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线性化求解器/参考文献/Kallesøe-Equations of motion for a rotor blade/auto/Kallesøe-Equations of motion for a rotor blade.md

@@ -98,7 +98,7 @@ To avoid unnecessary complications of the equations of motion, relatively small
 Before deriving the equations of motion, a transformation between the rotating $(x,\,y,\,z)$ -frame in which the blade deflection is described and the inertial $(X,\,Y,\,Z)$ -frame is found:  
 在推导运动方程之前，首先找到一个在其中描述叶片变形的旋转 $(x,\,y,\,z)$ 坐标系与惯性 $(X,\,Y,\,Z)$ 坐标系之间的变换：
 $$
-[\mathbf{i},\mathbf{j},\mathbf{k}]^{\mathrm{T}}\,{=}\,\mathbf{T}_{\beta}\mathbf{T}_{\phi}[\mathbf{I},\mathbf{J},\mathbf{K}]^{T}
+[\mathbf{i},\mathbf{j},\mathbf{k}]^{\mathrm{T}}\,{=}\,\mathbf{T}_{\beta}\mathbf{T}_{\phi}[\mathbf{I},\mathbf{J},\mathbf{K}]^{T}\tag 1
 $$  
 
 where $[\mathbf{i},\mathbf{j},\mathbf{k}]^{\mathrm{T}}$ and $[\mathbf{I},\mathbf{J},\mathbf{K}]^{\mathrm{T}}$ are the unit vectors in the $(x,y,z)$ and $(X,Y,Z)$ -frames, respectively. The matrices ${\bf{T}}_{\beta}$ and $\mathbf{T}_{\phi}$ are the transformations from the $(\hat{x},\,\hat{y},\,\hat{z})$ -frame to the $(x,\,y,\,z)$ -frame and from the $\left(X,\,Y,\,Z\right)$ - frame to the $\left(\hat{x},\hat{y},\hat{z}\right)$ -frame, respectively. Both matrices are given in Appendix A.  
@@ -113,20 +113,20 @@ The transformation between the principle axis and the $(x,y,z)$ -frame is given
 The strain in the blade is measured by Green’s strain tensor (cf. Hodges and Dowell3):  
 叶片应变由格林应变张量测量（参阅 Hodges and Dowell3）：
 $$
-2[\mathrm{d}s,\mathrm{d}\eta,\mathrm{d}\xi][\varepsilon_{i j}][\mathrm{d}s,\mathrm{d}\eta,\mathrm{d}\xi]^{\mathrm{T}}=\mathrm{d}\mathbf{r}_{\mathrm{l}}\cdot\mathrm{d}\mathbf{r}_{\mathrm{l}}-\mathrm{d}\mathbf{r}_{\mathrm{0}}\cdot\mathrm{d}\mathbf{r}_{\mathrm{0}}
+2[\mathrm{d}s,\mathrm{d}\eta,\mathrm{d}\xi][\varepsilon_{i j}][\mathrm{d}s,\mathrm{d}\eta,\mathrm{d}\xi]^{\mathrm{T}}=\mathrm{d}\mathbf{r}_{\mathrm{l}}\cdot\mathrm{d}\mathbf{r}_{\mathrm{l}}-\mathrm{d}\mathbf{r}_{\mathrm{0}}\cdot\mathrm{d}\mathbf{r}_{\mathrm{0}}\tag 2
 $$  
 
 where d denotes the differential, $\varepsilon_{i j}$ is the strain tensor and  
 其中 d 表示微分，$\varepsilon_{i j}$ 为应变张量，且
 $$
-\mathbf{r}_{0}\!=\![\mathbf{I},\mathbf{J},\mathbf{K}]\mathbf{T}_{\phi}^{\mathrm{T}}\mathbf{T}_{\beta}^{\mathrm{T}}\!\left[[l_{p i},0,w_{0}]^{\mathrm{T}}+(\mathbf{T}_{e}|_{u=\nu=\theta=0})^{\mathrm{T}}[\eta_{0},\xi_{0},0]^{\mathrm{T}}\right]
+\mathbf{r}_{0}\!=\![\mathbf{I},\mathbf{J},\mathbf{K}]\mathbf{T}_{\phi}^{\mathrm{T}}\mathbf{T}_{\beta}^{\mathrm{T}}\!\left[[l_{p i},0,w_{0}]^{\mathrm{T}}+(\mathbf{T}_{e}|_{u=\nu=\theta=0})^{\mathrm{T}}[\eta_{0},\xi_{0},0]^{\mathrm{T}}\right]\tag 3
 $$  
 
 is a position vector describing a point in the undeformed blade, where $(\eta_{0},\,\xi_{0})$ is the position of the point in the undeformed blade section. The same point in the deformed blade is given by  
 
