# The Modeling and Calculation of the Heading Machine Based on Differential Geometry

^{1}

^{2}

^{*}

## Abstract

**:**

## 1. Introduction

## 2. The Mechanism Analysis of the Horizontal Axis Heading Machine

## 3. The Differential Geometry Modeling from SO(2) to SO(3)

_{0}y

_{0}and the body frame which can rotate with the rod, as in Figure 2. Supposing that the axises from these two frames concide together at the initial moment, then the rod anticlockwise rotate $\mathsf{\theta}$ along the point O, the projection of the axises of the body frame on the inertail frame can be expressed as the matrix type [22] as

_{1}and Oy

_{1}in the inertial frame, the rotation matrix satisfies the following property

_{1}y

_{1}is ${\mathit{l}}_{A}$, then the projection of point A in Ox

_{0}y

_{0}after the rotation can be expressed as

_{0}y

_{0}is

## 4. The Pose and Attitude Analysis

_{1}, the cutting arm rotates along the frame O

_{2}, the cutting head rotates along the frame O

_{3}, the drive cylinder of the cutting arm is hinged with the turntable and the cutting arm by the point O

_{4}and O

_{5}, respectively. The distance between the points O

_{6}and O

_{7}represents the length of the slip sleeve’s extending and shrinking. The driving cylinder of the turntable is fixed with the main engine by the point O

_{8}, the gear of the turntable is engaged with the rack by O

_{9}. Constructing the body frame and the inertial frame at these points respectively, the rotation direction of the body frames is all along the anticlockwise direction [23].

_{3}to be ${\mathit{r}}_{1}$, and the rotation angle of the cutting head along the z axis of the frame O

_{3}to be ${\mathsf{\theta}}_{3}$, then the rotation matrix is as Equation (15). So the position vector of the cutting teeth in the inertial frame which is at the point of O

_{3}is

**R**

_{3}

**r**

_{1}.

_{3}relative to the O

_{7}to be

**r**

_{2}, then the coordinate value of the cutting tooth relative to the frame O

_{7}to be

**r**

_{2}+

**R**

_{3}

**r**

_{1}. Define the distance between O

_{6}and O

_{7}is ${l}_{S}$, and the x axis of the frame O

_{6}and O

_{7}are coincide with the line O

_{6}O

_{7}, so the coordinate of O

_{7}relative to point O

_{6}to be ${l}_{S}{\mathit{e}}_{1}$, then the position of the cutting tooth relative to the frame O

_{6}is

_{6}and O

_{2}is ${l}_{J}$, so the coordinate of O

_{6}in the frame of O

_{2}is ${l}_{J}{\mathit{e}}_{1}$, the coordinate of the cutting tooth in the frame O

_{2}is

_{2}, when the cutting arm rotates ${\mathsf{\theta}}_{2}$ along the z axis of the frame O

_{2}, the rotation matrix is

_{2}is

_{2}and O

_{1}to be ${l}_{Z}$, the position vector of O

_{2}relative to frame O

_{1}is ${l}_{Z}{\mathit{e}}_{1}$, then the position vector of the cutting tooth in frame O

_{1}is

_{5}in the body frame at point O

_{2}to be ${\mathit{r}}_{3}$, the point O

_{5}rotate with the body frame at point O

_{2}, so the position vector of point O

_{5}is ${\mathit{R}}_{2}{\mathit{r}}_{3}$, then the position of O

_{5}in the frame at O

_{1}is

_{4}in the frame O

_{1}to be ${\mathit{r}}_{4}$, then the distance between the two points O

_{4}and O

_{5}can be expressed as

_{4}O

_{5}and the rotation angle of the cutting arm. Supposing that the rotation angle of the rod rotate along the point is ${\mathsf{\theta}}_{4}$, the rotation matrix is

_{4}O

_{5}to coincide with the x axis of the frame O

_{4}, the coordinate of the point O

_{5}in the frame O

_{1}after the rotation can be expressed as

_{4}O

_{5}. When the turntable rotates ${\mathsf{\theta}}_{1}$ along the y axis of frame O

_{1}, the rotation matrix is

## 5. The Velocity Analysis

_{4}O

_{5}respectively. Differentiate Equation (24), replace ${\dot{\mathit{R}}}_{2}$ by the Equation (29), the relation between ${\dot{l}}_{Y}$ and ${\mathsf{\omega}}_{2}$ is

_{4}O

_{5}have a nonlinear relation. Differentiate Equation (26), the following relation can be obtained.

