text:
\section{Introduction}

It is common to encounter celestial bodies rotating about their axis of largest moment of inertia.
This naturally stable configuration describes the planets in our Solar System as well as many small bodies (e.g., asteroids, minor planets). Indeed, principal axis rotation is the norm for asteroids, with significant departures from this state primarily occurring for slowly rotating bodies \cite{Pravec:2000}.
Therefore, given the prevalence of principal axis rotators in the Solar System, it is interesting to consider how these dynamics might be exploited for shape modeling, navigation, and other practical purposes. Although principal axis rotation leads to a variety of helpful geometric properties, this work focuses on just one particular insight: the inertial path of any single point on the rotating body's surface must trace out a circle relative to the body's center. This is true regardless of the body's actual three-dimensioanl (3-D) shape. Such circles are commonly referred to as a \emph{circle of latitude} (CoL). The body's axis of rotation (i.e., the \emph{pole}) is perpendicular to the plane of each CoL and passes through each circle's center (see Fig.~\ref{fig:PoleGeom}). These CoLs project to ellipses in images collected by cameras (or telescopes), and the shape and orientation of these image ellipses may be used to infer the orientation of the observed body's pole. Moreover, the relative arrangement and sizes of many elliptical CoL projections may be used to constrain the body's shape to an unknown global scale.

\begin{figure}[b!]
    	\centering
    	\includegraphics[width=0.25\textwidth]{PoleGeom.png}
    	\caption{A circle of latitude (CoL) on a rotating celestial body lies in a plane perpendicular to the pole.}
    	\label{fig:PoleGeom}
\end{figure}

This work was inspired by a number of recent studies seeking to estimate the pole of an unknown spinning asteroid using the elliptical projection of CoLs \cite{Kuppa:2022a,Kuppa:2022b}. The related problem of estimating the latitude of a spacecraft from an image of constant-latitude bands in Jupiter's atmosphere has also been of recent interest \cite{Rivera:2020,Rivera:2022,Rivera:2023}. As compared to these prior works, this manuscript aims to (1) generalize the scenarios under which the pole may be estimated from the perspective projection of CoLs and (2) considerably expand the geometric information that might be extracted from such features.

The recent work of Kuppa, McMahon, and  Dietrich \cite{Kuppa:2022a,Kuppa:2022b} and of Danas Rivera and Peck \cite{Rivera:2020,Rivera:2022,Rivera:2023} provide excellent insights into the image processing required to localize the elliptical features in an image, and this piece of the problem is not discussed further here. These prior works, however, all make a number of limiting assumptions in the development of their navigation or reconstruction algorithms. In particular, they assume that the camera is pointed towards the center of the body and that the distance to the body is known. Neither of these assumptions are necessary to determine pole orientation or recover shape to scale. Although knowledge of the body's distance would remove the scale ambiguity, it is otherwise not needed. In addition to the assumptions on pointing and distance, the algorithms in these earlier works also require other pieces of information (or make other assumptions) that are not necessary---though the specifics vary from algorithm to algorithm. As is shown herein, all that is needed to determine the pole are the elliptical projections of CoLs in images from a calibrated camera. It does not matter where the observed CoLs lie on the body's surface nor does it matter where the camera is pointed or located relative to the body. Moreover, there is no need to correspond the observed CoLs with any known point, feature, or latitude band on the observed body.

In addition to enhanced algorithmic generality, this work introduces a number of new capabilities, including an analytic expression for estimated pole covariance, a procedure for shape reconstruction (to scale), and a means for resolving the unknown global scale in many practical cases. The advancements described here are made possible by transferring classic computer vision results on pose from circles \cite{Shiu:1989,Dhome:1990,Forsyth:1991,Kanatani:1993,Christian:2017AAS} to the problem of pole estimation, along with some novel application-specific improvements.

\section{Perspective Projection of Circles}
Conics project to conics under the action of perspective projection \cite{Hartley:2003}. In this case, the observed conic is a circle of latitude that (in practice) almost always projects to an ellipse in an image. The geometry of circle projections is now briefly reviewed.

Consider a CoL with radius $R$. Define a coordinate frame $L$ with its origin at the circle center and the $Z$-axis perpendicular to the plane of the circle (i.e., along the pole). Thus, the $X$-axis and $Y$-axis span the plane of the circle and any two-dimensional (2-D) point $\{X,Y\}$ within the $Z=0$ plane belonging to the circle must satisfy the implicit equation
\begin{equation}
    \label{eq:CircleSimple0}
    X^2 + Y^2 - R^2 = 0
\end{equation}
Defining the $3 \times 3$ matrix $\textbf{\emph{C}}_L$ of arbitrary scale,
\begin{equation}
   \textbf{\emph{C}}_L \propto \begin{bmatrix}
        1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -R^2
   \end{bmatrix}
\end{equation}
one may rewrite Eq.~\eqref{eq:CircleSimple0} as
\begin{equation}
    \label{eq:CircleConicImplicit}
    \bar{\textbf{\emph{p}}}^T_L \textbf{\emph{C}}_L \bar{\textbf{\emph{p}}}_L = 0
\end{equation}
where $\bar{\textbf{\emph{p}}}_L \in {\mathbb P}^2$ is given by
\begin{equation}
    \bar{\textbf{\emph{p}}}_L \propto \begin{bmatrix}
        X \\ Y \\ 1
   \end{bmatrix}
\end{equation}

Now suppose this CoL is viewed by a camera that produces images obeying perspective projection.
Following the camera frame conventions of Ref.~\cite{Christian:2021d} that places the camera's $+z$-axis out of the camera, a homogeneous point $\bar{\textbf{\emph{p}}}$ lying on the circle appears in a 2-D image at homogeneous coordinates $\bar{\textbf{\emph{x}}} \in {\mathbb P}^2$
\begin{equation}
    \begin{bmatrix}
        x \\ y \\ 1
   \end{bmatrix} = \bar{\textbf{\emph{x}}} \propto \textbf{\emph{T}}^L_C \textbf{\emph{S}} \bar{\textbf{\emph{p}}}_L + \textbf{\emph{r}}_C
\end{equation}
where $\textbf{\emph{r}}_C \in {\mathbb R}^3$ is the vector from the camera to the center of the CoL (expressed in the camera frame $C$), $\textbf{\emph{T}}^L_C = [\textbf{\emph{t}}_1, \textbf{\emph{t}}_2, \textbf{\emph{t}}_3]$ is the rotation matrix from frame $L$ to $C$ with columns given by $\{\textbf{\emph{t}}_i\}_{i=1}^3$, and $\textbf{\emph{S}}$ is the matrix
\begin{equation}
    \textbf{\emph{S}} =  \begin{bmatrix}
        1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 0
   \end{bmatrix}
\end{equation}
Which, simplifies to
\begin{equation}
   \bar{\textbf{\emph{x}}} \propto \textbf{\emph{H}} \bar{\textbf{\emph{p}}}
\end{equation}
where $\textbf{\emph{H}}$ is the homography matrix
\begin{equation}
   \textbf{\emph{H}} \propto \begin{bmatrix}
        \textbf{\emph{t}}_1 & \textbf{\emph{t}}_2 & \textbf{\emph{r}}_C
   \end{bmatrix}
\end{equation}
and it follows that
\begin{equation}
    \bar{\textbf{\emph{p}}} \propto \textbf{\emph{H}}^{-1} \bar{\textbf{\emph{x}}}
\end{equation}
Substituting this result into Eq.~\eqref{eq:CircleConicImplicit},
\begin{equation}
    \bar{\textbf{\emph{p}}}^T \textbf{\emph{C}} \bar{\textbf{\emph{p}}} = \bar{\textbf{\emph{x}}}^T\textbf{\emph{H}}^{-T} \textbf{\emph{C}} \textbf{\emph{H}}^{-1} \bar{\textbf{\emph{x}}} = 0
\end{equation}
Thus, defining $\textbf{\emph{A}}$ to be the symmetric $3 \times 3$ matrix of arbitrary scale defining the conic resulting from projection of the CoL into the image plane,
\begin{equation}
    \textbf{\emph{A}} \propto \textbf{\emph{H}}^{-T} \textbf{\emph{C}} \textbf{\emph{H}}^{-1}
\end{equation}
and one observes an image plane conic  of
\begin{equation}
    \label{eq:ImagePlaneConicQuad}
    \bar{\textbf{\emph{x}}}^T \textbf{\emph{A}} \bar{\textbf{\emph{x}}} = 0
\end{equation}
Defining the elements of $\textbf{\emph{A}}$ to be,
\begin{equation}
    \label{eq:DefA}
    \textbf{\emph{A}} = \begin{bmatrix}
        A & B/2 & D/2 \\
        B/2 & C & F/2 \\
        D/2 & F/2 & G
    \end{bmatrix}
\end{equation}
then it follows that Eq.~\eqref{eq:ImagePlaneConicQuad} describes the implicit equation for a conic
\begin{equation}
    \label{eq:ImplicitConic}
    A x^2 + Bxy + Cy^2 + Dx + Fy + G = 0
\end{equation}
which is an ellipse (or circle) when $B^2 < 4AC$. Equations relating the coefficients $A,\hdots,G$ to geometric parameters (e.g., ellipse center, semi-major axis) are given in the appendix. This will become important in Section~\ref{Sec:PoleCovariance} for the development of analytic covariance expressions.

Points on the image plane $[x,y]$ are related to the pixel coordinates $[u,v]$ in the corresponding digital image through an affine transformation,
\begin{equation}
    \begin{bmatrix}
        u \\ v \\ 1
   \end{bmatrix} = \bar{\textbf{\emph{u}}} =  \textbf{\emph{K}} \bar{\textbf{\emph{x}}}
\end{equation}
where $\textbf{\emph{K}}$ is the so-called camera calibration matrix (see Refs.~\cite{Hartley:2003} and \cite{Christian:2021d}). In almost all space applications the camera is calibrated and the matrix $\textbf{\emph{K}}$ is known (e.g., Refs.~\cite{Christian:2016} and \cite{Golish:2020}). Thus, a measured pixel coordinate $\bar{\textbf{\emph{u}}}$ may always be transformed into the image plane according to $\bar{\textbf{\emph{x}}} = \textbf{\emph{K}}^{-1} \bar{\textbf{\emph{u}}}$. Moreover, suppose an ellipse is fit to observed image points $[u,v]$ using a standard ellipse fitting method \cite{Fitzgibbon:1999,Chojnacki:2000,AlSharadqah:2009,Kanatani:2011,Szpak:2015} (with the ``semihyper'' least squares method from Ref.~\cite{Kanatani:2011} being especially recommended). If the image ellipse in pixel coordinates is described by the matrix $\textbf{\emph{B}}$ such that,
\begin{equation}
   \bar{\textbf{\emph{u}}}^T \textbf{\emph{B}} \bar{\textbf{\emph{u}}} = 0
\end{equation}
then the camera calibration matrix may be used to transform image ellipse $\textbf{\emph{B}}$ to image plane ellipse $\textbf{\emph{A}}$ according to
\begin{equation}
   \textbf{\emph{A}} \propto \textbf{\emph{K}}^T \textbf{\emph{B}} \textbf{\emph{K}}
\end{equation}
Consequently, the remainder of this work operates entirely in image plane coordinates $\bar{\textbf{\emph{x}}}$ and image plane ellipses $\textbf{\emph{A}}$---both of which may be readily obtained from a digital image collected by a calibrated camera.

\section{Estimating Pole Direction from Circle of Latitude Projections}
\label{Sec:PoleDirection}
The orientation of the pole for a rotating celestial body coincides with the normal direction to the plane of every CoL. Since each CoL projects to an ellipse in an image, the apparent image ellipse may be used to determine the pole orientation as expressed in the camera frame. This may be achieved by realizing that an image ellipse back-projects to a cone. If the observed image ellipse is known to originate from a circular feature (e.g., a CoL), then the objective is to find the two plane orientations whose intersection with this cone create a circle. This is a classical and well-studied result. While the solutions to this problem vary slightly \cite{Shiu:1989,Dhome:1990,Forsyth:1991,Kanatani:1993,Christian:2017AAS}, most fundamentally follow a two step procedure: (1) back-project the image ellipse to get a cone and (2) find the two plane orientations that cut a circle in this cone. These two steps are now discussed, followed by the development of an analytic covariance model.

