frumpy Posted January 5, 2017 Report Share Posted January 5, 2017 Hi guys, once I learned an easy rule of thumb for calculating the wind correction angle. It was so easy, that I completly forgot about it. Assume you land on runway 27 (270 deg), wind is 10 kts from 230 deg. So wind is 40° from the side. Sinus(40)*10 = 6.4, that means I aim for ~264°? There are many formulas out there, like CWC/ nm/min and stuff, but I am looking for one which incorporates a factor estimation based on sinus... something easy to calculate. Quote Link to comment Share on other sites More sharing options...
mmerelles Posted January 13, 2017 Report Share Posted January 13, 2017 On 5/1/2017 at 4:06 AM, frumpy said: Hi guys, once I learned an easy rule of thumb for calculating the wind correction angle. It was so easy, that I completly forgot about it. Assume you land on runway 27 (270 deg), wind is 10 kts from 230 deg. So wind is 40° from the side. Sinus(40)*10 = 6.4, that means I aim for ~264°? There are many formulas out there, like CWC/ nm/min and stuff, but I am looking for one which incorporates a factor estimation based on sinus... something easy to calculate. i do not know the formula you are trying to get, but any correction formula that does not include aircraft air speed will be incorrect. The amount of correction to stay on track requires wind intensity and wind direction BUT ALSO the aircraft airspeed. The slower the aircraft is flying the more correction that will be required not to deviate from desired trajectory. Quote Link to comment Share on other sites More sharing options...
frumpy Posted January 16, 2017 Author Report Share Posted January 16, 2017 Thanks for stepping in. You are right, I have to redefine my question - I got confused and mixed things up. So... Wind Correction Angle = Crosswindcomponent / (True Airspeed/60) For calculating CWC, sinus is approximated: 90° = 1 (direct crosswind, no headwind) 80° = 1 70° = .9 60° = .8 50° = .7 40° = .6 30° = .5 20° = .4 10° = .2 0° = 0 (direct headwind) Lets say we approach runway 26 with a Cessna (60 kts!), wind is 230/16: Wind angle is 30° off, which is .5*16 = 8, so final course is 252°. Pretty easy, CWC=WCA. WCA for 120kts: We are faster than the Cessna, so wind don't bother us that much. TAS/60 = 2, so WCA is .5*8/2 = 4, final course is 256°. All I need to remember is, that the sinus approximation is +.2 for windangle/10, except for the upper and lower two of the angles (90, 80 and 0, 10°) - which is 1, 1 and .2, 0. I guess that comes with practice. Quote Link to comment Share on other sites More sharing options...
poodster Posted January 16, 2017 Report Share Posted January 16, 2017 You can also download the e6b app and use the true and actual track tool. It incorporates everything you are using in your formula, but makes it easier by getting rid of the math. It's also quicker to use Quote Link to comment Share on other sites More sharing options...
frumpy Posted January 17, 2017 Author Report Share Posted January 17, 2017 I don't have a smartphone and I believe too much reliance on automation will diminish my ability to do simple math. Quote Link to comment Share on other sites More sharing options...
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