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I want to know the actual area of the entire area covered by an aerial photograph. _The Photo is 4000x3000 pixels _The Focal Length of the camera is 20mm. _The Picture height is 51 meters.

I researched and got the scale of the photo. In this case it is 1 mm / 2550 mm.

I understand the scale tells me that if I move 1 mm inside of the photo , in the real terrain I would be moving 2550 mm. I want to know how to get the real width and height of the picture. ¿? cm. I need this data to know the area of the photo and therefore know the terrain area that is pictured.

Thank you for your help.

dany luna
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    This seems to be arithmetic, not programming. – Teepeemm Aug 31 '15 at 21:35
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    Do you care about the difference between the 2D photo projection and spherical coordinates? It matters to NASA; less so to Google Maps. – duffymo Aug 31 '15 at 23:15
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    By “height” you mean the height from which the picture was taken? You have the size of the photo in pixels, but the scale in mm, can you relate these two quantities? In any case, I agree that this is not about programming. – MvG Aug 31 '15 at 23:34
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    I'm voting to close this question as off-topic because it is not about programming. – High Performance Mark Sep 01 '15 at 05:49

2 Answers2

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If you ignore distortion and curvature, the physics is qute simple. There are two similar triangles in the picture below:

enter image description here

You know the focal length (20mm), and the distance from the focal plane to the scene (51m). So you need the width of your sensor's CCD to put into the equation:

sensor width       focal length
------------    =  ------------
scene width        camera height

Make sure all the units are the same, i.e. don't mix millimetres and metres.

Mark Setchell
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If I understand your question correctly, consider the following comments:

Firstly, you must consider the loss of information in rendering a landscape as a 2-d photo: Precise information about depth is probably lost, unless there is a pronounced loss of resolution beyond and preceding the plane of focus.

That being said, you're right to consider the ratio of pixels to real distance (e.g., meters).

I suggest limiting your approach to making inferences about the length of certain objects in the photo on the same orthogonal (to viewer) plane as other objects in view. For instance, if I am looking head on at a wall, sufficiently close to ignore perspective's distortion, I can infer that a 10 pixel wall height translates to 5 meters if I know a window measuring 1 meter to be 2 pixels. This inference uses the technique of dimensional analysis (http://www.chem.tamu.edu/class/fyp/mathrev/mr-da.html).

David Shaked
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