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Mentally Calculating the Wind Correction Angle (cont.)

III - Applying the method

Example:

If your typical cruising true airspeed is 130 kts for your aircraft. Estimate the wind correction angle (WCA) for the following conditions:

     a) WS = 10 knots; AWA = 40o
     b) WS = 20 knots; AWA = 40o
     c) WS = 5 knots; AWA = 40o
     d) WS = 10 knots; AWA = 70o
     e) WS = 30 knots; AWA = 90o
     f) WS = 20 knots; AWA = 10o
     g) WS = 27 knots; AWA = 60o
     h) WS = 3 knots; AWA = 45o
     i) WS = 12 knots; AWA = 20o
     j) WS = 48 knots; AWA = 70o
     k) WS = 63 knots; AWA = 25o

Solution:

  Step 1: We know the typical cruising speed is 130 kts TAS.
 
  Step 2: Calculate the maximum wind correction angle using the formula below for a wind speed (WS) of 10 knots:

substituting 130 knots for TAS and 10 knots for WS we have:

This maximum correction angle result (WCAmax) of 4o will be the basis for all our estimates of wind correction angles (WCA) when the aircraft's true airspeed is around 130 knots.

Estimating each part using the mental calculation method:

a) WS = 10 knots; AWA = 40o

  Step 3: Since the WS of 10 knots is the same as the 10 knots we used to calculate the maximum wind correction angle (WCAmax), our maximum wind correction angle for this wind speed is going to be the same: 4o.
 
  Step 4: Next we look at the table in figure 2b and see what value is closest to 40o. If we use the table on the right, 40o is right in the middle between 30o and 50o. If it's in the middle, choose the value for the next highest one which will be for 50o. The factor is 3/4 or 0.75. Therefore our estimated wind correction will be 4o x 0.75 = 3o.


b) WS = 20 knots; AWA = 40o

  Step 3: Since the WS of 20 knots is double the 10 knots we used to calculate the maximum wind correction angle (WCAmax), our maximum wind correction angle for this wind speed is going to be doubled: 4o x 2 = 8o.
 
  Step 4: Next we look at the table on figure 2b and see what value is closest to 40o. If we use the table on the right, 40o is right in the middle between 30o and 50o. If it's in the middle, choose the value for the next highest one which will be for 50o. The factor is 3/4 or 0.75. Therefore our estimated wind correction will be 8o x 0.75 = 6o.

c) WS = 5 knots; AWA = 40o

  Step 3: Since the WS of 5 knots is half of the 10 knots we used to calculate the maximum wind correction angle (WCAmax), our maximum wind correction angle for this wind speed is going to be halved: 4o / 2 = 2o.
 
  Step 4: Next we look at the table on figure 2b and see what value is closest to 40o. If we use the table on the right, 40o is right in the middle between 30o and 50o. If it's in the middle, choose the value for the next highest one which will be for 50o. The factor is 3/4 or 0.75. Therefore our estimated wind correction will be 2o x 0.75 = 1.5o. You can round that up to 2o.


d) WS = 10 knots; AWA = 70o

  Step 3: Since the WS of 10 knots is the same as the 10 knots we used to calculate the maximum wind correction angle (WCAmax), our maximum wind correction angle for this wind speed is going to be the same: 4o.
 
  Step 4: Next we look at the table on figure 2b and see what value is closest to 70o. If we use the table on the right, 70o is going to be a factor of 1. Therefore our estimated wind correction will be 4o x 1 = 4o.


e) WS = 30 knots; AWA = 90o

  Step 3: Since the WS of 30 knots is three times the 10 knots we used to calculate the maximum wind correction angle (WCAmax), our maximum wind correction angle for this wind speed is going to be tripled: 4o x 3 = 12o.
 
  Step 4: Since the wind direction is 90o in relation to our course the wind correction angle will be the maximum. Therefore WCA = 12o.


f) WS = 20 knots; AWA = 10o

  Step 3: Since the WS of 20 knots is double the 10 knots we used to calculate the maximum wind correction angle (WCAmax), our maximum wind correction angle for this wind speed is going to be doubled: 4o x 2 = 8o.
 
  Step 4: Next we look at the table on figure 2b and see what value is closest to 10o. In this case it will be for 0o. The factor is 0. Therefore our estimated wind correction will be 0o.

 

g) WS = 27 knots; AWA = 63o

  Step 3: Round the WS to the nearest multiple of 5, so for a WS of 27 knots, we will calculate as if the WS was 25 knots. Since the WS of 25 knots is 2.5 or 2 1/2 times the 10 knots we used to calculate the maximum wind correction angle (WCAmax), our maximum wind correction angle for this wind speed is going to be 2 1/2 times that: 4o x (2 + 1/2) = 8o + 2o = 10o.
 
  Step 4: Next we look at the table on figure 2b and see what value is closest to 63o. If we use the table on the right, 60o is going to be a factor of 1. Therefore our estimated wind correction will be 10o x 1 = 10o.

 

h) WS = 3 knots; AWA = 45

  Step 3: For wind speed less the 5 knots I would probably just ignore them and assume the WCA is 0o. However, we were to use this method, round the WS to the nearest multiple of 5, so for a WS of 3 knots, we will calculate as if the WS was 5 knots. Since the WS of 5 knots is half of the 10 knots we used to calculate the maximum wind correction angle (WCAmax), our maximum wind correction angle for this wind speed is going to be halved: 4o / 2 = 2o.
 
  Step 4: Next we look at the table on figure 2b and see what value is closest to 45o. If we use the table on the right, 50o is going to be a factor of 3/4 or 0.75. Therefore our estimated wind correction will be 2o x 0.75 = 1.5o. You can round that up to 2o.


i) WS = 12 knots; AWA = 20o

  Step 3: Round the WS to the nearest multiple of 5, so for a WS of 12 knots, we will calculate as if the WS was 10 knots. Since the WS of 10 knots is the same as the 10 knots we used to calculate the maximum wind correction angle (WCAmax), our maximum wind correction angle for this wind speed is going to be the same: 4o.
 
  Step 4: Next we look at the table on figure 2b and see what value is closest to 20o. If we use the table on the right, 30o is going to be a factor of 1/2 or 0.5. Therefore our estimated wind correction will be 4o x 1/2 = 2o.

 

j) WS = 48 knots; AWA = 70o

  Step 3: Round the WS to the nearest multiple of 5, so for a WS of 48 knots, we will calculate as if the WS was 50 knots. Since the WS of 50 knots is five times the 10 knots we used to calculate the maximum wind correction angle (WCAmax), our maximum wind correction angle for this wind speed is going to be five times greater: 4o x 5 = 20o.
 
  Step 4: Next we look at the table on figure 2b and see what value is closest to 70o. If we use the table on the right, 60o is going to be a factor of 1. Therefore our estimated wind correction will be 20o x 1 = 20o.

 

k) WS = 63 knots; AWA = 25o

  Step 3: Round the WS to the nearest multiple of 5, so for a WS of 63 knots, we will calculate as if the WS was 65 knots. Since the WS of 65 knots is 6 and 1/2 times the 10 knots we used to calculate the maximum wind correction angle (WCAmax), our maximum wind correction angle for this wind speed is going to be 6 and 1/2 more: 4o x (6 + 1/2)= 24o+2o = 26o.
 
  Step 4: Next we look at the table on figure 2b and see what value is closest to 25o. If we use the table on the right, 30o is going to be a factor of 1/2 or 0.5. Therefore our estimated wind correction will be 26o x 1/2 = 13o.

 

  
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