text:
\section{Introduction}

In $\Bbb R^3,$  together with the plane, Catenoid is the only minimal surface of revolution.
If not counting the plane, it is the first minimal surface  discovered by Leonhard Euler in 1744.
The counterpart of minimal surfaces in the Lorentz-Minkowski space $\Bbb R^3_1,$ are  (spacelike or timelike) surfaces with zero mean curvature.
Since the metric in $\Bbb R^3_1$ is not positive definite, there are three types of vectors (spacelike, lightlike or timelike). Therefore, more complicated than rotations in Euclidean space, in $\Bbb R^3_1,$ there are three types of Lorentzian rotations depending on the causal of the rotation axises. Maximal surfaces of revolution in $\Bbb R^3_1$ have been classified in \cite{ko}.  Spacelike and timelike surfaces of revolution with constant mean curvature in $\Bbb R^3_1$ have been studied in \cite{leva}, \cite{leva1} and  \cite{lema}.  Recently, maximal surfaces in Lorentzian product spaces have been also studied by some authors (see, for example, \cite{al}, \cite{alal1}, \cite{alal2}, \cite{li} and \cite{li1}).

It is natural to study (spacelike or timelike) surfaces of revolution with zero weighted mean curvature, also called $f$-mean curvature, in $\Bbb R^3_1$ endowed with a density, i.e.,  a positive function defined on $\Bbb R^3_1$ used to weight the area  (the length) of surfaces (curves).

In this paper, such a density that we considered is the Gaussian-Euclidean density, i.e., the space is the Lorentzian product $\Bbb G^2\times \Bbb R_1,$  where $\Bbb G^2$ is the Gauss plane.
The space $\Bbb G^2\times\Bbb R_1$ is a special case of $n$-dimensional spacetime with a density that does not affect ``time''. It should be mentioned  that the space we are living can be seen as a  4-dimensional spacetime with density, the gravity,  that affect  ``space'' and does not affect ``time''.

It is showed that the axis of an $f$-maximal surface of revolution in $\Bbb G^2\times\Bbb R_1$ must be the $z$-axis. Then, solving the $f$-Maximal Surface Equation for surfaces of revolution we obtain new non-trivial examples,  called spacelike $f$-Catenoids. Beside horizontal planes, they are the only $f$-maximal surfaces of revolution.  This is the first result of the paper.

For the timelike case, by a similar proof, it is proved that  the axis of a timelike $f$-minimal surface of revolution must be the $x$-axis or the  $z$-axis. If the rotation axis is the $x$-axis, the only timelike $f$-minimal surfaces of revolution are vertical planes containing the $z$-axis.
If the rotation axis  is the $z$-axis, there are a family of  timelike $f$-minimal surfaces of revolution, called timelike $f$-Catenoids, that convergences to another timelike $f$-minimal surface, the cylinder $x^2+y^2=1.$   The second main result of the paper is that a timelike $f$-minimal surface of revolution is either the cylinder $x^2+y^2=1,$ a vertical plane containing the $z$-axis or a timelike $f$-Catenoid.

\section{Preliminaries}
For simplicity, all concepts as well as results in this section are introduced in 3-dimensional space. For more details about Lorentz-Minkowski
spaces, manifolds with density or the Gauss space we refer the reader to \cite{lo1}, \cite{mo1}, \cite{mo2}, \cite{nei}, \cite{we} and references therein.

Let $\Bbb R^{3}_1$ be the Lorentz-Minkowski 3-space endowed with the Lorentzian scalar product
$$\langle \ , \ \rangle=dx^2+dy^2-dz^2.$$
A nonzero
vector $\textbf x\in \Bbb R^{3}_1$ is called spacelike, lightlike  or timelike    if  $\langle \textbf x, \textbf
x\rangle>0$, \ $\langle \textbf x, \textbf x\rangle=0,$ or  $\langle \textbf x, \textbf x\rangle<0,$ respectively.

