Doubts of wings of fire
Q1) Apply 1 in 60 rule. 5.5=[(5x60)/d]
where 5.5 = error in vor
5=airway width in uk
d=half the max. distance of the 2 beacons.
Solving which we get d=54.5nm.
Therefore max. spacing=2d=109nm.
Q2) Apply 1 in 60 rule again.
Error=[(5x60)/100]=3 degrees.
Where error is the angle deviated from centreline, the aircraft being on the air way boundary which is 5nm i.e. airway width in uk.
100 is the distance of the beacon from the aircraft.
Given or even generally known that in OBS 1 dot is 2 degrees.
Therefore 3 degrees equals 1.5 dots.
Q3) Same as Q1)
Q4) Range=1.25(Square root of h1 + Square root of h2)
h1 is height of transmitter above msl in feet.
h2 is height of the aircraft above msl in feet.
Hence Range=1.25(Square root of 400 + Square root of 42000)=281nm.