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[求助]
多元非線性方程組求解,急求 已有1人參與
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這個(gè)4元非線性方程組怎么解啊?急求各位大神 function f=fun(x0) syms x1 x2 x3 x4 x0=[x1 x2 x3 x4]; fi=0.515; fe=0.53; B=fi+fe-1; d=pi/180; a0=45*d; Db=7.144; Dm=40; n=1200; L=Db/Dm; Z=atan(10/(pi*40)); p=7.801e-6; E1=2.07e5; E2=1.9e5; u1=0.29; u2=0.305; Ro=17.15; Ri=24.294; ro=3.872; ri=3.872; g=x2/sin(x4)+x1/sin(x3); A1=B*Db*sin(a0)+g; A2=B*Db*cos(a0); w1=sin(x3)*(0.03*Db+x1); w2=A2-cos(x4)*(0.015*Db+x2); v=atan(sin(x3)/(L+cos(x3))); Wm=2*pi*n/(1+(1+L*cos(x3))*(cos(x4)+tan(v)*sin(x4))/((1-L*cos(x4))*(cos(x3)+tan(v)*sin(x3)))); Wr=-Wm*(1+L*cos(x3))*cos(Z)/(L*(sin(v)*sin(x3)+cos(v)*cos(x3))); Fc=(pi/12)*p*(Db^3)*Dm*(Wm^2); Mg=(1/60)*p*pi*(Db^5)*Wm*Wr*sin(v); Ps11=2/Db; Ps12=2/Db; Pn11=2/Db; Pn12=2/Db; Ps21=1/Ro; Pn21=-1/Ri; Ps22=-ro; Pn22=-ri; S1=Ps11+Ps12+Ps21+Ps22; S2=Pn11+Pn12+Pn21+Pn22; Rs1=Ps11*Ps21/(Ps11+Ps21); Rs2=Ps12*Ps22/(Ps12+Ps22); Rn1=Pn11*Pn21/(Pn11+Pn21); Rn2=Pn12*Pn22/(Pn12+Pn22); K1=1.0339*(Rs2/Rs1)^0.636; K2=1.0339*(Rn2/Rn1)^0.636; Ke1=1.5277+0.6023*log(Rs2/Rs1); Ke2=1.5277+0.6023*log(Rn2/Rn1); Ee1=1.0003+0.5968*(Rs1/Rs2); Ee2=1.0003+0.5968*(Rn1/Rn2); e1=(1-1/(K1^2))^0.5; e2=(1-1/(K2^2))^0.5; y1=(2*Ke1/pi)*((1-(e1)^2)*pi/(2*Ee1))^(1/3); y2=(2*Ke2/pi)*((1-(e2)^2)*pi/(2*Ee2))^(1/3); E=0.5*((1-u1^2)/E1+(1-u2^2)/E2); Q1=(8*(x1/y2)^3/((1.5*E)^2*S2))^0.5; Q2=(8*(x2/y1)^3/((1.5*E)^2*S1))^0.5; f1=(A1-w1)^2+(A2-w2)^2-(0.015*Db+x4)^2; f2=w1^2+w2^2-(0.03*Db+x3); f3=Q2*sin(x4)-Q1*sin(x3)-Mg/Db*(cos(x4)-cos(x3)); f4=Q2*cos(x4)-Q1*cos(x3)+Mg/Db*(sin(x4)-sin(x3))+Fc; f=[f1; f2; f3 ;f4]; % format long; % f=vpa(f,2); % class(f); end |
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