|May 02, 2002, 04:31 PM|
Head Speed vs. Pitch, which is more efficient?
Ok, I think this one is for the more experienced heli flyer (compared to me that would be anyone ).
It takes a given amount of thrust the hover or fly and it seems that there are two ways to get it, low head speed and higher pitch or high head speed and lower pitch. So the question is which is more efficient? Anyone do any testing in this area or have an explanation for a hungry mind?
Soon to be heli pilot
|May 02, 2002, 05:21 PM|
Joined Apr 2000
Normally, drag varies as the square of the pitch angle but this is offset by the drag decrease due to higher Reynold's number of higher speed blades. This is true especially for small helis (24" rotors or so) but becomes less important for Logo 20 size helis.
|May 02, 2002, 05:25 PM|
I'd say a higher head speed would offer more advantage. Higher head speed means more control inputs per second, so you have decreased response time to directional controls (cyclic). Plus, as Hank pointed out, less drag.
I've used lower head speeds on my Eco and when the response gets sluggish, bad things are known to happen.
|May 02, 2002, 05:52 PM|
It all depends on the KV of your motor, your gearing, and the most efficient point on the torque speed curve for your motor. There is usually that happy medium that you want to operate on in order to achieve the most efficient performance.
|May 02, 2002, 06:15 PM|
Alot depends on the size of the heli too and the control setup.
A logo 10 will fly with a low HS (1200), but as posted above control can be sluggish. It is also happy at 1700+ with faster response times.
Disc loading will play a big factor too. A light disc loading will yield longer flight too.
Then take a voyager which won't like to fly at all unless it has a HS of over 1500.
Lager helis will tolerate slower HS, and small helis tend to like higher HS.
|May 02, 2002, 07:20 PM|
Matsuyama city, Japan
Joined May 2002
Ignoring the motor performance and considering only the aerodynamics of main rotor, large pitch and low rpm are defenitely higher efficient. It means to need lower input power for the same thrust power, so that, leading to longer flight duration.
This is not applicable for extreme low rpm (less than 900-1100rpm), where the main rotor fails to catch the air. In addition, low rpm leads to less (milder) cyclic pitch response.
|May 02, 2002, 10:49 PM|
Joined Apr 2000
If my math is correct, it goes like this:
Case 1: high blade angle, low rpm
Case 2: low blade angle, high rpm
Rotor lift = k*B*N^2*D^3
k is a constant, B is blade pitch angle, N is rotor rpm and D is the rotor diameter.
The lift in both case 1 and 2 must be equal. Then since the same rotor and constant is used in both cases:
B(1)*N(1)^2 = B(2)*N(2)^2
Assume B(1) = 2*B(2) (a) and substituting
2*B(2)*N(1)^2 = B(2)*N(2)^2 or
2*N(1)^2 = N(2)^2 and Finally
N(1) = N(2)/(2^(1/2)) (b)
Rotor drag = k*B^2*N^2*D^3/R for R less than 10^5.
Where the Reynold's number, R = kN
D = k*B^2*N*D^3
The ratio of the drags = (B(1)^2*N(1))/(B(2)^2*N(2)).
Substituting from (a) and (b) above
Drag ratio = (2*B(2)^2*N(2)/2^(1/2))/(B(2)^2*N(2))
or drag ratio = 2/(2^(1/2)) = 2^(1/2) = 1.414
THE BOTTOM LINE: If you double the blade angle you increase the drag by 1.414 where the Reynold's number is less than 10^5.
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