### 数学代写|matlab仿真代写simulation代做|Contour Analysis of a Mechanism with One Dyad

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## 数学代写|matlab仿真代写simulation代做|Closed Contour Equations

This study aims at providing an algebraic method to compute the velocities of a closed kinematic chain $[3,18,71]$. The method of independent contour (loop) equations is very efficient and can be applied to planar and spatial mechanical systems.

Consider two rigid bodies $(j)$ and $(k)$ connected by a joint or kinematic pair at $A$. The following relation exists between the velocity $\mathbf{v}{A}$ of the point $A{j}$ and the velocity $\mathbf{v}{A{k}}$ of the point $A_{k}$
$$\mathbf{v}{A{j}}=\mathbf{v}{A{k}}+\mathbf{v}{A{j k},}^{r}$$
where $\mathbf{v}{A{j k}}^{r}=\mathbf{v}{A{j} A_{k}}^{r}=\mathbf{v}{A{j} A_{k}}$ is the velocity of $A_{j}$ as seen by an observer at $A_{k}$ attached to body $(k)$ or the relative velocity of $A_{j}$ with respect to $A_{k}$, allowed at the joint $A$. The accelerations of $A_{j}$ and $A_{k}$ are expressed as
$$\mathbf{a}{A{j}}=\mathbf{a}{A{k}}+\mathbf{a}{A{j k}}^{r}+\mathbf{a}{A{j k}}^{c},$$
where $\mathbf{a}{A{j k}}^{r}=\mathbf{a}{A{j} A_{k}}^{r}=\mathbf{a}{A{j} A_{k}}$ is the relative acceleration of $A_{j}$ with respect to $A_{k}$ and $\mathbf{a}{A{j k}}^{c}=\mathbf{a}{A{j} A_{k}}^{c_{k}}$ is the Coriolis acceleration given by
$$\mathbf{a}{A{j k}}^{c}=2 \boldsymbol{\omega}{k} \times \mathbf{v}{A j k}^{r},$$
where $\omega_{k}$ is the angular velocity of the body $(k)$. Equations (3.1) and (3.2) are useful even for coincident points belonging to two rigid bodies that may not be directly

connected. Figure $3.1$ shows a monoloop closed kinematic chain with $n$ rigid links. The joint $A_{i}$ where $i=1,2, \ldots, n$ is the connection between the links $(i)$ and $(i-1)$. At the joint $A_{i}$ there are two instantaneously coincident points. The point $A_{i, i}$ belongs to link $(i), A_{i, i} \in(i)$ and the point $A_{i, i-1}$ belongs to body $(i-1), A_{i, i-1} \in(i-1)$. The absolute angular velocity of the rigid body ( $i$ ) is
$$\omega_{i}=\omega_{i-1}+\omega_{i, i-1}$$
where $\boldsymbol{\omega}{i, i-1}$ is the relative angular velocity of the rigid body (i) with respect to the rigid body $(i-1)$. For the $n$ link closed kinematic chain the equations for the angular velocities are $$\boldsymbol{\omega}{1}=\boldsymbol{\omega}{n}+\boldsymbol{\omega}{1, n}: \boldsymbol{\omega}{2}=\boldsymbol{\omega}{1}+\boldsymbol{\omega}{2,1}: \ldots \boldsymbol{\omega}{i}=\boldsymbol{\omega}{i-1}+\boldsymbol{\omega}{i, i-1}: \ldots \boldsymbol{\omega}{n}=\boldsymbol{\omega}{n-1}+\boldsymbol{\omega}{n, n-1} \text { (3.5) }$$ Summing the expressions in Eq. (3.5) the following relation is obtained $$\omega{1, n}+\omega_{2,1}+\ldots+\omega_{n, n-1}=\mathbf{0} \text { or } \sum_{(i)} \omega_{i, i-1}=\mathbf{0}$$
The first vectorial equation for the angular velocities of a simple closed kinematic chain is given by Eq. (3.6).

## 数学代写|matlab仿真代写simulation代做|Closed Contour Equations for R-RTR Mechanism

The planar R-RTR mechanism considered in Chap. 2 is shown in Fig. $3.3$ with the attached contour diagram. The input data and the position data are:
$\mathrm{AB}=0.10 ;$ \&ै ${\mathrm{m})$
$\mathrm{AC}=0.05 ;$ \& $(\mathrm{m})$
$C D=0.15 ;$ \& $;(\mathrm{m})$
$\mathrm{phi}=\mathrm{pi} / 6_{i}$ \& \& {rad $\rangle$
$x A=0 ; \quad y A=0 ;$
$r A-\left[\begin{array}{lll}x A & Y A & 0\end{array}\right] ;$
$x C=A C ; Y C=0 ;$
$r C_{-}=[x C y C 0] i$

