A helicopter has a main rotor diameter of 35 feet and a gross weight of 2,000 pounds. What is the main rotor disk loading (in pounds per square foot), and what does a higher disk loading generally indicate?
- A57.1 lb/ft²; higher disk loading always improves autorotation performance
- B4.16 lb/ft²; higher disk loading reduces the power required to hover
- C2.08 lb/ft²; higher disk loading generally requires more power to hover and produces a higher rate of descent in autorotation
- D2.08 lb/ft²; disk loading has no effect on power required or autorotation descent rate
Show answer & explanationAnswer & explanation
Correct answer: C. 2.08 lb/ft²; higher disk loading generally requires more power to hover and produces a higher rate of descent in autorotation
Disk area = π × r² = π × (17.5 ft)² = 3.1416 × 306.25 = 962.1 ft². Disk loading = gross weight ÷ disk area = 2,000 lb ÷ 962.1 ft² ≈ 2.08 lb/ft². Higher disk loading (heavier weight relative to disk area) requires more power to hover and results in a higher rate of descent during autorotation, since less rotor area is available to generate lift per pound of aircraft weight.
Why the other options are wrong
- A. 57.1 lb/ft² is an incorrect calculation, and higher disk loading worsens, not improves, autorotation.
- B. 4.16 lb/ft² miscalculates the disk area, and higher disk loading increases, not reduces, power required.
- D. The math is correct, but disk loading does significantly affect power and autorotation performance.
Rotor Disk Loading
Gross weight divided by main rotor disk area (lb/ft²); higher disk loading means more power is required to hover and a higher autorotational descent rate.
- Disk area = π × r² (r = half of rotor diameter)
- Disk loading = gross weight ÷ disk area
- Higher disk loading requires more power to hover
- Higher disk loading increases autorotation rate of descent
Memory trick: More weight per circle of rotor blades = more power to lift, faster fall if engine quits.