FAA Instrument Flight Instructor (FII)WeatherHard
A winds and temperatures aloft forecast (FD) for a station shows: 9,000 ft — 2721+03; 12,000 ft — 2734−04. Based on a linear interpolation between these two levels, approximately where should the pilot expect the freezing level?
- A9,300 feet
- B12,000 feet
- C10,300 feet
- D11,300 feet
Show answer & explanationAnswer & explanation
Correct answer: C. 10,300 feet
The temperature drops from +3°C at 9,000 ft to −4°C at 12,000 ft, a change of 7°C over 3,000 ft (about 2.33°C per 1,000 ft). To reach 0°C from +3°C requires a drop of 3°C, which occurs over 3/2.33 ≈ 1,286 ft. Adding this to 9,000 ft gives approximately 10,300 ft, the estimated freezing level.
Why the other options are wrong
- A. This underestimates the distance needed to cool 3°C at the computed lapse rate.
- B. 12,000 ft is the upper reporting level, already colder than freezing (−4°C), well above the actual freezing level.
- D. This overestimates the altitude; it assumes too little cooling occurs before 0°C is reached.
Freezing Level from Winds/Temps Aloft Forecast
The freezing level can be estimated by linear interpolation between two reported temperatures at different altitudes in an FD forecast, finding the altitude where temperature crosses 0°C.
- Negative temperatures are prefixed with a minus sign; positive temps have no sign shown above certain levels
- Interpolate proportionally between two known temperature/altitude pairs
- Useful for anticipating icing risk and planning altitude changes
- Temperatures/winds are omitted near station elevation on FD forecasts
Memory trick: Find where the temp line crosses zero — that's your ice line