The following are derivations of the output loads available in FAST for a 2-bladed turbine configuration. The loads for a 3-bladed turbine are very similar. Note that some of the loads are given multiple names in order to support variation among the user’s preferences. Along with most of the loads are associated partial loads. These partial loads will be used at the end of this document to redevelop portions of the equations of motion to speed up the computations. The definition of these partial loads is as follows: $$ F_{_{S o u r c e}}^{X_{i}}\left(\ddot{q},\dot{q},q,t\right)\!=\!\!\left(\sum_{r=I}^{22}F_{S o u r c e_{r}}^{X_{i}}\left(q,t\right)\ddot{q}_{r}\right)\!\!+\!F_{_{S o u r c e_{t}}}^{X_{i}}\left(\dot{q},q,t\right) $$ where FSXoiur are the partial forces and $F_{S o u r c e_{t}}^{X_{i}}$ is all components of $F_{S o u r c e}^{X_{i}}$ that are not of this form. $$ M_{S o u r c e}^{N_{i}@X_{i}}\left(\ddot{q},\dot{q},q,t\right)\!=\!\!\left(\sum_{r=I}^{22}{M_{S o u r c e_{r}}^{N_{i}@X_{i}}\left(q,t\right)\ddot{q}_{r}}\right)\!+\!{M_{S o u r c e_{t}}^{N_{i}@X_{i}}\left(\dot{q},q,t\right)} $$ where MSNoi u@rcXei are the partial moments and $M_{S o u r c e_{t}}^{N_{i}(\omega X_{i}}$ is all components of $M_{S o u r c e}^{N_{i}(\varpi X_{i}}$ that are not of this form. To find the loads characterizing the constraint forces between two bodies, say A and B, all that is needed is to remove body B from the equations of motion and determine what equivalent load applied on A would give the same effect that body B had on A originally. # Blade 1 Root Loads: There are 10 output loads at the root of blade 1. 5 of them are the 3 components of the root force $F_{B I}^{s t}(\boldsymbol{O})$ (2 components are expressed in both the coned and blade reference frames). The other 5 are the 3 components of the root bending moments, $M_{B I}^{H}({\boldsymbol{O}})$ (again, 2 components are expressed in both the coned and blade reference frames). If blade 1 is to be removed from the turbine, loads $F_{B I}^{s t}(\boldsymbol{O})$ and $M_{B I}^{H}({\boldsymbol{O}})$ applied to the hub at the blade 1 root $(r=0)$ must give the equivalent effect of blade 1 in the resulting equations of motion. The new generalized active force for the equations of motion resulting from these new loads is: $$ \begin{array}{r l}{F_{r}\big|_{B I}={^{E}\pmb{\nu}_{r}^{C}\cdot^{C}\cdot^{C}\pmb{F}_{B I}^{C}}+{^{E}\pmb{\omega}_{r}^{H}\cdot^{M}\pmb{{\cal H}}_{B I}^{H}}}&{{}\left(r=I,2,\ldots,22\right)}\end{array} $$ where the equivalent loads acting at the hub’s center of mass (point C) are related to $F_{B I}^{s t}(\boldsymbol{O})$ and $M_{B I}^{H}({\boldsymbol{O}})$ because the hub is rigid as follows: FBC1=FBS11 $(\!\!\!\begin{array}{c c c c c c c c c}{{(\!\!0)}}&{{\mathrm{and}}}&{{}}&{{M_{B I}^{\!\scriptscriptstyle H}=M_{B I}^{\!\scriptscriptstyle H}(\!\!0)+r^{c s I}(\!\!0)\times F_{B I}^{s\!\scriptscriptstyle I}(\!\!0)}}&{{\mathrm{or}}}&{{}}&{{M_{B I}^{\!\scriptscriptstyle H}=M_{B I}^{\!\scriptscriptstyle H}\left(\!\!0\right)+\left[r^{Q s I}\left(\!\!0\right)+\frac{1}{2}r^{Q s}\left(\!\!0\right)\right]^{\!\!\scriptscriptstyle T}.}}\end{array}$ −rQC×FBS11(0 ) But since ${}^{E}{\pmb{\nu}}_{r}^{c}={}^{E}{\pmb{\nu}}_{r}^{Q}+{}^{E}{\pmb{\omega}}_{r}^{H}\times{\pmb{r}}^{Q C}$ , this generalized active force can be expanded to: Fr B1=(E $\begin{array}{r l}{\mathbf{\nabla}_{r}^{\rho}+^{E}\boldsymbol{\omega}_{r}^{H}\times\mathbf{\boldsymbol{r}}^{Q C}\Big)\cdot F_{B I}^{S I}\left(\theta\right)+^{E}\boldsymbol{\omega}_{r}^{H}\cdot\left\{{M}_{B I}^{H}\left(\theta\right)+\left[{\boldsymbol{r}}^{Q S I}\left(\theta\right)-{\boldsymbol{r}}^{Q C}\right]\times{F}_{B I}^{S I}\left(\theta\right)\right\}}&{\left(r=I,2,\ldots,K\right)^{T},}\end{array}$ 22) Now applying the cyclic permutation law of the scalar triple product: $$ =^{E}\nu_{r}^{Q}\cdot F_{B I}^{S I}\left(\theta\right)+^{E}\omega_{r}^{H}\cdot\left\{r^{Q C}\times F_{B I}^{S I}\left(\theta\right)\right\}+^{E}\omega_{r}^{H}\cdot\left\{M_{B I}^{H}\left(\theta\right)+\left[r^{Q S I}\left(\theta\right)-r^{Q C}\right]\times F_{B I}^{S I}\left(\theta\right)\right\}+^{E}\omega_{r}^{H}\cdot\left[M_{B I}^{H}\left(\theta\right)+\left[r^{Q S I}\left(\theta\right)-r^{Q C}\right]\times F_{B I}^{S I}\left(\theta\right)\right\} $$ which simplifies to: $$ F_{r}\Big|_{B I}=^{E}\nu_{r}^{Q}\cdot F_{B I}^{S I}\left(0\right)+^{E}\omega_{r}^{H}\cdot\Big[M_{B I}^{H}\left(0\right)+r^{Q S I}\left(0\right)\times F_{B I}^{S I}\left(0\right)\Big]\;\;\;\left(r=I,2,\ldots,22\right) $$ [This can also be simplified to $F_{r}\big|_{B I}={}^{E}\nu_{r}^{S I}\left(0\right)\cdot F_{B I}^{S I}\left(0\right)+{}^{E}\omega_{r}^{H}\cdot M_{B I}^{H}\left(0\right)\quad\left(r=I,2,...,22\right)$ , which will be used later in the ensuing analysis.] This generalized active force must produce the same effects as the generalized active and inertia forces associated with blade 1. Thus, $$ F_{r}\big|_{B I}=F_{r}^{*}\big|_{B I}+F_{r}\big|_{A e r o B I}+F_{r}\big|_{G r a v B I}+F_{r}\big|_{E l a s t i c B I}+F_{r}\big|_{D a m p B I}\quad\big(r=I,2,\dots,22\big) $$ Since ${}^{E}{\nu}_{r}^{c}$ and $^E{\omega}_{r}^{H}$ (and $^E_{\nu_{r}^{Q}}$ and $^E_{\pmb{\nu}_{r}^{S I}}({\cal O}))$ are equal to zero unless $r=1,\!2$ ,…,14;Teet, the generalized active forces associated blade elasticity and damping do not contribute to the root loads (since also, $F_{r}\big|_{E l a s t i c B I}$ and $F_{r}|_{D a m p B I}$ are equal to zero within this range of $r\,\mathbf{\dot{s}}$ ). So, $$ \left.F_{r}\right|_{B I}=F_{r}^{*}\Big|_{B I}+F_{r}\Big|_{A e r o B I}+F_{r}\Big|_{G r a v B I}\quad\left(r=I,2,...,I\mathcal{I};T e e t\right) $$ Thus, Fr B1= $\begin{array}{l}{{\displaystyle\int_{0}^{B I d F l e c L}\varepsilon_{\nu_{r}^{S I}}(r)\cdot\left[F_{A e r o B I}^{S I}\left(r\right)-\mu^{B I}\left(r\right)g\varepsilon_{2}-\mu^{B I}\left(r\right)\varepsilon_{a}^{\phantom{S I I}}\varepsilon_{\phantom{S I I}}^{\phantom{S I I}}\left(r\right)\right]d r+\lefteqn{\displaystyle\int_{0}^{B I d F l e c L}\varepsilon_{\omega_{r}^{M I}}^{{M I}}\left(r\right)\cdot M_{A e r o B I}^{M I}\left(r\right)}=}}\\ {{\displaystyle+\,^{E}\nu_{r}^{S I}\left(B I d F l e x L\right)\cdot\left\{F_{T p D r a g B I}^{S I}\left(B I d F l e x L\right)-m^{B I T i p}\left[g z_{2}+\varepsilon_{a}^{\phantom{S I I}}\left(B I d F l e x L\right)\right]\right\}}}\end{array}$ oB1(r ) dr (r=1,2,,14;Teet) $$ ^{E}\nu_{r}^{s l}(r){=}^{E}\nu_{r}^{g}{+}^{H}\nu_{r}^{s l}(r){+}^{E}\omega_{r}^{H}\times{r}^{g s l}(r) $$ Fr $\begin{array}{c}{{\biggl|_{B I}=\displaystyle\int\displaystyle\sum_{\theta}^{B I d F e x t}\left[^{E}\nu_{\nu}^{Q}+^{H}\nu_{\nu}^{S I}\left(r\right)\right]\cdot\left[F_{A e r o B I}^{S I}\left(r\right)-\mu^{B I}\left(r\right)g z_{2}-\mu^{B I}\left(r\right)^{E}a^{S I}\left(r\right)\right]d r}}\\ {{+\left[^{E}\nu_{\nu}^{Q}+^{H}\nu_{r}^{S I}\left(B I d F l e x L\right)\right]\cdot\left\{F_{T i p D r a g B I}^{S I}\left(B I d F l e x L\right)-m^{B I T i p}\left[g z_{2}+^{E}a^{S I}\left(B I d F l e x L\right)\right]-m^{B I d F i c e n t}\right.}}\\ {{+\left.\displaystyle\int\displaystyle\sum_{\theta}^{B I d F e x L}\left[^{E}\omega_{r}^{H}\times r^{Q S I}\left(r\right)\right]\cdot\left[F_{A e r o B I}^{S I}\left(r\right)-\mu^{B I}\left(r\right)g z_{2}-\mu^{B I}\left(r\right)^{E}a^{S I}\left(r\right)\right]d r+\displaystyle\int\displaylimits_{0}^{B I d F i c e n t}}}\\ {{+\left[^{E}\omega_{r}^{H}\times r^{Q S I}\left(B I d F l e x L\right)\right]\cdot\left\{F_{T i p D r a g B I}^{S I}\left(B I d F l e x L\right)-m^{B I T i p}\left[g z_{2}+^{E}a^{S I}\left(B I d F l e x L\right)\right]\right\}.}}\end{array}$ L)} (r=1,2,,14;Teet) EωrM1(r )⋅MAMe1roB1(r ) dr L)} However, since $r$ is constrained to be between 1,2,…,14;Teet and since ${}^{H}\nu_{r}^{s t}(r)$ is equal to zero and $^{E}\pmb{\omega}_{r}^{M I}\left(r\right)$ equals $^E\pmb{\omega}_{r}^{H}$ with this constraint, this can be simplified as follows: $$ \begin{array}{c}{{\cdot\rfloor_{B I}=\displaystyle\int_{0}^{B I F l e c L}\varepsilon_{\nu_{r}^{Q}}\cdot\left[F_{_{A e r o B I}}^{S I}\left(r\right)-\mu^{B I}\left(r\right)g z_{2}-\mu^{B I}\left(r\right)\varepsilon_{a}^{S}a^{S I}\left(r\right)\right]d r+}}\\ {{\overset{B I d F e c L}{{+}}}}\\ {{+}}\\ {{\overset{B}{\lefteqnrightarrow}}}\\ {{+\left[\begin{array}{c}{{\displaystyle E\,\omega_{r}^{H}\times r^{Q S I}\left(r\right)-\left[F_{A e r o B I}^{S I}\left(r\right)-\mu^{B I}\left(r\right)-\mu^{B I}\left(r\right)g z_{2}-\mu^{B I}\left(r\right)^{E}a^{S I}\left(r\right)\right]d r+\displaystyle\int_{0}^{B I d F l e c s}\int_{0}^{1}\left(r\right)^{2}\left(r\right)^{2}\varepsilon_{a}^{B I}\left(r\right)^{2}d r}}\end{array}\right]_{a}}}\\ {{+\left[\begin{array}{c}{{\displaystyle E\,\omega_{r}^{H}\times r^{Q S I}\left(B I d F l e x L\right)}\right]\cdot\left\{F_{_{T p r o B I}}^{S I}\left(B I d F l e x L\right)-m^{B I T i p}\left[g z_{2}+\varepsilon_{a}^{E}a^{S I}\left(B I d F l e x L\right)\right]\right\}\left[g z_{2}+\varepsilon_{a}^{B I}\left(B I d F l e x L\right)-m^{B I d F i}\left(r\right)^{2}\left(r\right)^{2}\right]\right\}.