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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 ~B? Wg!  
    function sjr=nfdre(~) ^SW9J^9  
    wp5H|ctl  
    %系统焦距及各镜间距输入,间距取负正负 2?v }w<Ydl  
    XHOS"o$y  
    f=input('f:'); BjA$^i|8  
    d1=input('d1:'); #&fu"W+D96  
    d2=input('d2:'); @,-D P41g  
    d3=input('d3:'); VE1j2=3+o  
    8j :=D!S  
    A=f^2/(d3*d2)-f/d1; z.?slYe[  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); @A<~bod  
    C=d3/d2-f/d1; ^dsj1#3z  
    EJQT\c  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 I_kA!^  
    a2=d3/(a1*f);%α2 L GVy4D  
    b2=a1*(1-a2)*f/d2;%β2 %Pj}  
    b1=(1-a1)*f/(d1*b2);%β1 Zb|a\z8?  
    ,nGQVb   
    ^]~!:Ej0  
    %曲率半径 ET 0(/Zz  
    jA[")RVG  
    R1=2*f/(b1*b2) Zm7, O8  
    R2=2*a1*f/(b2*(1+b1)) g5u4|+70  
    R3=2*a1*a2*f/(1+b2) D*?LcxX  
    JNJ6HyCU  
    A1=b2^3*(a1-1)*(1+b1)^3; mEkYT  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; }$r]\v  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 4HX;9HPHE<  
    =dQ/^C_hj  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); hE$3l+  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); x25zk4-  
    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); Df:/r%  
    zK{}   
    CB=[C1 B1;C2 B2];  Zy8tI#  
    AB=[A1 B1;A2 B2]; <h}x7y?  
    AC=[A1 C1;A2 C2]; mZmEE2h  
    s.n:;8RibP  
    %非球面系数 `&xdSH  
    k2=-(det(CB)/det(AB)); +Ar4X-A{y  
    k3=-(det(AC)/det(AB)); @Y>PtA&w*  
    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 n2Mpo\2  
    k2=k2 }gB^C3b6  
    k3=k3 %y*'bS  
    $b2~H+u(  
    end V0&7MY*  
    Y6d~hLC  
    %有中间像,焦距输入为正数 LnN:;h  
    $#3[Z;\  
    function sjr=yfdre(~)  H{Lt,#  
    7Kb&BF|Q  
    f=input('f:'); w"#rwV&  
    d1=input('d1:'); -S&9"=v  
    d2=input('d2:'); _#+l?\u  
    d3=input('d3:'); |W@Ko%om  
    wL^x9O|`p9  
    A=f^2/(d3*d2)-f/d1; CdPQhv)m  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 0uPcEpIA  
    C=d3/d2-f/d1; @ 'N $5  
    SW+;%+`  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); p9mGiK4!  
    a2=d3/(a1*f); &0:Gj3`  
    b2=a1*(1-a2)*f/d2; UvB\kIH  
    b1=(1-a1)*f/(d1*b2); >i.$s  
    dLwP7#r  
    %曲率半径 (n jTS+?  
    pv&iJ7RN  
    R1=2*f/(b1*b2) #F2DEo^0  
    R2=2*a1*f/(b2*(1+b1)) QZa^Cng~  
    R3=2*a1*a2*f/(1+b2) d(R8^v/L  
    h4MBw=Tz~  
    A1=b2^3*(a1-1)*(1+b1)^3; @~N"MsF3  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; )1R[X!KQ7  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; I,,SR"  
    A )CsF  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); X*d!A >s  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); :?m"kh ~  
    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); Eb63O  
    WX_g  
    CB=[C1 B1;C2 B2]; "{H{-`Ni  
    AB=[A1 B1;A2 B2]; Yl$ @/xAa  
    AC=[A1 C1;A2 C2]; 1p&=tN  
    >r,z^]-  
    %二次系数 m39.j:BG5  
    W$J.B!O  
    k2=-(det(CB)/det(AB)); KV9~L`=]i  
    k3=-(det(AC)/det(AB)); a>,_o(]cW  
    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 /Dt:4{aTOC  
    k2=k2 [Fk|m1i!  
    k3=k3 >TawJ"q-6R  
    u(? U[pe[  
    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 \ :q@I]2