MATLAB是一个编程和数值计算平台，被数百万工程师和科学家用来分析数据、开发算法和创建模型。

MATLAB主要用于数值运算，但利用为数众多的附加工具箱，它也适合不同领域的应用，例如控制系统设计与分析、影像处理、深度学习、信号处理与通讯、金融建模和分析等。另外还有配套软件包提供可视化开发环境，常用于系统模拟、动态嵌入式系统开发等方面。

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

The dyad (binary group) is a fundamental kinematic chain with two links $(n=2)$ and three one degree of freedom joints $\left(c_{5}=3\right.$ ). Figure $1.7$ depicts different types of dyads: rotation rotation rotation (dyad RRR) or dyad of type one, Fig. 1.7a; rotation rotation translation (dyad RRT) or dyad of type two, Fig.1.7b; rotation translation rotation (dyad RTR) or dyad of type three, Fig. $1.7 \mathrm{c}$; translation rotation translation (dyad TRT) or dyad of type four, Fig. $1.7 \mathrm{~d}$; rotation translation translation (dyad RTT) or dyad of type five, Fig. 1.7e. The advantage of the group classification of a mechanical system is in its simplicity. The solution of the whole mechanical system can be obtained by adding partial solutions of different fundamental kinematic chains [55-57].

The number of DOF for the mechanism in Fig. $1.8$ is $M=1$. If $M=1$, there is one driver link (one actuator). The rotational link 1 can be selected as the driver link. If the driver link is separated from the mechanism the remaining moving kinematic chain (links $2,3,4,5$ ) has the number of DOF equal to zero. The dyad is the simplest system group with two links and three joints. On the contour diagram, the links 2 and 3 form a dyad and the links 4 and 5 represent another dyad. The mechanism has been decomposed into a driver link (link 1) and two dyads (links 2 and 3, and links 4 and

5). The dyad with the links 2 and 3 is a RTR dyad and the dyad with the links 4 and 5 is also a RTR dyad. The whole mechanism can be symbolized as a R-RTR-RTR mechanism.

For planar mechanisms the two degrees of freedom joints can be substituted and mechanisms with one degree of freedom joints are obtained. The transformed mechanism has to be equivalent with the initial mechanism from a kinematical point of view. The number of degrees of freedom of the transformed mechanism has to be equal to the number of degrees of freedom of the initial mechanism. The relative motion of the links of the transformed mechanism has to be the same as the relative motion of the links of the initial mechanism.

A planar straight link with the end nodes at $A$ and $B$ is considered. The coordinates of the joint $A$ are $\left(x_{A}, y_{A}\right)$ and the coordinates of the joint $B$ are $\left(x_{B}, y_{B}\right)$. The length $A B=l_{A B}$ is constant and the following relation can be written
$$\left(x_{B}-x_{A}\right)^{2}+\left(y_{B}-y_{A}\right)^{2}=A B^{2},$$

