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Aviation History
1931
1931 - 0391.PDF
24, 1931 THE AIRCRAFT ENGINEER SCPPLXMXMT TO FLIGHT Actually, the initial oil pressure is slightly greater than at the end of the stroke, but the difference is negligible when a reasonable factor of safety is allowed. Air Volumes and Pressures. If the static load on the tail skid is 10 per cent, of the weight of the aircraft, the static load on one wheel will be 4,500 X 8-284,500 lb. Static load on oleo is, therefore, = 4,6501b. Let— Pai = Air pressure with leg fully extended (F.E.) Pa2 = „ „ „ „ in standing position while stationary. Pas = t» » with leg fully compressed (F.C.) assuming a static condition. P«4 = ,i » with leg fully compressed in action. Pas = „ ,, with leg in standing position during action. Vs = Air volume with leg fully extended. V, = „ „ „ „ in standing position. V3 = „ „ „ „ fully compressed. (See Fig. 2.) Then Pal Vl = Pa8 Va = Pa3 V3 (Isothermal compression) and pal Vtn = pai Vs» = p,A V3» (Adiabatic compression) 4,650p at = -—- = 600 lb. p.s.i. At F.C., since velocity is zero, pa ( = pressure difference — Po — Pa) = 0. Also, Pai Y1 = pai V2 = Pai (Vx — Agt) where t is the travel of the leg from F.E. to standing position (S.T.) and Pai Vi" = pai V3» = pai (Vj — Ag T) where T is the total travel of the leg. Hence, pal = pai Therefore, —'• Pai \ V, 'Vt - A Pai V, f - AgT\ V, or Let and ^ I- ?*! (i - ;-?:) = (i _ Pai\ V, Pat Pat = z, n= 1-3 = y Then values of a^and yean be plotted for various values of Vlf and where the two curves intersect will be the true value of 0-30 0-2B 0-26 0-24 0-22 0-20 • 018 '<& 0-16 .X 0-1* fe 0-12 ^ 0-10 $0-08 0-06 0-04 0-02 TRUFVMUeOF l /S 86-S cu mcfies y ,—"•— <• y \ • y y _ — FIG. 3. GRA PLOT PHS OF"X"&"U" FED AGAIN5TY," y^ —-— 70 75 80 85 90 AIR VOLUME (CU. INCHES) 95 100 2000 1500 1000 0 —^ Pa "Alf PRESSUF \ PR hr- Pc -0 y FIG ESSUR IL PRESJ 7.4. i CUR\ 1 >UR£ /L \\ 'ES 1 / / k \ 3 4 TRAVEL (INCHES) -> 8 HOUrHOCA Vx for the particular case under consideration (see Fig. 3). With regard to the ratio —, this will be fixed automatically in cases where the oiTand air function consecutively ; but where the actions are concurrent—as in this example—it must be determined arbitrarily. The factors to bear in mind in fixing it are that the greater it is (i) the smaller the air chamber and, consequently, the lighter the unit and (ii) the harsher the action in taxying. So that, as is usual in aircraft design, a compromise must be made. Generally speaking, a value of 0-5 is sound and this figure will be taken in this example. All pressures and volumes can now be deduced as follows :— Vj = volume at F.E. = 86-5 cubic in. V2 = „ „ S.T. =V,-A^ = 86-5-31 = 55-5 cubic in. V3 = „ „ F.C. = Vx-AST= 86-5-62 = 24-5 cubic in. iT1_, Pa%Vt 600x55-5p ai = pressure at F.E. = —— = —— = „ S.T. (Isothermal) = /VA13(Adiabatic) = p ai (— ) = 385 1b. per sq. in. 4,650 = 7-75 600 lb. per sq. in. Pas Pas, = 385 —— = 686 1b. per sq. in. \55-5, p,n — ,, „ F.C. (Isothermal) = —* = V, 600 X 55-5 n 1L—— = 1,358 Jb. per sq. in. 24-5 Pat = „ „ „ (Adiabatic) = 2,000 lb. per sq. in. The Oil Valve. The frictional resistance of the oil in passing through the valve and also the friction of the mechanical parts, will be neglected. It is not possible to calculate these and, in any 362*
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