A formula is derived for the curvature of the horizon's image in photos of a sphere of radius R , taken by a camera with horizontal view angle alpha from height h above the sphere's surface. The formula is validated by means of an interactive GeoGebra construction: a key angle calculated from the formula derived here is compared to the angle actually present in the construction. Using the validated formula, the amount of curvature expected to be present in a photo of the Earth's horizon from an altitude of 3 m is calculated. The result is an order of magnitude smaller than typical degrees of barrel distortion present in consumers' digital cameras. Therefore, claims that "flat horizons in photos of waterscapes prove that the Earth is flat" are untenable.

1. Calculation of the Curvature Expected in
Photographs of a Sphere’s Horizon
Jim Smith
QueLaMateNoTeMate.webs.com
April 9, 2016
Abstract
A formula is derived for the curvature of the horizon’s image in
photos of a sphere of radius R, taken by a camera with horizontal
view angle α from height h above the sphere’s surface. The formula
is validated by means of an interactive GeoGebra construction: a
key angle calculated from the formula derived here is compared to
the angle actually present in the construction. Using the validated
formula, the amount of curvature expected to be present in a photo
of the Earth’s horizon from an altitude of 3 m is calculated. The
result is an order of magnitude smaller than typical degrees of
barrel distortion present in consumers’ digital cameras. Therefore,
claims that “flat horizons in photos of waterscapes prove that the
Earth is flat” are untenable.
Contents
1 Relation between the position of an object’s image in
a photo and the object’s position with respect to the
camera 2
2 Formulation of the problem 3
3 Relation between features on the sphere and the image
of those features on a photo of the sphere 3
4 Strategy for solving the problem 5
1

2. 5 Solution 5
5.1 Deriving expressions for key distances . . . . . . . . . . . 5
5.2 Deriving expressions for tan θ and d . . . . . . . . . . . . 6
6 Formula for the “drop” of the horizon from the center of
the photo to the edge 7
7 Validation of the results 8
7.1 Validation of the results . . . . . . . . . . . . . . . . . . . 8
8 Sample calculation of d/w for a photo of the Earth’s hori-
zon, and some comments 9
9 References 10
1 Relation between the position of an object’s
image in a photo and the object’s position
with respect to the camera
In a camera, the image of an object is formed on a film or sensor. For
sake of convenience, we’ll say that it’s a film. If the optical axis of the
camera passes through the center of the film, and is perpendicular to it,
then the distance from the center of a photo to the image of a distant
object is proportional to the tangent of the angle γ shown below:
2

3. 2 Formulation of the problem
Every point on the line of the horizon in a photograph is the image of a
point of tangency drawn from the camera to the sphere’s surface. We’ll
specify that the camera is aimed directly at the horizon. Thus, the line
of the horizon on the photo will have its highest point (T) in the center,
and will curve downward symmetrically to left and right.
We’ll quantify the curvature of the horizon’s image via the the ration
of the drop (d) of the horizon’s image to the photo’s width (w). We’ll ex-
press that ration in terms of the sphere’s radius, the camera’s horizontal
view angle, and the height of the camera above the sphere.
3 Relation between features on the sphere and
the image of those features on a photo of the
sphere
The two points of intersection (P1, P2) of the horizon with the edges
of the photo are images of points of tangency (Q1, Q2) drawn from the
camera to the surface of the sphere. Those two points of tangency, plus
the camera (C), define the “Blue Plane” that cuts the sphere as in the
two figures shown below:
3

4. The blue plane forms an angle θ with the tangent drawn from C to the
point, T, that is imaged at the center of the photo.
4

5. From the foregoing, we can deduce that
d
w
=
tan θ
2 tan α
2
. (1)
4 Strategy for solving the problem
We’ll express tan θ in terms of the sphere’s radius R, the camera’s height
h, and the camera’s horizontal view angle α. Having done so, we’ll be
able to write Eq. (1) in terms of those quantities.
5 Solution
We’ll begin by deriving expressions for the key distances t and CS.
5.1 Deriving expressions for key distances
The distance t can be expressed in many ways. We’ll apply the Pythagorean
Theorem to the triangle whose right angle is at T, thereby finding that
t = (R + h)2
− R2. (2)
We can find an expression for the distance CS by examining the
“blue plane”:
CS =
t
cos α
2
=
(R + h)2
− R2
cos α
2
(3)
5

