Photographing the horizon

Suppose you are on an island, looking out to see, how far away is the horizon?


We can approximate the earth as a sphere and apply pythagorous' theorem, then we find: \[ d^2 + r^2 = (h + r )^2 \] So the distance to the horizon is: \[d = \sqrt{h(h + 2r)}\] and the distance along the surface at sea level is: \[d_s = r \hspace{2 mm} sin^{-1} \left( \frac{d}{h+r} \right) \] Try out different heights to see the distance of the horizon. Also set the photo angle ( defined below ) to see the curvature of the horizon in a photo.

Height above sea level		meters
Distance to horizon		km
Distance along surface		km
Theta (photo angle)		degrees
Horizon curvature		%

If we apply pythagorous a couple of times to the diagram above and then do a bit of algebra, we find: \[g = \frac{d^2 - h^2}{2(r+h)}\] and \[ f = \sqrt{g ( 2r - g)}\]
Now suppose we take a photo of the horizon, what shape will the horizon appear to be when we have rendered the image? We will use the same ideas we used in the previous post Straight Lines in photos
The horizon will be a section of a cirle of radius f in a plane a distance h+g below the observer. When we point a camera horizontally and take a photo of the horizon, we assume that an image of that photo can be rendered on a screen a distance \(\nu\) from the observer, for whom all objects in the photo perfectly align with the real objects. Consider the cone which consists of the horizon circle at the base and with a vertex at the observer. The image of the horizon will be the intersection of the plane containing the screen and the cone. This is known as a conic section.

On the cone we have the equation: \[x^2_1 + x^2_3 = \rho^2\] The height of the cone is h+g and so, comparing similar triangles: \[ \frac{\rho}{x_2} = \frac{f}{h+g}\] Combining those two equations we find: \[ x^2_1 + x^2_3 = \frac{x^2_2 f^2 }{(h+g)^2} \] Now we are interested in intersection of that cone with the screen in the plane \(x_3 = \nu\)
We find \[x_2 = - \frac{h+g}{f} \sqrt{x^2_1 + \nu^2}\] That is an equation for a hyperbola. So the horizon will be a hyperbola.
We now have an expression for \(x_2\) in terms of \(x_1\), we could say that \(x_2\) is a function of \(x_1\) which we could write \(x_2(x_1)\)


We'll define \(\theta\) to be the angle in the horizontal plane from the centre of the screen to either vertical edge ( for the observer ).
We find: \[x_2( \nu tan(\theta)) = - \frac{h+g}{f} \nu \sqrt{1+tan^2(\theta)} \] and thus \[x_2( \nu tan(\theta)) = - \frac{(h+g) \nu}{f cos(\theta)} \] We define: \[\epsilon = x_2( 0 ) - x_2( \nu tan( \theta )) \] and then we measure the curvature of the horizon in the photo to be: \[\kappa = \frac{\epsilon}{\nu tan(\theta)}\] and we find: \[\kappa = \frac{(h+g)}{f} \frac{ (1-cos(\theta))}{sin(\theta)}\]

Stright lines in photos

In this post, I'll give a mathematical justification for the following statements about photos of line segments:
1: The photo of a straight line segment will be a straight line segment.
2: If the line segment is in a plane perpendicular to the direction that the camera is pointing, then the slope of the line in the photo will be the same as it is in reality.

As a result, if a camera is pointing in a horizontal direction, then vertical lines ( and vertical edges of objects ) will be vertical in the photo. Also a horizontal line which is in the plane perpendicular to the direction the camera is pointing will be horizontal in the photo.

I will show that those statements are true for regular photos, i.e. for photos which, when rendered as an image, maintain the original orientation, for an observer located in a particular location.

Consider a static image that is taken with a camera. We'll label the direction the camera is pointing as the \( x_3 \) axis. Suppose we render the image on a screen perpendicular to the \( x_3 \) asix, a distance \( \nu \) from the observer. The image of a point will the be rendered on the screen precisely in the spot that is covering the original point.
With 3 dimensional coordinates, we'll specify the position of the observer to be at the origin. The screen is in the pane \(x_3 = \nu \)
The \(x_1 \) axis is parallel to the ground and the \( x_2 \) axis is perpendicular to both \( x_1 \) and \( x_3 \).
When the camera is pointing in the horizontal plane, \( x_2 \) is vertical.
If the object is at \( \underline{x}^a = ( x_1^a , x_2^a, x_3^a ) \)
then the image should be somewhere along the line from the origin 0 to \( \underline{x}^a \)
In other words, for some real number \( \alpha \), the location of the images is given by: \[ Image(\underline{x}^a) =\underline{x}^a \alpha = ( x_1^a , x_2^a , x_3^a ) \alpha \] The image is on the screen so \[ \nu = x_3^a \alpha \] and so \[ \alpha = \frac{\nu}{x_3^a} \] hence \[ Image(\underline{x}^a) = ( \frac{x_1^a}{x_3^a} , \frac{x_2^a}{x_3^a} , 1 ) \nu \] Now suppose we had a line segment running from \( \underline{x}^a \) to \( \underline{x}^b \)