}}\end{array} $$$$ t^{B I}\left(r\right)g z_{2}-\mu^{B I}\left(r\right)^{E}a^{S I}\left(r\right)\right]d r+{}^{E}\nu_{r}^{Q}\cdot\left\{F_{T p r a g B I}^{S I}\left(B l d F l e x L\right)-m^{B I T i p}\left[g z_{2}+{}^{E}a^{S I}\left(B l d F l e x L\right)\right]\right\}^{R H}\,. $$ $$ \left(r=l,2,...,l\boldsymbol{4};T e e t\right) $$ Or by engaging the cyclic permutation law of the scalar triple product, $$ \begin{array}{r l r}{\lefteqn{\cdot\Big\vert_{B I}=\int\displaylimits_{0}^{B I d k f e n L}\varepsilon_{\nu}^{p}\cdot\left[F_{A e r o B I}^{S I}\left(r\right)-\mu^{B I}\left(r\right)g z_{2}-\mu^{B I}\left(r\right)^{E}a^{S I}\left(r\right)\right]d r+}}&{\varepsilon_{\nu}^{S I}\cdot\left\{F_{T p D r o g B I}^{S I}\left(B I d k e r n L\right)\right\}d r}\\ &{}&{+\left.\begin{array}{c}{B I d F l e c L}\\ {\varepsilon_{\quad}^{B I d l e c L}}\varepsilon_{\quad}^{R e}\cdot\left\{r^{g S I}\left(r\right)\times\left[F_{A e r o B I}^{S I}\left(r\right)-\mu^{B I}\left(r\right)g z_{2}-\mu^{B I}\left(r\right)^{E}a^{S I}\left(r\right)\right]\right\}d r+}\end{array}\right.\int\displaylimits_{0}^{B I d F l e c L}\left(B I d k e r n L\right)d r}\\ &{}&{+\left.\begin{array}{c}{\varepsilon_{\nu}^{G}}\quad cdot\left\{r^{g S I}\left(B I d F l e x L\right)\times\left\{F_{T p D r o g B I}^{S I}\left(B I d F l e x L\right)-m^{B I T i p}\left[g z_{2}+\varepsilon_{\quad}^{B I}a^{S I}\left(B I d F l e x L\right)\right]\right\}}\end{array}\right.}\end{array} $$ $$ \left(r=l,2,...,l\boldsymbol{4};T e e t\right) $$ Thus it is seen that, ${\cal F}_{g l}^{S I}\left({\cal O}\right)=\int_{0}^{B l d F l e c L}\Bigl[{\cal F}_{4e r o B I}^{S I}\left(r\right)-\mu^{B I}\left(r\right)g z_{2}-\mu^{B I}\left(r\right){}^{E}a^{S I}\left(r\right)\Bigr]d r+{\cal F}_{T p o r a g B I}^{S I}\left(B l d F l e x L\right)_{A g r o B I}^{E}\quad,$ )−mB1Tipgz2+EaS1(BldFlexL) and MBH1 $\begin{array}{c}{{\displaystyle(\theta)\!+\!r^{Q S I}\left(\theta\right)\!\times\!F_{B I}^{S I}\left(\theta\right)=\int_{0}^{B I d F l e c L}M_{A e r o B I}^{M I}\left(r\right)d r+\int_{0}^{B I d F l e c L}r^{Q S I}\left(r\right)\!\times\!\left[F_{A e r o B I}^{S I}\left(r\right)\!-\!\mu^{B I}\left(r\right)g z\right]}}\\ {{\displaystyle+\,r^{Q S I}\left(B I d F l e x L\right)\!\times\!\left\{F_{T p r o g B I}^{S I}\left(B I d F l e x L\right)\!-\!m^{B I T i p}\left[g z_{2}+{}^{E}a\left(1-r\right)\right]\right\}}}\end{array}$ z2−µB1(r )EaS1(r )dr S1(BldFlexL)} or MBH1(0 ) $\begin{array}{c}{{{\,\,-\,\,}}}\\ {{{\,\,\,}}}\\ {{{\,\,\,}}}\\ {{{\,\,\,}}}\\ {{{\,\,\,}}}\\ {{{\,\,\,}}}\\ {{{\,\,\,}}}\\ {{{\,\,\,}}}\end{array}\!\!\!\!-\,\,\frac{B l d F l e x L}{\theta}\left(r\right)d r+\displaystyle\int_{0}^{B l d F l e x L}\left(r\right)\times\!\left[F_{A e r o B I}^{S I}\left(r\right)\!-\!\mu^{B I}\left(r\right)g z_{2}-\mu^{B I}\left(r\right)^{E}{\,\pmb{a}}^{S I}\left(r\right)\right]d r}\\ {{\,\,}}\\ {{\,\,}}&{{+\,r^{Q S I}\left(B l d F l e x L\right)\times\left\{\!\frac{C S I}{\pi p_{D r a g B I}}\left(B l d F l e x L\right)\!-\!m^{B I T i p}\left[g\Xi_{2}+{\,}^{E}{\,\pmb{a}}^{S I}\left(B l d F l e x L\right)\right]\!\right\}}}\\ {{\,\,}}\\ {{\,\,}}&{{-\,r^{Q S I}\left(0\right)\times\displaystyle\left\{\!\!\!\begin{array}{c}{{\,\,\,}}\\ {{\,\,}}\\ {{\,\,\,}}\end{array}\!\!\!\left[F_{A e r o B I}^{S I F l e x L}\left(r\right)\!-\!\mu^{B I}\left(r\right)g z_{2}-\mu^{B I}\left(r\right)^{E}{\,\pmb{a}}^{S I}\left(r\right)\!\right]d r+F_{T i p n o g B I}^{S I}\left(B l d r\right)}}\end{array}$ r FlexL)−mB1Tipgz2+EaS1(BldFlexL) or # MBH1(0 )= AMe1roB1(r ) dr+ rQS1(r )−rQS1(0 )×FASe1roB1(r )−µB1(r ) gz2−µB1(r )EaS1(r )dr +rQS1(BldFlexL)−rQS1(0 )×{FTSip1DragB1(BldFlexL)−mB1Tipgz2+EaS1(BldFlexL)} $$ \begin{array}{r}{\stackrel{\mathrm{tora}}{\left\{F_{A e r o B l}^{S I}\left(r\right)-\mu^{B I}\left(r\right)\left\{\begin{array}{l}{g z_{2}+{^{E}}\nu_{_{I}}^{S I}\left(r\right)\vec{q}_{I}+\left[\sum_{i=l}^{H}E_{\nu_{i}}^{S I}\left(r\right)\vec{q}_{i}\right]+\left[\sum_{i=l}^{l,B}E_{\nu_{i}}^{S I}\left(r\right)\vec{q}_{i}\right]+\frac{E}{\nu_{_{I}}}\nu_{_{I}\infty}^{S I}}\\ {\qquad-\left[\displaystyle\sum_{i=l}^{H}\frac{d}{d t}\left({^{E}}\nu_{_{i}}^{S I}\left(r\right)\right)\vec{q}_{i}\right]+\left[\sum_{i=l\infty}^{[B}\frac{d}{d t}\left({^{E}}\nu_{_{i}}^{S I}\left(r\right)\right)\vec{q}_{i}\right]+\frac{d}{d t}\left({^{E}}\nu_{_{I e r}}^{S I}\left({^{\phantom{-}}\vec{q}_{i}\right)\vec{q}_{i}\right]+}\end{array}\right.}}\\ {\stackrel{\mathrm{tor~}}{\left.+\left.F_{r\mu\rho\rho\mu\lambda}^{S I}\left(B I d F l e x L\right)-m^{B I\pi}\right]\left\{\mathcal{G}z_{2}+\left[\sum_{i=l}^{\mathcal{M}}E_{\nu_{i}}^{S I}\left(B I d F l e x L\right)\vec{q}_{i}\right]+\left[\sum_{i=l\infty}^{[B}E_{\nu_{i}}^{S I}\left(B I d F l e x L\right)\vec{q}_{i}\right]\right]+\left[\sum_{i=l\infty}^{B}E_{\nu_{i}}^{S I}\left(B I d F l e x L\right)\vec{q}_{i}\right]\right\}}_{\alpha}}\end{array}\right. $$ $$ M_{B I}^{H}\left(0\right)=\int_{0}^{B l d F l e x L}M_{A e r o B I}^{M I}\left(r\right)d r $$ ![](images/f94847c77122a3b73ec487fb8cfdd59f3c1b34e6d2350069aef3bfff567fcb13.jpg) FBS11t(0 ) $\begin{array}{c}{{^{L}\mu^{B I}\left(r\right)^{E}\nu_{r}^{S I}\left(r\right)d r-m^{B I T p}~\nu_{r}^{S I}\left(B I d F l e x L\right)}}\\ {{\mathrm{}}}\\ {{\displaystyle\left\{F_{A e e n I}^{S I}\left(r\right)-\mu^{B I}\left(r\right)\left\{g z_{2}+\left[\sum_{i=4}^{M}\frac{d}{d t}\left(^{\varepsilon}\nu_{i}^{S I}\left(r\right)\right)\dot{q}_{i}\right]+\left[\sum_{i=l6}^{l8}\frac{d}{d t}\left(^{\varepsilon}\nu_{i}^{S I}\left(r\right)\right)\dot{q}_{i}\right]+\frac{d}{d t}\left(^{\varepsilon}\nu_{{r e e}}^{S I}\left(r\right)\right)\right\}\left(^{\varepsilon}\mu^{B}\right)\right\}~}}\\ {{\displaystyle B I T p\left\{g z_{2}+\left[\sum_{i=d}^{M}\frac{d}{d t}\left(^{\varepsilon}\nu_{i}^{S I}\left(B I d F l e x L\right)\right)\dot{q}_{i}\right]+\left[\sum_{i=l6}^{l8}\frac{d}{d t}\left(^{\varepsilon}\nu_{{b}}^{S I}\left(B I d F l e x L\right)\right)\dot{q}_{i}\right]+\frac{d}{d t}\left(^{\varepsilon}\nu_{{r e e}}^{S I}\left(1-{\cal E}\right)\right)\right\}~}}\\ {{\displaystyle\left\{s!I\right.}}\end{array}$ r ))qTeet dr −mB1Tip BldFlexL))qTee + FS FTSip1DragB1(BldFlexL) MBH1r(0 )= − ∫ rQS1(r )−rQS1(0 )×µB1(r )EvrS1(r )dr−mB1TiprQS1(BldFlexL)−rQS1(0 )×EvrS1(BldFlexL) (r=1,2,,14;16,17,18;Teet) MBH1t(0 )= ∫ rQS1(r )−rQS1(0 )×FASe1roB1(r )−µB1(r )gz2+1∑4 ddt(EviS1(r ))qi+ $\sum_{i=l6}^{l\delta}\frac{d}{d t}\Big(^{E}\nu_{i}^{s t}\left(r\right)\Big)\dot{q}_{i}$ +ddt(EvTSe1et(r ))qTeetdr gz2+1∑4 ddt(EviS1(BldFlexL))qi+1∑8 ddt(EviS1(BldFlexL))qi +rQS1(BldFlexL)−rQS1(0 )×FTSip1DragB1(BldFlexL)−mB1Tip + d (EvTSe1et(BldFlexL))qTeet BldFlexL MAMe1roB1(r ) dr The output loads are as follows: RootFxc1=FBS11(0 )⋅i1B1/ 1,000 $R o o t F y c l=F_{B I}^{S I}\left(0\right)\cdot\dot{\pmb{t}}_{2}^{B I}\,/\,l,000$ $R o o t F x b I=F_{B I}^{S I}\left(0\right)\cdot\pmb{j}_{I}^{B I}\,/\,I,000$ $R o o t F y b I=F_{B I}^{S I}\left(0\right)\cdot\pmb{j}_{2}^{B I}\,/\,I,0O0$ Blade 1 OoP shear force at the blade root (directed along the xc1-axis), (kN) Blade $1\;\mathrm{IP}$ shear force at the blade root (directed along the yc1-axis), (kN) Blade 1 flapwise shear force at the blade root (directed along the xb1-axis), (kN) Blade 1 edgewise shear force at the blade root (directed along the yb1-axis), (kN) $R o o t F z c l=R o o t F z b I=F_{B I}^{S I}\left(O\right)\cdot\dot{\imath}_{3}^{B I}\mathrm{~/~}l,O O O=F_{B I}^{S I}\left(O\right)\cdot\dot{\jmath}_{3}^{B I}\mathrm{~/~}l,O O O$ Blade 1 axial force at the blade root (directed along the zc1-/zb1-axis), (kN) $R o o t M x c l=R o o t M I P l=M_{B I}^{H}\left(\boldsymbol{\theta}\right)\cdot\dot{\boldsymbol{\imath}}_{I}^{B I}\mathrm{~/~}l,000$ Blade 1 IP moment (i.e., the moment caused by IP forces) at the blade root (about the xc1-axis), $$ R o o t M y c I=R o o t M O o P I=M_{B I}^{H}\left(0\right)\cdot\dot{\pmb{i}}_{2}^{B I}\mathrm{~/~}I,000 $$ Blade 1 OoP moment (i.e., the moment caused by OoP forces) at the blade root (about the yc1- $$ R o o t M x b I=R o o t M E d g I=M_{B I}^{H}\left(\boldsymbol{\theta}\right)\cdot\dot{\boldsymbol{j}}_{I}^{B I}\mathrm{\Delta}/\mathit{I},000 $$ Blade 1 edgewise moment (i.e., the moment caused by edgewise forces) at the blade root (about $R o o t M y b I=R o o t M F l p I=M_{B I}^{H}\left(\boldsymbol{\theta}\right)\cdot\dot{\boldsymbol{J}}_{2}^{B I}\mathrm{\textit{/}}I,000$ Blade 1 flapwise moment (i.e., the moment caused by flapwise forces)at the blade root (about the $R o o t M z c I=R o o t M z b I=M_{B I}^{H}\left({\boldsymbol{\partial}}\right)\cdot{\dot{\boldsymbol{i}}}_{3}^{B I}\left/I,000=M_{B I}^{H}\left({\boldsymbol{\partial}}\right)\cdot{\dot{\boldsymbol{j}}}_{3}^{B I}\left/I,000\right.