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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 \<|a>{`7]i  
    function sjr=nfdre(~) ,]Yjo>`tW  
    K4SR`Q  
    %系统焦距及各镜间距输入,间距取负正负 hJ4==ILx  
    JfKhYRl  
    f=input('f:'); OdgfvHDgW  
    d1=input('d1:'); RyD2LAf)J  
    d2=input('d2:'); WhE5u&`  
    d3=input('d3:'); j)Kk:BFFY  
    7}Z.g9<  
    A=f^2/(d3*d2)-f/d1; Q nZR  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); $>'}6?C.  
    C=d3/d2-f/d1; Z]$yuM  
    %s$_KG!&  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 \F,?ptu  
    a2=d3/(a1*f);%α2 nhk +9  
    b2=a1*(1-a2)*f/d2;%β2 z QoMHFL3  
    b1=(1-a1)*f/(d1*b2);%β1 \^EjE  
    C JiMg'K  
    Bx E1Ky8@A  
    %曲率半径 lO%Z4V_Mj  
    [=e61Z  
    R1=2*f/(b1*b2) uI%h$  
    R2=2*a1*f/(b2*(1+b1)) ; 5my(J*b  
    R3=2*a1*a2*f/(1+b2) /.'1i4Xa1P  
    8v1asFxs.  
    A1=b2^3*(a1-1)*(1+b1)^3; cgYMo{R3  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; [HF)d#A  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; la)f\Nk  
    r-]R4#z>  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); C,]Q/6'>  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Ul@ZCv+  
    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); 9foQ0#R  
    i/O!bq[o  
    CB=[C1 B1;C2 B2]; )%X\5]w`  
    AB=[A1 B1;A2 B2]; )~d2`1zGS  
    AC=[A1 C1;A2 C2]; C'n 9n!hR  
    3I:DL#f  
    %非球面系数 TW3:Y\p  
    k2=-(det(CB)/det(AB)); PG<N\  
    k3=-(det(AC)/det(AB)); H=X>o.iVqi  
    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 VmBLNM?  
    k2=k2 V$q%=Sip  
    k3=k3 Y141Twjvd  
    $ }B"u;:SU  
    end 6J%SkuxR  
    68I4MZK>4  
    %有中间像,焦距输入为正数 dUb(C1h  
    6ap,XFRMh  
    function sjr=yfdre(~) uO[4 WZ  
    k% In   
    f=input('f:'); M*c\=(  
    d1=input('d1:'); L@a-"(TN+  
    d2=input('d2:'); "45BOw&72G  
    d3=input('d3:'); i":-g"d  
    M?nnpO  
    A=f^2/(d3*d2)-f/d1; GrM~ %ng  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); #Q'i/|g   
    C=d3/d2-f/d1; ~PlwPvWo  
    MIR17%G  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); }ZYK3F  
    a2=d3/(a1*f); 5V?1/  
    b2=a1*(1-a2)*f/d2; oR-_=U^  
    b1=(1-a1)*f/(d1*b2); ,\=u(Y\I[  
    }FM<uBKW  
    %曲率半径 (O`=$e  
    u'32nf?  
    R1=2*f/(b1*b2) -3 W 4  
    R2=2*a1*f/(b2*(1+b1)) l}O`cC  
    R3=2*a1*a2*f/(1+b2) uxh4nyE  
    YRYrR|I  
    A1=b2^3*(a1-1)*(1+b1)^3; b7>;UX  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 7#g C(&\A  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; er qm=)  
    ZV Gw@3  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 6]rrj  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); h M8G"b  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); Q'|cOQX  
    U{O\  
    CB=[C1 B1;C2 B2]; (g4g-"rc  
    AB=[A1 B1;A2 B2]; vfy- ;R(  
    AC=[A1 C1;A2 C2]; V_ ]4UE  
    %^5$=w  
    %二次系数 3BSeZ:j7  
    3Q;^X(Ml*  
    k2=-(det(CB)/det(AB)); ]&r/H17  
    k3=-(det(AC)/det(AB)); ($cu!$lY~  
    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 F2:7UNy,  
    k2=k2 n `n3[  
    k3=k3 +s S*EvF  
    tNUcmiY  
    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 /f6]XP\'`+