%MATLAB程序：已知三个位置设计平面四杆机构求解程序（位移矩阵法）clear;clc; %凡是变量名前带v的为数值变量，不带的是符号变量vxp1=0; vyp1=0; vsita1=0*pi/180;vxp2=-2; vyp2=6; vsita2=40*pi/180;vxp3=-10; vyp3=8; vsita3=90*pi/180; %精确位置P1，P2，P3及各角度vsita12=vsita2-vsita1;vsita13=vsita3-vsita1;vxa=-10; vya=-2;vxd=-5; vyd=-2; %选定A,D点%所有数值均在此确定，更改此处即可解出不同数值的四杆机构位移矩阵方程syms xp1 yp1 xp2 yp2 xp3 yp3 sita12 sita13;syms xa ya xb1 yb1 xb2 yb2 xb3 yb3;f1=(xb2-xa)2+(yb2-ya)2=(xb1-xa)2+(yb1-ya)2;f2=(xb3-xa)2+(yb3-ya)2=(xb1-xa)2+(yb1-ya)2; %前两个机构方程f3=xb2=cos(sita12)*xb1-sin(sita12)*yb1+xp2-xp1*cos(sita12)+yp1*sin(sita12);f4=yb2=sin(sita12)*xb1+cos(sita12)*yb1+yp2-xp1*sin(sita12)-yp1*cos(sita12); %由第一个位移矩阵方程得出f5=xb3=cos(sita13)*xb1-sin(sita13)*yb1+xp3-xp1*cos(sita13)+yp1*sin(sita13);f6=yb3=sin(sita13)*xb1+cos(sita13)*yb1+yp3-xp1*sin(sita13)-yp1*cos(sita13); %由第二个位移矩阵方程得出f1=subs(f1,xa,ya,vxa,vya);f2=subs(f2,xa,ya,vxa,vya);f3=subs(f3,xp1,xp2,yp1,sita12,vxp1,vxp2,vyp1,vsita12);f4=subs(f4,xp1,yp1,yp2,sita12,vxp1,vyp1,vyp2,vsita12);f5=subs(f5,xp1,xp3,yp1,sita13,vxp1,vxp3,vyp1,vsita13);f6=subs(f6,xp1,yp1,yp3,sita13,vxp1,vyp1,vyp3,vsita13); %代入具体数值xb1,xb2,xb3,yb1,yb2,yb3=solve(f1,f2,f3,f4,f5,f6); %解方程vxb1=vpa(xb1);vyb1=vpa(yb1);vxb2=vpa(xb2);vyb2=vpa(yb2);vxb3=vpa(xb3);vyb3=vpa(yb3);(vxb1-vxa)2+(vyb1-vya)2;(vxb2-vxa)2+(vyb2-vya)2;(vxb3-vxa)2+(vyb3-vya)2; %去掉这三行分号可验证B点三个位置是否距离A点相等syms xd yd xc1 yc1 xc2 yc2 xc3 yc3;f7=(xc2-xd)2+(yc2-yd)2=(xc1-xd)2+(yc1-yd)2;f8=(xc3-xd)2+(yc3-yd)2=(xc1-xd)2+(yc1-yd)2; %前两个机构方程f9=xc2=cos(sita12)*xc1-sin(sita12)*yc1+xp2-xp1*cos(sita12)+yp1*sin(sita12);f10=yc2=sin(sita12)*xc1+cos(sita12)*yc1+yp2-xp1*sin(sita12)-yp1*cos(sita12); %由第一个位移矩阵方程得出f11=xc3=cos(sita13)*xc1-sin(sita13)*yc1+xp3-xp1*cos(sita13)+yp1*sin(sita13);f12=yc3=sin(sita13)*xc1+cos(sita13)*yc1+yp3-xp1*sin(sita13)-yp1*cos(sita13); %由第二个位移矩阵方程得出f7=subs(f7,xd,yd,vxd,vyd);f8=subs(f8,xd,yd,vxd,vyd);f9=subs(f9,xp1,xp2,yp1,sita12,vxp1,vxp2,vyp1,vsita12);f10=subs(f10,xp1,yp1,yp2,sita12,vxp1,vyp1,vyp2,vsita12);f11=subs(f11,xp1,xp3,yp1,sita13,vxp1,vxp3,vyp1,vsita13);f12=subs(f12,xp1,yp1,yp3,sita13,vxp1,vyp1,vyp3,vsita13); %代入具体数值xc1,xc2,xc3,yc1,yc2,yc3=solve(f7,f8,f9,f10,f11,f12); %解方程vxc1=vpa(xc1);vyc1=vpa(yc1);vxc2=vpa(xc2);vyc2=vpa(yc2);vxc3=vpa(xc3);vyc3=vpa(yc3);(vxc1-vxd)2+(vyc1-vyd)2;(vxc2-vxd)2+(vyc2-vyd)2;(vxc3-vxd)2+(vyc3-vyd)2; %去掉这三行分号可验证C点三个位置是否距离D点相等%最终答案xb1,yb1,xc1,yc1Lab=sqrt(vxb1-vxa)2+(vyb1-vya)2)Lbc=sqrt(vxb1-vxc1)2+(vyb1-vyc1)2)Lcd=sqrt(vxc1-vxd)2+(vyc1-vyd)2)Lad=sqrt(vxa-vxd)2+(vya-vyd)2) %得到四杆长曲柄存在条件：%得出四杆长后计算得到可靠到位条件：vxc1-vxb1,vyc1-vyb1*vxc1-vxd,vyc1-vydvxc2-vxb2,vyc2-vyb2*vxc2-vxd,vyc2-vydvxc3-vxb3,vyc3-vyb3*vxc3-vxd,vyc3-vyd顺序到位条件：%未完成输出结果：xb1 = (-7-4*sin(2/9*pi)+4*cos(2/9*pi)/(4*cos(2/9*pi)+4*sin(2/9*pi)-5) yb1 = -1 xc1 = -6*(27+24*sin(2/9*pi)-64*cos(2/9*pi)/(-31*cos(2/9*pi)-5+39*sin(2/9*pi) yc1 = -2*(72*cos(2/9*pi)-175+192*sin(2/9*pi)/(-31*cos(2/9*pi)-5+39*sin(2/9*pi) Lab = 1.0288436025165976748172169832223 Lbc = 2.9872531417317691216303250912289 Lcd = 6.9831476545729886023199865357226 Lad =5ans =曲柄存在条件：ans =可靠到位条件：ans = 14.605219997928496422368168445525 ans = 19.799913716084881287517588922012 ans = 20.814756669957613005391246805307 ans =顺序到位条件：