\subsection{Cone Back-Projection}
The back-projection of rays from the camera through the image ellipse forms a conical surface \cite{Hartley:2003}. To see this, define a 3-D point $\textbf{\emph{y}}\in {\mathbb R}^3$ lying on the ray described by image plane point $\bar{\textbf{\emph{x}}}$ such that $\textbf{\emph{y}} = a \bar{\textbf{\emph{x}}}$ for $a > 0$. Substituting this into Eq.~\eqref{eq:ImagePlaneConicQuad}, any point $\textbf{\emph{y}}$ belonging to the cone must satisfy
\begin{equation}
    \textbf{\emph{y}}^T \textbf{\emph{A}} \textbf{\emph{y}} = 0
\end{equation}
The observed image plane ellipse described by $\textbf{\emph{A}}$ is the conic section of this cone formed by an intersection with the camera frame's $z=1$ plane. Likewise, in the present application, the circle of latitude is a different conic section of this same cone.

Since $\textbf{\emph{A}}$ is a symmetric $3 \times 3$ matrix, an eigendecomposition may be used to diagonalize the system. Moreover, since $\textbf{\emph{A}}$ is guaranteed to be an indefinite matrix when describing an ellipse, the result is three eigenvalues $\{\lambda_i\}^3_{i=1}$ of mixed sign. Thus, recalling that the scale (and sign) of $\textbf{\emph{A}}$ is arbitrary, one can write without loss of generality
\begin{equation}
    \label{eq:ConeEigenvalueOrder}
    \lambda_1 \geq \lambda_2 > 0 > \lambda_3
\end{equation}
which will always be the case if $\textbf{\emph{A}}$ is scaled such that $\text{det}(\textbf{\emph{A}})=-1$. This is the particular scaling of $\textbf{\emph{A}}$ used in all numerical examples within this work, though different scalings are equally reasonable (so long as one is careful with the eigenvalue sign conventions).

The three corresponding unit eigenvectors $\{\textbf{\emph{u}}_i\}_{i=1}^3$ form an orthonormal basis (since $\textbf{\emph{A}}$ is symmetric) and describe the principal axes of the cone (as thoroughly discussed in Ref.~\cite{Christian:2017AAS}). This is denoted as frame $P$. The eigenvector corresponding to the eigenvalue of unique sign describes the cone's centerline, which will be $\textbf{\emph{u}}_3$ given the convention from Eq.~\eqref{eq:ConeEigenvalueOrder}.
Moreover, since the signs of $\pm \textbf{\emph{u}}_i$ are ambiguous, choose the sign of $\pm \textbf{\emph{u}}_3$ to be positive out of the camera: $\textbf{\emph{k}}^T \textbf{\emph{u}}_3 > 0$, where $\textbf{\emph{k}}^T = [0, 0, 1]$. Also make the system right-handed by adjusting the sign of $\pm \textbf{\emph{u}}_2$ to ensure that $\textbf{\emph{u}}_1 \times \textbf{\emph{u}}_2 = \textbf{\emph{u}}_3$. Therefore, one may construct a rotation matrix to transform vectors from the cone principal axis frame $P$ to the camera frame $C$ as
\begin{equation}
    \label{eq:RotationConeToCamera}
    \textbf{\emph{T}}^P_C = \begin{bmatrix}
        \textbf{\emph{u}}_1 & \textbf{\emph{u}}_2 & \textbf{\emph{u}}_3
    \end{bmatrix}
\end{equation}

\subsection{Estimating Pole Direction from a Single Circle Projection}
Given a cone, it is always possible to find two plane orientations whose conic sections are a circle. Different planes of this orientation will produce circles of different radius, which will become important later. There are a few different ways to describe these two planes found in the literature and they will be briefly reviewed here (and shown to be equivalent).

It has been shown by numerous authors \cite{Shiu:1989,Christian:2017AAS} that the two normal directions $\textbf{\emph{n}}_P \in {\mathbb P}^2$ producing circular conic sections in the cone principal axis frame are given by
\begin{equation}
    \label{eq:PlaneNormalCone0}
    \textbf{\emph{n}}_{P} \propto
    \begin{bmatrix}
        \pm \sqrt{\alpha} \\ 0 \\ -\sqrt{ -\lambda_3 / \lambda_1}
    \end{bmatrix}
\end{equation}
where
\begin{equation}
    \label{eq:DefAlpha}
    \alpha = \frac{1/\lambda_2 - 1/\lambda_1}{1/\lambda_2 - 1/\lambda_3} = \frac{\lambda_3}{\lambda_1}\frac{\lambda_1 - \lambda_2}{\lambda_3 - \lambda_2}
\end{equation}
Note the proportional sign in Eq.~\eqref{eq:PlaneNormalCone0} which indicates that $\textbf{\emph{n}}_P \in {\mathbb P}^2$ is not a unit vector and simply defines a direction in ${\mathbb R}^3$. Since the physical interpretation of this can be somewhat opaque, substitute for $\alpha$ in Eq.~\eqref{eq:PlaneNormalCone0} and expand
\begin{equation}
    \label{eq:NormalInConeFrameSimple}
    \textbf{\emph{n}}_{P} \propto
    \begin{bmatrix}
        \pm \sqrt{(\lambda_1 - \lambda_2)/(\lambda_2 - \lambda_3)} \\ 0 \\ 1
    \end{bmatrix} \propto
    \begin{bmatrix}
        \pm \sqrt{\lambda_1 - \lambda_2} \\ 0 \\ \sqrt{\lambda_2 - \lambda_3}
    \end{bmatrix}
\end{equation}
which represents a rotation of the vector $\textbf{\emph{k}} = [0,0,1]^T$ about the $P$ frame's $y$-axis by the angle $\pm \theta$,
\begin{equation}
    \label{eq:CirclePlaneAngle}
   \tan \pm \theta = \sqrt{ \frac{\lambda_1 - \lambda_2}{\lambda_2 - \lambda_3} }
\end{equation}
which is exactly the angle described by Forsyth, et al. \cite{Forsyth:1991}. Thus, one could equivaletnly write $\textbf{\emph{n}}_P \propto \textbf{\emph{T}}_2(\pm \theta) \, \textbf{\emph{k}}$, where $\textbf{\emph{T}}_2( \, \cdot \, )$ is the elementary rotation matrix about the second coordinate direction (i.e., $y_P$-axis).

It is sometimes desirable to write $\textbf{\emph{n}}_P$ explicitly as the unit vector $\hat{\textbf{\emph{n}}}_P = \textbf{\emph{n}}_P / \| \textbf{\emph{n}}_P \|$. Therefore, observe that the right-hand side of Eq.~\eqref{eq:NormalInConeFrameSimple} has a norm of
\begin{equation}
    \sqrt{ (\lambda_1 - \lambda_2) + (\lambda_2 - \lambda_3) } = \sqrt{ \lambda_1 - \lambda_3 }
\end{equation}
and so the unit normal in the cone's principal axis frame becomes
\begin{equation}
    \label{eq:UnitNormalInConeFrame}
    \hat{\textbf{\emph{n}}}_{P} =
    \begin{bmatrix}
        \pm \sqrt{(\lambda_1 - \lambda_2)/(\lambda_1 - \lambda_3 )} \\ 0 \\ \sqrt{(\lambda_2 - \lambda_3)/(\lambda_1 - \lambda_3 )}
    \end{bmatrix}
\end{equation}
Finally, since the rotation back into the camera frame is given by $\textbf{\emph{T}}^P_C$ from Eq.~\eqref{eq:RotationConeToCamera}
\begin{equation}
    \textbf{\emph{n}}_C \propto \textbf{\emph{T}}^P_C \textbf{\emph{n}}_P
\end{equation}
which may be written out as as a linear combination of the eigenvectors
\begin{equation}
    \textbf{\emph{n}}_C \propto \sqrt{\lambda_1 - \lambda_2} \; \textbf{\emph{u}}_1 \pm \sqrt{\lambda_2 - \lambda_3} \; \textbf{\emph{u}}_3
\end{equation}
or, if desired, explicitly as a unit vector
\begin{equation}
    \label{eq:UnitNormalFinal}
    \hat{\textbf{\emph{n}}}_C =  \sqrt{\frac{\lambda_1 - \lambda_2}{\lambda_1 - \lambda_3 }} \; \textbf{\emph{u}}_1 \pm \sqrt{\frac{\lambda_2 - \lambda_3}{\lambda_1 - \lambda_3 }} \; \textbf{\emph{u}}_3
\end{equation}
which agrees with the result presented in \cite{Kanatani:1993}. Note that the eigenvalue convention from Eq.~\eqref{eq:ConeEigenvalueOrder} guarantees that $\lambda_1 - \lambda_3 > 0$ and there is no risk of dividing by zero.  In summary, the different formulations of Refs.~\cite{Shiu:1989,Dhome:1990,Forsyth:1991,Kanatani:1993,Christian:2017AAS} all give the exact same result and there is no compelling reason to prefer one over the other. The reader interested in a detailed derivation is directed to these references.

It is insightful to consider what happens in the case of orthographic projection---which one approaches in the limit when the observed body is along the camera boresight and the distance becomes very large. Images from narrow field of view cameras or telescopes often appear nearly orthographic, and so this limiting case helps build geometric intuition. The reader is strongly cautioned, however, to not use this limiting geometry for pole estimation as its performance does not deteriorate gracefully. The analytic result from Eq.~\eqref{eq:UnitNormalFinal} should always be used to actually solve for the pole.  Therefore, considering the limiting case of orthographic projection, observe that
\begin{equation}
    \sqrt{\frac{\lambda_1 - \lambda_2}{\lambda_1 - \lambda_3 }} \; \rightarrow \; \sqrt{1 - b^2/a^2} = e
\end{equation}
\begin{equation}
    \sqrt{\frac{\lambda_2 - \lambda_3}{\lambda_1 - \lambda_3 }} \; \rightarrow \;  b/a = \epsilon
\end{equation}
where $e$ is the ellipse eccentricity, $\epsilon$ is the ellipse ellipticity, and $a$ and $b$ are the image ellipse semi-major and semi-minor distances.
Recalling the observation from Ref.~\cite{Forsyth:1991} discussed in Eq.~\eqref{eq:CirclePlaneAngle}, if the angle $\theta$ is the angle between the cone principal axis $\textbf{\emph{u}}_3$ and the normal vector of the circular feature $\hat{\textbf{\emph{n}}}_C$, then $\sin \theta \rightarrow e$ and $\cos \theta \rightarrow \epsilon$ as one approaches orthographic projection and it is readily apparent that
\begin{equation}
    1 = \sin^2 \theta + \cos^2 \theta  \rightarrow e^2 + \epsilon^2 = 1
\end{equation}
It follows, then, that the unit normal $\hat{\textbf{\emph{n}}}_C$ tends towards
\begin{equation}
    \label{eq:NormalLimit0}
    \hat{\textbf{\emph{n}}}_C \; \rightarrow \; e \; \textbf{\emph{u}}_1 \pm \epsilon \; \textbf{\emph{u}}_3
\end{equation}
which remains exactly a unit vector, even if $e$ and $\epsilon$ do not exactly correspond to the correct normal orientation $\theta$. Again, while helpful for geometric intuition, the reader is urged to not use Eq.~\eqref{eq:NormalLimit0} for estimating $\hat{\textbf{\emph{n}}}_C$.
The approximation of Eq.~\eqref{eq:NormalLimit0} does not appear especially robust as the geometry departs from orthographic projection. The practical computation of $\hat{\textbf{\emph{n}}}_C$ should always be accomplished with Eq.~\eqref{eq:UnitNormalFinal} or one of it's equivalents.

\subsection{Approximate Covariance of Pole Direction for a Single Circle Projection}
\label{Sec:PoleCovariance}
It is often desirable to know the error covariance for the estimated unit vector $\hat{\textbf{\emph{n}}}_C$ from Eq.~\eqref{eq:UnitNormalFinal}. This task is made more difficult than usual since Eq.~\eqref{eq:UnitNormalFinal} depends on the eigenvectors and eigenvalues of $\textbf{\emph{A}}$ which change in a complicated manner with perturbations in the image ellipse's apparent shape. Fortunately, however, the error covariance of $\hat{\textbf{\emph{n}}}_C$ may still be found analytically using well-understood procedures for computing the Jacobian of eigenvalues and eigenvectors \cite{Magnus:1985,Liounis:2019}. The methods developed here work well everywhere except when camera's position is nearly along the 3-D line formed by the observed body's pole. Such a vantage point leads to the condition $\lambda_1 \approx \lambda_2$ where Eq.~\eqref{eq:UnitNormalFinal} is not differentiable. Although the estimate of $\hat{\textbf{\emph{n}}}_C$ from Eq.~\eqref{eq:UnitNormalFinal} is still valid, a different means of developing the covariance in this special scenario is needed---and this task is left to future work. Therefore, assuming the camera is not perfectly along the pole (i.e., that $\lambda_1 \neq \lambda_2$), the development is briefly summarized here and the full partial derivatives are provided in the appendix.