$ $(\mathrm{kN}\!\cdot\!\mathrm{m})$ Blade 1 pitching moment at the blade root (about the zc1-/zb1-axis), Blade 1 Local Moment Outputs: There are 3 output loads at any of the selected span stations $i$ $\mathbf{\zeta}(r=\,R^{s p a n\;i}\,\rightleftharpoons\,R^{s p a n\;i}\,\rightleftharpoons\,R^{s p a n\;i}\,,$ ) of blade 1 $(i{=}1,\!2,\!...,\!5)$ . These are the 3 components of the bending moment $M_{B I}^{M I}\left(R^{S p a n\;i}\right)$ expressed in the local blade coordinate system (principal structural axes). Examining the results for the blade 1 root loads, it follows that: ![](images/c2a077c10ce590a223033998754939009a99543553f12a468ae951e8ef30dde3.jpg) The output loads are as follows: $S p n i M L x b I=M_{B I}^{M I}\left(R^{S p a n\;i}\right)\cdot n_{I}^{B I}\left(R^{S p a n\;i}\right)/\,I,000$ $S p n i M L y b I=M_{B I}^{M I}\left(R^{S p a n\;i}\right)\cdot n_{2}^{B I}\left(R^{S p a n\;i}\right)/\,I,000$ $\mathit{S p n i M L z b l}=M_{\mathit{B I}}^{\mathit{M I}}\left(R^{\mathit{S p a n\;i}}\right)\cdot n_{3}^{\mathit{B I}}\left(R^{\mathit{S p a n\;i}}\right)/\mathit{I,O00}$ Blade 1 local edgewise moment at span station $i$ (about the local xb1-structural axis), $(\mathrm{kN}\!\cdot\!\mathrm{m})$ Blade 1 local flapwise moment at span station $i$ (about the local yb1-structural axis), $(\mathrm{kN}\!\cdot\!\mathrm{m})$ Blade 1 pitching moment at span station $i$ (about the zc1-/zb1-/local zb1-axis), $(\mathrm{kN\cdotm})$ Blade 2 Root Loads: The equations for $F_{B2}^{S2}\left(0\right),\;M_{B2}^{H}\left(0\right),\;F_{B2_{r}}^{S2}\left(0\right),\;F_{B2_{t}}^{S2}\left(0\right),\;M_{B2_{r}}^{H}\left(0\right),\;M_{B2_{t}}^{H}\left(0\right)$ , and all 10 output loads are similar to blade 1. # Hub and Rotor Loads: There are 14 output loads at the hub end of the low-speed shaft. 5 of them are the 3 components of the thrust and shear force $F_{R o t o r}^{P}$ (2 components are expressed in a nonrotating frame, 2 components are expressed in a rotating frame, and 1 component is independent of rotation). 5 other loads are the 3 components of the shaft bending moments, $M_{R o t o r}^{L@P}$ (again, 2 components are expressed in a nonrotating frame, 2 components are expressed in a rotating frame, and 1 component is independent of rotation). The $11^{\mathrm{th}}$ and $12^{\mathrm{th}}$ loads are the rotor power and rotor power coefficient, respectively. The $13^{\mathrm{th}}$ and $14^{\mathrm{th}}$ loads are the rotor thrust and rotor torque coefficients, respectively. For a 2-blader, all these loads are given relative to the teeter pin (point P) as indicated. For the 3-blader, all of these loads are given relative to the apex of rotation (point Q, which is coincident with point P). The new generalized active force for the equations of motion resulting from these new loads is: $$ F_{r}\big|_{R o t o r}={^{E}\pmb{\nu}}_{r}^{P}\cdot F_{R o t o r}^{P}+{^{E}\pmb{\omega}}_{r}^{L}\cdot M_{R o t o r}^{L@P}\quad\left(r=l,2,...,22\right) $$ This generalized active force must produce the same effects as the generalized active and inertia forces associated with blade 1, blade 2, the hub, and the teeter springs and dampers. Thus, $$ \begin{array}{r l}&{F_{r}\big|_{R o t o r}=F_{r}^{*}\big|_{B I}+F_{r}\big|_{A e r o B I}+F_{r}\big|_{G r a v B I}+F_{r}\big|_{E l a s t i c B I}+F_{r}\big|_{D a m p B I}}\\ &{~+F_{r}^{*}\big|_{B_{2}}+F_{r}\big|_{A e r o B2}+F_{r}\big|_{G r a v B2}+F_{r}\big|_{E l a s t i c B2}+F_{r}\big|_{D a m p B2}}&{\big(r=I,2,\dots,22\big)}\\ &{~+F_{r}^{*}\big|_{H}+F_{r}\big|_{G r a v H}+F_{r}\big|_{S p r i n g T e e t}+F_{r}\big|_{D a m p T e e t}}\end{array} $$ Since $^E_{\nu_{r}^{P}}$ and $\varepsilon_{\pmb{\omega}_{r}^{L}}$ are equal to zero unless $r=1,2,...,14$ , the generalized active forces associated with blade and teeter elasticity and damping do not contribute to the hub and rotor loads (since also, $F_{r}|_{E l a s t i c B I}\,,\;F_{r}|_{D a m p B I}\,,\;F_{r}|_{E l a s t i c B2}\,,\;F_{r}|_{D a m p B2}\,,\;F_{r}|_{S p r i n g T e e}\,.$ , and $F_{r}|_{D a m p T e e t}$ are equal to zero if $r=$ 1,2,…,14). So, $$ F_{r}\Big|_{R o t o r}=F_{r}^{*}\Big|_{B I}+F_{r}\Big|_{A e r o B I}+F_{r}\Big|_{G r o w B I}+F_{r}^{*}\Big|_{B2}+F_{r}\Big|_{A e r o B2}+F_{r}\Big|_{G r o w B2}+F_{r}^{*}\Big|_{H}+F_{r}\Big|_{G r o w B}\quad(r=0,1) $$ When using the results for the blade 1 and blade 2 root loads, this equation can be simplified as follows: $$ F_{r}\big|_{R o t o r}=F_{r}\big|_{B I}+F_{r}\big|_{B2}+F_{r}^{*}\big|_{H}+F_{r}\big|_{G r a v H}\quad\big(r=I,2,...,I\mathcal{I}\big) $$ Thus, $$ \begin{array}{r l}&{\mathbf{\Xi}_{B I}^{S I}\left(\theta\right)+}\\ &{\mathbf{\Xi}_{B I}^{E}\left(\theta\right)+}\\ &{\varepsilon_{\nu}^{C}\cdot\left({\vphantom{\int_{0}^{\tau}}}\varepsilon_{a}^{C}+g\varepsilon_{2}\right)-}\end{array}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\left({\cal M}_{B I}^{H}\left(\theta\right)+\left[r^{\underline{{{\theta}}}S I}\left(\theta\right)-r^{\underline{{{\theta}}}C}+\right]\times{\cal F}_{B I}^{S I}\left(\theta\right)\right\}+{\vphantom{\int_{0}^{\tau}}}{\varepsilon_{\nu}^{C}}\cdot{\cal F}_{B2}^{S2}\left(\theta\right)+}\\ &{\varepsilon_{\nu}^{C}\cdot\left({\vphantom{\int_{0}^{\tau}}}\varepsilon_{a}^{C}+g\varepsilon_{2}\right)-}\varepsilon_{\omega_{r}^{H}}\cdot\left(\overline{{{\cal I}}}^{H}\cdot{\vphantom{\int_{0}^{\tau}}}\alpha_{\nu}^{H}+{\vphantom{\int_{0}^{\tau}}}\omega^{H}\times\overline{{{\cal I}}}^{H}\cdot{\vphantom{\int_{0}^{\tau}}}\omega^{H}\right)}\end{array} $$ or when grouping like terms: $$ \begin{array}{l}{{\overline{{{F}}}_{B I}^{S I}\left(\boldsymbol{\theta}\right)+F_{B2}^{S2}\left(\boldsymbol{\theta}\right)-m^{H}\left(\frac{\varepsilon}{\varepsilon}{a}^{c}+g\varepsilon_{2}\right)\right]}}\\ {{{\mathbf{\nabla}^{\prime}}\cdot\left\{M_{B I}^{H}\left(\boldsymbol{\theta}\right)+M_{B2}^{H}\left(\boldsymbol{\theta}\right)+\left[r^{Q S I}\left(\boldsymbol{\theta}\right)-r^{Q C}\right]\times F_{B I}^{S I}\left(\boldsymbol{\theta}\right)+\left[r^{Q S2}\left(\boldsymbol{\theta}\right)-r^{Q C}\right]\times F_{B2}^{S2}\left(\boldsymbol{\theta}\right)-\overline{{{\overline{{{I}}}}}}^{H}\mathbf{\nabla}^{E}\right\}}}\end{array} $$ Recognizing that EvrC =EvrP EωH×(rPQ+rQC) , this generalized force can be expanded to: $$ \begin{array}{r l}&{^{E}\nu_{r}^{P}+^{E}\omega_{r}^{H}\times\left(r^{P Q}+r^{Q C}\right)\Biggr]\cdot\left[F_{B I}^{S I}\left(\theta\right)+F_{B2}^{S2}\left(\theta\right)-m^{H}\left(^{E}a^{C}+g z_{2}\right)\Biggr]}\\ &{+^{E}\omega_{r}^{H}\cdot\Bigl\{M_{B I}^{H}\left(\theta\right)+M_{B2}^{H}\left(\theta\right)+\left[r^{Q S I}\left(\theta\right)-r^{Q C}\right]\times F_{B I}^{S I}\left(\theta\right)+\Bigl[r^{Q S2}\left(\theta\right)-r^{Q C}\Bigr]\times F_{B2}^{S2}\left(\theta\right)-^{\frac{1}{2}}}\end{array} $$ Now applying the cyclic permutation law of the scalar triple product: $$ \begin{array}{r l}&{{\nu_{r}^{P}}\cdot\left[F_{B I}^{S I}\left(\theta\right)+F_{B2}^{S2}\left(\theta\right)-m^{H}\left({}^{E}a^{C}+{g}{z_{2}}\right)\right]+{}^{E}\omega_{r}^{H}\cdot\left\{\left(r^{P Q}+r^{Q C}\right)\times\left[F_{B I}^{S I}\left(\theta\right)+F_{B2}^{S2}\left(\theta\right)-F_{B I}^{S I}\left(\theta\right)\right]\right\}}\\ &{+{}^{E}\omega_{r}^{H}\cdot\left\{M_{B I}^{H}\left(\theta\right)+M_{B2}^{H}\left(\theta\right)+\left[r^{Q S I}\left(\theta\right)-r^{Q C}\right]\times F_{B I}^{S I}\left(\theta\right)+\left[r^{Q S2}\left(\theta\right)-r^{Q C}\right]\times F_{B2}^{S2}\left(\theta\right)-\frac{1}{L}\omega_{r}^{H}\left(\theta\right)\right\}}\end{array} $$ which simplifies to: $$ \left.