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

Let’s pick this up right where the geometry left off. We just worked out that with a 10‑minute leg, a distance of 14.14 nautical miles gives you a ground speed of 85 knots — that was the answer to that earlier problem. Now I want to walk you through something that trips up a lot of pilots, and it’s a real trap in Flight Planning. Here’s the misconception: people assume that if the wind is at right angles to your heading or your track, then there is no headwind or tailwind component at all. In other words, they think a beam wind — a wind blowing exactly across your path — has zero effect on your ground speed. For many problems, especially in Flight Planning, you’re actually expected to make that assumption and treat the wind component as zero. But strictly speaking, that is not true. Let me show you why. Consider this question. An aircraft is tracking 090° true at a true airspeed of 180 knots. The wind is from due North, and the wind component is 5 knots head. So you’re flying east, the wind is coming from the north, and it’s giving you a 5‑knot headwind component. Now the question asks: what will the wind component be if the aircraft flies a track of 270° true — that is, the reciprocal, due west? The options are: 5 knots tail, zero, 5 knots head, or not possible to tell. Now, the intuitive but wrong answer is “zero,” because you think, “well, the wind is still at right angles to the track, so no head or tail component.” But that’s the trap. The correct reasoning is that the wind component is still 5 knots, but now it’s a tailwind. Why? Because the wind direction hasn’t changed — it’s still from due North. When you reverse your track from east to west, the same wind that was opposing you is now helping you. So the answer is 5 knots tail. Let me show you how you’d actually solve this on your navigation computer, because it’s not one of the conventional methods. Put the wind direction, 360°, at the 12 o’clock position on the wind face. Then draw a straight line vertically downwards from the blue circle — about 3 to 4 centimetres long. Now 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 — that is, to the right. Now rotate the wind face to the right, and 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 try it practically and you’ll see that it occurs at a heading of 077°, using 13° of starboard drift. That gives you a wind of 360° / 42 — meaning the wind is from 360° at 42 knots. Now apply that same wind, 360° / 42, to a track of 270° true, and you’ll find you need to fly a heading of 283° true. That gives you a ground speed of 175 knots. So here’s the key takeaway: the wind component is not simply zero just because the wind is at right angles to your track. The geometry of the triangle of velocities means the wind can still have a head or tail component depending on the exact relationship between your heading, your track, and the wind direction. In this case, the same wind that gave you a 5‑knot headwind on an easterly track gives you a 5‑knot tailwind on a westerly track. That’s the implication of geometry on the triangle of velocities.