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HW6Pr2soln

# HW6Pr2soln - radius(ft velocity(ft/sec Maximum turn rate...

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Correction to Problem 6.1 solution: Start with the equation below “Squaring both sides…” (corrections shown in red, in most cases) (4) This yields, for BD-J4, Obviously, other results, given in the solution, involving this velocity would be incorrect too. The recalculated values are (by my calculations): , One could observe this graphically as illustrated below. As shown in text, obtain Figure 6.2 for this aircraft.

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150 200 250 300 350 400 1 2 3 4 5 6 7 8 9 10 load factor .vs. velocity velocity (ft/sec) load factor stall thrust Clearly, the maximum load factor occurs at the intersection of these curves, corresponding to the computed velocity of about 190 ft/sec. (The actual velocity is 189.83 ft/sec, both analytically and numerically). This is also displayed in the radius.vs.velocity plot given below, on a different scale.
150 160 170 180 190 200 210 220 230 240 250 1 1.5 2 2.5 3 3.5 4 load factor .vs. velocity load factor stall thrust 150 160 170 180 190 200 210 220 230 240 250 400 500 600 700 800 turn radius .vs. velocity

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Unformatted text preview: radius (ft) velocity (ft/sec) Maximum turn rate: The thrust limited maximum turn-rate velocity is 194 ft/sec., as computed in the author’s solution. This point (denoted by point D in Figure 6.2) lies to the right of the intersection point A, and is on the thrust-limit curve. To observe that the best turn-rate velocity, generate plot for the turn rate, using equations 6.18 and 6.11. This is given below. 150 160 170 180 190 200 210 220 230 240 250 0.2 0.25 0.3 0.35 0.4 0.45 0.5 turn-rate .vs. velocity turn-rate (rad/sec) velocity (ft/sec) stall thrust The maximum turn-rate at any given velocity is the minimum of the two curves, implying that for velocities above the interection point A, the max. turn-rate is determined by the thrust-limit. The maximum of the thrust limit curve lies at the computed velocity of 194 ft/sec . And hence the maximum turn-rate is 0.331 rad/sec ....
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