Defining $\boldsymbol{\lambda}^T = [\lambda_1, \lambda_2, \lambda_3]$ to be a vector of the three eigenvalues of $\textbf{\emph{A}}$, compute the $3 \times 9$  Jacobian matrix $\textbf{\emph{N}}$ as
\begin{equation}
    \label{eq:PartialsN}
    \textbf{\emph{N}} = \begin{bmatrix} \partial \hat{\textbf{\emph{n}}}_C / \partial \boldsymbol{\lambda} & \partial \hat{\textbf{\emph{n}}}_C / \partial \textbf{\emph{u}}_1 & \partial \hat{\textbf{\emph{n}}}_C / \partial \textbf{\emph{u}}_3 \end{bmatrix}
\end{equation}
Then compute the $9 \times 9$ Jacobian of the eigenvalues and relevant eigenvectors with respect to the vectorized matrix $\textbf{\emph{A}}$
\begin{equation}
    \label{eq:PartialsM}
    \textbf{\emph{M}} = \begin{bmatrix} \partial \boldsymbol{\lambda} / \partial \text{vec}(\textbf{\emph{A}}) \\
    \partial \textbf{\emph{u}}_1 / \partial \text{vec}(\textbf{\emph{A}}) \\
    \partial \textbf{\emph{u}}_3 / \partial \text{vec}(\textbf{\emph{A}})
     \end{bmatrix}
\end{equation}
Now, since $\textbf{\emph{A}}$ is symmetric [see Eq.~\eqref{eq:DefA}], define the $9 \times 6$ matrix $\textbf{\emph{P}}$ (see appendix) such that $\text{vec}(\textbf{\emph{A}}) = \textbf{\emph{P}} \textbf{\emph{a}}$ and where $\textbf{\emph{a}} = [A,B,C,D,F,G]^T$ is a $6 \times 1$ vector of the coefficients of the implicit equation for an ellipse [see Eq.~\eqref{eq:ImplicitConic}]. Consequently, one may write the Jacobian of interest as
\begin{equation}
    \partial \hat{\textbf{\emph{n}}}_C / \partial \textbf{\emph{a}} = \textbf{\emph{N}} \textbf{\emph{M}} \textbf{\emph{P}}
\end{equation}
such that
\begin{equation}
    \delta \hat{\textbf{\emph{n}}}_C = \textbf{\emph{N}} \textbf{\emph{M}} \textbf{\emph{P}} \delta \textbf{\emph{a}}
\end{equation}
Thus, the covariance of the unit pole direction $\hat{\textbf{\emph{n}}}_C$ is
\begin{equation}
    \label{eq:PoleCovariance}
    \textbf{\emph{R}}_{\hat{\textbf{\emph{n}}}_C} = E[\delta \hat{\textbf{\emph{n}}}_C \, \delta \hat{\textbf{\emph{n}}}^T_C] = \textbf{\emph{N}} \textbf{\emph{M}} \textbf{\emph{P}} \textbf{\emph{R}}_{\textbf{\emph{a}}} \textbf{\emph{P}}^T \textbf{\emph{M}}^T \textbf{\emph{N}}^T
\end{equation}
where $\textbf{\emph{R}}_a = E[\delta \hat{\textbf{\emph{a}}} \, \delta \hat{\textbf{\emph{a}}}^T]$ is the known error covariance of the ellipse fit. The covariance $\textbf{\emph{R}}_a$ is a $6 \times 6$ matrix with $\text{rank}[\textbf{\emph{R}}_a]=5$ since $\textbf{\emph{a}}$ has arbitrary scale.  Note that this covariance was intentionally developed in terms of errors in $\textbf{\emph{a}}$ because most modern ellipse fitting algorithms estimate $\textbf{\emph{a}}$ directly (and not things like ellipse center location, semi-major axis, and so on), and so statistics on an ellipse fit are most readily available in terms of $\textbf{\emph{a}}$. If statistics are instead known for the five geometric parameters $\textbf{\emph{g}} = [a,b,x_c,y_c,\phi]^T$, then one may compute $\textbf{\emph{R}}_a$ as
\begin{equation}
    \textbf{\emph{R}}_{\textbf{\emph{a}}} = \left( \frac{\partial \textbf{\emph{a}}}{\partial \textbf{\emph{g}}} \right) \textbf{\emph{R}}_\textbf{\emph{g}} \left( \frac{\partial \textbf{\emph{a}}}{\partial \textbf{\emph{g}}} \right)^T
\end{equation}
where $\textbf{\emph{R}}_\textbf{\emph{g}} = E[\delta \hat{\textbf{\emph{g}}} \, \delta \hat{\textbf{\emph{g}}}^T]$ is the error covariance of $\textbf{\emph{g}}$.

\subsection{Estimating the Pole from Many Circle Projections}
Each CoL produces a pair of possible solutions for $\hat{\textbf{\emph{n}}}_C$ from Eq.~\eqref{eq:UnitNormalFinal}. Except when looking nearly down the pole, these two directions are distinct. Thus, if one has many CoL projections, it is simple to unambiguously separate the estimates of $\hat{\textbf{\emph{n}}}_C$ into two groups. In the idealized (and unrealistic) case of perfect CoL projections with no measurement noise, one group of solutions for $\hat{\textbf{\emph{n}}}_C$ will be exactly the same for every projection and the other group of solutions will be slightly different from one another. However, with realistic measurement noise, the differences in the incorrect choice of $\hat{\textbf{\emph{n}}}_C$ are so small that they aren't useful in distinguishing which of the two choices for $\hat{\textbf{\emph{n}}}_C$ is correct. Thus, without additional contextual information, it is often difficult to determine which of the two pole orientation estimates corresponds to the true pole orientation. Usually, however, a simple inspection of the image or an occlusion check will allow the analyst to select the correct pole solution.

Proceeding under the assumption that the pole ambiguity has been resolved, it is possible to construct a maximum likelihood estimate (MLE) of the pole orientation from many CoL projections. Since an analytic pole covariance is available for each individual pole estimate, the MLE solution takes the form of a weighted least squares problem \cite{Tapley:2004}. The solution to this is, therefore, given by
\begin{equation}
    \label{eq:PoleMLE}
    \hat{\textbf{\emph{n}}}_C = \left[ \sum_{i=1}^n \textbf{\emph{R}}^{-1}_{\hat{\textbf{\emph{n}}}_{C,i}} \right]^{-1} \sum_{i=1}^n \textbf{\emph{R}}^{-1}_{\hat{\textbf{\emph{n}}}_{C,i}} \hat{\textbf{\emph{n}}}_{C,i}
\end{equation}
with covariance
\begin{equation}
    \label{eq:PoleMLEcovariance}
    \textbf{\emph{R}} = \left[ \sum_{i=1}^n \textbf{\emph{R}}^{-1}_{\hat{\textbf{\emph{n}}}_{C,i}} \right]^{-1}
\end{equation}
When no covariance is available (e.g, when no estimate of ellipse fit quality is available) then each observation is weighted equally and Eq.~\eqref{eq:PoleMLE} collapses to a simple average
\begin{equation}
    \hat{\textbf{\emph{n}}}_C = \frac{1}{n} \sum_{i=1}^n \hat{\textbf{\emph{n}}}_{C,i}
\end{equation}

\section{Recovering Three-Dimensional Circle of Latitude Structure}
When more than one CoL projection is observed, it is possible to infer more information about the observed body than just the pole orientation.
If $\hat{\textbf{\emph{n}}}_{C}$ is as shown in Eq.~\eqref{eq:UnitNormalFinal}, then it was shown in Refs.\cite{Shiu:1989} and \cite{Christian:2017AAS} that vector from the camera to the circle center is
\begin{equation}
    \textbf{\emph{r}}_{C} = \frac{R}{\sqrt{\alpha - \lambda_3 / \lambda_1}} \textbf{\emph{T}}^P_C
    \begin{bmatrix}
        \mp \sqrt{-\alpha \lambda_3 / \lambda_1} \\ 0 \\ 1
    \end{bmatrix}
\end{equation}
where $R$ is the CoL radius, $\textbf{\emph{T}}^P_C$ is from Eq.~\eqref{eq:RotationConeToCamera}, and $\alpha$ is from Eq.~\eqref{eq:DefAlpha}.
If the radius $R$ is not (yet) known, then one may estimate the relative position as normalized by the circle radius. Therefore, defining $\boldsymbol{\rho}_{C} = \textbf{\emph{r}}_{C} / R$, it is always possible compute $\boldsymbol{\rho}_{C}$ as
\begin{equation}
    \label{eq:DefRhoC}
    \boldsymbol{\rho}_{C} = \frac{\textbf{\emph{r}}_{C}}{R}= \frac{1}{\sqrt{\alpha - \lambda_3 / \lambda_1}} \textbf{\emph{T}}^P_C
    \begin{bmatrix}
        \mp \sqrt{-\alpha \lambda_3 / \lambda_1} \\ 0 \\ 1
    \end{bmatrix}
\end{equation}
where everything on the right-hand side is known from the eigenvectors and eigenvalues of the ellipse shape matrix $\textbf{\emph{A}}$. This result may be used to determine the 3-D structure to an unknown global scale.

\subsection{Scaled Three-Dimensional Structure}
\label{Sec:Scaled3DStructure}
Suppose there are $n \geq 2$ image ellipses formed by the projection of $n$ unknown CoLs. Denote the circle radii as $\{R_i\}_{i=1}^n$ and the circle relative positions as $\{ \textbf{\emph{r}}_{C_i} \}_{i=1}^n$. If the CoLs all come from the same spinning celestial object then they must have the same normal, hence $\hat{\textbf{\emph{n}}}_{C_i} = \hat{\textbf{\emph{n}}}_C$ for $i=1,\hdots,n$. Choose one of the elliptical projections as the reference feature and define the corresponding CoL's radius and relative position as $R_r$ and $\textbf{\emph{r}}_{C_r}$.

Since every CoL center lies along the pole, they may all be written as
\begin{equation}
    \label{eq:RelativePositionWithScale}
    \textbf{\emph{r}}_{C_i} = \textbf{\emph{r}}_{C_r} + \delta Z_{i} \hat{\textbf{\emph{n}}}_C
\end{equation}
where $\delta Z_i = Z_i - Z_r$ is the distance (positive in the direction of $\textbf{\emph{n}}_C$) from the reference CoL to the $i$-th CoL. This geometry is illustrated in Fig.~\ref{fig:SpheroidCoL} (where frame subscripts have been removed).

\begin{figure}[b!]
    	\centering
    	\includegraphics[width=0.7\textwidth]{SpheroidCoL.png}
    	\caption{Illustration of multiple CoL geometry that may be used for 3-D reconstriction.}
    	\label{fig:SpheroidCoL}
\end{figure}

However, since the relative positions $\textbf{\emph{r}}_{C_r}$ and $\textbf{\emph{r}}_{C_i}$ may not be directly obtained from the image,
one may instead rewrite Eq.~\eqref{eq:RelativePositionWithScale} as
\begin{equation}
    R_i \boldsymbol{\rho}_{C_i} = R_r \boldsymbol{\rho}_{C_r} + \delta Z_{i} \hat{\textbf{\emph{n}}}_C
\end{equation}
where $\hat{\textbf{\emph{n}}}_C$, $\boldsymbol{\rho}_{C_r}$, and $\boldsymbol{\rho}_{C_i}$ are all known from the methods above [see Eqs.~\eqref{eq:UnitNormalFinal} and \eqref{eq:DefRhoC}]. Clearly, $R_r$, $R_i$, and $\delta Z_{i}$ may not be simultaneously determined in an absolute sense. Therefore, normalize by $R_r$ to obtain
\begin{equation}
   (R_i/R_r) \boldsymbol{\rho}_{C_i} = \boldsymbol{\rho}_{C_r} + (\delta Z_{i}/R_r) \hat{\textbf{\emph{n}}}_C
\end{equation}
Letting the normalized distances $R'_i$ and $\delta Z'_{i}$ be
\begin{subequations}
\label{Eq:ScaledCoL}
\begin{equation}
    R'_i = R_i/R_r
\end{equation}
\begin{equation}
    \delta Z'_{i} = \delta Z_{i}/R_r
\end{equation}
\end{subequations}
one obtains the $3 \times 2$ linear system in these two unknowns
\begin{equation}
    \label{eq:LinSysReconstruct}
   \boldsymbol{\rho}_{C_r} = \begin{bmatrix} \boldsymbol{\rho}_{C_i} & -\hat{\textbf{\emph{n}}}_C  \end{bmatrix}
   \begin{bmatrix} R'_i \\ \delta Z'_{i}  \end{bmatrix}
\end{equation}
This system may be directly solved for $R'_i$ and $\delta Z'_{i}$. In the case of the reference CoL itself (i.e., when $i=r$), this produces the expected result of $R'_r = 1$ and $\delta Z'_r = 0$. Observability into $R'_i $ and $\delta Z'_i$ disappears when $\boldsymbol{\rho}_{C_i}$ and $\hat{\textbf{\emph{n}}}_C$ are parallel (or nearly parallel), which occurs when the camera lies on the line defining the poles (i.e., is located directly above/below the poles). This happens to be the same degenerate configuration in which Eq.~\eqref{eq:UnitNormalFinal} for the normal direction $\textbf{\emph{n}}_C$ is not differentiable, leading to difficulties with computing the pole direction covariance. Regardless, outside of this single degenerate viewpoint, the result of Eq.~\eqref{eq:LinSysReconstruct} may be used to reconstruct the 3-D system of CoLs to an unknown global scale---with this unknown scale being the radius $R_r$ of the reference CoL.
The unknown global scale may be resolved in some special cases, as discussed in Section~\ref{Sec:ScaleForSpheroid}.

