So we write it as: departure equals change of longitude times cos latitude — Page 565, Lesson 562

Let’s start with the departure formula, because that’s the heart of this page. Departure is the distance along a parallel of latitude, measured in nautical miles, and it equals the change of longitude multiplied by the cosine of the latitude. So we write it as: departure equals change of longitude times cos latitude. The change of longitude must be in minutes of arc, not degrees, so if you have 10 degrees, you multiply by 60 to get 600 minutes, then multiply by the cosine of the latitude. For example, at latitude 45 degrees, cosine is 0.7071, so 600 times 0.7071 gives 424.3 nautical miles. That’s the departure. Now, the excerpt also shows a shortcut: change of longitude times 0.80. That’s just a specific case where the cosine of the latitude is 0.8, so you can use that multiplier directly. But remember, the general formula is departure equals change of longitude times cos latitude. Next, we have a worked example about resolving a track into change of latitude and change of longitude. You’re flying 70 nautical miles along a track of 225 degrees. To resolve that into components, you use sine and cosine. The change of latitude is 70 times sine of 45 degrees, and the change of longitude is 70 times cosine of 45 degrees. Both come out to 49 nautical miles. Since you’re at the Equator, the 49 nautical miles west can be taken directly as 49 minutes of change of longitude, because at the Equator, one minute of longitude equals one nautical mile. So the final position is 49 minutes south and 49 minutes west of the original position. Now, there’s a question about the Earth’s shape. The polar diameter and equatorial diameter are in the ratio 296 to 297. The semi-major axis is half the equatorial diameter, and the semi-minor axis is half the polar diameter. We need the polar radius. So we set up the ratio: polar radius divided by 6378.4 equals 296 divided by 297. Solving that gives a polar radius of 6356.9 kilometers. Alternatively, you can remember that the polar diameter is 43 kilometers shorter than the equatorial diameter, so the radius is 21.5 kilometers shorter. Subtract that from 6378.4 to get 6356.9 kilometers. Either method works. Then there’s a combined problem involving departure and scale. We already found the departure as 424.3 nautical miles. Now, scale is chart length divided by earth distance. The chart length is 14 centimeters, and the earth distance is 424.3 nautical miles converted to centimeters. Since one nautical mile is 1852 meters, and there are 100 centimeters in a meter, you multiply 424.3 by 1852 by 100 to get the distance in centimeters. That gives a scale of approximately 1 in 5.6 million. Finally, there’s a descent profile problem. You’re flying at FL370, which is 37,000 feet, and you need to descend to FL80, which is 8,000 feet. The difference is 29,000 feet. At a rate of 1800 feet per minute, that descent takes 16.1 minutes. Your mean ground speed in the descent is 232 knots, so in 16.1 minutes you cover 62 nautical miles. Your total distance to run is 185 nautical miles. If 62 of those are in the descent, the high-level distance to the top of descent is 123 nautical miles. At your level ground speed of 320 knots, 123 nautical miles takes 23 minutes. Add that to the time of 0422 to get 0445 as the ETA for the top of descent. And there’s a bearing problem. True heading is 140, true track is 150. At point A, the relative bearing to the NDB is 35 degrees left of the nose, so the true bearing is 105. At point B, the NDB is 80 degrees left of the nose, giving a true bearing of 060. That covers the key concepts on this page. Let me know if you want to go deeper into any of these.