机械原理大作业22题2凸轮Word文档格式.docx
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二.凸轮推杆运动规律
1.运动规律(等加速等减速运动)
推程
2.运动规律(等加速等减速运动)
回程
三.推杆位移、速度、加速度线图及凸轮
线图
采用VB编程,其源程序及图像如下:
1.位移:
PrivateSubCommand1_Click()
Timer1.Enabled=True'
开启计时器
EndSub
PrivateSubTimer1_Timer()
StaticiAsSingle
DimsAsSingle,qAsSingle'
i作为静态变量,控制流程;
s代表位移;
q代表角度
Picture1.CurrentX=0
Picture1.CurrentY=0
i=i+0.1
Ifi<
=45Then
q=i
s=240*(q/90)^2
Picture1.PSetStep(q,-s),vbRed
ElseIfi>
=45Andi<
=90Then
s=120-240*((90-q)^2)/(90^2)
Picture1.PSetStep(q,-s),vbGreen
=90Andi<
=150Then
s=120
Picture1.PSetStep(q,-s),vbBlack
=150Andi<
=190Then
s=120-240*(q-150)^2/6400
Picture1.PSetStep(q,-s),vbBlue
=190Andi<
=230Then
s=240*(230-q)^2/6400
=230Andi<
=360Then
s=0
Else
EndIf
2.速度
PrivateSubCommand2_Click()
Timer2.Enabled=True'
PrivateSubTimer2_Timer()
DimvAsSingle,qAsSingle,wAsSingle'
i为静态变量,控制流程;
q代表角度;
w代表角速度,此处被赋予50
Picture1.CurrentX=0
w=50
v=480*w*q/8100
Picture1.PSetStep(q,-v),vbRed
v=480*w*(90-q)/8100
Picture1.PSetStep(q,-v),vbBlack
v=0
Picture1.PSetStep(q,-v),vbGreen
v=-480*w*(q-150)/6400
Picture1.PSetStep(q,-v),vbBlue
v=-480*w*(230-q)/6400
Else
3.加速度
PrivateSubCommand3_Click()
Timer3.Enabled=True'
PrivateSubTimer3_Timer()
DimaAsSingle,wAsSingle,qAsSingle'
a代表加速度;
w代表角速度
i=i+0.1
a=480*w^2/8100
Picture1.PSetStep(q,-a),vbRed
a=-480*w^2/8100
Picture1.PSetStep(q,-a),vbBlack
a=0
Picture1.PSetStep(q,-a),vbGreen
a=-480*w^2/6400
Picture1.PSetStep(q,-a),vbBlue
a=480*w^2/6400
4.ds/dq---ds
PrivateSubCommand4_Click()
Timer4.Enabled=True'
开启计时器;
建立坐标系
Picture1.Scale(-400,-400)-(400,400)
PrivateSubTimer4_Timer()
DimxAsSingle,sAsSingle,qAsSingle,scaleyAsSingle,tAsSingle'
x代表位移;
s代表纵坐标ds/dq;
Picture1.CurrentY=0
scaley=1
t=3.14/180
q=i*t
x=194.734*q
s=240*(2*q/3.14)^2
Picture1.PSetStep(x,-s),vbRed
x=194.734*(3.14/2-q)
s=120-97.367*(3.14/2-q)^2
x=0
s=120*scaley
x=-246.46*(q-5*3.14/6)
s=120-123.23*(q-5*3.14/6)^2
x=-246.46*(23*3.14/18-q)
s=123.23*(23*3.14/18-q)^2
四.确定凸轮基圆半径和偏距
1.求切点转角
在图中,右侧曲线为升程阶段的类速度-位移图,作直线Dtdt与其相切,且位移轴正方向呈夹角[
1]=300,则切点处的斜率与直线Dtdt的斜率相等,因为kDtdt=tan600,右侧曲线斜率可以表示为:
q;
q=tan600继而求出切点坐标(337.272,292.084)。
同理可求出另一切点(-89.766,16.347)
2.确定直线方程
y-292.084=1.732*(x-337.272)
y-16.347=-0.364*(x+89.766)
3.绘图确定基圆半径和偏距
程序如下:
PrivateSubCommand5_Click()'
画出ds/dq-s图的切线
Timer5.Enabled=True
Picture1.Scale(-400,-400)-(400,400)
Picture1.Line(-50,378.671)-(200,-54.329),vbRed
Picture1.Line(-200,-65.683)-(40,-481.363),vbBlue
Picture1.Line(-200,-209.127)-(200,-63.527),vbBlack
