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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 JKM(fX+  
    function sjr=nfdre(~) 5 qfvHQ ~M  
    "i;.>  
    %系统焦距及各镜间距输入,间距取负正负 ]LC4rS  
    G)?*BH  
    f=input('f:'); b}R_@_<u  
    d1=input('d1:'); s0?'mC+p  
    d2=input('d2:'); rS BI'op  
    d3=input('d3:'); 4@-tT;$  
    -pYmM d,  
    A=f^2/(d3*d2)-f/d1; ~{Iw[,MJ  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); >+dS PI  
    C=d3/d2-f/d1; L K #A  
    '\#q7YjaL  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 =IMmtOvJ  
    a2=d3/(a1*f);%α2 ?l9sj]^w  
    b2=a1*(1-a2)*f/d2;%β2 #Zm`*s`  
    b1=(1-a1)*f/(d1*b2);%β1 A`3KE9ED  
    HqWWWCWal  
    L[2qCxB'^  
    %曲率半径 a20w.6F  
    w"9h_;'C_  
    R1=2*f/(b1*b2) KY  
    R2=2*a1*f/(b2*(1+b1)) JI!1 .]&  
    R3=2*a1*a2*f/(1+b2) 5qnei\~  
    ,H7_eVLWR  
    A1=b2^3*(a1-1)*(1+b1)^3; 89&9VX^A  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; qB$-H' j:;  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 8?nn4]P  
    Z2]0brV  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); FFw(`[A_  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); fS4foMI63)  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); K]m#~J3d>  
    ` 7iA?;  
    CB=[C1 B1;C2 B2]; :g|.x  
    AB=[A1 B1;A2 B2]; X <xM '  
    AC=[A1 C1;A2 C2]; 8`*5[ L~~/  
    }'P|A  
    %非球面系数 ^s6~*n<fH  
    k2=-(det(CB)/det(AB)); (sKg*G2  
    k3=-(det(AC)/det(AB)); LG,?,%_s  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 wMCMrv:  
    k2=k2 ">Qxb.Y}  
    k3=k3 vX }iA|`#  
    pqO3(2F9  
    end 5>9Q<*   
    i\IpS@/{-v  
    %有中间像,焦距输入为正数 bKS/T^UQ  
    */K[B(G  
    function sjr=yfdre(~) 2@a'n@-  
    )isS^O$qH  
    f=input('f:'); HAO-|=c4  
    d1=input('d1:'); _ooHB>sH  
    d2=input('d2:'); VzSkqWF/"  
    d3=input('d3:'); @TALZk'%  
    Lmjd,t  
    A=f^2/(d3*d2)-f/d1; J8~hIy6]  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); w~B1TfqNo  
    C=d3/d2-f/d1; F4\:9ws  
    aZ~e;}w.Zq  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); Q I";[  
    a2=d3/(a1*f); hXI[FICQU{  
    b2=a1*(1-a2)*f/d2; ODu/B'*  
    b1=(1-a1)*f/(d1*b2); c(hC'Cp  
    &;ddnxFI  
    %曲率半径 -btNwE6[.  
    &pI\VIx ?  
    R1=2*f/(b1*b2) |5;,]lbt  
    R2=2*a1*f/(b2*(1+b1)) v>K|hH  
    R3=2*a1*a2*f/(1+b2) Tr;.%/4Q  
    dwB#k$VIOw  
    A1=b2^3*(a1-1)*(1+b1)^3; ^r}Uu~A>  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; x}a?B  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; %t_'rv  
    qsp3G7\'=  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); TgV-U  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); A&1EOQ=N  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); 0i[t[_sce  
    1R-0b{w[  
    CB=[C1 B1;C2 B2]; ypM,i  
    AB=[A1 B1;A2 B2]; E*)A!2rlK  
    AC=[A1 C1;A2 C2]; iOa<=  
    9*iVv)jd  
    %二次系数 VT>-*  
    n2*Ua/J-8  
    k2=-(det(CB)/det(AB)); E7h@c>IK  
    k3=-(det(AC)/det(AB)); jR1^e$  
    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 qX5]\nX&G  
    k2=k2 (1S9+H>g  
    k3=k3 0 F8xS8vK+  
    w:B&8I(n}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 T:Q+ Z }v+