Class Exercise 9/22/00
 
 

You want to fly due North over the ground. Your airspeed is 100 kts and the wind is from the SE at 25 kts. What heading must you fly and what is your groundspeed?

Here's the vector diagram:

The red vector has known magnitude (25 kts) and known direction, as shown. The blue vector has known magnitude (100 kts) but unknown direction- that is one of the desired answers. We can call the angle marked with the black arc q . The black vector is the velocity over the ground. Its length (call it v) is also one of our desired answers.

Now from the sine rule:
 
 

sinq /25 kts = sin 45° /100 kts

sinq =1/(4Ö 2)

q = 10°



Then from the cosine rule:

v2 = (100 kt)2 + (25 kt)2 - 2(100 kt)(25 kt)cos(180° -45° -10° )

= 1350 (kt)2 and thus

v = 116 kt



So you must head 10° East of North and your groundspeed will be 116 kts.