What is the track error overhead Cranfield? — Page 208, Lesson 190

I want to walk you through a navigation exercise that uses the 1-in-60 rule — a fundamental mental tool we use in the cockpit to estimate track errors and heading corrections without a calculator. This excerpt gives us two scenarios, and I'll teach the first one fully, then the second. Let's start with the first scenario, which is set up as a series of questions about a flight from Oxford to Cranfield to Cambridge. The questions are: a. What is the track error overhead Cranfield? b. What is the Track Made Good (TMG) from Oxford? c. What was the expected drift? d. What has the actual drift been? e. What alteration of heading should be made over Cranfield to fly direct to Cambridge? f. What is the new heading to be flown from overhead Cranfield? Now, the excerpt doesn't give us the raw numbers for this first scenario — it only lists the questions. That tells me the numbers were probably on a diagram or in a preceding paragraph. But the structure itself teaches us the workflow. Let me define each term as we go, because these are the exact terms you'll use in the cockpit. Track error is the angular difference between your planned track and your Track Made Good (TMG). It's measured at your current position. So question (a) asks: at the moment you're overhead Cranfield, how many degrees off track are you, and to which side? Track Made Good (TMG) is the actual path you've flown over the ground, expressed as a bearing from your departure point to your present position. So question (b) asks: from Oxford to your position overhead Cranfield, what was the actual bearing you flew? Expected drift is the crosswind correction you planned for — the difference between your planned heading and your planned track. If you planned a heading of, say, 260°(M) and a track of 250°(M), your expected drift would be 10° left or right depending on the wind. Question (c) asks what that planned value was. Actual drift is what really happened — the difference between the heading you actually flew and the TMG you actually achieved. Question (d) asks for that real-world value. Alteration of heading is the change you need to make to your current heading to achieve a new goal. Question (e) asks: from overhead Cranfield, how many degrees should you turn to fly directly to Cambridge? That's a direct tracking correction — you're not trying to regain the old track; you're pointing straight at the destination. New heading is the actual compass heading you'll fly after making that alteration. Question (f) asks for that final heading value. Now let's move to the second scenario, which has actual numbers. This one is a flight from Norwich to Oxford. The planned track is 250°(M), distance 96 NM. The heading being flown is 260°(M), and the ground speed is 180 kt. The aircraft departs Norwich at 1000 hrs. At 1012 hrs, the aircraft is overhead Ely, which is 3 NM right of planned track. Let's work through each question using the 1-in-60 rule. The 1-in-60 rule says that at 60 NM from your starting point, 1 NM off track equals 1° of track error. More generally: track error in degrees = (distance off track in NM ÷ distance flown in NM) × 60. First, we need the distance flown. From 1000 to 1012 is 12 minutes. At 180 kt ground speed, in 12 minutes you cover 180 × (12 ÷ 60) = 36 NM. So at 1012, you've flown 36 NM. You're 3 NM right of planned track. Using 1-in-60: track error = (3 ÷ 36) × 60 = 5°. So you're 5° to the right of your planned track. Now let's answer each question. a. What was the planned drift? Planned drift is the difference between planned heading and planned track. Planned heading is 260°(M), planned track is 250°(M). That's 10° of drift. Since heading is greater than track, the planned drift is 10° to the right — meaning you planned for a wind from the left. b. What is the track error at 1012 hrs? We just calculated it: 5° to the right of planned track. c. What TMG has been flown between 1000 hrs and 1012 hrs? TMG is the actual bearing from Norwich to your position at Ely. You planned track 250°(M), but you're 3 NM right after 36 NM. That 5° right means your TMG is 250° + 5° = 255°(M). d. What has the actual drift been between 1000 hrs and 1012 hrs? Actual drift is the difference between the heading you flew (260°(M)) and the TMG (255°(M)). That's 5° of drift — but to the left, because your heading is greater than your TMG. So actual drift is 5° left, which is less than the planned 10° right. That tells you the wind is weaker than forecast, or from a different direction. e. What alteration of heading should be made to track directly to Oxford? This is a direct tracking correction. You're at Ely, and you want to fly straight to Oxford. The 1-in-60 rule for direct tracking uses the distance to go. The total distance is 96 NM, you've flown 36 NM, so distance to go is 60 NM. You're 3 NM right of the planned track, but the direct track to Oxford from Ely is a new line. However, since Oxford is on the planned track, the direct track from Ely to Oxford is parallel to the planned track but offset. The correction needed is: track error angle (5°) plus the angle to converge. For direct tracking, the formula is: alteration = (distance off track ÷ distance to go) × 60. That's (3 ÷ 60) × 60 = 3°. But you also need to remove the existing track error. The standard method: to track directly to the destination, you turn by the track error angle plus the closing angle. The closing angle is (distance off track ÷ distance to go) × 60 = 3°. So total alteration = 5° + 3° = 8° to the left. f. What heading is required to fly directly to Oxford? Your current heading was 260°(M). You alter left by 8°, so new heading = 260° - 8° = 252°(M). g. What alteration of heading should be made to regain track at 1024 hrs? Now we're doing a regain track maneuver. You want to be back on the original planned track by 1024 hrs. From 1012 to 1024 is 12 minutes. At 180 kt, that's another 36 NM. So distance to go to the regain point is 36 NM. You're 3 NM right. To regain track, you need a closing angle of (3 ÷ 36) × 60 = 5°. Plus you need to correct the existing track error of 5°. Total alteration = 5° + 5° = 10° to the left. h. What heading should be flown between 1012 and 1024 hrs to regain track at 1024 hrs? Current heading 260°(M), alter left 10°, so heading = 250°(M). i. Given the situation in g. and h. above, what heading change should be made at 1024 hrs and what heading should be flown from 1024 hrs onwards? At 1024 hrs, you're back on track. Now you need to fly the original planned track of 250°(M). But you also need to apply the actual drift you've observed, which was 5° left. So to maintain track 250°(M) with 5° left drift, you need to steer 250° + 5° = 255°(M). So at 1024, you turn right by 5° (from 250° to 255°). j. Estimate the ETA at Oxford. You departed at 1000. You flew 36 NM to Ely by 1012. Then you fly another 36 NM to the regain point by 1024. From the regain point, distance to Oxford is total 96 minus 72 = 24 NM. At 180 kt, 24 NM takes 8 minutes. So ETA = 1024 + 8 = 1032 hrs. That's the full worked example. The key takeaway: the 1-in-60 rule lets you convert a linear off-track distance into an angular correction, and you choose between direct tracking (using distance to go) or regaining track (using distance to the regain point).