切换到宽版
  • 广告投放
  • 稿件投递
  • 繁體中文
    • 1951阅读
    • 1回复

    [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]

    上一主题 下一主题
    离线songshaoman
     
    发帖
    652
    光币
    2557
    光券
    0
    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 %y+j~]^:  
    function sjr=nfdre(~) /2cOZ1G;  
    W78-'c  
    %系统焦距及各镜间距输入,间距取负正负 jzMGRN/67  
    Q3Lqj2r  
    f=input('f:'); @$G K<jl  
    d1=input('d1:'); "ZW*O{  
    d2=input('d2:'); !W^II>Y  
    d3=input('d3:'); GefgOlg5"  
    }cmL{S  
    A=f^2/(d3*d2)-f/d1; ]!w52kF7  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); I =pdjD  
    C=d3/d2-f/d1; qChPT:a  
    H/f= 2b  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 )dcGV$4t[  
    a2=d3/(a1*f);%α2 DLZ63'  
    b2=a1*(1-a2)*f/d2;%β2 Mla,"~4D5  
    b1=(1-a1)*f/(d1*b2);%β1 !<ucwWY,  
    ']x]X ,  
    0@;E8^pa  
    %曲率半径 :Fl:bRH+  
    v<4zcMv  
    R1=2*f/(b1*b2) 4V5*6O9(u  
    R2=2*a1*f/(b2*(1+b1)) Ye\%o[X  
    R3=2*a1*a2*f/(1+b2) q~C6+  
    [ ]NAV  
    A1=b2^3*(a1-1)*(1+b1)^3; |kD?^Nx  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Emo]I[<&q  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; M Cz3RZK  
    /hHD\+0({  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 7 m%|TwJN  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); f,*e?9@;s  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); "]% L{a P  
    Sw~<W%! ?  
    CB=[C1 B1;C2 B2]; 3? "GH1e  
    AB=[A1 B1;A2 B2]; }|;n[+}  
    AC=[A1 C1;A2 C2]; EP @=i  
    BScysoeD  
    %非球面系数 )xU70:X  
    k2=-(det(CB)/det(AB)); &E~7ty'  
    k3=-(det(AC)/det(AB)); 8u4]@tJH  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 uG<+IT|x  
    k2=k2 VrDvd  
    k3=k3 wTLHg2'y^  
    |Ia46YS  
    end !&X}? NK  
    p 7sYgz  
    %有中间像,焦距输入为正数 (jXgJ" m  
    X~x]VKr/  
    function sjr=yfdre(~) F4@h} T5)  
    G[jCmkK  
    f=input('f:'); nBQG.3  
    d1=input('d1:'); [KJm&\evp  
    d2=input('d2:'); mu(S 9  
    d3=input('d3:'); n '&WIf3  
    sUA)I%Q!  
    A=f^2/(d3*d2)-f/d1;  ,}bC  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 4aGHks8Z,\  
    C=d3/d2-f/d1; *Owq_)_ (|  
    sI`Lsd'V  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); V&gUxS]*  
    a2=d3/(a1*f); -.xs=NwB.|  
    b2=a1*(1-a2)*f/d2; YoXXelO&  
    b1=(1-a1)*f/(d1*b2); ^ c:(HUo#  
    5}1cNp6@  
    %曲率半径 X)Kd'6zg  
    >2_J(vm>  
    R1=2*f/(b1*b2) [z?q -$#  
    R2=2*a1*f/(b2*(1+b1)) H -`7T;t~  
    R3=2*a1*a2*f/(1+b2) :.@gd7T  
    h:l\kr|9  
    A1=b2^3*(a1-1)*(1+b1)^3; m'Wz0b^BO  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; =3sBWDB[  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 7Rq;V=2YV  
    Mk7#qiPo  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); RXh/[t+  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); S&9{kt|BI  
    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); .aF+>#V=Q  
    2 rne=L  
    CB=[C1 B1;C2 B2]; U$qSMkj6RK  
    AB=[A1 B1;A2 B2]; 1}9@aKM  
    AC=[A1 C1;A2 C2]; 'Exj|Y&  
    J-hP4t&x  
    %二次系数 |0dmdrKD  
    x# 0(CcKK  
    k2=-(det(CB)/det(AB)); Ly lw('zZ  
    k3=-(det(AC)/det(AB)); ?HttqK)  
    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 2?#IwT'  
    k2=k2 _yumUk-QW  
    k3=k3 0d~>zKho  
    j $a,93P5  
    end
     
    分享到
    离线doushan
    发帖
    14
    光币
    0
    光券
    0
    只看该作者 1楼 发表于: 2023-03-01
    谢谢分享,学习一下 e}D#vPaSY