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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 %)1y AdG 8  
    function sjr=nfdre(~) 9jM}~XvV  
    C5o#i*|  
    %系统焦距及各镜间距输入,间距取负正负 'jWr<]3  
    *4\:8  
    f=input('f:'); xF!,IKlBBp  
    d1=input('d1:'); LBYMCY  
    d2=input('d2:'); +r2+X:#~T  
    d3=input('d3:'); f6hnTbJ  
    d,k!qjf=r  
    A=f^2/(d3*d2)-f/d1; (ZUHvvL  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); lXW%FH6c+  
    C=d3/d2-f/d1; gb[5&> (#  
    6m}Ev95  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 {$0mwAOH "  
    a2=d3/(a1*f);%α2 %O|iE M  
    b2=a1*(1-a2)*f/d2;%β2 C0T;![/4A  
    b1=(1-a1)*f/(d1*b2);%β1 Ni9/}bb  
    1m4$p2j  
    5-:?&|JK;  
    %曲率半径 -_=nDH  
    f,U.7E  
    R1=2*f/(b1*b2) \V;F/Zy(  
    R2=2*a1*f/(b2*(1+b1)) ZgTW.<.%2  
    R3=2*a1*a2*f/(1+b2) ` Fa~  
    $*^7iT4q_t  
    A1=b2^3*(a1-1)*(1+b1)^3; ]E5o1eeg  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; D+TD 95t  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 03$mYS_?  
    `V}q-Zdy  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); g];!&R-  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); p $S*dr  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); l!D}3jD  
    5'OrHk;u  
    CB=[C1 B1;C2 B2]; c[0}AG J  
    AB=[A1 B1;A2 B2]; Z@4Ar fl  
    AC=[A1 C1;A2 C2]; Vr3Zu{&2  
    lU8l}Ndz"  
    %非球面系数 =x/X:;)>  
    k2=-(det(CB)/det(AB)); g9 .Q<JwO  
    k3=-(det(AC)/det(AB)); \j$&DCv   
    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 8SMxw~9$  
    k2=k2 T^zXt?  
    k3=k3 X]ipI$'+C  
    /:cd\A}  
    end /2&c$9=1  
    9SX +  
    %有中间像,焦距输入为正数 #|uCgdi  
    \[;0 KV_  
    function sjr=yfdre(~) O %\*@4zM  
    nGC/R&  
    f=input('f:'); \;,_S+Fz8  
    d1=input('d1:'); pIc#L>{E  
    d2=input('d2:'); 3/P1!:g9  
    d3=input('d3:'); /4yo`  
    (Lbbc+1m  
    A=f^2/(d3*d2)-f/d1; &sl0W-;0  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ]=\].% >  
    C=d3/d2-f/d1; GV1pn) 4  
    lt/1f{v[:  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); #NQMy:JHD)  
    a2=d3/(a1*f); (^ J I%>  
    b2=a1*(1-a2)*f/d2; Pd8![Z3  
    b1=(1-a1)*f/(d1*b2); B`EJb71^Xy  
    x[cL Bc<  
    %曲率半径 A[B<~  
    kXViWOXU^  
    R1=2*f/(b1*b2) QFA8N  
    R2=2*a1*f/(b2*(1+b1)) :a!^   
    R3=2*a1*a2*f/(1+b2) pJ>P[  
    49eD1h3'X[  
    A1=b2^3*(a1-1)*(1+b1)^3; xw.A #Zb\_  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; aEB_#1  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; zOJ%}  
    \P[Y`LYL  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Kf3"Wf^q   
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); &w\{TZ{  
    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); Lw1Yvtn  
    <]ox;-56  
    CB=[C1 B1;C2 B2]; wH*-(*N "  
    AB=[A1 B1;A2 B2]; d z|or9&  
    AC=[A1 C1;A2 C2]; BT !^~S%w  
    1Yq!~8  
    %二次系数 b 1c y$I  
    9i:L&dN  
    k2=-(det(CB)/det(AB)); F~-(:7j  
    k3=-(det(AC)/det(AB)); @@f"%2ZR[  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 e1yt9@k,  
    k2=k2 +L$Xv  
    k3=k3 KR} ?H#%  
    /'SNw?&  
    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 <sbu;dQ`