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Where should a pilot start descent?

Where should a pilot start descent?. By: Alex, Chelsea, Gio , Ian, Jake, and Tessa . The Problem. An approach path for an aircraft landing in shown in the figure and satisfies the following conditions…. Conditions.

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Where should a pilot start descent?

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  1. Where should a pilot start descent? By: Alex, Chelsea, Gio, Ian, Jake, and Tessa

  2. The Problem An approach path for an aircraft landing in shown in the figure and satisfies the following conditions…

  3. Conditions The cruising altitude is h when descent starts at a horizontal distance l from touch-down at the origin The pilot must maintain a constant horizontal speed v through the descent The absolute value of the vertical acceleration should not exceed a constant k (which is much less than the acceleration due to gravity)

  4. Question #1 Find a cubic polynomial p(x)= ax3+bx2+cx+d that satisfies condition 1 , by imposing suitable conditions on p(x) and p’(x) at the start of descent and at touchdown. Objective Find constants a, b, c, and d that justify an equation for the plane’s flight path.

  5. Question #1 Work Rates Distance: p(x)= ax3+bx2+cx+d Velocity: p’(x)= 3ax2+bx+c Acceleration: p’’(x)= 6ax+2b We found… • d=0, by plugging 0 into the distance equation • c = 0, by setting the velocity equation (the tangent line) • a in terms of band L • Knowing that p(L)=h, we plugged a back into the distance equation to findb in terms of h and L • We had an equation for a in terms of b and L, so we plugged our value for b back in to find a in terms of h and L 12. Slide 12 13. Slide 13 Solution p(x) = -2hx3/L3 + 3hx2/L2

  6. Question #2 Use the conditions 2 and 3 to show that 6hv2/L2<k Given • a =-2h/L2 • b = 3h/L3 • p(x) = ax3+bx2 • x = L

  7. Question #2 Work Steps • Find the first implicit derivative of p(L) dp/dt = 3aL2(dx/dt)+2b(dx/dt) note: dx/dt = -v • Find the second implicit derivative of p(x) d2p/dt2 = -6avL(dx/dt)-2bv(dx/dt) d2p/dt2 = -6avL(-v)-2bv(-v) d2p/dt2 = -6Lav2-2bv2 note: d2p/dt2<k • Plug in a and b -6Lav2 -2bv2< K -6Lv2(-2h/L3) - 2(3h/L2)v2 < K 12hLv2/L3 – 6hv2/L2< K 12hv2/L2 - 6hv2/L2< K 6hv2/L2<k

  8. Question #3 Suppose that an airline decides not to allow vertical acceleration of a plane to exceed k = 860 mi/h2. If the cruising altitude of a plane is 35,000 ft. and the speed is 300 mi. how far away from the airport should the pilot start descent? Objective Find L, which is the point at which the pilot should start the descent

  9. Question #3 Work • Given: h=35,000 ft converted to 6.62878 mi v= 300 mi/hk=860 mi/h2 • 6hv2/L2<k we now need to plug in our values • 6(6.62878)(300) 2/ L2 < 860 • 3579541.2 /L2< 860 multiply l2 on both sides and divide by 860 to isolate l2 on the right side. • 33579541.2/860 = 4162.257209 • √4162.257209 < √L2 L=64.515 miles • When the plane is 64.515 miles away the pilot should start the descent to the airport

  10. Question #4 Graph the approach path if the conditions stated in Problem 3 are satisfied. • In order to graph we use the original equation which was ax3 + bx2 + cx+d. Because in question one we found that c and d equaled 0, the equation is ax3 + bx2. • In problem 2 we found what a and b equaled in terms of h and l, now that we have what h equals and l equals from number 3, we plug it in for a and b.

  11. Question #4 Work Given a= -2h/l3 b=3h/l2 Height=6.629 Length=64.517 Steps A= -2(6.629)/(64.517)3 B=3(6.629)/(64.517)2 A=-.00004937 B=.004777 Now you plug in A and B to the original equation Solution p(x)=-.00004937x3+ .004777x2 Windowxmin=0, xmax=64.5(length), ymin=0, ymax=6.6 (height)

  12. Walk away…

  13. \ P(0) = 0 P’(0) = 0 P’(L) = 0 P(L) = h Question

  14. 5. Question

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