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Find the actual distances in an image.

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Chathu
Chathu 2016 年 4 月 15 日
コメント済み: Chathu 2016 年 4 月 30 日
I want to find the minimum x(1) value which touches the image,A. Then maximum x(2) value which touches the image,A.(ps:image is attached). I made the axis ON. SO i want to let the program identify the x(min), x(max),y(min) and y(max), next draw straight lines(as shown in the image) along above said points(eg:x(1),x(2),etc) and then perform the spatial calibration (note; image is not symmetric). Can anyone give me a hint how to resolve this issue.
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Adam
Adam 2016 年 4 月 15 日
Where is your image coming from? And is it actually an image or a line plot? This kind of thing would usually just be calculated on the raw data that produces the image, but without knowing what format that is in it is hard to suggest clear solutions.
Chathu
Chathu 2016 年 4 月 15 日
it is an image. Let me add it here now. Sorry for the confusion.

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採用された回答

Mohammad Abouali
Mohammad Abouali 2016 年 4 月 15 日
Similar approach to this should work:
I=imread('P3290337.JPG');
I=padarray(I,[1 1 0],0);
thresh = 15;
mask = ~(I(:,:,1) < thresh & I(:,:,2) < thresh & I(:,:,1) < thresh);
xProj = any(mask);
xMin = find(xProj,1,'first')-0.5;
xMax = find(xProj,1,'last');
yProj = any(mask,2);
yMin = find(yProj,1,'first');
yMax = find(yProj,1,'last');
imshow(I);
hold on
plot([xMin xMin],ylim,'r','LineWidth',2)
plot([xMax xMax],ylim,'r','LineWidth',2)
plot(xlim,[yMin yMin],'r','LineWidth',2)
plot(xlim,[yMax yMax],'r','LineWidth',2)
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Mohammad Abouali
Mohammad Abouali 2016 年 4 月 19 日
編集済み: Mohammad Abouali 2016 年 4 月 19 日
If you need to know the real world coordinates then you need your image to be geo-referenced. Have a look at pix2map() function.
another easy solution is that if you know how wide are each pixel then you can multiply the Euclidiean distance in pixel unit by the size of each pixel and you get the distance in your requested unit. However, this is only good for short distances, If your image is covering large areas you can not use this method to compute the distance and you should use other methods. For example, if you are measuring the distance between to cities located in two different state you should not do this. Also this requires your image to be corrected for distortions so that all pixels do have same size.
Chathu
Chathu 2016 年 4 月 30 日
@ Mohammad, i saw your response just now. Your approach is perfect for my purpose, as it is short distance.
Thank you so much for your response. Highly appreciate it:)

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その他の回答 (1 件)

Image Analyst
Image Analyst 2016 年 4 月 16 日
See my spatial calibration demo.
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Chathu
Chathu 2016 年 4 月 16 日
Thank you so much Image Analyst. Highly appreciated.

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