a pilot wishes to fly on a course bearing 45 and with a ground speed of 600 km per hour. If a wind is blowing from the southwest (bearing 315) at 80 km per hour. What must be the heading and air speed of the aircraft?
3 answers
since the wind and the desired course are in the same direction, the plane just needs to have an air speed of 520 km/hr at 45
Vp + 80km/h[315] = 600[45o].
Vp + 80*sin315 + i80*Cos315 = 600*sin45 + i600*Cos45,
Vp - 56.6 + i56.6 = 424.3 + i424.3,
Vp = 480.9 + i367.7 = 605.4km/h[52.6o] = Air speed and heading of the plane.
Vp + 80*sin315 + i80*Cos315 = 600*sin45 + i600*Cos45,
Vp - 56.6 + i56.6 = 424.3 + i424.3,
Vp = 480.9 + i367.7 = 605.4km/h[52.6o] = Air speed and heading of the plane.
Vp + 80[45o] = 600[45o].
Vp + 80*sin45 +i80*Cos45 = 600*sin45 + i600*Cos45,
Vp + 56.6 + i56.6 = 424.3 + i424.3,
Vp = 367.7 + i367.7 = 520 km/h[45o]. = Air speed and heading of the plane.
Note: If the wind is blowing from southwest, the bearing is 225o Not 315o.
Vp + 80*sin45 +i80*Cos45 = 600*sin45 + i600*Cos45,
Vp + 56.6 + i56.6 = 424.3 + i424.3,
Vp = 367.7 + i367.7 = 520 km/h[45o]. = Air speed and heading of the plane.
Note: If the wind is blowing from southwest, the bearing is 225o Not 315o.
1 answer