gaoxj3000 发表于 2008-3-5 21:50

看看谁会画这个系统的分岔图

文章参看函数中的Reference,文章中说是用的Kaas-Petersen 的 Continuation routine PATH 做分岔图, 可我找不到有关这个程序的资料,且一直没搞清楚文章中的分岔图是如何画出来的,请高手指点,如果能帮我写一下分岔程序,哪再好不过了,后面两幅图是文章中的分岔图,是行车速度与某个部件位移的分岔图

function F = True1( t, X )
% Reference:
% Preben Isaksen and Hans True. On the ultimate transition to chaos in the dynamics of Cooperrider's bogie. Chaos,Solitons & Fractals,1997,8(4):559-581.
global Speed;
% ******************** Parameter Definition ******************** %
Mw = 1022;% The mass of the wheel axle
Mf = 2918;%               frame
Iwy = 678;% The moment of inertia for the yaw motion of the wheel axle
Ify = 6780; %                                                 frame
Ifr = 6780; % The roll motion innertia of the frame
K1 = 1.823e6;K2 = 3.646e6;K3 = 3.646e6;K4 = 0.1823e6;K5 = 0.3333e6;K6 = 2.710e6;
D1 = 20.0e3;   D2 = 29.2e3;
a = 0.716;% Half of the track gauge
b = 1.074;%             axle distance
d1 = 0.620;
d2 = 0.680;
h1 = 0.0762;
h2 = 0.6584;
Lambda = 0.05;delta = 0.0091;R0 = 0.4572;
% ***** Coordinate Definition ***** %
Dq1 = X(1);Dq2 = X(2);Dq3 = X( 3);Dq4 = X( 4);Dq5 = X( 5);Dq6 = X( 6);Dq7 = X( 7);
   q1 = X(8);   q2 = X(9);   q3 = X(10);   q4 = X(11);   q5 = X(12);   q6 = X(13);   q7 = X(14);
% ******************** Computation of creep force ******************** %
Psi1 = 0.54219;Phi = 0.60252;Gpiab = 6.563e6;Nmiu = 10000;
xi_xf = Dq1/Speed - q2;xi_yf = a*Dq2/Speed + Lambda*q1/R0;% Creepage for front axle
xi_xr = Dq3/Speed - q4;xi_yr = a*Dq4/Speed + Lambda*q3/R0;%               rear
xi_Rf = sqrt( xi_xf^2/Psi1^2 + xi_yf^2/Phi^2 );
xi_Rr = sqrt( xi_xr^2/Psi1^2 + xi_yr^2/Phi^2 );
Uf = Gpiab*xi_Rf/Nmiu;Ur = Gpiab*xi_Rr/Nmiu;
ifUf<3
      F_Rf = Nmiu*( Uf - Uf^2/3 + Uf^3/27 );
elseif Uf>=3
       F_Rf = Nmiu;
end
ifUr<3
      F_Rr = Nmiu*( Ur - Ur^2/3 + Ur^3/27 );
elseif Ur>=3
       F_Rr = Nmiu;
end
Fxf = xi_xf*F_Rf/(Psi1*xi_Rf);Fyf = xi_yf*F_Rf/(Phi*xi_Rf);% The creep forces for front axle
Fxr = xi_xr*F_Rr/(Psi1*xi_Rr);Fyr = xi_yr*F_Rr/(Phi*xi_Rr);%                      rear
% ******************** Computation of flange force ******************** %
K0 = 14.6e6;
ifabs(q1)<=delta
      Fq1 = 0.0;
elseif q1>delta
      Fq1 = K0*( q1 - delta );
elseif q1<-delta
      Fq1 = K0*( q1 + delta );
end
ifabs(q3)<=delta
      Fq3 = 0.0;
elseif q3>delta
      Fq3 = K0*( q3 - delta );
elseif q3<-delta
      Fq3 = K0*( q3 + delta );
end
% ******************** Definition of other variable ******************** %
A1 = 2*K1*( q1 - q5 - b*q6 - h1*q7 );
A2 = 2*K1*( q3 - q5 + b*q6 - h1*q7 );
A3 = 2*K2*d1^2*( q2 - q6 );
A4 = 2*K2*d1^2*( q4 - q6 );
A5 = 2*D2*( Dq5 - h2*Dq7 ) + 2*K4*( q5 - h2*q7 );
A6 = K6*q6;
A7 = 2*D1*d2^2*Dq7 + 2*K5*d2^2*q7 + 4*K3*d1^2*q7;
% ******************** The state equation ******************** %
F(1) = -( Fq1 + 2*Fxf + A1 )/Mw;
F(2) = -( 2*a*Fyf + A3 )/Iwy;
F(3) = -( Fq3 + 2*Fxr + A2 )/Mw;
F(4) = -( 2*a*Fyr + A4 )/Iwy;
F(5) = -( A5 - A1 - A2 )/Mf;
F(6) = -( A6 - A4 - A3 + b*A2 - b*A1 )/Ify;
F(7) = -( A7 - h2*A5 - h1*A2 - h1*A1 )/Ifr;
F(8)=Dq1;F(9)=Dq2;F(10)=Dq3;F(11)=Dq4;F(12)=Dq5;F(13)=Dq6;F(14)=Dq7;
% ******************** End of the state equation ******************** %
X = X';
F = F';

[ 本帖最后由 gaoxj3000 于 2008-3-5 22:02 编辑 ]
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