbs and the tail skid 10 lbs., a total of 107 lbs. for the landing gear of a machine of 3,230 lbs. weight, or -0331 of the total weight. I have assumed that weight of landing gear varies directly as total weight of aeroplane and is of this value, -0331. I am not certain that this assumption is entirely sound, but it possesses at least the merit of simplicity. " We have now got the following equations for the total
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