Monday, November 27, 2017

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: \[\kappa = \frac{(h+g)}{f} \frac{ (1-cos(\theta))}{sin(\theta)}\]

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