Resistant moment of the rotor and of the electric generator


I’d like to know if you have the values of the rotational frequency of the rotor blades (rot_freq) in function of the mean wind speed at the hub (Um_hub) and of the aerogenerators that you test for exemples in FAST.

Is it possible also to have the trend of the resistant angular momentum in function of the Um_hub or of the rot_freq ?

In the analysis of the dynamic response of the aerogenerator with the rot_freq varing in time (FAST makes it, i believe), noting the moment around the rotor axis due to the wind speed, how FAST calculates the instant rot_freq ? Is not necessary to know the resistant moment of the rotor and of the electric generator rispect to the rotor axis ? How can i calculate it with the input qunatities known by FAST ?

I really thank you for supports, best regards

Lorenzo Montanari

Dear Lorenzo,

I’m not exactly sure what you are asking. For example: What do you mean by the “resistant angular momentum”? Do you mean the rotational inertia?

The average rotor rotational speed as a function of mean wind speed for the NREL 5-MW turbine, as computed by FAST simulations, can be found in Figure 9-1 of the specifications document: No such specifications documents have been published for the other models supplied in the FAST CertTest. You may be able to find reports that contain some information on these models (altough I can’t think of specific reports at the moment). To get exactly what you need, however, you will probably need to run the simulations yourself.

The rotor rotational speed is dictated by the overall system-dynamics equations of motions, including the effects of:
*Steady, periodic, and stochastic aerodynamic torque
*Rotor dynamics, including blade vibrations and hub inertia
*Generator inertia and generator reaction torque
*Gearbox friction
*Stiffness and damping of the drivetrain torsional spring

While the overall system-dynamics equations of motions would include rotor-rotation equations that couple with many other system DOFs, to clarify how the drivetrain model in FAST works, here are some points to note:

  1. HSShsftTq and GenTq are different. GenTq is the torque applied to the high-speed shaft from the generator, which is equally and oppositely applied as a reaction to the nacelle. HSShftTq is the torque transmitted through the HSS, which includes the inertia torque from generator acceleration or deceleration. (HSShftTq is the torque one would measure from a strain gage mounted on the HSS.) From a free-body diagram of a rigid drivetrain with no gearbox inefficiencies and the mass lumped in the rotor and generator, the equations are:

LSShftTq = T_Aero – J_RotorAlpha = HSShftTqGBRatio = T_GenGBRatio + J_GenAlpha*GBRatio^2

LSShftTq = torque transmitted through the low-speed shaft (LSS) = RotTorq - positive in the direction of positive Alpha
HSShftTq = torque transmitted through the high-speed shaft (HSS) - positive in the direction of positive Alpha
T_Aero = aerodynamic torque applied to the LSS - positive in the direction of positive Alpha
T_Gen = generator torque applied to the HSS - negative in the direction of positive Alpha
GBRatio = HSS to LSS gearbox ratio
J_Rotor = rotor inertia relative to the LSS
J_Gen = generator inertia relative to the HSS
Alpha = LSS rotational acceleration

So: LSShftTq = T_Aero = HSShftTqGBRatio = T_GenGBRatio only when the drivetrain is not accelerating or decelerating (Alpha = 0). The problem gets more complicated if you have a gearbox efficiency less than 100%, and even more complicated if you enable other system DOFs, such as flexibility in the drivetrain or blades. The system must then be represented by a series of differential equations (the equations of motion).

  1. A gearbox efficiency (GBoxEff) less than 100% causes the power of the output shaft to be less than the power of the input shaft by the efficiency. So, HSShftTq = LSShftTqGBoxEff/GBRatio (for positive torque) or HSShftTq = LSShftTq/(GBoxEffGBRatio) (for negative torque).

  2. A generator efficiency (GenEff) less than 100% causes the electrical power to be less than the mechanical power while generating or the mechanical power to be less than the electrical power while motoring. So, GenPwr = T_GenGenSpeedGenEff (for positive torque) or GenPwr = T_Gen*GenSpeed/GenEff (for negative torque).

