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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 ?0 KiR?  
    function sjr=nfdre(~) 'qD'PLV  
    C}M0XW  
    %系统焦距及各镜间距输入,间距取负正负 !;xf>API  
    %8rr*l5  
    f=input('f:'); ,6y-.m7>  
    d1=input('d1:'); Mo:!jS~a(Z  
    d2=input('d2:'); @'k,\$/  
    d3=input('d3:'); : 9djMsd  
    k'WS"<-  
    A=f^2/(d3*d2)-f/d1; +j)-L \  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); |,M#8NOp:  
    C=d3/d2-f/d1; JO<gN= [  
    tW Cv]*  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 S?,KgMVM  
    a2=d3/(a1*f);%α2 DRKc&F6Qy  
    b2=a1*(1-a2)*f/d2;%β2 o}r!qL0c  
    b1=(1-a1)*f/(d1*b2);%β1 n;S0fg  
    Sh~ 8jEk  
    S+Y y  
    %曲率半径 WNF=NNO-R  
    )Bm^aMVl3  
    R1=2*f/(b1*b2) vMW-gk  
    R2=2*a1*f/(b2*(1+b1)) z$8e6*  
    R3=2*a1*a2*f/(1+b2) }R(0[0NQe-  
    sTYuwna~   
    A1=b2^3*(a1-1)*(1+b1)^3; dZ;~b(CA  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; gOES2 4$2  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ~PH1|h6  
    7Dx .;  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); g O\f:Pg  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); RJ`/qXL  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); pO7{3%  
    f4aD0.K.g|  
    CB=[C1 B1;C2 B2]; t<EX#_i,  
    AB=[A1 B1;A2 B2]; 7Da^Jv k  
    AC=[A1 C1;A2 C2]; gl(6m`a>  
    8YJqM,t5)  
    %非球面系数 [w#x5Xsn  
    k2=-(det(CB)/det(AB)); *fuGVA  
    k3=-(det(AC)/det(AB)); I;|5C=!  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 ,uqbS  
    k2=k2 /K Jx n6  
    k3=k3 ^JF_;~C  
    Um0<I)  
    end 7K5o" "  
    pFv[z':&Q  
    %有中间像,焦距输入为正数 |0vHy7CE  
    'k(~XA}X:  
    function sjr=yfdre(~) &oK/ ]lub  
    /iJcy:J  
    f=input('f:'); J?,!1V=  
    d1=input('d1:'); PUFW^"LV  
    d2=input('d2:'); exrt|A] _[  
    d3=input('d3:'); o"+ &^  
    {$QF*j  
    A=f^2/(d3*d2)-f/d1; IG3K Pmu  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); % &Q7;?  
    C=d3/d2-f/d1; 2M( PH]D  
    VkP:%-*#v  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); C6=;(=?C  
    a2=d3/(a1*f); krnk%ug  
    b2=a1*(1-a2)*f/d2; oe_[h]Hgl  
    b1=(1-a1)*f/(d1*b2); 8Q)mmkI\=  
    !A^w6Q;`V  
    %曲率半径 ?PxYS%D_L  
    *mhw5Z=!  
    R1=2*f/(b1*b2) r@@eC['  
    R2=2*a1*f/(b2*(1+b1)) KlX |PQ  
    R3=2*a1*a2*f/(1+b2) S bqM=I+  
    Jv{"R!e"P  
    A1=b2^3*(a1-1)*(1+b1)^3; ]2s Zu7  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; .Mft+,"  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; "62Ysapq+  
    z1KC$~{O  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); s? \9i6  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ^[?+=1 k  
    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); $X\` 7`v  
    )b2E/G@X&  
    CB=[C1 B1;C2 B2]; e !x-:F#4j  
    AB=[A1 B1;A2 B2]; n~>CE"q  
    AC=[A1 C1;A2 C2]; )1yUV*6  
    R O3e  
    %二次系数 &8YI)G%  
    yLa5tv/  
    k2=-(det(CB)/det(AB)); ,["|wqM  
    k3=-(det(AC)/det(AB)); SIBIh-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 -0J<R;cVs  
    k2=k2 BMsy}08dQ  
    k3=k3 T+`GOFx  
    Va[dZeoy  
    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 ( =/L#Yg_