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I am not sure if this is a telemetry question or simple geometry but came across to similar questions in this forum, but I think this is a bit different; and actually "simpler" (compared to "autofocus" solutions) since I move the camera itself by a given amonut of dX and somehow "measure" the image height and observe a dI amount of image size change. Hence I believe make it camera independent. I am not sure if it depends on focal length. Does it help using more than one object image ratios for a given dX?

Image below is a hypothetic one with E0, E1 being camera positions with a difference of dX. O1 , O2 are different objects at positions x1, x2; and I10, I 11, I 20, I21 are image sizes for O1, O2 for E0, E1. Then the question comes down to, if there is any relationship for dX and dI1 and/or dI2?

ImageRatiosVsdX

FOLLOW-UP:

Yes; mathematically speaking I have ended up with such an equation:

x1**2 + x1*dX = O1**2/(I10*I11)

Here the O1 represents the object height, which we need but do not have.

On the other hand again in this forum I had seen a piece of information which said the image size is inversly proportional with distance, i.e.,

O1 = a/x1

But my original drawing was not drawn with this logic, to be able to display a larger image for a nearer object (for x1- dX case). Hence I can not use these to equations as one set to solve for x1. I just could not manage to combine the two (I can not comment to an answer to my question, so I had to answer my question again to ask a question).

So I have managed to put up a better drawing, which I believe is correct:

enter image description here

Here we have one object only. Camera moves forward to take image I2. Normally, I believe f1 and f2 are different; but compared to x1 distance, if we assume f1 = f2, then we get a set of two equations, to solve for x1 in the form of:

a*x1**2 + b*x1 + c = 0

where 

a = (I2/I1)
b = - a*dX
c = -1

and then we can solve this quadratic equation and find the roots. Is this a viable solution? You may well argue that I2 >> I1 as f2 >> f1, and we gain nothing out of these equations.

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  • I don't know the correct equations, but I think that a prime lens used with the aperture stopped down may help, taking focal length out of the equation and increasing depth of field to avoid effects of re-focusing. – Jahaziel Mar 2 '17 at 18:48
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    I'm voting to close this question as off-topic because this is about using a camera as a tool for a purpose other than photography. – Please Read My Profile May 1 '17 at 18:57
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Yes, it's possible to do this. It's called photogrammetry. Depending on what you're trying to calculate, you may need to already know other pieces of information. For example, if you want to know the distance between 2 objects, you need to know their sizes.

I believe that Photoshop has some options for photogrammetry, though I've not used them myself.

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    Hypothetically, yes. But things like focus breathing might alter the results enough to make them unreliable. I would use scientific surveying equipment and not consumer cameras for this kind of work. – Lumigraphics Jan 1 '17 at 17:39
  • @Lumigraphics not just focus breathing but lens distortion too. A fisheye lens wouldn't be my first choice for this task. – Mark Ransom Sep 28 '17 at 22:30

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