\subsection{Three-Dimensional Pole Representations with Pl\"{u}cker Coordinates}
It may sometimes be useful to describe the 3-D line in space describing the pole of the spinning body. A 3-D line may be represented as a point in ${\mathbb P}^5$ and may often be efficiently described with Pl\"{u}cker coordinates \cite{Plucker:1865,Hartley:2003}. The location of the center of circle $\textbf{\emph{r}}_{C_r}$ and the pole direction $\textbf{\emph{n}}_C$ may be combined to describe a 3-D line in the camera frame. Specifically, following Ref.~\cite{Christian:2021streak}, one finds that
\begin{equation}
   \label{eq:PluckerMat}
    \textbf{\emph{L}}_C \propto \begin{bmatrix}
        \left[(\textbf{\emph{n}}_C \times \textbf{\emph{r}}_C) \times \right] & -\textbf{\emph{n}}_C\\
        \textbf{\emph{n}}_C^T & 0
    \end{bmatrix}
\end{equation}
which projects to the line in the image according to
\begin{equation}
   \boldsymbol{\ell} \propto \textbf{\emph{n}}_C \times \textbf{\emph{r}}_C
\end{equation}
where image points belonging to the projected 2-D line $\boldsymbol{\ell}$ satisfy the constraint $\boldsymbol{\ell}^T \bar{\textbf{\emph{x}}} = 0$. The 2-D image line $\boldsymbol{\ell}$ passes through the projected points corresponding to the centers of each of the 3-D circles (which are points along the pole). Note, however, that the line does \emph{not} necessarily pass through the center of the projected ellipses since the circle centers to not project to the ellipse centers. In the event that scale is not known, $\boldsymbol{\rho}_C$ may be used in place of $\textbf{\emph{r}}_C$ in Eq.~\eqref{eq:PluckerMat} to obtain the line for the scaled 3-D geometry.

\section{Camera Localization for an Oblate Spheroid}
\label{Sec:ScaleForSpheroid}

Most of the large celestial bodies in our Solar System may be well-modeled as an oblate spheroid. This occurs since rotating and self-gravitating bodies in hydrostatic equilibrium (e.g., Jupiter) take the shape of an oblate spheroid, as was shown by Newton in the \emph{Principia} \cite{Newton:1687}. Thus, since many of the large celestial bodies in our Solar System have a well-known spheroidal shape, the scale-ambiguous CoL structure from Section~\ref{Sec:Scaled3DStructure} may be adjusted to tightly lie on the spheroidal surface.

If the celestial body is an oblate spheroid, then its shape is defined by the known equatorial radius $R_E$ and polar radius $R_P$. Thus, any CoL must satisfy the constraint
\begin{equation}
   \frac{R^2_i}{R_E^2} + \frac{Z_i^2}{R_P^2} = 1
\end{equation}
which is the same as
\begin{equation}
   \frac{R^2_i}{R_E^2} + \frac{(Z_r + \delta Z_i)^2}{R_P^2} = 1
\end{equation}
Proceed by substituting from Eq.~\eqref{Eq:ScaledCoL}
\begin{equation}
   \frac{R^2_r R'^2_i}{R_E^2} + \frac{R^2_r(Z'_r + \delta Z'_i)^2}{R_P^2} = 1
\end{equation}
which, defining $\epsilon = R_P/R_E \leq 1$, may be rearranged to find
\begin{equation}
   (\epsilon^2 R'^2_i + \delta Z'^2_i) R^2_r + (2 \delta Z'^2_i) R^2_r Z'_r + (R_r^2 Z'^2_r - R_P^2) = 0
\end{equation}
Here, it is noted that $\epsilon$ is known since $R_E$ and $R_P$ are known. Finally, rearranging leads to the linear system
\begin{equation}
   \begin{bmatrix}
       (\epsilon^2 R'^2_1 + \delta Z'^2_1)& (2 \delta Z'^2_1) & 1 \\
       \vdots & \vdots & \vdots \\
       (\epsilon^2 R'^2_n + \delta Z'^2_n)& (2 \delta Z'^2_n) & 1
   \end{bmatrix}
   \boldsymbol{\xi} =
   \begin{bmatrix}
       0 \\
       \vdots  \\
       0
   \end{bmatrix}
\end{equation}
\begin{equation}
    \boldsymbol{\xi} = \begin{bmatrix}
        \xi_1 \\ \xi_2 \\ \xi_3
   \end{bmatrix}  \propto
   \begin{bmatrix}
        R^2_r \\ R^2_r Z'_r \\ R_r^2 Z'^2_r - R_P^2
   \end{bmatrix}
\end{equation}
This is a null space problem, so all that is important are the ratios
$\xi_1:\xi_2:\xi_3$.
The optimal solution for $\boldsymbol{\xi}$ is found via the singular value decomposition. With an estimate of $\boldsymbol{\xi}$ in hand,
one can immediately recover the scaled offset as $Z'_r = \xi_2 / \xi_1$. Likewise, one can recover the reference CoL radius $R_r$ as
\begin{equation}
    \label{Eq:RefCoLRadiusSpheroid}
    R_r = R_P \, \sqrt{ \frac{ \xi_1^2}{\xi_2^2 - \xi_1 \xi_3}  }
\end{equation}
Therefore, since $Z_r = R_r Z'_r$, one finds that
\begin{equation}
    \label{Eq:RefCoLOffsetSpheroid}
    Z_r = R_P \, \sqrt{ \frac{ \xi_2^2}{\xi_2^2 - \xi_1 \xi_3}  }
\end{equation}
Consequently, from the geometry in Fig.~\ref{fig:SpheroidCoL}, the position from the spacecraft to the center of the oblate spheroid as expressed in the camera frame
\begin{equation}
    \textbf{\emph{p}}_C = R_r \boldsymbol{\rho}_{C_r} - Z_r \hat{\textbf{\emph{n}}}_C
\end{equation}
where $\hat{\textbf{\emph{n}}}_C$ is from Eq.~\eqref{eq:PoleMLE}, $\boldsymbol{\rho}_{C_r}$ is from Eq.~\eqref{eq:DefRhoC}, $R_r$ is from Eq.~\eqref{Eq:RefCoLRadiusSpheroid}, and $Z_r$ is from Eq.~\eqref{Eq:RefCoLOffsetSpheroid}.

\section{Numerical Results}

\subsection{Example with a Spinning Asteroid}
\label{Section:BennuExample}
As the first numerical example, consider the scenario where the motion of surface points on a spinning asteroid create elliptical paths in a sequence of images. The asteroid Bennu is used as the reference model to allow for easy comparison with the prior work of Kuppa, McMahon, and Dietrich \cite{Kuppa:2022a,Kuppa:2022b}. Suppose that ellipses are fit to the projections of two CoLs. Each of these two CoLs pass through the surface point that generated them. However, since Bennu is not a body of revolution (e.g., an oblate spheroid), these CoLs do not necessarily lie on any other point on Bennu's surface---this can be seen in Fig.~\ref{fig:BennuCoL}.

\begin{figure}[t!]
    	\centering
    	\includegraphics[width=0.5\textwidth]{BennuCoL.png}
    	\caption{Mesh model of Bennu with two CoLs (red and cyan circles) produced by two reference surface points (red and cyan dots).}
    	\label{fig:BennuCoL}
\end{figure}

Within this scenario, two specific configurations are considered: (1) a spacecraft at a latitude of 60 deg and (2) a spacecraft at a latitude of 30 deg. In both configurations, the camera is not pointed directly at the center of Bennu or at the center of either of the observed CoLs. These projections produce the image plane ellipses shown in Fig.~\ref{fig:BennuProjEllipse}. To process these image ellipses, the algorithm is not provided any information about the camera-relative position or orientation of Bennu---only the image-plane ellipse fits [as parameterized by the matrix $\textbf{\emph{A}}$ from Eq.~\eqref{eq:DefA}] are available. Note that the method from Ref.~\cite{Kuppa:2022a,Kuppa:2022b} does not work in such a situation, since it requires the camera to be pointed at the asteroid center and that the distance is known.
\begin{figure}[t!]
    	\centering
    	\includegraphics[width=0.5\textwidth]{BennuProjEllipse.png}
    	\caption{Projection of CoLs from Fig.~\ref{fig:BennuCoL} onto the image plane for a spacecraft at a latitude of 60 degrees (top) and 30 degrees (bottom).}
    	\label{fig:BennuProjEllipse}
\end{figure}

Begin by considering the noise free case. The pole directions are computed using Eq.~\eqref{eq:UnitNormalFinal} and are shown in Tables~\ref{tab:AsteroidPerfectPole60} and \ref{tab:AsteroidPerfectPole30}. The correct pole direction is bolded in these tables, and it is observed that these match exactly for the two CoL projections. The bolded pole directions in Tables~\ref{tab:AsteroidPerfectPole60} and \ref{tab:AsteroidPerfectPole30} agree with the simulation truth to machine precision. Also note that the correct (bolded) and incorrect (not bolded) are different enough from each other to allow for easy separation into two groups (i.e., it is easy to match the corresponding pole solutions from the different CoL projections).

\begin{table}[b!]
    \centering
    \caption{Noise-free estimates of pole direction for spacecraft at a latitude of 60 degrees.}
    \begin{tabular}{cccc}
    \hline
     \multicolumn{2}{c}{CoL Projection 1 (red)} & \multicolumn{2}{c}{CoL Projection 2 (cyan)}\\
     $\hat{\textbf{\emph{n}}}_C$ Sol. 1 & $\hat{\textbf{\emph{n}}}_C$ Sol. 2 & $\hat{\textbf{\emph{n}}}_C$ Sol. 1 & $\hat{\textbf{\emph{n}}}_C$ Sol. 2 \\
    \hline
    \textbf{-0.143087} &  0.098228 & \textbf{-0.143087} &  0.097531 \\
    \textbf{-0.555760} &  0.421967 & \textbf{-0.555760} &  0.419187 \\
    \textbf{-0.818937} & -0.901274 & \textbf{-0.818937} & -0.902646 \\
    \hline
    \end{tabular}
    \label{tab:AsteroidPerfectPole60}
\end{table}

\begin{table}[b!]
    \centering
    \caption{Noise-free estimates of pole direction for spacecraft at a latitude of 30 degrees.}
    \begin{tabular}{cccc}
    \hline
     \multicolumn{2}{c}{CoL Projection 1 (red)} & \multicolumn{2}{c}{CoL Projection 2 (cyan)}\\
     $\hat{\textbf{\emph{n}}}_C$ Sol. 1 & $\hat{\textbf{\emph{n}}}_C$ Sol. 2 & $\hat{\textbf{\emph{n}}}_C$ Sol. 1 & $\hat{\textbf{\emph{n}}}_C$ Sol. 2 \\
    \hline
     -0.327999 & \textbf{ 0.392413} & -0.326582 & \textbf{ 0.392413} \\
      0.764579 & \textbf{-0.816434} &  0.761901 & \textbf{-0.816434} \\
     -0.554829 & \textbf{-0.423612} & -0.559331 & \textbf{-0.423612} \\
    \hline
    \end{tabular}
    \label{tab:AsteroidPerfectPole30}
\end{table}

While the correct pole directions in Tables~\ref{tab:AsteroidPerfectPole60} and \ref{tab:AsteroidPerfectPole30} match exactly for each configuration, observe that the incorrect pole directions are slightly different. Unfortunately, however, the incorrect pole directions are usually not different enough for this mismatch to be a reliable means of identifying the incorrect pole in the presence of noise. Additional contextual information (e.g., occlusion check) is necessary to identify which pole solution to use.

