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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 @B9O*x+n:  
    function sjr=nfdre(~) <Gj]XAoe%  
    fRK=y+gl@  
    %系统焦距及各镜间距输入,间距取负正负 @ S)p{T5G  
    <tgfbY^nL  
    f=input('f:'); 1k!$#1d<  
    d1=input('d1:'); D:0?u_[W  
    d2=input('d2:'); ge|Cv v  
    d3=input('d3:'); Vo(>K34  
    ![ @i+hl  
    A=f^2/(d3*d2)-f/d1; q85 4k+C  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 3B5 `Y  
    C=d3/d2-f/d1; Z> <,t~o}  
    R47tg&k6[  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 /E^j}H{  
    a2=d3/(a1*f);%α2 m+3]RIr&A  
    b2=a1*(1-a2)*f/d2;%β2 HE@P<  
    b1=(1-a1)*f/(d1*b2);%β1 =VGRM#+D  
    PMZ*ECIJU  
    bo[[<j!"I  
    %曲率半径 `P jS  
    |vY|jaV}  
    R1=2*f/(b1*b2) C j:  
    R2=2*a1*f/(b2*(1+b1)) :Fdk`aC  
    R3=2*a1*a2*f/(1+b2) "O<TNSbrC  
    d=/a{lP\  
    A1=b2^3*(a1-1)*(1+b1)^3; #C+7~ns'  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; !yu-MpeG  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; e_dsBmTh  
    nmoC(| r  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); q(Zu;ecBN  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); k(dNHT  
    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); FB n . 4  
    ~QU\kZ7Z  
    CB=[C1 B1;C2 B2]; }0 =gP?.kE  
    AB=[A1 B1;A2 B2]; oB%j3aAH  
    AC=[A1 C1;A2 C2]; .83z =  
     EHda  
    %非球面系数 2o1 RJk9  
    k2=-(det(CB)/det(AB)); M [6WcH0/T  
    k3=-(det(AC)/det(AB)); !CLL{\F  
    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 tNYCyw{K  
    k2=k2 >/7[HhBT  
    k3=k3 85#+_}#  
    &NK6U  
    end g&O!w!T  
    4n/CS AT1  
    %有中间像,焦距输入为正数 o_?A^u  
    M~-jPY,+  
    function sjr=yfdre(~) ;xjw'%n,  
    *7K)J8kq  
    f=input('f:'); gF&HJF 0x  
    d1=input('d1:'); H~~>ut6`  
    d2=input('d2:'); e`;U9Z  
    d3=input('d3:'); e! 0Y`lQ  
    #RP7?yGM,  
    A=f^2/(d3*d2)-f/d1; -_fh=}.n+"  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); B8 R&Q8Q  
    C=d3/d2-f/d1; Jl{g"N{2u'  
    fe7DS)U  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); ]`\~(*;[W9  
    a2=d3/(a1*f); #& &  
    b2=a1*(1-a2)*f/d2; d5 U+]g  
    b1=(1-a1)*f/(d1*b2); F/U38[  
    eG%Q 3h  
    %曲率半径 ;(;{~1~  
    ="s>lI-1a  
    R1=2*f/(b1*b2) #i| AE`  
    R2=2*a1*f/(b2*(1+b1)) e18}`<tW-  
    R3=2*a1*a2*f/(1+b2) FWuk@t[<O  
    *!L it:H  
    A1=b2^3*(a1-1)*(1+b1)^3; k>!A~gfP~  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; P~u~`eH*  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; a'YK1QX  
    {KM5pK?,BJ  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); _rfGn,@BH  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ( 2i{8  
    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); I] 0 D*z  
    ~\[\S!"  
    CB=[C1 B1;C2 B2]; Q5/BEUkC  
    AB=[A1 B1;A2 B2]; T |ZJ$E0  
    AC=[A1 C1;A2 C2]; {%XDr,myd  
    Ib0@,yS[  
    %二次系数 992cy2,Fb  
    KU)~p"0[6]  
    k2=-(det(CB)/det(AB)); TW~9<c  
    k3=-(det(AC)/det(AB)); Qj(|uGqm3  
    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 <#No t1R  
    k2=k2 FdM xw*}  
    k3=k3 !F~*Q2PZ9  
    _J -3{a  
    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 H?W8_XiN