切换到宽版
  • 广告投放
  • 稿件投递
  • 繁體中文
    • 3041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
    661
    光币
    2655
    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 _I8-0DnOM  
    function sjr=nfdre(~) Fn[~5/  
    s &.Z;X  
    %系统焦距及各镜间距输入,间距取负正负 qL'3MY.!  
    0I#<-9&d-  
    f=input('f:'); 8vT:icl  
    d1=input('d1:'); r p @=  
    d2=input('d2:'); #5H@/o8!s=  
    d3=input('d3:'); ;JZXSM-3  
    D>|:f-Z6Z  
    A=f^2/(d3*d2)-f/d1; PN'8"8`{  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); &a|oJ'clz  
    C=d3/d2-f/d1; #-@{rgH  
    {'b8;x8h  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 SHGO;  
    a2=d3/(a1*f);%α2 K[ \z'9Q  
    b2=a1*(1-a2)*f/d2;%β2 kqyMrZ#  
    b1=(1-a1)*f/(d1*b2);%β1 |SxEJ  
    kEXcEF_9P  
    "(xS  
    %曲率半径 9 V"j=1B}  
    mi-\PD>X  
    R1=2*f/(b1*b2) "~[Rwh?  
    R2=2*a1*f/(b2*(1+b1)) Qb|dp~K.M  
    R3=2*a1*a2*f/(1+b2) c3}}cFe  
    sbs"26IE  
    A1=b2^3*(a1-1)*(1+b1)^3; S1+#qs {5a  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; i]YQq!B  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; b9Y pUm7#  
    }Sh-4:-D  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 2Z97Tq  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); tS9m8(Hr%Q  
    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); $` oA$E3  
    srSTQ\l4  
    CB=[C1 B1;C2 B2]; 1]<!Xuk^f  
    AB=[A1 B1;A2 B2]; B.WJ6.DkS  
    AC=[A1 C1;A2 C2]; {c1qC zM4  
    oF>GWst TR  
    %非球面系数 q-RGplx  
    k2=-(det(CB)/det(AB)); %*gO<U4L]  
    k3=-(det(AC)/det(AB)); zm"\D vN)  
    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 [yyV`&  
    k2=k2 vr  vzV  
    k3=k3 68)^i"DM<  
    3K{8sFDO  
    end &Ch)SD  
    9=o b:  
    %有中间像,焦距输入为正数 7^iAc6QSy3  
    l<HRD  
    function sjr=yfdre(~) ' EDi6  
    b1#=q0Zl  
    f=input('f:'); $"i690  
    d1=input('d1:'); K+}Z6_:  
    d2=input('d2:'); toWmm(7v  
    d3=input('d3:'); saT9%?4-  
     n=&c5!  
    A=f^2/(d3*d2)-f/d1; [v`4OQF/  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); cDx^}N!  
    C=d3/d2-f/d1; :9#`| #uh  
    |W <:rT  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); zfZDtKq  
    a2=d3/(a1*f); n1t(ns|  
    b2=a1*(1-a2)*f/d2; ypd?mw&1}  
    b1=(1-a1)*f/(d1*b2); !BX62j\?  
    3wE8y&  
    %曲率半径 `#f=&S?k  
    =l(JJ  
    R1=2*f/(b1*b2) cOb%SC[A{  
    R2=2*a1*f/(b2*(1+b1)) c{Kl?0#[  
    R3=2*a1*a2*f/(1+b2) eTc`FXw`  
    E8i:ER $$7  
    A1=b2^3*(a1-1)*(1+b1)^3; Wa(S20y F  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; CwvNxH#LVu  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; =#tQIhX`  
    kut|A  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); TJpv"V  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); u\xm8}A  
    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); Q.1ohj0)  
    X2[cR;;'  
    CB=[C1 B1;C2 B2]; f 99PwE(=  
    AB=[A1 B1;A2 B2]; F4Uk+|]Bu  
    AC=[A1 C1;A2 C2]; {wP|b@(1t  
    As|/ O7%  
    %二次系数 Z-|C{1}A  
    .LhmYbQ2WE  
    k2=-(det(CB)/det(AB)); }^[@m#  
    k3=-(det(AC)/det(AB)); CK(ev*@\D,  
    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 L_mqC(vn  
    k2=k2 2-0cB$W+  
    k3=k3 }NCvaO  
    ?vFh)U  
    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 f9UaAdJ(