Aviation Mathematics & Calculations Flashcards

Using the formula WCA ≈ (wind speed ÷ TAS) × sin(wind angle) × 60, if wind is 20 knots, TAS is 120 knots, and the wind angle is 90°, what is the wind correction angle?

Answer: 10°

WCA = (20 ÷ 120) × sin(90°) × 60 = 0.167 × 1 × 60 = 10°.

An aircraft weighs 4,000 lbs and enters a 60° bank level turn where the load factor is 2.0. What is the effective structural load on the aircraft?

Answer: 8,000 lbs

Effective load = aircraft weight × load factor = 4,000 × 2.0 = 8,000 lbs.

A glider has a best glide ratio of 20:1. At 6,000 ft AGL (using 6,000 ft ≈ 1 nm), what is the maximum glide distance?

Answer: 20 nm

Glide distance = altitude × glide ratio = 6,000 ft × 20 = 120,000 ft ÷ 6,000 ft/nm = 20 nm.

Pressure altitude is 4,000 ft with an ISA temperature deviation of +20°C. Using 120 ft per degree of deviation, what is the density altitude?

Answer: 6,400 ft

Density altitude = 4,000 + (20 × 120) = 4,000 + 2,400 = 6,400 ft.

An aircraft flies at 150 knots TAS heading 090°. Wind is from 180° at 30 knots (a pure crosswind). What is the approximate groundspeed?

Answer: 153 knots

With a pure 90° crosswind, GS = √(TAS² + wind²) = √(150² + 30²) = √23,400 ≈ 153 knots.

An aircraft requires 2,400 ft of runway at sea level. At density altitude 5,000 ft, the correction factor is 1.4. What is the new required takeoff distance?

Answer: 3,360 ft

New takeoff distance = 2,400 × 1.4 = 3,360 ft.

An aircraft has a total moment of 264,000 in-lbs and a CG located 88 inches aft of the datum. What is the total weight of the aircraft?

Answer: 3,000 lbs

Weight = total moment ÷ CG arm = 264,000 ÷ 88 = 3,000 lbs.