The norm of the vector $\textbf x$ is then defined by $\|\textbf x\|=\sqrt{|\langle\textbf x, \textbf x\rangle|}.$
Two vectors $\textbf x_1=(x_1,y_1, z_1),\  \textbf x_2=(x_2, y_2, z_2)\in \Bbb R^{3}_1$  are said to be orthogonal if $\langle \textbf x_1, \textbf x_2\rangle=0,$ i.e., $x_1x_2+y_1y_2-z_1z_2=0.$ The
 Lorentzian vector product of $\textbf x_1$ and $\textbf x_2,$  denoted by $\textbf x_1\wedge\textbf x_2,$  is defined by
$$\textbf x_1\wedge\textbf x_2=\begin{vmatrix} \textbf e_1&\textbf e_2&-\textbf e_3\\
x_1&y_1&z_1\\
x_2&y_2&z_2\end{vmatrix}, $$
where  $\{\textbf e_1,\textbf e_2,\textbf e_3\}$ is the canonical basis of $\Bbb R^{3}_1.$
For every $\textbf x\in \Bbb R^{3}_1,$
$$\langle \textbf x, \textbf x_1\wedge\textbf x_2\rangle=\det( \textbf x, \textbf x_1,\textbf x_2).$$
It follows that $\textbf x_1\wedge\textbf x_2$ is orthogonal to both
$\textbf x_1$ and $\textbf x_2.$

A surface  in  $\Bbb R^{3}_1$ is called spacelike (timelike) if its induced metric
from $\Bbb R^{3}_1$ is Riemannian (Lorentzian) or equivalently,  every normal vector of the surface is timelike (spacelike).

For example, let $\alpha$ be a plane whose general equation is $Ax+By+Cz+D=0, \ A^2+B^2-C^2\ne 0.$ It is easy to see that, the vector
${\bf n}=(A,B,-C)$ is a  normal vector of $\alpha.$ The plane $\alpha$ is spacelike or timelike if and only if $\bf n$ is timelike or spacelike, respectively.

The following formula for computing the mean curvature of a (spacelike or timelike) surface in $\Bbb R^3_1$  is well-known (see \cite{lo2}, for instance)
\begin{align}\label{mean}
H=\epsilon\frac{Eg-2 Ff+Ge}{2(EG-F^2)},
\end{align}
where $\epsilon=-1,$ if the surface is spacelike; $\epsilon=1,$ if the surface is timelike; $E, F, G$ are the coefficients of the first fundamental form and $e,f,g$  are the coefficients of the second fundamental form.

A spacelike (timelike) surface is called maximal (timelike minimal) if its mean curvature $H$ is zero everywhere.

There are three kinds of rotations in $\Bbb R^3_1$: rotations about a spacelike axis, rotations about a timelike axis and rotations about a lightlike axis  (see \cite{leva}, for instance).  Below are the matrices of some typical kinds of rotations that will be used in the proof of Lemma \ref{sf} and Lemma \ref{tf}.
\begin{enumerate}
\item The matrix corresponding to a rotation about the $y$-axis is
$$	\begin{pmatrix}
	\cosh \theta & 0 & \sinh\theta\\
	0&1&0\\
\sinh\theta & 0& \cosh \theta
\end{pmatrix}. $$
\item The matrix corresponding to a rotation about the lightlike axis  $x=z,\ y=0$  is
\begin{align*}
	\begin{pmatrix}
	1-\dfrac{v^2}{2}&-v&\dfrac{v^2}{2}\\
	v&1&-v\\
	\dfrac{-v^2}{2}&-v&1+\dfrac{v^2}{2}
	\end{pmatrix}.
\end{align*}
\item The matrix corresponding to a rotation about the $z$-axis is
$$	\begin{pmatrix}
	\cos\theta &-\sin\theta &0\\
\sin\theta &  \cos \theta &0\\
0&0&1\\
\end{pmatrix}. $$
\end{enumerate}

A surface of revolution is a surface in $\Bbb R^3_1$ obtained by rotating a  curve $\gamma,$ the generatrix, around an axis of rotation $l,$  assuming that $\gamma$ and $l$ are in a plane.

A density on $\Bbb R^{3}_1$ is a positive function, denoted by $e^{-f},$  used to weight area (length) of surfaces (curves).
The weighted mean curvature or the $f$-mean curvature of a (spacelike or timelike) surface,  denoted by $H_f,$ is defined by
$$H_f=H+\frac 12\langle  \nabla f, N\rangle.$$
 A spacelike (timelike) surface $\Sigma$ is called $f$-maximal (timelike $f$-minimal) if  $H_f=0$ everywhere, i.e., $H=-\frac 12\langle \nabla f, N\rangle.$

Gauss space $\Bbb G^2,$ is  just $\Bbb R^2$ with the Gaussian probability density
$$e^{-f(x,y)}=\frac 1{2\pi}e^{-\frac{x^2+y^2}2},$$
where $ (x,y)\in\Bbb G^2.$

Therefore, the Lorentzian product $\Bbb G^2\times\Bbb R_1$ can be seen as  $\Bbb R^{3}_1=\Bbb R^2\times\Bbb R_1$ endowed with the Gaussian-Euclidean density
$$e^{-f(x,y,z)}=\frac 1{2\pi}e^{-\frac{x^2+y^2}2},$$
where $ (x,y,z)\in\Bbb G^2\times \Bbb R_1.$ It should be noted that the last coordinate is not dependent on the density.

