Roses are Red, Violets are Blue. What colour is fuel 100LL?

How well do you know your fuel colours?

Now, you will need to. For the PPL exam you need to be able to identify fuel colours. In practice, the idea is you need to ensure the right fuel is used to fill your aircraft's tanks. Using the wrong fuel can prove disastrous.

In case you didn't know: Avgas is aviation gasoline and Mogas (or Autogas) is motor gasoline

Here they are (sorted by octane ratings except for jet fuel) :

RED - Grade 80 or 80/87 is a used primarily in low compression ratio engines..
GREEN - 100/130 and P87-90 is also known as MOGAS or automobile gasoline.
BLUE - Grade 100LL is a medium power output fuel with low lead content (LL).
PURPLE - 115/145 - purple, primarily military use
STRAW (light yellow or colourless) - is jet fuel or also known as Kerosene. That's right, jets and rockets run on paraffin:)

If you didn't know most fuel is colourless, but they use fuel dyes to help differentiate them.

Hey, now you won't have a problem answering the following sample PPL exam question:


What grade of fuel is in your tanks when upon your review the fuel colour is blue?

• Grade 80 or 80/87
• Grade 100
• Grade 100LL
• Kerosene

Checkout more info about avgas here.

Upper Winds (FDs) - Prelude to Nav 101

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
3421

Read: at 3000 feet, you can expect winds from 340 degrees true at 21 knots

6000
3526-09

Read: 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 degrees

Similarly, 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 be

• 150/15
• 150/8
• 120/15
• 140/8

In 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.