Let’s pick this up right where the problem left off — Page 514, Lesson 515

Let’s pick this up right where the problem left off. We had just worked out the ground speed for that triangle-of-velocities question, and the answer was 85 knots. Now I want to walk you through the next idea, because it’s a trap that catches a lot of people. Here’s the trap. It is often assumed, wrongly, that if the wind is at right angles to your heading or your track, there is no head or tail wind component at all. For many problems, especially in Flight Planning, you are expected to assume zero wind component. But strictly speaking, that assumption is not true. The geometry of the triangle of velocities means that a wind that is exactly abeam still produces a small head or tail component, because your heading and your track are not the same thing when there’s drift. Let me show you with a question. An aircraft is tracking 090°(T) at a TAS of 180 knots. The wind is from due North, and the wind component is 5 knots head. The question is: what will the wind component be if the aircraft flies a track of 270°(T)? The options are 5 knots tail, zero, 5 knots head, or not possible to tell. Now, the answer is not zero, and it’s not simply 5 knots tail either. Let me explain how you actually solve it, because it’s not one of the conventional methods on the navigation computer. You put the wind direction, 360°, at the 12 o’clock position, and you draw a straight line vertically downwards from the blue circle, about 3 to 4 centimetres long. Then you put 180 knots under the blue circle. Bring the track, 090°, up to the heading index to start. Your initial drift will be to starboard. Then you rotate the wind face to the right, and you select a heading off to the left, until the amount of difference between heading and track is the same as the drift-line which crosses the wind direction at 175 knots ground speed. It’s a little difficult to explain without a diagram, but if you try it practically, you will see that it occurs at a heading of 077°, using 13°S drift. That gives you a wind of 360° / 42. Now apply that wind, 360° / 42, to a track of 270°(T), and you will find you need to fly a heading of 283°(T). That gives you a ground speed of 175 knots. So the key point here is that the wind component is not zero, even though the wind is at right angles to the track. The geometry of the triangle of velocities means that the wind component changes with the track, and it’s not simply a matter of the wind being abeam. The correct answer to that question is that it is not possible to tell, because the wind component depends on the full geometry, not just the wind direction relative to track. Let me make sure you’ve got the numbers straight. The aircraft is tracking 090°(T) at 180 knots TAS. The wind is from due North, 360°, and the component is 5 knots head. When you solve it on the computer, you find the wind is actually 360° / 42, meaning the wind speed is 42 knots. For a track of 090°, you fly a heading of 077° with 13°S drift, and your ground speed is 175 knots. Then for a track of 270°, you fly a heading of 283°(T), and your ground speed is still 175 knots. So the lesson here is that you cannot assume zero wind component just because the wind is at right angles to the track. The triangle of velocities is a geometric reality, and the wind component is a function of the full vector relationship between heading, track, TAS, and wind. That’s why the answer is not zero, and not simply 5 knots tail. It’s not possible to tell without doing the full calculation. Let me also connect this back to the earlier ground speed problem. There, we used Pythagoras to find the distance BC, which was 14.14 nautical miles, and then we used the navigation computer to find that 14.14 NM in 10 minutes gives 85 knots. That was a straightforward triangle of velocities. But this new problem shows that the geometry can be more subtle, and that’s why you need to be careful with assumptions about wind components. So, to summarise: the wind component is not zero when the wind is abeam, and you cannot assume it is. The correct approach is to solve the full triangle of velocities, as we did here, and the answer to that question is that it is not possible to tell without the full calculation.