Let $\Sigma$ be an oriented  (spacelike or timelike) surface in $\mathbb{G}^2\times\mathbb{R}_1,$ \ $N$ be a unit normal vector field  on $\Sigma$ and
 $\rho$  be the projection onto  the $z$-axis. Then  at any point $p\in \Sigma,$  we have the following.
\begin{lem}\label{lem1}(Geometric meaning of the quantity $\langle\nabla f,\mathbf{N}\rangle$)
\begin{align*}
\left| \langle\nabla f,\mathbf{N}\rangle (p)\right|=\operatorname{d_E}\big(\rho(p),T_p\Sigma\big).|\mathbf{N}|_{\operatorname{E}},
\end{align*}
where $d_E$ and $| \ \  |_E$ denote the Euclidean distance the Euclidean length, respectively.
\end{lem}
\begin{proof}
Suppose that $p=(x_0, y_0, z_0)$ and $\mathbf{N}(p)=(a,b,c),\ a^2+b^2-c^2=\pm 1.$ An equation of $T_p\Sigma$ is of the form $ax+by-cz+d=0.$
We have $\nabla f(p)=(x_0,y_0,0)$ and $\rho(p)=(0,0,z_0).$ Therefore
$$\left| \langle\nabla f,\mathbf{N}\rangle (p)\right|=|ax_0+by_0|=|cz_0-d|=\operatorname{d_E}\big(\rho(p),T_p\Sigma\big).|\mathbf{N}|_{\operatorname{E}}.$$
\end{proof}
By Lemma \ref{lem1},  it is not hard to prove the followings.
\begin{cor} \label{co1} In  $\mathbb{G}^2\times\mathbb{R}_1,$
\begin{enumerate}
\item horizontal planes  are $f$-maximal surfaces;
\item vertical planes have constant $f$-mean curvature, such a plane containing the $z$-axis is timelike $f$-minimal;
\item circular cylinders about the $z$-axis  have constant $f$-mean curvature, such a cylinder is timelike $f$-minimal if and only if the radius is 1.
\end{enumerate}
\end{cor}
\section{Spacelike $f$-maximal surfaces of revolution in $\Bbb G^2~\times~\Bbb R_1$}
\subsection{Spacelike $f$-Catenoids in $\Bbb G^2\times\Bbb R_1$}

 In the $xz$-plane, consider the curve $\gamma_S$ that is the graph of the function (see Figure 1).
$$h(u)= \int_{0}^{u}\sqrt{\dfrac{1}{1+\tau^2e^{\tau^2+C}}}d\tau,\ \ u\in\Bbb R.$$

 Rotating  $\gamma_S$ the about the $z$-axis, we obtain a surface of revolution (see Figure 2), denoted by  $\Sigma_S,$  that can be parametrized as follows.
$$	X(u,v)=\left(u\cos v, u\sin v, \int_{0}^{u}\sqrt{\dfrac{1}{1+\tau^2e^{\tau^2+C}}}d\tau\right),$$
where $C$ is a constant. It is easy to verify that the curve is spacelike and therefore the surface $\Sigma_S$ is spacelike. The surface $\Sigma_S$ has a singular point, that is the origin.
By a direct computation, it follows that the $f$-mean curvature  of  $\Sigma_S$ is zero, i.e., $\Sigma_S$ is $f$-maximal. We call $\Sigma_S$  a spacelike $f$-Catenoid.
\vskip 1cm
\begin{center}
	\includegraphics{DuongSinhSpacelike-1}\\
	\textit{Figure 1.\ The generatrix $\gamma$}
\end{center}

\begin{center}
	\includegraphics{DuongSinhSpacelike-2}\\
	\textit{Figure 2.\ Spacelike $f$-Catenoid.}
\end{center}

\subsection{Classification of $f$-maximal surfaces of revolution in $\Bbb G^2\times\Bbb R_1$}
In the Lorentz-Minkowski space $\Bbb R^3_1,$ because the mean curvature of a (spacelike or timelike) surfaces is invariant under Lorentzian transformation, when studying surfaces of revolution of constant mean curvature, if the rotation axis is timelike, spacelike or lightlike we can suppose it is the $z$-axis, the $x$-axis, or the lightlike axis $x=z, y=0,$ respectively. In the space $\Bbb G^2\times \Bbb R_1,$ with the appearance of the density, we can not do this because the $f$-mean curvature is not invariant under some Lorentzian transformations. Since the density is dependent on the distance from points to the $z$-axis and not dependent on the last coordinate, the $f$-mean curvature of a surface does not change under rotations about as well as  translations along the $z$-axis (see Lemma (\ref{lem1}). This observation is useful for the rest of the paper to simplify some calculations.