6. 5.2 Deriving expressions for tan θ and d
From the figure shown below, θ = φ − ψ. Therefore,
tan θ =
sin θ
cos θ
=
sin φ cos ψ − cos φ sin ψ
cos φ cos ψ + sin φ sin ψ
. (4)
6

7. From that same figure, we obtain the following expressions for sin ψ,
cos ψ, sin φ, and cos φ:
sin φ =
CS
R + h
=
(R + h)2
− R2
(R + h) cos α
2
∴ cos φ = 1 − sin2
φ =
R2 − (R + h)2
sin2 α
2
(R + h) cos α
2
sin ψ =
t
R + h
=
(R + h)2
− R2
(R + h)
cos ψ =
R
R + h
.
Substituting those expressions in Eq. (4), then simplifying and rearrang-
ing, we obtain
tan θ =
(R + h)2
− R2 R − R2 − (R + h)2
sin2 α
2
(R + h)2
− R2 + R R2 − (R + h)2
sin2 α
2
. (5)
6 Formula for the “drop” of the horizon from
the center of the photo to the edge
From Eqs. (1) and (5),
7

8. d =
w
2 tan α
2



(R + h)2
− R2 R − R2 − (R + h)2
sin2 α
2
(R + h)2
− R2 + R R2 − (R + h)2
sin2 α
2



. (6)
7 Validation of the results
7.1 Validation of the results
The figures shown in this document are portions of screen shots from the
interactive GeoGebra worksheet Simulate Low-Altitude Photo of Horizon
of Sphere. In addition to simulating the image of a horizon, that spread-
sheet calculated tan θ in two ways: from Eq. (5), and from the lengths
of segments t and CS that resulted when figures like those shown in this
document were constructed according to user-selected values of R, h,
and α. The results of those calculations agree, as shown in the following
detail from a screen shot of an example of the worksheet’s output:
8

9. 8 Sample calculation of d/w for a photo of the
Earth’s horizon, and some comments
It’s interesting to calculate the ratio d/w for photos of the Earth’s horizon
taken from low elevations. Either analytically or by using the GeoGebra
construction “Simulate Low-Altitude Photo of Horizon of Sphere” , we
can find that d/w increases with α for given values of R and h. Therefore,
to estimate an upper bound on d/w, we’ll assume that α = 90◦. (We’ll
comment on that choice shortly.) The Earth’s radius is 6370 km, so for
a camera mounted 3 m above the ground, d/w = 0.0002:
For an image 300 mm wide, the “drop” would be only 0.6 mm.
Our derivation assumed that the camera is optically perfect. How-
ever, wide-angle lenses (such as the one we’ve assumed) would almost cer-
tainly suffer from barrel distortion. For a discussion of this phenomenon,
see this introduction to distortions, and this review of a high-quality
zoom lens. Those references indicate in typical consumer-grade digital
cameras, the distortions are equal to about 1% of the photo’s height.
In other words, the distortions are an order of magnitude larger than
9

10. the value of d/w that we calculated. Therefore, claims that “flat horizons
in photos of waterscapes prove the Earth is flat!” are untenable.
9 References
Bockaert Vincent. (Date unknown; retrieved 8 April 2016) ”Barrel Dis-
tortion” Digital Photography Review.
http://www.dpreview.com/glossary/optical/barrel-distortion
GeoGebra worksheet, “Simulate Low-Altitude Photo of Horizon of Sphere”.
http://www.geogebra.org/material/simple/id/3130935. 2016.
Westlake Andy. 2008 (Retrieved 8 April 2016)) “Olympus Zuiko Digital
12-60mm 1:2.8-4.0 review”. Digital Photography Review
http://www.dpreview.com/reviews/olympus-12-60-2p8-4-o20/3.
10