In the presence of measurement noise the ellipse fits will not be perfect, and so the estimate of $\hat{\textbf{\emph{n}}}_C$ from Eq.~\eqref{eq:UnitNormalFinal} will not be perfect. In this case, the analytic covariance from Section~\ref{Sec:PoleCovariance} may be used to quantify the errors in $\hat{\textbf{\emph{n}}}_C$ arising from the fit to the red CoL from Figs.~\ref{fig:BennuCoL} and \ref{fig:BennuProjEllipse}. The ellipse fit is performed using the semi-hyper ellipse fit method from Ref.~\cite{Kanatani:2011}.
Points around image ellipse are sampled with a measurement error of about 15 arcsec.
Since the semi-hyper ellipse fit directly estimates the implicit coefficient vector $\textbf{\emph{a}}$, the the covariance of the ellipse fit is $\textbf{\emph{R}}_a = E[\delta \textbf{\emph{a}} \, \delta \textbf{\emph{a}}^T]$. The geometric ellipse parameters (e.g., semi-major and semi-minor axis, center coordinates) need not be computed at any point in the process.
Since the pole estimate covariance $\textbf{\emph{R}}_{\hat{\textbf{\emph{n}}}_C}$ is rank deficient (errors in pole direction lie in the plane perpendicular to $\hat{\textbf{\emph{n}}}_C$), it makes sense to plot the covariance in this plane. This approach is validated for the two configurations shown here via a 10,000-run Monte Carlo analysis and the results are shown in Fig.~\ref{fig:BennuCovPDF}. Excellent agreement is seen between the analytic error covariance (red line) and the Monte Carlo sample covariance (dashed cyan line).

\begin{figure}[b!]
    	\centering
    	\includegraphics[width=0.5\textwidth]{BennuCovPDF.png}
    	\caption{Pole direction errors in the plane normal to $\hat{\textbf{\emph{n}}}_C$ for a spacecraft at a latitude of 60 degrees (top) and 30 degrees (bottom).}
    	\label{fig:BennuCovPDF}
\end{figure}

Pole estimate errors remain approximately Gaussian and the analytic error covariance remains a consistent with Monte Carlo results everywhere except for very high latitudes (i.e., when the camera position is near the line coinciding with the pole). In this case, the projected ellipses become nearly circles and small perturbations in the measured ellipse produce complicated changes in the apparent pole direction. This limitation was discussed in Section~\ref{Sec:PoleCovariance}. To quantify what happens in such a situation, consider a spacecraft at a latitude of 88 degrees. As before, a 10,000-run Monte Carlo analysis is performed. The resulting normal direction errors are as shown in Fig.~\ref{fig:BennuPolePDF88}, which is clearly non-Gaussian. Also note that the magnitude of the errors in the normal direction become larger as the latitude increases, with the worst performance occurring at (or near) a latitude of 90 deg. Consequently, it is desirable to avoid high-latitude observations when using CoL projections to determine the orientation of a spinning body's pole.

\begin{figure}[b!]
    	\centering
    	\includegraphics[width=0.4\textwidth]{BennuPolePDF88.png}
    	\caption{Pole direction errors in the plane normal to $\hat{\textbf{\emph{n}}}_C$ for a spacecraft at a latitude of 88 degrees.}
    	\label{fig:BennuPolePDF88}
\end{figure}

\subsection{Example with Jupiter Latitutde Bands}
As the second numerical example, consider the scenario of image-based navigation relative to Jupiter---as previously explored by Danas Rivera and Peck \cite{Rivera:2020,Rivera:2022,Rivera:2023}. As a gas giant, Jupiter's shape is very well-approximated as an oblate spheroid. This simulation uses the 2015 shape model from the International Astronautical Union (IAU) \cite{Archinal:2018}, which defines Jupiter's equatorial radius as $R_E = 71,492 \pm 4$ km and polar radius as $R_P = 66,854 \pm 10$ km. This equates to an ellipticity of $\epsilon = R_P/R_E \approx 0.935$.

Suppose now that a spacecraft collects a single image of Jupiter. In that image, two latitude bands are observed at latitudes of 7 deg and 45 deg. Further suppose that no onboard ''map'' of the latitude bands exists, and so the spacecraft has no internal knowledge of the actual latitude of the observed bands. Instead, the spacecraft only knows that each of these bands correspond to a CoL lying on the surface of an oblate spheroid of known shape (e.g., the 2015 IAU Jupiter shape model). The existing algorithm of Danas Rivera and Peck \cite{Rivera:2020,Rivera:2022,Rivera:2023} could not navigate since it requires correspondence between observed CoL projections and known latitude bands (assumption \#1 in Ref.~\cite{Rivera:2023}).

To create some representative numerical results, let the spacecraft observation occur from a latitude of 60 deg. Suppose that the camera is pointing in the general direction of Jupiter (so that it appears in the image), but that the camera boresight is not perfectly pointed at Jupiter's center or at the center of either of the CoLs. In such a situation, the algorithm of Danas Rivera and Peck \cite{Rivera:2020,Rivera:2022,Rivera:2023} would not work since the camera is not perfectly pointed at the planet center (assumption \#3 in Ref.~\cite{Rivera:2023}) and the planetocentric distance is not known (assumption \#5 in Ref.~\cite{Rivera:2023}).

In this case, the orientation of Jupiter's pole within the camera frame may be estimated as described in Section~\ref{Sec:PoleDirection} and similar behavior to Section~\ref{Section:BennuExample} is seen. Now, since Jupiter is an oblate spheroid, the relative position may be found using the results from Section~\ref{Sec:ScaleForSpheroid}. As in the prior example, assume an imaging system producing LOS errors to points around the observed CoL with an error of about 15 arcsec. If the spacecraft views the two specified CoLs from a range of about $50 R_E \approx 3.57 \times 10^6$ km, then Jupiter-relative localization results similar to Fig.~\ref{fig:JupiterError} are obtained. Note that the results in Fig.~\ref{fig:JupiterError} provide the position vector from the spacecraft to Jupiter's center as expressed in the camera frame. As with most vision-based OPNAV techniques, the error is largest in the direction of the observed object. Here, the $1\sigma$ error corresponds to about 1.6\% of the range to Jupiter.

\begin{figure}[b!]
    	\centering
    	\includegraphics[width=0.75\textwidth]{JupiterError.png}
    	\caption{Scatter plot of Jupiter localization error in the camera frame ($+z$ along boresight).}
    	\label{fig:JupiterError}
\end{figure}

Since the inertial orientation of navigation cameras is usually well known, this relative position may often be converted to the inertial frame. Thus, assuming the spacecraft inertial orientation is known (e.g., from a star tracker), the CoL projections can be used to find the absolute relative position between the spacecraft and planet---not just the latitude. In this regard, navigation by latitude bands (i.e., by CoLs) produce a very similar measurement as the classical horizon-based OPNAV approach \cite{Christian:2021d}. This flexibility in measurement source (horizon or latitude band) may provide valuable operational flexibility in some situations. In most cases, however, horizon localization will correspond to a crisper feature and have lower image processing errors---and, therefore, horizon-based OPNAV will often provide better navigation results that latitude band-based OPNAV.

\section{Conclusions}
Inspired by recent work on pole estimation \cite{Kuppa:2022a,Kuppa:2022b} and latitude band-based navigation \cite{Rivera:2020,Rivera:2022,Rivera:2023} with images, this work introduces a new framework for navigation with the perspective projection of circles of latitude (CoLs). These advancements begin by transferring established algorithms for image-based localization with circles, including a comparison of different formulations and a discussion of their equivalence. These historical methods are then expanded by introduction of an analytic covariance expression and development of a localization algorithm for oblate spheroids. The result is an algorithmic framework that more accurately captures the projective geometry and allows for more opereational flexiblity (i.e., has fewer constraints) than existing methods.

\section*{Appendix}
The pole covaraince requires computation of a number of measurement partials. These may all be computed analytically. To begin, consider Eq.~\eqref{eq:PartialsN} for $\textbf{\emph{N}}$,
\begin{equation}
    \textbf{\emph{N}} = \begin{bmatrix} \partial \hat{\textbf{\emph{n}}}_C / \partial \boldsymbol{\lambda} & \partial \hat{\textbf{\emph{n}}}_C / \partial \textbf{\emph{u}}_1 & \partial \hat{\textbf{\emph{n}}}_C / \partial \textbf{\emph{u}}_3 \end{bmatrix}
\end{equation}
which consists of the individual partials
\begin{subequations}
\begin{equation}
   \frac{\partial \hat{\textbf{\emph{n}}}_C}{ \partial \textbf{\emph{u}}_1 } = \sqrt{\frac{\lambda_1 - \lambda_2}{\lambda_1 - \lambda_3 }} \textbf{\emph{I}}_{3\times 3}
\end{equation}
\begin{equation}
   \frac{\partial \hat{\textbf{\emph{n}}}_C}{ \partial \textbf{\emph{u}}_3 } = \pm \sqrt{\frac{\lambda_2 - \lambda_3}{\lambda_1 - \lambda_3 }}\textbf{\emph{I}}_{3\times 3}
\end{equation}
\begin{equation}
    \frac{\partial \hat{\textbf{\emph{n}}}_C}{\partial \boldsymbol{\lambda}} = \textbf{\emph{g}}^T \otimes \textbf{\emph{m}}_C
\end{equation}
\end{subequations}
where
\begin{equation}
    \textbf{\emph{g}} = \frac{1}{2(\lambda_1-\lambda_3)^2}
    \begin{bmatrix}
        \lambda_2 - \lambda_3 \\
        \lambda_3 - \lambda_1 \\
        \lambda_1 - \lambda_2
    \end{bmatrix}
\end{equation}
\begin{equation}
    \textbf{\emph{m}}_C =  \sqrt{\frac{\lambda_1 - \lambda_3 }{\lambda_1 - \lambda_2}} \; \textbf{\emph{u}}_1 \mp \sqrt{\frac{\lambda_1 - \lambda_3 }{\lambda_2 - \lambda_3}} \; \textbf{\emph{u}}_3
\end{equation}

Then compute the $9 \times 9$ Jacobian of the eigenvalues and relevant eigenvectors with respect to the vectorized matrix $\textbf{\emph{A}}$, as given by the Eq.~\eqref{eq:PartialsM} for $\textbf{\emph{M}}$
\begin{equation}
    \textbf{\emph{M}} = \begin{bmatrix} \partial \boldsymbol{\lambda} / \partial \text{vec}(\textbf{\emph{A}}) \\
    \partial \textbf{\emph{u}}_1 / \partial \text{vec}(\textbf{\emph{A}}) \\
    \partial \textbf{\emph{u}}_3 / \partial \text{vec}(\textbf{\emph{A}})
     \end{bmatrix}
\end{equation}
which consists of the individual partials (see Refs.~\cite{Magnus:1985} and \cite{Liounis:2019}),
\begin{equation}
    \frac{\partial \boldsymbol{\lambda}}{\partial \text{vec}(\textbf{\emph{A}})}  = \begin{bmatrix} \textbf{\emph{u}}^T_1 \otimes \textbf{\emph{u}}^T_1   \\
    \textbf{\emph{u}}^T_2 \otimes \textbf{\emph{u}}^T_2  \\
    \textbf{\emph{u}}^T_3 \otimes \textbf{\emph{u}}^T_3
     \end{bmatrix}
\end{equation}
\begin{equation}
    \frac{\partial \textbf{\emph{u}}_i}{\partial \text{vec}(\textbf{\emph{A}})}  =
    \textbf{\emph{u}}^T_i \otimes \left[ \left( \textbf{\emph{A}} - \lambda_i \textbf{\emph{I}}_{3 \times 3} - \textbf{\emph{u}}_i \textbf{\emph{u}}_i^T \textbf{\emph{A}} \right)^{-1} \left(\textbf{\emph{u}}_i \textbf{\emph{u}}_i^T - \textbf{\emph{A}} \right) \right]
\end{equation}
Since $\textbf{\emph{A}}$ is symmetric, it only has six unique entries that depend on the $6 \times 1$ vector of coefficients $\textbf{\emph{a}} = [A,B,C,D,F,G]^T$. Given the entries in $\textbf{\emph{A}}$ from Eq.~\eqref{eq:DefA}, it follows that $\text{vec}(\textbf{\emph{A}}) = \textbf{\emph{P}} \textbf{\emph{a}}$, where $\textbf{\emph{P}}$ is
\begin{equation}
\textbf{\emph{P}} = \begin{bmatrix}
     1&   0& 0&   0&   0& 0 \\
     0& 1/2& 0&   0&   0& 0 \\
     0&   0& 0& 1/2&   0& 0 \\
     0& 1/2& 0&   0&   0& 0 \\
     0&   0& 1&   0&   0& 0 \\
     0&   0& 0&   0& 1/2& 0 \\
     0&   0& 0& 1/2&   0& 0 \\
     0&   0& 0&   0& 1/2& 0 \\
     0&   0& 0&   0&   0& 1
    \end{bmatrix}
\end{equation}
Thus, one finds that
\begin{equation}
    \frac{\partial \hat{\textbf{\emph{n}}}_C}{\partial \textbf{\emph{a}}} = \textbf{\emph{N}} \textbf{\emph{M}} \textbf{\emph{P}}
\end{equation}
with the explicit expressions for $\textbf{\emph{N}}$, $\textbf{\emph{M}}$, and $\textbf{\emph{P}}$ given above.

