# distance between pixels = a fixed measurement (inch, foot etc)?

Let's say the distance from my camera to my object is 3 feet. The object is a 6 foot long pole. The pole has a mark every inch. Will the number of pixels between the bottom 1 inch be the same number of pixels of the top 1 inch? Will I be able to say that at a 3 foot distance from my camera and any object that for example 100 vertical pixels is an inch, across the entire vertical view of the image?

I did a simple test. I took a picture of some graph paper, then converted to jpg to a pbm. At the bottom if the image, 30 pixels was 1/10 of an inch. But at the to of the image, it was more like 37 pixels for the 1/10 of an inch.

Was there a problem in the way i did my simple experiment. The jpg was taken with my cellphone. Would I see similar results with my Canon G11 Powershot?

• What does trigonometry say? – Michael C Aug 5 at 5:24
• trig shows that the angles between increasing in height but same "units" is not uniform. My question was about cameras. Are you implying my camera follows trigonometric functions. I dont know much about camera's, lens and sensors. The picture of GUM department store in Red Square is thus flawed due to these imperceptible,but actual, trigonometric functions? – Walter L. Preuninger II Aug 5 at 5:49
• In a way yes. Plus the effect gets even more pronounced when using wide angle lenses. And you end up with something completly different, when you have an anamorphic lens which skews the image towards the right and left borders. – Kai Mattern Aug 5 at 7:18
• @MichaelC Trigonometry has nothing to say. Whether cameras employ rectilinear lenses is an empirical question, not a mathematical one. A camera is a device for projecting an image onto a sensor. There is no "correct" projection function, any more than there is a "correct" map projection. – Acccumulation Aug 6 at 0:07
• Technically, there's distance per pixel, not distance between pixels. – Acccumulation Aug 6 at 0:08

In a camera with the usual lenses, the size in pixels in the image corresponds to an angle. Things that are far away elicit a smaller angle of view and so correspond to less pixels in the image. Take a picture of a building from the street, the windows in the upper floors will appear smaller than the ones at street level. This is also happens what happens with your pole, assuming the pole is parallel to the sensor (or perpendicular to the lens) the part of the pole at the center of the photo is closer to your camera than the extremities, so there are more pixels between marks at the center.

However, if you step back with the lens and zoom in (or take a lens with a longer focal lenth) to compensate, then the distance difference between the camera on one side and the center and the ends of the pole on the other will be less and you will see closer sizes in pixels (but still no totally equal)

• If we are counting pixels here one must pay attention to lens distortion too. – Arjihad Aug 5 at 10:10
• With a rectilinear lens, which is what most lenses are approximations of, distance is measured along the optical axis. Objects that are in a plane perpendicular to the optical axis (and thus parallel to the sensor plane), are considered to be at the same distance. Equal sized objects on such a parallel plane should result in the same size on the sensor. – Roel Schroeven Aug 5 at 12:49
• Re “the size in pixels in the image corresponds to an angle”: no, it doesn't. Not unless you are using an equidistant fisheye. And even then, only in the sagittal direction. – Edgar Bonet Aug 5 at 14:26
• @RoelSchroeven, except, at some point, designers of digital photography systems figured out that they don't have to work so hard to make the lens rectilinear all by its own self. When the sensor has enough spatial resolution, they can get away with applying geometric transformations to the image after it's been captured in order to correct any geometric distortions caused by the lens. When I open RAW images from my own camera in RawTherapee, it gives me the option to view the image with or without correcting for the lens geometry, and the difference is significant. – Solomon Slow Aug 5 at 19:02
• The theoretical field of focus of a highly corrected rectilinear lens is shaped more like a lasagna noodle than a flat plane. Fun with Field of Focus Part 1 and Fun with Field of Focus II: Copy-to-Copy Variation and Lens Testing – Michael C Aug 6 at 6:27

In theory, when you have different objects all on the same plane that's parallel to your sensor plane (such as the marked inches on your pole, assuming that pole is perfectly parallel to your sensor) and you are using a rectilinear lens (most lenses are, or try to be, rectilinear; fish eye lenses are decidedly not rectilinear) then equal-sized objects result in equal-sized images on the sensor plane.

In practice, lenses are not perfectly rectilinear; they have some distortion causing differences. With wide lenses the distortions are more pronounced.

Try to put the zoom of your G11 at the largest optical magnification, and put the camera further from the pole (or graph paper or any other object) until the pole nicely fits into the view again. That should give much better results; Digital Photography Review says about the G11 lens: "The distortion is fairly well controlled, with a noticeable 1.3% barrel distortion at the wide-angle setting and a perfectly rectilinear projection at the other end".

And don't forget to make sure your camera is perfectly parallel to the pole or graph paper.

• Yes, the right hand side image here is relevant: en.wikipedia.org/wiki/Intercept_theorem#/media/… S would be the focal point of an idealized lens (or the hole in a pin-hole camera), the photographed object would contain B, E, D, and C, F, A would be on the sensor. (see en.wikipedia.org/wiki/Intercept_theorem) – Carsten S Aug 5 at 13:48
• In theory, even a perfectly rectilinear lens does not perform as you describe. The center point of the pole will be closer to the optical center of the lens than the end points of the pole. No lens can, even in theory, be perfectly corrected for field curvature at various wavelengths of light. Please see does focal length change across an image? and What is the shape of the focal plane? – Michael C Aug 6 at 6:20
• In the above comment, "No lens can, even in theory..." should read, "No lens with real thickness can, even in theory..." What you describe can only be achieved in theory with an idealized thin lens that has zero thickness. – Michael C Aug 6 at 6:34

Fundamentals of optics & Ray Tracing

With ideal optics, which no one actually has:

(1) A ray leaving the object parallel to the lens axis is refracted and passes through the far focal point.

(2) A ray passing through the near focal point is refracted and proceeds parallel to the lens axis.

(3) A ray that leaves the object and passes through the center of the lens is not deviated. The relative sizes and positions of the image is LINEAR. There is no angular effect of size, that is it is not distorted.

Now of course the reality is that we are not dealing with ideal thin lenses so there will be a distortion effect as a function of the lens, but lens designers aim for this ideal.

``````--- Edit to address the Geometry misconceptions ---
`````` P/D = Tan Θ

V/F = Tan Θ

P/D = V/F

P/V = D/F

Both D & F are constants, so let D/F = K

P/V = K ; V = P/K

The angle factors out. The linear distance “V” on the sensor is a constant ratio of the linear distance "P" of the object.

Not angle dependent!