The 1 in 60 Rule — Page 199, Lesson 173

Let’s start with the heart of this lesson: the 1 in 60 rule. This is a mental shortcut we use in navigation to convert a small distance off track into an angular error, called the track error angle. The rule gets its name from the fact that at a distance of 60 units along track, a 1-unit offset off track equals roughly 1 degree of angular error. That’s the geometric basis. Now, when the numbers don’t lend themselves to simple mental proportioning, we use the formula. I want you to write this down exactly as it appears: z = (distance off / distance gone) × 60 Here, z is the track error angle in degrees. Distance off is how far you are laterally from your required track, in nautical miles. Distance gone is how far you have travelled along the track from your departure point, also in nautical miles. The 60 is the constant that converts the ratio into degrees. Let me give you the worked example from the text. You find yourself 6 miles right of track after 40 miles along track. Plug those numbers in: z = (6 / 40) × 60 That gives you 0.15 × 60, which equals 9 degrees. So your track error angle is 9 degrees to the right. Now, a few important points about how to use this. The formula gives you the angular error between your required track and your actual track made good. If you are right of track, your track made good is to the right of your required track, so you would subtract that angle from your required track to correct back. If you are left of track, you add the angle. That’s the basic correction logic. Let’s also look at the questions that follow, because they test exactly this. Question 1: you are 60 NM outbound from A and 7 NM left of track. So distance gone is 60, distance off is 7. z = (7/60) × 60 = 7 degrees. Since you are left of track, your track error angle is 7 degrees left. Question 2: 120 NM outbound, 8 NM right. z = (8/120) × 60 = 4 degrees right. Question 3: 90 NM outbound, 6 NM right. z = (6/90) × 60 = 4 degrees right. Question 4: 30 NM outbound, 4 NM left. z = (4/30) × 60 = 8 degrees left. Now, questions 5 through 7 ask for the track made good, not just the error angle. For question 5, required track is 045°(T), you are 80 NM outbound and 4 NM left. First find z: (4/80) × 60 = 3 degrees. Since you are left of track, your track made good is 045° minus 3°, which is 042°(T). Question 6: required track 220°(T), 45 NM outbound, 3 NM right. z = (3/45) × 60 = 4 degrees. Right of track means track made good is 220° plus 4°, so 224°(T). Question 7: required track 315°(T), 40 NM outbound, 6 NM left. z = (6/40) × 60 = 9 degrees. Left of track, so track made good is 315° minus 9°, which is 306°(T). Now, the rule also applies to vertical navigation, which is what questions 8 through 10 are about. Question 8 is a surveyor problem: he is 660 metres from a mast and measures an elevation angle of 4° to the top. Using the 1 in 60 rule, distance off is the height of the mast, distance gone is 660 metres. So height = (4/60) × 660 = 44 metres. That’s the height of the mast. Question 9: instrument approach, glide slope angle 3.00°, you are exactly 2 NM from touchdown. Assume 1 NM = 6000 feet. So distance gone is 2 NM, which is 12,000 feet. The height you should be passing is (3/60) × 12,000 = 600 feet. So you should be at 600 feet above the touchdown point. Question 10: glide slope 2.5°, you are at 1000 feet QFE. QFE means your altimeter is set to airfield pressure, so 1000 feet is your height above the airfield. We need the range from touchdown. Rearranging the formula: distance gone = (distance off / z) × 60. Distance off is 1000 feet, z is 2.5 degrees. So distance gone = (1000 / 2.5) × 60 = 24,000 feet. Convert to NM: 24,000 / 6000 = 4 NM. So your range from touchdown is 4 NM. Finally, questions 11 and 12 ask what track you must fly to arrive overhead the destination. This is a slightly different use. Here, distance gone is your distance from the destination, not from your departure. Question 11: required track 125°(T), you are 40 NM from R and 2 NM left. z = (2/40) × 60 = 3 degrees. Since you are left of track, to arrive overhead R you must fly a track that is 125° plus 3°, which is 128°(T). Question 12: required track 272°(T), 50 NM from T, 5 NM right. z = (5/50) × 60 = 6 degrees. Right of track, so you must fly 272° minus 6°, which is 266°(T). So the key distinction: when you are outbound from your departure, you correct by subtracting if right, adding if left. When you are inbound to your destination, you do the opposite — you add if left, subtract if right, because you are turning toward the track to intercept it. That’s the full 1 in 60 rule as presented here. The formula, the units, the direction logic, and how it applies both laterally and vertically.