The angle of the link $A B$ with the horizontal axis $O x$ is $\phi$ and the slope $m$ of $A B$ is
$$m=\tan \phi=\frac{y_{B}-y_{A}}{x_{B}-x_{A}} .$$
The equation of the straight link is $y=m x+n$, where $x$ and $y$ are the coordinates of any point on this link and $n$ is the intercept of $A B$ with the vertical axis $O y$.
The R-RRT (slider-crank) mechanism shown in Fig. 1.10a has the dimensions: $A B=0.5 \mathrm{~m}$ and $B C=1 \mathrm{~m}$. The driver link 1 makes an angle $\phi=\phi_{1}=60^{\circ}$ with the horizontal axis. The positions of the joints and the angles of the links with the horizontal axis will be calculated. A Cartesian reference frame $x y$ is selected. The joint $A$ is in the origin of the reference frame, $A \equiv O$,
$$x_{A}=0, y_{A}=0 .$$
The coordinates of the joint $B$ are
\begin{aligned} &x_{B}=A B \cos \phi=(0.5) \cos 60^{\circ}=0.250 \mathrm{~m} \ &y_{B}=A B \sin \phi=(0.5) \sin 60^{\circ}=0.433 \mathrm{~m} \end{aligned}
The coordinates of the joint $C$ are $x_{C}$ and $y_{C}$. The joint $C$ is located on the horizontal axis $y_{C}=0$. The length of the segment $B C$ is constant
$$\left(x_{B}-x_{C}\right)^{2}+\left(y_{B}-y_{C}\right)^{2}=B C^{2},$$
or
$$\left(0.250-x_{C}\right)^{2}+(0.433-0)^{2}=1^{2} .$$
The two solutions for $x_{C}$ are:
$$x_{C_{1}}=1.151 \mathrm{~m} \text { and } x_{C_{2}}=-0.651 \mathrm{~m}$$
To determine the position of the joint $C$, for the given angle of the link 1 , an additional condition is needed. For the first quadrant, $0 \leq \phi \leq 90^{\circ}$, one condition can be $x_{C}>$ $x_{B}$. The $x$-coordinate of the joint $C$ is $x_{C}=x_{C_{1}}=1.151 \mathrm{~m}$. The angle of the link 2 (link $B C$ ) with the horizontal is
$$\phi_{2}=\arctan \frac{y_{B}-y_{C}}{x_{B}-x_{C}} .$$
The numerical solutions for $\phi_{2}$ is $154.341$ degrees. The MATLAB code is:
clear; clc; close all;
뭉 Input data
clear; clc; close
ㅇㅎㅇ Input data
$\mathrm{AB}=0.5 ;$ 황 $(\mathrm{m}}$
$A B=0.5 ;$ 핳 $(\mathrm{m})$

## 数学代写|matlab仿真代写simulation代做|Velocity and Acceleration Analysis for Rigid Body

The motion of a rigid body $(R B)$, with respect to a fixed reference frame, is defined by the position, velocity and acceleration of all points of the rigid body. A fixed orthogonal Cartesian reference frame $x_{0} y_{0} z_{0}$ with the constant unit vectors $\mathbf{1}{0}, \mathbf{J}{0}$, and $\mathbf{k}_{0}$, is shown in Fig. 1.12. The body fixed (mobile or rotating) orthogonal Cartesian

