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I am working on a virtual camera inside a game engine, which provides me with an FoV control, but no Focal length control (and no filming sensor dimensions short of aspect ratio), and I am less than a beginner in photography.

Now when I position the Camera in a specific location in the world, and I position an object in front of the camera such that the object covers 50 degrees of the camera field horizontally. The camera FoV is 100 degrees horizontally.[see image please]. enter image description here Then I measure the object in the image and I find it covering around 40% of the image width instead of 50%.

What is the most expected mistake that I am making here?

Relevant information: I don't know the position of the virtual camera is the position in the middle of the lens itself, the position of the sensory retina, or the position in between.

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@ Physician --- We can use trigonometry and set-up a condition whereby the horizontal angle of view will be 100°

On a full frame 35mm camera format demotions 24mm by 36mm, mount a 15mm lens and horizontal angle of view will be 100°

On an APS-C compact, the frame size is 16mm by 24mm. Mount a 10mm lens and the horizontal angle of view is 100°

In other words, the angle of view of camera system intertwines focal length and format dimension.

For the full frame camera above: The image plane falls 15mm downstream from the lens. The Image triangle has a height of 15mm. The size of the angle at the apex of this triangle is 50°. We bisect this triangle creating two right triangles. The angle at the apex is 25°. The height of this triangle is 15mm. If you draw this triangle, the height is the adjacent side and the base is the opposite size. The tan of 25° = 0.4663. Thus Opposite / Adjacent = 0.4663. Thus Opposite side = 15mm X 0.4663 = 6.9946mm. The length of the image of the object will be 2X this value = 14mm.

For the APS-C format: The image plane falls 10mm downstream. Such as lash-up delivers a 100° horizontal angle of view. The image triangle angles are the same as for the full frame. Bisecting the image triangle, we get two right triangles each with a 25° apex. The tan of this angle = 0.4663. The height of this triangle is the focal length of the lens = 10mm. 10mm X 0.4663 = 4.66mm X 2 = 9.33mm. The image of the object will be 9.33mm long.

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  • $\begingroup$ So, I will say exactly what I understood and correct me if I am wrong. Using an APS-C format, the sensor placed at 10mm behind the lens, the object image width will be 9.33mm out of a full image of width 24mm. Which means 9.33/24 * 100, which equals 38.8 %. Meaning an objects covering 50 degree of 100 degree FoV will give a less than 40 % of image representation. That's fascinating. $\endgroup$
    – Physician
    Dec 14 '20 at 17:18
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You are probably assuming that the apparent screen size of an object is proportional to the angle, while in fact it is proportional to the tangent of half angle.

The difference is small for small angles but big enough at your 100°, try sketching your example for 179° FOV camera and 89.5° (half FOV) object to see how deep the rabbit hole goes.

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  • $\begingroup$ Yup, that what I was assuming exactly. $\endgroup$
    – Physician
    Dec 13 '20 at 21:06
  • $\begingroup$ @szulat Execuse me, now let's say that my camera has horizontal FoV of 100 degrees. And I have an object in front of the camera with an angular diameter (I calculate the angle using 2*arctan(d/2D)) of 50 degrees horizontally. What is the image size of the object in the photo. Is it 50 % of the image width? $\endgroup$
    – Physician
    Dec 14 '20 at 16:49
  • $\begingroup$ it will be 39% of the image width (tan(50°/2) divided by tan(100°/2)) $\endgroup$
    – szulat
    Dec 14 '20 at 17:20
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When we calculate the field of view of a lens mounted on a camera, we can evaluate three different unique angles. These are height, width, and diagonal. As an example, if we mount a 46mm lens on a 35mm full frame camera with a format of 24mm by 36mm, the angles of view are 29° height, by 42° length, and 50° diagonal. While likely the last value, the diagonal is value most often given. If you think this is bizarre, consider that TV sets are sized by their diagonal measure.

We can trace an object triangle, a trace from its boundary points to a point at the center of the lens. We can also trace an image triangle, inside the camera, beginning from its boundaries to the center of the lens.

These two triangles will be “similar”. Their angles will be identical; their sides will be in ratio. Likely the field of view you are using is in reality the diagonal angle of view, and the image length you are referencing corresponds to the hormonal angle of view.

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  • $\begingroup$ Hi Alan. Now Let's say that my camera has horizontal FoV of 100 degrees. And I have an object in front of the camera with an angular diameter of 50 degrees horizontally. What is the image size of the object in the photo. Is it 50 % of the image width? $\endgroup$
    – Physician
    Dec 14 '20 at 13:21
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@ Physician --- There are three flies in the ointment, all miniscule: The object distance is a measurement taken from a point within the optical system called the front nodal. The image distance is the focal length, a measurement taken from a point in the optical system called the rear nodal. The focal length is a measurement taken from the rear nodal to the image when the camera is focused on an object at infinity.

In a multi element lens system the distance between front and rear nodal points will be unknown unless we run some tests. This could change the value we are using for the height of the image triangle. Additionally, if the object being imaged is closer than infinity, the height of the image triangle will be elongated. We are using the focal length as this value.

Suppose a 10mm focal length lens is deployed. Suppose the object being imaged is 1 meter distant. What is the height of the image triangle? 1/10 x1000 = power of camera lens in diopters = 0.1 x 1000 = 100d. The object distance can be express in diopter units as 1/1000 x 1000 = 1d.

We can calculate the image distance as 100d – 1d = 99d. Convert to millimeters = 1/99 x 1000 = 10.1mm. In other words, the fact that the object is not at infinity, elongates the image distance. If the object is 1 meter from the lens, using a 10mm lens, the image forms 10.1mm downstream from the lens. We were using 10mm as the height of the image triangle. In actually this should be 10.1mm Nothing is easy!

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  • $\begingroup$ Thanks for your contribution. Mmm, What do you suggest as introduction to this subject, I would like to read more about it. By the way, you can comment instead of adding new answer every time. ( I think it is not allowed to add more than one answer to one question, but I don't really know, I am new here). $\endgroup$
    – Physician
    Dec 14 '20 at 22:15

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