# Application of Absolute Nodal Coordinate Formulation in Calculation of Space Elevator System
## 2 Tether Modelling
\begin{split} S_{1}(x)&=1-3\xi_{e}^{2}+2\xi^{3}\\ S_{2}(x)&=L(\xi-2\xi^{2}+\xi^{3})\\ S_{3}(x)&=3\xi^{2}-2\xi^{3}\\ S_{4}(x)&=L(-\xi^{2}+\xi^{3})\end{split}  \tag{4}\]  
where \(\xi=x/L\). As shown in Equation (2), the shape function and node coordinates only depend on the element coordinates and time variables, respectively, so the element displacement derivative, velocity and acceleration are expressed as follows:  
\[ \begin{split}\vec{r}_{x}(x,t)&=\mathbf{S}_{x}(x)\vec{q }_{e}(t)\\ \vec{r}_{xx}(x,t)&=\mathbf{S}_{xx}(x)\vec{q}_{e}(t)\\ \vec{\dot{\tau}}(x,t)&=\mathbf{S}(x)\vec{q}_{e}(t)\\ \vec{\ddot{\tau}}(x,t)&=\mathbf{S}(x)\vec{\dot{q}}_{e}(t) \end{split}  \tag{5}\]  
For a rope element with cross-sectional area \(A\) and density \(\rho\), its kinetic energy is:  
\[T^{e}=\frac{1}{2}\int_{0}^{L}\rho\int_{A}\vec{r}\dot{\tau}\dot{\tau}\mathrm{d} A\mathrm{d}x=\frac{1}{2}\vec{q}_{e}^{\intercal}\mathbf{M}^{e}\dot{q}_{e} \tag{6}\]  
where \(M^{e}\) is the element mass matrix, which can be presented as:  
\[\mathbf{M}^{e}=\int_{0}^{L}\rho AS(x)^{\intercal}S(x)\mathrm{d}x \tag{7}\]  
It can be seen from the displacement field function of the element that the axial and bending deformation are considered in the element. The axial strain \(\varepsilon\) and curvature \(\kappa\) can be presented as:  
\[\varepsilon=\frac{1}{2}(\vec{r}_{x}^{\intercal}\vec{r}_{x}-1) \tag{8}\]  
\[\kappa=\frac{\left\|\vec{r}_{x}\times\vec{r}_{xx}\right\|}{\left\|\vec{r}_{x} \right\|^{3}} \tag{9}\]  
According to Euler Bernoulli beam theory, the element strain energy can be presented as:  
\[U_{e}^{e}=\frac{1}{2}\int_{0}^{L}(EA\varepsilon^{2})\mathrm{d}x+\frac{1}{2} \int_{0}^{L}(EJ\kappa^{2})\mathrm{d}x \tag{10}\]where \(E\) and \(J\) are the Young's modulus and the section moment of inertia of the element material. The generalized elastic force of the element can be obtained by calculating the partial derivative of the strain energy in Equation (10) with respect to the generalized coordinate:  
\[\vec{Q}_{\varepsilon}^{\varepsilon}=\int_{0}^{L}EA\frac{\partial\varepsilon}{ \partial\vec{q}}\mathrm{d}x+\int_{0}^{L}EJ\kappa\frac{\partial\kappa}{ \partial\vec{q}}\mathrm{d}x \tag{11}\]  
The generalized structural force is composed of universal gravitation and centrifugal force. As shown in Figure 1, there is no difference between the perturbation of universal gravitation in the \(XY\) plane or \(XZ\) plane, but the centrifugal force is different in the \(XY\) plane and the \(XZ\) plane. This is because the calculation of universal gravitation depends on the distance between the particle and the earth's center, while centrifugal force depends on the distance between the particle and the earth's rotation axis. As a result, the \(z\) component in \(\vec{r}\) has no contribution to centrifugal force, a transformation matrix \(\mathbf{C}\) is set to calculate centrifugal force, which can be presented as:  
\[\mathbf{C}= \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&0\end{pmatrix}  \tag{12}\]  
The gravitational force \(\vec{Q}_{g}^{x}\) and centrifugal force \(\vec{Q}_{\varepsilon}^{x}\) with the element coordinate \(x\) can be expressed as:  
\[\vec{Q}_{g}^{x}=-\frac{\mu m_{x}^{\varepsilon}}{\vec{r}(x)^{2}} \tag{13}\]  
\[\vec{Q}_{\varepsilon}^{x}=m_{x}^{\varepsilon}\omega^{2}\mathbf{C}\vec{r}(x) \tag{14}\]  
where \(\mu\) is the gravitational constant, and the point mass \(m_{x}^{\varepsilon}=\rho A\). For objects with mass \(m_{o}\) in the geostationary orbit, the gravitational force is equal to the centrifugal force, and we can obtain:  
\[\frac{\mu m_{o}}{R_{g}^{2}}=m_{o}\omega^{2}R_{g} \tag{15}\]  
where \(R_{g}\) is the geosynchronous orbit radius.