\begin{lem}\label{sf} A spacelike surface of revolution $\Sigma$ in $\Bbb G^2\times\Bbb R_1$ can be parametrized as follows.
\begin{enumerate}
\item If the rotation axis is spacelike
\begin{align}\label{ss} X(u,v)=(u\cosh\theta+g(u)\sinh \theta\cosh v, u\sinh v+a, u\sinh\theta+g(u)\cosh\theta\cosh v).
\end{align}
\item If the rotation axis is lightlike
\begin{align}\label{ls}
X(u,v)=\left(u-\left[u-g(u)\right]\dfrac{v^2}{2}+a,v[u-g(u)],g(u)-\left[u-g(u)\right]\dfrac{v^2}{2}\right).
\end{align}
\item If the rotation axis is timelike
\begin{align}\label{ts}
X(u,v)=(u\cos v\cosh\theta+g(u)\sinh\theta, u\sin v +a,u\cos v\sinh\theta+g(u)\cosh\theta).
\end{align}
\end{enumerate}
\end{lem}

\begin{proof}
\begin{enumerate}
\item {\bf The case $l$ is spacelike}.
Under a suitable rotation about the $z$-axis,  we can assume that the plane containing the generatrix $\gamma$ and the rotation axis $l$ are parallel to or coincident with the $xz$-plane.  If $l$ and the $z$-axis are not intersect, we assume that the common perpendicular line of $l$ and the $z$-axis is the $y$-axis. If  $l$ and the $z$-axis are intersect, we assume that the intersection point is the origin $O.$ Let $\{H\}=(0,a,0)$ be the intersection point of $l$ and $xy$-plane and let  $\theta$ be the angle between $l$ and $Ox.$

There exist a Lorentz transformation that maps $\Sigma$ to a surface of revolution $\Sigma_1$ obtained by rotating a spacelike $\gamma_1,$ that lies in the $xz$-plane, about the $x$-axis.
This transformation is a composition of a translation along $y$-axis by a vector $v=(0, a, 0)$ and a rotation about $y$-axis of angle $\theta.$
Because the curve $\gamma_1$ is spacelike, it can be parametrized as $\gamma_1(u)=(u,0,g(u)),\ u\in I\subset\Bbb R,\ g\ne 0,\ 1-g'^2>0.$ Then, a parametrization  of $\Sigma_1$ is $X(u,v)= (u, g(u)\sinh, g(u)\cosh v)$ and therefore a parametrization of $\Sigma$ is
\begin{align*}
	X(u,v)&=\begin{pmatrix}
	\cosh \theta & 0 & \sinh\theta\\
	0&1&0\\
\sinh\theta & 0& \cosh \theta
\end{pmatrix}
\begin{pmatrix}
u\\ g(u)\sinh v\\
  g(u)\cosh v
\end{pmatrix} + \begin{pmatrix}
 0\\
a\\
 0
\end{pmatrix}\\
&=(u\cosh\theta+g(u)\sinh \theta\cosh v, g(u)\sinh v+a, g(u)\sinh\theta+g(u)\cosh\theta\cosh v)
\end{align*}
\item {\bf The case $l$ is lightlike}.

By a suitable rotation about the $z$-axis, we can assume that the plane containing the generatrix $\gamma$ and the rotation axis $l$ is the $xz$-plane and $l$ is parallel to $e_1+e_3.$   Let $\{H\}=(a,0,0)$ be the intersection point of $l$ and the $xy$-plane and suppose that $\gamma(u)=(u, 0, g(u)).$ Then,  a parametrization of $\Sigma$ is

\begin{align*}
X(u,v)&=\begin{pmatrix}
	1-\dfrac{v^2}{2}&-v&\dfrac{v^2}{2}\\
	v&1&-v\\
	\dfrac{-v^2}{2}&-v&1+\dfrac{v^2}{2}
	\end{pmatrix}\begin{pmatrix}
u\\ 0\\ g(u)
\end{pmatrix} + \begin{pmatrix}
a\\
0\\
 0
\end{pmatrix}\\
&=\left(u-\left[u-g(u)\right]\dfrac{v^2}{2}+a,v[u-g(u)],g(u)-\left[u-g(u)\right]\dfrac{v^2}{2}\right).
\end{align*}
\item {\bf The case $l$ is timelike}.

