
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 = 40^{o}
b) WS = 20 knots; AWA = 40^{o}
c) WS = 5 knots; AWA = 40^{o}
d) WS = 10 knots; AWA = 70^{o}
e) WS = 30 knots; AWA = 90^{o}
f) WS = 20 knots; AWA = 10^{o}
g) WS = 27 knots; AWA = 60^{o}
h) WS = 3 knots; AWA = 45^{o}
i) WS = 12 knots; AWA = 20^{o}
j) WS = 48 knots; AWA = 70^{o}
k) WS = 63 knots; AWA = 25^{o}
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 (WCA_{max}) of 4^{o} 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 = 40^{o}

Step 3: Since the WS of 10 knots is the same as the 10 knots we used to calculate the maximum wind correction angle (WCA_{max}), our maximum wind correction angle for this wind speed is going to be the same: 4^{o}.


Step 4: Next we look at the table in figure 2b and see what value is closest to 40^{o}. If we use the table on the right, 40^{o} is right in the middle between 30^{o} and 50^{o}. If it's in the middle, choose the value for the next highest one which will be for 50^{o}. The factor is 3/4 or 0.75. Therefore our estimated wind correction will be 4^{o} x 0.75 = 3^{o}. 
b) WS = 20 knots; AWA = 40^{o}

Step 3: Since the WS of 20 knots is double the 10 knots we used to calculate the maximum wind correction angle (WCA_{max}), our maximum wind correction angle for this wind speed is going to be doubled: 4^{o} x 2 = 8^{o}.


Step 4: Next we look at the table on figure 2b and see what value is closest to 40^{o}. If we use the table on the right, 40^{o} is right in the middle between 30^{o} and 50^{o}. If it's in the middle, choose the value for the next highest one which will be for 50^{o}. The factor is 3/4 or 0.75. Therefore our estimated wind correction will be 8^{o} x 0.75 = 6^{o}. 
c) WS = 5 knots; AWA = 40^{o}

Step 3: Since the WS of 5 knots is half of the 10 knots we used to calculate the maximum wind correction angle (WCA_{max}), our maximum wind correction angle for this wind speed is going to be halved: 4^{o} / 2 = 2^{o}.


Step 4: Next we look at the table on figure 2b and see what value is closest to 40^{o}. If we use the table on the right, 40^{o} is right in the middle between 30^{o} and 50^{o}. If it's in the middle, choose the value for the next highest one which will be for 50^{o}. The factor is 3/4 or 0.75. Therefore our estimated wind correction will be 2^{o} x 0.75 = 1.5^{o}. You can round that up to 2^{o}. 
d) WS = 10 knots; AWA = 70^{o}

Step 3: Since the WS of 10 knots is the same as the 10 knots we used to calculate the maximum wind correction angle (WCA_{max}), our maximum wind correction angle for this wind speed is going to be the same: 4^{o}.


Step 4: Next we look at the table on figure 2b and see what value is closest to 70^{o}. If we use the table on the right, 70^{o} is going to be a factor of 1. Therefore our estimated wind correction will be 4^{o} x 1 = 4^{o}. 
e) WS = 30 knots; AWA = 90^{o}

Step 3: Since the WS of 30 knots is three times the 10 knots we used to calculate the maximum wind correction angle (WCA_{max}), our maximum wind correction angle for this wind speed is going to be tripled: 4^{o} x 3 = 12^{o}.


Step 4: Since the wind direction is 90^{o} in relation to our course the wind correction angle will be the maximum. Therefore WCA = 12^{o}. 
f) WS = 20 knots; AWA = 10^{o}

Step 3: Since the WS of 20 knots is double the 10 knots we used to calculate the maximum wind correction angle (WCA_{max}), our maximum wind correction angle for this wind speed is going to be doubled: 4^{o} x 2 = 8^{o}.


Step 4: Next we look at the table on figure 2b and see what value is closest to 10^{o}. In this case it will be for 0^{o}. The factor is 0. Therefore our estimated wind correction will be 0^{o}. 
g) WS = 27 knots; AWA = 63^{o}

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 (WCA_{max}), our maximum wind correction angle for this wind speed is going to be 2 1/2 times that: 4^{o} x (2 + 1/2) = 8^{o} + 2^{o} = 10^{o}.


Step 4: Next we look at the table on figure 2b and see what value is closest to 63^{o}. If we use the table on the right, 60^{o} is going to be a factor of 1. Therefore our estimated wind correction will be 10^{o} x 1 = 10^{o}. 
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 0^{o}. 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 (WCA_{max}), our maximum wind correction angle for this wind speed is going to be halved: 4^{o} / 2 = 2^{o}.


Step 4: Next we look at the table on figure 2b and see what value is closest to 45^{o}. If we use the table on the right, 50^{o} is going to be a factor of 3/4 or 0.75. Therefore our estimated wind correction will be 2^{o} x 0.75 = 1.5^{o}. You can round that up to 2^{o}. 
i) WS = 12 knots; AWA = 20^{o}

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 (WCA_{max}), our maximum wind correction angle for this wind speed is going to be the same: 4^{o}.


Step 4: Next we look at the table on figure 2b and see what value is closest to 20^{o}. If we use the table on the right, 30^{o} is going to be a factor of 1/2 or 0.5. Therefore our estimated wind correction will be 4^{o} x 1/2 = 2^{o}. 
j) WS = 48 knots; AWA = 70^{o}

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 (WCA_{max}), our maximum wind correction angle for this wind speed is going to be five times greater: 4^{o} x 5 = 20^{o}.


Step 4: Next we look at the table on figure 2b and see what value is closest to 70^{o}. If we use the table on the right, 60^{o} is going to be a factor of 1. Therefore our estimated wind correction will be 20^{o} x 1 = 20^{o}. 
k) WS = 63 knots; AWA = 25^{o}

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 (WCA_{max}), our maximum wind correction angle for this wind speed is going to be 6 and 1/2 more: 4^{o} x (6 + 1/2)= 24^{o}+2^{o } = 26^{o}.


Step 4: Next we look at the table on figure 2b and see what value is closest to 25^{o}. If we use the table on the right, 30^{o} is going to be a factor of 1/2 or 0.5. Therefore our estimated wind correction will be 26^{o} x 1/2 = 13^{o}. 

