I am preparing to my ATPL theory exams now, and, of course, I am making records. Here I will put some of them, probably somebody will find it helpful.

First of all, I’d like to leave a note about a number of questions and each subject exam duration. Here they are:

Air Law – 44 questions, 1 hour;

Aircraft General Knowledge (Airframe, Systems Engines and Electrics) – 80 questions, 2 hours;

Instrumentation – 60 questions, 1 hour 30 minutes;

Mass and Balance – 25 questions, 1 hour;

Performance – 35 questions, 1 hour;

Flight Planning and Monitoring – 43 questions, 2 hours;

Human Performance and Limitations – 48 questions, 1 hour;

Meteorology – 84 questions, 2 hours;

General Navigation – 60 questions, 2 hours;

Radio Navigation – 66 questions, 1 hour 30 minutes;

Operational Procedures – 45 questions, 1 hour 15 minutes;

Principles of Flight – 44 questions, 1 hour;

VFR Communication – 24 questions, 30 minutes;

IFR Communication – 24 questions, 30 minutes.

Each one has some difficulties, but some of them became my personal nightmares. For example, Air Law, Operation Procedures and General Navigation. The first two did because of huge amount of information to memorize, and the latter is because of timing: a lot of computations with only about 2 minutes per question.

A lot of subjects require mainly knowledge. The only way to pass is reading the books. But some of them, for example, navigation, allows to solve a huge part of questions only by knowing a proper method. That’s why I decided to start this page with navigation problems.

Imagine that we have the following problem:

Course required: 085° (T)

Forecast W/V: 030/100 kt

TAS: 470 kt

Distance: 265 NM

Probably we can use a flight computer, but I do not prefer it in this kind of tasks because it can be solved with a simple sketch and 2-3 formulae. So, firstly let’s draw it:

I did not do it in scale, but it does not matter for computation. We have our final track α (black angle around A, 85°). Wind angle is γ (green angle around point C, 30°), and wind speed is CB (green arrow). We also know AC (470 knots, it’s our speed vector without wind correction). We have to find AB (final speed value after wind correction) and ε, angle between AC and vertical axis (which is our required HDG).

We can see that δ = β = 90° – 85° = 5° (CM ⊥ AM, CE ⊥ AE so δ = β).

CE = wind * cos(γ + δ)

sin(α – ε) = CE / AC, so

ε = α – arcsin(CE / AC) = α – arcsin(wind * cos(γ + δ) / AC)

ε = 85° – arcsin(100° * cos(30° + 5°) / 470) = 75°

Now let’s find AB:

AB = AE – BE

AE = AC * cos(α – ε)

BE = wind * sin(γ + δ)

AB = AE – BE = AC * cos(α – ε) – wind * sin(γ + δ) =

470 * cos(85° – 75°) – 100 * sin(30° + 5°) = 405

Flight time is dist / speed = 265 / 405 = 0.654 h = 39 minutes.