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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 lf8xL9v  
    function sjr=nfdre(~) ,xj3w#`zaf  
    MOyT< $  
    %系统焦距及各镜间距输入,间距取负正负 kr{)  
    o PaZ  
    f=input('f:'); @,YlmX}  
    d1=input('d1:'); JmjxGcG  
    d2=input('d2:'); u0 BMyH  
    d3=input('d3:'); .\)k+ R  
    !2tw,QM  
    A=f^2/(d3*d2)-f/d1; "!a`ygqpT  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ?{j@6,  
    C=d3/d2-f/d1; *')Q {8`  
    iIB9j8  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 Oc^m_U8>^  
    a2=d3/(a1*f);%α2 XSl!T/d  
    b2=a1*(1-a2)*f/d2;%β2 /p}{#DLB  
    b1=(1-a1)*f/(d1*b2);%β1 F8 ?uQP8  
    +',^((o  
    L1F###c  
    %曲率半径 :Q=z=`*2w  
    :K;T Q  
    R1=2*f/(b1*b2) ?k::tNv0  
    R2=2*a1*f/(b2*(1+b1)) T\cR2ZT~  
    R3=2*a1*a2*f/(1+b2) .d e  
    @9<S*  
    A1=b2^3*(a1-1)*(1+b1)^3; 7g-$oO  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; .}~$1QKS  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; L^jaBl  
    1XGG.+D  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Uf^RLdoDn  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); shy  
    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); u x#. :C|  
    N)P((>S;  
    CB=[C1 B1;C2 B2]; J& )#G@fRX  
    AB=[A1 B1;A2 B2]; w`0)x5 TGR  
    AC=[A1 C1;A2 C2]; + lP5XY{  
    EFwL.'Fh  
    %非球面系数 gjQ=8&i  
    k2=-(det(CB)/det(AB)); $^K]&Mft  
    k3=-(det(AC)/det(AB)); 4XD)E&   
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 /_HwifRQ  
    k2=k2 /4^G34  
    k3=k3 >j) w\i  
    FXeV6zfrE  
    end m]1!-`(*  
    7:h<`_HT(X  
    %有中间像,焦距输入为正数 ||`qIElAW,  
    .E#<fz  
    function sjr=yfdre(~) 0FTRm2(  
    Y=3X9%v9g  
    f=input('f:'); 0Ux<16#  
    d1=input('d1:'); U|9U(il  
    d2=input('d2:'); "NJ ,0A  
    d3=input('d3:'); 'qdg:_L"  
    mZ~mf->%  
    A=f^2/(d3*d2)-f/d1; )&XnM69~b  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); kAY@^vi  
    C=d3/d2-f/d1; A"0wvk)UcY  
    jzMhJ  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); \Oz,Qzr|  
    a2=d3/(a1*f); K/Sq2:  
    b2=a1*(1-a2)*f/d2; .r7D )xNa@  
    b1=(1-a1)*f/(d1*b2); C?{D"f`[]  
    cJSVT8  
    %曲率半径 Gee~>:_Q{J  
    "$]ls9-%n  
    R1=2*f/(b1*b2) T.J`S(oI  
    R2=2*a1*f/(b2*(1+b1)) 2rF?Q?$,B  
    R3=2*a1*a2*f/(1+b2) Sy4 mZ}:  
    ^@M [t<  
    A1=b2^3*(a1-1)*(1+b1)^3; lfXH7jL2~  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Go-wAJ>  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 4 U}zJP(L  
    lt{lHat1  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); >'eB2  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); lj4%(rB=  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); *Yj~]E0`1  
    1]_?$)$T  
    CB=[C1 B1;C2 B2]; C:rRK*  
    AB=[A1 B1;A2 B2]; U?JiVxE^  
    AC=[A1 C1;A2 C2]; ]cn/(U`  
    +{5JDyh0  
    %二次系数 x(rd$oZO  
    *Kp}B}}J  
    k2=-(det(CB)/det(AB)); Hde]DK,d  
    k3=-(det(AC)/det(AB)); h U 9\y  
    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 J 5Wz4`'  
    k2=k2 N$C{f;xV  
    k3=k3 oG+K '(BB  
    lTx Y6vi  
    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 $?[1#%