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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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use the angle that makes you move parallel to the runway... –  ratchet freak May 13 '14 at 11:11
@ratchetfreak, that is fbf? fly by feels :) –  CGCampbell May 13 '14 at 12:31
@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. –  Lnafziger May 13 '14 at 12:49
@DeltaLima - Death Reckoning. Is that done by listening to really heavy metal music while hoping you're going the right direction? –  Jamiec May 13 '14 at 16:23
@Jamiec, it's flying a lead zeppelin. –  Phil Perry May 13 '14 at 16:55

2 Answers 2

up vote 2 down vote accepted

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

Now take you XWC (crosswind component). Devide 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 devided by 60 is 2, so our speed number is 2. 18kts wind devided 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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and my answer came to that: $$60*\frac{XWC}{airspeed}=\frac{XWC}{airspeed/60}$$ –  ratchet freak May 14 '14 at 17:39

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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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? –  CGCampbell May 13 '14 at 12:30
Woaw! MathJaX equations –  menjaraz May 13 '14 at 13:43
@CGCampbell, the 3.75 is your crab angle in degrees. If the crosswind is from the north, then your heading would be 064. –  fooot May 13 '14 at 14:24
Bleh. Trig. I liked your shorter answer better! –  voretaq7 May 13 '14 at 15:54
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 –  reirab May 13 '14 at 18:59

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