effect of GBRevers


I’m just going over some of the parameters that FAST requires.

GBRevers - indicates if the output shaft of the gearbox is the same as the input shaft (T). The variable HSShftTq does not seem to be affected by GBRevers if it has the same sign as LSShftTq (I think it should be affected, but if it did it might confuse people that hadn’t thought about it and since it doesn’t, it might confuse people that have thought about it… maybe).

I think the effect is that with a gearbox with a ratio of 67:1 (for example) and rotation in the same direction (torque in opposite directions), the torque imparted by the gearbox to its mounting would be 66/67 * LSShftTq and when the shafts are rotating in the opposite direction (torque in the same direction) the torque imparted by the gearbox on its mounting would be 68/67 * LSShftTq.

The manual isn’t that clear about the effect of GBRevers. Is the above accurate?

There doesn’t seem to be an output for Gearbox torque resisted by gearbox mounting. Is that correct?

Would it be worth adding one?



Dear Alec,

Its been a long time since I’ve thought about FAST input parameter GBRevers. As far as I know, it is very rarely (if ever) used (GBRevers is typically set to False). FAST is set up such that the torque outputs are independent of the setting of GBRevers. That is, for example, the generatorq torque output by FAST, GenTq, is positve when producing power and negative when motoring regardless of the value of GBRevers. If I recall correctly, the setting of GBRevers really only impacts some of the generator inertia terms, which tend to be small (e.g., the impact of the generator inertia on the gyroscopic yaw moment).

I’m not sure I really understand your example regarding the torque imparted by the gearbox.

While most of the rotor torque is transferred into the tower through the generator torque (and perhaps the high-speed shaft (HSS) brake torque, if engaged), some of the torque will be absorbed through the gearbox if the gearbox has friction. Gearbox friction can be added to a FAST model by setting the gearbox efficiency less than 100% (GBoxEff < 100 (%)). In this case, a gearbox friction torque is calculated inside FAST, but you are right that it is not currently output. However, you can easily calculate the gearbox friction torque relative to the low-speed shaft (LSS), GBoxTq, from knowledge of the LSS torque, LSShftTq, HSS torque, HSShftTq, and gearbox ratio, GBRatio. That is:

GBoxTq = LSShftTq - HSShftTq*GBRatio

An equivalent relationship is:

GBoxTq = LSShftTq*( 1 - (GBoxEff/100) ) when the generating producing power, or:
GBoxTq = LSShftTq*( 1 - 1/(GBoxEff/100) ) when the generator is motoring.

I hope that helps.

Best regards,

Thanks Jason

Maybe another way to describe it is that with one direction of HSS rotation, the torque that the gearbox mounting needs to react is LSShftTq + HSShftTq and with the other it would be LSShftTq - HSShftTq, the latter having slightly lower loads.

The component that is reacting the torque is the torque tube in the image below (courtesy of Nordic Windpower). I’m not sure about putting other people’s pictures on your site. Please remove if it is inappropriate. I think anything on the web is for public consumption if you quote the source.

Most of the torque reacted by the tower is therefore from the LS Shaft, not the HSS.

I take the point about the gyroscopic effects and it is easy to add/subtract the torques manually if you’re interested in designing the gearbox mounting using FAST.


Dear Alec,

OK, now I understand what you are saying and agree with it. I’ll try to clarify things a bit:

Let T_LSS represent the torque in the low-speed shaft at one end of the gearbox and T_HSS represent the torque in the high-speed shaft at the other end of the gearbox. Also, let T_LSS and T_HSS be defined as positive-valued when the torque is such that right-hand rule for the torque has your thumb pointing downwind regardless of the direction of shaft rotation. Then, the the torque transferred through the gearbox mounting, T_Mount, would be:

T_Mount = T_LSS + T_HSS

In FAST, however, output HSShftTq doesn’t equal T_HSS because the sign of HSShftTq is tied to the sign of LSShftTq. That is, the following relationships are used in FAST:

LSShftTq = T_LSS

But you can use GBRevers to determine the proper sign of T_HSS based on the values of LSShftTq and HSShftTq output by FAST:

T_LSS = LSShftTq
T_HSS = HSShftTq when GBRevers = False
T_HSS = -HSShftTq when GBRevers = True

Thus you can write the torque transferred through the gearbox mounting based on the available FAST outputs as follows:

T_Mount = LSShftTq + HSShftTq when GBRevers = False
T_Mount = LSShftTq - HSShftTq when GBRevers = True

which agrees with the equations you gave.

Best regards,