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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 r4<As`&  
    function sjr=nfdre(~) :X:s'I4J D  
    L=fy!R  
    %系统焦距及各镜间距输入,间距取负正负 9JX@c k  
    U| ?68B3  
    f=input('f:'); Xfbr;Jt"<  
    d1=input('d1:'); FnI}N;"  
    d2=input('d2:'); oE6`]^^  
    d3=input('d3:'); bC98<if  
    t0r0{:  
    A=f^2/(d3*d2)-f/d1; =f [/Pv  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); wwQ2\2w>Hm  
    C=d3/d2-f/d1; X Q CE`m  
    @vXXf/  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 "6U@e0ht  
    a2=d3/(a1*f);%α2 l9OpaOVfJ  
    b2=a1*(1-a2)*f/d2;%β2 ;yyR_N S  
    b1=(1-a1)*f/(d1*b2);%β1 DVMdRfA  
    ,zr,>^ v  
    r:$*pC&{  
    %曲率半径 !]Qk?T~9-  
    MkjB4:"  
    R1=2*f/(b1*b2) ]$ Nhy8-  
    R2=2*a1*f/(b2*(1+b1)) kFY2VPP~  
    R3=2*a1*a2*f/(1+b2) uv8k ea .(  
    HFTeG4R  
    A1=b2^3*(a1-1)*(1+b1)^3; \8m9^Z7IfK  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; "Nb2[R  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; NZ&ZK@h}.  
    C.":2F;-e  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); kB=B?V~#  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); jkta]#O  
    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); Otx>S' 5  
    N*6~$zl&  
    CB=[C1 B1;C2 B2]; FG+pR8aA$  
    AB=[A1 B1;A2 B2]; ujkWVE'  
    AC=[A1 C1;A2 C2]; WagL8BpLx  
    `OgT"FdL!  
    %非球面系数 1?^ P=^8   
    k2=-(det(CB)/det(AB)); Lu:*nJ%1[  
    k3=-(det(AC)/det(AB)); [KHlApL  
    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 !iNwJ|0  
    k2=k2 ]!P8{xmb@  
    k3=k3 ks6iy}f7  
    >5Lp;  
    end ~@ PD\  
    F[%k ;aJ  
    %有中间像,焦距输入为正数 |<,0*2  
    [>pBz3fn,  
    function sjr=yfdre(~) r-xP 6  
    vqQ)Pu?T  
    f=input('f:'); J?X{NARt  
    d1=input('d1:'); u/S>*E  
    d2=input('d2:'); Ab/JCZNn  
    d3=input('d3:'); H~vrCi~t"  
     $RRX-  
    A=f^2/(d3*d2)-f/d1; 3@Fa  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2);  <)~-]  
    C=d3/d2-f/d1; \xl$z *zI  
    f.^|2T I1g  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); ^,^MW  
    a2=d3/(a1*f); 3h=kn@I  
    b2=a1*(1-a2)*f/d2; M&29J  
    b1=(1-a1)*f/(d1*b2); pg7~%E4  
    y9R%%i  
    %曲率半径 w:deQ:k  
    Fp@>(M#3  
    R1=2*f/(b1*b2) o8<~zeI  
    R2=2*a1*f/(b2*(1+b1)) {5X,xdzR  
    R3=2*a1*a2*f/(1+b2) R-NM ~gp  
    Cm%I/4  
    A1=b2^3*(a1-1)*(1+b1)^3; R6fkc^  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Z:N;>.3i  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Lp(`m=;O  
    =j- ,yxBvJ  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); \ HUDZ2 s  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); W]7<PL*u  
    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); =JfwHFHd#  
    ;Bz| hB{  
    CB=[C1 B1;C2 B2]; yn]Sc<uK  
    AB=[A1 B1;A2 B2]; j|Vl\Z&o)  
    AC=[A1 C1;A2 C2]; &uO-h  
    SZ[?2z  
    %二次系数 [Z:P{yr  
    -`\^_nVC  
    k2=-(det(CB)/det(AB)); #'> )?]tn  
    k3=-(det(AC)/det(AB)); E8#aE\'t  
    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 ~>9G\/u j  
    k2=k2 9,_~qWw  
    k3=k3 I)` +:+P  
    62'9lriQ  
    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 A. tGr(r