Departure — Page 266, Lesson 235

We're starting a brand-new topic: Departure. This is one of the core ideas in general navigation, and it's all about the relationship between longitude and actual distance on the Earth's surface. Let me give you the precise definition first, because everything else builds on it. Departure is the distance between two meridians along a specified parallel of latitude, usually in nautical miles. Let me unpack that. A meridian is a line of longitude running from pole to pole. A parallel of latitude is a line running east-west. So if I pick two meridians—say, the 5°W meridian and the 6°W meridian—and I measure the distance between them along a specific parallel, that distance is the departure. And we measure it in nautical miles. Now, here's the crucial point. Two meridians always represent the same change of longitude, no matter where you are. But the actual east-west distance between them changes with latitude. Why? Because the meridians converge—they get closer together—as you move from the Equator toward the poles. So the same change of longitude does not represent the same east-west distance in nautical miles at different latitudes. Think about it this way. If I take the Equator as my specified parallel, the distance between two meridians will be greater than at some higher latitude. At the Equator, the meridians are at their maximum separation. In fact, departure is maximum at the Equator, where 1° change of longitude—which we abbreviate as ch.long—equals 60 minutes of arc of a Great Circle. That's 60 nautical miles. At the other extreme, departure is zero at both poles, because the meridians converge and meet at these two points. They all come together at a single point, so the distance between them is nothing. So departure varies smoothly between these two extremes. And here's the beautiful part: departure varies as the cosine of the latitude. That gives us our fundamental formula: Departure (NM) = ch.long (in minutes) × cos lat Let me make sure you understand each piece. "Departure" is in nautical miles. "ch.long" is the change of longitude, expressed in minutes of arc—not degrees. And "cos lat" is the cosine of the latitude. So if I'm at 60°N and I change longitude by 1°, my departure is 60 minutes × cos 60°, which is 60 × 0.5, or 30 nautical miles. At the Equator, cos 0° is 1, so 1° of longitude gives you the full 60 nautical miles. At the pole, cos 90° is 0, so departure is zero. It all fits together. One more important point before we calculate. Since this distance is always measured along a parallel of latitude, it represents a Rhumb Line distance. A rhumb line is a line that crosses all meridians at the same angle—and a parallel of latitude does exactly that. So departure is a rhumb line distance, not a great circle distance. Now let's actually calculate one. Consider two meridians joined by a parallel of latitude. Let's say the change of longitude is 20 degrees, and we're at latitude 52°. First, we multiply the change of longitude by 60 to convert it into minutes. So 20 × 60 = 1200 minutes. Then we multiply by the cosine of 52°. So: Departure (NM) = ch.long (min) × cos lat = 20 × 60 × cos 52° = 738.8 nautical miles. That's the basic calculation. Now, in the exam, there are two main types of departure questions. The first is variations on the basic departure formula—where you're given some of the pieces and you have to solve for the unknown. The second is given departure at one latitude, calculate it at another. Let me walk you through the first type with two worked examples. Example 1: An aircraft at position 60°00'N 005°22'W flies 165 km due East. What is the new position? First, we need to convert 165 km to nautical miles. We divide by 1.852, because there are 1.852 kilometers in a nautical mile. So 165 / 1.852 = 89 NM. Now we substitute into the departure formula. We know departure is 89 NM, and we're at latitude 60°N. So: 89 = ch.long × cos 60° The cosine of 60° is 0.5. So: ch.long = 89 / 0.5 = 178 minutes of longitude. Now we convert that back to degrees and minutes. 178 minutes is 2°58'. Since the aircraft flew due East, we add this to the initial longitude. The initial longitude was 005°22'W. Adding 2°58' East gives us 002°24'W. So the new position is 60°00'N 002°24'W. Example 2: In which latitude is a difference of longitude of 44°11' equivalent to a departure of 2000 NM? First, convert 44°11' to minutes of arc. 44 degrees is 2640 minutes, plus 11 minutes, gives us 2651 minutes. Now we use the formula. We know departure is 2000 NM, and ch.long is 2651 minutes. So: 2000 = 2651 × cos lat Rearranging: cos lat = Departure / ch.long = 2000 / 2651 = 0.7544 Now we find the angle whose cosine is 0.7544. That's approximately 41°. So the latitude is 41° North or South—because the cosine is the same for a given latitude in either hemisphere. That's the core of departure. The formula, the maximum at the Equator, the zero at the poles, and how to rearrange it to solve for any unknown. Take a moment to let that sink in, and we can move on to the second type of question when you're ready.