Question about Sum of the out-of-plane moments

Dear all,

Recently, I am trying to sum up the blade 1,2,3 out-of-plane moments using FAST. The results suggest me that the sum of the blade out-of-plane moments tends to a constant, and has a prefered constant angle (in a stationaty coordinate system). Is this result reasonable? If yes, is the fact caused by the teeter? Or I had some bug in my proceedure, which is the following:

My=M1cos(A)+M2cos(A+2pi/3)+M3cos(A+4pi/3);
Mz=M1
sin(A)+M2sin(A+2pi/3)+M3sin(A+4pi/3);
M(out-of-plane)=(Mz^2+My^2)^0.5;
Direction=atan2(Mz/My,1)/pi*180;

Where M1, M2, M3 are Blade 1, 2, 3 out of plane moments(RootMyc1,RootMyc2,RootMyc3)
A is the Azimuth angle(Azimuth)

I appreciate any discussion and consideration. Thank you.

Sincerely,

Zhichao Yu
8/3/13

Dear Zhichao,

Your equations look correct for a 3-bladed rotor when the azimuth angle “A” is defined as zero degres for blade 1 pointing up. I don’t really understand your point about teeter because a 3-bladed turbine would not have a teeter degree of freedom. A constant rotor-total out-of-plane moment and direction would suggest that the rotor thrust and center-of-thrust remain stationary. I would expect this to be the case when simulating with a steady wind speed, but not if the wind is time-varying.

Best regards,

Thank you so much, Dr. Jonkman. Now I will continue the work about the out-of-plane moments.