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Aviation History
1911
1911 - 0362.PDF
|p]GHf] APRIL 22, 1911. sufficient plane area ahead of the main planes at a greater angle than that of the main planes, so that, neglecting the effect of inertia, the machine is stable and cannot, consequently, be in balance at more than one attitude when in gliding flight. In gliding the only two forces in action are the resultant of all the air pressures, P, acting at p, the centre of pressure, and the weight, IV, acting at g, the centre of gravity. These forces must be equal and in the same straight line. Let Fig. 2 represent the same aeroplane in horizontal flight, flying at the same attitude as it glides at, for it is desirable that the aeroplane should fly at a constant attitude, whether the direction of flight be downward, horizontal, or upward. If this is so, variations of thrust will not affect the attitude. Then the position and the direction of P relative to the machine remains unaltered, and g lies in P produced downwards. W, acting vertically downwards at g, must be balanced by making AH) the vertical component of P, equal to it. Also A5H, is the horizontal com ponent of/". Now D, the resultant of the thrust and the weight, must be equal to P and lie in the same straight line as P, and consequently must pass through g. By completing the lower parallelogram we see that T, the thrust, must pass horizontally through f, and must, of course, equal AH. So the thrust line should pass through the centre of gravity. Since P produced must pass through g, in order that D, wnich passes through g, can be in the same straight line sin £ sin e sin e -^ n r, Wsin/8_ Wsin(-y + a) _Wsin(> + a) 1 sin e sin(ao - 7) cos y ^sin(7 + o) Thrust required for ascending at angle o = IV. Speed on inclined trajectory (Va). P _ L _ W Now —.—,—5 7. > : :— • sin(i8o-j8-f) sine sine cosy, P = Wsin(l8o°-0-e)_ ^sin l8o°-(y + a) -(90*-7) sin(9O0-7) Wsin(90° — o)_ ffcoss eos 7 cos 7 .-. Component of PaX right angles to trajectory Wcosa _, = • • cos 7 = Wcos o. cos 7 But this as K F2o and W = K V"H. . •. K V-a — Wcos a. — K V2H cos a. .". Va= VuVcosa. V. Now power varies as thrust x velocity. ~LetP=cTV. DIRECTION OF FLIGHT FIG i as P, and P only passes through g at the attitude shown (see proviso of first paragraph), the attitude will not be affected by variations of thrust, and the thrust for any chosen trajectory must bear such a relation to the weight that D and P are in the same straight line. Gliding Speed and Angle see Fig. 1). I. 7=angle between Pana Lg (since P is vertical and Lg per pendicular to the direction of flight). _ . Pg AH resistance . , _, . g*.\ tan y— r — r = --• . , - (as given by Chanute). ' Lg AH weight v & * II. Gliding speed Vg (equal to VH, the horizontal flight speed wjien 7 is small). VZV+A^=-W=P-- J&P - L?g tan 2 y = LgJi + tan2 y = Lg •\/sec27 = Lg sec 7. But IV= LH = K V^a, and Lg= K V-g. . •. K F*H = K V"g sec 7. . •. fV= = F2H cos7. 0 sec 7 *. Vg= FH-/COS7. Altitude Flying (angle of trajectory with horizontal = a). (See Fig. 3.) III. j3=7 + o, 8 = angle between L and A' + 90°Ja. .•. «- = 5-£ = ao0 + a-(7-ra) = 900-7. DlRECTlOI- 3F FLI6HT FIG + Then power for ascending at o (say, Pa) _J-Tsin(7 + a) — = c • KHVCOSO, cos 7 ' and power for horizontal flight (say, Pn) = c • Wisiay . VH Pa _ sin(7 + g) A/COS a __ sin (7 + a) -/cos a PH tan 7. cos y sin 7 PH sin(7 + a) Vcos a P*- siny By substituting in this formula we find that a machine whose gliding angle = 50, will need 3 "86 times the power required for horizontal flight to rise at 150, 6*12 times bo rise at 300, and 7*38 times to rise at 450, the attitude for the upward flight being the same as the attitude for horizontal flight. Thus a 9-h. p. aeroplane, whose thrust line passed through its centre of gravity, would require about 66J-h.p to rise at 450. From formula IV we see that, if the attitude remain unchanged, increase of thrust results, not in increase of speed, but in decrease of speed. Consequently the fitting of a motor or a more efficient propeller to an aeroplane, whose thrust line passes through the centre of gravity, will not increase its speed unless at the same time the angle of the planes, or their area, be decreased. If we consider an aeroplane whose thrust line passes above the centre of gravity (Fig. 4) we find that in order that D and P may be in the same straight line when the thrust line is horizontal, the centre of gravity g, must be aft of D, and therefore behind the position required for gliding at the same attitude. Now, because of the proviso of our first paragraph, the centre of pressure moves back as the angle of incidence increases, and because of the 364
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