# Analysis of a Tracked In-Pipe Robot's Obstacle-Crossing Performance
## 3 Results  
### Force Analysis of Pipe Diameter Adaptation Mechanism  
As shown in Figures 2 and 3, \(X\) is the horizontal distance between points \(A\) and \(O\), and \(D\) represents the radius of the robot. The distance from the support point to the lead screw is denoted as \(U\), while the distance from the tail of the connecting rod to the traveling mechanism is represented by \(H.\)\(R\) is the radius of the pipe. \(AB,BO,\) and \(AC\) are denoted as \(L_{3},L_{2},\) and \(L_{1}\), respectively. The angle between \(OB\) and the lead screw is denoted as \(\alpha\), while the angle between \(AB\) and the \(X\)-axis is denoted as \(\beta\). The angle between the connection line from the driving wheel to the center of the pipe and the center line of the pipe robot is \(\varphi\), the arc length is represented by \(s\), the attitude angle is \(\gamma\), and \(\theta\) is set to 60\({}^{\circ}\).  
The following conclusions can be drawn:  
\[\left\{\begin{aligned} & X=L_{3}\text{cos}\ \beta+L_{2}\text{cos}\ \alpha\\ & L_{3}\text{sin}\ \beta+U=L_{2}\text{sin}\ \alpha\\ & s=R\varphi\\ & D=L_{1}\text{sin}\ \beta+U+H\\ & D\text{sin}\ \theta=R\text{sin}\ \varphi\\ & Y=L_{1}\text{sin}\ \beta+U\\ \end{aligned}\right. \tag{1}\]  
\[\text{cos}\ \varphi=\frac{\sqrt{R^{2}-D^{2}sin^{2}\theta}}{R} \tag{2}\]  
According to the formula above, it follows that  
\[F=\left(\frac{(1-\mu\sin\theta)R^{2}-D^{2}\sin^{2}\theta}{R^{2}-D^{2}\sin^{2} \theta}\right) \tag{3}\]  
Figure 1: Screw and nut diameter-changing mechanism.  
As derived from the virtual displacement theorem,  
\[F\delta_{s}+T\delta_{\omega}\eta=0 \tag{4}\]  
\(T\) is the output torque of the lead-screw motor and \(\eta\) is the transmission efficiency of the lead screw. After arrangement, we obtain:  
\[\begin{array}{l}T=\frac{P}{2\pi\eta}\left(\frac{(1-\mu\sin\theta)R^{2}-D^{2} \sin^{2}\theta}{R^{2}-D^{2}\sin^{2}\theta}\right)\\ \frac{L_{1}m\cos\theta\cos\gamma}{L_{3}(\tan\alpha+\tan\beta)}\end{array} \tag{5}\]  
Here, \(P\) denotes the lead of the lead screw, and \(\omega\) represents the rotation angle between the lead screw and the nut.  
When the robot is in equilibrium, both the attitude angle and the inclination angle are \(0^{\circ}\). According to Equation (5), when the pipe robot adapts to the maximum pipe diameter of 340 mm, \(\alpha\) is 69.7\({}^{\circ}\) and \(\beta\) is 76.2\({}^{\circ}\). When adapting to the minimum pipe diameter of 260 mm, \(\alpha\) is 15.8\({}^{\circ}\) and \(\beta\) is 23.4\({}^{\circ}\). From the output torque formula of the lead-screw motor,  
Figure 3: In-pipe pose of pipeline robot.  
Figure 2: Mechanical model of pipe diameter adaptation mechanism.  
it is evident that the output torque of the lead screw is influenced by the values of \(\alpha\) and \(\beta\). Specifically, the output torque of the lead-screw motor is inversely proportional to these angles. As the length of AB increases, the required torque decreases. Conversely, as AB decreases, the output torque of the lead screw gradually increases, along with the thrust force \(F\) required for the operation of the variable-diameter mechanism. Moreover, the output torque of the lead-screw motor is inversely proportional to both the attitude angle and the inclination angle. When these angles deviate from \(0^{\circ}\), a larger angle results in a smaller output torque of the lead screw and a greater required power. Therefore, the robot should endeavor to maintain balance within the pipe, align with the central axis of the pipe, and enhance the performance and efficiency of robotic operations.