 是一个位置矢量，描述了未变形叶片中的一个点，其中 $(\eta_{0},\,\xi_{0})$ 是该点在未变形叶片截面中的位置。变形叶片中的同一点由下式给出
 $$
-\mathbf{r}_{\mathrm{1}}\!=\![\mathbf{I},\mathbf{J},\mathbf{K}]\mathbf{T}_{\phi}^{\mathrm{T}}\mathbf{T}_{\beta}^{\mathrm{T}}\big[[l_{p i}+u,\nu,w_{0}+w_{\mathrm{1}}]^{\mathrm{T}}+\mathbf{T}_{\mathrm{c}}^{\mathrm{T}}{[\eta_{\mathrm{1}},\xi_{\mathrm{1}},0]}^{\mathrm{T}}\big]
+\mathbf{r}_{\mathrm{1}}\!=\![\mathbf{I},\mathbf{J},\mathbf{K}]\mathbf{T}_{\phi}^{\mathrm{T}}\mathbf{T}_{\beta}^{\mathrm{T}}\big[[l_{p i}+u,\nu,w_{0}+w_{\mathrm{1}}]^{\mathrm{T}}+\mathbf{T}_{\mathrm{c}}^{\mathrm{T}}{[\eta_{\mathrm{1}},\xi_{\mathrm{1}},0]}^{\mathrm{T}}\big]\tag 4
 $$  
 
 where $(\eta_{1},\,\xi_{1})$ is the position of the point in the deformed blade section.  
@@ -135,13 +135,13 @@ where $(\eta_{1},\,\xi_{1})$ is the position of the point in the deformed blade
 Assuming uniaxial stress $\sigma_{22}=\sigma_{33}=\sigma_{23}=0$ , where $\sigma_{i j}$ is the stress tensor. Applying Hook’s law gives $\varepsilon_{22}$ $=\varepsilon_{33}=-\nu\varepsilon_{11}$ , where $\nu$ is Poisson’s ratio. By expanding these relations to second order of the bookkeeping parameter $\varepsilon$ , it can be shown that $\eta_{1}=\eta_{0}$ and $\xi_{1}=\xi_{0}$ to second order. Expanding the remanding strain tensor components to second order of $\varepsilon$ gives  
 假设单轴应力 $\sigma_{22}=\sigma_{33}=\sigma_{23}=0$，其中 $\sigma_{i j}$ 是应力张量。应用胡克定律得到 $\varepsilon_{22}$ $=\varepsilon_{33}=-\nu\varepsilon_{11}$，其中 $\nu$ 是泊松比。将这些关系式展开到记账参数 $\varepsilon$ 的二阶，可以证明 $\eta_{1}=\eta_{0}$ 和 $\xi_{1}=\xi_{0}$ 到二阶。将剩余的应变张量分量展开到 $\varepsilon$ 的二阶，得到
 $$
-\begin{array}{l l}{{\displaystyle\varepsilon_{11}\!=\!-u^{\prime\prime}\big(\eta\cos(\overline{{\theta}})-\xi\sin\!\big(\hat{\theta}\big)\big)-\nu^{\prime\prime}\big(\eta\sin\!\big(\hat{\theta}\big)+\xi\cos\!\big(\overline{{\theta}}\big)\big)}}\\ {{\displaystyle\varepsilon_{12}\!=\!-\frac{1}{2}\xi\theta^{\prime},\quad\!\varepsilon_{13}\!=\!\frac{1}{2}\eta\theta^{\prime}}}\end{array}
+\begin{array}{l l}{{\displaystyle\varepsilon_{11}\!=\!-u^{\prime\prime}\big(\eta\cos(\overline{{\theta}})-\xi\sin\!\big(\hat{\theta}\big)\big)-\nu^{\prime\prime}\big(\eta\sin\!\big(\hat{\theta}\big)+\xi\cos\!\big(\overline{{\theta}}\big)\big)}}\\ {{\displaystyle\varepsilon_{12}\!=\!-\frac{1}{2}\xi\theta^{\prime},\quad\!\varepsilon_{13}\!=\!\frac{1}{2}\eta\theta^{\prime}}}\end{array}\tag 5
 $$  
 
