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At my workplace there is a large aerial photo that is over 10 years old on the wall showing our company site and a partner company's site. A colleague of mine who is also an amateur pilot would like to take an updated photo to show a then-and-now depiction. To do this he wants to take the photo from the same location (coordinates and altitude), but the source of the original photo has gone in the mists of time so we are left with making the determination by mathematical means.

Conveniently, in the photo there is a landmark in the distance that is vertically aligned with the two sites so we can work in two dimensions. The scenario is pictured below.

P   Plane (as in aeroplane/airplane)
G   Ground-point directly below plane
A   Our site
B   Partner's site
C   Landmark
a/b/c   Equivalent locations to A/B/C on the photo viewed from the perspective of P


P
|
|
|
|
|
|
|
|
G---------------------A---------------B----------------------------------C-----

The following values are known. (We are working in metric so I've provided some conversion factors if you require them.)

AB = 800 metres
BC = 8750 metres
ab = 290 millimetres
bc = 415 millimetres

1 metre = 1000 millimetres
1 metre = 3.28084 feet
1 inch = 25.4 millimetres

The following assumptions are made.

The Earth is flat!
There is no refraction.
The viewpoint is with the naked eye.

Question: What is the location of "P", i.e. the values for GA and GP?

Although there is obviously a relationship between the mathematical plane of the photograph and that of the ground, I've been unable to figure out how to represent it.

Previous questions I found on here of a similar theme (below) were unanswered so I've tried to provide more information, especially by way of the pictorial representation.

How can I calculate the GPS coordinate of an object in an image using the distance, angle and GPS coordinate of the center of the image in python? Ask

Locating Geographic Coordinates in an Oblique Photograph


And for the adventurous...

I'm willing to believe it's possible that the location of P has multiple solutions, i.e. it's defined by a curve. In this case the question would be, What is the function for the curve?

And probably related (but possibly a separate issue), it's likely the photo was taken using a telephoto lens from higher and further away. Again in this case the question would be, What is the function for the curve?

Thanks

Wayne

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    $\begingroup$ Are A, B and C in a straight line? Is G assumed to also be in this straight line? (I.e. are a, b and c in a vertical line in the photo?) $\endgroup$ – Arthur Sep 11 at 7:26
  • $\begingroup$ Hi @Wayne, and welcome to the Math.se. The problem you have posed is interesting but seems quite difficult: however, it reminds me of a technique called Photogrammetry, which allows for the measure of real dimensions of an object from one of its photo or its perspective drawing: it probably works also in the inverse sense, i.e. if you have a photo and the real dimensions of an object, you could probably find the location of the point of view (including orientation of the camera) from which it was taken. I hope this can help you. $\endgroup$ – Daniele Tampieri Sep 11 at 8:15
  • $\begingroup$ @Arthur, yes to all your questions. $\endgroup$ – Wayne Ivory Sep 11 at 9:40
  • $\begingroup$ I think you are right that there is a curve of solutions. The photo could have been printed at any scale, so from the two photo measurements you really only get one data point, the ratio between the two lengths. This corresponds to the ratio between the angles APB and BPC (approximately). $\endgroup$ – Jaap Scherphuis Sep 11 at 11:57
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I think as posed the problem is not solvable, you are missing one bit of information. That is for any position of the air plane I can hold the camera in a way so that the distances on the photo match what you described. If you have a second land mark that should work. Knowing the height of one of the buildings both in real life and on the picture should also work.

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  • $\begingroup$ Unless you're including the telephoto lens from my "advanced question" I don't think holding the camera a different way would maintain the distances on the photo as you describe. Taking the extreme case, if I reversed the plane 100 kilometres (or 100 miles) obviously the distances ab and bc would tend to zero. But if I've misunderstood you please explain more. $\endgroup$ – Wayne Ivory Sep 11 at 9:50

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