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

1) The cruising altitude is h when descent starts at a horizontal distance l from touch-down at the origin

2) The pilot must maintain a constant horizontal speed v through the descent

3) The absolute value of the vertical acceleration should not exceed a constant k (which is much less than the acceleration due to gravity)

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.

Question #1 WorkRatesDistance: p(x)= ax3+bx2+cx+d Velocity: p’(x)= 3ax2+bx+cAcceleration: 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 b and L • Knowing that p(L)=h, we plugged a back into the distance equation to find b 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

p(x) = -2hx3/L3 + 3hx2/L2

Solution

12. Slide 12

13. Slide 13

Question #2

Use the conditions 2 and 3 to show that6hv2/L2 < k

Given• a = -2h/L2

• b = 3h/L3

• p(x) = ax3+bx2

• x = L

Question #2 WorkSteps1. Find the first implicit derivative of p(L)

dp/dt = 3aL2(dx/dt)+2b(dx/dt)note: dx/dt = -v

2. 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

3. Plug in a and b-6Lav2 -2bv2 < K-6Lv2(-2h/L3) - 2(3h/L2)v2 < K12hLv2/L3 – 6hv2/L2 < K12hv2/L2 - 6hv2/L2 < K6hv2/L2 < k

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?

ObjectiveFind L, which is the point at which the pilot

should start the descent

Question #3 Work• Given: h=35,000 ft converted to 6.62878 mi

v= 300 mi/h k=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

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.

Question #4 WorkGivena= -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

Solutionp(x)=-.00004937x3+ .004777x2

Window xmin=0, xmax=64.5(length), ymin=0, ymax=6.6 (height)

Walk away…

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Question

P(0) = 0

P’(0) = 0

P’(L) = 0

P(L) = h

5. Question