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\documentclass[letterpaper,10pt]{article}
\usepackage{amsmath}
\usepackage[empty,cm]{fullpage}
\pagenumbering{gobble}
\begin{document}
\title{Pannellum Equirectangular Projection Reference}
\author{Matthew Petroff}
\date{January 2013}
\maketitle
Let a coordinate system be defined with the origin at the focal point, the
x-axis parallel to the camera's horizontal axis with positive to the right as
viewed from behind the camera, the y-axis parallel to the camera's vertical
axis with positive upward as viewed from behind the camera, and the z-axis be
perpendicular to these two axes with positive extending away from the
camera.\footnote{Note that this is a left-handed coordinate system.}
Let \(\lambda\) represent the panorama's longitude, \(\phi\) the panorama's
latitude, \(\theta\) the latitudinal offset, \(\psi\) the longitudinal offset,
and \(f\) the focal length. Let \(\mathbf{a}\) represent a point's position
vector, \(\mathrm{R}_x(\theta)\) the latitudinal rotation matrix,
\(\mathbf{d}\) a point's position vector in the camera's reference frame, \(x\)
and \(y\) a point's position on the image plane, and \(p_x\) and \(p_y\) a
point's position on an equirectangular projection.
\begin{equation} \label{eq:posvec}
\mathbf{a} = \begin{bmatrix}
\mathbf{a}_x \\ \mathbf{a}_y \\ \mathbf{a}_z
\end{bmatrix} = \begin{bmatrix}
\sin(\lambda)\cos(\phi) \\ \sin(\phi) \\ \cos(\phi)\cos(\lambda) \end{bmatrix}
\end{equation}
\begin{equation} \label{eq:rotmat}
\mathrm{R}_x(\theta) = \begin{bmatrix}
1 & 0 & 0 \\
0 & \cos(\theta) & -\sin(\theta) \\
0 & \sin(\theta) & \cos(\theta)
\end{bmatrix}
\end{equation}
With a point's position vector \eqref{eq:posvec} and the latitudinal rotation
matrix \eqref{eq:rotmat}, one can calculate a point's position in the camera's
reference frame.\nopagebreak
\begin{equation}
\mathbf{d} = \begin{bmatrix}
\mathbf{d}_x \\ \mathbf{d}_y \\ \mathbf{d}_z
\end{bmatrix} = \mathrm{R}_x(\theta)\mathbf{a} = \begin{bmatrix}
1 & 0 & 0 \\
0 & \cos(\theta) & -\sin(\theta) \\
0 & \sin(\theta) & \cos(\theta)
\end{bmatrix}
\begin{bmatrix}
\sin(\lambda)\cos(\phi) \\ \sin(\phi) \\ \cos(\phi)\cos(\lambda)
\end{bmatrix} = \begin{bmatrix}
\sin(\lambda)\cos(\phi) \\
\sin(\phi)\cos(\theta) - \cos(\phi)\cos(\lambda)\sin(\theta) \\
\sin(\phi)\sin(\theta) + \cos(\phi)\cos(\lambda)\cos(\theta)
\end{bmatrix}
\end{equation}
Using the pinhole camera model, one can then calculate positions on the
image plane.\nopagebreak
\begin{equation}
\begin{bmatrix} x \\ y \end{bmatrix} = \frac{f}{\mathbf{d}_z}
\begin{bmatrix} \mathbf{d}_x \\ \mathbf{d}_y \end{bmatrix}
= \frac{f}{\sin(\phi)\sin(\theta) + \cos(\phi)\cos(\lambda)\cos(\theta)}
\begin{bmatrix}
\sin(\lambda)\cos(\phi) \\
\sin(\phi)\cos(\theta) - \cos(\phi)\cos(\lambda)\sin(\theta)
\end{bmatrix}
\end{equation}
\begin{equation} \label{eq:xf}
\frac{x}{f} = \frac{\sin(\lambda)\cos(\phi)}{\sin(\phi)\sin(\theta) +
\cos(\phi)\cos(\lambda)\cos(\theta)}
\end{equation}
\begin{equation} \label{eq:yf}
\frac{y}{f} = \frac{\sin(\phi)\cos(\theta) -
\cos(\phi)\cos(\lambda)\sin(\theta)}{\sin(\phi)\sin(\theta) +
\cos(\phi)\cos(\lambda)\cos(\theta)}
\end{equation}
Solving the system of \eqref{eq:xf} and \eqref{eq:yf} for \(\lambda\) and
\(\phi\) and adding the longitudinal offset, one is able to find a point's
position on the unit sphere from its position on the image plane.\nopagebreak
\begin{equation}
\lambda = \tan ^{-1}\left(\frac{x}{\sqrt{x^2 + \left(f \cos(\theta) -
y \sin(\theta) \right)^2}},\frac{f \cos(\theta) - y \sin(\theta)}{\sqrt{x^2 +
\left(f \cos(\theta) - y \sin(\theta) \right)^2}}\right) + \psi
\end{equation}
\begin{equation}
\phi = \tan ^{-1}\left(\frac{y \cos(\theta) + f \sin(\theta)}{\sqrt{x^2 +
\left(f \cos(\theta) - y \sin(\theta) \right)^2}}\right)
\end{equation}
Once a point's position on the unit sphere is known, it is trivial to find its
position on an equirectangular projection.\nopagebreak
\begin{equation}
p_x = \frac{\lambda}{\pi}
\end{equation}
\begin{equation}
p_y = \frac{\phi}{\frac{\pi}{2}}
\end{equation}
\end{document}