General Navigation Problems — Page 514, Lesson 514

I want to walk you through a classic General Navigation problem — one that combines a VOR/DME fix with a bit of geometry to find track and ground speed. This is the kind of question you'll see in the exam, and the method I'll show you is the one you should use every time. Let's set the scene. At 1000 hours, the aircraft is on the 310° radial from a VOR/DME, at 10 nautical miles range. Ten minutes later, at 1010, the radial and range are 040/10 NM. So the aircraft has moved from one fix to another, and we need to find its track and ground speed. First, let's define our terms. A radial is a magnetic bearing from the VOR station — so if you're on the 310° radial, you're somewhere along the line that extends 310° magnetic from the VOR. The DME gives you the slant range in nautical miles. So "310°/10 NM" means you're 10 nautical miles from the station, on that 310° bearing. Now, the first step — and this is the key habit — is to draw a diagram. It's just a sketch, not a scale drawing. You could solve this by drawing it accurately to scale, but that's not the ideal method. A rough sketch with the geometry marked is what we want. So let's draw a Magnetic North reference arrow. Then we plot the two fixes. Let's call the first fix, at 1000 hours, point B — that's the 310°/10 NM position. And the second fix, at 1010, point C — that's the 040°/10 NM position. The VOR/DME station itself is point A, at the origin. Now here's where the geometry comes in. If radial B is 310°(M), then the angle between Magnetic North and the line AB is 50°. Why? Because 360° minus 310° is 50° — the line AB sits 50° west of north. Similarly, the angle up to AC is 40° — because the 040° radial is 40° east of north. So at corner A, we have a right-angled triangle: 50° plus 40° equals 90°. Now, AB and AC are both 10 nautical miles. So we have an isosceles triangle ABC — two sides equal. In any triangle, the internal angles must add up to 180°. We already know 90 of those degrees are at corner A. So the remaining 90° are split equally between corners B and C. Therefore both B and C are 45° each. Now we need to find the track angle. There are a couple of ways to do this. The easiest is to look at the left-hand of the two smaller triangles. The bottom angle there is 50°, and angle B is 45°. These add up to 95°. So the remaining angle, which we'll call angle x, must be 85° — because 180° minus 95° is 85°. And if angle x is 85°, then so is angle y, because they're corresponding angles in the symmetric triangle. Angle y is the track angle. There's another way to arrive at the same answer. If direction AB is 310°(M), then direction BA — that's the reciprocal, the opposite direction — must be 130°(M). We know angle B is 45°. So the track direction is 130° minus 45°, which equals 085°(M). So the track is 085°(M). Now for ground speed. The aircraft covered the distance from B to C in 10 minutes — from 1000 to 1010. We need to know that distance. In an isosceles triangle with two sides of 10 NM and an included angle of 90° at A, the third side BC is the hypotenuse of a right-angled isosceles triangle. Using Pythagoras, that's the square root of (10² + 10²), which is the square root of 200, which is approximately 14.14 nautical miles. Now, ground speed is distance divided by time. In 10 minutes — that's one-sixth of an hour — the aircraft covered 14.14 NM. So ground speed is 14.14 divided by (1/6), which is 14.14 times 6, which equals approximately 84.8 knots. That rounds to 85 knots. So our answer is track 085°(M), ground speed 85 knots. Looking at the options, that matches option b: 085°(M) / 85 knots. Let me just recap the method, because this is the transferable skill. Draw the sketch with the Magnetic North arrow. Plot both fixes. Identify the angles from the radials. Recognise the isosceles triangle. Split the remaining angles. Find the track angle. Then use the time interval and the distance to get ground speed. That's the complete solution. The key insight is that the radials give you the angles, the equal ranges give you the isosceles triangle, and the 10-minute interval converts distance into speed.