Sometimes it is more convenient to compute the partials with respect to the geometric parameters $\textbf{\emph{g}} = [a,b,x_c,y_c,\phi]^T$. The coefficients $\textbf{\emph{a}} = [A,B,C,D,F,G]^T$ may be written directly in terms of $\textbf{\emph{g}}$ according to \cite{Christian:2021crater}
\begin{subequations}
\begin{equation}
A = a^2 \sin^2 \psi + b^2 \cos^2 \psi \label{eq:ConicImplicitA}
\end{equation}
\begin{equation}
B = 2 (b^2 - a^2) \cos \psi \, \sin \psi \label{eq:ConicImplicitB}  \end{equation}
\begin{equation}
C = a^2 \cos^2 \psi + b^2 \sin^2 \psi \label{eq:ConicImplicitC}
\end{equation}
\begin{equation}
D = -2 A x_c - B y_c
\end{equation}
\begin{equation}
F = -B x_c - 2 C y_c
\end{equation}
\begin{equation}
G = A x_c^2 + B x_c y_c + C y_c^2 - a^2 b^2
\end{equation}
\end{subequations}
which allows one to compute the partials $\partial \textbf{\emph{a}} / \partial \textbf{\emph{g}}$. Explicit computation of these partials is tedious, but straightforward.


You might have already heard about Lily, the aerial video camera that tracks you as it flies above and around you recording video from afar of you and your surroundings. The shots and angels that Lily captures is truly stunning.
Navigate to their homepage and watch the feature video on to get an idea of just how cool Lily's video can be. One other thing that stands out on Lily's homepage is the proper and effective use of animations to enhance a website's design.
One of the key ways that animations on any site can contribute to a great user experience is when they are used to communicate information. Lily's homepage effectively uses animations to convey concepts and ideas about how Lily is used, how Lily functions and how Lily is assembled and put together. Go to Lily's site and scroll through the page to see the animations in action.
Animations provide a sense of interaction with the website, which I think this is one of the key benefits of using animations like the ones on Lily's website. Lily's homepage for example interacts with the user base on their input. Lily's animations are driven through scrolling. The animations draws the user's attention, engaging them as they scroll down the page and highlighting key points of interest. Instead of displaying static content such as flat images, the animations give depth to a page by adding another dimension to the animated elements. Animations simply give life to a page. The web design of the site is enhanced via the animations.
Just because you can doesn't mean you should. Too many animations to your a page can have a negative effect on users. Having to many animations can make the page seem very busy and distracting. In addition to that, too many animations can make a website seem slow, feel jittery and somewhat unresponsive. Lily's doesn't over do the animations, it has the proper amount and they are simple, not hugely complicated. We love to see sites that use animation effectively as many right now are over-doing it with gratuitous animation effects that add nothing to the experience.
Go check out Lily and the animations on their homepage.
Starting in the last few months there have been many new TLD options to put your website on something different than the traditional .com, .net, or .org TLD. TLD's (or top level domains) simply are the end of the URL that is registered for your web address. The team at Lily opted to use the new .CAMERA TLD for their site, therein emphasizing how this drone is intended to be used (.drone itself is also a new TLD).


A judge has cleared the MV Marathassa of charges arising from an April 2015 oil spill in English Bay.
The cargo ship MV Marathassa has been acquitted of the last charges it faced relating to the April 8, 2015, spill that saw 2,700 litres of bunker fuel dumped into Vancouver's English Bay.
On Thursday, Judge Kathryn Denhoff acquitted the ship of charges under the Canada Shipping Act, ruling that while the ship discharged a pollutant, she accepted the ship's defence that it used due diligence in the case.
The ship built in Japan was only three weeks old. The judge said that "the hazards in this case of defective alarms and a defective valve were not foreseeable. The Marathassa, the crew and Transport Canada all expected that the brand new vessel would be free of those defects."
She also ruled that the Crown had failed to prove the ship and its crew didn't implement an oil pollution emergency plan. She also noted that "given the extensive training, verification and ongoing monitoring of the crew, the Marathassa took all reasonable steps to ensure that the crew followed the requirements" of the emergency plan.
She also acquitted the ship on one charge under the Migratory Birds Convention Act.
The spill took more than two weeks to clean up, and outraged environmentalists and some residents.


text:
Q: Get Timestamp in ML.NET Anomaly detection Im following this tutorial: anomaly-detection with ML.NET where the results looks like this:
Alert   Score   P-Value
0       271.00  0.50
0       124.30  0.13
1       341.50  0.00 <-- Spike detected
0       197.80  0.48
0       127.90  0.13

but in my data i also have a timestamp. Is it possible to see the timestamp in the results?
Alert   Score   P-Value Timestamp
0       271.00  0.50    2022-06-01: 22:30:15
0       124.30  0.13    2022-06-01: 22:30:20
1       341.50  0.00    2022-06-01: 22:30:25 <-- Spike detected
0       197.80  0.48    2022-06-01: 22:30:30
0       127.90  0.13    2022-06-01: 22:30:35

If it is not possible, how do i use the results?

A: If that's a column in your dataset, you can define it in your Input / Output classes.
https://learn.microsoft.com/dotnet/machine-learning/tutorials/sales-anomaly-detection#create-classes-and-define-paths
Add a property to the classes for your TimeStamp column. i.e.
public class ProductSalesData
{
    [LoadColumn(0)]
    public string Month;

    [LoadColumn(1)]
    public float numSales;

    [LoadColumn(2)]
    public string TimeStamp;
}

public class ProductSalesPrediction
{
    public string TimeStamp;

    //vector to hold alert,score,p-value values
    [VectorType(3)]
    public double[] Prediction { get; set; }
}

