# Is there a formula to calculate the physical size of an image?

I'm trying to figure out the size of the physical space an image represents (IE, how many meters long/wide the real world in the image is), but I can't seem to find a formula that works with only the data I have. I know that the picture was taken from a UAV, facing straight down. I know the altitude where the picture was taken, and the model of the digital camera it was taken from. I don't have access to the camera, or to the places the pictures were taken of, so I can't use test images. Does a formula exist to calculate this information?

Sorry if I'm missing something obvious, photography isn't my strong suit.

Edit - Specifics

The camera in question is a Sigma DP1 digital camera, images were taken around 230 meters above the ground.

• Please post the information that you have, not just the fact that you have it =) – null Jun 17 '15 at 15:46

This can be derived from the basic lens formulas. The lens divides the entire system in two halves:

• the subject side, with two variables
• S the real size of the subject, this is what you are looking for
• s the distance between the subject and the lens, this is the altitude
• the image side, also with two variables
• I the size of the subject in the image, this is what you can measure
• i the distance between the sensor and the lens, this is a property of the camera. this is not the flange distance.
• and of course there's the focal length f

Dividing sizes and distances yields a constant value.

``````s/i = S/I = constant
``````

From how the rays go through a lens, one can derive a formula including the focal length and without the i which is unknown:

``````S/I = (s-f) / f
``````

By multiplying with I, you get a formula for S:

``````S = (s-f)*I / f
``````

Is I measured in proportions?

No. All the variables mentioned: i,I,s,S and f are lengths. They have a dimension of length and therefore have a unit of length. The focal length is usually given in mm, but any unit that describes a length will do.

Number of pixels?

Yes and no. As mentioned above, it is a length. It is the size of the object as it is optically projected onto the sensor. Of course, the sensor captures that projected image and you end up with a digital image made out of pixels. You have to convert the pixels back to a length value, e.g. you have a sensor that is 100pixels wide. The physical width of the sensor is 20mm (roughly 4/5 of an inch if you don't like metric). If the object is 39px wide in the image, how wide was it when being projected onto the sensor?

First find how big one pixel is:

``````20mm / 100px = 0.2mm/px (or 200µm/px)
``````

now you can multiply this with the pixel size of your object to find I:

``````I = 39px * 0.2mm/px = 7.8mm
``````

And is there a name for the property i? Or a method of calculating it?

It is usually referred to as "the distance between image and lens". The term used in optics is "image distance" Please take a look at the following image from here: Attention: it uses different symbols. What I call s is what they call S1 and what I call i is what they call S2. The thing is that in optics, a lot of calculations use a certain reference for their calculations. (where "0" is) This reference point is not simply the "middle" of the lens. It doesn't have to be "within" the lens at all. It is a mathematical construct to describe the lens, based on its shape. For any real world lenses, like those used on cameras, there are actually two such reference points. (basically speaking)

As you can see, this is not a trivial thing to measure. And it isn't a practical thing to do either, because it isn't necessarily constant. Yes, you read that right, it's a variable property. Essentially, this is the variable that your camera changes when it focuses on something. That's why I suggested the formula that involves the focal length, which is a (more) constant value.

I tried to keep my answer simple and not too optics/physics/math heavy or just as much as necessary.

• Thanks! I'm still unsure about I and i though. Is I measured in proportions? Number of pixels? And is there a name for the property i? Or a method of calculating it? – Turniper Jun 17 '15 at 19:07
• @Turniper I added some more information while trying to keep it simple. Does this help? – null Jun 17 '15 at 20:50

I guess it depends how accurately you want to measure, but this is how I would do it:

Wikipedia has a formula for calculating the horizontal/vertical angles of view of a lens from the lens' focal length and sensor dimensions:

https://en.wikipedia.org/wiki/Angle_of_view#Calculating_a_camera.27s_angle_of_view

In the case of the Sigma DP1, the focal length is 16.6mm and the sensor measures 20.7 x 13.8mm. (I got those specs from Wikipedia also.)

Once you know the horizontal/vertical angles of view, and the altitude, you should be able to use basic trigonometry (tan A = opposite/adjacent => opposite = tan A * adjacent) to reasonably approximately calculate the distances represented by the sides of the image.