[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 WQ<J<$$uu  
function sjr=nfdre(~) ~(aQ!!H6  
j` 9pZAF  
%系统焦距及各镜间距输入，间距取负正负 {v}f/cu  
Aj#CB.y  
f=input('f:'); S "R]i  
d1=input('d1:'); 5*xk8*  
d2=input('d2:'); V{p*N*  
d3=input('d3:'); r]sv50Fy  
SG2s!Ht  
A=f^2/(d3*d2)-f/d1; -LJbx<'  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); Igt:M[ /  
C=d3/d2-f/d1; ".O+";wk  
m:59f9WXA  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 2K'3ry)[y  
a2=d3/(a1*f);%α2 \C5YVl#  
b2=a1*(1-a2)*f/d2;%β2 8C#R  
b1=(1-a1)*f/(d1*b2);%β1 r9sW:cM:e  
?1K|.lr  
,`v)nwP  
%曲率半径 F=U3o=-:  
8 sZ~3  
R1=2*f/(b1*b2) UM(tM9  
R2=2*a1*f/(b2*(1+b1)) jxU1u"WU  
R3=2*a1*a2*f/(1+b2) O6NgI2[O  
"~0m_brf  
A1=b2^3*(a1-1)*(1+b1)^3; xAw$bJj~s  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; im_WTZz2P  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; U+F?b\  
Sg(fZ' -  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2);  iUJqAi1o  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); eEePK~%c  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2);  d!5C$C/x  
L}*:,&Y/  
CB=[C1 B1;C2 B2]; j-8v$0'  
AB=[A1 B1;A2 B2]; ml<X92Y  
AC=[A1 C1;A2 C2]; wN\%b}pp  
2`tdH|Z`  
%非球面系数 lG/M%i  
k2=-(det(CB)/det(AB)); 2F[smUL
 
k3=-(det(AC)/det(AB)); @,F8gv*  
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 9>\P]:  
k2=k2 "Kx2k>ym  
k3=k3 @.;] $N&J  
hqIYo .<  
end !'o5X]s  
0)`{]&  
%有中间像，焦距输入为正数 [`nY/g:  
o4,fwPkB  
function sjr=yfdre(~) YjN2 ,Xi  
wYQTG*&h  
f=input('f:'); s+&Ts|c#  
d1=input('d1:'); WL$nchS9  
d2=input('d2:'); P,r9<  
d3=input('d3:'); M9'Qs m  
~G|u
n}g=  
A=f^2/(d3*d2)-f/d1; 99w;Q 2k  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); LW<DhMV  
C=d3/d2-f/d1; S*-n%D0q5  
WJbdsPs  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); 64;F g/t  
a2=d3/(a1*f); V#["Z}  
b2=a1*(1-a2)*f/d2; z/6/   
b1=(1-a1)*f/(d1*b2); xP%`QTl\  
J0CEZ  
%曲率半径 l!CWE  
Bf33%I~  
R1=2*f/(b1*b2) }_93}e  
R2=2*a1*f/(b2*(1+b1)) 6REv(E]  
R3=2*a1*a2*f/(1+b2) F4'g}yOLd  
}1E'a>^|  
A1=b2^3*(a1-1)*(1+b1)^3; g"vg {Q  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 7OY<*ny  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ^HYmi\`  
[ fvip_Pt  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); jiQJ{yY  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ,_;+H*H>
"  
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); 'zCJ
K~x`x  
"D0:Y(\  
CB=[C1 B1;C2 B2]; n!=%MgF'*p  
AB=[A1 B1;A2 B2]; [5K&J-W  
AC=[A1 C1;A2 C2]; e=K2]Y Q{  
5N "fD{v{  
%二次系数 ?&$??r^i  
Y2709LWmP  
k2=-(det(CB)/det(AB)); F F<xsoZJ  
k3=-(det(AC)/det(AB)); VeK^hz R^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 q] g'rO'  
k2=k2 }%(e`[?1  
k3=k3 dYEF,\Z'  
<tkxE!xF`J  
end

谢谢分享，学习一下 {TT@Mkz_QC