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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 j:T/iH!YF  
    function sjr=nfdre(~) hRI"y":zD  
    G|w=ez  
    %系统焦距及各镜间距输入,间距取负正负 <:/&&@2  
    V-I(WzR9y  
    f=input('f:'); 9//+Bh  
    d1=input('d1:'); .o1^Oh  
    d2=input('d2:'); Ab%;Z5$fr  
    d3=input('d3:'); @RFs/'  
    ev0oO+u  
    A=f^2/(d3*d2)-f/d1; n~V4nj&_T  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 87)zCq  
    C=d3/d2-f/d1; /yz=Cjoz  
    {Sl57!U5  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 s5.AW8X=?*  
    a2=d3/(a1*f);%α2 _I`,Br:N  
    b2=a1*(1-a2)*f/d2;%β2 Ok7t@l$  
    b1=(1-a1)*f/(d1*b2);%β1 +MbIB&fRCB  
    ,:fl?x.X  
    \;-fi.Hrf$  
    %曲率半径 QVF]Ci_=  
    bPD`+: A_  
    R1=2*f/(b1*b2) M/?KV9Xk2  
    R2=2*a1*f/(b2*(1+b1)) tt?58dm|  
    R3=2*a1*a2*f/(1+b2) I KtB;  
    YCe7<3>J4  
    A1=b2^3*(a1-1)*(1+b1)^3; G2LK]  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; I1X /Lj=  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ^J Z^>E~  
    =cN&A_L(  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); L%v^s4@  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); .6O"| Mqb  
    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); v5;I]?72l~  
    {U 'd}Q  
    CB=[C1 B1;C2 B2]; ZvYLL{>}w  
    AB=[A1 B1;A2 B2]; Q9d`zR]  
    AC=[A1 C1;A2 C2]; ms($9Lv/  
    =.]l*6W V  
    %非球面系数 %p^.\ch9  
    k2=-(det(CB)/det(AB)); i,V;xB2  
    k3=-(det(AC)/det(AB)); wxm:7$4C  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 :+{ ?  
    k2=k2 H/M Au7  
    k3=k3 kyAXRwzI  
    "G-1>:   
    end 'Y$R~e^Y?  
    4`Q3v4fOF  
    %有中间像,焦距输入为正数 {QBB^px  
    ;!o]wHmA  
    function sjr=yfdre(~) 2f U$J>Y  
    xD&^j$Em  
    f=input('f:'); ]0;864X0  
    d1=input('d1:'); |/g W_;(  
    d2=input('d2:'); IchCACK  
    d3=input('d3:'); =. y*_Ja  
    | K?#$~  
    A=f^2/(d3*d2)-f/d1; WwC 5!kZ  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); UA[,2MBp  
    C=d3/d2-f/d1; L,d LE-L  
    7r|(}S  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); bX.ja;;   
    a2=d3/(a1*f); *A}cL  
    b2=a1*(1-a2)*f/d2; QKN<+,h!z>  
    b1=(1-a1)*f/(d1*b2); v7%X@j]ji  
    b}T6v  
    %曲率半径  tvXW  
    T!wo2EzE  
    R1=2*f/(b1*b2) UgWs{y2SE.  
    R2=2*a1*f/(b2*(1+b1)) IHgeQ F ~  
    R3=2*a1*a2*f/(1+b2) }SIGPVM  
    }M1sksk5  
    A1=b2^3*(a1-1)*(1+b1)^3; fzjU<?}  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; e*+F pW@  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; o!:8nXw  
    p8s:g~ W  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); [^8n0{JiN  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); vP7K9K x  
    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); e4~>G?rM_  
    }HE6aF62O  
    CB=[C1 B1;C2 B2]; :'aAZegQY  
    AB=[A1 B1;A2 B2]; LZ@|9!KDw  
    AC=[A1 C1;A2 C2]; {0! ~C=P  
    `mye}L2I  
    %二次系数 Cf B.ZT  
    H+ h07\? %  
    k2=-(det(CB)/det(AB)); GE>[*zN  
    k3=-(det(AC)/det(AB)); 9N%JP+<89  
    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 mMMQ|ea  
    k2=k2 pZ#ap<|>I  
    k3=k3 j3q~E[Mz\  
    sH[ -W-  
    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 6Es? MW=