\begin{array}{l}{{\cdot\left[F_{B I}^{S I}\left(\theta\right)+F_{B^{2}}^{S2}\left(\theta\right)-m^{H}\left(^{E}{\boldsymbol{a}}^{C}+g z_{2}\right)\right]}}\\ {{\omega_{r}^{H}\cdot\left\{\begin{array}{l}{{M_{B I}^{H}\left(\theta\right)+M_{B^{2}}^{H}\left(\theta\right)+\left[r^{P Q}+r^{\vartheta S I}\left(\theta\right)\right]\times F_{B I}^{S I}\left(\theta\right)+\left[r^{P Q}+r^{\vartheta S2}\left(\theta\right)\right]\times F_{B^{2}}^{S2}\left(\theta\right)}}\\ {{-m^{H}\left(r^{P Q}+r^{Q C}\right)\times\left(^{E}{\boldsymbol{a}}^{C}+g z_{2}\right)-\overline{{{\overline{{{I}}}}}}\cdot{\boldsymbol{\varepsilon}}_{\alpha}^{E}{\boldsymbol{a}}^{H}-{^{E}}\omega^{H}\times\overline{{{\overline{{{I}}}}}}^{H}\cdot{^{E}}\omega^{H}}}\end{array}\right\}}}\end{array}\left(r\right)} $$ However, $^E_{\omega_{r}^{H}}$ equals $^E\pmb{\omega}_{r}^{L}$ when $r$ is not equal to Teet. Thus the generalized active force associated with the rotor can be expressed as follows: $$ \begin{array}{r l}&{=^{E}\nu_{r}^{P}\cdot\left[F_{B I}^{S I}\left(\theta\right)+F_{B2}^{S2}\left(\theta\right)-m^{H}\left(^{E}a^{C}+g z_{2}\right)\right]}\\ &{\phantom{=}+^{E}\omega_{r}^{L}\cdot\left\{\begin{array}{l l}{M_{B I}^{H}\left(\theta\right)+M_{B2}^{H}\left(\theta\right)+\left[r^{P Q}+r^{Q S I}\left(\theta\right)\right]\times F_{B I}^{S I}\left(\theta\right)+\left[r^{P Q}+r^{Q S2}\left(\theta\right)\right]\times F_{B2}^{S2}\left(\theta\right)}\\ {\phantom{=}-m^{H}\left(r^{P Q}+r^{Q C}\right)\times\left(^{E}a^{C}+g z_{2}\right)-\overline{{\bar{I}}}^{H}\cdot^{E}\alpha^{H}-^{E}\omega^{H}\times\overline{{\bar{I}}}^{H}\cdot^{E}\omega^{H}}\end{array}\right.}\end{array} $$ Thus it is seen that, $$ F_{R o t o r}^{P}=F_{B I}^{S I}\left(0\right)+F_{B2}^{S2}\left(0\right)-m^{H}\left({}^{E}a^{C}+g z_{2}\right) $$ MRLo@toPr=MBH1(0 )+MBH2(0 )+rPQ+rQS1(0 )×FBS11(0 )+rPQ+rQS2(0 )×FBS22(0 )−mH(rPQ+rQC) ( a +gz2)−I a −EωH×IH⋅Eω Thus, $$ \mathbf{\Psi}_{R o t o r}^{P}=F_{B I}^{S I}\left(\boldsymbol{\theta}\right)+F_{B2}^{S2}\left(\boldsymbol{\theta}\right)-m^{H}\left\{\left(\sum_{i=I}^{H}E_{\nu_{i}^{C}}\hat{q}_{i}\right)+\;E_{\nu_{T e t}^{C}}\hat{q}_{T e e t}+\left[\sum_{i=I}^{H}\frac{d}{d t}\Big(\boldsymbol{\varepsilon}\nu_{i}^{C}\Big)\hat{q}_{i}\right]+\frac{d}{d t}\Big(\boldsymbol{\varepsilon}\nu_{T e t}^{C}\hat{q}_{i}\Big)\right\}, $$ $$ \begin{array}{c}{{\displaystyle\Gamma_{R o t o r}^{A\bar{\alpha}P}=M_{B I}^{H}\left(\theta\right)+M_{B2}^{H}\left(\theta\right)+\left[r^{P Q}+r^{Q S I}\left(\theta\right)\right]\times F_{B I}^{S I}\left(\theta\right)+\left[r^{P Q}+r^{Q S2}\left(\theta\right)\right]\times F_{B2}^{S2}\left(\theta\right)}}\\ {{\displaystyle-m^{H}\left(r^{P Q}+r^{Q C}\right)\times\left\{\left(\sum_{i=I}^{I\!\!\!/}^{k}{\varepsilon_{\nu}}_{i}^{C}\ddot{q}_{i}\right)\!\!+{}^{E}{\nu}_{{\nu}\!\!\!/e c t}^{C}\ddot{q}_{{\mathrm{ref}}}\!\!+\!\left[\sum_{i=I}^{I\!\!\!/}\frac{d}{d t}\!\left({}^{E}\nu_{{\nu}\!\!\!/}^{C}\right)\dot{q}_{i}\right]\!\!+\!\frac{d}{d t}\!\left({}^{E}\nu_{{\nu}\!\!\!/e c t}^{C}\right)\dot{q}_{i}\right.}}\\ {{\displaystyle-\overline{{{\cal F}}}^{H}\cdot\left\{\left(\sum_{i=I}^{I\!\!\!/}{}^{E}{\varepsilon_{\nu}}_{i}^{H}\ddot{q}_{i}\right)\!\!+{}^{E}{\omega}_{I e r t}^{H}\ddot{q}_{{\mathrm{ref}}}\!\left.+\!\left[\sum_{i=I}^{I\!\!\!/}\frac{d}{d t}\!\left({}^{E}{\omega}_{i}^{H}\right)\dot{q}_{i}\right]\!\!+\!\frac{d}{d t}\!\left({}^{E}{\omega}_{{\nu}\!\!\!/e c t}^{H}\right)\dot{q}_{{\mathrm{ref}}}\right\}-{}^{E}{\omega}_{I}}}\end{array} $$ Or, $$ \begin{array}{l}{{\displaystyle F_{R o t o r_{r}}^{P}=F_{B I_{r}}^{S I}\left(\boldsymbol{\theta}\right)+F_{B2_{r}}^{S2}\left(\boldsymbol{\theta}\right)-m^{H}^{\textit{E}}\pmb{\nu}_{r}^{c}\quad\left(r=I,2,\ldots,I\mathcal{I};I6,I7,\ldots,22\right)}}\\ {{\displaystyle F_{R o t o r_{t}}^{P}=F_{B I_{t}}^{S I}\left(\boldsymbol{\theta}\right)+F_{B2_{t}}^{S2}\left(\boldsymbol{\theta}\right)-m^{H}\left\{\left[\sum_{i=d}^{M}\frac{d}{d t}\!\left(^{\varepsilon}\nu_{i}^{c}\right)\boldsymbol{\dot{q}}_{i}\right]\!+\!\frac{d}{d t}\!\left(^{\varepsilon}\nu_{{I}\!e e t}^{c}\right)\boldsymbol{\dot{q}}_{{T e e t}}+g\boldsymbol{z}_{2}\right\}}}\end{array} $$ $\begin{array}{l}{{{\cal M}_{R o t o r_{r}}^{L@P}={\cal M}_{B_{I_{r}}}^{H}\left(0\right)+{\cal M}_{B_{2r}}^{H}\left(\theta\right)\mathrm{+}\left[r^{P Q}+r^{Q S I}\left(0\right)\right]\times F_{B{I_{r}}}^{S I}\left(\theta\right)\mathrm{+}\left[r^{P Q}+r^{Q S2}\left(\theta\right)\right]\times F_{B{I_{r}}}^{S2}\left(\theta\right)}}\\ {{{\cal M}_{R o t o r_{t}}^{L@P}={\cal M}_{B_{I_{r}}}^{H}\left(\theta\right)\mathrm{+}{\cal M}_{B_{2\mathrm{t}}}^{H}\left(\theta\right)\mathrm{+}\left[r^{P Q}+r^{Q S I}\left(\theta\right)\right]\times F_{B{I_{r}}}^{S I}\left(\theta\right)\mathrm{+}\left[r^{P Q}+r^{Q S2}\left(\theta\right)\right]\times F_{B{I_{r}}}^{S2}\left(\theta\right)}}\\ {{\mathrm{\quad~}}}\\ {{{\displaystyle~~-\,\overline{{\cal I}}^{H}\cdot\left\{\left[\sum_{i=7}^{M}\frac{d}{d t}\Big({}^{E}\omega_{i}^{H}\Big)\dot{q}_{i}\right]\!\!+\!\frac{d}{d t}\Big({}^{E}\omega_{T e t}^{H}\Big)\dot{q}_{T e t}\Big\}\!\!-{}^{E}\omega^{H}\times\overline{{\cal I}}^{H}\cdot{}^{E}\omega^{H}}}}\end{array}$ −mH(rPQ+rQC) EωH (r=1,2,,14;16,17,, −mH(rPQ+rQC)×1∑4 ddt(EviC)qi+ddt(EvTCeet)qTeet+gz2 The output loads are as follows, $R o t T h r u s t=L S S h f t F x s=L S S h f t F x a=F_{R o t o r}^{P}\cdot e_{I}\;/\;I,000=F_{R o t o r}^{P}\cdot c_{I}\;/\;I,000$ is constant along the shaft and is equivalent to the rotor thrust force), $(\mathrm{kN})$ Low-speed shaft thrust force (directed along the xs-/xa-axis) (this $L S S h f t F y a=F_{R o t o r}^{P}\cdot e_{2}\ ⁄I,O O O$ $L S S h f t F z a=F_{R o t o r}^{P}\cdot e_{3}\,/\,l,000$ $L S S h f t F y s=-F_{R o t o r}^{P}\cdot c_{s}\,/\,l,000$ $L S S h f t F z s=F_{R o t o r}^{P}\cdot c_{2}\mathit{\Omega}/{I,O O O}$ Rotating low-speed shaft shear force (directed along the ya-axis) (this is constant along the shaft), (kN) Rotating low-speed shaft shear force (directed along the za-axis) (this is constant along the shaft), (kN) Nonrotating low-speed shaft shear force (directed along the ys-axis) (this is constant along the shaft), (kN) Nonrotating low-speed shaft shear force (directed along the zs-axis) (this is constant along the shaft), (kN) RotT $\diamond r q=L S S h f t T q=L S S h f t M x s=L S S h f t M x a=M_{\cal R o t o r}^{L a P}\cdot e_{I}\;/\;I,\theta\theta0=M_{\cal R o t o r}^{L a P}\cdot e_{I}\;/\;I,\theta\theta0=0,\forall01\theta\in\cal R^{3}:\cal R^{3}.$ Low-speed shaft torque (about the xs-/xa-axis) (this is constant along the shaft and is equivalent to the rotor torque), $(\mathrm{kN\cdotm})$ $L S S T i p M y a=M_{R o t o r}^{L@P}\cdot e_{2}\;/\;l,O O O$ (about the ya-axis), $(\mathrm{kN\cdotm})$ $L S S T i p M z a=M_{R o t o r}^{L@P}\cdot e_{3}\,/\,l,000$ (about the za-axis), $(\mathrm{kN}\!\cdot\!\mathrm{m})$ Rotating low-speed shaft bending moment at the shaft tip [teeter pin for 2 blades] [apex of rotation for 3 blades] Rotating low-speed shaft bending moment at the shaft tip [teeter pin for 2 blades] [apex of rotation for 3 blades] $L S S T i p M y s=-M_{R o t o r}^{L@P}\cdot c_{s}\,/\,I,O O O$ blades] (about the ys-axis), $(\mathrm{kN\cdotm})$ Nonrotating low-speed shaft bending moment at the shaft tip [teeter pin for 2 blades] [apex of rotation for 3 # $L S S T i p M z s=M_{R o t o r}^{L@P}\cdot c_{2}\;/\;l,000$ ![](images/0c89b69f3ec2d36217613615b0156cfa3e52ee6367691ddd11433722d6d39506.jpg) The rotor torque is equal to low-speed shaft torque as seen above. It is noted that this torque can be computed differently using the drivetrain flexibility and damping, though the load summation method and this other constraint method are equivalent. This can be demonstrated as follows. First of all, the equation above is equivalent to saying: $$ L S S h f t T q={}^{E}\omega_{D r T r}^{L}\cdot M_{R o t o r}^{L@P}\slash{I},000 $$ However, since $^E_{\nu_{D r T r}^{P}}$ is equal to zero, it is also equivalent to say: $$ \begin{array}{l}{{^{E}\nu_{_{D r T}}^{P}\cdot F_{_{R o t o r}}^{P}+^{\varepsilon}\omega_{D r T}^{L}\cdot M_{R o t o r}^{L\ @P}}}\\ {\mathrm{~}}\\ {{^{\sum_{\substack{J r\sim\Gamma}}}\left|_{R o t o r}/\mathit{I},000\right.\mathrm{~or}\qquad L S S h/t{\Gamma q}=\left(F_{_{D r T}}^{\ast}\Big|_{B I}+F_{D r T}\Big|_{A e r o B I}+F_{D r T}\Big|_{G r a v B I}+F_{D r T}^{\ast}\Big|_{B J}+F_{D r r}\Big|_{B I}\right)}}\end{array} $$ From the equations of motion, it is easily seen that this is equivalent to saying: $$ \begin{array}{r}{L S S h f t T q=\left(-F_{D r T r}\right|_{E l a s t i c D r i v e}-F_{D r T r}\Big|_{D a m p D r i v e}\right)/{\mathit{I}},000}\end{array} $$ and thus, $\begin{array}{r l}{L S S h f t T q=\left(D T T o r S p r\cdot q_{D r T}+D T T o r D m p\cdot\dot{q}_{D r T r}\right)/\,I,000}&{\left(=\,M_{R o t o r}^{L a P}\cdot c_{I}\,/\,I,000\right)}\end{array}$ and is equivalent to the rotor torque) Thus, both the load summation method and the constraint method are equivalent. However, if the drivetrain DOF is disabled, then $q_{D r T r}$ will equal zero and $\dot{q}_{D r T r}$ will equal zero, which implies that, at least, DTTorSpr is equal to infinity (since the product of DTTorSpr and $q_{D r T r}$ is, in general, nonzero). Thus, to avoid using 2 different methods to calculate LSShftTq , it is best just to use $M_{R o t o r}^{L@P}\cdot c_{I}\,/\,I,O O O$ , which will always work, regardless of the number of DOFs disabled. Like the LSShftTq , it is noted that LSSTipMya can also be computed differently using the teeter springs and dampers, though the load summation method and this other constraint method are equivalent. This also can be demonstrated as follows. First of all, the equation above is equivalent to saying: $$ L S S T i p M y a=\mathit{\Pi}^{E}\omega_{T e e t}^{H}\cdot M_{R o t o r}^{L@P}\mathit{\Pi}/\mathit{I},000 $$ Or, $$ M\!y a={}^{E}\!