practice problems

practice problems - this isn't a very large velocity...

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rho 0 slug/ft^3 W 18000 lbs S 320 ft^2 CL,max 2.4 CLg 1.8 CD = .4+.05CL^2 mu_g 0.4 carrier runway 700 ft using drag polar -> CDg 0.56 Vs = (2*W/(rho*S*Clmax))^.5 Vs 140.34 VTD (1.15*Vs) V_TD 161.39 0.7 VTD used for average accel… v_avg 112.97 q_avg 15.19 g*(-mu_g -(CD - mu_g*CL)*q/(W/S)) a_avg -11.51 d_nowind = V_TD^2/(-2*a_avg) d_nowind 1131.86 we need to find wind for distance of 700 relative to carrier, what is the velocity ^2 that leads to 700? Vg^2 = 2*a_avg*700 16108.81 Vg at TD 126.92 Vw = V_TD - 127 34.47 (fps) 23.5 (mph) Let's check our answer…. VTD^2 - Vw^2 24858.9 d 1080.23 (ft) t 11.03 (sec) carrier will travel. . 380.23 ft allowed distance (700 + 380) 1080.23
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if there was no wind, we would need this much runway to decelerate from V_TD to 0 Vg goes from ( V_TD - Vw) at touchdown to 0 at the end of the runway Carrier needs be going approximately 24 mph
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Unformatted text preview: this isn't a very large velocity relative to V_TD, so using 0.7 V_TD as the average velocity seems pretty reasonable If the carrier is moving 24 mph, then the airspeed of the aircraft will go from VTD to Vw what distance is needed to decelerate from VTD to Vw how long it takes to decelerate: (VTD-Vw)/a_avg carrier will travel this distance during the ground roll. . . S rho V W 320 2.38E-03 400 18000 Radius CD_0 K 1500 0.02 0.05 n tan(phi) 3.31 --> phi = 1.28 73.2 3.46 L_turn/L_level 3.46 CL_level = 2.96E-01 CD_level = 2.44E-02 --> D_level = 1.48E+03 CL_turn = 1.02E+00 CD_turn = 7.24E-02 --> D_turn = 4.40E+03 delta thrust required: 2.92E+03 when back to level, thrust excess: 2.92E+03 lbs R/C 6.49E+01 3.90E+03 ft/sec fpm...
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This note was uploaded on 10/17/2010 for the course MAE 154s taught by Professor Tooney during the Spring '09 term at UCLA.

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practice problems - this isn't a very large velocity...

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