Density Altitude at Matsumoto
Pilot plans a Cessna 172N (max gross 2,300 lb) from Matsumoto Airport (elevation 2,200 ft, runway 17/35 1,600 m) to Toyama. OAT is 32°C, QNH 1004 hPa. Actual takeoff weight is 2,250 lb. The POH takeoff table (flaps 10°, full throttle, paved, level runway) gives: at 2,500 ft pressure altitude and 30°C, ground roll 520 m and total over 50 ft 980 m; at 5,000 ft and 30°C, ground roll 780 m and total 1,450 m. The runway has a 150 m displaced threshold for landing only, so takeoff run available is 1,600 m. Rising terrain begins 1.5 km beyond the departure end, and the pilot's planned climb gradient is 350 ft/NM. Fuel is 40 gal usable, burn 9.5 gph, trip time 1.4 h, reserve 45 min. Load: two adults 340 lb, bags 80 lb, full fuel 240 lb.
1. Calculate pressure altitude and density altitude. Which table entry applies?
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Pressure altitude = airport elevation + 30 ft per hPa below 1013: 2,200 + (1013 - 1004) × 30 = 2,200 + 270 = 2,470 ft. ISA temperature at 2,470 ft is about 15 - 2 × 2.47 ≈ 10°C. The OAT of 32°C is therefore about ISA +22°C. Density altitude ≈ pressure altitude + 120 × (ISA deviation) = 2,470 + 120 × 22 ≈ 5,110 ft. Use the 5,000 ft/30°C table entry, but because the actual DA is slightly higher and OAT is 32°C, add a margin.
2. Using the table, estimate the total takeoff distance at 2,250 lb and decide whether the takeoff is acceptable with the 1,600 m runway and rising terrain.
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At 5,000 ft and 30°C the table gives 1,450 m total over 50 ft at 2,300 lb. At 2,250 lb, takeoff distance is roughly 3-5% lower, say 1,380-1,410 m. Correcting for 5,110 ft DA and 32°C adds about 5-10%, giving roughly 1,450-1,550 m. The runway offers 1,600 m TORA, so the margin is only 50-150 m, under 10%. With rising terrain 1.5 km beyond the end and a planned 350 ft/NM climb gradient, that margin is not adequate for a PPL operation. The takeoff should be rejected or the weight reduced and/or departure delayed to cooler conditions. A POH table is not a guarantee for a marginal runway with rising terrain.
3. How much fuel would need to be offloaded to reach 2,200 lb, and would that solve the problem?
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To lose 50 lb: 50 lb ÷ 6 lb/gal = 8.3 gal, about 31 L. At 9.5 gph that is 0.87 h of fuel. Current fuel is 40 gal; trip plus 45-minute reserve requires 1.4 + 0.75 = 2.15 h × 9.5 = 20.4 gal, so 8.3 gal can be offloaded without violating fuel requirements. But 50 lb is only about 2% of gross weight and may reduce takeoff distance by only 30-40 m, so it does not solve the terrain and margin problem. Better to offload bags/passenger or wait for a cooler part of the day.