_{4}O

_{5}rotate along O

_{4}, which can be expressed as

## 6. The Acceleration Analysis of the Motion

_{4}O

_{5}and the angular acceleration of the cutting arm is derived. Differentiate Equation (30), the angular acceleration response of the cutting arm is

## 7. The Numerical Calculation Flow

## 8. The Simulation Results

^{2}, the initial and the end positions are 0 and 1500 mm respectively, the acceleration of the slip sleeve is 1 mm/s

^{2}, the initial and the end positions are 1538 mm and 1938 mm, respectively, the acceleration of the drive cylinder of the cutting arm is 0.5 mm/s

^{2}, the initial and the end positions are 2096 mm and 2296 mm, respectively. In the first 10 s, the turntable, the slip sleeve and the drive cylinder of the cutting arm are accelerated with their constant accelerations. Then they move with a constant velocity with the next 20 s, and decelerate within 10 s. The whole time of the simulation is 40 s. The simulation results of the kinematic parameters of the heading machine are shown in Figure 4, Figure 5, Figure 6 and Figure 7.

## 9. Summary and Prospects

## Acknowledgments

## Author Contributions

## Conflicts of Interest

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The calculation flow |

Set the initial values: ${l}_{Y}^{0},{l}_{S}^{0},{\mathit{R}}_{1}^{0},{\mathit{R}}_{3}^{0},{\mathsf{\omega}}_{1}^{0},{\mathsf{\omega}}_{3}^{0},{\dot{l}}_{Y}^{0},{\dot{l}}_{S}^{0},{\dot{\mathsf{\omega}}}_{1}^{0},{\dot{\mathsf{\omega}}}_{3}^{0},{\ddot{l}}_{Y}^{0},{\ddot{l}}_{S}^{0}$ |

For loop |

1 The pose and attitude responses: ${\mathit{R}}_{2}=f\left({l}_{Y}\right),{\mathit{R}}_{4}=f\left({l}_{Y},{\mathit{R}}_{2}\right),{\mathit{r}}_{j}=f\left({\mathit{R}}_{1},{\mathit{R}}_{2},{\mathit{R}}_{3},{l}_{S}\right)$ |

2 The velocities responses: ${\mathsf{\omega}}_{2}=f\left({l}_{Y},{\mathit{R}}_{2},{\dot{l}}_{Y}\right)$ ${\mathsf{\omega}}_{4}=f\left({l}_{Y},{\mathit{R}}_{2},{\mathit{R}}_{4},{\mathsf{\omega}}_{2}\right)$ |

${\dot{\mathit{r}}}_{j}=f\left({\mathsf{\omega}}_{1},{\mathsf{\omega}}_{2},{\mathsf{\omega}}_{3},{\dot{l}}_{S},{\mathit{R}}_{1},{\mathit{R}}_{2},{\mathit{R}}_{3},{l}_{S}\right)$ |

3 The acceleration responses: ${\dot{\mathsf{\omega}}}_{2}=f\left({l}_{Y},{\mathit{R}}_{2},{\mathsf{\omega}}_{2},{\dot{l}}_{Y},{\ddot{l}}_{Y}\right)$, ${\dot{\mathsf{\omega}}}_{4}=f\left({l}_{Y},{\mathit{R}}_{2},{\mathit{R}}_{4},{\dot{l}}_{Y},{\mathsf{\omega}}_{2},{\mathsf{\omega}}_{4},{\dot{\mathsf{\omega}}}_{2}\right)$ |

${\ddot{\mathit{r}}}_{j}=f\left({l}_{Y},{l}_{S},{\mathit{R}}_{1},{\mathit{R}}_{2},{\mathit{R}}_{3},{\mathsf{\omega}}_{1},{\mathsf{\omega}}_{2},{\mathsf{\omega}}_{3},{\dot{l}}_{S},{\dot{\mathsf{\omega}}}_{1},{\dot{\mathsf{\omega}}}_{2},{\dot{\mathsf{\omega}}}_{3},{\ddot{l}}_{S}\right)$ |

4 The update of the initial values: ${l}_{Y},{l}_{S},{\mathit{R}}_{1},{\mathit{R}}_{3},{\mathsf{\omega}}_{1},{\mathsf{\omega}}_{3},{\dot{l}}_{Y},{\dot{l}}_{S},{\dot{\mathsf{\omega}}}_{1},{\dot{\mathsf{\omega}}}_{2},{\dot{\mathsf{\omega}}}_{3},{\ddot{l}}_{Y},{\ddot{l}}_{S}$ |

End |

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**MDPI and ACS Style**

Bai, L.; Ma, L.-h.; Ge, X.-s.
The Modeling and Calculation of the Heading Machine Based on Differential Geometry. *Math. Comput. Appl.* **2016**, *21*, 40.
https://doi.org/10.3390/mca21040040

**AMA Style**

Bai L, Ma L-h, Ge X-s.
The Modeling and Calculation of the Heading Machine Based on Differential Geometry. *Mathematical and Computational Applications*. 2016; 21(4):40.
https://doi.org/10.3390/mca21040040

**Chicago/Turabian Style**

Bai, Long, Lu-han Ma, and Xin-sheng Ge.
2016. "The Modeling and Calculation of the Heading Machine Based on Differential Geometry" *Mathematical and Computational Applications* 21, no. 4: 40.
https://doi.org/10.3390/mca21040040