\omega_{T e e t}^{H}\cdot\left\{\!M_{B I}^{H}\left(\theta\right)\!+\!M_{B2}^{H}\left(\theta\right)\!+\!\left[{r}^{P Q}+{r}^{Q S I}\left(\theta\right)\right]\!\times\!F_{B I}^{S I}\left(\theta\right)\!+\!\left[{r}^{P Q}+{r}^{Q S2}\left(\theta\right)\right]\!\times\!F_{B2}^{S I}\left(\theta\right)\!+\!\left[{r}^{P Q}\right]\!+\!{r}^{Q S2}\left(\theta\right)\!\right]\!\times\!F_{B2}^{S I}\left(\theta\right)\!+\!{r}^{P Q}\left(\theta\right)^{2}, $$ Now applying the cyclic permutation law of the scalar triple product: $$ \begin{array}{l}{{^{E}\omega_{T e t}^{H}\times\left[r^{P Q}+r^{Q S I}\left(0\right)\right]\cdot F_{B I}^{S I}\left(0\right)+^{E}\omega_{3}^{H}\times\left[r^{P Q}+r^{Q S2}\left(0\right)\right]\cdot F_{B2}^{S2}\left(0\right)-m^{H}\,^{E}\omega_{T e t}^{H}\times\left(r^{P Q}+r^{Q S2}\left(0\right)\right)}}\\ {{\mathrm{}\,\,\,\,\,+\,\,^{E}\omega_{T e t}^{H}\cdot\left[M_{B I}^{H}\left(0\right)+M_{B2}^{H}\left(0\right)-\overline{{{\overline{{{I}}}}}}\,^{H}\cdot^{E}\alpha^{H}-^{E}\omega^{H}\times\overline{{{\overline{{{I}}}}}}\,^{H}\cdot^{E}\omega^{H}\right]}}\end{array} $$ Recognizing also that $E_{\psi_{I e e t}}^{\phantom{\dagger}}\left(\theta\right)={}^{E}\omega_{I_{e e t}}^{H}\times\left[r^{P Q}+r^{Q S I}\left(\theta\right)\right]\quad,\qquad\quad{}^{E}\nu_{I_{e e t}}^{S2}\left(\theta\right)={}^{E}\omega_{I_{e e t}}^{H}\times\left[r^{P Q}+r^{Q S2}\left(\theta\right)\right],\quad\forall\theta\in\mathbb{R}^{3}.$ nd ${^E}{\nu}_{T e e t}^{C}={^E}\omega_{T e e t}^{H}\times\left(r^{P Q}+r^{Q C}\right)$ , this can be simplified as follows: $$ \begin{array}{r l}&{\frac{\nu I}{\nu e t}\left(\partial\right)\cdot F_{s t l}^{S I}\left(\partial\right)+^{E}\omega_{2r e t}^{H}\cdot M_{B I}^{H}\left(\partial\right)+^{E}\nu_{T e e t}^{S2}\left(\partial\right)\cdot F_{s t2}^{S2}\left(\partial\right)+^{E}\omega_{T e t}^{H}\cdot M_{B2}^{H}\left(\partial\right)-m^{H}^{E}\nu_{T e t}^{C}\cdot\left(^{L}\nu_{T e e t}^{C}\cdot M_{B2}^{H}\left(\partial\right)\right)}\\ &{-^{E}\omega_{T e t}^{H}\cdot\left(\overline{{\overline{{I}}}}^{H}\cdot^{E}\alpha^{H}+^{E}\omega^{H}\times\overline{{\overline{{I}}}}^{H}\cdot^{E}\omega^{H}\right)}\\ &{\qquad\qquad\times\left(\left|_{B I}+F_{T e e t}\right|_{B2}+F_{T e e t}^{*}\right|_{H}+F_{T e e t}\right|_{G r a b l}\right)/I,000}\\ &{\left.\left|_{t}\right|_{H}+F_{T e e t}^{*}\right|_{B I}+F_{T e e t}^{*}\right|_{B2}+F_{T e e t}^{*}\right|_{A e r o B l}+F_{T e e t}\Big|_{G r a b l}+F_{T e e t}\Big|_{G r a b l}+F_{T e e t}\Big|_{G r a b l}+F_{T e e t}\Big|_{G r a b l}}\end{array} $$ From the equations of motion, it is easily seen that this is equivalent to saying: $$ \begin{array}{r}{L S S T i p M y a=\left(-F_{T e e t}\vert_{S p r i n g T e e t}-F_{T e e t}\vert_{D a m p T e e t}\right)/\,I,000}\end{array} $$ and thus, $$ r i p M y a=\left\{\begin{array}{l l}{I F\Big[\Big|q_{T e e l}\Big|>T e e t S S t P,T e e t S S P\cdot S I G N\big(q_{T e e l}\Big)\Big(\Big|q_{T e e l}\Big|-T e e t S S t P\Big),0\Big]}\\ {+I F\Big[\Big|q_{T e e l}\Big|>T e e t H S t P,T e e t H S S p\cdot S I G N\big(q_{T e e l}\Big)\Big(\Big|q_{T e e l}\Big|-T e e t H S t P\Big),0\Big]}\\ {+I F\Big[\dot{q}_{T e e l}\bigtriangleup(0,T e e t C D m p\cdot S I G N\big(\dot{q}_{T e e l}\big),0\Big]+I F\Big[\Big|q_{T e e l}\Big|>T e e t D m p,T e e t C D m p\cdot}\end{array}\right].}\end{array}\right. $$ Thus, both the load summation method and the constraint method are equivalent. Thus, to avoid using 2 different methods to calculate LSSTipMya if various DOFs are disabled, it is best just to use $M_{R o t o r}^{L@P}\cdot e_{2}\,/\,I,O O O$ , which will always work. Shaft Strain Gage Loads: There are 4 output loads at point SG on the low-speed shaft [which is a point on the shaft a distance ShftGagL towards the nacelle from point P (or point Q for a 3-blader since point $\mathrm{P}$ does not exist)]. These are 2 of the 3 components of the shaft bending moments, $M_{R o t o r}^{L@S G}$ (2 components are expressed in a nonrotating frame, 2 components are expressed in a rotating frame, and third component which is directed in the $c_{I}$ direction is not used because it is the same as the rotor torque). Since the low-speed shaft is assumed to be rigid and massless between points $\mathrm{P}$ and SG, it is easily seen that: $$ M_{R o t o r}^{L@S G}=M_{R o t o r}^{L@P}-r^{P S G}\times F_{R o t o r}^{P} $$ since $r^{P S G}$ equals $-r^{S G P}$ . Thus, $L S S G a g M y a=M_{R o t o r}^{L@F}\cdot e_{2}\,/\,I,O00=L S S T i p M y a+S h f t G a g L\cdot L S S h f t F z a$ (about the ya-axis), $(\mathrm{kN\cdotm})$ $\ L S S G a g M z a=M_{\scriptscriptstyle R o t o r}^{L@F}\cdot e_{_3}\cdot/\,I,O00=L S S T i p M z a-S h f t G a g L\cdot L S S h f t F y a$ (about the za-axis), $(\mathrm{kN}\!\cdot\!\mathrm{m})$ $L S S G a g M y s=-M_{R o t o r}^{L@F}\cdot c_{3}\ /\ I,O O O=L S S T i p M y s+S h f t G a g L\cdot L S S h f t F z s$ (about the ys-axis), $(\mathrm{kN}\!\cdot\!\mathrm{m})$ $L S S G a g M z s=M_{\scriptscriptstyle R o t o r}^{L@F}\cdot c_{_2}/I,000=L S S T i p M z s-S h f t G a g L\cdot L S S h f t F y s$ Rotating low-speed shaft bending moment at the shaft’s strain gages Rotating low-speed shaft bending moment at the shaft’s strain gages Nonrotating low-speed shaft bending moment at the shaft’s strain gages Nonrotating low-speed shaft bending moment at the shaft’s strain gages Note that no shear or thrust forces need be output at point SG since these would be the same as the shear and thrust forces at point P. Note also that $$ {c_{I}}\cdot{M_{R o t o r}^{L\left(\varpi P\right)}}={c_{I}}\cdot{M_{R o t o r}^{L\left(\varpi S G\right)}} $$ and thus the low-speed shaft torque or rotor torque are constant along the shaft. Generator and High-Speed Shaft Loads: There are 9 output loads on the high-speed shaft. The first and second are the high-speed shaft torque, HSShftTq , and high-speed shaft torque coefficient, HSShftCq , whose convention is that it has a positive value when the LSShftTq is positive. The third and fourth are the high-speed shaft power, HSShftPwr , and high-speed shaft power coefficient, $H S S h f t C p$ . The fifth and sixth are the generator electrical torque, GenTq , and generator electrical torque coefficient, GenCq . The seventh is the high-speed shaft braking torque, HSSBrTq . The eighth is the generator electrical power, GenPwr . The ninth is the electrical generator power coefficient, GenCp . From a simple free-body diagram of a black-box gearbox, ![](images/f1f02883fd1f61f1fb7826ddd0f0d8b199cbab6b0a36a06fb3ea9b0105a76f05.jpg) High-speed shaft torque (this is constant along the shaft and has the convention that it is positive when the LSShftTq is positive), (kN·m) This can alternatively be written in terms of the high-speed shaft motions and torques through use of the equation for the GeAz DOF as follows. From earlier work, $$ H S S h j t T q=\frac{\frac{E}{\omega_{D r T r}^{L}}\cdot M_{R o t o r}^{L@P}G B o x E\!f\!\!f^{S I G N\left(L S S h\!\!f t T q\right)}}{I,000\cdot G B R a t i o} $$ From the equations of motion for the GeAz DOF, it is seen that this is equivalent to saying: $$ H S S h f t T q=\frac{\left(-F_{G e d z}^{*}\Big|_{G}-F_{G e d z}\Big|_{G e n}-F_{G e d z}\Big|_{B r a k e}-F_{G e d z}\Big|_{G B F r i c}\right)G B o x E f f^{S I G N}(L S S h\beta t T q)}{I,O\theta O\cdot G B R a t i o} $$ ![](images/074b15b551123781bc9b4b291cda94c4624a89d0c711cba32a2d75349bd95295.jpg) or, HSShftTq $=$ or, HSShftTq=GenIner ⋅GBRatio⋅ qGeAz+GenDir ⋅GenInerEαR⋅c1+TGen(GBRatio ⋅qGeAz,t)+TBrake(t )/ 1,000 $H S S h f t C q=\frac{I,O O O\cdot H S S h f t T q}{\frac{I}{2}R h o\cdot P r\,o j A r e a\cdot V_{o}^{2}\cdot T i p R a d}$ $$ H S S h j t P w r=H S S h j t T q\cdot G B R a t i o\cdot\dot{q}_{G e d z} $$ High-speed shaft power, (kW) High-speed shaft power coefficient, (-) $H S S B r T q=T^{B r a k e}\left(t\right)/I,O O O$ $G e n T q=T^{G e n}\left(G B R a t i o\cdot\dot{q}_{G e d z},t\right)/\mathit{I},000$ situation or power input), $(\mathrm{kN\cdotm})$ Electrical generator torque (positive reflects power extracted and negative represents a motoring-up $$ G e n C q\!