XXavier
Member
- Joined
- Nov 13, 2006
- Messages
- 1,481
- Location
- Madrid, Spain
- Aircraft
- ELA R-100 and Magni M24 autogyros
- Total Flight Time
- 913 gyro (June 2023)
Chatting with a friend, he touched the issue of the minimum power required by a gyro to stay aloft. I didn't know an answer, but thinking about that question, I have found the following (tentative) answer. Any comments will be welcome...
I remember having read somewhere that an autogyro, in vertical autorotation, has a sink rate more or less like a parachute having the diameter of the rotor, and loaded with the same mass of the gyro.
I've found, in Wikipedia, the formula for the terminal velocity of a body of mass m, area s, falling with a drag coefficient Cd in air of density rho:
v = 1,41 · m^0,5 · g^0,5 · rho^–0,5 · s^–0,5 · Cd^–0,5
I get v = 9,3 m/s for an ELA gyro, with a mass of 400 kg and a rotor diameter of 8,5 m and assuming Cd = 1,3
That vertical autorotation speed sounds plausible... It's 1830 fpm... Now, for a descending parachute at a constant speed (or for a gyro in vertical autorotation), the drag (or the rotor lift) is equal in magnitude to the weight. The power associated with that moving force is m · g · v. For the ELA, that's 400 · 9,8 · 9,3 = 36456 W = 49,5 hp. It looks a bit high for minimum power required, since the efficiency of the prop may be around 0,6. That would mean a shaft power of 83 hp...
As stated above, any comments (or corrections) will be welcome...
I remember having read somewhere that an autogyro, in vertical autorotation, has a sink rate more or less like a parachute having the diameter of the rotor, and loaded with the same mass of the gyro.
I've found, in Wikipedia, the formula for the terminal velocity of a body of mass m, area s, falling with a drag coefficient Cd in air of density rho:
v = 1,41 · m^0,5 · g^0,5 · rho^–0,5 · s^–0,5 · Cd^–0,5
I get v = 9,3 m/s for an ELA gyro, with a mass of 400 kg and a rotor diameter of 8,5 m and assuming Cd = 1,3
That vertical autorotation speed sounds plausible... It's 1830 fpm... Now, for a descending parachute at a constant speed (or for a gyro in vertical autorotation), the drag (or the rotor lift) is equal in magnitude to the weight. The power associated with that moving force is m · g · v. For the ELA, that's 400 · 9,8 · 9,3 = 36456 W = 49,5 hp. It looks a bit high for minimum power required, since the efficiency of the prop may be around 0,6. That would mean a shaft power of 83 hp...
As stated above, any comments (or corrections) will be welcome...