用matlab分析凸轮机构

Matlab作业

一、用matlab求从动件的运动规律，设计凸轮外形。

设推杆的行程h为10，凸轮角速度为5rad/s，推程角为2π/3，回程角为2π/3，近休止为2π/3，基圆半径为20。

1、位移：

h=10;

w=5;

th1=2*pi/3;

th2=4*pi/3;

th3=2*pi;

j1=linspace(0,th1,120);

s1=h*(1-cos(pi.*j1/th1))/2;

j2=linspace(th1,th2,120);

s2=h*(1+cos(pi.*(j2-th1)/(th2-th1)))/2;

j3=linspace(th2,th3,120);

s3=0+0.*j3;

plot(j1,s1,j2,s2,j3,s3)

grid on

title('位移线图')

h=10;

w=5;

th1=2*pi/3;

th2=4*pi/3;

th3=2*pi;

j1=linspace(0,th1,120);

v1=pi*w*h*sin(pi.*j1/th1)/(2*th1);

j2=linspace(th1,th2,120);

v2=-pi*h*w*sin(pi*(j2-th1)/(th2-th1))/(2*(th2-th1)); j3=linspace(th2,th3,120);

v3=0+0.*j3;

plot(j1,v1,j2,v2,j3,v3)

grid on

title('速度线图')

3、加速度：

h=10;

w=5;

th1=2*pi/3;

th2=4*pi/3;

j1=linspace(0,th1,120);

a1=pi^2*h*w^2*cos(pi.*j1/th1)/(2*th1^2);

j2=linspace(th1,th2,120);

a2=-pi^2*h*w^2*cos(pi.*(j2-th1)/th2)/(2*(th2-th1)^2); j3=linspace(th2,th3,120);

a3=0+0.*j3;

plot(j1,a1,j2,a2,j3,a3)

grid on

title('加速度线图')

h=10;

w=5;

th1=2*pi/3;

th2=4*pi/3;

th3=2*pi;

j1=linspace(0,th1,120);

s1=h*(1-cos(pi.*j1/th1))/2;

j2=linspace(th1,th2,120);

s2=h*(1+cos(pi.*(j2-th1)/(th2-th1)))/2;

j3=linspace(th2,th3,120);

r0=20;

x1=(r0+s1).*cos(j1);

y1=(r0+s1).*sin(j1);

x2=(r0+s2).*cos(j2);

y2=(r0+s2).*sin(j2);

x3=(r0+s3).*cos(j3);

y3=(r0+s3).*sin(j3); plot(x1,y1,x2,y2,x3,y3) grid on

title('凸轮外形') xlabel('x轴')

ylabel('y轴')