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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 \LB =_W$  
    function sjr=nfdre(~) x;NCW  
    [V`j@dV  
    %系统焦距及各镜间距输入,间距取负正负 A3%s5`vNvH  
    Ij>x3L\-  
    f=input('f:'); *JXiOs  
    d1=input('d1:'); y4`<$gL   
    d2=input('d2:'); SJ1 1LF3)  
    d3=input('d3:'); [T', ZLR|  
    pfW0)V1t  
    A=f^2/(d3*d2)-f/d1; )f4D2c&VE  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 't=\YFQ*v  
    C=d3/d2-f/d1; M-KjRl  
    rkA0v-N6v  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 nf!RB-orF  
    a2=d3/(a1*f);%α2 4cK6B)X  
    b2=a1*(1-a2)*f/d2;%β2 P[PBoRd2  
    b1=(1-a1)*f/(d1*b2);%β1 ,)A^3Q*  
    y wlN4=  
    b7>^w<ki  
    %曲率半径 R}4o{l6  
    h d1H  
    R1=2*f/(b1*b2) +M%i3A  
    R2=2*a1*f/(b2*(1+b1)) XJnDx 09h  
    R3=2*a1*a2*f/(1+b2) K^AX=B  
    dwks"5l  
    A1=b2^3*(a1-1)*(1+b1)^3; O4FW/)gq  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ann!"s_  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; )F 6#n&2  
    vTYI ez`g  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 8Dpf{9Y-E  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); MJ[#Gq\0R  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); FQ?,&s$Bmd  
    :qy`!QPUm  
    CB=[C1 B1;C2 B2]; V#!ihL/>  
    AB=[A1 B1;A2 B2]; B+|E|8"  
    AC=[A1 C1;A2 C2]; RsU=fe,  
    "/hM&  
    %非球面系数 )NZ6!3[@  
    k2=-(det(CB)/det(AB)); CtVY;eG  
    k3=-(det(AC)/det(AB)); cH6ie?KvAo  
    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 k%#`{#n i  
    k2=k2 _GK^7}u  
    k3=k3 -i|qk`Y  
    hNUAwTH6  
    end N]: "3?%  
    xEaRuH c  
    %有中间像,焦距输入为正数 +4ax~fuU  
    )'\Jp 7*3  
    function sjr=yfdre(~) w.J[3m/  
    qVC_K/w 7  
    f=input('f:'); `(1em%}  
    d1=input('d1:'); EDvK9J  
    d2=input('d2:'); tA$,4B?  
    d3=input('d3:'); NAhV8  
    La? q>  
    A=f^2/(d3*d2)-f/d1; { yU1db^  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); @lAOi1m,,  
    C=d3/d2-f/d1; =If% m9  
    nL@ "FZ`(  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); i0,{*LD%^  
    a2=d3/(a1*f); ?UQVmE&  
    b2=a1*(1-a2)*f/d2; 8YraW|H  
    b1=(1-a1)*f/(d1*b2); oM-{)rvQd  
    0.O pgv2K  
    %曲率半径 )gV+BHK  
    wNDLN`,^H  
    R1=2*f/(b1*b2) `|wH=  
    R2=2*a1*f/(b2*(1+b1)) ,T"pUeVJ  
    R3=2*a1*a2*f/(1+b2) -2|D( sO  
    ;_K+b,  
    A1=b2^3*(a1-1)*(1+b1)^3; sl|s#+Z  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; q#v.-013r  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; zc]F  
    v83uGEq(  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); hxx,E>k  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); (}O)pqZ>  
    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); [e3|yE6  
    |K'{R'A  
    CB=[C1 B1;C2 B2]; UA{sUj+?  
    AB=[A1 B1;A2 B2]; [6 wI22  
    AC=[A1 C1;A2 C2];  ?1r@r  
    <qZXpQ#  
    %二次系数 B P"PUl:  
    ]l+Bg;F#V  
    k2=-(det(CB)/det(AB)); %9[GP7?  
    k3=-(det(AC)/det(AB)); wc)[r~On(5  
    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 y 4,2Xs9,  
    k2=k2 51.F,uY  
    k3=k3 Y+iC/pd  
    <?52Svi}}  
    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 10}oaL S