GenDOF simple question...or is it?

a new user to FAST here.

If I set GenDOF to FALSE and then define the the rotor speed, what torque is applied to the rotor shaft (besides the aerodynamic torque), if any at all?

I have taken the SWRT Burgey style model and switched the GenDOF to False, then set the rotor speed to 25 and had a wind input file running from 1 m/s to 18 m/s over 1500 seconds. I then repeated this for 50 rpm, 75 rpm etc. with the intention of creating a family of torque and power curves which would then allow me to plot a surface plot.

I am using RotTroq and Rotpwr which I have verified against RotCq 0.51.225windspeed^2swept arearadius and RotCp 0.512…5windspeed^3swept arearadius.

I changed the torque on the torque speed look up table written by KP and this had no effeect on the performance characteristics of the model leading me to conclude that when GenDOF is False, there is no torque applied to the rotor shaft by the generator. Therefore the RotCq and RotCp values I am using are a direct reflection of the rotor characteristics when unloaded.

Any comments on this would be greatly appreciated. I am also willing to accept that I have a weak grasp of this and am barking up the wrong tree.



Dear Phllip,

Disabling a degree-of-freedom (DOF) in FAST (or the ElastoDyn module FAST v8) really means that the acceleration will be zero, velocity will be constant, and displacement will vary linearly (if the constant velocity is nonzero), regardless of the loads applied. Inside FAST, the equation of motion for the disabled DOFs are not formulated, resulting in no acceleration of those DOF.

Thus, disabling GenDOF will force the generator to rotate at a constant speed regardless of the torques applied. With GenDOF enabled, the generator will accelerate/decelerate if the applied torques are not balanced.

As discussed here, the RotTorq output of FAST (equivalent to LSShftTq) will equal the applied aerodynamic torque if the shaft acceleration is zero.

Best regards,