Let's assume that I have the following camera parameters confirmed:

  • Focal Length
  • F/#
  • Coc
  • And a Lens Focusing Distance (the focus distance it was optimized to in the manufacturing process of the lens).

I am trying to quantify how the HFOV and VFOV changes with changes to object distance. My knowledge gap at the moment is as follows:

I have a lens, which was manufactured to a specific focusing distance. This also has an associated near and far DOF. I would like to understand how the horizontal and vertical fields of view change as my object distance changes. My current calculation for HFOV and VFOV is:

Horizontal FOV (mm) = Focus Distance(mm) * TAN(Hor FOV (in degrees)/2) * 2

How do I incorporate a changing object distance into my calculation for the changing Horizontal and Vertical FOV?

  • If the lens was manufactured for a certain focus distance, does this mean you can't change the focus? Is the focal length fixed? Because then your (angular) FOV will remain constant as long as you don't change the sensor size.
    – Grebu
    Oct 10 '17 at 15:04
  • Yes, the focal length is fixed. So I'm thinking in terms of the "Pyramid" that is created from the lens -> outward. The horizontal FOV and vertical FOV will reduce as I get closer to the lens; and I am trying to quantify this.
    – Gary
    Oct 10 '17 at 15:06
  • The "pyramid" is usually considered to be a "frustum."
    – dav1dsm1th
    Oct 11 '17 at 11:34

As the focus distance and the focal lenght are fixed, the angular FOV is fixed. Assuming here that the (used) sensor size or the distance to the sensor won't change either.

Basically the formula of the question can be used, only the Focus distance has to be replaced by the object distance. Horizontal "linear" FOV = Object distance * tan(horizontal angular FOV/2)*2 This eventually leaves you with the problem of measuring the object distance. More accurately this formula calculates the horizontal width of the plane captured at the given object distance.

The result of tan(horizontal angular FOV/2)* 2 is a constant C in this configuration. Therefore increasing the object distance by x will increase the horizontal plane width by C*x. C is the change (or slope of the first equation). The same can be done with the vertical calculations.

  • This is the confirmation that I was looking for. Thanks!
    – Gary
    Oct 12 '17 at 13:25

Angle of View and Dimension of Angular Field: As you have discovered, our cameras sport a rectangular format. We can trace imaginary lines outward from our camera lens to determine the angle of view and the dimensions of the angular field. Such a trace reveals that the pyramid has a rectangular base. The size of this base expands with distance. The trace of the sides of this pyramid reveals that each of the four sides is in the shape of an iisosceles triangle.

We can use the trigonometry of right triangles to most easily solve. If we bisect an iisosceles triangle, we create two “right” triangles. A “right” triangle has one angle that equals 90°. Now we can use the trigonometric functions of “right” triangles (the easiest trig) and solve for the size angles and the dimensions of the base. Our solution will be for ½ the triangle; we multiply by 2 to discover these values for our iisosceles triangle.

For the angle of view – solve for horizontal angle of view APS-C format 24mm long with 30mm lens mounted.

Angle of view = (ArcTan (d÷ f÷ 2))X2

d = dimension

f = focal length

Angle of View = (ArcTan (24÷30÷2) X 2

Angle of View = (ArcTan (0.40) X 2

Angle of View = 21.8 X 2

Angle of View = 43.6°

Solve for vertical Angle of View

Angle of View = (ArcTan (16÷30÷2) X 2

Angle of View = (ArcTan (0.27) X 2

Angle of View = 14.9 X 2

Angle of View = 29.9°

Solve for the dimensions of the Angular Field. We bisect the angle to construct a “right” triangle from the side of the pyramid which is an iisosceles triangle. The height of this triangle is the distance lens to subject. The Tan function of the apex angle multiplied by the Tan value is the dimension sought.

Distance = 5 meters Lens mounted is 30mm

For Angular Field Vertical Divide 29.9° by 2 = 14.95° Tan 14.95 = 0.267 Multiply Tan by distance = 0.267 X 5 = 1.335
This is ½ of Vertical field thus 1.335 X 2 = 2.67 meters = 2,670mm = 105 inches

For Angular Field Horizontal Divide 43.6° by 2 = 21.8° Tan 21.8 = 0.40 Multiply Tan by distance = 0.40 X 5 = 1.99 This is ½ of Vertical field thus 1.99 X 2 = 3.99 meters = 3,999mm = 157 inches

For any other distance, multiply tan of ½ the angle by distance and then multiply by 2. Answer will be in the same unit of measure as used for distance value.


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