Let \( \underline{x}^{\lambda} \) be a point on that line segment such that: \[ \underline{x}^{\lambda} = \underline{x}^a ( 1 - \lambda) + \underline{x}^b \lambda \] where \( 0 \le \lambda \le 1\)
So \[ Image(\underline{x}^{\lambda}) = \left( \frac{x^a_1 (1 - \lambda ) + x^b_1 \lambda }{x^a_3 (1 - \lambda ) + x^b_3 \lambda} , \frac{x^a_2 (1 - \lambda ) + x^b_2 \lambda }{x^a_3 (1 - \lambda ) + x^b_3 \lambda} , 1 \right) \nu \] If we let \[ \beta =\frac{ x^b_3 \lambda}{x^a_3 (1 - \lambda ) + x^b_3 \lambda} \] then after doing some algebra we find: \[ Image(\underline{x}^{\lambda}) = \left( \frac{x^a_1}{x^a_3} ( 1 - \beta) + \frac{x^b_1}{x^b_3} \beta , \frac{x^a_2}{x^a_3} ( 1 - \beta) + \frac{x^b_2}{x^b_3} \beta , 1 \right) \nu \] And as we vary \( \lambda \), \( \beta \) changes and the image traces out a line segment on the screen from \( ( \frac{x^a_1}{x^a_3} , \frac{x^a_2}{x^a_3}, 1 ) \nu \) to \( ( \frac{x^b_1}{x^b_3} , \frac{x^b_2}{x^b_3}, 1 ) \nu \)
So we have shown that in our model, the image of a straight line segment is a straight line segment.
It took some effort using algebra to get to that statement. It turns out we could have arrived at that statement using some geometry. The image of the line segment from \( \underline{x}^a\) to \( \underline{x}^b\) will be along the intersection of the plain \( x_3 = \nu \) and the plane containing \( \underline{x}^a\), \( \underline{x}^b\) and the origin. The intersection of two planes is a line and so the image of the straight line segment will be a straight line segment.

Again let us consider the image of a line segment from \( \underline{x}^a\) to \( \underline{x}^b\) but this time let's assume that both \( \underline{x}^a\) and \( \underline{x}^b\) are equidistant from the plane containing the screen at \( x_3 = \nu \).
So \[ x^a_3 = x^b_3 \] In this case, the image will be rendered along the line segment from \( \left( \frac{x_1^a}{x^a_3}, \frac{x_2^a}{x^a_3} , 1 \right) \nu \)
to \( \left( \frac{x_1^b}{x^a_3}, \frac{x_2^b}{x^a_3} , 1 \right) \nu \)
Ignoring for a moment the \( x_3 \) axis, we find the slope of the line segment from \( \underline{x}^a\) to \( \underline{x}^b\), in the \( x_1 \) , \(x_2 \) plane is: \[ slope_{12}( \underline{x}^a, \underline{x}^b) = \frac{x^b_2 - x^a_2}{x^b_1 - x^a_1} \] which is the same as the slope of the line segment from \( Image( \underline{x}^a ) \) to \( Image( \underline{x}^b ) \).
So in this case the slope is preserved.
From a practical point of view, if a camera is pointing in the horizontal plane and it is used to take a photo of a building with some vertical lines, then those vertical lines will be preserved in the photo.

The image of a horizontal line segment which is parallel to the \( x_3 = \nu \) pane ( parallel to the screen ), will be rendered as a horizontal line in the photo.
So if you take a photo with the camera pointing perpendicularly at a wall, then horizontal lines on that wall will be horizontal lines in the photo.
If you are perpendicular to a flat straight road, then a line at the edge of a road will appear as horizontal, going across a photo.

On the other hand, if the photographer is very close to a building and points the camera upwards, then the real vertical axis will not be parallel to the \( x_2 \) axis and so a real-life vertical line, may not appear to be vertical in the photo. In fact the building may appear to lean back in the photo.

All bets are off when a fish-eye lens is used and then the image is rendered on a flat screen. The objects in the image will not be aligned with the real objects and you may observe straight lines to appear to become curved in the photo.