Where Should A Pilot Start Descent ?

By : Mardhatillah, Visca Amelia S, Sary Widrafebi, Murtia Zaili, Dwi Ratna Dian Sari, Zalfa Ahmad, Rani Febrian, Elita Putri

(i) The cruising altitude is h when descent starts at a horizontal distance l from touchdown at the origin.

(ii) The pilot must maintain a constant horizontal speed v throughout descent.

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

Conditions ...

Problem 1 . . .

Find a cubic polynomial P(x) = ax³+bx²+cx+d that satisfies conditions 1 by imposing suitable conditions on P(x) and P'(x) at the start of descent and at touchdown

Solve . . .

We have P(0) = 0, P'(0) = 0, P(L) = h, P'(L) = 0

By that conditions we can get :

From P(0) we get d = 0

From P'(0) we get c = 0

We know that p(L)=h, we can put a to the distance equation to find b in terms of h and L .

So, we get :a =

b = l

b

3

2

l

b

3

2

2

3

l

h2

3

l

h

2

3

l

h

- 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

We get :

2

2

3

3 32

L

hx

L

hx

2

2

3

3 32

L

hx

L

hx

2) Use conditions 2 and 3 to show that

kl

hv

2

26

kl

hv

2

26

AnswerCondition (ii) dx =-v dt so, x(t)=l –vtCondition (iii) | d²y | ≤ k dt²

P(x)= ax³ +bx²P’(x) = 3ax² (dx) + 2bx (dx)

dt dt = 3x² (2h) (-v) + 2x (3h ) (-v) -l³ l² = -6hvx² - 6hvx -l³ l² |d²y| = 6hv (2x) dx -6hv dx dt² l³ dt l² dt = 12hv(-v) x – 6hv (-v)

l³ l² =-12hv² l + 6hv² when t=0, x=l

l³ l²

-12hv² + 6hv² = -6hv² l² l² l²

So 6hv² ≤ k l²

Process:

Look condition ii and iii

Take the derivative of the

equation P(x)=ax3+bx2 with

respect to time (t)

•(dx/dt) being with respect to

time, replace (dx/dt) with (-v).

Where (-v) stands the horizontal

velocity

Find the derivative again

Substitute a, and b that were

found in the previous problem

SOLVE

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 35000 ft and the speed is 300 mi how far away from the airport should the pilot start descent ?

Problem 3 . . .

Work:known:k =860 mi/h²

h= 35000 ftchange ft to mih=35000ft X 1 mi

5280 fth= 6,63 miv= 300 mi/h

Answer:6hv²/l² ≤ k6 [6,63 mi . (300mi/h)²] ≤ 860 mi/h²

l²6 [6,63 mi . 90000mi²/h²]≤860 mi/h². l²

3580200 mi³/h² ≤ 860 mi/h². l² 3580200 mi³/h² ≤ l² 860 mi/h²

4163,02356 ≤ l²64,5 miles ≤ l

Process:Take the equation that was

found in part 2 and apply it to this problem.

Take the equation and the variables that are known and start plugging into the equation. We already know that: k = 860 mi/h² v = 300 mi h = 35000 ft but we must find l

Since h is in feet, we must change it to miles so we must divide 35000 by 5280 getting an answer of 6.63 miles.

We must now plug in the variables into the equation.

Find the square root of l² and get l

Graph the approach path if the condition stated in problem 3 are satisfied .

Problem 4 . . .

From number 1 we known a= 2h -l³And we known :h= 6,63 mi

l= 64,5 mi

Substitute the value of h and l a= 2( 6,63 mi) (-64,5mi)³a= 13,26 mi -268336,125

a= - 4,9 . 10 ^-5

b=3h l²Subtitute the value of h and l

b= 3( 6,63 mi) (64,5 mi) ²b= 19,89 mi 4160,25 b= 0,00478

b= 4,78 .10^-3

We can find P(x) ,with subtitute the value of a and bP(x)= ax³+ bx²P(x)= (-4,9 .10^-5) x³ + (4,78 .10^-3) x²

64,5 mi

6,63 mi

P(x)= (-4,9 .10^-5) x³ + (4,78 .10^-3) x²

Graph

In conclusion to our project, we found l in the equation.

Thank You . . .