I started a post on navigation and flight planning. After a few pages of content I decided I needed to break it down into smaller more manageable parts and to explain a few critical components.
Understanding of upper wind forecasts is one of them.
Here is a typical chart you can expect to see on Navcanada site. You will use this data in your flight planning to calculate headings, fuel burn, time, airspeed, drift etc.
| STN VBI - 49.5N 94.1W | for use | 3000 | 6000 | 9000 | 12000 | 18000 |
| FDCN01 CWAO FCST BASED ON 141200 DATA VALID 141800 | 17-21 | 3421 | 3526-09 | 3317-13 | 3217-18 | 3318-31 |
| FDCN02 CWAO FCST BASED ON 141200 DATA VALID 150000 | 21-06 | 3424 | 3616-12 | 3419-15 | 3324-18 | 3530-30 |
| FDCN03 CWAO FCST BASED ON 141200 DATA VALID 151200 | 06-17 | 3416 | 3420-12 | 3527-14 | 3425-18 | 3226-30 |
The first two columns are very intuitive so you should be able to figure them out...
Here is the important stuff:
3000
3421Read: at 3000 feet, you can expect winds from
340 degrees true at
21 knots
6000
3526-09Read: at 6000 feet, you can expect winds from
350 at
26 knots and temperature of minus
9 degrees (temperature is not given for altitudes at or below 3000 feet)
So far so good. The challenge?
What if you chose your flight altitude to be at 5500 feet? How will you find the wind direction and wind velocity...
Before you start the calculations apply some other knowledge namely about nature of
veering and
backing winds.
The rule is:
- Veering is when the wind direction changes in a clockwise direction and with increasing altitude velocity of the wind also increases (due to lowering of surface friction)
- Backing is when the wind direction changes in a counter-clockwise direction and with decreasing altitude velocity of the wind also decreases (due to increasing surface friction)
In this case, you can expect the wind direction to veer from 3000 to 6000 feet, with the wind direction to turn clockwise and velocity to increase. The answer will be anywhere between 340 and 350 degrees with speeds of 21 to 26 knots. Anything outside of this range signals an error in the calculation.
Here is a practical calculation method to get your info for 5500 feet. First, pick an altitude right in the middle between those forecasts in this case 4500 feet.
The new wind direction becomes 345 = (350-340)/2, the new wind velocity becomes 24= 26+21/2 (rounded up)
Now you are still 1500 feet away from 6000 and 1000 from your desired altitude of 5500.
What you should do now is calculate for the additional 1500 feet first.
5 degrees for additional 1500 feet (350-345)
divide by 3 to get to
500 feet= 1.6 degrees
multiply by 2 to get to
1000 feet=3.2 degrees (4 degrees rounded up)The wind direction at 5500 feet is 345+4=
349 degreesSimilarly, you can do the same for the wind velocity
2 kt divided by 3= 0.67
multiply by 2 = 1.34 (2 rounded up)Therefore, rounded to the nearest integer the new wind velocity becomes 24+2=
26 kt
The final answer is : at 5500 feet, you can expect winds from
349 degrees true at
26 knots
Here is another table to help you interpret the data.
| EXAMPLE | DECODED |
| 9900 + 00 | Wind light and variable, temperature 0˚C |
| 2625 | 260˚ true at 25KT |
| *791159 | 290˚ true (79 - 50 = 29) at 111 KT (11 + 100 = 111), temperature -59˚C |
| *859950 | 350˚ true (85 - 50 = 35) at 199 KT or greater, temperature -50˚C |
*Encoded wind speeds from 100 to 199 kt have 50 added to the direction code and 100 subtracted from the speed.
For more info about FD's go here...
You are now ready to answer the following two sample PPL exam questions.
At 1200Z one could expect the winds at 5,500 feet to be
- 340 degrees magnetic at 24 knots.
- 349 degrees true at 26 knots.
- 350 degrees true at 29 knots.
- 349 degrees magnetic at 26 knots.
- 340 degrees magnetic at 24 knots.
With the wind at the surface being reported at 120/10, one could reasonably expect the wind at 3,000 feet AGL to beIn the upcoming posts I will cover other areas important for your Navigation section of the exam.
Let me know if you have any specific requests.