=\!\frac{I,\!O00\cdot G e n T q}{\frac{I}{2}R h o\cdot P r\,o j\!A r e a\cdot V_{o}^{2}\cdot T i p R a d} $$ Electrical generator torque coefficient, (-) Though the HSShftTq is calculated the same regardless of the generator model employed, GenPwr is not. Similar to how power is transmitted through the gearbox with a simple efficiency, for the simple generator or simple variable-speed generator control models, the electrical generator power is as follows: $$ G e n P w r=G B R a t i o\cdot\dot{q}_{G e,i z}\cdot G e n T q\cdot G e n E j f^{S I G N\left[T^{G e n}\left(G B R a t i o\cdot\dot{q}_{G e,i z},t\right)\right]}\;/\;I,0O0 $$ Electrical generator power (positive reflects power extracted and negative represents a motoring-up situation or power input), (kW) And for the Thevenin-Equivalent induction generator model, Electrical generator power (positive reflects power extracted and negative represents a motoring-up situation or power input), (kW) where, $\begin{array}{r l}&{P w r_{_{M e c h a n i c a l}}=G B R a t i o\cdot\dot{q}_{_{G e d z}}\cdot T^{G e n}\left(G B R a t i o\cdot\dot{q}_{_{G e d z}},t\right)}\\ &{P w r_{_{S t a t o r L o s s}}=T E C_{-}N P h a\left|\overline{{I}}_{I}\right|^{2}T E C_{-}S\,R e s}\end{array}$ ![](images/21058b4b11f8e9e916b6011546863a021d1b2b479898ba91140dfb2c7d560aec.jpg) ![](images/69c115517884fc61e62994dd36aa05afefffe25bfcde7a77df92cee85ebb85ba.jpg) $$ \overline{{{I}}}_{I}=\overline{{{I}}}_{2}+\frac{V_{I A}}{T E C\_M R\overline{{{j}}}} $$ (the sign of this is governed by $T^{G e n}$ ) (always positive) (always positive) where the definition of $V_{I a},R_{e I},X_{e I}$ , and Slip are given elsewhere and $\overline{{\boldsymbol{j}}}=\sqrt{-\boldsymbol{l}}$ . Otherwise, the electrical generator power, GenPwr, is a user-defined function of the high-speed shaft speed, GBRatio $\dot{q}_{G e A z}$ , and time $t$ . Electrical generator power coefficient, (-) # Rotor-Furl Axis Loads: There is 1 output load on the rotor-furl axis. This is the rotor-furl moment about the rotor-furl axis. Of course, we could also output all 6 components of the force $F_{G e n,R o t}^{V}\ /$ moment $M_{G e n,R o t}^{N\!(a V}$ acting on the rotor-furl axis at point V on the nacelle. Following the analysis for finding the blade root loads, the new generalized active force for the equations of motion resulting from these new loads is: $$ F_{r}|_{G e n,R o t}={}^{E}\nu_{r}^{V}\cdot F_{G e n,R o t}^{V}+{}^{E}\omega_{r}^{N}\cdot M_{G e n,R o t}^{N\overline{{\omega}}V}\quad\left(r=l,2,...,22\right) $$ This generalized active force must produce the same effects as the generalized active and inertia forces associated with blade 1, blade 2, the hub, the drivetrain, and the structure that furls with the rotor. Thus, $$ \begin{array}{r l}{\lefteqn{\mathrm{\partial_{r}=\}F_{r}^{*}\Big|_{B I}+F_{r}\Big|_{A e r o s h I}+F_{r}\Big|_{G r o w B I}+F_{r}\Big|_{E l a s t i c B I}+F_{r}\Big|_{D a m p B I}+F_{r}^{*}\Big|_{B2}+F_{r}\Big|_{A e r o B2}+F_{r}\Big|_{G r o w B2}+F_{r}\Big|_{E l a s t i c B}}\\ &{\quad+F_{r}^{*}\Big|_{H}+F_{r}\Big|_{G r o w H}+F_{r}\Big|_{S p r i n g T e e t}+F_{r}\Big|_{D a m p T e e t}+F_{r}^{*}\Big|_{G}+F_{r}\Big|_{G e n}+F_{r}\Big|_{B r a k e}+F_{r}\Big|_{G B F r i c}+F_{r}\Big|_{E l a s t i c}}\\ &{\quad+F_{r}^{*}\Big|_{R}+F_{r}\Big|_{G r o w B}+F_{r}\Big|_{S p r i n g R F}+F_{r}\Big|_{D a m p R F}}\end{array} $$ Since $^E\nu_{r}^{\nu}$ and $^E\pmb{\omega}_{r}^{N}$ are equal to zero unless $r=1,2,...,11$ , the generalized active forces associated with blade, drivetrain, rotor-furl, and teeter elasticity and damping, as well as the generator torque, HSS braking torque, and gearbox friction do not contribute to the rotor-furl loads (since also, Fr Ela $_{s t i c B I},\;F_{r}\big|_{D a m p B I},\;F_{r}\big|_{E l a s t i c B},\;F_{r}\big|_{D a m p B2},\;F_{r}\big|_{S p r i n g T e e}\,,\;F_{r}\big|_{D a m p T e e t},\;F_{r}\big|_{S p r i n g R F}\,,\;F_{r}\big|_{D a m p R F}\,,$ ElasticDrive, Fr DampDrive, Fr Gen, Fr Brake, and $F_{r}|_{G B F r i c}$ are equal to zero if $r=1,2,...,11)$ . So, FrGen,Ro $,={F_{r}^{*}}\Big|_{B I}+{F_{r}}\Big|_{A e r o B I}+{F_{r}}\Big|_{G r a v B I}+{F_{r}^{*}}\Big|_{B2}+{F_{r}}\Big|_{A e r o B2}+{F_{r}}\Big|_{G r a v B2}+{F_{r}^{*}}\Big|_{H}+{F_{r}}\Big|_{G r a v H}+{F_{r}^{*}}\Big|_{R}+{F_{r}}\Big|_{G r a v B}$ ravR+Fr\*G (r=1,2,,11) When using the results for hub and rotor loads, this equation can be simplified as follows: $$ F_{r}\big|_{G e n,R o t}=F_{r}\big|_{R o t o r}+F_{r}^{*}\big|_{R}+F_{r}\big|_{G r a v R}+F_{r}^{*}\big|_{G}\quad\big(r=I,2,...,I I\big) $$ Thus, Fr Gen,Ro $\v_{\tau}={}^{E}\nu_{r}^{P}\cdot F_{R o t o r}^{P}+{}^{E}\omega_{r}^{L}\cdot M_{R o t o r}^{L@P}-m^{R}{}^{E}\nu_{r}^{D}\cdot\left({}^{E}a^{D}+g z_{2}\right)-{}^{E}\omega_{r}^{R}\cdot\left({\overline{{{\overline{{I}}}}}}^{R}\cdot{}^{E}a^{R}+{}^{E}\omega^{R}\times{\overline{{{\overline{{I}}}}}}^{R}\cdot{}^{E}a^{R}\right)$ ωR)−EωrG⋅(IG⋅EαG+EωG×IG⋅EωG) (r=1,2,,11) However, ${}^{E}{\pmb\omega}_{r}^{L}\,,\ {}^{E}{\pmb\omega}_{r}^{G}\,,\ {}^{E}{\pmb\omega}_{r}^{R}$ , and $^E\pmb{\omega}_{r}^{N}$ are all equal when $r$ is constrained to be between 1 and 11. Thus, when grouping like terms: $$ {\bf\Pi}_{\prime}={}^{E}{\nu}_{r}^{P}\cdot F_{R o t o r}^{P}-m^{R}{}^{E}{\nu}_{r}{\nu}_{\prime}^{D}\cdot\left({}^{E}a^{D}+g{\bar{z}}_{2}\right)+{}^{E}\omega_{r}^{N}\cdot\left(M_{R o t o r}^{L a P}-{\overline{{{\bar{I}}}}}{}^{R}\cdot{}^{E}a^{R}-{}^{E}{\omega}^{R}\times{\overline{{{\bar{I}}}}}{}^{R}\cdot{}^{E}{}\omega^{R}-{\overline{{{\bar{I}}}}}{}^{R}\cdot{}^{E}{}\omega^{R}\right){\bf\Pi}_{\prime}^{\mathrm{~a~c~o~n~d~}}{\bf\Pi}_{\mathrm{~I~p~o~t~o~r~}}^{E}=0 $$$$ +\,g\!z_{2}\!\left)+\,^{E}\!\omega_{r}^{N}\cdot\left(M_{R o t o r}^{L\bar{\alpha}P}-\overline{{{\overline{{{I}}}}}}\,^{R}\cdot^{E}\!\alpha^{R}-^{E}\!\omega^{R}\times\overline{{{\overline{{{I}}}}}}\,^{R}\cdot^{E}\!\omega^{R}-\overline{{{\overline{{{I}}}}}}\,^{G}\cdot^{E}\!\alpha^{G}-^{E}\!\omega^{G}\times\overline{{{\overline{{{I}}}}}}\,^{G}\cdot^{E}\!\omega^{G}\right)\quad(r\to\infty)^{N} $$ Recognizing also that ${}^{E}{\pmb{\nu}}_{r}^{P}={}^{E}{\pmb{\nu}}_{r}^{V}+{}^{E}{\pmb{\omega}}_{r}^{N}\times{\pmb{r}}^{V P}$ and ${}^{E}{\pmb{\nu}}_{r}^{D}={}^{E}{\pmb{\nu}}_{r}^{V}+{}^{E}{\pmb{\omega}}_{r}^{N}\times{\pmb{r}}^{V D}$ , when $r=1,2,...,11$ , this generalized force can be expanded to: Fr Gen,Ro $\begin{array}{l}{{\_=\left(^{E}\nu_{_{r}}^{V}+^{E}\omega_{_{r}}^{N}\times r^{V P}\right)\cdot F_{\_{R o t o r}}^{P}-m^{R}\left(^{E}\nu_{_{r}}^{V}+^{E}\omega_{_{r}}^{N}\times r^{V D}\right)\cdot\left(^{E}a^{D}+g\mathrm{z}_{_{2}}\right)}}\\ {{\_=\ F_{\_}^{E}\cdot^{E}\left(M_{_{R o t o r}}^{L@P}-\overline{{{I}}}^{R}\cdot^{E}a^{R}-^{E}\omega^{R}\times\overline{{{I}}}^{R}\cdot^{E}\omega^{R}-\overline{{{I}}}^{G}\cdot^{E}a^{G}-^{E}\omega^{G}\times\overline{{{I}}}^{G}\cdot^{E}\omega^{G}\right)}}\end{array}\left(r=F_{\rightarrow}\right)$ ,2,,11) Now applying the cyclic permutation law of the scalar triple product: $$ \begin{array}{c}{{\ \ \ _{_{I,R o t}}={}^{E}\nu_{r}^{V}\cdot\left[F_{_{R o t o r}}^{P}-m^{R}\left({}^{E}a^{P}+{}g z_{2}\right)\right]+{}^{E}\omega_{r}^{N}\cdot\left[r^{V P}\times F_{_{R o t o r}}^{P}-m^{R}r^{V D}\times\left({}^{E}a^{P}+{}g z_{2}\right)\right]}}\\ {{+\ {}^{E}\omega_{r}^{N}\cdot\left(M_{R o t o r}^{L@P}-\overline{{{I}}}^{R}\cdot{}^{E}a^{R}-{}^{E}\omega^{R}\times\overline{{{I}}}^{R}\cdot{}^{E}\omega^{R}-\overline{{{I}}}^{G}\cdot{}^{E}a^{G}-{}^{E}\omega^{G}\times\overline{{{I}}}^{G}\cdot{}^{E}\omega^{G}\right)}}\end{array}\ \left(\begin{array}{c}{{F}}\\ {{A_{R o t o r}^{E}\times\left({}^{E}a^{P}+{}^{E}z_{2}\right)}}\\ {{+\ {}^{E}\omega_{r}^{N}\cdot\left({}^{E}a^{E}\cdot{}^{E}a^{E}\right)}}\end{array}\right) $$ which simplifies to: $$ \begin{array}{r l}{\lefteqn{\mathrm{\Large~\stackrel~:~}{\varepsilon}_{\nu}V_{r}^{\star}\,\biggl\lbrack F_{R o t o r}^{P}-m^{R}\biggl(\varepsilon_{a}^{\,b}+g z_{2}\biggr)\biggr\rbrack}}\\ &{+\mathrm{\Large~\stackrel~{\varepsilon}{\omega}^{N}\!\!~}_{r}\,\biggl\langle M_{R o t o r}^{L\bar{\alpha}P}+r^{V P}\times{\cal F}_{R o t o r}^{P}-m^{R}{\cal r}_{r}^{V D}\times\Bigl({\cal{E}}_{a}^{\,b}+g z_{2}\Bigr)-\overline{{\cal{I}}}^{R}\cdot{\cal{E}}_{a}^{\,b}\!\!\!\!