How to calculate the distance of an object

Question

I have two screenshots (1920x1080) of a game, one with a 348-pixel-tall object that is 1 meter distant from the camera, and the other with a 138-pixel-tall version of the same thing. Given that the camera's field of vision is 90 degrees in the second screenshot, how can I precisely measure the object's distance from the camera?

I tried using a formula to determine the object's distance from the camera based on the object's height and camera distance, but the results were inaccurate.

Can you include pictures in your post? I would be tempted to use Thales' theorem or something... But without seeing the picture it's hard to know whether that's appropriate or not. — Stef, Jan 21 '23 at 10:53
So, determining the equation linking the size and distance of the object and the angular-size is relatively straightforward with trigonometry. The question that remains is, how does this angular size translate to a pixel size. I don't know enough about photography to write an answer I'm confident in. — Stef, Jan 21 '23 at 11:40
Rel: Pixel size at long distance, What is the relationship between size of object with distance?, Calculate object size when I have physical pixel size, Calculate the distance of an object in a picture? — Stef, Jan 21 '23 at 11:42
Your two images are inconsistent. Is Distance supposed to be measured from the camera or the 348pixel object? — , Jan 21 '23 at 12:34
@SimonGoater I don't see any inconsistency between the two pictures, unless you take "Distance" to be a variable name, as opposed to the well-defined word "distance". — Stef, Jan 21 '23 at 12:36
On the left picture, you have Distance= and a double-headed arrow between the pink and green boxes, but the double-headed arrow on the right goes to the camera. — , Jan 21 '23 at 12:38
I just realized what was wrong, the problem was the game camera itself, after moving away from some object the fov changed which caused me to get the wrong result. And yes, I gave the arrow wrong in the first picture, it was about the distance of the object from the camera and not from the second object sorry. the formula I used to calculate the distance: object_distance = object_size_1/object_size_2 — , Jan 21 '23 at 13:18

score 4 · Accepted Answer · answered Jan 21 '23 at 13:21

Precise answer: Without knowing the projection model, this is not possible.

An ad-hoc engineer's answer: One may assume the projection model is a pinhole camera model. This is probably a good assumption, because it is the model with least strange distortions and with geometrically beneficial properties. For a virtual game camera with horizontal* 90° opening angle this assumption seems to be reasonable.

Now you can translate your question into simple equations based on intercept and trigonometric theorems:

  (I)  w_1 / f = s / d_1
 (II)  w_2 / f = s / d_2
(III)  w_45 / f = tan(45°)
where
  f = focal length (unknown)
  s = object width (unknown)
  d_1 = object distance in image 1 = 1 m
  d_2 = object distance in image 2 (unknown)
  w_1 = object width in image coordinates in image 1 = 348 px
  w_2 = object width in image coordinates in image 2 = 138 px
  w_45 = 1920 px / 2

What remains is to solve this equation system:

(I) => f * s = w_1 * d_1
(II) => f * s = w_2 * d_2
___
=> w_1 * d_1 = w_2 * d_2
=> d_2 = w_1 * d_1 / w_2 = 348 px / (138 px) * 1 m = 2.52... m

So you do not even need equation (III). You would need it in case you are interested in focal length f = 960 px or object width s = 0.3625 m.

*I assume it is the horizontal opening angle from what is common and from what you drew in your question.

score 0 · Answer 2 · answered Jan 21 '23 at 14:22

This might be of interest. I don't think this approach is entirely accurate due to perspective distortions, but in this case, I think a fair estimate of Distance D from the camera can be obtained by assuming constant pixels-per-unit-angle near the centre of the picture. Assume the object are poles of height h and the camera is pointing at their centre. Suppose the angle from the centre of the camera view to the top of the pole at 1m is Theta_1 and for the other object it is Theta_2. Then

tan Theta_1 = h/2
tan Theta_2 = h/2D
(tan Theta_1) / (tan Theta_2) = D

Taking a first order approximation,

D approx1.= Theta_1 / Theta_2

If we also assume 348 Theta_2 = 138 Theta_1, then we get the answer posted. For a 2nd order approximation, which hopefully should be more accurate, particularly for more distended cases,...

D = (sin Theta_1)(cos Theta_2) / ((sin Theta_2)(cos Theta_1))
approx2.= Theta_1(1 - Theta_2^2/2) / (Theta_2(1 - Theta_1^2/2))
approx2.= (Theta_1/Theta_2)(1 - Theta_2^2/2)(1 + Theta_1^2/2)
approx2.= (Theta_1/Theta_2)(1 - Theta_2^2/2 + Theta_1^2/2)

So if 348 Theta_2 = 138 Theta_1

D approx2.= (348/138)(1 + Theta_1^2(1/2 - (138/348)^2/2))

If 1920 pixels is pi/2 radians, then estimate

Theta_1 approx.= 348*pi/(2*2*1920) = 0.142353417

giving

D approx2.= (348/138)(1.008538919)

which suggests a less than 1 percent difference from the 1st order approximation in this case.

How to calculate the distance of an object

2 Answers2