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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 3Ul*QN{6  
    function sjr=nfdre(~) R5D1w+  
    8Wx=p#_  
    %系统焦距及各镜间距输入,间距取负正负 DrR@n~  
    r" y.KD^  
    f=input('f:'); L#J1b!D&<6  
    d1=input('d1:'); >j/w@Fj  
    d2=input('d2:'); paK2 xX8E  
    d3=input('d3:'); ]`K2 N  
    2 nCA<&  
    A=f^2/(d3*d2)-f/d1; wz%-%39q%  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); rH-23S  
    C=d3/d2-f/d1; \85i+q:LuA  
     )2.Si#  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 N['  .BN  
    a2=d3/(a1*f);%α2 yAt ^;  
    b2=a1*(1-a2)*f/d2;%β2 f8~_E  
    b1=(1-a1)*f/(d1*b2);%β1 ,prf;|e?  
    Xhm c6?  
    M  >u_4AY  
    %曲率半径 \dVOwr  
    CrLrw T  
    R1=2*f/(b1*b2) HtFDlvdy]  
    R2=2*a1*f/(b2*(1+b1)) DVA:Cmh\  
    R3=2*a1*a2*f/(1+b2) ~Y;*u]^  
    &8H'eAA  
    A1=b2^3*(a1-1)*(1+b1)^3; Cy e.gsCT  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 6Oq 7#3]  
    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{aN+  
    Da|z"I x  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); AH^/V}9H  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 80I#TA6C  
    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); rp$'L7lrX  
    Y4-t7UlS;  
    CB=[C1 B1;C2 B2]; hb$Ce'}N  
    AB=[A1 B1;A2 B2]; qPNR`%}Q  
    AC=[A1 C1;A2 C2]; _f83-':W6  
    Nb\4 /;#  
    %非球面系数 V0@=^Bls  
    k2=-(det(CB)/det(AB)); 2G7Wi!J  
    k3=-(det(AC)/det(AB)); .A|udZ,  
    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 'L'R9&o<X  
    k2=k2 qGo.WZ$  
    k3=k3 W1~0_;  
    c{|p.hd  
    end M%HU4pTW#o  
    23PGq%R  
    %有中间像,焦距输入为正数 dPlV>IM$z  
    @JMiO^  
    function sjr=yfdre(~) 3fj4%P"  
    ?,mmYW6TjB  
    f=input('f:'); o-5TC  
    d1=input('d1:'); b6bHTH0  
    d2=input('d2:'); /v{I  
    d3=input('d3:'); pBHRa?Y5  
    01]f2.5  
    A=f^2/(d3*d2)-f/d1; _6Sp QW  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); j#|ZP-=1_  
    C=d3/d2-f/d1; 4?kcv59  
    K;?+8(H  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); pK*TE5]  
    a2=d3/(a1*f); N~Jda o  
    b2=a1*(1-a2)*f/d2; Txu/{ M,  
    b1=(1-a1)*f/(d1*b2); $Sq:q0  
    |yCMt:Hk  
    %曲率半径 -?\D\\+t  
    O- wzz  
    R1=2*f/(b1*b2) )=+|i3]U  
    R2=2*a1*f/(b2*(1+b1)) Gc?a+T  
    R3=2*a1*a2*f/(1+b2) i-1op> Y  
    MgZ/(X E  
    A1=b2^3*(a1-1)*(1+b1)^3; 1 MFbQs^  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3;  wwqEl(  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 81F9uM0  
    =;L|gtH"  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Z,gk|M3.  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); M\j.8jG  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); /vt3>d%B;  
    z{q`GwW  
    CB=[C1 B1;C2 B2]; zLQx%Yg!  
    AB=[A1 B1;A2 B2]; *. t^MP  
    AC=[A1 C1;A2 C2]; ~%oR[B7=|  
    k$VlfQ'+  
    %二次系数 7V>M]  
    kh<2BOV  
    k2=-(det(CB)/det(AB)); C!gZN9-  
    k3=-(det(AC)/det(AB)); kJU2C=m@e2  
    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 P}iE+Z 3  
    k2=k2 !WlH'y-I  
    k3=k3  *CMx-_  
    ;uW FHc5@B  
    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 x-c"%Z|