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Aviation History
1911
1911 - 1084.PDF
DECEMBER 16, 1911. Conducted by The relative Areas of Main Plane and Elevator. Let the figure be supposed to represent a part plan of a self- rising model in so far as the fuselage and planes are concerned. In Figure (1) let P be the position of the combined central vertical mast (or spar) and chassis. We assume also that the model is supported by the wheels of this chassis and rear skid at K. Then the centre of gravity G will require to be slightly in the rear of P. If it be not so placed then the model, instead of rising, is just as likely (under Ct- V. E. JOHNSON, M.A. gravity (in steady flight the upward pressure, R, obviously acts upwards through the centre of gravity); then, if we place at the point where the centre of pressure of the elevator will act a weight equal to the elevator (the weight of the elevator in, and balance the complete machine about a knife-edge, we shall know G, this case being a predetermined one), i.e., we know b and c and x can be found. In formula ii if we increase A (the other factors remaining constant) x increases—similarly if b be increased x is also increased —in other words, as we shift the main plane nearer tc we must increase the relative size of the elevator, and conversely as we bring the main plane further forward x diminishes. As c is increased x diminishes, and also as the factor 2 (ii) or 1 + -(i) becomes greater. All these results arc borne out in actual practice. The general tendency among model aero-modelists undoubtedly r.P 6 P :'-%%- .-S91 .--r.-:;s.. LJL. Fig. 1. the thrust of the twin propellers) to dip its nose down and run along the ground like a dog following a scent. The centre of gravity meant is of the model complete with propellers and motor. The position of the main plane A is now fixed, i.e., slightly behind G. The position of the elevator is also supposed determined, also the area of the main plane is supposed known. To determine the size or area of the elevator. In Fig. 2 let A denote the area of the main plane and a its inclination to the line of flight, let x denote the area of the elevator and yS its inclination to the Ike of flight; / and p1 the centres of pressure ; and y the angle elevator makes with the main plane : now since a and /3 are small angles we may take the lift as proportional to the angle. The air pressure on A and x may therefore be taken as Aa and x$ respectively, acting upwards at their centres of pressure p and p1. By the laws of elementary mechanics the resultant of these two pressures is a force (say R) acting upwards and dividing//1 into two parts b and c proportional to xfi and Aa respectively. Whence - • P) <<•*$ Now A is supposed known, also b and c can be measured, because (in the case of flat planes) the distance of the centre of pressure from the leading edge is known at various angles of inclination from Rateau's diagram, and the position of the elevator is supposed fixed. In the case of a cambered surface or aerocurve with a curvature T^th the span (say) the same is also known, and can be determined with a sufficient degree of approximation in other cases. But we have still left the factor r-f—*. a Now, if 7 =- a, i.e., if the inclination or the elevator were twice m that of the main plane, then 1 + - would equal 2, and our formula a. Ab would become x = — (ii) Now, if such an assumption be a fair one, and such as is found to work out in practice—with respect to this, or any other proposed substitution, I shall be glad to hear from correspondents—then we have a very simple and easily-applied formula. Suppose we know A (position and area), but not G, the centre of Fig. 2. (primarily) to use elevators of much too large an area. Applying formula i to one of my own models to determine I + , I find it gives a value 1*219. Next week- I shall bs glad to hear what value others arrive at. " Plan Forms of Supporting Surfaces." To Calculate the PitchTof a Propeller. Take any point on one of the blades and measure carefully the inclination of the blade at that point to the plane of rotation. -1 1 Say the angle so formed is 15° (taken at the tip, say). ' Now 14*48°, i.e., approx. I5J, is 1 in 4 (see any table of equivalent inclinations), say this point is 10" from the centre, then every revolution this point will travel a distance 22 27rr=2X — XIO= 62*8'* '. 7 •* Now, since the inclination is 1 in 4, the propeller will travel forward theoretically one-fourth of this distance 62*85 = • • • • = 1571 = ISl ms. approx. Similarly, any other case may be dealt with. The pitch is sup posed uniform. If the blade be hollow-faced, take the mean effective pitch—such a blade has, of course, an increasing pitch from the leading to the trailing edge—considering any particular section. An 8 in. propeller would want an inclination of about 37° at the tip to have a pitch of 2 ft. allowing for slip. Propellers of different diameters, both having the same pitch, are not similar. They are similar when the pitch angle is the same. Petrol Motors. In the only successful small petrol-engined model that we know of the motor was a four-cylindered one and weighed complete 5J lbs. and developed ij-h.p. at 1,300 r.p.m. The propeller was 29 ins. in diameter, and of 36-in. pitch, with a static thrust of 7 lbs. The machine had a spread of 8 ft. 2 ins., and a length of 6 ft. 10 ins. Total weight, 21 lbs. It soared at 16 m.p.h. The engine was air- cooled, and a double float-feed carburettor and single-coil ignition and distributor were employed. In another model, made by a well-known aeronautical engineer, the total supporting-area was 13*3 sq. ft. Total weight, 16 lbs. ; engine, inclusive, 8 lbs. 140ZS. ; length of model, 5 ft. ; span, 7 ft. 8 ins. Engine developed 1 h. p. at 1,500 r.p.m. Diameter of propeller, 25 ins. So far as we know, however, this model has never been tested in actual flight. 1092
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