By the same arguments as in the case $l$ is spacelike, but in this case, $\theta$  is the angle between $l$ and the $z$-axis and $\Sigma_1$ is the surface of revolution obtained by rotating $\gamma_1$ about the $z$-axis.

A parametrization of $\Sigma_1$ is  $(u\cos v, u\sin v, g(u)).$ Therefore, a parametrization of $\Sigma$ is
\begin{align*}X(u,v)&=	\begin{pmatrix}
	\cosh \theta & 0 & \sinh\theta\\
	0&1&0\\
\sinh\theta & 0& \cosh \theta
\end{pmatrix}\begin{pmatrix}
u\cos v\\  u\sin v\\ g(u)
\end{pmatrix} + \begin{pmatrix}
 0\\
a\\
 0
\end{pmatrix}\\
&=(u\cos v\cosh\theta+g(u)\sinh\theta, u\sin v +a, u\cos v\sinh\theta+g(u)\cosh\theta).
\end{align*}
\end{enumerate}
\end{proof}
\begin{thm}\label{thm1}
 An $f$-maximal surface of revolution $\Sigma$  in $\Bbb G^2\times\Bbb R_1$ is either a horizontal plane or a spacelike $f$-Catenoid.
 \end{thm}
\begin{proof}
 Since Lorentz transformations preserve the mean  curvature of surfaces,  along a coordinate curve $u=u_0,$  the mean curvature $H$ of $\Sigma$ is constant.
Therefore if the $f$-mean curvature $H_f$ of $\Sigma$ is zero,  along any coordinate curve $u=u_0,$
$\langle \nabla f, N\rangle$ must be constant. This fact will be used to eliminate the case that the rotation axis $l$ is spacelike or lightlike.
\end{proof}
The followings are obtained by straightforward computations.
\begin{itemize}
\item If $l$  is spacelike and (\ref{ss}) is a parametrization of $\Sigma,$ then
\begin{align*
\langle \nabla f, N\rangle&=
\frac{1}{\sqrt{1-[g'(u)]^2}}[u\sinh\theta\cosh\theta\cosh v + g(u)(\sinh^2\theta\cosh^2v+\sinh^2v\\
&\;\;+g'(u)\sinh\theta\cosh v\cosh\theta)+ug'(u)\cosh^2\theta+a\sinh v{\sqrt{1-[g'(u)]^2}}].
\end{align*}
The condition ``$\langle \nabla f, N\rangle$ is not constant'' is equvalent to that ``for any $u$
$$Q:=\displaystyle\frac{\partial}{\partial v}\sqrt{1-[g'(u)]^2}\langle \nabla f, N\rangle,$$
must be zero for every $v.$''
By a straightforward computation, we obtain
\begin{align*}
Q
&=\dfrac{u\sinh2\theta}{2}\sinh v+ g(u)\left(\sinh^2\theta\sinh 2v+\sinh 2v+g'(u)\dfrac{\sinh2 \theta}{2}\sinh v \right) + a\cosh v\\
	&=g(u)\sinh 2v\cosh^2\theta+[u+g(u)g'(u)]\dfrac{\sinh2 \theta}{2}\sinh v+a\cosh v.
\end{align*}
It is not hard to see that  if  for any $u\in I,$\ $Q$  vanishes for every $v$ then $g(u)=0.$ This is impossible because $g\ne 0.$
\item If $l$  is lightlike  and (\ref{ls}) is a parametrization of $\Sigma$, then
\begin{align*}
	\langle \nabla f, N\rangle&=\frac 1{g'^2(u)-1}\left[u-\left[u-g(u)\right]\dfrac{v^2}{2}\right]\left[g'(u)+\dfrac{v^2}{2}\left[1-g'(u)\right]\right]+v^2[u-g(u)][g'(u)-1]\\
	&=\frac 1{{4}(g'^2(u)-1)}\left[(g(u)-u)(1-g'(u))v^4+2(u+g(u)g'(u)-2g(u))v^2\right]+ug'(u).
\end{align*}
	We can verify that $\langle \nabla f, N\rangle$ is not constant.
\item The case $l$ is timelike  and (\ref{ts}) is a parametrization of $\Sigma,$
A direct computation shows that
\begin{align}
\label{eq2}
\langle \nabla f, N\rangle&=\dfrac{1}{\sqrt{1-[g'(u)]^2}}\left[ ug'(u)(1+\cos^2v\sinh^2\theta)+[u+g(u)g'(u)]\dfrac{\sinh2\theta}{2}\cos v+g(u)\sinh^2\theta\right]\nonumber\\
&\quad+\dfrac{ag'(u)\sin v}{\sqrt{1-[g'(u)]^2}}.
\end{align}
\end{itemize}
We can see that, for any $u,$
$\langle \nabla f, N\rangle$ is constant if and only if $\theta=a=0,$ i.e.,  $l$ must be the $z$-axis.
The parametrization of  $\Sigma$ is now become
\begin{align} 	X(u,v)=(u\cos v, u\sin v, g(u)).
\end{align}