redpajama_set_name: RedPajamaStackExchange


text:
Russia-Ukraine struggle stay: a whole bunch of our bodies present in Kharkiv after Russian retreat; nuclear menace is worst in 60 years, Biden says
A whole bunch of our bodies present in Kharkiv after Russians left
Within the northeastern Kharkiv area the place Ukrainian forces regained a big swathe of floor in September, the our bodies of 534 civilians together with 19 youngsters have been discovered after Russian troops left, Serhiy Bolvinov of the Nationwide Police in Kharkiv informed a briefing posted on-line Thursday.
The entire included 447 our bodies present in Izium, Reuters experiences. He additionally stated that investigators had discovered proof of twenty-two websites getting used as "torture rooms". There was no instant remark from Russia.
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North Korean chief Kim Jong Un has congratulated Vladimir Putin on his birthday, applauding him for his "distinguished management and powerful will" within the newest signal of deepening ties between the 2 international locations.
Within the birthday message, Kim spoke of Putin's achievements in "constructing highly effective Russia" and stated the Russian chief was "having fun with excessive respects and assist from the broad plenty of individuals".
Russian President Vladimir Putin and North Korea's chief Kim Jong Un in Vladivostok, Russia, 2019. {Photograph}: Reuters
Russia is "reliably defending the dignity of the state and its basic pursuits from the challenges and threats by the US and its vassal forces", he added. "Such actuality is unthinkable with out your distinguished management and powerful will," he stated.
Cooperation between North Korea and Russia had been strengthened "as by no means earlier than", Kim stated, including that he hoped their private ties would play a much bigger function in additional creating the pleasant relationship between the 2 international locations.
Abstract of the day up to now …
Within the northeastern Kharkiv area the place Ukrainian forces regained a big swathe of floor in September, the our bodies of 534 civilians together with 19 youngsters have been discovered after Russian troops left, Serhiy Bolvinov of the Nationwide Police in Kharkiv informed a briefing. The entire included 447 our bodies present in Izium. He additionally stated that investigators had discovered proof of twenty-two websites getting used as "torture rooms".
Joe Biden has warned the world might face "Armageddon" if Vladimir Putin makes use of a tactical nuclear weapon to attempt to win the struggle in Ukraine. The US president made his most outspoken remarks to this point about the specter of nuclear struggle, saying it was the closest the world had come to nuclear disaster for sixty years. "We now have not confronted the prospect of Armageddon since Kennedy and the Cuban missile disaster," he stated. "We've received a man I do know pretty nicely," Biden stated, referring to the Russian president. "He's not joking when he talks about potential use of tactical nuclear weapons or organic or chemical weapons as a result of his navy is, you would possibly say, considerably underperforming."
The 2022 Nobel Peace Prize has been awarded to human rights advocate Ales Bialiatski from Belarus, the Russian human rights organisation Memorial and the Ukrainian human rights organisation Middle for Civil Liberties.
Ukrainian sources try to make clear what President Volodymyr Zelenskiy meant yesterday when he talked about "preventive strikes" being essential to cease Russia utilizing tactical nuclear weapons in Ukraine. Russia's international minister Sergey Lavrov seized on the feedback, decoding them as a name for Nato to make use of nuclear weapons in opposition to Russia, and stated the feedback confirmed why Russia's "particular navy operation" inside Ukraine's borders had been essential. Advisor to Zelenskiy, Serhii Nykyforov, stated "Colleagues, you have got gone a bit of too far along with your nuclear hysteria and now you hear nuclear strikes even the place there are none. The president spoke in regards to the interval earlier than 24 February. Then it was essential to use preventive measures to stop Russia from beginning the struggle. Let me remind you that the one measures that have been about then have been preventive sanctions."
The headquarters of the armed forces of the self-proclaimed Donetsk Individuals's Republic (DPR) has claimed to have captured three settlements from Ukrainian forces in Donetsk.
No less than 5 individuals have been killed and as many injured after Ukrainian forces struck a bus whereas shelling a strategically vital bridge within the Russian-controlled a part of Ukraine's southern Kherson area, Russia's Tass information company has reported.
The workplace of Turkey's president, Recep Tayyip Erdoğan, has issued a quick read-out after he held a name together with his Russian counterpart. The pair mentioned the most recent developments in Ukraine, and Erdoğan repeated Ankara's willingness to do its half to peacefully resolve the struggle.
That's it from me, Martin Belam, for now. Léonie Chao-Fong can be with you shortly.
Ukrainian sources try to make clear what President Volodymyr Zelenskiy meant yesterday when he talked about "preventive strikes" being essential to cease Russia utilizing tactical nuclear weapons in Ukraine.
Russia's international minister Sergey Lavrov earlier seized on the feedback, decoding them as a name for Nato to make use of nuclear weapons in opposition to Russia, and stated the feedback confirmed why Russia's "particular navy operation" inside Ukraine's borders had been essential.
Reuters experiences that adviser to Zelenskiy, Serhii Nykyforov, has written on Fb:
Colleagues, you have got gone a bit of too far along with your nuclear hysteria and now you hear nuclear strikes even the place there are none. The president spoke in regards to the interval earlier than 24 February. Then it was essential to use preventive measures to stop Russia from beginning the struggle. Let me remind you that the one measures that have been about then have been preventive sanctions.
Nykyforov went on to say that hints at the usage of nuclear weapons have been "afforded solely by the terrorist state Russia. You'll by no means hear such calls from Ukraine."
Earlier US president Joe Biden gave his starkest warning but in regards to the prospect of the usage of nuclear weapons within the Ukraine battle, saying the world might face "Armageddon" if Vladimir Putin elects to make use of the weapons.
Russian and Ukraine human rights organisations amongst Nobel Peace Prize winners
The 2022 Nobel Peace Prize has been awarded to human rights advocate Ales Bialiatski from Belarus, the Russian human rights organisation Memorial and the Ukrainian human rights organisation Middle for Civil Liberties. You'll be able to comply with the most recent on that with my colleague Sam Jones.
Valentyn Reznichenko, governor of the Dnipropetrovsk oblast, has posted to Telegram to say that the town of Nikopol "shuddered all night time from the explosions". He wrote:
Virtually 40 Russian shells arrived in Nikopol. Within the metropolis, a number of high-rise buildings, greater than ten non-public homes, farm buildings, a transport infrastructure enterprise, fuel pipelines and electrical energy networks have been broken.
The claims haven't been independently verified.
The workplace of Turkey's president, Recep Tayyip Erdoğan, has issued a quick read-out after he held a name together with his Russian counterpart Vladimir Putin. Reuters experiences the pair mentioned the most recent developments in Ukraine, and that Erdoğan repeated Ankara's willingness to do its half to peacefully resolve the struggle.
The headquarters of the armed forces of the self-proclaimed Donetsk Individuals's Republic (DPR) has claimed to have captured three settlements from Ukrainian forces in Donetsk. The RIA Novosti information company quotes an official supply saying:
As of seven October, on the territory of the Donetsk Individuals's Republic, a bunch of troops of the DPR and Luhansk Individuals's Republic (LPR) , with hearth assist from the Russian Federation armed forces , liberated Odradivka, Vesela Dolyna and Zaitseve
The three settlements are near Bakhmut. The claims haven't been independently verified. The DPR and LPR are solely recognised as professional authorities by three UN member states: Russia, Syria and North Korea. Donetsk is among the areas of occupied Ukraine that Russia has claimed to annex, regardless of not having full management over the territory.
Reuters has a fast snap {that a} staff of 4 specialists from the Worldwide Atomic Power Company (IEAE) is predicted be arrive on the Zaporizhzhia nuclear energy plant later at present.
5 killed in Ukrainian shelling of key bridge in Kherson area – experiences
Russian-installed authorities within the area stated the strike occurred because the bus drove civilians throughout the bridge close to the village of Darivka. The claims of the assault haven't been independently verified.
Reuters experiences movies shared by the Russian armed forces information outlet Zvezda confirmed the burnt out wreckage of what gave the impression to be the chassis of a bus, and a badly-damaged van behind it with smoke pouring out its entrance compartment.
"Ambulance crews promptly arrived on the scene and supplied emergency medical therapy to the victims," a Telegram channel run by the Russian-installed well being authorities stated.
The Daryivskiy bridge, which spans practically 100 metres, is among the solely Russian-controlled crossings throughout the Inhulets river, a tributary of the huge Dnipro.
The bridge is strategically vital because it joins two Russian-occupied areas of the area and is simply 20km (12 miles) north-east of the town of Kherson. It's in one of many areas of Ukraine which Russia has claimed to annex.
Russia's international minister, Sergey Lavrov, has stated this morning that Volodymyr Zelenskiy's feedback about Nato making "preventive strikes" confirmed that Russia had wanted to hold out what it phrases its "particular navy operation" inside Ukraine's borders.
Yesterday, talking through video hyperlink to an viewers in Australia, Ukraine's president had stated "What ought to Nato do? Make it inconceivable for Russia to make use of nuclear weapons. What's vital, I as soon as once more attraction to the worldwide group … preventive strikes in order that they know what is going to occur to them in the event that they use them [nuclear weapons]."
As we speak marks Russian President Vladimir Putin's birthday, an occasion that has not gone unnoticed by Arvydas Anušauskas, Lithuania's defence minister. He has sarcastically tweeted, in a reference to pictures of Putin assembly world leaders at a distance, that the Russian president has been gifted "a brand new desk and binoculars", and stated that in his honour "Lithuanian persons are elevating cash for some 'fireworks'".
As we speak Putin on the event of his birthday was given a brand new desk and binoculars for speaking with guests. Lithuanian persons are elevating cash for some "fireworks".
— Arvydas Anušauskas (@a_anusauskas) October 7, 2022
Russian President Vladimir Putin attends a gathering with French President Emmanuel Macron in Moscow in February. {Photograph}: SPUTNIK/Reuters
Oleh Synyehubov, governor of Kharkiv, has posted to Telegram to say that within the final 24 hours, eight individuals have been hospitalised within the area because of Russian shelling. He additionally stated {that a} 47-year-old lady was injured by stepping on a mine within the Izium area. He cautioned residents once more in regards to the dangers, saying "the minefields left by the enemy are enormous". The claims haven't been independently verified.
Ukraine's president, Volodymyr Zelenskiy, appealed to Australia for assist in a essential UN vote subsequent week, asking Australian officers to affix in diplomatic efforts to make sure it was 'as unanimous as attainable'. 'I'm asking Australia to make use of all of its affect to persuade as many international locations as attainable to not stay impartial and to vote for worldwide regulation and in opposition to a Russian annexation,' he stated.
Addressing the Sydney-based Lowy Institute by video hyperlink, Zelenskiy revealed Australia was providing heavy arms to Ukraine in its subsequent package deal of navy assist and thanked Australia for its 'very significant defence' help, saying the Bushmaster protected-mobility autos which the nation had beforehand contributed had 'carried out masterfully'.
Zelenskiy calls on Australia to again UN vote condemning Russia's purported annexation – video
Julian Borger
Joe Biden has warned the world might face "Armageddon" if Vladimir Putin makes use of a tactical nuclear weapon to attempt to win the struggle in Ukraine.
The US president made his most outspoken remarks to this point about the specter of nuclear struggle, at a Democratic fundraiser in New York, saying it was the closest the world had come to nuclear disaster for sixty years.
"We now have not confronted the prospect of Armageddon since Kennedy and the Cuban missile disaster," he stated.
"We've received a man I do know pretty nicely," Biden stated, referring to the Russian president. "He's not joking when he talks about potential use of tactical nuclear weapons or organic or chemical weapons as a result of his navy is, you would possibly say, considerably underperforming."
Putin and his officers have repeatedly threatened to make use of Russia's nuclear arsenal in an effort to discourage the US and its allies from supporting Ukraine and serving to it resist the all-out Russian invasion launched in February. One concern is that he might use a brief vary "tactical" nuclear weapon to attempt to cease Ukraine's counter-offensive in its tracks and pressure Kyiv to barter and cede territory:
Ukraine takes again 500 sq km of territory in every week, Zelenskiy says
Ukraine's proclaimed wins within the southern area of Kherson are the most recent in a sequence of Russian defeats undermining the Kremlin's declare to have annexed round 20% of Ukraine.
"Greater than (500 sq. kilometres) have been liberated from Russian occupiers within the Kherson area alone" because the begin of October, President Volodymyr Zelensky introduced late Thursday in his nightly deal with.
The recaptured territory was house to dozens of cities and villages that had been occupied by Russian forces for months, southern military command spokeswoman Natalia Gumeniuk stated.
Kherson, a area with an estimated pre-war inhabitants of round a million individuals, was captured early and simply by Moscow's troops after their invasion launched on 24 February.
Welcome and Abstract
Good day and welcome to our stay protection of the Russia-Ukraine struggle. My title is Helen Sullivan and I'll be bringing you the most recent for the following few hours.
My colleague Julian Borger experiences from Washington that Joe Biden has warned the world might face "Armageddon" if Vladimir Putin makes use of a tactical nuclear weapon to attempt to win the struggle in Ukraine.
In the meantime listed here are the opposite key latest developments:
The EU has imposed a brand new spherical of sanctions on Russia, increasing import and export bans and blacklisting people over Moscow's annexation of 4 Ukrainian areas.
Volodymyr Zelenskiy, the Ukrainian president, informed European heads of state gathered in Prague that Ukraine should win in order that Russia doesn't "advance on Warsaw or once more on Prague".
The Russian opposition activist Vladimir Kara-Murza, who has been imprisoned in Moscow since April, is being investigated for "excessive treason", because the authorities step up their case in opposition to him for his criticism of the struggle in Ukraine.
The US has accused Russian mercenaries of exploiting pure sources in Central African Republic, Mali, Sudan and elsewhere to assist fund Moscow's struggle in Ukraine, a cost Russia rejected as "anti-Russian rage". The US ambassador to the United Nations, Linda Thomas-Greenfield, has stated the Wagner Group of mercenaries are exploiting pure sources and "these ill-gotten good points are used to fund Moscow's struggle machine in Africa, the Center East and Ukraine".
Ukrainian emergency providers stated three our bodies have been pulled from rubble after a Russian rocket strike destroyed a five-storey condo block within the southern Ukrainian metropolis of Zaporizhzhia.
The Kremlin denied experiences that 700,000 Russians had fled the nation since Moscow introduced a mobilisation drive it stated would name up a whole bunch of 1000's to struggle in Ukraine.
Two Russians who stated they fled their nation to keep away from obligatory navy service have requested asylum within the US after touchdown in a small boat on a distant Alaska island within the Bering Sea, Alaska Senator Lisa Murkowski's workplace stated Thursday.
Categories Top 10 Tags Biden, bodies, Hundreds, Kharkiv, Live, nuclear, retreat, Russian, RussiaUkraine, threat, war, Worst, years Post navigation
Musk's Twitter Deal, Oath Keepers Trial, Illinois Baker's Campaign
Nobel Peace Prize 2022 winners introduced

redpajama_set_name: RedPajamaCommonCrawl


text:
HomeNews DeskLocal NewsPolice release more information in Tuesday evening shooting of 17-year-old
Police release more information in Tuesday evening shooting of 17-year-old
June 2, 2017 Local News, News Desk, Police/Crime News 0
Cedartown, Ga. – June 2, 2017: Police are releasing additional information in what they are calling a "gang-related drug deal gone bad."
Officials have obtained an aggravated assault arrest warrant for Derian Bona Ventura-Flores of a Lawrenceville, Georgia, address.
The incident took place Tuesday, May 30, at 8:30 p.m. in the parking lot of the Polk School Central District Office on South College Street.
Upon arriving at the scene, police found a 17-year-old white male with a gunshot wound lying on the lawn in front of school district office.
The teen was transported to Floyd Medical Center where he underwent surgery and is recovering.
Officials are not releasing his name.
"We are not releasing the name of the victim because as of this point in time, he has not been charged with a crime, he is also a juvenile and is currently in DFCS custody," said Cedartown Police Chief Jamie Newsome.
Chief Newsome did say that the 17-year-old and Ventura-Flores had engaged in some form of cellular communication prior to the incident.
"They did have communication that brought them both to the location on South College Street," Newsome said.
"The incident had nothing to do with Polk School District, it just happened to be the location."
As to the label of a gang-related drug deal gone bad, Newsome said two factors contributed to that conclusion: the cell phone communication between the alleged perpetrator and the shooting victim, as well as $2,300 in fake currency located near the victim at the time officers responded.
Rome Braves rally to beat West Virginia 5-4 to open home stand
One Person Confirmed Dead In Drowning At Little River Falls
Miami bar shooting leaves 4 wounded, gunman dead: police January 18, 2022
The gunman who opened fire outside the Chicagoan Bar in Miami late Monday had returned after security kicked him out for assaulting his female companion, authorities said.
Stephen Sorace
Trisha Yearwood does Betty White Challenge, raises more than $30,000 for animal rescue charity January 18, 2022
Trisha Yearwood celebrated what would have been Betty White's 100th birthday by raising money for animals in the late star's honor.
Concha rips Jen Psaki for criticism of Youngkin's school mask order: 'He was elected, she was selected' January 18, 2022
Fox News contributor Joe Concha slammed Jen Psaki for criticizing Gov. Glenn Youngkin on his first day in officer over revoking the mask mandate in schools.
Fox News Staff
Rooftop Revelations: From selling drugs to 'Wings, Greens and Things' January 18, 2022
Tina Marie Jackson went from running a 10-man crew for the Gangster Disciples to running her own soul food takeout restaurant on the South Side of Chicago.
Eli Steele
Blinken to visit Ukraine amid rising tensions with Russia January 18, 2022


