Eye of the eclipse...

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A lot of people have asked if we can measure the distance of an object in photo. But I was wondering if we can measure the size of an object in photo. Of course assuming we know everything from sensor size to height of the photographer etc.

Example: If I take a photo of a bird in flight with only sky as background, there would be no objects in the frame to give an idea of size of the bird. So if I could calculate the size to some approximate level, it would be really nice to know.

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2 Answers 2

It would never be more than an estimate, without some external reference or accurate distance to the subject. If you have estimated distance to the subject, FOV (/length of the lens) and the sensor size it's not too hard to calculate. The biggest uncertainty would be the distance to the subject.

To find the spatial resolution (how large (wide or high) a pixel is in the subject) use:

(distance*(sensor size / focal length)) / number of pixels

Make sure the sensor size and focal length are using the same scale (both are in millimetres or inches).

Then you can count the number of pixels for the subject, and multiply that with the spatial resolution for one pixle.

I'm assuming a linear change in FOV over the whole area of the lens, which isn't 100% accurate, but quite close for normal and longer lenses.

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Too bad we don't have the LaTeX-math working here at photo. – Håkon K. Olafsen May 15 '12 at 7:22
The equation is wrong. Why 2x? You are calculating an area, therefore, it is squared. It should be distance^2. See photo.stackexchange.com/a/12437/7292 – Rado Jun 7 '14 at 13:09
Oh .. and the focal length should also be squared – Rado Jun 7 '14 at 13:49
@HåkonK.Olafsen: Observe: sensor_x/focal_length = picture_x/distance; sensor_y/focal_length=picture_y/distance (distance is calculated from the focal point). Considering a square pixel, you can take any of these, and get: picture_x = distance*sensor_x/focal_length. Divide it by horizontal pixel count pixel_cnt_x you get: distance(sensor_x/focal_length)/pixel_cnt_x. This is the horizontal spatial resolution. The "2x" multiplier is not correct. If you have non-square pixels, you will have to do the math for both horizontal and vertical resolution. – TFuto Jun 7 '14 at 15:45

My camera stores the focus distance within the picture. With this you know how far away your bird was. It won't be too hard to calculate the size of the bird. For instance, you can take an object of known size and take a picture with the same lens from the same distance and compare the size of this object in the image, with the size of the bird in the image.

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What camera is this? – ysap May 15 '12 at 6:47
It's a Nikon D300 – Rene May 15 '12 at 8:15
But doesn't the focus distance go only up to around 3m and then it becomes infinity? When I use the Magic Lantern firmware on my Canon 550D it does at least. – Bart Arondson May 15 '12 at 8:16
I would be wary of trusting the distance provided by the camera at anything remotely approaching a focus distance of infinity, as the smallest calibration error would yield errors that could be several orders of magnitude of the provided distance. – Michael Clark Jul 5 '13 at 11:44
@BartArondson It depends on the focal length of the lens. A longer focal length, such as often used when birding, will focus much further before it reaches ∞ than a wide angle lens. That is why the last measure before ∞ for distance scales on WA lenses is usually only around 1 meter, but a long telephoto lens will have a marking of over 50 meters or even 100 meters just before the ∞ mark. – Michael Clark Jul 5 '13 at 11:54

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