A direct computation shows that
\begin{align*}
	& H=\dfrac{-1}{2}\dfrac{(1-g'^2)g'+ug''}{u(1-g'^2)^{3/2}},\\
	&\langle \nabla f, N\rangle=\dfrac{-g'u}{\sqrt{1-g'^2}}.
\end{align*}
Therefore, $\Sigma$ is $f$-maximal if and only if $g$ satisfies the following equation
\begin{align}
\label{revolution1}
	(1-g'^2)g'+ug''+u^2g'(1-g'^2)=0.
\end{align}

We solve equation \eqref{revolution1}.
\begin{itemize}
\item It is clear that $g(u)=a,$  where $a$ is constant, is a solution of \eqref{revolution1}, i.e., $\Sigma$ is a vertical plane.
\item Now locally we can suppose that  $g'(u)\ne0$ for every $u\in J,$ where $J\subset I.$  Multiply both sides of \eqref{revolution1} by $g'$ and set $h=g'^2$, we get
\begin{align*}
- \dfrac{dh}{du}=2\dfrac{u^2+1}{u}h(1-h).
\end{align*}
Solving this equation, we obtain
\begin{align*}
 \ln\dfrac{1-h}{h}=u^2+\ln u^2 + C,\; C\in\mathbb{R},
\end{align*}
or
\begin{align*}
h=\dfrac{1}{1+u^2e^{u^2+C}}.
\end{align*}
Therefore,
\begin{align*}
 g'(u)=\pm \sqrt{\dfrac{1}{1+u^2e^{u^2+C}}},
\end{align*}
and
\begin{align*}
 g(u)=\pm\int_{u_0}^u\sqrt{\dfrac{1}{1+\tau^2e^{\tau^2+C}}}d\tau,\ \ u_0\in J.
\end{align*}
The function $g$ is defined over $\Bbb R,$  therefore we can assume that $I=\Bbb R, \  u_0=0,$ i.e.,
$\gamma\equiv\gamma_s$ or $\gamma\equiv\overline{\gamma_s},$ where $\overline{\gamma_s}$ is the graph of the function
$$\overline{h}(u)=-\int_{0}^u\sqrt{\dfrac{1}{1+\tau^2e^{\tau^2+C}}}d\tau,\ \ u\in \Bbb R.$$
It is clear that $\gamma_s$ and   $\overline{\gamma_s}$ generate the same surface of revolution $\Sigma_S.$
\end{itemize}

\section{Timelike $f$-minimal surfaces of revolution in $\Bbb G^2~\times~\Bbb R_1$}

\subsection {Timelike $f$-Catenoids in $\Bbb G^2\times \Bbb R_1$}\label{4.1}

In the $xz$-plane consider the curve $\gamma_T$ that is the graph of the function
$$h(u)=\int_{u_0}^u {\sqrt{\dfrac{e^{\tau^2}}{e^{\tau^2}-C\tau^2}}d\tau },$$
where $C$ is a positive constant and $u_0$ belongs to the domain $D$ of the function.
The domain $D$ is determined by the following lemma.
\begin{lem}\label{domain}
Consider the function $h:\Bbb R\longrightarrow\Bbb R$ defined by $h(u)=e^{u^2}-Cu^2,$ where $ C>0. $ Then
\begin{enumerate}
\item If $0<C<e,$ then $h(u)>0,\ \forall u\in \Bbb R.$
\item If $C=e,$ then $h(u)>0,\ \forall u\ne-1,1.$
\item If $C>e,$ then there exist $0<u_1<1< u_2,$ such that $h(u)>0,\ \forall u\in (-\infty, -u_2)\cup(-u_1,u_1)\cup(u_2,+\infty).$
\end{enumerate}
\end{lem}
\begin{proof}
Because the function $h$ is even, we just consider the case $u\ge 0.$ Taking the derivative of the function, we obtain
$$ h'(u)=2u\left(e^{u^2}-C\right).$$

