The inertia of the drivetrain brings about generalized inertia forces and the load in the generator, high-speed shaft brake, gearbox (friction forces resulting from nonzero GBoxEff ) and the flexibility of the low speed shaft bring about generalized active forces. Note that all of these equations assume that the rotor is spinning about the positive $c_{1}$ axis (they assume that the rotor can’t be forced to rotate in the opposite direction). This model works for any gearbox arrangement (including no gearbox, single stage, or multi-stage) as long as the generator rotates about the shaft axis (it may not be skewed relative to the shaft, even though it may rotate in the opposite direction of the low-speed shaft due to the gearbox stages). If there is no gearbox, simply set $G B R a t i o\,=\,G B o x E f f=G e n D i r=1$ (GB R $e\nu e r s e={\mathrm{False}}$ ).
传动系惯性带来广义惯性力和发电机负载、高速轴制动器、齿轮箱（由于GBoxEff非零导致的摩擦力）以及低速轴挠曲性，都会产生广义主动力。需要注意的是，这些方程都假设转子绕正$c_{1}$轴旋转（即假设转子不能被强制旋转到相反方向）。该模型适用于任何齿轮箱布置（包括无齿轮箱、单级或多级），只要发电机绕轴旋转即可（即使它可能由于齿轮箱级数而以低速轴相反的方向旋转，但不能相对于轴倾斜）。如果不存在齿轮箱，只需将$G B R a t i o\,=\,G B o x E f f=G e n D i r=1$ (GB $Rev e r s e={\mathrm{False}}$ )。
The mechanical torque within the generator is applied to the high speed shaft and equally and oppositely to the structure that furls with the rotor as follows:
\left.F_{r}\right|_{G e n}=\left\{\begin{array}{l l}{{^{E}\pmb{\omega}_{G e A z}^{G}\cdot M_{G e n}^{G}}}&{{f o r~~r=G e A z}}\\ {{0}}&{{o t h e r w i s e}}\end{array}\ \right.\tag{2}
\boldsymbol{M_{G e n}^{G}}=-G e n D i\boldsymbol{r}\cdot\boldsymbol{T^{G e n}}\left(G B R a t i o\cdot\dot{q}_{G e d z},t\right)\boldsymbol{c}_{1}\tag{3}
Note that a positive $T^{G e n}$ represents a load (positive power extracted) whereas a negative $T^{G e n}$ represents a motoring-up situation (negative power extracted, or power input). Thus,
F_{r}\Big|_{G e n}=\left\{\begin{array}{l l}{\left(G e n D i r\cdot G B R a t i o\cdot c_{1}\right)\cdot\left[-G e n D i r\cdot T^{G e n}\left(G B R a t i o\cdot\dot{q}_{G e d z},t\right)c_{1}\right]}&{f o r\;\;\;r=G e Az}\\ {0}&{o t h e r w i s e}\end{array}\right.\tag{4}
Similarly, the mechanical torque applied to the high-speed shaft from the high-speed shaft brake is applied equally and oppositely to the structure that furls with the rotor. Thus,
F_{r}\big|_{B r a k e}=\left\{\begin{array}{l l}{{^{E}\pmb{\omega}_{G e A z}^{G}\cdot M_{B r a k e}^{G}}}&{{f o r\;\;\;r=G e A z}}\\ {{0}}&{{o t h e r w i s e}}\end{array}\!\!\right.\tag{8}
$$
where
$$M_{B r a k e}^{G}=-G e n D i r\cdot T^{B r a k e}\left(t\right)c_{1}\qquad\mathrm{~and~where}\qquad T^{B r a k e}\left(t\right)=H S S B r k T\left(t\right)\tag{9}
F_{r}\bigr|_{B r a k e}=\left\{\begin{array}{l l}{-G B R a t i o\cdot T^{B r a k e}\left(t\right)}&{f o r\;\;\;r=G e A z}\\ {O}&{o t h e r w i s e}\end{array}\right.\tag{10}
If the translational inertia of the drivetrain is assumed to be incorporated into that of the structure that furls with the rotor, then the high-speed shaft generator inertia generalized force is as follows:
However, since $c_{1}\cdot\frac{d}{d t}(^{E}\omega_{G e A z}^{G}\Big)\propto c_{1}\cdot(^{E}\omega^{R}\times c_{1}\Big)={^{E}\omega^{R}}\cdot\Big(c_{1}\times c_{1}\Big)=0$ (the first $c_{1}$ coming from $\overline{\overline{{I}}}^{\text{G}}$ ), this simplifies as follows:
However since $c_{1}\cdot\left({\,^{E}}{\omega^{G}}\times c_{1}\,\right)={^{E}}{\omega^{G}}\cdot\left(c_{1}\times c_{1}\right)=0$ (the first $c_{1}$ coming from $\overline{{\overline{{I}}}}^{G}$ ), this simplifies again as follows:
The terms associated with DOFs 4,5,…,12 represent the fact that the rate of change of angular momentum of the generator can be considered as an additional torque on the structure that furls with the rotor (i.e., in addition to the torques on the structure transmitted directly from the low-speed shaft).
Similar to the generator and HSS brake, the mechanical friction torque applied to the high speed shaft is applied equally and oppositely to the structure that furls with the rotor. Thus,
where, from a free-body diagram of the high and low-speed shafts, it is easily seen that the friction torque applied on the LSS upon the gearbox entrance, $T^{G B F r i c}\left(\ddot{q},\dot{q},q,t\right)$ , is always positive in value and equal to:
从高速轴和低速轴的受力图上可以清楚地看到，作用于低速轴进入变速箱处的摩擦转矩，$T^{G B F r i c}\left(\ddot{q},\dot{q},q,t\right)$，其值始终为正，且等于：
It is noted that the gearbox friction generalized active force effects both the mass matrix and the forcing function. Its effect on the mass matrix can be thought of as an apparent mass in the system (i.e., an active friction force which is a function of the generator acceleration). The gearbox friction causes the mass matrix to become unsymmetric. Note that the equation for DOF GeAz is an implicit equation (since the gearbox friction is a function of DOF GeAz), which has no closed-form solution. To avoid the complications resulting from this implicit characterization, the value of the LSShftTq from the previous time step is used in the SIGN() function in place of the value of the LSShftTq in the current time step. This may be done since it may be assumed that LSShftTq will not change much between time steps if the time step is considered small enough. Note that gearbox friction is the only term in the equations of motion that cause the mass matrix to be unsymmetrical. The mass matrix will only be unsymmetric if GBoxEff is not $100\%$ —if GBoxEff is $100\%$ , then the mass matrix is completely symmetric.
