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I am wondering what is the rule of thumb with correction of the bat?

Is it 2° for every 5kts?

Now i know you can use a E6B. But im talking about landing ILS or visuals without bracketing the approach.

So lets say you landing on runway 06 the magnetic heading is 060, the wind is 080/05kts. So would the correction be 062?

The aircraft is a C172

Apologies if this is not a valid question, getting back into aviation

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    $\begingroup$ use the angle that makes you move parallel to the runway... $\endgroup$
    – ratchet freak
    Commented May 13, 2014 at 11:11
  • $\begingroup$ @ratchetfreak, that is fbf? fly by feels :) $\endgroup$
    – CGCampbell
    Commented May 13, 2014 at 12:31
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    $\begingroup$ @CGCampbell No, it is trial and error. The wind is constantly changing anyway, so you need to make constant corrections to compensate for the wind. $\endgroup$
    – Lnafziger
    Commented May 13, 2014 at 12:49
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    $\begingroup$ the calculation of crosswind correction angles is useful for dead reckoning in cruise. For landing you would fly whatever keeps you aligned with the runway / ILS. The wind you encounter during the approach will typically be different from the reported surface wind. $\endgroup$
    – DeltaLima ♦
    Commented May 13, 2014 at 12:50
  • $\begingroup$ I thought as much. Remember chatting to some one and they mentioned the 2° to 5kts. Thank every one $\endgroup$
    – chaos505
    Commented May 13, 2014 at 13:05

6 Answers 6

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What I have learned during my commercial flight training is making use of the speed number. Take your TAS, divide it by 60. This is your speed number.

Now take you XWC (crosswind component). Divide the XWC by your speed number. This is the amount of degrees you should crab to stay on track (wind correction angle)

Lets use an example:

We are flying in a C172 at 120kts TAS. XWC is 18kts from the left. 120 divided by 60 is 2, so our speed number is 2. 18kts wind divided by 2 is 9. Now adjust your heading by 9 degrees to the left (into the wind), and you should stay on track.

Worked perfectly fine for me so far. Hope it helps! Cheers

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  • $\begingroup$ and my answer came to that: $$60*\frac{XWC}{airspeed}=\frac{XWC}{airspeed/60}$$ $\endgroup$
    – ratchet freak
    Commented May 14, 2014 at 17:39
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    $\begingroup$ It helps explain that "divide by 60" if you call "speed number" as "number of miles per minute". ie 120 kts is about 2 miles a minute (ie. 120/60 => 2). makes it easier to remember as "cross wind divided by number of miles a minute" $\endgroup$
    – Radu094
    Commented Feb 21, 2015 at 12:40
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    $\begingroup$ @Radu094: This may help remember the rule, but it doesn't actually explain anything. (Mathematically the number shouldn't be 60, but 57.3 which is the number of degrees in a radian -- but that much precision is not warranted here). $\endgroup$
    – hmakholm left over Monica
    Commented Feb 21, 2015 at 19:00
  • $\begingroup$ Well, these days, when every pilot has at least a hand-held GPS unit at a minimum, wouldn't you just steer whatever heading gives you the desired ground track? $\endgroup$
    – quiet flyer
    Commented May 8, 2020 at 20:36
  • $\begingroup$ @quietflyer Even better, since it gives you your current Track, the difference between it and your heading is the WCA, which you can directly apply to your desired course. $\endgroup$
    – Steven
    Commented Nov 13, 2022 at 6:20
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If we set our runway to be aligned to a $x$ axis so the angle is $0°$ and we have an airspeed of $\vec{v_a}$ and a wind of $\vec{v_w}$, this means that ground speed is $\vec{v_g} = \vec{v_a} - \vec{v_w}$.

We want the $y$ component of $\vec{v_g}$ to be 0 so this means that the $y$ components of $\vec{v_g}$ and $\vec{v_w}$ must cancel each other out.