/^{\,b}-{\cal{E}}_{\mathrm{\Large~\omega}}{^{R}}\times\overline{{\cal{I}}}^{R}\cdot{\cal{E}}_{\omega}{^{R}}-m^{R}{\cal{E}}_{a}^{\,b}\biggr\rbrack\,.}\end{array} $$ Thus it is seen that, and $$ \begin{array}{l}{{\displaystyle{\sf R}_{\theta\alpha}={\cal F}_{R o\omega}^{P}-m^{R}\left({}^{E}a^{D}+g z_{2}\right)}}\\ {{\displaystyle{}}}\\ {{\hphantom{\displaystyle{}}_{|R\alpha\theta}^{\gamma}={\cal M}_{R\alpha\theta r}^{L\bar{\alpha}P}+{\cal F}_{R o\omega r}^{V P}-m^{R}{\cal r}_{|{}}^{V D}\times\left({}^{E}a^{D}+g z_{2}\right)-\overline{{{\cal I}}}^{R}\cdot{}^{E}a^{R}-{}^{E}{\omega}^{R}\times\overline{{{\cal I}}}^{R}\cdot{}^{E}{\omega}^{R}-\overline{{{\cal I}}}^{G}\cdot{}^{E}}}\end{array} $$ MGNe@n αG−EωG×IG⋅EωG Thus, $$ F_{G e n,R o t}^{V}=F_{R o t o r}^{P}-m^{R}\left\{\left(\sum_{i=I}^{I2}\varepsilon_{\pmb{\nu}_{i}^{D}\ddot{q}_{i}}\right)+\left[\sum_{i=J}^{I2}\frac{d}{d t}\Big({\bf\nabla}^{E}\nu_{i}^{D}\Big)\dot{q}_{i}\right]+g z_{2}\right\}\;. $$ and = ML@P $\begin{array}{l}{{{\displaystyle{\cal L}_{\theta\omega\sigma}^{\alpha\boldsymbol{P}}+{\cal P}^{\nu\boldsymbol{P}}\times{\cal F}_{R o\omega\sigma}^{\boldsymbol{P}}-m^{R}r^{\nu\boldsymbol{D}}\times\left\{\left(\sum_{i=I}^{l2}\varepsilon_{\nu_{i}^{D}}^{\phantom{\dagger}}\tilde{q}_{i}\right)+\left[\sum_{i=d}^{l2}\frac{d}{d t}\left(^{E}\nu_{\nu_{i}^{D}}^{D}\right)\dot{q}_{i}\right]+g z_{2}\right\}}}\\ {{{\displaystyle{^*{R}}\cdot\left\{\left(\sum_{i=d}^{l2}E_{\omega_{i}^{R}}^{\phantom{\dagger}}\tilde{q}_{i}\right)+\left[\sum_{i=7}^{l2}\frac{d}{d t}\left(^{E}\omega_{i}^{R}\right)\dot{q}_{i}\right]\right\}-{^{E}\omega^{R}}\times\overline{{{{\cal I}}}}^{R}\cdot{^{E}}\omega^{R}-\overline{{{{\cal I}}}}^{G}\cdot\left\{\left(\sum_{i=d}^{l3}E_{\omega_{i}^{G}}^{\phantom{\dagger}}\tilde{q}_{i}\right)+\left[\sum_{i=7}^{l3}\frac{d}{d t}\left(^{E}\omega_{i}^{R}\right)\right]\right\}}}\end{array}$ EωiG)qi−EωG×IG⋅EωG Or, FGen,Rotr 二 F Rotorr P mR E v (r=1,2,,14;16,17,,22) Gen,Rott =FP Rotort mR ∑ d (EviD)qi +gz2 and MGNe@n,VRotr = = ML@P Rotorr +r VP ×FRotorr m (r=1,2,,14;16,17,,22) MN@V Gen,Rott = MRLo@toPr +rVP ×FP Rotort mRrVD × ∑dt(EviD)qi +gz2 −IR⋅1∑2 ddt(EωiR)qi−EωR×IR⋅EωR−IG⋅1∑3 ddt(EωiG)qi−EωG×IG⋅EωG The output loads is as follows, RFrlBrM=MGNe@n,VRo ⋅rfa/ 1,000 Rotor-furl bearing furl moment (about the rotor-furl axis), $(\mathrm{kN\cdotm})$ Like the LSShftTq and LSSTipMza, it is noted that the rotor-furling furl moment can be computed differently using the rotor-furl springs and dampers, though the load summation method and this other constraint method are equivalent. This can be demonstrated as follows. First of all, the equation above is equivalent to saying: $$ R F r l B r M=}^{E}\omega_{R F r l}^{R}\cdot M_{G e n,R o t}^{N@V}/{I,O O O} $$ Or, RFrlBrM $=^{E}\omega_{R F r}^{R}\cdot\Bigg[M_{R o t o r}^{L@P}+\boldsymbol{r}^{V P}\times\boldsymbol{F}_{R o t o r}^{P}-m^{R}\boldsymbol{r}^{V D}\times\left(^{E}\boldsymbol{a}^{D}+g\boldsymbol{z}_{2}\right)-\overline{{\boldsymbol{I}}}^{R}\cdot^{E}\boldsymbol{a}^{R}-^{E}\omega^{R}\times\overline{{\boldsymbol{I}}}^{R}\cdot^{E}\boldsymbol{\omega}^{R}-^{E}\boldsymbol{a}^{R}\cdot^{E}\boldsymbol{a}^{R}\Bigg],$ IG⋅EαG−EωG×IG⋅EωG/ 1,000 Now applying the cyclic permutation law of the scalar triple product: $$ =\left[\begin{array}{l}{{\varepsilon_{\omega_{R F H}^{R}}\times r^{V P}\cdot F_{R o t o r}^{P}-m^{R}\,\varepsilon_{\omega_{R F H}^{R}}\times r^{V D}\cdot\left({\varepsilon_{a}}^{P}+{g}{z_{2}}\right)}}\\ {{\;+\;{^{E}\omega}_{R F H}^{R}\cdot\left(M_{R o t o r}^{L@P}-\overline{{{I}}}^{R}\cdot{^{E}{\alpha}^{R}}-{^{E}{\omega}^{R}}\times\overline{{{I}}}^{R}\cdot{^{E}{\omega}^{R}}-\overline{{{I}}}^{G}\cdot{^{E}{\alpha}^{G}}-{^{E}{\omega}^{G}}\times\overline{{{I}}}^{G}\cdot{^{E}{\omega}^{G}}\right)\right]/\left[\begin{array}{l}{{\varepsilon_{a}}^{R}}\\ {{\varepsilon_{b}}^{R}}\end{array}\right]}}\end{array} $$ Recognizing also that ${}^{E}{\nu}_{R F r l}^{P}={}^{E}{\omega}_{R F r l}^{R}\times{r}^{V P}$ and ${}^{E}{\nu}_{R F r l}^{D}={}^{E}{\omega}_{R F r l}^{R}\times{r}^{V D}$ , and also that $^E{\pmb{\omega}}_{R F r l}^{L}\,,\ {^E\pmb{\omega}}_{R F r l}^{R}$ , and $^E_{\omega_{R F r l}^{G}}$ , are equal, this can be expanded as follows: $$ B r M=\left[{}^{E}\nu_{R F r i}^{P}\cdot F_{R o t o r}^{P}+{}^{E}\omega_{R F r i}^{L}\cdot M_{R o t o r}^{L@P}-m^{R\ E}\nu_{R F r i}^{D}\cdot\left({}^{E}a^{D}+g z_{2}\right)-{}^{E}\omega_{R F r i}^{R}\cdot\left({\overline{{{I}}}}^{R}\cdot{}^{E}a^{R}+g z_{1}\right)\right]. $$ $$ ^{R\;E}\nu_{R F H}^{\;D}\cdot\left(^{E}a^{D}+g\bar{z}_{2}\right)-^{E}\omega_{R F H}^{R}\cdot\left(\overline{{{\overline{{{I}}}}}}\bar{}^{R}\cdot^{E}a^{R}+^{E}\omega^{R}\times\overline{{{\overline{{{I}}}}}}^{R}\cdot^{E}\omega^{R}\right)-^{E}\omega_{R F H}^{G}\cdot\left(\overline{{{\overline{{{I}}}}}}\bar{}^{G}\cdot^{E}a^{G}+^{E}\omega^{R}\times\overline{{{\overline{{{I}}}}}}\omega^{R}\right) $$ $$ \begin{array}{l}{{F_{R F r l}}{\Big|}_{R o t o r}+{F_{R F r l}^{*}}{\Big|}_{R}+{F_{R F r l}}{\Big|}_{G r a v R}+{F_{R F r l}^{*}}{\Big|}_{G}\Big)\thinspace/\thinspace I,000}\\ {\mathrm{~}}\\ {{F_{R F r l}^{*}}{\Big|}_{R}+{F_{R F r l}^{*}}{\Big|}_{H}+{F_{R F r l}^{*}}{\Big|}_{B I}+{F_{R F r l}^{*}}{\Big|}_{B2}+{F_{R F r l}^{*}}{\Big|}_{G}+{F_{R F r l}}{\Big|}_{A e r o B I}+{F_{R F r l}}{\Big|}_{A e r o B2}+{F_{R F r l}}{\Big|}_{G r a v R}+{F_{R F l}}{\Big|}_{G r a v R}+{F_{R F r l}}}\end{array} $$ $\left.\gamma\right|_{G}+F_{R F r}\right|_{A e r o B I}+F_{R F r}\right|_{A e r o B2}+F_{R F r}\right|_{G r a v e}+F_{R F r}\right|_{G r a v e H}+F_{R F r}\right|_{G r a v e B I}+F_{R F r}\right|_{G r a v e B I}=F_{R F r}\Big|_{G r a v e B2}\Big//\left.I,000$ From the equations of motion, it is easily seen that this is equivalent to saying: $$ R F r l B r M=\left(-\left.F_{R F r l}\right|_{S p r i n g R F}-\left.F_{R F r l}\right|_{D a m p R F}\right)/\,l,000 $$ and thus, ![](images/69e0fdf47d8fcc449007e1620543938173b306a662ac01c9b1b4f627f9962fc5.jpg) Thus, both the load summation method and the constraint method are equivalent. Thus, to avoid using 2 different methods to calculate RFrlBrM if various DOFs are disabled, it is best just to use $M_{G e n,R o t}^{N@V}\cdot r\mathbf{f}a\left/I,O O O\right.$ , which will always work. # Tail-Furl Axis Loads: There is 1 output load on the tail-furl axis. This is the tail-furl moment about the tail-furl axis. Of course, we could also output all 6 components of the force $F_{T a i l}^{\mu}$ / moment $M_{T a i l}^{N\!\alpha\mu}$ acting on the tail-furl axis at point W on the nacelle. Following the analysis for finding the rotor-furl loads, the new generalized active force for the equations of motion resulting from these new loads is: $$ F_{r}|_{T a i l}={^{E}\nu_{r}^{W}}\cdot F_{T a i l}^{W}+{^{E}\omega_{r}^{N}}\cdot M_{T a i l}^{N\overline{{{\omega}}}W}\quad\left(r=I,2,...,22\right) $$ This generalized active force must produce the same effects as the generalized active and inertia forces associated with the tail and tail fin. Thus, $$ F_{r}|_{T a i l}=F_{r}^{*}|_{A}+F_{r}|_{G r a v d}+F_{r}|_{A e r o A}+F_{r}|_{S p r i n g T F}+F_{r}|_{D a m p T F}\quad(r=I,2,\dots,22) $$ Since $^E_{\nu_{r}^{\mu}}$ and $^E\pmb{\omega}_{r}^{N}$ are equal to zero unless $r=1,2,...,11$ , the generalized active forces associated with tail-furl elasticity and damping do not contribute to the tail-furl loads (since also, $F_{r}|_{S p r i n g T F}$ and $F_{r}|_{D a m p T F}$ are equal to zero if $r\!=\!1,\!2$ ,…,11). So, $$ F_{r}\big|_{T a i l}=F_{r}^{*}\big|_{A}+F_{r}\big|_{G r a v A}+F_{r}\big|_{A e r o A}\quad\big(r=I,2,...,I I\big) $$ Thus, $$ {\v{r}}_{I}=-m^{B\ E}\nu_{\nu_{r}}^{I}\cdot\left({\v{\overline{{\varepsilon}}}}_{a}^{\ E}+g\Xi_{2}\right)-m^{F\ E}\nu_{\nu_{r}}^{J}\cdot\left({\v{\overline{{\varepsilon}}}}_{a}^{\ E}+g\Xi_{2}\right)+{\v{}}^{E}\nu_{r}^{K}\cdot F_{A e r o t}^{K}+{\v{}}^{E}\omega_{r}^{A\cdot}\cdot\left(M_{A e r o t}^{A\quad}-\overline{{{\bar{I}}}}^{\ A}\cdot F_{A e r o t}^{B\quad}\right), $$ However, $^E{\omega}_{r}^{A}$ and $^E{\pmb\omega}_{r}^{N}$ are all equal when $r$ is con $\begin{array}{r}{^{E}\nu_{r}^{J}={}^{E}\nu_{r}^{W}+{}^{E}\omega_{r}^{N}\times{r}^{W J}\,,\qquad\mathrm{and}\qquad^{E}\nu_{r}^{K}={}^{E}\nu_{r}^{W}+{}^{E}\omega_{r}^{N}\times{r}^{W K}\,,}\end{array}$ when $r=1,2,...,11$ , this generalized force can be expanded to: $$ \begin{array}{r l}&{+\phantom{}^{E}\nu_{r}^{K}\cdot F_{A e r o A}^{K}+\phantom{}^{E}\omega_{r}^{A}\cdot\left(M_{A e r o A}^{A}-\overline{{\overline{{I}}}}\,^{A}\cdot{}^{E}\alpha^{A}+{}^{E}\omega^{A}\times\overline{{\overline{{I}}}}\,^{A}\cdot{}^{E}\omega^{A}\right)\quad\left(r=I,2,\ldots,I I\right)}\\ &{\mathrm{strained~to~be~between~1~and~11.