Oswald in the doorway: the blog of the Oswald Innocence Campaign, by Ralph Cinque

Saturday, June 7, 2014

I have reading about the angle of view of a photograph, and there is Math involved in determining it. It involves a horizontal, vertical, and diagonal component.

Calculating a camera's angle of view[edit]

For lenses projecting rectilinear (non-spatially-distorted) images of distant objects, the effective focal length and the image format dimensions completely define the angle of view. Calculations for lenses producing non-rectilinear images are much more complex and in the end not very useful in most practical applications. (In the case of a lens with distortion, e.g., a fisheye lens, a longer lens with distortion can have a wider angle of view than a shorter lens with low distortion)^[2] Angle of view may be measured horizontally (from the left to right edge of the frame), vertically (from the top to bottom of the frame), or diagonally (from one corner of the frame to its opposite corner).
For a lens projecting a rectilinear image (focused at infinity, see derivation), the angle of view (α) can be calculated from the chosen dimension (d), and effective focal length (f) as follows:^[3]

\alpha = 2 \arctan \frac {d} {2 f}

d

represents the size of the film (or sensor) in the direction measured. For example, for film that is 36 mm wide,

d = 36

mm would be used to obtain the horizontal angle of view.
Because this is a trigonometric function, the angle of view does not vary quite linearly with the reciprocal of the focal length. However, except for wide-angle lenses, it is reasonable to approximate

\alpha\approx \frac{d}{f}

radians or

\frac{180d}{\pi f}

degrees.
The effective focal length is nearly equal to the stated focal length of the lens (F), except in macro photography where the lens-to-object distance is comparable to the focal length. In this case, the magnification factor (m) must be taken into account:

f = F \cdot ( 1 + m )

(In photography

m

is usually defined to be positive, despite the inverted image.) For example, with a magnification ratio of 1:2, we find

f = 1.5 \cdot F

and thus the angle of view is reduced by 33% compared to focusing on a distant object with the same lens.
A second effect which comes into play in macro photography is lens asymmetry (an asymmetric lens is a lens where the aperture appears to have different dimensions when viewed from the front and from the back). The lens asymmetry causes an offset between the nodal plane and pupil positions. The effect can be quantified using the ratio (P) between apparent exit pupil diameter and entrance pupil diameter. The full formula for angle of view now becomes:^[4]

\alpha = 2 \arctan \frac {d} {2 F\cdot ( 1 + m/P )}

Angle of view can also be determined using FOV tables or paper or software lens calculators.^[5]

Example[edit]

Consider a 35 mm camera with a lens having a focal length of F = 50 mm. The dimensions of the 35 mm image format are 24 mm (vertically) × 36 mm (horizontal), giving a diagonal of about 43.3 mm.
At infinity focus, f = F, and the angles of view are:

horizontally, $\alpha_h = 2\arctan\frac{h}{2f} = 2\arctan\frac{36}{2 \times 50}\approx 39.6^\circ$
vertically, $\alpha_v = 2\arctan\frac{v}{2f} = 2\arctan\frac{24}{2 \times 50}\approx 27.0^\circ$
diagonally, $\alpha_d = 2\arctan\frac{d}{2f} = 2\arctan\frac{43.3}{2 \times 50}\approx 46.8^\circ$

I am not going to attempt to critique the lines of Martin Hinrichs because to do that, I would have to apply the above formulas to Altgens' conditions, and I am not prepared to do that. But, I will state again that I have my doubts about his lines. This is how they compare to the example that they gave, which looks more like what I would expect. But again, I am not going to debate it.

Saturday, June 7, 2014

Calculating a camera's angle of view[edit]

Example[edit]

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