Right, let's pick this up — Page 486, Lesson 484

Right, let's pick this up. We've just dealt with the true bearing to and from the NDB, and the reciprocal relationship there. Now I want to move on to a piece of equipment you'll use constantly for fixing your position: Distance Measuring Equipment, or DME. Now, the key thing to understand about DME is that it doesn't give you a flat distance on a chart. It gives you what we call slant range. That's the straight-line distance from the aircraft, up in the air, directly to the ground facility. Think of it as the hypotenuse of a triangle. The vertical side is your height, and the horizontal side is your true distance over the ground, which we call plan range. Strictly speaking, you should convert that slant range into plan range before you plot it on your chart. But here's the practical reality: unless the question actually gives you the aircraft's height, you simply cannot do that conversion. And there's a useful rule of thumb here. If the range you're reading is more than about 10 nautical miles, the difference between slant range and plan range is insignificant—unless you happen to be at a very high altitude. So for most practical plotting, you just use the DME reading as if it were the plan range. Now, how do we use this in a fix? The straightforward case is combining a VOR bearing with a DME range. You draw the bearing line from the VOR, and you measure the range along it, and there's your fix. That's simple. But sometimes the question asks you to derive your position from two DME ranges. And this creates a problem: ambiguity. If you have two ground stations and you know your distance from each, you draw a circle around each station. Those two circles will intersect at two points. Both points satisfy the ranges, so you have two possible positions, labelled A and B. In a real-world situation, you'd probably know roughly where you are, so you could pick the correct intersection. But in an exam, they have to give you a way to disambiguate. So the question will normally tell you your heading and whether the ranges are increasing or decreasing. Let me give you the example from the text. Suppose you're told the aircraft is heading 270°(T)—that's due west—and the ranges are decreasing. If you were at position A, flying west, you'd be flying away from the stations, so the ranges would be increasing. Since they're decreasing, you must be at position B, where flying west takes you toward the stations. That's how you resolve the ambiguity. Now, let's talk about what happens when you have three position lines instead of two. If you plot three lines—say, three bearings—they rarely all cross at a single perfect point. Instead, they form a small triangle. We call this a 'cocked hat'. The fix is then taken to be at the point where the bisectors of the angles of the triangle would meet. That's the centre point of the triangle, effectively. And here's the important operational point: the size of the cocked hat is an indication of the probable accuracy of your fix. A big cocked hat means your fix is less reliable; a small one means you're more confident. Now, let's shift gears completely and look at climb and descent. These aren't really plotting problems in the same way, but it's convenient to group them together. They can come as two types of question, but they're essentially the same problem. The core idea is this: you want to make the time taken in the vertical direction—the climb or descent—exactly the same as the time taken in the horizontal direction. So the relationship is: the vertical distance divided by the rate of climb or descent must equal the horizontal distance divided by the ground speed. Let me show you with the first example. Example 1: You're 65 NM from a VOR, and you commence a descent from FL330 in order to arrive over the VOR at FL100. Your mean ground speed in the descent is 240 knots. What rate of descent is required? Let's work through it. First, the horizontal time: 65 NM at 240 knots. 65 divided by 240 gives you 0.2708 hours, which is 16.25 minutes. That's how long you have to make the descent. Now the vertical distance. FL330 is 33,000 feet, and FL100 is 10,000 feet. So you need to lose 23,000 feet. And you must lose that in the same 16.25 minutes. So the rate of descent, or ROD, is 23,000 divided by 16.25, which gives you 1415 feet per minute. And looking at the options, that's answer (a), 1420 feet per minute—the closest value. Now Example 2. You're homing to overhead a VORTAC—that's a VOR and TACAN combined at the same site. You will descend from 7500 QNH to be 1000 AMSL by 6 NM DME. Your ground speed is 156 knots, and the ROD will be 800 feet per minute. At what range from the VORTAC do you commence the descent? So here, we work it the other way. First, the vertical time: you're descending from 7500 to 1000 feet, so that's a loss of 6,500 feet. At 800 feet per minute, that takes 6500 divided by 800, which is 8.125 minutes. Now, in that same 8.125 minutes, at 156 knots ground speed, how far do you travel horizontally? 156 knots is 2.6 NM per minute. Multiply that by 8.125 minutes, and you get 21.125 NM. But remember, you want to be at 1000 feet AMSL by 6 NM DME from the VORTAC. So you must start your descent 21.125 NM before that point. That means you commence at 6 plus 21.125, which is 27.1 NM from the VORTAC. So the answer is (a), 27.1 NM. So you see, in both cases, it's the same principle: match the vertical time to the horizontal time.