Let’s pick this up right where the bearings left off — Page 486, Lesson 484

Let’s pick this up right where the bearings left off. You’ve just worked out a true bearing of 153°(T) TO the NDB. To get the bearing FROM the NDB, you take the reciprocal — add or subtract 180 — and that gives you 333°(T). That’s the bearing you’d plot from the NDB to your aircraft. Keep that in your head, because it’s the foundation for everything we’re about to do with position lines. Now I want to introduce you to Distance Measuring Equipment — DME. DME tells you the slant range from the ground DME facility to your aircraft. Slant range is the straight-line distance through the air, from the antenna on the ground up to you. Strictly speaking, you should convert that slant range into plan range — the horizontal distance you’d plot on a chart — before you plot it. But here’s the practical catch: unless the question gives you the height, you can’t do that conversion. And there’s a useful rule of thumb: if the range is more than about 10 nautical miles, the difference between slant range and plan range is insignificant — unless you’re at high altitude. So for most exam work, you just plot the DME range as it is. Now, how do you use DME to get a fix? Often the question simply asks you to combine a VOR bearing with a DME range. That’s straightforward — one bearing line and one range circle, and where they cross is your fix. But sometimes the question wants you to derive your position from two DME ranges. And that creates a problem — ambiguity. Two DME ranges give you two circles, and those circles intersect at two positions. Both are mathematically valid, so which one are you actually at? In a real situation, you’d probably know your approximate position and could work out which intersection is correct. But in an exam question, they’ll normally tell you your heading and whether the ranges are increasing or decreasing. Let me make that concrete. Suppose you’re told the aircraft is heading 270°(T) — that’s due west — and the ranges are decreasing. Then you must be at position B. Why? Because if you were at position A, flying that same heading, the ranges would be increasing. The trend of the ranges — increasing or decreasing — combined with your heading, disambiguates the two intersections. That’s the whole trick. Now let’s move to something called the ‘Cocked Hat’. When you plot three position lines — say three bearings — they rarely meet at a single point. Instead, they form a triangle, and that triangle is called a cocked hat. The fix is then taken to be at the point where the bisectors of the angles of the triangle would meet. So you take each angle of the triangle, draw the bisector — the line that splits the angle in half — and where those bisectors converge is your fix. And here’s the key interpretation: the size of the cocked hat is an indication of the probable accuracy of your fix. A small triangle means your position lines agree closely, so your fix is probably accurate. A big triangle means they disagree, so your fix is less reliable. Let’s shift gears now to climb and descent. These aren’t really plotting problems, but it’s convenient to group them together. They come as two types of question, but they’re essentially the same problem. The idea is to make the time taken in the vertical direction — in the climb or descent — the same as the time taken in the horizontal direction. So the vertical distance divided by the rate of climb or descent must equal the horizontal distance divided by the ground speed. That’s the core relationship: vertical time equals horizontal time. Let me walk you through Example 1 to see it in action. You’re 65 nautical miles from a VOR, and you commence a descent from FL330 — that’s flight level 330, or 33,000 feet — in order to arrive over the VOR at FL100, which is 10,000 feet. Your mean ground speed in the descent is 240 knots. The question asks: what rate of descent is required? First, work out the horizontal time. 65 nautical miles at 240 knots takes 16.25 minutes. Now, the vertical distance: you need to lose 33,000 minus 10,000 feet, which is 23,000 feet. And you need to lose that in the same 16.25 minutes. So the required rate of descent — ROD — is 23,000 divided by 16.25, which gives you 1415 feet per minute. That matches answer (a), 1420 feet per minute — the closest option. Now Example 2. You’re homing to overhead a VORTAC — that’s a facility combining VOR and TACAN, so it gives you both bearing and DME. You’ll descend from 7500 QNH to be 1000 feet AMSL by 6 nautical miles DME. Your ground speed is 156 knots, and the rate of descent will be 800 feet per minute. The question asks: at what range from the VORTAC do you commence the descent? So here, the vertical distance is 7500 minus 1000, which is 6500 feet. At 800 feet per minute, the time to descend is 6500 divided by 800, which is 8.125 minutes. Now, in that same time, at 156 knots ground speed, you cover 156 times 8.125 divided by 60 — that gives you about 21.1 nautical miles. But you need to be at 1000 feet by 6 nautical miles DME, so you commence the descent 21.1 nautical miles before that — which means at about 27.1 nautical miles from the VORTAC. That matches answer (a), 27.1 NM. So the whole climb-and-descent idea is just matching vertical time to horizontal time. Once you’ve got that, both question types fall out the same way.