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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 s`8M%ZLu  
    function sjr=nfdre(~) <Dt /Rad  
    XBfiaj  
    %系统焦距及各镜间距输入,间距取负正负 es.\e.HK  
    "TBQNWZ  
    f=input('f:'); 33#7U+~]@  
    d1=input('d1:'); Ft%TnEp  
    d2=input('d2:'); uPv;y!Lsa@  
    d3=input('d3:'); 5XSxQG@k^z  
    W2r6jm!  
    A=f^2/(d3*d2)-f/d1; !1a|5 xrn  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); eZN3H"H  
    C=d3/d2-f/d1; A@@)lD.  
    O?C-nw6kP  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 RE`J"&  
    a2=d3/(a1*f);%α2 `}k&HRn  
    b2=a1*(1-a2)*f/d2;%β2 q G :jnl  
    b1=(1-a1)*f/(d1*b2);%β1 ^*cMry  
    Q.pEUDq/  
    X/`#5<x  
    %曲率半径 `_J^g&y~  
    .R$+#_  
    R1=2*f/(b1*b2) WTV3p,;6a  
    R2=2*a1*f/(b2*(1+b1))  Vq .!(x  
    R3=2*a1*a2*f/(1+b2) <FcPxZ  
    66^1&D"  
    A1=b2^3*(a1-1)*(1+b1)^3; s3MMICRT.  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; zJG x5JC  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; gCk y(4  
    F(KH-  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); .q_uJ_qu-  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); dPH! V6r  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); dVsAX(  
    >?G|Yz*kEJ  
    CB=[C1 B1;C2 B2]; .UT,lqEkv  
    AB=[A1 B1;A2 B2]; JGSk4  
    AC=[A1 C1;A2 C2]; /+<%,c$n  
    }QWTPRn  
    %非球面系数 L!8 -:)0b  
    k2=-(det(CB)/det(AB)); ,IT)zCpaBP  
    k3=-(det(AC)/det(AB)); I9*BENkR  
    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 U<g UX07  
    k2=k2 Y6;0khp  
    k3=k3 A<YZBR_  
    D)O6| DiO  
    end Zeme`/aBb  
    ]df9'\  
    %有中间像,焦距输入为正数 {x&jh|f`g  
    282+1X  
    function sjr=yfdre(~) 80s~ae;  
    B(en5|  
    f=input('f:'); I7G\X#,iz  
    d1=input('d1:'); ohc/.5Kl  
    d2=input('d2:'); wCq)w=,  
    d3=input('d3:'); d5sGkR`(  
    ziLr }/tg  
    A=f^2/(d3*d2)-f/d1; 3h D2C'KD  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); U&w 5&W{F}  
    C=d3/d2-f/d1; MOqA$b  
    CJ}@R.Zy  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); ?9('o\N:  
    a2=d3/(a1*f); OO !S w  
    b2=a1*(1-a2)*f/d2; \6`%NhkM_  
    b1=(1-a1)*f/(d1*b2); DETajf/<F  
    Iu1Sj`A  
    %曲率半径 }lNuf u  
    ,8J*S  
    R1=2*f/(b1*b2)  ,3@15j  
    R2=2*a1*f/(b2*(1+b1)) #o r7T^  
    R3=2*a1*a2*f/(1+b2) 7u`}t83a  
    *v:,rh  
    A1=b2^3*(a1-1)*(1+b1)^3; ?^yh5   
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; jC/JiI  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; od5w9E.  
    Y8`))MeD  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); E? m#S  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Cj4b]*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); #o"HD6e  
    Z;~E+dXC  
    CB=[C1 B1;C2 B2]; |{ /O)3  
    AB=[A1 B1;A2 B2];  Sj{rvW  
    AC=[A1 C1;A2 C2]; p\]LEP\z,  
    9Pob|UA  
    %二次系数 yz2oS|0'  
    li_pM!dWU_  
    k2=-(det(CB)/det(AB)); m"|(w`n]E+  
    k3=-(det(AC)/det(AB)); zIYr0k*%  
    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_g/e1A3  
    k2=k2 X_sG6Q@  
    k3=k3 {u_k\m[Y  
    #`#aSqGmc  
    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 Q0"F> %Cn