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    [原创]在框架结构确定的情况下,基于matlab的消四种像差的三反系统初始结构的求解 [复制链接]

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    离线songshaoman
     
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    只看楼主 倒序阅读 楼主  发表于: 2020-05-25
    %无中间像,焦距输入为负数 U bYEEY#  
    function sjr=nfdre(~) ?r2#.W  
    aBzszp]l+  
    %系统焦距及各镜间距输入,间距取负正负 P(a.iu5   
    *;XWLd#  
    f=input('f:'); n\ Hs@.  
    d1=input('d1:'); > MH(0+B*  
    d2=input('d2:'); A?*o0I  
    d3=input('d3:'); ZY56\qcY  
    )=DGdI Et  
    A=f^2/(d3*d2)-f/d1; HQ9X7[3  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); )H}#A#ovj7  
    C=d3/d2-f/d1; ]0r|_)s  
    YQ0)5}  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 W8g' lqc|  
    a2=d3/(a1*f);%α2 S{K0.<,E  
    b2=a1*(1-a2)*f/d2;%β2 \`w4|T  
    b1=(1-a1)*f/(d1*b2);%β1 c )7j QA  
    wP/A^Rs  
    99EXo+g  
    %曲率半径 jp+_@S>  
    K]xa/G(  
    R1=2*f/(b1*b2) vs j3  
    R2=2*a1*f/(b2*(1+b1)) ,]5Ic.};p  
    R3=2*a1*a2*f/(1+b2) zT ZVehEe  
    >5 b/or  
    A1=b2^3*(a1-1)*(1+b1)^3; Y+qQIMZ  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; .6~`Ubr}E  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; OD=!&LM  
    m~'? /!!  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); _Zc%z@}  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); tV/Z)fpyH  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); CD0VfA>Z  
    T%Pp*1/m7  
    CB=[C1 B1;C2 B2]; 9GdB#k6W`  
    AB=[A1 B1;A2 B2]; )J(q49  
    AC=[A1 C1;A2 C2]; auWXgkwZs/  
    ]P[%Mhg^  
    %非球面系数 yE} dj)wd  
    k2=-(det(CB)/det(AB)); %/.a]j!  
    k3=-(det(AC)/det(AB)); ^JR;epVJ  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 /b;K  
    k2=k2 PNeh#PI 6)  
    k3=k3 \x >65;  
    X _$a,"'~)  
    end eb|i 3.  
    w-$[>R[hw  
    %有中间像,焦距输入为正数 G9g6.8*&  
    +([!A6:  
    function sjr=yfdre(~) +!0eu>~_&  
    s4H2/EC  
    f=input('f:'); j 6ut}Uq  
    d1=input('d1:'); l =IeJh  
    d2=input('d2:'); e &9F\e  
    d3=input('d3:'); ZlKw_Sq:  
    FP"$tt(  
    A=f^2/(d3*d2)-f/d1; ;PyZ?Z;  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); gX5&d\y  
    C=d3/d2-f/d1; n +1y  
    X%9*O[6{  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); <a le$[  
    a2=d3/(a1*f); rgcWRt  
    b2=a1*(1-a2)*f/d2; M*pRv  
    b1=(1-a1)*f/(d1*b2); :1 )DqoAJ  
    +[ .Yy  
    %曲率半径 `3q;~ 9  
    T{vR,  
    R1=2*f/(b1*b2) =wq;@'U  
    R2=2*a1*f/(b2*(1+b1)) 'YSuQP>  
    R3=2*a1*a2*f/(1+b2) PKq-@F%X  
    _U)%kY8  
    A1=b2^3*(a1-1)*(1+b1)^3; v$w++3H  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; "zZI S6j  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; KbxR Lx]w  
    H' J|U|  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); o'%e I  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 7k=fZ$+O  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); Z$KV&.=+  
    s* j fMY  
    CB=[C1 B1;C2 B2]; ;Pb8YvG1$  
    AB=[A1 B1;A2 B2]; F#+.>!  
    AC=[A1 C1;A2 C2]; /2NSZO  
    B>TSdn={>  
    %二次系数 DHfB@/q#  
    u2sR.%2U<  
    k2=-(det(CB)/det(AB)); /owO@~G  
    k3=-(det(AC)/det(AB)); vi {uy  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 19d6]pJ5  
    k2=k2 q9}m!*8e  
    k3=k3 v@u<Ww;=@  
    K3p@$3hQ  
    end
     
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    离线doushan
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    只看该作者 1楼 发表于: 2023-03-01
    谢谢分享,学习一下 !E& MBAKy