\begin{enumerate}
 \item If  $ C\le 1,$ then $ h'(u)>0, \forall u>0.$ The function $h$ is  monotonically increasing  and therefore $ h(u)>0, \forall u\ge 0.$ Note that  $h(0)=1.$
\item If $C>1,$  the function has the only minimum point at $u=\sqrt{\ln C}$ and $h(\sqrt{\ln C})=C-C\ln C.$ We consider the following subcases.
\begin{itemize}
\item The case $1<C<e.$ Because $h(\sqrt{\ln C})=C-\ln C>0,$ \ $ h(u)>0, \forall u\ge 0.$
\item  The case $C=e.$ 	We can see that  $ h(u)>0, \forall u\ne 1$ and $h(1)=0.$
\item The case $C>e.$ Because $h(\sqrt{\ln C})=C-C\ln C<0,$ there exist two values  $0<u_1<1 <u_2$ such that \ $h(u_1)=h(u_2)=0,$ \ $h(u)>0, \ \forall u \notin [u_1, u_2]$ and $h(u)\le0, \ \forall u \in [u_1, u_2].$
\end{itemize}
\end{enumerate}

\begin{center}
	\includegraphics{BienThienLorent-1}
\end{center}
\end{proof}
By Lemma \ref{domain}, the domain $D$   and $u_0$ are chosen as follows.
\begin{enumerate}
\item If $0<C<e,$ then $D=\Bbb R$ and $u_0=0.$
\item If $C=e,$ then $D=(-\infty, -1), \ u_0=-1$ or $D=(-1,1),\ u_0=0$ or or $D=(1, +\infty), u_0=1.$
\item If $C>e,$ then $D=(-\infty, -u_2), \ u_0=-u_2$ or $D=(-u_1,u_1), \ u_0=0;$ or $D=(u_2, +\infty), u_0=u_2.$
\end {enumerate}
 Rotate the curve about the $z$-axis, we obtain a surface of revolution, denoted by  $\Sigma_T,$ that can be parametrized as follows.
$$	X(u,v)=\left(u\cos v, u\sin v, \int_{u_0}^u {\sqrt{\dfrac{e^{\tau^2}}{e^{\tau^2}-C\tau^2}}d\tau }\right).$$
 By a direct computation, it follows that the curve is timelike, $\Sigma_T$ is timelike. Moreover
 $\Sigma_T$ is timelike $f$-minimal. We call $\Sigma_T$ a timelike $f$-Catenoid.

\begin{center}
	\includegraphics[width=7cm]{DuongsinhLorentz-1}\\
\vskip.3cm
	\textit{Figure 3. Generatrices corresponding to $ C=2, 1.5, 1, 0.5$ and the line $x=z,$ respectively}
\end{center}

\begin{center}
		\includegraphics[width=4cm]{C2P5-1}\hspace*{2cm}
		 \includegraphics[width=4cm]{C2P5-2}\\
\vskip .1cm
	\textit{Figure 4. The generatrix and the corresponding timelike $f$-minimal surface ($C<e$)}

	\end{center}

\begin{center}
	\includegraphics[width=4cm]{CBangE_e-1}\hspace*{1cm}	\includegraphics[width=3cm]{CBangE_e-2}\\
	\vskip .5cm
	\textit{Figure 5. The generatrix and the corresponding timelike $f$-minimal surface ($C=e,\ D=(-1,1)$)}
		\end{center}

\vskip 1cm
\begin{center}
	\includegraphics[width=6cm]{CBangE_e-3}\hskip1cm
	\includegraphics[width=6cm]{CBangE_e-4
	\vskip .5cm
	\textit{Figure 6. The generatrix and the corresponding timelike $f$-minimal surface ($C=e,\ D=(-\infty,-1)\cup(1, \infty)$)}
\end{center}

\begin{center}
	\includegraphics[width=6cm]{C3P1-1}\hspace*{0.5cm}
	\includegraphics[width=4cm]{C3P1-2}\\
		\vskip .5cm
	\textit{Figure 7. The generatrix and the corresponding timelike $f$-minimal surface ($C=3.1,\ D=(-0.75,0.75)$)}
\end{center}
	\vskip .3cm
\begin{center}
	\includegraphics[width=5cm]{C3P1-8}\hspace*{0.5cm}
	\includegraphics[width=5cm]{C3P1-5}\\
		\vskip .5cm
	\textit{Figure 8. The generatrix and the corresponding timelike $f$-minimal surface ($C=3.1,\ D=(1.267,3)$)}
\end{center}

