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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 Q(@IK&v  
    function sjr=nfdre(~) I9y.e++/  
    ;</Lf=+Vm  
    %系统焦距及各镜间距输入,间距取负正负 4Yjx{5QSAG  
    m <k!^jp  
    f=input('f:'); F} DUEDND*  
    d1=input('d1:'); b"j|Bb  
    d2=input('d2:'); <:;^'x>!  
    d3=input('d3:'); *~VxC{  
    JBX[bx52<r  
    A=f^2/(d3*d2)-f/d1; &Ral+J  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); &jQ?v@|1c  
    C=d3/d2-f/d1; (?&=T.*^  
    ;[0&G6g  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 =hH.zrI6e  
    a2=d3/(a1*f);%α2 -8Ti*:  
    b2=a1*(1-a2)*f/d2;%β2 lC?Icn|o  
    b1=(1-a1)*f/(d1*b2);%β1 \FzM4-  
    & eZfQ27$  
    mxG]kqi  
    %曲率半径 h3G.EM:eG  
    7^e +  
    R1=2*f/(b1*b2) (!K_Fy@  
    R2=2*a1*f/(b2*(1+b1)) CnF |LTi  
    R3=2*a1*a2*f/(1+b2) MXh "Y*}  
    i76 Yo5  
    A1=b2^3*(a1-1)*(1+b1)^3; c+' =hR[  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; D&:yMp(  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Bx~[F  
    }xb=<  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 12`_;[37  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); udqS'g&  
    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); Sr.;GS5i  
    x8#ODuH  
    CB=[C1 B1;C2 B2]; &XLD S=j  
    AB=[A1 B1;A2 B2]; pd@;b5T  
    AC=[A1 C1;A2 C2]; 5F $V`kYT  
    ra7uU*  
    %非球面系数 ] Uc`J8p,  
    k2=-(det(CB)/det(AB)); R4's7k  
    k3=-(det(AC)/det(AB)); x%> e)L<  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 :Ao!ls' =  
    k2=k2 R MYP"  
    k3=k3 Kp iF0K  
    W0`Gc {  
    end -M5=r>1;  
    p='-\M74K  
    %有中间像,焦距输入为正数 *wbZ;rfF  
    A7XnHPIw  
    function sjr=yfdre(~) b3$k9dmxV+  
    0Fr1Ku!  
    f=input('f:'); ,d,\-x-+/  
    d1=input('d1:'); ] gb=  
    d2=input('d2:'); B0UJq./`  
    d3=input('d3:'); gQWd&)'muf  
    2(YPz|~W  
    A=f^2/(d3*d2)-f/d1; JcO08n  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); |1= !;.#  
    C=d3/d2-f/d1; "dh:-x6  
    q!,zq  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); d.<~&.-$  
    a2=d3/(a1*f); 4/> Our 5  
    b2=a1*(1-a2)*f/d2; D$e B ,~  
    b1=(1-a1)*f/(d1*b2); F1azZ (  
    <&!]K?Q9i  
    %曲率半径 ,K9f_bv  
    p#d+>7  
    R1=2*f/(b1*b2) 4g _"ku  
    R2=2*a1*f/(b2*(1+b1)) u V6g[J  
    R3=2*a1*a2*f/(1+b2) ,2[ra9n  
    Yn51U6_S  
    A1=b2^3*(a1-1)*(1+b1)^3; ffDc 6*.Q  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; i^z`"3#LE  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; !E\[SjY@J  
    Bo_ym36N  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); @!tVr3;N$  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); f>Td)s1 M  
    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,Z{ w"  
    Mio>{%/  
    CB=[C1 B1;C2 B2]; 1'w:`/_  
    AB=[A1 B1;A2 B2]; n(Y%Vmy  
    AC=[A1 C1;A2 C2]; " BTE  
    ak:v3cQR  
    %二次系数 WPuz]Ty  
    YhKZ|@  
    k2=-(det(CB)/det(AB)); y&T&1o  
    k3=-(det(AC)/det(AB)); ]n1dp2aH  
    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 US&B!Q:v  
    k2=k2 6)RbPPeE  
    k3=k3 ;]D(33) (  
    Tt# bg1  
    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 !F4;_A`X