 Using engineering strain $\varepsilon_{s s}=\varepsilon_{11}$ , $\varepsilon_{s\eta}=2\varepsilon_{12}$ , $\varepsilon_{s\xi}=2\varepsilon_{13}$ and stresses $\sigma_{\mathrm{ss}}=E\varepsilon_{s s},\,\sigma_{\mathrm{s}\eta}=G\varepsilon_{s\eta},\,\sigma_{s\xi}=G\varepsilon_{s\xi}$ where $E$ is the tensile modulus of elasticity (Young’s modulus) and $G$ is the shear modulus of elasticity, the elastic energy becomes  
 使用工程应变 $\varepsilon_{s s}=\varepsilon_{11}$ 、$\varepsilon_{s\eta}=2\varepsilon_{12}$ 、$\varepsilon_{s\xi}=2\varepsilon_{13}$ 和应力 $\sigma_{\mathrm{ss}}=E\varepsilon_{s s}$、$\sigma_{\mathrm{s}\eta}=G\varepsilon_{s\eta}$、$\sigma_{s\xi}=G\varepsilon_{s\xi}$，其中 $E$ 为拉伸弹性模量（杨氏模量），$G$ 为剪切弹性模量，弹性应变能变为
 $$
-\delta V_{e l a}=\int_{r}^{R}\iint_{A}{(\sigma_{s s}\delta\varepsilon_{s s}+\sigma_{s\eta}\delta\varepsilon_{s\eta}+\sigma_{s\xi}\delta\varepsilon_{s\xi})\mathrm{d}\eta\mathrm{d}\xi\mathrm{d}s}
+\delta V_{e l a}=\int_{r}^{R}\iint_{A}{(\sigma_{s s}\delta\varepsilon_{s s}+\sigma_{s\eta}\delta\varepsilon_{s\eta}+\sigma_{s\xi}\delta\varepsilon_{s\xi})\mathrm{d}\eta\mathrm{d}\xi\mathrm{d}s}\tag 6
 $$  
 
 The potential energy associated with the gravity field measured from the inertial frame $\left(X,\,Y,\,Z\right)$ is described by  
@@ -149,13 +149,13 @@ The potential energy associated with the gravity field measured from the inertia
 
 
 $$
-V_{g r a}=\int_{r}^{R}\mathbf{r}_{c g}^{T}\cdot\mathbf{g}\mathrm{d}s
+V_{g r a}=\int_{r}^{R}\mathbf{r}_{c g}^{T}\cdot\mathbf{g}\mathrm{d}s\tag 7
 $$  
 
 where $\mathbf{g}=[0,\,0,\,-g]^{\mathrm{T}}$ is the gravity field and  
 
 $$
-\mathbf{r}_{c g}\!=\![\mathbf{I},\mathbf{J},\mathbf{K}]\mathbf{T}_{\phi}^{\mathrm{T}}\mathbf{T}_{\beta}^{\mathrm{T}}\big[[l_{p i}+u,\nu,w_{0}+w_{1}]^{\mathrm{T}}+\mathbf{T}_{\mathrm{c}}^{\mathrm{T}}[l_{c g},0,0]^{\mathrm{T}}\big]
+\mathbf{r}_{c g}\!=\![\mathbf{I},\mathbf{J},\mathbf{K}]\mathbf{T}_{\phi}^{\mathrm{T}}\mathbf{T}_{\beta}^{\mathrm{T}}\big[[l_{p i}+u,\nu,w_{0}+w_{1}]^{\mathrm{T}}+\mathbf{T}_{\mathrm{c}}^{\mathrm{T}}[l_{c g},0,0]^{\mathrm{T}}\big]\tag 8
 $$  
 