  3. The drivetrain torsion DOF is modeled as part of the low-speed shaft. With appropriate units applied, in equation form: LSShftTq = DTTorSpr*(Azimuth-LSSGagPxs) + DTTorDmp*( RotSpeed- LSSGagVxs). This equation only applies when the drivetrain DOF is enabled. When the drivetrain is rigid, (Azimuth-LSSGagPxs) is zero, DTTorSpr is infinite, and their product is equal to the finite value of LSShftTq, as calculated in the first equation above.

I hope that helps.

Best regards,


To ensure that I understand the drivetrain dynamics correctly from your post, let me make a following point:

In case of your forth point, when torsion DOF is present in the drivetrain, the equation in point 1) would apply as

T_Aero – J_RotorAlpha = LSShftTq = DTTorSpr(Azimuth-LSSGagPxs) + DTTorDmp*( RotSpeed- LSSGagVxs)

Now, since the torsion DOF is only modeled (or transferred to) on LSS side of the gearbox and not on HSS side, the dynamics on HSS side would stay as
LSShftTq = HSShftTqGBRatio = T_GenGBRatio + J_GenAlphaGBRatio^2

Is this how FAST solves for alpha when torsion is present?


Dear Dee Doge,

Yes, your understanding is correct. However, when the drivetrain torsion is enabled, the “Alpha” multiplying J_Rotor does not equal the “Alpha” multiplying J_Gen. Instead, I would write:

LSShftTq = T_Aero – J_RotorRotAccel = DTTorSpr(Azimuth-LSSGagPxs) + DTTorDmp*( RotSpeed- LSSGagVxs) = HSShftTqGBRatio = T_GenGBRatio + J_GenLSSGagAxsGBRatio^2

where “Alpha” has been replaced with “RotAccel” for the rotor-side of the shaft and “LSSGagAxs” for the generator-side of the shaft, respectively (with appropriate units applied). RotAccel and LSSGagAxs are outputs available from FAST and will be equal when drivetrain torsion is disabled, but will differ when drivetrain torsion is enabled.

Best regards,

Thank you Jason . It was a decisif explanation for me .It helped me very much .


Not to revive an old thread but if I wanted to calculate the aerodynamic rotor torque I just need to:

T_Aero = LSShftTq + J_Rotor*RotAccel

where does J_Rotor get calculated in FAST?

Dear Andre,

I agree with your equation in the case of a rigid rotor (all blade and/or teeter degrees-of-freedom disabled). The rotor inertia, “J_Rotor”, is computed and output to the FAST summary (.fsm) file. However, if the rotor is not rigid, “J_Rotor” will change with blade deflection and/or rotor teeter during the course of a FAST simulation, so this calculation will only be exact for a rigid rotor (for a flexible rotor, the calculation will work when averaged over some length of time). More discussion on this topic is available in the following forum topic:

Best regards,

To further this question, what if T_Aero is solved via the full drivetrain equation of motion from the “5MW Reference Tower Offshore Definition” paper?

T_Aero = I_DrivetrainRotAccel + N_GearGenTrq

Is anything missing from this equation?

Dear Sean,

Your equation is also correct for a rigid rotor and drivetrain. Actually, your equation can be derived from the first equation in my Sep 29, 2011 post above (with a little notation change):

LSShftTq = T_Aero – J_RotorAlpha = HSShftTqGBRatio = T_GenGBRatio + J_GenAlphaGBRatio^2
→ T_Aero = ( J_Rotor + J_Gen
GBRatio^2 )Alpha + T_GenGBratio
→ T_Aero = J_DrivetrainAlpha + T_GenGBratio

where J_Drivetrain = J_Rotor + J_Gen*GBRatio^2.

Best regards,

Thanks for the response Jason. Admittedly, I was verifying my work. :slight_smile: