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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 T$<'ZC  
    function sjr=nfdre(~) :9x]5;ma  
    -byaV;T?"  
    %系统焦距及各镜间距输入,间距取负正负 SdMLO6-  
    z?kE((Ey  
    f=input('f:'); pEN`6*  
    d1=input('d1:'); %1{O  
    d2=input('d2:'); vo)W ziHh  
    d3=input('d3:'); Lc]hwMGR*  
    ;p <BiC$b  
    A=f^2/(d3*d2)-f/d1; <HS{A$]  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); Vu4LC&q  
    C=d3/d2-f/d1; =,qY\@fq  
    EKN<KnU%  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 ]-a/)8  
    a2=d3/(a1*f);%α2 'gD./|Z0  
    b2=a1*(1-a2)*f/d2;%β2 ,VUOsNN4\  
    b1=(1-a1)*f/(d1*b2);%β1 ni )G  
    +<V$G/"  
    d|~'#:y@  
    %曲率半径 on5\rY<I:@  
    Iue=\qUK^  
    R1=2*f/(b1*b2) #}~?8/h!  
    R2=2*a1*f/(b2*(1+b1)) ~d,$ nZ"z  
    R3=2*a1*a2*f/(1+b2) /M3;~sx  
    -!M>;M@  
    A1=b2^3*(a1-1)*(1+b1)^3; r9b(d]  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 9U3}_  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Uqj$itqUQ  
    K*1]P ar;  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 87)/dHc  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); | "M1+(k7  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); 9oP  
    M\JAB ;A  
    CB=[C1 B1;C2 B2]; Pd `~#!  
    AB=[A1 B1;A2 B2]; !mwMSkkq  
    AC=[A1 C1;A2 C2]; 8 K)GH:a  
    zJUT<%[U  
    %非球面系数 pj3H4yCM:  
    k2=-(det(CB)/det(AB)); gOE ?  
    k3=-(det(AC)/det(AB)); meThjCC  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 ~% `hh9]  
    k2=k2 e~,+rM  
    k3=k3 P+_1*lOG  
    Wap\J7NY  
    end XMxm2-%olP  
    T0b/txS  
    %有中间像,焦距输入为正数 Z3u6m0!  
    A%&lW9z7  
    function sjr=yfdre(~) Zm6jF  
    od,,2pwK+  
    f=input('f:'); J5Q.v;  
    d1=input('d1:'); Klu0m~X@  
    d2=input('d2:'); 30sA\TZ  
    d3=input('d3:'); WigTNg4  
    h+YPyeAs  
    A=f^2/(d3*d2)-f/d1;  ggfCfn  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); dg+"G|nr  
    C=d3/d2-f/d1; W>b\O">  
    >*+n`"6  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); c0X1})q$  
    a2=d3/(a1*f); Zba<|C  
    b2=a1*(1-a2)*f/d2; W+s3rS2  
    b1=(1-a1)*f/(d1*b2); L$,Kdpj  
    zpNt[F?~1  
    %曲率半径 5;XU6Rz!  
    c7tO'`q$e  
    R1=2*f/(b1*b2) $0~1;@`rQ6  
    R2=2*a1*f/(b2*(1+b1)) N>sHT =_  
    R3=2*a1*a2*f/(1+b2) \t&8J+%  
    KO[T&#y'  
    A1=b2^3*(a1-1)*(1+b1)^3; o##!S6:A  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; IkGM~3e  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 9c0  
    DwTVoCC  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Gsm.a  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); -y$<fu9 e  
    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); 4T){z^"  
    XN3'k[  
    CB=[C1 B1;C2 B2]; XF@34b5(  
    AB=[A1 B1;A2 B2]; 0juP"v$C>  
    AC=[A1 C1;A2 C2]; |a'$v4dCF  
    Fd%JF#Hk  
    %二次系数 ~eiD(04^r*  
    7o7FW=^  
    k2=-(det(CB)/det(AB)); Y%(8'Ch  
    k3=-(det(AC)/det(AB)); qR%as0;  
    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 7Fzr\&  
    k2=k2 mMCd   
    k3=k3 {t]8#[lo  
    ?+{_x^  
    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 4i"fHVp8