 is a position vector describing the center of gravity.  
@@ -165,7 +165,7 @@ is a position vector describing the center of gravity.
 The inertia of the system is described by a mass pr. length $m$ , a moment of rotational inertia pr. length $I_{c g}$ of the blade and a moment of rotational inertia $J_{g e n}$ that describes the hub, gear box and generator. The use of concentrated mass description of the blade inertia, instead of a more general description integration over the cross section, leads much to less complexity in the derivation. A general description will lead to extra terms, such as rotational inertiae about $x-$ and $y_{\mathrm{~\,~}}$ -axis, but these terms turn out to be relatively small anyway. The kinetic energy of the system is given by  
 系统的惯性由单位长度质量 $m$、叶片的单位长度转动惯量 $I_{c g}$ 以及描述轮毂、齿轮箱和发电机的转动惯量 $J_{g e n}$ 描述。使用叶片惯性的集中质量描述，而不是更一般的截面积分描述，大大降低了推导的复杂性。一般描述会引入额外的项，例如绕 $x$ 轴和 $y$ 轴的转动惯量，但这些项无论如何都相对较小。系统的动能由下式给出
 $$
-T\!=\!\frac{1}{2}J_{g c n}\dot{\phi}^{2}+\int_{r}^{R}\!\Big(\frac{1}{2}m\mathbf{r}_{c g}^{\mathrm{T}}\cdot\dot{\mathbf{r}}_{c g}+\frac{1}{2}I_{c g}\big(\dot{\beta}+\dot{\theta}\big)^{2}\Big)\mathrm{d}s
+T\!=\!\frac{1}{2}J_{g c n}\dot{\phi}^{2}+\int_{r}^{R}\!\Big(\frac{1}{2}m\mathbf{r}_{c g}^{\mathrm{T}}\cdot\dot{\mathbf{r}}_{c g}+\frac{1}{2}I_{c g}\big(\dot{\beta}+\dot{\theta}\big)^{2}\Big)\mathrm{d}s\tag 9
 $$  
 
 where ${\dot{\beta}}+{\dot{\theta}}$ is the angular velocity of the blade section around the elastic axis.  
@@ -176,13 +176,13 @@ where ${\dot{\beta}}+{\dot{\theta}}$ is the angular velocity of the blade sectio
 The nonconservative forces are taken into account by describing the variational work done by them for any admissible variation:  
 非保守力通过描述它们对任何容许变分所做的变分功来加以考虑：
 $$
-\delta Q=T_{g c n}\delta\phi+M_{p i t c h}\delta\beta+\int_{r}^{R}(\mathbf{f}^{\mathrm{{T}}}\cdot\delta\mathbf{r}_{e a}+M\delta(\theta+\beta))\mathrm{d}s
+\delta Q=T_{g c n}\delta\phi+M_{p i t c h}\delta\beta+\int_{r}^{R}(\mathbf{f}^{\mathrm{{T}}}\cdot\delta\mathbf{r}_{e a}+M\delta(\theta+\beta))\mathrm{d}s\tag 10
 $$  
 
 where $\mathbf{f}=\mathbf{T}_{\phi}^{\mathrm{T}}\mathbf{T}_{\beta}^{\mathrm{T}}[f_{u},f_{\nu},f_{w}]^{\mathrm{T}}$ and  
 
 $$
-\mathbf{r}_{e a}\!=\![\mathbf{I},\mathbf{J},\mathbf{K}]\mathbf{T}_{\phi}^{\mathrm{T}}\mathbf{T}_{\beta}^{\mathrm{T}}[l_{p i}+u,\nu,w_{0}+w_{1}]^{\mathrm{T}}
+\mathbf{r}_{e a}\!=\![\mathbf{I},\mathbf{J},\mathbf{K}]\mathbf{T}_{\phi}^{\mathrm{T}}\mathbf{T}_{\beta}^{\mathrm{T}}[l_{p i}+u,\nu,w_{0}+w_{1}]^{\mathrm{T}}\tag 11
 $$  
 