reference frame $x y z$ with unit vectors $\mathbf{1}, \mathbf{J}$ and $\mathbf{k}$ is attached to the moving rigid body. The unit vectors $\mathbf{1}{0}, \mathbf{J}{0}$, and $\mathbf{k}{0}$ of the primary reference frame are constant with respect to time and the unit vectors $\mathbf{1}, \mathbf{J}$, and $\mathbf{k}$ are functions of time, $t$. The unit vectors $\mathbf{1}, \mathbf{J}$, and $\mathbf{k}$ of the body-fixed frame of reference rotate with the body-fixed reference frame. The origin $O$ is arbitrary. The position vector of any point $M, \forall M \in(R B)$, with respect to the fixed reference frame $x{0} y_{0} z_{0}$ is $\mathbf{r}{M}=\mathbf{r}{O_{0} M}$ and with respect to the rotating reference frame $O x y z$ is denoted by $\mathbf{r}=\mathbf{r}{O M}$. The position vector of the origin $O$ of the rotating reference frame with respect to the fixed point $O{0}$ is $\mathbf{r}{o}=\mathbf{r}{{ }{0}} o$. The position vector of $M$ is $$\mathbf{r}{M}=\mathbf{r}{O}+\mathbf{r}=\mathbf{r}{O}+x \mathbf{1}+y \mathbf{J}+z \mathbf{k}$$
where $x, y$, and $z$ represent the projections of the vector $\mathbf{r}=\mathbf{r}{O M}$ on the rotating reference frame $$\mathbf{r}=x \mathbf{1}+y \mathbf{j}+z \mathbf{k}$$ The distance between two points of the rigid body $O$ and $M$ is constant, $O \in(R B)$, and $M \in(R B)$. The components $x, y$ and $z$ of the vector $\mathbf{r}$ with respect to the rotating reference frame are constant. The unit vectors $\mathbf{1}, \mathbf{J}$, and $\mathbf{k}$ are time-dependent vector functions. For the unit vectors of an orthogonal Cartesian reference frame, 1, Jand $\mathbf{k}$, there are the following relations $$\mathbf{1} \cdot \mathbf{1}=1, \quad \mathbf{J} \cdot \mathbf{J}=1, \quad \mathbf{k} \cdot \mathbf{k}=1, \quad \mathbf{1} \cdot \mathbf{J}=0, \quad \mathbf{j} \cdot \mathbf{k}=0, \quad \mathbf{k} \cdot \mathbf{1}=0 .$$ The velocity of an arbitrary point $M$ of the rigid body with respect to the fixed reference frame $x{0} y_{0} z_{0}$, is the derivative with respect to time of the position vector $\mathbf{r}{M}$ \begin{aligned} \mathbf{v}{M} &=\frac{d \mathbf{r}{M}}{d t}=\frac{d \mathbf{r}{O_{o} M}}{d t}=\frac{d \mathbf{r}{O}}{d t}+\frac{d \mathbf{r}}{d t} \ &=\mathbf{v}{O}+x \frac{d \mathbf{1}}{d t}+y \frac{d \mathbf{j}}{d t}+z \frac{d \mathbf{k}}{d t}+\frac{d x}{d t} \mathbf{1}+\frac{d y}{d t} \mathbf{j}+\frac{d z}{d t} \mathbf{k} \end{aligned}
where $\mathbf{v}{O}=\dot{\mathbf{r}}{O}$ is the velocity of the origin of the rotating reference frame $O x y z$ with respect to the fixed reference frame. Because all the points in the rigid body maintain their relative position, their velocity relative to the rotating reference frame $x y z$ is zero, i.e., $\dot{x}=\dot{y}=\dot{z}=0$.
The velocity of point $M$ is
$$\mathbf{v}{M}=\mathbf{v}{O}+x \frac{d \mathbf{1}}{d t}+y \frac{d \mathbf{j}}{d t}+z \frac{d \mathbf{k}}{d t}=\mathbf{v}_{O}+x \mathbf{i}+y \mathbf{j}+z \dot{\mathbf{k}}$$
The derivative of the Eq. (1.11) with respect to time gives
$$\frac{d \mathbf{1}}{d t} \cdot \mathbf{1}=0, \quad \frac{d \mathbf{J}}{d t} \cdot \mathbf{J}=0, \quad \frac{d \mathbf{k}}{d t} \cdot \mathbf{k}=0$$

## matlab代写

5）。具有链接 2 和 3 的对子是 RTR 对子，具有链接 4 和 5 的对子也是 RTR 对子。整个机制可以用R-RTR-RTR机制来表示。

(X乙−X一种)2+(是乙−是一种)2=一种乙2,

X一种=0,是一种=0.

X乙=一种乙因⁡φ=(0.5)因⁡60∘=0.250 米 是乙=一种乙罪⁡φ=(0.5)罪⁡60∘=0.433 米

(X乙−XC)2+(是乙−是C)2=乙C2,

(0.250−XC)2+(0.433−0)2=12.

XC1=1.151 米 和 XC2=−0.651 米

φ2=反正切⁡是乙−是CX乙−XC.

clear; cl；关闭所有;
뭉 输入数据

ㅇㅎㅇ 输入数据

## 数学代写|matlab仿真代写simulation代做|Velocity and Acceleration Analysis for Rigid Body

d1d吨⋅1=0,dĴd吨⋅Ĵ=0,dķd吨⋅ķ=0

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。