資源簡介
完成對圖像的逆透視映射變換,需要給出攝像頭的高度,視角和最終圖片的橫向?qū)挾龋祝R坏┙o定參數(shù)準(zhǔn)確,即可得到效果很好的逆透視映射變換圖片。
代碼片段和文件信息
A=imread(‘IMAG0652.jpg‘);
figure;
if?isgray(A)==0;
????A=rgb2gray(A);
end
imshow(A);
[m?n]=size(A);
B=A;
tem=(m/3/n);%保留角度
for?i=1:n
????for?j=1:m
????????if?(i<(n/2+1)?&&?(-j+m-m/2)>(tem*(i-n/2)))?||?(i>(n/2)?&&?(-j+m-m/2)>(-tem*(i-n/2)))
%?????????if?j ????????????B(ji)=0;
????????end
????end
end
figure
imshow(B);
after_row=2000;
after_col=1000;
C=zeros(after_rowafter_col);
h=1;%垂直高度
d=0;%橫向偏移
l=0;%縱向偏移
alpha=25*pi/180;%垂直視角
beta=20*pi/180;%水平視角
theta=pi/2-atan(0.45/4);%垂直俯角?
gama=0;%水平傾角
row=1:m;
col=1:n;
zoom=400;
for?j=round(m/2):m
????for?i=1:n
????????if(B(ji)~=0)
????????????v_inv=(i-1-n/2);
????????????u_inv=(-(j-1)+m/2);%coordinate?trans
????????????
????????????x=h*(m*tan(theta)+2*u_inv*tan(alpha))/(m-2*tan(theta)*tan(alpha)*u_i
?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----
?????文件???????1280??2012-05-22?16:08??fun3_ipm2.m
?????文件???????1079??2012-04-12?09:28??fun_ipm.m
?????文件????2076381??2011-12-29?07:23??IMAG0582.jpg
?????文件????1194096??2012-03-06?20:00??IMAG0652.jpg
?????文件???????1123??2012-04-12?09:30??fun_ipm_2.m
?????文件???????1317??2010-03-25?21:33??facefinding.asv
?????文件???????1219??2012-04-12?09:32??fun2_ipm1.m
?????文件???????1081??2012-04-12?09:33??fun2_ipm2.m
?????文件???????1332??2012-04-16?10:16??fun3_ipm1.m
-----------?---------??----------?-----??----
??????????????3278908????????????????????9
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