~Recognizing~also~that}\qquad\phantom{.}^{E}\nu_{r}^{I}={}^{E}\nu_{r}^{W}+{}^{E}\omega_{r}^{N}\times{}r^{W I}}\end{array} $$ $$ \begin{array}{r l}&{\_=\Big(^{E}\pmb{\nu}_{r}^{W}+^{E}\omega_{r}^{N}\times\pmb{r}^{W K}\Big)\cdot\pmb{F}_{\_{A e r o A}}^{K}-m^{B}\left(^{E}\nu_{r}^{W}+^{E}\omega_{r}^{N}\times\pmb{r}^{W I}\right)\cdot\left(^{E}\pmb{a}^{I}+g\Xi_{2}\right)-m^{F}\left(^{E}\nu_{r}^{W}+^{E}\omega_{r}^{A}\times^{A}\right)}\\ &{\quad+\;^{E}\omega_{r}^{N}\cdot\left(\pmb{M}_{\_{A e r o A}}^{A}-\overline{{\overline{{I}}}}\cdot^{A}\cdot^{E}\alpha^{A}-^{E}\omega^{A}\times\overline{{\overline{{I}}}}\mathcal{A}\cdot^{E}\omega^{A}\right)}\end{array} $$ Now applying the cyclic permutation law of the scalar triple product: $$ \begin{array}{r l}&{\mathbf{\nabla}_{r}^{\mathcal{W}}\cdot\left[F_{A e r o A}^{K}-m^{B}\left(\mathbf{\nabla}^{E}a^{I}+g z_{2}\right)-m^{F}\left(\mathbf{\nabla}^{E}a^{J}+g z_{2}\right)\right]}\\ &{+\mathbf{\nabla}^{E}\omega_{r}^{N}\cdot\left(M_{A e r o A}^{A}+r^{\mathcal{W}K}\times F_{A e r o A}^{K}-m^{B}r^{\mathcal{W}I}\times\left(\mathbf{\nabla}^{E}a^{I}+g z_{2}\right)-m^{F}r^{\mathcal{W}J}\times\left(\mathbf{\mathcal{E}}a^{J}+g z_{2}\right)-\overline{{\mathbf{\nabla}}}^{A}\cdot\mathbf{\nabla}^{E}a^{J}\right),}\end{array} $$ Thus it is seen that, $$ \begin{array}{l}{{\displaystyle-m^{B}\left({\vphantom{\int}}^{E}a^{I}+g z_{2}\right)-m^{F}\left({\vphantom{\int}}^{E}a^{J}+g z_{2}\right)}}\\ {{\displaystyle}}\\ {{\displaystyle-m^{A}+r^{W\kappa}\times{\cal F}_{A e n d}^{\kappa}-m^{B}r^{W}\times\left({\vphantom{\int}}^{E}a^{I}+g z_{2}\right)-m^{F}r^{W\kappa}\times\left({\vphantom{\int}}^{E}a^{J}+g z_{2}\right)-\overline{{\overline{{I}}}}^{A}\cdot{\vphantom{\int}}\alpha^{A}-{\vphantom{\int}}_{\omega}^{4}\omega^{4}\times\overline{{\overline{{I}}}}^{J}\times\left(-g z_{2}\right)}}\\ {{\displaystyle}}\\ {{\displaystyle-m^{B}\left\{\left(\sum_{i=I}^{I}E_{\nu_{i}^{I}}{\vphantom{\int}}^{\vec{\dagger}}\vec{q}_{i}\right)+{\vphantom{\int}}^{E}{\nu}_{T r{\vec{U}}}^{I}{\ddot{q}}_{T r i}\right.+\left[\sum_{i=d}^{I}\frac{d}{d t}\left({\vphantom{\int}}^{\varepsilon}\nu_{i}^{I}\right){\dot{q}}_{i}\right]+\frac{d}{d t}\left({\vphantom{\int}}^{\varepsilon}\nu_{T r{\vec{U}}}^{I}\right){\dot{q}}_{T r i}+g z_{2}\right\}}}\\ {{\displaystyle}}\\ {{\displaystyle-m^{F}\left\{\left(\sum_{i=I}^{I}\varepsilon_{\nu_{i}^{\prime}}{\vec{q}}_{i}\right)+{\vphantom{\int}}^{E}{\nu}_{T r{\vec{U}}}^{J}{\ddot{q}}_{T r i}+\left[\sum_{i=d}^{I}\frac{d}{d t}\left({\vphantom{\int}}^{\varepsilon}\nu_{i}^{\prime}\right){\dot{q}}_{i}\right]+\frac{d}{d t}\left({\vphantom{\cal F}}_{\nu T r}^{J}\right){\dot{q}}_{T r i}+g z_{2}\right\}}}\end{array} $$ $$ \begin{array}{c}{{\displaystyle r_{T a l}^{N\alpha W}=M_{~A e r o A}^{4}+r^{W K}\times F_{_{A e r o A}}^{K}-m^{B}r^{W}\times\left\{\left(\displaystyle\sum_{i=I}^{I I}\varepsilon_{\nu_{i}^{I}}\ddot{q}_{i}\right)+{}^{E}\nu_{T e r i}^{I}\ddot{q}_{T r r i}+\left[\displaystyle\sum_{i=I}^{I I}\frac{d}{d t}\Big({}^{E}\nu_{{\nu}_{i}^{I}}^{I}\Big)\dot{q}_{i}\right]+:}}\\ {{\displaystyle-m^{F}r^{W J}\times\left\{\left(\displaystyle\sum_{i=I}^{I I}\varepsilon_{\nu_{i}^{J}}\ddot{q}_{i}\right)+{}^{E}\nu_{T e r i}^{J}\ddot{q}_{T r r i}+\left[\displaystyle\sum_{i=I}^{I I}\frac{d}{d t}\Big({}^{E}\nu_{{\nu}_{i}^{J}}^{J}\Big)\dot{q}_{i}\right]+\frac{d}{d t}\Big({}^{E}\nu_{T r r i}^{J}\Big)\dot{q}_{T r r i}+g z_{i},}}\\ {{\displaystyle-\overline{{\bar{I}}}^{\,4}\cdot\left\{\left(\displaystyle\sum_{i=I}^{I I}\varepsilon_{\omega_{i}^{A}}\ddot{q}_{i}\right)+{}^{E}\omega_{T r r i}^{A}\ddot{q}_{T r r i}+\left[\displaystyle\sum_{i=I}^{I I}\frac{d}{d t}\Big({}^{E}\omega_{i}^{A}\Big)\dot{q}_{i}\right]+\frac{d}{d t}\Big({}^{E}\omega_{T r r i}^{A}\Big)\dot{q}_{T r r i}\Big\}-{}^{E}\omega_{T r i}^{J}\;,}}\end{array} $$ Or, $\begin{array}{l}{{\displaystyle F_{r a i_{r}}^{\scriptscriptstyle{W}}=-m^{B}\,^{\cal E}{}_{\nu}^{\scriptscriptstyle{E}}-m^{\cal F}\,^{\scriptscriptstyle{E}}\nu_{\nu}^{\scriptscriptstyle{J}}\quad\left(r=l,2,\ldots,l{l};l5\right)\nonumber}}\\ {{\displaystyle F_{r a i_{t}}^{\scriptscriptstyle{W}}=F_{\scriptscriptstyle{A e r o}}^{\scriptscriptstyle{K}}-m^{B}\,\biggl\{\biggl[\sum_{i=d}^{\underline{{{l}}}}\frac{d}{d t}\Bigl(^{\varepsilon}\nu_{i}^{\prime}\Bigr)\dot{q}_{i}\biggr]+\frac{d}{d t}\Bigl(^{\varepsilon}\nu_{\scriptscriptstyle{F H}}^{\prime}\Bigr)\dot{q}_{\scriptscriptstyle{T F r l}}+g z_{2}\biggr\}-m^{\cal F}\,\biggl\{\biggl[\sum_{i=d}^{\underline{{{l}}}}\frac{d}{d t}\Bigl(^{\varepsilon}\nu_{i}^{\prime}\Bigr)\dot{q}_{i}\biggr]\nonumber}}\end{array}$ +ddt(EvTJFrl)qTFrl+gz2 and MTNai@lrW $\begin{array}{l}{{\displaystyle-m^{B}r^{W I}\times^{E}\nu_{r}^{I}-m^{F}r^{W J}\times^{E}\nu_{r}^{J}-\overline{{{\bar{I}}}}{^{A}}\cdot^{E}\omega_{r}^{A}}}&{{\left(r=I,2,\ldots,I I;I5\right)}}\\ {{\displaystyle M_{_{A e r o A}}^{4}+r^{W K}\times F_{_{A e r o A}}^{K}-m^{B}r^{W J}\times\left\{\left[\sum_{i=4}^{I I}\frac{d}{d t}\Big(^{E}\nu_{i}^{I}\Big)\dot{q}_{i}\right]+\frac{d}{d t}\Big(^{E}\nu_{_{T R I}}^{I}\Big)\dot{q}_{T e r t}+g z_{2}\Bigg\}-m^{F}r^{W J}\times\left\{\begin{array}{l}{{\displaystyle-m^{B}}}\\ {{\displaystyle-\frac{1}{d t}\Big(^{E}\omega_{_{I}}^{A}\Big)\dot{q}_{T e r t}\Bigg\}-^{E}\omega^{A}\times\overline{{{\bar{I}}}}{^{A}}\cdot^{E}\omega^{A}}}\end{array}\right.}}\\ {{\displaystyle-\,\overline{{{\bar{I}}}}^{A}\cdot\left\{\sum_{i=7}^{I I}\frac{d}{d t}\Big(^{E}\omega_{i}^{A}\Big)\dot{q}_{i}\right\}+\frac{d}{d t}\Big(^{E}\omega_{_{T F I}}^{A}\Big)\dot{q}_{T e r t}\Bigg\}-^{E}\omega^{A}\times\overline{{{\bar{I}}}}^{A}\cdot^{E}\omega^{A}}}\end{array}$ M t TNai@lW = ∑ddt(EviJ)qi+ddt(EvTJFrl)qTFrl+gz2 The output loads is as follows, # TFrlBrM=MTNai@lW ⋅tfa/ 1,000 Tail-furl bearing furl moment (about the tail-furl axis), $(\mathrm{kN}\!\cdot\!\mathrm{m})$ Like the LSShftTq, LSSTipMza, and RFrlBrM, it is noted that the tail-furling furl moment can be computed differently using the tail-furl springs and dampers, though the load summation method and this other constraint method are equivalent. This can be demonstrated as follows. First of all, the equation above is equivalent to saying: $$ T F r l B r M=\mathit{\Omega}^{E}\omega_{T F r l}^{A}\cdot M_{T a i l}^{N\ @W}\mathit{\Omega}/\mathit{I},000 $$ Or, TFrlBrM $=^{E}\omega_{T H}^{A}\cdot\left[M_{A e r o A}^{A}+r^{W K}\times F_{A e r o A}^{K}-m^{B}r^{W I}\times\left(^{E}a^{I}+g z_{2}\right)-m^{F}r^{W J}\times\left(^{E}a^{J}+g z_{2}\right)-\overline{{{\overline{{{I}}}}}}^{A}\cdot^{E}\omega_{S}\right]$ αA−EωA×IA⋅EωA/ 1,000 Now applying the cyclic permutation law of the scalar triple product: $$ \begin{array}{r l}&{^{E}\omega_{T F r i}^{A}\times r^{W\kappa}\cdot F_{^{A_{e r o d.}}}^{K}-m^{B\;E}\omega_{T F r i}^{A}\times r^{W H}\cdot\left(^{E}a^{I}+g z_{2}\right)-m^{F\;E}\omega_{T F r i}^{A}\times r^{W J}\cdot\left(^{E}a^{J}+g z_{2}\right)\Biggr]/I_{^{\prime}F r i}}\\ &{\quad+\;^{E}\omega_{T F r i}^{A}\cdot\left(M_{^{A_{e r o d.}}}^{A}-\overline{{\overline{{I}}}}^{\;a}\cdot^{E}\alpha^{A}-^{E}\omega^{A}\times\overline{{\overline{{I}}}}^{\;a}\cdot^{E}\omega^{A}\right)}\end{array} $$ Recognizing also that ${}^{E}\nu_{T F r l}^{I}={}^{E}\omega_{T F r l}^{A}\times r^{W I}$ ${}^{E}\nu_{T F r l}^{J}={}^{E}\omega_{T F r l}^{A}\times r^{W J}$ , and ${}^{E}\nu_{T F r l}^{K}={}^{E}\omega_{T F r l}^{A}\times r^{W K}$ , this can be expanded as follows: $T F r/B r M=\left[{\ \varepsilon_{\nu_{T H r M}}^{\kappa}\cdot F_{A e r o A}^{\kappa}+\varepsilon_{\omega_{T H r i}^{A}}^{\kappa}\cdot M_{A e r o A}^{A}-m^{B\;E}\nu_{\nu_{T H}}^{I}\cdot\left({\varepsilon_{a}^{I}}+{\varepsilon_{\bar{z}}}_{2}\right)-m^{F\;E}\nu_{T H}^{J}\cdot\left({\varepsilon_{a}^{A}}+{\varepsilon_{b}}\right)}\right].$ $_{\!\gamma}\cdot\left(^{\varepsilon}a^{I}+g z_{2}\right)-m^{F~E}\nu_{_{I F H}}^{J}\cdot\left(^{E}a^{J}+g z_{2}\right)-^{\varepsilon}\omega_{_{I F H}}^{A}\cdot\left(\overline{{{I}}}^{\,4}\cdot^{E}\alpha^{4}+^{E}\omega^{A}\times\overline{{{I}}}^{\,4}\cdot^{E}\omega^{A}\right)\right]/I,00$ 0 or, $T F r l B r M=\left(F_{T F r l}^{*}\vert_{A}+F_{T F r l}\vert_{G r a v A}+F_{T F r l}\vert_{A e r o A}\right)/\,I,000$ From the equations of motion, it is easily seen that this is equivalent to saying: $$ T F r l B r M=\left(-\left.F_{T F r l}\right\vert_{S p r i n g T F}-F_{T F r l}\right\vert_{D a m p T F}\right)/\left.I,000 $$ and thus, $$ M=\left\{\begin{array}{l l}{{T F r l S p r cdot q_{T r r l}+I F\left[q_{T r r l}>T F r/U S S P,T F r/U S S p r\left(q_{T r r l}-T F r/U S S P\right),0\right]}}\\ {{+I F\left[q_{T r l}T F r/U S D P,T F r/U S D m p\cdot\dot{q}_{T r r l},0\right]+I F\left[q_{T r l}