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hw06_solutions

# hw06_solutions - AA311 Atmospheric Flight Mechanics Autumn...

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AA311 - Atmospheric Flight Mechanics Autumn 2010 University of Washington Homework 6 Solutions Problem 1 Part a. We can first convert to consistent, metric units F-14 S = 54.5 m 2 W = 52 000 lbf 4.44822162 N 1 lbf = 231 308 N F E i = 27 800 lbf 4.44822162 N 1 lbf = 123 661 N i = 1, 2(thrust from engines) Runway d = 300 ft 1 m 3.2808399 ft = 91.44 m AA311 - Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students

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S = 54.5 W = 52 000 H 4.44822162 L FE1 = 27 800 H 4.44822162 L FE2 = FE1; d = 300 1 3.2808399 20 54.5 231 308. 123 661. 91.44 Part b. If we assume that there is no difference between infinite and finite wings, we can use Appendix D with the c l vs. a curve for flaps of standard roughness to see that at a = 0 c l = C L = 1.4 CL = 1.4; We require the lift to equal the weight in order to not lose altitude at the end of the runway W = L = C L q S recall: q = Å 1 2 r V 2 = C L I Å 1 2 r V 2 M S V = J 2 W r S C L N 1 ê 2 where r = 1.225 kg ë m 3 (standard sea-level conditions) 2 hw06_solutions.nb Printed by Mathematica for Students
ρ infty = 1.225; Vinfty = 2 W ρ infty S CL 1 ê 2 70.3525 So we see that we require V = 70.35 m ê s º 157 mi ê hr Part c. We can draw a free body diagram of the aircraft So the forces on the aircraft are / F = F E 1 + F E 2 + F C where F E 1 = thrust from engine 1 (123661 N) F E 1 = thrust from engine 2 (123661 N) F C = thrustfrom catapult So the acceleration experienced is given by / F = m a a = F E 1 + F E 2 + F C m note: m = W ê g a = F E 1 + F E 2 + F C W ê g (Eq.1) We note that the acceleration is constant because all terms on the right hand side are assumed to be constant We can obtain the velocity as a function of time using V H t L = Ÿ 0 t a H t L t + V H 0 L note: a H t L = a hw06_solutions.nb 3 AA311 - Atmospheric Flight Mechanics Christopher Lum Printed by Mathematica for Students

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V H t L = a t + V H 0 L (Eq.2) We can find the position as a function of time by integrating Eq.2 p H t L = Ÿ 0 t V H t L t + p H 0 L = Ÿ 0 t a t + V H 0 L t + p H 0 L p H t L = Å 1 2 a t 2 + V H 0 L t + p H 0 L (Eq.3) We now would like to find the time T when the aircraft reaches the end of the runway. We assume that the aircraft starts at rest V H 0 L = 0 and at the back of the runway p H 0 L = 0 p H T L = d Å 1 2 a T 2 + V H 0 L T + p H 0 L = d recall: V H 0 L = p H 0 L = 0 Å 1 2 a T 2 = d T = I 2 d a M 1 ê 2 (Eq.4) We know that at t = T , we require that V H T L = V V H T L = V a T + V H 0 L = V recall: V H 0 L = 0 a T = V recall: T = I 2 d a M 1 ê 2 from Eq.4 a I 2 d a M 1 ê 2 = V H 2 a d L 1 ê 2 = V recall: a = F E 1 + F E 2 + F C W ê g from Eq.1 J 2 J F E 1 + F E 2 + F C W ê g N d N 1 ê 2 = V 2 J F E 1 + F E 2 + F C W ê g N d = V 2 F E 1 + F E 2 + F C = W g V 2 2 d F C = W g V 2 2 d - F E 1 - F E 2 4 hw06_solutions.nb Printed by Mathematica for Students
where g = 9.81 m s 2 g = 9.81; FC = W g Vinfty 2 2 d FE1 FE2 390 816. So we see that F C = 390816 N º 87859 lbf Part d. We can compare the catapult thrust to the force provided by the two engines h = F C F E 1 + F E 2 η = FC FE1 + FE2 1.58019 We see that the catapult must provide approximately 1.5 times the thrust that both engines are capable of producing.

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