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

Let’s start with the first part, the Cranfield–Cambridge problem, because it sets up the whole method. I’m going to teach you the 1-in-60 rule as we go, because that’s the tool you’ll use for every one of these questions. The 1-in-60 rule — here’s the core idea. If you fly for 60 NM and you’re 1 NM off track, your track error is 1°. That’s the whole rule. It’s a ratio: track error in degrees equals the distance off track divided by the distance flown, times 60. So if you’re 3 NM off after 60 NM, that’s 3°. If you’re 3 NM off after only 30 NM, that’s 6°, because the error is building faster. Now, the first question: What is the track error overhead Cranfield? You need the distance flown from Oxford to Cranfield and how far off track you are at that point. The track error is the angle between your planned track and your actual track — the TMG. Using 1-in-60, you take the distance off track, divide by the distance flown, multiply by 60. That gives you the track error in degrees. Second: What is the Track Made Good (TMG) from Oxford? TMG is the actual path you’ve flown over the ground, measured as a bearing from your departure point. It’s not the planned track — it’s what you actually did, including the wind effect. So if your planned track was, say, 045°(M) and you’ve drifted right, your TMG will be a few degrees greater than 045°(M). Third: What was the expected drift? Expected drift is the wind correction you planned for. It’s the difference between your heading and your planned track. If you planned a heading of 050°(M) to make good a track of 045°(M), your expected drift is 5° starboard — that’s the drift you anticipated from the forecast wind. Fourth: What has the actual drift been? This is the difference between your heading and your TMG. If your heading was 050°(M) but your TMG was 048°(M), your actual drift is only 2° starboard. The difference between expected and actual drift tells you the wind isn’t doing what you forecast — that’s why you’re off track. Fifth: What alteration of heading should be made over Cranfield to fly direct to Cambridge? This is the opening correction. You’re overhead Cranfield, you know your track error, and you want to fly straight to Cambridge. The alteration is the track error plus the closing angle to Cambridge. The closing angle is the angle between your TMG and the bearing to Cambridge, again using 1-in-60: distance off track divided by distance to go, times 60. Sixth: What is the new heading to be flown from overhead Cranfield? This is your original heading, plus the alteration you just calculated. If you were on 050°(M) and you need to alter 8° to the right, your new heading is 058°(M). Now let’s move to the second problem — Norwich to Oxford. This is the same method, but with more steps because you’re also regaining track and estimating ETA. Planned track 250°(M), distance 96 NM, heading 260°(M), ground speed 180 kt. Depart Norwich at 1000 hrs. At 1012 you’re overhead Ely, 3 NM right of planned track. a. What was the planned drift? Planned drift is heading minus planned track: 260° minus 250° is 10° starboard. That’s what you expected from the wind. b. What is the track error at 1012 hrs? You’ve flown from 1000 to 1012 — that’s 12 minutes. At 180 kt, in 12 minutes you cover 36 NM. You’re 3 NM off track. Using 1-in-60: 3 divided by 36, times 60, gives you 5°. So your track error is 5°. c. What TMG has been flown between 1000 and 1012? Your planned track was 250°(M), and you’re 5° right of it, so your TMG is 255°(M). d. What has the actual drift been between 1000 and 1012? Actual drift is heading minus TMG: 260° minus 255° is 5° starboard. So the wind gave you only 5° of drift, not the 10° you planned for. e. What alteration of heading should be made to track directly to Oxford? You’re at Ely, 3 NM right of track, with 60 NM to go to Oxford. The closing angle is 3 divided by 60, times 60 — that’s 3°. You add your track error of 5° plus the closing angle of 3°, giving an alteration of 8° to the left. f. What heading is required to fly directly to Oxford? Your original heading was 260°(M). Alter 8° left, and your new heading is 252°(M). g. What alteration of heading should be made to regain track at 1024 hrs? This is the double-the-track-error method. You want to regain the planned track, not just fly direct. From 1012 to 1024 is another 12 minutes — another 36 NM. To regain track, you double the track error: 5° times 2 is 10°. That’s your alteration to the left. h. What heading should be flown between 1012 and 1024 to regain track at 1024? Original heading 260°(M) minus 10° gives you 250°(M). You fly this for 12 minutes to converge back onto the planned track. i. Given the situation in g and h, what heading change should be made at 1024 and what heading should be flown from 1024 onwards? At 1024 you’re back on track. Now you need to fly parallel to the planned track, so you remove the correction. You add back the 10° you took out, returning to your original heading of 260°(M). From 1024 onwards, you fly 260°(M). j. Estimate the ETA at Oxford. You’ve flown 36 NM to Ely, then another 36 NM to 1024 — that’s 72 NM. Total distance is 96 NM, so you have 24 NM to go. At 180 kt, 24 NM takes 8 minutes. 1024 plus 8 minutes gives you an ETA of 1032 hrs. Let me check that logic once more. From 1000 to 1012 you covered 36 NM. From 1012 to 1024 you covered another 36 NM. That’s 72 NM flown. 96 minus 72 is 24 NM remaining. At 180 kt, that’s 8 minutes. So yes, ETA 1032 hrs. That’s the full method — 1-in-60 for track error, closing angle for direct-to, double-the-error for regaining track, and time-speed-distance for ETA. You’ve now got the complete toolkit for these navigation problems.