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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 M\!z='Fi  
    function sjr=nfdre(~) =}xH6^It  
    Ty5}5)CRZ  
    %系统焦距及各镜间距输入,间距取负正负 mNDd>4%H_  
    z<jH{AU  
    f=input('f:'); J;#7dRW{  
    d1=input('d1:'); H]<@\g*l@P  
    d2=input('d2:'); sqE? U*8.-  
    d3=input('d3:'); g?1! /+  
    ^)i1b:4  
    A=f^2/(d3*d2)-f/d1; [6}>?  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 277Am*2  
    C=d3/d2-f/d1; 1c19$KHu  
    CGCI3Z'  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 %1UdG6&J_  
    a2=d3/(a1*f);%α2 +hL%8CVU M  
    b2=a1*(1-a2)*f/d2;%β2 P7|x=Ew;`  
    b1=(1-a1)*f/(d1*b2);%β1 V")u y&Ob  
    V 3yt{3Or  
    S @tpd'  
    %曲率半径 wxW\L!@  
    }\oy%]_mY  
    R1=2*f/(b1*b2) xL#UMvZ>;h  
    R2=2*a1*f/(b2*(1+b1)) /xh/M@G3  
    R3=2*a1*a2*f/(1+b2) Re]7G.y  
    qf?X:9Wt  
    A1=b2^3*(a1-1)*(1+b1)^3; 3?Tk[m1b  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; fb;y*-?#  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 6Sr}I,DG  
    cms9]  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 7 yp}  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); eUZvJTE  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); L>>Cx`ASi  
    wu`P=-  
    CB=[C1 B1;C2 B2]; oJln"-M1nx  
    AB=[A1 B1;A2 B2]; -j"]1JLQ  
    AC=[A1 C1;A2 C2]; G Z~W#*|V  
    d7i 0'R  
    %非球面系数 Qk#`e  
    k2=-(det(CB)/det(AB)); "+k^8ki  
    k3=-(det(AC)/det(AB)); m)oGeD( !  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 kY.3x# w  
    k2=k2 c_dg/ !Iu  
    k3=k3 Lju)q6  
    %[J|n~8_Z  
    end k*"FMJG_  
    fXfO9{E  
    %有中间像,焦距输入为正数 ")i>-1_H  
    i-E/#zni  
    function sjr=yfdre(~) rFl6xM;F  
    `zjbyY  
    f=input('f:'); 'A)r)z {X  
    d1=input('d1:'); DB>.Uf"  
    d2=input('d2:'); /+g)J0u  
    d3=input('d3:'); KXvBJA$  
    >xK!J?!K  
    A=f^2/(d3*d2)-f/d1; SM1[)jZ-  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); +L>?kr[i[  
    C=d3/d2-f/d1; h&O8e;S#  
    SQ0t28N3h  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); pj/w9j G6  
    a2=d3/(a1*f); i?D KKjN$  
    b2=a1*(1-a2)*f/d2; ai@hQJ*  
    b1=(1-a1)*f/(d1*b2); 'pQ\BH  
    6>R|B?I%  
    %曲率半径 d^W1;0  
    o{I]c#W  
    R1=2*f/(b1*b2) H%^j yGS  
    R2=2*a1*f/(b2*(1+b1)) `S+B-I0  
    R3=2*a1*a2*f/(1+b2) .Mz'h 9@  
    gK+/wTQ%  
    A1=b2^3*(a1-1)*(1+b1)^3; '%\FT-{  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Yj/[I\I"m  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; [r f.&  
    rUOl+p_47  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); TF7~eyLg  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); lk`,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); _DAj$$ Ru4  
    9W+RUh^W  
    CB=[C1 B1;C2 B2]; [Z$E^QAP  
    AB=[A1 B1;A2 B2]; 6&KvT2?tA`  
    AC=[A1 C1;A2 C2]; NR|t~C+  
    e3!0<A[X  
    %二次系数 hUO&rov3@  
    Ka|, qkb  
    k2=-(det(CB)/det(AB)); Z(0sMOaX  
    k3=-(det(AC)/det(AB)); 4ht+u  
    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 3qNLosm#M  
    k2=k2 1Bhd-  
    k3=k3 qrxn%#\XP  
    hCFgZiH2  
    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 $7 FT0?kG