The $y$ component of the wind is our crosswind ($v_c$). To get the $y$ component of our airspeed we take $|\vec{v_a}|\cdot\tan \theta$ where $\theta$ is our heading (0 is parallel to the runway).

This means that $|\vec{v_a}|\tan \theta - v_c=0$ or that $\tan \theta = v_c/|\vec{v_a}|$.

At low crosswind speeds this means that your crab angle in degrees is $\sim 60*\frac{crosswind}{airspeed}$.

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    $\begingroup$ Bad at math... if crosswind is 5kts and airspeed is 80kts, 5/80 is .0625, heading is 060, so 60 * .0625 is 3.75, so my crab angle should be ~ 64 degrees? $\endgroup$
    – CGCampbell
    Commented May 13, 2014 at 12:30
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    $\begingroup$ Woaw! MathJaX equations $\endgroup$
    – menjaraz
    Commented May 13, 2014 at 13:43
  • $\begingroup$ @CGCampbell, the 3.75 is your crab angle in degrees. If the crosswind is from the north, then your heading would be 064. $\endgroup$
    – fooot
    Commented May 13, 2014 at 14:24
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    $\begingroup$ Bleh. Trig. I liked your shorter answer better! $\endgroup$
    – voretaq7
    Commented May 13, 2014 at 15:54
  • $\begingroup$ My flight instructor once told me about an incident where an examiner asked him what the correct angle would be for a given crosswind scenario. He simply stated the answer without hesitation. Little did the examiner know that he's a math professor. - lol $\endgroup$
    – reirab
    Commented May 13, 2014 at 18:59
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For VMC approaches, just fly whatever tracks the extended centerline. There should be no need to look at your HSI, heading bug, etc. other than to make sure you're landing on the correct runway.

For IMC approaches, take the crosswind and divide it by the number of miles per minute you're traveling. You find this by dividing your TAS by 60 or just using your mach number.

Miles/Minute = MachNumber * 10
OR
Miles/Minute = TAS / 60

Drift correction = Crosswind / (MilesPerMinute)

This will get you in the general ballpark. What you should do is then bug this heading and see how it's working for you. If the localizer is swinging one way or the other, then make a 1-2 degree correction to avoid chasing. Rebug this heading and see how that adjustment works.

Or just use the flight director...

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Turn into the wind so you continue on a course parallel to (better directly over) landing runway. Eyeball. No computation required. As you flare, kick rudder to swing nose over runway centerline as you descend. If you time it correctly, you'll land rolling forward down the runway rather than taxiing off across it.

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  • $\begingroup$ This is the method preferred by pilots of airliners, while pilots of light planes often prefer to use the wing-down method to have a stabilized no-drift situation all the way to touchdown. Especially pilots of taildraggers. $\endgroup$
    – quiet flyer
    Commented May 8, 2020 at 20:38
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Just look out of the window, crab down the centre line and kick off the drift at the flare. No need for numbers. I gather this doesn’t work so well in airliners.

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  • $\begingroup$ Welcome to aviation.SE. I fail to see what your answer adds to already existing answers. Could you please expand it so that it is not a simple repetition of already existing answers. $\endgroup$
    – Manu H
    Commented Feb 16, 2019 at 11:30
  • $\begingroup$ @Graham no it is the opposite, this is the method preferred by pilots of airliners, while pilots of light planes often prefer to use the wing-down method to have a stabilized no-drift situation all the way to touchdown. Especially pilots of taildraggers. $\endgroup$
    – quiet flyer
    Commented May 8, 2020 at 20:37
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People are getting way too technical here. You’ll do this kind of the computations for wind correction in cruise flights, but on final approach, that’s all done by visual reference and visually tracking the aircraft down the centerline of the runway during the approach. Once you enter the round out it’s only enough rubber pressure to keep the nose pointed at the far end of the runway, and enough aileron pressure to hold the aircraft on the center line on the runway. That’s it.

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