# Find the actual distances in an image.

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Chathu il 15 Apr 2016
Commentato: Chathu il 30 Apr 2016
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 il 15 Apr 2016
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 il 15 Apr 2016
it is an image. Let me add it here now. Sorry for the confusion.

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### Risposta accettata

Mohammad Abouali il 15 Apr 2016
Similar approach to this should work:
thresh = 15;
mask = ~(I(:,:,1) < thresh & I(:,:,2) < thresh & I(:,:,1) < thresh);
xMin = find(xProj,1,'first')-0.5;
xMax = find(xProj,1,'last');
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 il 19 Apr 2016
Modificato: Mohammad Abouali il 19 Apr 2016
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 il 30 Apr 2016
@ 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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### Più risposte (1)

Image Analyst il 16 Apr 2016
See my spatial calibration demo.
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Chathu il 16 Apr 2016
Thank you so much Image Analyst. Highly appreciated.

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