wind components of speed:
north -35 cos 40 = -26.8
east -35 sin 40 = -22.5
heading angle clockwise from N = h
north air speed = 220 cos h
east air speed = 220 sin h
speed made good = v
direction = 110
cos 110 = -.342
sin 110 = .940
north = v cos 110 = -.342 v
east = v sin 110 = .940 v
so
-26.8 + 220 cos h = -.342 v
-22.5 + 220 sin h = +.940 v
eliminate v to get heading h
multiply first by 2.75
-73.7 + 605 cos h = -.940 v
-22.5 + 220 sin h = +.940 v add
--------------------------
- 96.2 + 605 cos h + 220 sin h = 0
The easy way is try angles until you converge but
605 cos h + 220 sin h = 96.2
let y = sin h
then cos h = sqrt(1-y^2)
605(1-y^2)^.5 + 220 y = 96.2
(1-y^2)^.5 + .364 y = .159
(1-y^2)^.5 = .159-.364 y
1 - y^2 = .0253 - .116 y + .132 y^2
1.132 y^2 - .116 y -.975 = 0
y = .983 or -.878
y = sin h so h = 79.4 or -61.4
since we want to end up going 20 deg south of east and we are being pushed south by wind we better pick that heading of 79.4 deg east of north
let's call it 80 degrees
then remember
-22.5 + 220 sin h = +.940 v
-22.5 + 217 = .940 v
v = 207 speed made good
600 km/207 km/h = 2.9 hours
1 pm + 2.9 hours ---> 4 pm about
An paris aeroplane left Benin airport at 1:00pm to fly to Yola airport 600km away on a bearing of 110• at an airspeed of 220km/hr. On that particular day, the meteorologists reported that there existed a steady wind of 35km/hr from a bearing of 40•. What is the course of the aeroplane and it's time of arrival.
1 answer