Picture1.Line(-200,-56.472)-(200,89.128),vbGreen
PrivateSubTimer5_Timer()'
scaley是图形比例系数;
t为角度,弧度转化系数
DimxAsSingle,sAsSingle,qAsSingle,scaleyAsSingle,tAsSingle
在轴心公共许用区内取轴心位置,能够满足压力角要求,现取直线红线与直线绿线的交点为轴心位置,通过解方程组:
x=131.557y=-64.214r0=146.392偏距e=131.557s0=64.013
五.绘制凸轮理论廓线和实际廓线
1.绘制凸轮理论轮廓
PrivateSubCommand6_Click()'
Picture1.Scale(-500,-500)-(500,500)
Timer6.Enabled=True
PrivateSubTimer6_Timer()'
e为偏距;
s为基圆半径平方减去偏距平方再开方所得;
q为角度;
x为横坐标;
y为纵坐标
DimeAsSingle,qAsSingle,sAsSingle,xAsSingle,yAsSingle
e=131.557
s=64.213
Picture1.CurrentX=12.764
Picture1.CurrentY=-95.125
i=i+0.001
=3.14/4Then
x=(s+240*(2*q/3.14)^2)*Cos(q)-e*Sin(q)
y=(s+240*(2*q/3.14)^2)*Sin(q)+e*Cos(q)
Picture1.PSetStep(x,-y),vbRed
=3.14/4Andi<
=3.14/2Then
x=(s+120-97.367*(3.14/2-q)^2)*Cos(q)-e*Sin(q)
y=(s+120-97.367*(3.14/2-q)^2)*Sin(q)+e*Cos(q)
=3.14/2Andi<
=5*3.14/6Then
x=(s+120)*Cos(q)-e*Sin(q)
y=(s+120)*Sin(q)+e*Cos(q)
=5*3.14/6Andi<
=19*3.14/18Then
x=(s+120-123.23*(q-5/6*3.14)^2)*Cos(q)-e*Sin(q)
y=(s+120-123.23*(q-5/6*3.14)^2)*Sin(q)+e*Cos(q)
=19/18*3.14Andi<
=23/18*3.14Then
x=(s+123.23*(23/18*3.14-q)^2)*Cos(q)-e*Sin(q)
y=(s+123.23*(23/18*3.14-q)^2)*Sin(q)+e*Cos(q)
=23/18*3.14Andi<
=2*3.14Then
x=s*Cos(q)-e*Sin(q)
y=s*Sin(q)+e*Cos(q)
EndSub
2.确定滚子半径并绘制曲率半径图
PrivateSubCommand7_Click()'
开启计时器,建立坐标系
Timer7.Enabled=True
Picture1.Scale(-10,-500)-(10,10)
PrivateSubTimer7_Timer()'
x代表dx/dq;
y代表dy/dq;
p代表曲率;
s为基圆半径平方减去偏距平方再开方所得
DimsAsSingle,qAsSingle,xAsSingle,yAsSingle,pAsSingle
s=64.213
e=131.557
Picture1.CurrentY=-Sqr(e^2+s^2)
i=i+0.01
x=(480*4*q/3.14^2)*Cos(q)+(s+240*(2*q/3.14)^2)*(-Sin(q))-e*Cos(q)
y=(480*4*q/3.14^2)*Sin(q)-e*Sin(q)+(s+240*(2*q/3.14)^2)*Cos(q)
p=Sqr(x^2+y^2)
Picture1.PSetStep(q,-p),vbRed
x=(2*97.367*(3.14/2-q))*Cos(q)+(s+120-97.367*(3.14/2-q)^2)*(-Sin(q))-e*Cos(q)
y=(2*97.367*(3.14/2-q))*Sin(q)+(s+120-97.367*(3.14/2-q)^2)*Cos(q)-e*Sin(q)
=5/6*3.14Then
x=(s+120)*(-Sin(q))-e*Cos(q)
y=(s+120)*Cos(q)-e*Sin(q)
=5/6*3.14Andi<
=19/18*3.14Then
x=(-2*123.23*(q-5/6*3.14))*Cos(q)+(s+120-123.23*(q-5/6*3.14)^2)*(-Sin(q))-e*Cos(q)
y=(-2*123.23*(q-5/6*3.14))*Sin(q)+(s+120-123.23*(q-5/6*3.14)^2)*Cos(q)-e*Sin(q)
x=(-2*123.23*(23/18*3.14-q))*Cos(q)+(s+123.23*(23/18*3.14-q)^2)*(-Sin(q))-e*Cos(q)
y=(-2*123.23*(23/18*3.14-q))*Sin(q)+(s+123.23*(23/18*3.14-q)^2)*Cos(q)-e*Sin(q)
x=s*(-Sin(q))-e*Cos(q)
y=s*Cos(q)-e*Sin(q)
其中曲率最小值为126.026mm可以得到滚子半径为63.013mm
3.绘制实际轮廓线
PrivateSubCommand8_Click()
Picture1.Scale(-500,-500)-(500,500)'
建立