ILS Approach with DME and ADF Cross-Check
You are flying an ILS approach to runway 27. The glideslope is 3.0°, the threshold elevation is 250 ft, and the ILS DME reads distance to the threshold. At 5.2 DME, with 140 kt groundspeed, your altimeter reads 2,100 ft. The localizer is centred, but the glideslope pointer is one dot above centre. The published approach uses a 3° path and requires a stabilized approach by 3 DME. An NDB is located 1 NM left of the final approach course at the threshold; with heading 270°M, the ADF relative bearing is 020°. Magnetic variation is 8°E. The aircraft is cleared for the approach.
1. Calculate the nominal glideslope altitude at 5.2 DME and 3.0 DME, determine the vertical deviation, and decide on the required action.
Read the worked answer
Nominal GP height at 5.2 DME = 5.2 × 6076 × tan 3° = 1,655 ft above threshold; add 250 ft = 1,905 ft. Actual 2,100 ft → 195 ft high. At 3 DME: 3 × 6076 × tan 3° = 955 ft + 250 = 1,205 ft. To lose 895 ft in 2.2 NM at 140 kt: time = 2.2/140 × 60 = 0.943 min; required ROD = 895/0.943 = 949 ft/min. Normal 3° ROD at 140 kt = 140 × 6076 × tan 3°/60 = 742 ft/min. The required correction is about 28% steeper than normal and close to 1,000 ft/min, so a stabilized approach by 3 DME is unlikely if the deviation is not corrected immediately. Correct at once; if not back on path by 4 DME, go around.
2. Compute the magnetic and true bearing to the NDB from the ADF indication and assess whether it is consistent with the aircraft position on the localizer.
Read the worked answer
MB to NDB = heading + relative bearing = 270 + 020 = 290°M. True bearing = 290 + 8 = 298°T. On the centreline at 5.2 DME, the NDB 1 NM left of the threshold should bear arctan(1/5.2) = 10.9° left of the 270°M course, i.e. about 259°M. The ADF shows 290°M, which is 31° right of the expected bearing and 20° right of the aircraft nose. With the localizer centred and DME stable, this is a gross inconsistency. Likely causes are the wrong NDB tuned, a bearing indicator error, or an incorrect database/plate position. Do not use the ADF for guidance; continue on the ILS and verify the NDB before using it for the missed approach.
3. The glideslope pointer is one dot above centre. If one dot corresponds to 0.35°, compute the vertical error at 5.2 DME and compare it with your answer.
Read the worked answer
One dot = 0.35°. At 5.2 DME, the vertical displacement is 5.2 × 6076 × tan 0.35° = 193 ft. This matches the 195 ft calculated from the altimeter and glideslope geometry. The glideslope receiver, DME and altimeter are therefore consistent: the aircraft really is about one dot high, not suffering a display or altimeter error. The decision should be based on the resulting 949 ft/min correction requirement, not on a hope that the indication is false.