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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 Oq*a4_R'YV  
    function sjr=nfdre(~) Kl(}s{YFn.  
    _=U XNr8S  
    %系统焦距及各镜间距输入,间距取负正负 OTN"XKa$  
    g\rujxHlH  
    f=input('f:'); A"vI6ud>  
    d1=input('d1:'); `"GD'Oa  
    d2=input('d2:'); wLkHU"'   
    d3=input('d3:'); x~Se-#$  
    #xYkG5`lm  
    A=f^2/(d3*d2)-f/d1; dMRwQejY{7  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); QHP^1W`  
    C=d3/d2-f/d1; aFIet55o  
    lCd^|E  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 =\CbX  
    a2=d3/(a1*f);%α2 wKk  
    b2=a1*(1-a2)*f/d2;%β2 h=`rZC  
    b1=(1-a1)*f/(d1*b2);%β1 [0/?(i|  
    )I1LBvfQ  
    <Y~V!9(~{Q  
    %曲率半径 rp=?4^(u  
    jG)>{D  
    R1=2*f/(b1*b2) G|3OB:  
    R2=2*a1*f/(b2*(1+b1)) JC>}(yQA  
    R3=2*a1*a2*f/(1+b2) C$ZY=UXz!T  
    EnwiE  
    A1=b2^3*(a1-1)*(1+b1)^3; (e F5?I  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; -BEPpwb<g  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; gMHH3^\VH)  
    QXXcJc~  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); I62Yg p$K  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); uA}asm  
    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); C$,S#n@  
    2GZUMXK  
    CB=[C1 B1;C2 B2]; Aqp3amW!  
    AB=[A1 B1;A2 B2]; u6(7#n02  
    AC=[A1 C1;A2 C2]; K VQZ  
    BOh&Db*  
    %非球面系数 fC~WuG 3  
    k2=-(det(CB)/det(AB)); w`!Yr:dU  
    k3=-(det(AC)/det(AB)); f3v/Y5)  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 >vP^l {SD  
    k2=k2 N3x}YHFF  
    k3=k3 K.X% Q,XD  
    k{@z87+&  
    end SxOM@A  
    vP,WV9Q1u  
    %有中间像,焦距输入为正数 [oKB1GkA  
    =#y&xWxL  
    function sjr=yfdre(~) |/p ^e  
    )'fIrBT  
    f=input('f:'); %>JqwMK  
    d1=input('d1:'); \u)s Zh  
    d2=input('d2:'); f5sk,Z  
    d3=input('d3:'); OW #pBeX99  
    r@ejU'uz  
    A=f^2/(d3*d2)-f/d1; P!]DV$o  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); D_, 2z  
    C=d3/d2-f/d1; _a.Q@A4'  
    j&44wuf  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); iqOd]H]v  
    a2=d3/(a1*f); Ge97e/ CY  
    b2=a1*(1-a2)*f/d2; aZBaIl6I  
    b1=(1-a1)*f/(d1*b2); y<;#*wB  
    }* BY!5  
    %曲率半径 nk-?$'i9q  
    i 5_g z>  
    R1=2*f/(b1*b2) xNm<` Y?  
    R2=2*a1*f/(b2*(1+b1)) B*?v`6  
    R3=2*a1*a2*f/(1+b2) 3J:!8Gmk  
    kM9E)uT>(<  
    A1=b2^3*(a1-1)*(1+b1)^3; Y$,]~Qzq  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; &}P62&  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; koAc-o  
    sS+9ly{9J  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); -M/ny-; `}  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 43P?f+IYrk  
    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); g(<@r2p  
    Xw![}L >  
    CB=[C1 B1;C2 B2]; *_^AK=i  
    AB=[A1 B1;A2 B2]; 0}w>8L7i{  
    AC=[A1 C1;A2 C2]; .|o7YTcR:  
    dc:|)bK M  
    %二次系数 o3uv"# C  
    P/ug'  
    k2=-(det(CB)/det(AB)); ?MN?.O9-  
    k3=-(det(AC)/det(AB)); "lUw{3  
    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 ? ZN8Ku  
    k2=k2 %=_ Iq\lC  
    k3=k3 *;4r|# LG  
    Nf3Kz#!B  
    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 JQ0KXS Nr