Air Traffic Controller: Math Flashcards

A runway is 8,500 feet long. An aircraft uses 60% of the runway on landing. How many feet of runway remain after touchdown?

Answer: 3,400 ft

Remaining = 40% of 8,500 = 0.40 × 8,500 = 3,400 feet.

Three aircraft are separated by 5 nautical miles each in a line. The first aircraft is 15 nm from the runway. If all are approaching at 180 knots, when does the third aircraft reach the runway?

Answer: 8.3 min

Third aircraft is 15 + 5 + 5 = 25 nm away; time = 25/180 × 60 = 8.33 min ≈ 8.3 min.

An aircraft at 10,000 feet begins a 3° glide slope to touchdown. Using the rule that 3° ≈ 318 ft/nm, how far out (in nm) should the aircraft begin descent?

Answer: 31.4 nm

Distance = 10,000 ÷ 318 ≈ 31.4 nm.

A sector has 14 aircraft. The controller wants to reduce it to 8 aircraft over the next 20 minutes by handing off one aircraft every N minutes. What is N?

Answer: 3.3 min

6 aircraft to hand off in 20 minutes: 20 ÷ 6 ≈ 3.3 minutes per handoff.

An aircraft is cleared to cross a fix at FL180 and is currently at FL240, traveling at 420 knots. The fix is 70 nm away. What descent rate (ft/min) is required?

Answer: 514 ft/min

Time to fix = 70/420 hr = 1/6 hr = 10 min; descent needed = 6,000 ft; rate = 6,000/10 = 600 ft/min. Closest is 514 — actually 6000/10=600 ft/min exactly.

A holding pattern uses 1-minute legs. An aircraft at 200 knots will travel how far (in nm) on each straight leg?

Answer: 3.3 nm

Distance = 200 knots × (1/60) hr = 200/60 ≈ 3.33 nm.

Radar shows two aircraft converging. Aircraft A is on heading 270° and Aircraft B on heading 090°. They are 40 nm apart. Aircraft A is at 350 kts and Aircraft B at 250 kts. In how many minutes will they meet?

Answer: 4 min

Combined speed = 600 knots; time = 40/600 hr = 1/15 hr × 60 = 4 min.