Find the dimensions (height, width) of the object using the camera

Question

I want to find a solution to get the size of an object using a camera, well it sounds like Duplicate one

How to measure the height, width and distance of an object using the camera?

But the solution does not help me. Start with the link above, I got some idea to find out the distance ( Measure Distance ).

Can someone suggest me how should I get the width and height of an object. Simple math or any idea would really be helpful.

Are there any opportunities to achieve the above solution using OpenCV. Height and width measurement

What I have tried so far:

Suppose we fix a distance, we can calculate the elevation angle

tan(α/2) = (l/2)/d, hence α = 2*atan(l/2d) 

But we still do not know the value of L (object length)

Another way to find the viewing angle:

  double thetaV = Math.toRadians(camera.getParameters().getVerticalViewAngle()); double thetaH = Math.toRadians(camera.getParameters().getHorizontalViewAngle()); 

Doesn't seem to work !!

Answer 1

The actual physics of the lens is explained, for example, on this website of the State University of Georgia .

See this illustration for an explanation of how you can use linear zoom or focal length ratios to determine the size of an object from the image size:

In particular, -i / h' = o / h , and this o / h ratio is valid for all such triangles (i.e. an object of size 2h at a distance of 2o has the same size h' in the figure). Thus, as you can see, even in the case of a complete equation, you cannot know the distance o and the size h object, but one will give you the other.

On the other hand, two objects at the same distance o will see that their sizes h1' and h2' in the image will be proportional to their sizes in real life h1 and h2 , since h1' / h1 = M = h2' / h2 >.

Therefore, if you know both o and h for one object, you know M , thereby knowing the size of the object on the film, you can subtract its size from a distance and vice versa.

The value of -i / h' naturally expressed for maximum h' . If the size of the object accurately fills the image, it fills the field of view, then the ratio of its distance to its size is tan(α/2) = (l / 2) / d (note that in the legend of the image below, d = o and l = 2 * h ).

This α is what you call theta in your example. Now, from the size of the image you can get at what angle you see the image - that is, what size l will the image have if it is at a distance d . From there, you can infer the size of an object from its distance and vice versa.

Algorithm Steps:

• get the coefficient r = size of object in image (in px) / total size of image (in px) .
  Do this along the axis for which you know or plan to get the real size of the object, of course.
• get the corresponding field of view and angle, multiply r by the tangent half of this angle
  r *= tan(camera.getParameters().getXXXXViewAngle() / 2)
• r - now the tangent of the polygon under which you see the object, so the following relationships are true: r = (l / 2) / d = h / o (with the corresponding designation of the drawings).
• • If you know the distance d to the object, its size l = 2 * r * d
  • If you know the size l object, it is at a distance d = l / (2 * r)

This works for objects that the camera actually points to, if they are not centered, math can be turned off.