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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 ~NYy@l   
    function sjr=nfdre(~) OF^:_%c/  
    9C?;'  
    %系统焦距及各镜间距输入,间距取负正负 *>x~`  
    g<,kV(_7  
    f=input('f:'); X2avo|6e  
    d1=input('d1:'); so~vnSQ!x  
    d2=input('d2:'); Rj&V~or  
    d3=input('d3:'); qd@x#"qT  
    :JBvCyj4PE  
    A=f^2/(d3*d2)-f/d1; (DzV3/+p^  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); Tk[`kmb  
    C=d3/d2-f/d1; s bf\;_!  
    "i+fO&LpZ  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 %kM|Hk3d  
    a2=d3/(a1*f);%α2 N1dp%b9W(  
    b2=a1*(1-a2)*f/d2;%β2 qA4w*{JN  
    b1=(1-a1)*f/(d1*b2);%β1 U%2[,c_  
    h{s- e.  
    }O+F#/6  
    %曲率半径 %i!&Fr  
    x.Sq2rw]V  
    R1=2*f/(b1*b2) EeW%5/;  
    R2=2*a1*f/(b2*(1+b1)) ET ;=o+\d  
    R3=2*a1*a2*f/(1+b2) Q fI =  
    |Qq_;x]  
    A1=b2^3*(a1-1)*(1+b1)^3; c< ke)@  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; lM1Y }  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 1aC ?*,e?  
    o $k1&hyH  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); M" |Mte  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); TCW[;d  
    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); P I gbeP  
    yFp8 >  
    CB=[C1 B1;C2 B2]; jl# )CEx  
    AB=[A1 B1;A2 B2]; %xE9vN;  
    AC=[A1 C1;A2 C2]; :Oz! M&Ov  
    F1skI _!  
    %非球面系数 #!,tId  
    k2=-(det(CB)/det(AB)); H-gq0+,yE  
    k3=-(det(AC)/det(AB)); Z4U8~i  
    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 W~ 6ii\  
    k2=k2 p 4k*vuu>  
    k3=k3 F\1{bN|3  
    a8K"Z-LlQ  
    end <^}{sdOyu  
    \ "193CW!  
    %有中间像,焦距输入为正数 ]=5nC)|  
    Z!Y ^iN  
    function sjr=yfdre(~) '5V2{k$4U  
    2=pVX  
    f=input('f:'); cwK 6$Ax  
    d1=input('d1:'); =;(wBj  
    d2=input('d2:'); KNtsz[#b  
    d3=input('d3:'); K8 Y/sHl  
    !^ko"^p  
    A=f^2/(d3*d2)-f/d1; 8 Zy`Z  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); u@v0I$  
    C=d3/d2-f/d1; Yrb[:;Y  
    Y }*[Krw  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); xviz{M9g  
    a2=d3/(a1*f); t\2Lo7[Pu  
    b2=a1*(1-a2)*f/d2; {}ks[%,_\  
    b1=(1-a1)*f/(d1*b2); HbWl:yU  
    SWujj,-[  
    %曲率半径 > <WR]`G  
    a%2r]:?^?  
    R1=2*f/(b1*b2) Fwn4c4-%  
    R2=2*a1*f/(b2*(1+b1)) #8.%YG  
    R3=2*a1*a2*f/(1+b2) 0( fN  
    !Kv.v7'N/k  
    A1=b2^3*(a1-1)*(1+b1)^3; "g7`Ytln  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; sMh3IL9(*  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; j'lfH6_')e  
    !2oe;q2X[G  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); OyVdQ".  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 3RpDIl`0  
    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); p;av63 i  
    ?`?"j<4e  
    CB=[C1 B1;C2 B2]; DJSSc  
    AB=[A1 B1;A2 B2]; yE3g0@*  
    AC=[A1 C1;A2 C2]; c!@g<<}[(  
    [G{{f  
    %二次系数 _iBNy   
    j>s> i  
    k2=-(det(CB)/det(AB)); !( xeDX  
    k3=-(det(AC)/det(AB)); >U@7xeK  
    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 ^` N+mlh  
    k2=k2 N_TWT&o4  
    k3=k3 cPe0o'`[  
    [4,=%ez  
    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 ks=j v: