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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4元非線性方程組怎么解。考鼻蟾魑淮笊 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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No solution found. fsolve stopped because the relative size of the current step is less than the default value of the step size tolerance squared, but the vector of function values is not near zero as measured by the default value of the function tolerance. <stopping criteria details> x1 = 0.264988629986718 - 0.227162298915769i 0.157276044576307 - 0.503604785301162i 53.759850087022613 - 0.866858320147245i 34.658568519774903 + 0.590132528775326i 得出的是這個,這不算求出解啊 |
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