## 3 Kinematic and Static Force Analysis
During the opening process of the landing gear, the tension spring between the main rods remains in a stretched state. Let \(x_{0}\) denote the original length of the spring and \(\Delta x=x-x_{0}\) the elongation. Here, \(x\) represents the length of the spring after stretching. The \(k\) is the spring stiffness. The tension in each spring is denoted as \(F_{\rm T}\). As the dual-function landing gear opens, the spring is gradually stretched. The angle \(\theta_{1i}\) and the spring elongation \(\Delta x\) have the following relationship:  
Figure 6:0  
The mass of the aux-rods is neglected. The combined weight of the main rod and its associated multiple components is denoted as G. The center of weight is assumed to lie in the rotational plane of the main rod. In the coordinate system \(x_{1i},y_{1i},z_{1i}\), the coordinates of the center of weight can be determined:  
\[\left\{ \begin{aligned} x_{\rm c}&=\frac{\sum m_{n}x_{ 1i_{n}}}{M}\cos\theta_{1i}\\ y_{\rm c}&=0\\ z_{\rm c}&=\frac{\sum m_{n}z_{1i_{n}}}{M}\sin\theta_{1i }\end{aligned} \right. \tag{10}\]  
During the process of landing gear deployment, with revolute joint \(A_{i}\) serving as the center of rotation, the relationship between the tensile force \(F_{\rm P}\) exerted on the main rod by the cable and the tensile force \(F_{\rm T}\) exerted by the spring can be determined based on the conditions of moment equilibrium:  
When \(5^{\circ}\leq\theta_{1i}\leq\pi/2-\arcsin{D_{\rm e}}/{\rm h}\),  
\[F_{\rm P}\cdot D_{\rm e}+2{\rm ka}^{2}\sin^{2}\theta_{1i}={\rm G}\cdot(x_{c}- {\rm r}) \tag{11}\]  
When \(\pi/2-\arcsin{D_{\rm e}}/{\rm h}<\theta_{1i}\leq 90^{\circ}\),  
\[F_{\rm P}(\sin\theta+D_{\rm e}\cos\theta)+2{\rm ka}^{2}\sin^{2}\theta_{1i}={ \rm G}\cdot(x_{c}-{\rm r}) \tag{12}\]The allowable safety factor for the quick-release mechanism is set to \(k=1.8\), and the transmission efficiency \(\eta\) is taken as 0.9024. Thus, the relationship between the torque \(T\) on the motor output shaft and the tensile force \(F_{\mathrm{P}}\) on the cable is as follows:  
\[T=3\frac{F_{\mathrm{P}}\cdot F_{\mathrm{w}}\cdot k}{i\cdot\eta} \tag{13}\]  
Compared to gripping mechanisms that rely solely on friction or clamping force, the dual-function landing gear can grasp heavier objects with the same clamping force. In an ideal scenario, after the landing gear captures the spherical object, a force \(T\) that gradually increases along the \(z\)-axis downward is applied. Due to the axisymmetric characteristic of the landing gear, the centroid of the sphere always remains on the \(z\)-axis at \((0,0,z_{\mathrm{s}})\). Initially, the sphere is in a stationary state, as illustrated in Figure 3a. Not only do points \(C_{i}\) and \(D_{i}\) on the landing gear simultaneously contact the sphere, but the main rod also contacts the sphere at point \(K_{i}\). According to geometric relationships,  
\[\left\{ \begin{array}{l}D_{i,x}^{2}+D_{i,y}^{2}+(z_{\mathrm{s}}-D_{i,z})^{2}= R^{2}\\ C_{i,x}^{2}+C_{i,y}^{2}+(z_{\mathrm{s}}-C_{i,z})^{2}=R^{2}\end{array} \right. \tag{14}\]  
the angles between the normal forces at points \(C_{i}\) and \(D_{i}\) and the \(z\)-axis are given by the following expressions:  
\[\alpha_{\mathrm{C}}=\arcsin\frac{\sqrt{C_{i,x}^{2}+C_{i,y}^{2}}}{R},\alpha_{ \mathrm{D}}=\arcsin\frac{\sqrt{D_{i,x}^{2}+D_{i,y}^{2}}}{R} \tag{15}\]  
Considering the auxiliary link as a two-force member, the forces \(F_{\mathrm{elastic}}\) and \(F_{\mathrm{NC}_{i}}\) acting at the hinge points on the main link are equal in magnitude and direction. For the target sphere, based on the conditions of static equilibrium, we have:  
\[F_{\mathrm{NC}_{i}}^{\prime}\cos\alpha_{\mathrm{C}}+F_{\mathrm{ND}_{i}}^{ \prime}\cos\alpha_{\mathrm{D}}=3\mathrm{G}+T+F_{\mathrm{NK}_{i}}^{\prime} \cos\theta_{1i} \tag{16}\]  
Taking the revolute joint \(A_{i}\) as the rotational center, and according to the principle of moment equilibrium, we obtain the following:  
\[