A commenter on a post regarding the PPL tests had trouble with the following question:

The average wind applicable to a direct flight from Winnipeg (CYWG) to Brandon (CYBR) at 5,500 ft would be

(1) 290°M at 30 kt.

(2) 290°T at 30 kt.

(3) 310°M at 31 kt.

(4) 310°T at 31 kt.FDCN CWAO 061920

ISSUED 1200Z 07 FEB 2007 FOR USE 6-17Z

3000 6000 9000 12000 18000 24000 YWG 2825 2728-07 2932-10 2935-15 2939-26 2841-38 YBR 3030 3132-06 3133-10 3135-15 3041-28 2948-40 YYQ 3529 3428-13 3229-14 3130-19 3032-32 2733-42 YYL 3327 3435-10 3338-14 3337-19 3136-31 3038-44

He got as far as interpolating the wind speeds and directions at 4,500 feet by averaging the respective values but did not know how to proceed after that. Here is how I solved it:

The pertinent information is as follows:

3000 | 6000 | |

YWG | 2825 | 2728-07 |

YBR | 3030 | 3132-06 |

Between 6,000’ and 3,000’ there is a difference of 3,000’. The altitude required is 5,500’ which is 2,500’ above 3,000’ (or 500’ below 6,000’ depending how you look at it. For the purposes of this example I will be using the 2,500’). There are two steps to this problem:

- Find wind speed and direction at the required altitude at the respective airports.
- Average them to find the the average wind.

#### Step 1: Finding wind speed and direction at the respective airports

I will show you the calculation for YWG. Follow the same steps to find the values at YBR.

Wind Direction Adjustment = (270 – 280) * 2500 / 3000 = – 8.333…

Wind Speed Adjustment = (28 – 25) * 2500 / 3000 = 2.5

Wind Direction = 280 + –8.333… = 271°

Wind Speed = 25 + 2.5 = 27.5 kt

At YBR the wind direction would be 308° and wind speed would be 32 kt.

** Explanation:** Finding the wind at non-reported levels is simply a matter of properly interpolating the information you are provided. In this case I used a weighted average to give me the values I needed. One thing to keep in mind with this calculation is that order is important. The order I followed was going from 6,000’

*(I took the values here)*to 3,000’

*(subtracted these values)*then back up to 5,500’

*(multiplied by the weight, 2,500/3,000 and added to the values at 3,000’)*.

Another way to think about it is you will be going 2,500 / 3,000 = 83% of the way from 3,000’ to 6,000’ so you add 83% of the difference between measurements at the different altitudes to the values at the lower altitude (or you subtract 17% of the difference from the values at the higher altitude).

#### Step 2: Average wind

Average wind direction = (308 + 271) / 2 = 289.5

Average wind speed = (32 + 27.5) / 2 = 29.75

With rounding, the applicable wind is from 290 at 30 kts. Keeping in mind that the FD reports directions in degrees TRUE leads to the correct answer being **(2) 290°T at 30 kt.**

Notice that these calculations assume the change from one altitude to another is linear which may not always be the case.

Frank Ch. EiglerWhat an unfortunate question. In practice, the inaccuracies

of the measurements and the unreality of the linear interpolation

assumption will render the difference between 290/30 and 310/31

absolutely moot.

MichelleThe question is testing if you can read the winds aloft correctly and if you understand that they are always reported in degrees true.

Answers 1 and 3 are wrong because the winds are reported as magnetic.

Use the 6000 feet values, which are close enough to the flight altitude 5,500 feet.

Average winds are approximately an average of 28 kts @ 270 degrees and 32 kts @ 310 degrees true

Answer 4 cannot be correct because it’s 310 degrees.

Answer 2 remains, and makes sense because 290 degrees is halfway between 270 degrees and 310 degrees and 30 kts halfway between 28 kts and 32 kts.