of $\mathrm{rC}{-}=0.050,0.000,0$ 와 $r \mathrm{~B}{-}=0.087,0.050,0$
of phi2 $=$ phi $3=53.794$ (degrees)
If $\mathrm{rD}{-}=[0.139,0.121,0] \quad(\mathrm{m})$ The mechanism has one contour. The contour is made of $0,1,2,3$, and 0 . A clockwise path is chosen for the contour. The contour has: a rotational joint $R$ between the links 0 and 1 (joint $A{\mathrm{R}}$ ); a rotational joint $\mathrm{R}$ between the links 1 and 2 (joint $B_{\mathrm{R}}$ ); a translational joint $\mathrm{T}$ between the links 2 and 3 (joint $B_{\mathrm{T}}$ ); a rotational joint $\mathrm{R}$ between the links 3 and 0 (joint $C_{\mathrm{R}}$ ). The angular velocity $\omega_{10}$ of the driver link is $\omega_{10}=\omega_{1}=$ $\omega=n \pi / 30=-50 \pi / 30 \mathrm{rad} / \mathrm{s}=-5 \pi / 3 \mathrm{rad} / \mathrm{s}$. The velocity equations for the contour are:
\begin{aligned} &\boldsymbol{\omega}{10}+\boldsymbol{\omega}{21}+\boldsymbol{\omega}{03}=\mathbf{0}, \ &\mathbf{r}{B} \times \boldsymbol{\omega}{21}+\mathbf{r}{C} \times \boldsymbol{\omega}{03}+\mathbf{v}{B_{3} B_{2}}^{\text {rel }}=\mathbf{0}, \end{aligned}
where $\mathbf{r}{B}=x{B} \mathbf{1}+y_{B} \mathbf{J}, \mathbf{r}{C}=x{C} \mathbf{1}+y_{C} \mathbf{J}$, and
\begin{aligned} &\boldsymbol{\omega}{10}=\omega{10} \mathbf{k}, \boldsymbol{\omega}{21}=\omega{21} \mathbf{k}, \boldsymbol{\omega}{03}=\omega{03} \mathbf{k} \ &\mathbf{v}{B{3} B_{2}}^{\text {rel }}=\mathbf{v}{B{22}}=v_{B_{12}} \cos \phi_{2} \mathbf{1}+v_{B_{32}} \sin \phi_{2} \mathbf{J} \end{aligned}
The relative angular velocities $\omega_{21}, \omega_{03}$, and relative linear velocity $v_{B_{22}}$ are the unknowns. Equation (3.18) can be written as
$$\left[\begin{array}{ccc} \mathbf{1} & \mathbf{J} & \mathbf{k} \ x_{B} & y_{B} & 0 \ 0 & 0 & \omega_{21} \end{array}\right]+\left[\begin{array}{ccc} \mathbf{1} & \mathbf{J} & \mathbf{k} \ x_{C} & y_{C} & 0 \ 0 & 0 & \omega_{03} \end{array}\right]+v_{B_{32}} \cos \phi_{2} \mathbf{1}+v_{B_{32}} \sin \phi_{2} \mathbf{J}=\mathbf{0} .$$
Equation (3.19) represents a system of 3 algebraic equations with 3 unknowns and can be written in MATLAB as:
\begin{aligned} &n=-50 \ldots ; \text { \& }(\mathrm{rpm}) \ &\text { omega1 }=\left[\begin{array}{ll} 0 & \left.0 \mathrm{pi}{ }^{} \mathrm{n} / 30\right] ; \text { 영 } \end{array} \quad(\mathrm{rad} / \mathrm{s})\right. \end{aligned} of symbolic unknowns syms omega2 $1 \mathrm{z}$ omega03z vB32 omega10_ = omega1_i omega21v_ $=\left[\begin{array}{ll}0 & 0 \text { omega21z}\end{array}\right]$; omega03v $=\left[\begin{array}{ll}0 & 0 \text { omega03z }\end{array}\right]$; $v 32 v_{-}=v B 32^{}[\cos (p h i 2) \sin (p h i 2) \quad 0] ;$
eqlomega_ $=$ omegalo $1_{-}+$omega $21 v_{-}+$omega03v_;

