Density Altitude and a Ridge: The 1500 Departure
An unpressurised, normally aspirated light aircraft departs an aerodrome at elevation 4,500 ft with QNH 1013 hPa. At 0900 the OAT is 28 °C and the dewpoint 12 °C. The route crosses a ridge at 9,000 ft, and your operator requires 1,000 ft of terrain clearance, so you plan to cross at 10,000 ft. The 1500 forecast gives a surface temperature of 34 °C, isolated cumulus with tops FL200, and the dewpoint unchanged. The aircraft has a service ceiling of 14,000 ft, a sea-level rate of climb of 700 fpm, and no supplemental oxygen. ISA temperature at 4,500 ft is about 6 °C.
1. Calculate the density altitude at the aerodrome at 0900 and at 1500, and explain what it means for the take-off.
Read the worked answer
Pressure altitude = 4,500 ft (QNH is standard). ISA temperature at 4,500 ft ≈ 15 − 1.98 × 4.5 ≈ 6 °C.
| Time | OAT | ISA deviation | Density altitude |
|---|---|---|---|
| 0900 | 28 °C | +22 °C | 4,500 + 120 × 22 ≈ 7,140 ft |
| 1500 | 34 °C | +28 °C | 4,500 + 120 × 28 ≈ 7,860 ft |
The 120 ft per °C rule is a rule of thumb, but the conclusion is robust: at 1500 the aeroplane is taking off from an aerodrome that behaves like 7,900 ft. The engine produces less power, the propeller produces less thrust, and true airspeed for the same IAS is roughly 18% higher, so the ground roll is close to double the sea-level ISA figure. Leaning for best power before take-off is also required at this density altitude; a full-rich mixture will cost further power and can foul plugs. The 0900 take-off is still a high-density-altitude take-off, not a normal one.
2. Estimate the density altitude at the 10,000 ft crossing level at 1500 and compare it with the aircraft's capability.
Read the worked answer
Free-air temperature at 10,000 ft ≈ 34 − 2 × 5.5 ≈ 23 °C. ISA at 10,000 ft ≈ 15 − 1.98 × 10 ≈ −5 °C. Deviation is therefore +28 °C, essentially unchanged from the surface.
Density altitude at the crossing level ≈ 10,000 + 120 × 28 ≈ 13,360 ft.
The service ceiling is 14,000 ft, and the service ceiling is defined at ISA - it is the level at which the aeroplane still climbs at about 100 fpm. At 13,360 ft density altitude, on a hot afternoon, in a normally aspirated aeroplane, you have roughly 600 ft of theoretical margin and a real climb rate measured in tens of feet per minute. The last 2,000 ft of the climb will take 20–30 minutes, during which the temperature may rise further and the convection build.
The practical rule is never to plan a crossing that needs more than 70–80% of the available ceiling, because sink, downdraughts on the lee side of a ridge, and a forecast error of 2 °C (240 ft of density altitude) all eat into the margin. At 1500, 10,000 ft is not a safe crossing level for this aeroplane.
3. Estimate the convective cloud base at 1500 and at 0900, and state your decision.
Read the worked answer
Cloud base (ft AGL) ≈ temperature–dewpoint spread × 400 ft/°C.
| Time | Spread | Base AGL | Base AMSL |
|---|---|---|---|
| 0900 | 16 °C | 6,400 ft | 10,900 ft |
| 1500 | 22 °C | 8,800 ft | 13,300 ft |
At 1500 the crossing level (10,000 ft) is 3,300 ft below the base, so cloud clearance is not the problem - performance is. At 0900 the crossing level is only about 900 ft below the forecast base, which is marginal for any below-cloud clearance requirement and leaves no room to climb if a Cu forms on the ridge.
Decision: the 1500 departure is not viable with this aeroplane and this route. The 0900 departure is achievable but only just: plan 10,000 ft as a hard ceiling, brief a turn-back point at the ridge for any climb rate below 200 fpm, and expect the crossing level to sit close to the cloud base. If the operator needs a comfortable margin, the answer is a different aeroplane, a lower route, or a different time of year - not a different technique. Note also that FL200 tops mean a build-up over the ridge can close the pass regardless of density altitude.