[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]

%无中间像，焦距输入为负数 xr7}@rq"U<  
function sjr=nfdre(~) (RV#piM  
s1[&WDedM  
%系统焦距及各镜间距输入，间距取负正负 iC
/*d  
_+NjfF|  
f=input('f:'); r)>3YM5  
d1=input('d1:'); At?|[%<`  
d2=input('d2:'); , ?U)mYhI  
d3=input('d3:'); {H)hoAenA  
PfRA\  
A=f^2/(d3*d2)-f/d1; @uCi0Pt  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 1n[)({OQ  
C=d3/d2-f/d1; Nr~!5XO  
z<%bNnSO  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 z!O;s ep?/  
a2=d3/(a1*f);%α2 <%Nf"p{K  
b2=a1*(1-a2)*f/d2;%β2 _,)_(R ,h  
b1=(1-a1)*f/(d1*b2);%β1 d"06 gp  
iD G&Muc  
H-+U^@w  
%曲率半径 'z AvQm  
G)%V 3h  
R1=2*f/(b1*b2) 3qf#NJN}  
R2=2*a1*f/(b2*(1+b1)) j?m(l,YD|*  
R3=2*a1*a2*f/(1+b2) S.~L[iLc  
BVNJas  
A1=b2^3*(a1-1)*(1+b1)^3; N6"sXwm  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; $f0u  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; {)lZfj}l  
; F'IS/ttX  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); z$R&u=J  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); j8p<HE51  
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); } &+]UGv  
-Pp{aFe  
CB=[C1 B1;C2 B2]; 5z,q~CU  
AB=[A1 B1;A2 B2]; VsL,t\67  
AC=[A1 C1;A2 C2]; jI7 x<=  
cB){b'WJ  
%非球面系数 :ig=zETM  
k2=-(det(CB)/det(AB)); [|oG}'Xz  
k3=-(det(AC)/det(AB)); ?
.Vuet  
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 /y{:N  
k2=k2 JX&]>#6|E  
k3=k3 v#&;z_I+  
~6n|GxR.[  
end A{T9-f@X  
@I?,!3`jS  
%有中间像，焦距输入为正数 7>y]uT@ar  
4"kc(J`c  
function sjr=yfdre(~) XjN=UhC  
Z9$pY=8^?  
f=input('f:'); JI]Lz1i  
d1=input('d1:'); ftTD-d  
d2=input('d2:'); 81x/bx@L%  
d3=input('d3:'); e:nByzdH0[  
hRX9Du`$  
A=f^2/(d3*d2)-f/d1; y,`n9[$K\  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); #~nXAs]Q  
C=d3/d2-f/d1; Ve%ua]qA  
~Ze!F"  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); yZ,pH1  
a2=d3/(a1*f); S8dfe~|7:  
b2=a1*(1-a2)*f/d2; .8^mA1fmX  
b1=(1-a1)*f/(d1*b2); J{dO0!7y  
8%m\J:eR  
%曲率半径 8iaMr278W  
lqOpADLS3  
R1=2*f/(b1*b2) wi7Br&bGi  
R2=2*a1*f/(b2*(1+b1)) T90O.]S  
R3=2*a1*a2*f/(1+b2) eUQmW^  
8A&N+sT  
A1=b2^3*(a1-1)*(1+b1)^3; 2[`n<R\  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; }||p#R@?  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; -@>]iBl  
;%2+Tc-7I  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); g]L8Jli  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); *uRDB9#9,  
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); 1gK^x^l*f  
5*Zz_ .  
CB=[C1 B1;C2 B2]; 'XKfKv >;  
AB=[A1 B1;A2 B2]; WuY#Kx~2  
AC=[A1 C1;A2 C2]; !Z2h?..O  
? bWc<]  
%二次系数 #Y7iJPO  
h SS9mQ  
k2=-(det(CB)/det(AB)); s3JzYDpy  
k3=-(det(AC)/det(AB)); ZrEou}z(*  
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 ]I,&Bme  
k2=k2 cI8\d 4/py  
k3=k3 ,^c-}`!K  
7=fNvES2  
end

谢谢分享，学习一下 +sn0bi/rG