## 数学代写|matlab仿真代写simulation代做|Force Analysis for R-RTR Mechanism

The inertia forces and moments and the gravity forces on links 2 and 3 are shown in Fig. 3.4a. The input data for joint force analysis are:
동 $r C_{-}=[0.050,0.000,0]$ (m)
뭉 $\mathrm{rB}=[0.087,0.050,0]$ (m)
뭉 phiz $=$ phis $=53.794$ (degrees)
뭉 $\mathrm{rD}{-}=[0.139,0.121,0] \mathrm{(m)}$ 당 $\mathrm{rCl}{-}=[0.043,0.025,0]$ (m)
당 $\mathrm{rC}{-}=[0.087,0.050,0] \mathrm{(m)}$ 당 $\mathrm{rC3}{-}=[0.094,0.061,0]$ (m)
당 $\mathrm{aCl}{-}=[-1.187,-0.685,0]\left(\mathrm{m} / \mathrm{s}^{A} 2\right)$ 당 $a C^{2}=[-2.374,-1.371,0]\left(\mathrm{m} / \mathrm{s}^{A} 2\right)$ 닿 $\mathrm{aC}{3}=[-0.538,-5.162,0]\left(\mathrm{m} / \mathrm{s}^{A} 2\right)$
웅 alpha1 $=[0,0,0.000]\left(\mathrm{rad} / \mathrm{s}^{4} 2\right)$
닿 alpha2 $=[0,0,-34.865]\left(\mathrm{rad} / \mathrm{s}^{4} 2\right)$
당 alpha3 ${ }^{3}=[0,0,-34.865]\left(\mathrm{rad} / \mathrm{s}^{\wedge} 2\right)$

당 $\mathrm{Gl}{-}=-\mathrm{m} 1 \mathrm{~g}{-}=[0,-0.078,0] \quad(\mathrm{N})$
닿 Fin1 $=-\mathrm{m} 1 \mathrm{aC1}=[0.009,0.005,0]$ (N)
망 Min1 $=-I C 1$ alpha1 $=[0,0,0](\mathrm{N} \mathrm{m})$

왛 $\mathrm{G}{-}=-\mathrm{m} 2 \mathrm{~g}{-}=[0,-0.063,0] \quad(\mathrm{N})$
of $\operatorname{Fin} 2_{-}=-\mathrm{m} 2 \mathrm{aC2}=[0.015,0.009,0] \quad$ (N)
of Min2 $=-I C 2$ alpha2 $2=[0,0,3.719 \mathrm{e}-05](\mathrm{N} \mathrm{m})$

왕 $\mathrm{G}{-}=-\mathrm{m} 3 \mathrm{~g}{-}=0,-0.118,0$
왛 $\mathrm{Fin}{3}=-\mathrm{m} 3 \mathrm{aC3}=[0.006,0.062,0]$ (N) 망 $\operatorname{Min} 3$ $=-\operatorname{IC3}$ alpha3 $=[0,0,7.880 \mathrm{e}-04](\mathrm{N} \mathrm{m})$
of $\mathrm{Me}=[0,0,100.000]$ (N m)

To calculate the joint reaction $\mathbf{F}{03}$ with the application point at the rotational joint $C$ the diagram shown in Fig. 3.4(b) is used. For the link 3, the joint reaction force $\mathbf{F}{23}$ is eliminated if the sum of the forces on link 3 are dot multiplied with the direction of the link $B C$ :
syms F03x F03y
F03_ $=[F 03 x, F 03 y, 0]$; 뫄 unknown joint force of 0 on 3 at $C$
돵 for joint B translation
뭉 $\left(\mathrm{FO} 3_{-}+\mathrm{G} 3_{-}+\mathrm{Fin} 3_{-}+\mathrm{F} 23_{-}\right) \cdot\left(\mathrm{rB}-\mathrm{rC}{-}\right)=0$ 망 $\mathrm{F} 23$ – perpendicular to $\mathrm{BC}$ : $\mathrm{F3} 2{-} \cdot \mathrm{BC}{-}=0$ => eqF3 = $\left(\mathrm{FO} 3{-}+\mathrm{G} 3_{-}+\mathrm{Fin} 3_{-}\right) *\left(\mathrm{rB}-{ }{-} \mathrm{C}{-}\right) \cdot \prime ;$ o $(\mathrm{F} 1)$

