This document discusses how to determine the optimal descent path for an airplane. It presents three problems: 1. Deriving a cubic polynomial equation to model the descent path based on conditions of starting altitude, touchdown point, and constant horizontal speed. 2. Showing that the vertical acceleration must be below a maximum value based on the horizontal speed and distance from touchdown. 3. Calculating the distance from the airport at which descent should begin, given constraints on maximum vertical acceleration, starting altitude, and airspeed. The document solves each problem, derives the relevant equations, and provides an example graph of a descent path satisfying the given conditions.

1. 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

2. (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 ...

3. 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

4. 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

5. 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
h
2
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

6. We get :
2
2
3
3
32
L
hx
L
hx
+−
2
2
3
3
32
L
hx
L
hx
+−

7. 2) Use conditions 2 and 3 to show that
k
l
hv
≤2
2
6
k
l
hv
≤2
2
6

8. Answer
Condition (ii)
dx =-v
dt
so, x(t)=l –vt
Condition (iii)
| d²y | ≤ k
dt²
P(x)= ax³ +bx²
P’(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

9. 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 . . .

10. Work:
known:k =860 mi/h²
h= 35000 ft
change ft to mi
h=35000ft X 1 mi
5280 ft
h= 6,63 mi
v= 300 mi/h
Answer:
6hv²/l² ≤ k
6 [6,63 mi . (300mi/h)²] ≤ 860 mi/h²
l²
6 [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 ≤ l²
64,5 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

11. Graph the approach path if the condition stated
in problem 3 are satisfied .
Problem 4 . . .

12. From number 1 we known
a= 2h
-l³
And we known :h= 6,63 mi
l= 64,5 mi
Substitute the value of h and l
a= 2( 6,63 mi)
(-64,5mi)³
a= 13,26 mi
-268336,125
a= - 4,9 . 10 ^-5
b=3h
l²
Subtitute the value of h and l
b= 3( 6,63 mi)
(64,5 mi) ²
b= 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 b
P(x)= ax³+ bx²
P(x)= (-4,9 .10^-5) x³ + (4,78 .10^-3) x²

13. 64,5 mi
6,63 mi
P(x)= (-4,9 .10^-5) x³ + (4,78 .10^-3) x²
Graph

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

15. Thank You . . .