Sprained Ankle at Dusk: Bearing, Scale and Route Choice on a 1:50,000 Sheet
A school trekking party of 12 cadets and one teacher is at grid 472 318 on a 1:50,000 sheet at 1515 in February. The planned camp is at grid 501 344. The bearing measured on the map between the two points is 048°. The sheet's magnetic declination is 2°30′ east. A ridge track runs from the party's position through 489 331 and joins the camp track; it is about 5.4 km long with roughly 2.5 km of climbing. A shorter valley track of about 4.2 km follows a stream through a gorge and crosses the stream four times. Contour interval is 20 m. One cadet has a sprained ankle and cannot move faster than 1.5 km/h on level ground or 1 km/h uphill. Sunset is 1830, with usable light until 1905. The party carries no radio, one mobile phone that works only on the ridge top, two torches and one first-aid kit.
1. Convert the 048° grid bearing to a magnetic bearing and calculate the straight-line distance between the two grid references. Show each step.
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Scale. 1:50,000 means 1 cm = 500 m, so one 2 cm grid square = 1 km and the last digit of a six-figure reference is a tenth of a square, i.e. 100 m.
Distance. Eastings: 501 − 472 = 29 tenths = 2.9 km. Northings: 344 − 318 = 26 tenths = 2.6 km. Straight line = √(2.9² + 2.6²) = √(8.41 + 6.76) = √15.17 ≈ 3.9 km.
Check the bearing. tan θ = 2.9 / 2.6 = 1.115, so θ ≈ 48°, which agrees with the 048° measured on the map. A bearing that does not match the grid differences means one of the two is wrong.
Declination. Magnetic north lies 2.5° east (clockwise) of grid north, so the needle already points 2.5° to the right of the target direction. 'East is least': magnetic bearing = grid bearing − declination = 48° − 2.5° = 045½°.
The party must therefore set 045½° on the compass to walk the grid line, not 048°, and a 2.5° error over 3.9 km is roughly 170 m of lateral drift.
2. Estimate the time to camp by each route and decide what the party should actually do at 1515.
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The party's speed is set by the slowest cadet, so use 1.5 km/h on the flat and 1 km/h uphill, plus halts.
Ridge track: 2.5 km uphill ÷ 1.0 = 2.5 h; 2.9 km level ÷ 1.5 = 1.93 h; total 4.43 h, add 30 min of halts = 4.9 h, arriving about 2030.
Valley track: 0.8 km uphill ÷ 1.0 = 0.8 h; 3.4 km level ÷ 1.5 = 2.27 h; total 3.07 h, add 40 min for four stream crossings and 20 min of halts = 4.07 h, arriving about 1937.
Neither route reaches camp in daylight: last usable light is 1905. The valley route is shorter but puts a hopping cadet into a gorge with four stream crossings in failing light, which is the worst combination available.
Decision: stop well short. Halt at 489 331 - about 2.4 km along the ridge, roughly 1.8 h of walking, arriving around 1700 - pitch camp, treat and elevate the ankle, and send two fit cadets to the ridge top with the phone. Arriving late at a planned camp is a minor failure; a night evacuation from a gorge is a serious one.
3. What drill should the teacher follow on halting, and what rules must govern the two cadets going for the phone signal?
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At the halt, in order: headcount by name against the roll; treat the ankle (rest, ice if available, compression, elevation) and re-check the other cadets' feet; water and food before any task; site the camp on dry, level ground away from the stream, with tents upwind and the cooking area separate; establish a sentry roster and a whistle signal convention; mark the position on the map and note the time; log the decision and the next intended move, so that a search party later has a written trail; keep everyone in windproof layers before anyone feels cold.
Rules for the two cadets: never fewer than two; map, compass set to 045½°, torch, whistle, water and a written note of the route and the time due back; follow the ridge track only and stay in sight of it; no stream crossing and no short cut down the spur; turn back at a fixed time - 1815 - regardless of whether the call connected; three whistle blasts means 'come back'. The teacher writes down their departure time before they leave.