## 数学代写|matlab仿真代写simulation代做|Closed Contour Equations

ω一世=ω一世−1+ω一世,一世−1

## 数学代写|matlab仿真代写simulation代做|Closed Contour Equations for R-RTR Mechanism

X一种=0;是一种=0;
r一种−[X一种是一种0];
XC=一种C;是C=0;
rC−=[XC是C0]一世

$\mathrm{rC}{-}= 0.050,0.000,0와哇r \ mathrm ~~ B {-} = 0.087,0.050,0这FpH一世2=pH一世3=53.794(d和Gr和和s)一世F\mathrm{rD}{-}=[0.139,0.121,0] \quad(\mathrm{m})吨H和米和CH一种n一世s米H一种s这n和C这n吨这在r.吨H和C这n吨这在r一世s米一种d和这F0,1,2,3,一种nd0.一种Cl这Cķ在一世s和p一种吨H一世sCH这s和nF这r吨H和C这n吨这在r.吨H和C这n吨这在rH一种s:一种r这吨一种吨一世这n一种lj这一世n吨Rb和吨在和和n吨H和l一世nķs0一种nd1(j这一世n吨A {\ mathrm {R});一种r这吨一种吨一世这n一种lj这一世n吨\数学{Rb和吨在和和n吨H和l一世nķs1一种nd2(j这一世n吨B_{\数学{R});一种吨r一种nsl一种吨一世这n一种lj这一世n吨\mathrm{T}b和吨在和和n吨H和l一世nķs2一种nd3(j这一世n吨B_{\mathrm{T}});一种r这吨一种吨一世这n一种lj这一世n吨\数学{Rb和吨在和和n吨H和l一世nķs3一种nd0(j这一世n吨C_{\数学{R}).吨H和一种nG在l一种r在和l这C一世吨是\omega_{10}这F吨H和dr一世在和rl一世nķ一世s\omega_{10}=\omega_{1}=\ omega = n \ pi / 30 = -50 \ pi / 30 \ mathrm {rad} / \ mathrm {s} = – 5 \ pi / 3 \ mathrm {rad} / \ mathrm {s.吨H和在和l这C一世吨是和q在一种吨一世这nsF这r吨H和C这n吨这在r一种r和:ω10+ω21+ω03=0, r乙×ω21+rC×ω03+在乙3乙2相对 =0,在H和r和\mathbf{r}{B}=x{B} \mathbf{1}+y_{B} \mathbf{J}, \mathbf{r}{C}=x{C} \mathbf{1}+y_{ C} \mathbf{J},一种ndω10=ω10ķ,ω21=ω21ķ,ω03=ω03ķ 在乙3乙2相对 =在乙22=在乙12因⁡φ21+在乙32罪⁡φ2Ĵ吨H和r和l一种吨一世在和一种nG在l一种r在和l这C一世吨一世和s\omega_{21}，\omega_{03},一种ndr和l一种吨一世在和l一世n和一种r在和l这C一世吨是v_{B_{22}}一种r和吨H和在nķn这在ns.和q在一种吨一世这n(3.18)C一种nb和在r一世吨吨和n一种s[1Ĵķ X乙是乙0 00ω21]+[1Ĵķ XC是C0 00ω03]+在乙32因⁡φ21+在乙32罪⁡φ2Ĵ=0.和q在一种吨一世这n(3.19)r和pr和s和n吨s一种s是s吨和米这F3一种lG和br一种一世C和q在一种吨一世这ns在一世吨H3在nķn这在ns一种ndC一种nb和在r一世吨吨和n一世n米一种吨大号一种乙一种s:영n=−50…; \& (rp米) 欧米茄1 =[00p一世n/30]; 零 (r一种d/s)这Fs是米b这l一世C在nķn这在nss是米s这米和G一种21 \mathrm{z}omega03z vB32 omega10_ = omega1_i omega21v_omega03z vB32 omega10_ = omega1_i omega21v_=\左[00 欧米茄21z\对];这米和G一种03在=\左[00 omega03z \对];v 32 v_{-}=v B 32^{}[\cos (phi 2) \sin (phi 2) \quad 0] ;eqlomega_eqlomega_=这米和G一种l这1_{-}+这米和G一种21 v _ {-} +$ omega03v_;

## 数学代写|matlab仿真代写simulation代做|Force Analysis for R-RTR Mechanism

동rC−=[0.050,0.000,0](m)

뭉 phiz=菲斯=53.794（度）
뭉rD−=[0.139,0.121,0](米)聚会rCl−=[0.043,0.025,0](m)

_Gl−=−米1 G−=[0,−0.078,0](ñ)

왕 $\ mathrm {G} {-} = – \ mathrm {m} 3 \ mathrm ~ g {-} = 0, -0.118,0왛哇\ mathrm {Fin} {3} = – \ mathrm {m} 3 \ mathrm {aC3} = [0.006,0.062,0]망(ñ)网\operatorname{Min} 3=-\运营商名称{IC3}一种lpH一种3= [0,0,7.880 \ mathrm {e} -04] (\ mathrm {N} \ mathrm {m})这F\mathrm{我}=[0,0,100.000]$ (N m)

syms F03x F03y
F03_=[F03X,F03是,0]; 뫄 0 对 3 的未知联合力C
돵 联合 B 翻译
뭉(F这3−+G3−+F一世n3−+F23−)⋅(r乙−rC−)=0网F23– 垂直于乙C : F32−⋅乙C−=0=> eqF3 =(F这3−+G3−+F一世n3−)∗(r乙−−C−)⋅′;这(F1)

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