Let’s pick this up right where the worked example left off — Page 247, Lesson 226

Let’s pick this up right where the worked example left off. We’ve just solved a convergency problem two different ways, and now I want to show you the two methods side by side, because you’ll meet both definitions in exams and in practice. First, the definition-based approach. If you define convergency as the change in great circle track between two meridians, then in our example the great circle track changed by 8.5 degrees between A and B. The aircraft left A on a track of 060°. Now, using the D-I-I-D rules from chapter 2 — that’s the mnemonic for whether track increases or decreases — travelling eastwards in the Northern hemisphere, the track will increase. So the track angle as it passes through B is 060° plus 8.5°, which is 068.5°. And if you want the track from B back to A, as it passes through B, you take the reciprocal of 068.5°, which is 248.5°. Now the second definition. If you prefer to think of convergency as the angle of inclination between two meridians, you start differently. You parallel the meridian at A over to B. That gives you an angle M, which is 8.5° — that’s the inclination of the meridians. Then angle N is 060°, which comes from corresponding angles between meridians in classical geometry. So the total of M plus N is 068.5°. Now, the meridian at B defines True North at that point. So the track angle of the continuation of line AB is 068.5° at B. Therefore the great circle bearing of A from B, measured at B, is 248.5°. Same answer, two routes. Now let’s try a fresh problem. Question 2: The initial great circle track from C, at 36°N 015°E, to D, at latitude 42°N, is 300°(T), and the final great circle track at D is 295°(T). Part (a) asks: what is the longitude of D? Part (b) asks: what is the approximate great circle track direction at longitude 011°E? Let’s work the answer. The track has changed from 300°(T) to 295°(T), so the convergence must be 5°. The question tells us C is at 36°N and D is at 42°N, so the mean latitude is 39°N. Now we substitute into the equation: convergency equals change of longitude times sine of mid latitude. So 5° equals change of longitude times sine of 39°. Rearranging, change of longitude equals 5° divided by sine of 39°, which gives 8°. If the change in longitude is 8° and C is at 015°E, and D is west of C, then the longitude of D is 007°E. For part (b), longitude 011°E is halfway between 015°E and 007°E. Therefore the great circle track will be halfway between 300°(T) and 295°(T), which is 297.5°(T). Now Question 3: The initial great circle track from H, at 40°S 170°W, to G, at 45°S 174°E, is 250°(T). What is the initial great circle track from G to H? This time we’re in the Southern hemisphere. Draw in an initial great circle track of 250°(T). It must be from the right-hand side of the diagram if it’s going to cut the other meridian. So the right-hand one must be H and the left-hand one must be G. Just to be certain, check the latitudes and longitudes. H is 40°S, G is 45°S, so G should be south of H on the diagram — and it is. H is 170°W and G is 174°E. As we’re crossing the Greenwich anti-meridian, which is 180°E/W, left and right are the right way round on this diagram — confusing, but correct. If you need clarification on that, check the explanation in chapter 1. Then you apply the same equation: convergency equals change of longitude times sine of mid latitude. That gives you the convergency, and from there you can find the initial great circle track from G to H. Take a moment to look at the figures — Figure 14.12 shows meridians converging northwards, Figure 14.13 shows paralleling the meridian, and Figure 14.14 shows the basic diagram for Question 2. They’ll help you visualise exactly what we just did.