Sunlight once shone continuously over the Kingdom of Everdawn. Not even the tiniest mouse knew what darkness truly was. Night had yet to find them.
High on the cliff over the Ocean of Trust, a castle proudly sat. Beings of great importance lived within those walls—some got to lead, others only were meant to follow. There was, of course, one exception to that. There was one meant for more.
When the kingdom was young, it was led by King Harold the Fantastic. He was a kind and honest man, who ruled with a fair tone and a wizard's staff. He married Princess Poppy from the Land of Truths, a collection of islands in the north ocean. This marriage bonded the land in light for a hundred more years. No one thought of the consequences. After all, what could be bad about constant light?
King Harold and the new Queen Poppy had three sons in sixteen seasons. Their hopes were that each boy carried a legacy with them. Sadly, the eldest, a strong and handsome lad, fell in the War of the Plains. The second, who was even stronger and better looking, fell in the War of the Mountains. But, their third son, Prince Elric was the most handsome of the three. Though, nothing about him was a warrior, and he did not share his brother's goals or strengths. No, Elric was a baker at heart. His views on royalty were different from most. He believed all should be equal in the land of forever daylight. So, when it became his turn to rule, he changed it all by marrying the sweet Sarilene, who worked the market every third day. When those two married, the sun began to shine even brighter, and the rain came even less frequently.
The ground started to crack deep under the kingdom's dungeons and sewers. The cracks were small and went unnoticed by the people above.
Fifteen season cycles had passed after King Elric, and Queen Sarilene had their one and only child, Dalfinia, a girl with bright yellow hair and a kind heart. The days had grown hotter, and the rain hadn't shown itself in two seasons. The ponds and lakes were drying up, and the next crop wasn't growing. The kingdom had begun to spend their riches on food from other plains. King Elric began to worry about the future of his lands.
Delfina's hair sparkled in the sunlight as she dipped her bare toes in the tide of the Forbidden Coast. All the other bodies of water were nearly dry, yet the Ocean Of Truths was rising. Her parents didn't trust this, so they forbid everyone from the kingdom from going on the coast. Dalfinia didn't like to break the rules very much, but she could not bear to stay away. She didn't care to be indoors or to be still. She spent her childhood running through rainbow meadows with the wood's wolf pups and hiding from ogres woken from their naps. Her father would say if the wind could be seen it would look a lot like Dalfinia. Every day seemed to involve some kind deed quest or another. For Dalfinia's path was lined with helping others. Her kindness was so pure and rare; tales began to travel about the princess who was only kind, nothing else.
On the day when her life as she knew it changed forever, her parents were unable to find her for lunch. The midday feast was set down, but the princess never showed. Now, it didn't take long for the Queen to find herself in a fit of worry, and before the lunch plates were gathered, guards and scouts were sent to find Princess Dalfinia. The King was calmer when it came to matters of the Princess; he was a bit more understanding of her free-spitted ways.
As the Queen was picturing her daughter being carried off by hungry ogres, Princess Dalfinia was scaling the Wall Of Might. Dalfinia's hair blew in wild circles around her slight body. She reached for a large, jagged stone above her head and to the left. The angry waters of the Forbidden Coast beat on the rocky shores beneath her. She had only one more foot to go, and she should be able to reach the Rockbirds' nest. She clutched a baby bird close to her chest. Dalfinia had found this bird washed up by her feet. Only Dalfinia would bring the baby back to its mother without thinking of herself. As Dalfinia carefully placed the bird back with its siblings, her eyes caught a glimpse of the OtherCliffs, the kingdom to the west. All that was once green was brown and yellow. Everything was dying. But, why? Sadness pawed at the princess. The mother bird flew in at this point; she's looking in the same direction as Dalfinia. They both knew that the sunlight had become too much.
"There was only one father," Dalfinia had answered this realizing then that there should have been many more nests on the cliffs. She wondered and worried about the missing birds.
Now Dalfinia hung upside down from her mother's throne in the empty ballroom. The King's birthday was after all a big day, and even though the Kingdom should have been rationing all of their supplies they didn't hold back for these festivities. The treats were lavish and the meat plentiful. Young children lined up with flutes and piccolos and were getting ready to perform a particular song for the king. Something deep in Dalfinia's stomach told her something was wrong. Brown grass and missing birds circled in her head. Since the troubling thoughts were circling rather furiously, she did not notice that her parents and the first-choice-guests entered the room. It wasn't until the Queen grabbed her foot, did Dalfinia realize she was no longer alone. She sat upright and quickly shifted into her own thrown. The princess chose that moment not to look at her mother in the eye.
Queen Sarilene sat down beside Dalfinia, her father beside her. A question sat on Queen Sarilene's tongue. Just as it was going to come to light, a trumpet sounded, and the rest of the guests entered the ballroom. The Queen decided it would have to wait until later.
The feast carried on much like they usually do. Good smells, healthy laughter, and voices on top of voices. Dalfinia sat there much quieter than usual, only picking at her turkey leg. The King was in a loud debate about what makes parsnip pie so much better than turnip pie. Across from the royal family, sat four large men that Dalfinia has never seen before. She cast her eyes at the guards, the other villagers, and then her parents. Was Dalfinia the only who noticed these strangers? Just as that question tunneled its way through her head, the main doors flew open. A swirl of sparkles, then a puff of smoke. Three small-winged fairies floated in the large entrance-way.
"Too much of one thing, just too much." The blue-haired fairy said to the other two.
Everyone in ballroom stood silent and stared.
"Everyone is staring at us you know. You talk to yourself, and people will think you're crazy. You suppose, what can we do?" The smallest fairy said this. Her hair was short, purple, and poking every-which-way on top of her head.
"The Dark Mountain needs a payment," Blue said looking at her feet, her voice grim.
"What payment?" The king asked, but it's possible some say that he already knew the answer.
"Something bright, something pure, something unlike anything else." The green-headed one said.
As quickly as they arrived, the fairies were gone.
That next day, Princess Dalfinia said goodbye to her grief-stricken parents and made her way across the bridge into the village lands. Everyone came out of their storefronts and homes to watch the princess leave, all knowing she will never return to them. The Princess never gave herself permission to look back. Looking back is a side effect of not truly moving forward. Dalfinia didn't want to do anything that would risk her failing any tests of any kind. She knew that she had to sacrifice herself to the Dark Mountains to save the kingdom from disappearance, and that was what she was going to do. Everyone has their path, destiny all their own written in the stars above. Bringing the night, was Dalfinia's. She had imagined a long and full life for herself, expecting to be Queen some day and have children of her own. But, she accepted it for it was. She loved her kingdom and the people in it.
The first few days quickly blurred together, and her feet grew sore. The sadness inevitably found her in moments when she was most tired. She wondered how her world was getting on without her.
The Dark Mountain grew closer as the princess entered a dense forest. As she entered the tree-covered world, she got a sense of what darkness was. The sun still found spaces between the branches, so some light still came in, but this was far more darkness than she had ever seen before. The darkness made her want to make her way through as quick as possible. Every ounce of Dalfinia feared the darkness corners of the forest around her. Fear of darkness was new to her. Then, she realized the real importance of her quest, her fate. The world without contrast won't thrive. Everyone needs the two sides; the world needs both good and bad.
Without hesitation Dalfinia reached into her satchel, pulled out her canteen, and offered it to the fox. "You may have the rest," she said softly.
"What will you drink?" asked the fox. His pointy ears twitched.
"I don't need to drink anymore," She replied as she extended her arm further, insisting that the fox take it. The fox did, but the fox was curious.
"Why don't you need water?" The fox asked. They were walking side by side now, still with no end to the woods in sight.
"This is what I hear." The princess's voice in fainter than it was.
"Thank you for the water, and good luck," the fox said before scampering away.
The princess carried on further when she finally saw a break in the woods and a bright meadow in the far distance, she also saw a wolf stalking something. Dalfinia continued slowly, keeping her eyes on the wolf. Finally, when she got right behind it, she was able to see that wolf had a rabbit in its sights. Just then, Dalfinia stepped on a stick, causing it to break. A loud snap traveled through the small patch of forest the three of them shared. The wolf's eyes fixed on the girl, and the rabbit ran off without a glance in their direction.
The large gray wolf turned to look at the young and exhausted princess. "You did that on purpose," the wolf snarled.
"I did no such thing." Princess Dalfinia continued to walk to the meadow that she was so pleased to see in front of her. She reached into her satchel again and pulled out some trail rations. "You may have the rest. I no longer need food," she said softly.
"What will you eat?" asked the wolf. His haunches relaxed.
"I don't need to eat anymore," She replied as she extended her arm further, insisting that the wolf take it. The wolf did, but the wolf was curious.
"Why don't you need food?" The wolf asked. They were walking side by side now, the open meadow getting closer to them.
"This is what I hear." The princess's voice was just above a whisper.
"Thank you for the food, and good luck," the wolf said before running off.
The bright, wide-open sky was a relief. Dalfinia even though she was at her weakest state, was happy to be out in the open again. The massive Dark Mountain to the North of her was even closer. Dalfinia knew that today was her last day. She isn't as scared as she thought she would be. By now, it appears that she's ready.
A child's cry filled the meadow. Dalfinia looked to the direction of the screams. Two children ran toward her. One girl was crying, and one girl was soaking wet.
Dalfinia without hesitation took her cloak and wrapped it around the drenched child.
"Thank you." The girl said. Her hair as red as roses and her eyes as green as the grass in October. She then looked to her right at the girl crying. This girl was a little younger and a little smaller. Her hair was red like her sisters, but her eyes were a dark brown. "I told you we'd find her. Mom never goes far." The older of the two spat, as she wrapped herself tightly with Princess Dalfinia's cloak.
"I'm not your mother." The princess said, her voice now was only a whisper.
"Sure you are. Maybe not yet, but you will be." The eldest girl said. Her voice squeaked like a mouse.
"This is what I hear." The princess thinks but isn't able to say. The girls were gone. A thick darkness cast an enormous shadow across the meadow. Dalfinia pulled in a deep breath as she turned to face the mountain.
The mountain said nothing back. Its darkness reached out like a helping hand to someone fallen. Clouds came down from the sky, wrapped themselves around her, and brought up to the top of the mountain. The Dark Mountain continued to extend its reach. Dalfinia disappeared into the mists. The sun sighed as it set above Everdawn. As the moon rose, King Elric and Queen Sarilene looked at it with tears. Princess Dalfinia had her legacy, and so did the king.
Day and night took turns and the beautiful kingdom and the lands that surrounded it carried on, as did the story of the one meant for more.


text:
Daniel Goleman est un psychologue américain né le  à Stockton (Californie).

Biographie
Il est diplômé de l'université Harvard et docteur en psychologie clinique et développement personnel, puis devient journaliste au New York Times, où il suit particulièrement les sciences du comportement.

Il publie en 1995 l'ouvrage Intelligence émotionnelle.

Il est membre du conseil d'administration du Mind and Life Institute, qui facilite les rencontres entre la science et le bouddhisme. Il fait partie de l'Association américaine pour le progrès de la science.

Travaux

Intelligence émotionnelle

Les styles de leadership
Pour lui, il y a six styles de leadership :
 Autoritaire
 Visionnaire
 Collaboratif
 Participatif
 Chef de file
 Coach

Ouvrages

 EspritScience : Dialogue Orient-Occident, avec le Dalai Lama, Herbert Benson, Howard E. Gardner, Robert A.F. Thurman, éd. Claire Lumière, 1993,  éd, 1999, ,

 Participe à Bonheur de la méditation, de Mingyour Dorje, avec Eric Swanson, et Matthieu Ricard (2007), Fayard,
 Focus. The Hidden Driver of Excellence, Harper, 2015.

Voir aussi

Bibliographie
 (Vidéo) .

Liens externes

Naissance à Stockton (Californie)
Naissance en mars 1946
Étudiant de l'université Harvard
Psychologue américain
Auteur en management

redpajama_set_name: RedPajamaWikipedia


Help your kid to identify colors with coloring pages. There are so many benefits to coloring the pages. When they start to color they enjoy themselves and their concentration will increase. Here are some of the coloring pages I uploaded. I will sure that they will enjoy with these worksheets.
Observe the worksheets and learn how to draw pictures. You have to apply the color lightly using different colors as you want to create and set background color with your own creativity.
Below are some interesting coloring pages with Halloween theme. If you are having a Halloween party or want to make the kids occupied with activity for Halloween, print and share the attached coloring pages.