The document discusses how air traffic control (ATC) calculates the minimum safe distance between two aircraft that are crossing paths at an angle. It explains that ATC uses Pythagorean theorem and calculus, setting the derivative of the distance function equal to zero, to find the point of minimum distance. The solution is that the minimum distance occurs when the two aircraft pass each other at the midpoint between their positions. Some key equations and the final solution of minimum separation distance are also presented.

1. Delhi Public School
Vasant Kunj

2. I

4. 1000 feet 2000 feet Longitudinal 2Lateral
Based on time Based on distance

6. How ATC calculates minimum distance
between two aircrafts when they are
crossing at an angle, suppose 90:
In this case we apply Pythagoras theorem and calculus
using the property that if the minima of function is at a point then its
differential gives value =0 at a
S(x) = [(D1-x)2 + (D2-x)2 ]1/2
For minimum distance
dS = 0
dx
= > d[(D1-x)2 + (D2-x)2 ]1/2 =0
dx
= > [-1/2.{(D -x)2 + (D -x)2 }] ′ * [{(D -x)2 + (D -x)2 )}(-1/2)] =

7. The expression [{(D1-x)2 + (D2-x)2 )}(-1/2)] can not be zero ∀ x R , so for ds/dx
=0 , it requires that :
[(D1-x)2 + (D2-x)2 ]’ = 0
=>( D12-2D1x+x2 + D22-2D2x+x2) ′ = 0
=> ( -2D1 +2x - 2D2 + 2x ) = 0
=> 4x = 2 ( D1+D2)
x = ( D1+D2)
2

8.
S(x=(D1+D2)/2)) = [{D1-(D1+D2)}2 + {D2-(D1+D2)}2 ]1/2
2 2
=[ (D1-D2)2 + (D2-D1)2]1/2
(2)2 (2)2
(D1-D2)2= (D2-D1)2
S = [2 . ( D1-D2)2 ](1/2)
4
S= [ (D1-D2)2]1/2=[(D1-D2)2]1/2=
2 2
S= (D1-D2)
√2

10.  If D1 and D2 are the distances of two planes
from a crossing point on tracks that form an
angle of α and we want to calculate how much
their future minimum separation Smin will be ,
then
 Smin = [0.7+0.03*(90-α)]*(D1-D2)

13. D = (60 nm)X
tan(a)= (60 nm) X[ 0.0174a + (0.0174a)2 + (0.0174a )3 + .....)] =
If x 60X0.017aNM approximately =1.044a NM 3 + ….
is in radian, then tan x = x + x 2 + x
But or for a=1 degree it is1 NM practically.
1 radian= 0.0174 degrees
“a”degrees = 0.0174 a radian
Therefore 1 degree at 60 NM is about 1 NM
In a distance of 60 nm ATC create a track deviation of 1 nm for
every 1 degree of vectoring angle. This is thumb rule used by the
controller.

21. Emirates flight is maintaining at 35000 ft. And Russian airbus is cleared by
Consequently, the pilotsft/min= 1200/60=20ft/sec. Therefore, -dR/dtto avert the
ATC to ascend at 1200 of both flights change their path in order = 20 ft./sec
collision.pilots mistakes it as clearance for 36000 ft. The closure separation of
But the Thanks to TCAS
the flights is recorded by TCAS as 700 ft.
As soon as the flight reaches FL 350, TCAS alarm starts blaring. Therefore,
TCAS calculates Tao = R/(-dR/dt) =700/20=35 seconds. Resolution Advisories
(RA) is issued to conflicting aircraft 35 sec before predicted collision.
TCAS
Intruder
35000 ft.
35 seconds
34000 ft.
January 27, 2013 21

23. "I bite… If the pilot tries to touch any control"

25.  Questi
ons ?

26.  PRESENTED BY:
 PULKIT JAISWAL
 AYUSH SINGH
 ADARSH BAKSHI
 MENTORED BY:
 MS. HIMANI ASIJA
 P.G.T. MATHEMATICS