 is a position vector describing the elastic axis.  
@@ -198,7 +198,7 @@ By demanding that any admissible variation of the action integral $\delta H\equi
 The equation of motion of the $x$ - and $y$ -directions becomes  
 $x$ 和 $y$ 方向的运动方程变为
 $$
-\begin{array}{r l}&{m\big(\ddot{u}-\ddot{\theta}l_{c g}\sin(\overline{{\theta}})\big)+F_{u,1}\big(\ddot{\beta},\dot{\beta},\dot{\phi},\dot{\nu},\dot{\theta},u^{\prime},u,\nu,\theta,\beta\big)+F_{u,2}\big(\dot{\phi},\dot{u},\dot{\nu},u^{\prime},\nu,\theta,\beta\big)}\\ &{\quad+F_{u,3}(\phi,\beta,\theta,u^{\prime},\nu^{\prime})+F_{u,4}(u^{\prime\prime},\nu^{\prime\prime},\theta)+F_{u,5}\big(\ddot{\phi},\beta\big)=f_{u}+\Big((u^{\prime}+l_{p i}^{\prime})\int_{s}^{R}f_{w}\mathrm{d}\rho\Big)^{\prime}}\\ &{\quad m\big(\ddot{\nu}+\ddot{\theta}l_{c g}\cos(\overline{{\theta}})\big)+F_{\nu,1}\big(\ddot{\beta},\dot{\beta},\dot{\phi},\dot{\theta},\nu^{\prime},u,\nu,\theta,\beta\big)+F_{\nu,2}\big(\dot{\phi},\dot{u},\dot{\nu},u^{\prime},\nu,\theta,\beta\big)}\\ &{\quad+F_{\nu,3}\big(\phi,\beta,\theta,u^{\prime},\nu^{\prime}\big)+F_{\nu,4}\big(u^{\prime\prime},\nu^{\prime\prime},\theta\big)+F_{\nu,5}\big(\ddot{\phi},\beta\big)=f_{\nu}+\Big(\nu^{\prime}\int_{s}^{R}f_{w}\mathrm{d}\rho\Big)^{\prime}}\end{array}
+\begin{array}{r l}&{m\big(\ddot{u}-\ddot{\theta}l_{c g}\sin(\overline{{\theta}})\big)+F_{u,1}\big(\ddot{\beta},\dot{\beta},\dot{\phi},\dot{\nu},\dot{\theta},u^{\prime},u,\nu,\theta,\beta\big)+F_{u,2}\big(\dot{\phi},\dot{u},\dot{\nu},u^{\prime},\nu,\theta,\beta\big)}\\ &{\quad+F_{u,3}(\phi,\beta,\theta,u^{\prime},\nu^{\prime})+F_{u,4}(u^{\prime\prime},\nu^{\prime\prime},\theta)+F_{u,5}\big(\ddot{\phi},\beta\big)=f_{u}+\Big((u^{\prime}+l_{p i}^{\prime})\int_{s}^{R}f_{w}\mathrm{d}\rho\Big)^{\prime}}\\ &{\quad m\big(\ddot{\nu}+\ddot{\theta}l_{c g}\cos(\overline{{\theta}})\big)+F_{\nu,1}\big(\ddot{\beta},\dot{\beta},\dot{\phi},\dot{\theta},\nu^{\prime},u,\nu,\theta,\beta\big)+F_{\nu,2}\big(\dot{\phi},\dot{u},\dot{\nu},u^{\prime},\nu,\theta,\beta\big)}\\ &{\quad+F_{\nu,3}\big(\phi,\beta,\theta,u^{\prime},\nu^{\prime}\big)+F_{\nu,4}\big(u^{\prime\prime},\nu^{\prime\prime},\theta\big)+F_{\nu,5}\big(\ddot{\phi},\beta\big)=f_{\nu}+\Big(\nu^{\prime}\int_{s}^{R}f_{w}\mathrm{d}\rho\Big)^{\prime}}\end{array}\tag 12
 $$  
 
 The direction of the $x_{\mathrm{{}}}$ - and $y$ -axis can be swapped by changing the $\beta$ angle, hence the only differences between the terms in equations (12a) and (12b) are the directions of projection of the forces. In the following, the individual terms in equation (12) are shown and the physical interpretation of them is discussed. Because of the similarity between the terms from equations (12a) and (12b), only the terms from equation (12a) will be discussed. The influence of pitch action is described by