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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 正序阅读 楼主  发表于: 2020-05-25
    %无中间像,焦距输入为负数 U~"Y8g#qgy  
    function sjr=nfdre(~) kQlwl9  
    [Y@>,B!V  
    %系统焦距及各镜间距输入,间距取负正负 rMZuiRz*  
    1(:!6PY  
    f=input('f:'); wD{c$TJ?{F  
    d1=input('d1:'); eMFxdtH  
    d2=input('d2:'); 3/ }  
    d3=input('d3:'); ^eO/?D8~h  
    ) < U9  
    A=f^2/(d3*d2)-f/d1; LGZa l&9AY  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); R)\^*tkz7  
    C=d3/d2-f/d1; #)]t4wa_W  
    C&*1H`n  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 ez*QP|F*9  
    a2=d3/(a1*f);%α2 2=0HQXXrq  
    b2=a1*(1-a2)*f/d2;%β2 2u!&Te(!9  
    b1=(1-a1)*f/(d1*b2);%β1 fxr#T'i  
    Myiv#rQ)  
    A%$~  
    %曲率半径 S >CKm:7  
    w( XZSE  
    R1=2*f/(b1*b2) +0UBP7kn  
    R2=2*a1*f/(b2*(1+b1)) G\;6n  
    R3=2*a1*a2*f/(1+b2) x(eX.>o\  
    c-Yd> 4+ 1  
    A1=b2^3*(a1-1)*(1+b1)^3; Rq[d\BN0.d  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ~eoM 2XlW  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; h ! R=t  
    7X/t2Vih@  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); p e+h8  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); fbOqxF"?we  
    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); lG94^|U  
    emnT;kJ>  
    CB=[C1 B1;C2 B2]; Km%L1Cd]  
    AB=[A1 B1;A2 B2]; X~JP 1  
    AC=[A1 C1;A2 C2]; 7>F{.\Z  
    q^EY?;Y  
    %非球面系数 !%('8-x%  
    k2=-(det(CB)/det(AB)); 6:Z8d%Z  
    k3=-(det(AC)/det(AB)); A!x&,<  
    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 W [K.|8ho  
    k2=k2 wT::b V{  
    k3=k3 )(-;H|]?  
    u YT$$'S  
    end bG@2f"  
    T:Klr=&V  
    %有中间像,焦距输入为正数 /YF:WKr2  
    e}d(.H%l0  
    function sjr=yfdre(~) X V;j6g  
    Kmx^\vDs  
    f=input('f:'); '=fk;AiQ  
    d1=input('d1:'); Op ?"G  
    d2=input('d2:'); B<m0YD?>~>  
    d3=input('d3:'); BrwC9:  
    x}?<9(nE c  
    A=f^2/(d3*d2)-f/d1; 3+>n!8x ;A  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ;t\h"K<,|  
    C=d3/d2-f/d1; p,$N-22a  
    &.*UVc2+Y  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); tMiIlf!>p  
    a2=d3/(a1*f); K=v:qY4Z  
    b2=a1*(1-a2)*f/d2; !!^z6jpvn  
    b1=(1-a1)*f/(d1*b2); =ZIT!B?4  
    AT~,  
    %曲率半径 d)Yl D]I  
    j"69uj` R  
    R1=2*f/(b1*b2) \BXzmok  
    R2=2*a1*f/(b2*(1+b1)) CG=c@-"n/  
    R3=2*a1*a2*f/(1+b2) ls]N&!/hq  
    3$k#bC  
    A1=b2^3*(a1-1)*(1+b1)^3; +)jll#}?  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; cZ?QI6|[  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 3Hom0g,V4  
    '|^:,@8P9  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 1I9v`eT4  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ]zSFX =~(S  
    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);  XTJD>  
    e}e8WR=B  
    CB=[C1 B1;C2 B2]; 8I#ir4z#<  
    AB=[A1 B1;A2 B2]; n_e'n|T  
    AC=[A1 C1;A2 C2]; UUJQc ~=  
    L9 D`hefz  
    %二次系数 kk3^m1  
    sV  
    k2=-(det(CB)/det(AB)); MCT1ZZpPr  
    k3=-(det(AC)/det(AB)); RU:Rt'  
    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 F`nQS&y  
    k2=k2 Iv6 q(c  
    k3=k3 `um,S  
    // o.+?S  
    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 smQ^(S^