PPRuNe Forums - View Single Post - Wind Triangle Vectors & Curved Trajectories
Old 10th September 2010 | 06:40
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Badmachine
 
Joined: Jul 2008
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From: Las Vegas, NV
One example would be:

P (plane): approximate compass bearing 209° (traveling approximately southwest) at 272 f/s (185 mph); W (wind): traveling south at 32 f/s (22 mph).

Plane and wind vector components represented by ordered pairs:

P = [272 f/s cos(241°), 272 f/s sin(241°)] = -131.8, -237.9
W = [32 f/s cos(270°), 32 f/s sin(270°)] = 0, -32

-131.8 + 0 = -136
-237.9 + (-32) = -269.9

Resolved components substituted into Pythagoreans theorem for resultant speed:

||P + W|| = 131.8² + 269.9² = 90,2171/2 = 300.4 f/s (205 mph)

Resolved components substituted for resultant bearing:

tan −1(269.9/131.8) = 64°; (90° - 64°) + 180° = 206°

Deflection angle = 209° - 206° = 3°

Ground track displacement = 3°tan(300.4 f/s) = 15.7 f/s
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