# What is the relationship between size of object with distance?

How does the size/length of an object vary with distance?

Is it a logarithmic relationship? exponential? linear?

I plotted a curve of the size/length of an object for different distances from the camera, and the curve looked exponential/logarithmic. I was trying to understand the reasoning behind that.

The relationship is a simple inverse, i.e.

`````` object size in image = Object size * focal length / object distance from camera
``````

If you keep the same object and the same focal length you get: `size = 1/ distance` (the =-sign should be proportional-sign).

• Thanks. How is that different when it's object size vs object location image in terms of pixel (x,y) coordinates – fmvpsenior Jul 17 '13 at 16:48
• It's not different, pixels are just a different unit than say meters. You can change units by knowing the size of one pixel. – Unapiedra Jul 17 '13 at 17:02

Inversely linear is a good approximation. Imagine a 1,7m tall girl at 1 m distance b. Her head is at point B.

How does the size/length of an object vary with distance?

Let the girl walk away from you. Her size a stays the same. She appears smaller, because she is appearing under a smaller angle. Her angular size changes. Try to imagine it with the picture attached. Using arctangent to calculate her angular size is the correct way. For small angles you can simplify:

Angular size is inversely proportional to its object distance, without using optical devices.

An object on full-field with focal length of 12 mm would be measured incorrectly. An error 2-5% in length measurement may be made. For fish-eye lenses this may be even worse. Hands-on rule: Use the inverse relationship if angular size is smaller than 10°.

I'm sure this is a duplicate, but I can't find a good answer to the question in the archives so here goes.

The relationship between object size and distance is an inverse linear relationship, i.e. size is 1 / distance. This makes sense when you think about it as if you double the distance the size halves.

This is why you appear to be observing an exponential: the exponent is -1, if you take the reciprocal of the size, your graph should be a straight line.

• Thanks. I am actually plotting the size of the object vs distance in pixels (or location). So I start with the object being on the far left of the image (pixel coord x = 0) and end with the object being on the far right of the image along the same line (pixel coord x=3000). I wonder if the relationship is the same in that case. – fmvpsenior Jul 17 '13 at 16:47
• This only holds for object with an angle of view smaller 10°. Since arctangent Alpha is not Alpha. An object on full-field with focal length of 12 mm would be measured incorrectly. See my answer. – Stefan Bischof Aug 12 '17 at 21:35

It depends on what 'size' means in the question.

1. Each linear dimension of an object will halve as distance from the camera is doubled and each linear dimension of an object will double as the distance from the camera is halved.

2. The area of the sensor of film upon which an object is projected will quarter when the distance to the camera is doubled and will quadruple when the distance to the camera is halved.

To put it another way, so long as the subject fits in the frame, doubling the focal length potentially allows recording four times more information with the sensor. Composition aside, that's really the important thing. Thus in terms of focal length:

• Because doubling the focal length halves the angular field of view in both dimensions the area of the sensor upon which the object is projected quadruples.

• Likewise halving the focal length, quarters the sensor area upon which the object is projected.

In practice this means that going from a 200mm lens to a 300mm lens more than doubles the degree to which a distant subject fills the frame. It's why a 18mm lens is a lot (rather than a little) wider than a 24mm. A 1.4x teleconverter doubles the area the subject projects on the sensor and a 2x teleconverter quadruples it.