\begin{rem}
\begin{enumerate}
\item If $C=e,$ the integral $\displaystyle\int_{0}^{1}\sqrt{\dfrac{e^{\tau^2}}{e^{\tau^2}-C\tau^2}}d\tau $ is divergence.
\begin{center}
	\includegraphics[width=10cm]{H1}
	\vskip .3cm
	\textit{The divergence of  the integral is  showed by WolframAlpha}
\end{center}
\item If $ C>e= 2.718281828...,$ let $ u_1<1<u_2$ are solutions of the equation $ e^{u^2}-Cu^2=0,$ $\displaystyle I_1=\int_{0}^{u_1}\sqrt{\dfrac{e^{\tau^2}}{e^{\tau^2}-Cu^2}}d\tau $
and $\displaystyle I_2=\int_{u_2}^{4}\sqrt{\dfrac{e^{\tau^2}}{e^{\tau^2}-Cu^2}}d\tau.$ The following table computed by Maple gives us some specific values. We can see that the intergal $I_1$ is convergence. When $C$ goes to $\infty$ both $u_1$ and $I_1$ goes to 0.
 \vskip 1cm
 \begin{tabular}{c|cccccccc}
 	$C$ &2.72&2.725 &2.73 &2.74 & 2.75\\
 	\hline
 	$u_1$&0.9822782644&0.9650767467 & 0.9539872965 &0.9375982937 &0.9248309636\\
 	\hline
 	$u_2$&1.017827051& 1.035334695 &1.046729750  & 1.063728407&1.077103331\\
 	\hline
 	$I_1$& 3.39646&2.913861676 &2.716714471 & 2.497786122 &2.363204279 \\
 	\hline
 	$I_2$&5.63324 &5.150340780 &4.952913186 & 4.733425412 &4.598285949
 \end{tabular}

\begin{tabular}{c|cccccccc}
	$C$ &2.8  & 2.9 & 3.0 & 3.1 & 3.2\\
	\hline
	$u_1$&0.8808758710 & 0.8258522408 & 0.7868044780 &0.7556136794 & 0.7293528965\\
	\hline
	$u_2$&1.124065962 & 1.184957908 & 1.229688803 &1.266389104 & 1.297996253\\
	\hline
	$I_1$& 2.026032087  & 1.740252676 & 1.583315220 & 1.474783161&1.391943189\\
	\hline
	$I_2$&4.258352255 & 3.967178286 & 3.805008572 & 3.691396684 &3.603620332
\end{tabular}

\begin{tabular}{c|cccccccc}
	$C$ &4 & 5 & 6 &7 &8 \\
	\hline
	$u_1$& 0.5978318795&0.5090885010 & 0.4521962510 & 0.4113302857 & 0.3800280951\\
	\hline
	$u_2$& 1.467410087 & 1.594566197 & 1.683195738 & 1.751120026&1.806013755\\
	\hline
	$I_1$&  1.051169565& 0.8629326647 & 0.7526390978 & 0.6770807650&0.6208959900\\
	\hline
	$I_2$& 3.227665154& 3.003422822 &2.863129355 &2.761819500 &2.683055126
\end{tabular}

\begin{tabular}{c|cccccccc}
	$C$ &10 &20 &  30 &40 &50\\
	\hline
	$u_1$& 0.3344137545&  0.2295778377 &0.1857512382 & 0.1601547153 & 0.1428721249\\
	\hline
	$u_2$& 1.891336053 & 2.121262664& 2.239037675 & 2.317248453 &2.375356389\\
	\hline
	$I_1$&  0.5412131026 & 0.3655508690 & 0.2943554575 &0.2532125983&0.2255845814\\
	\hline
	$I_2$&  2.565108172 & 2.266933975 & 2.122055409 &2.027988230 &1.959035117
\end{tabular}

\begin{tabular}{c|cccccccc}
	$C$ &100 &200 &  300 &400 &500\\
	\hline
	$u_1$&  0.1005063540 & 0.07088856878 & 0.05783165509&0.05006269611&0.04476619308\\
	\hline
	$u_2$&  2.544164917 & 2.698888129 &2.784186390&2.842705995 &2.887054100\\
	\hline
	$I_1$& 0.1582764960 & 0.1114918791 & 0.09091788154&0.07868765650 & 0.07035385008\\
	\hline
	$I_2$&   1.762485425 &1.586306978 & 1.490482830 & 1.425201168 &1.375955282
\end{tabular}
\end{enumerate}
\end{rem}

\subsection{Classification of timelike $f$-minimal surfaces of revolution in $\Bbb G^2\times \Bbb R_1$}
Let $\Sigma$ be a timelike surface of revolution with the generatrix $\gamma$ and the rotation axis $l.$
Since $\gamma$ is timelike, it can be expressed as below
$$\gamma(u)=(g(u), 0,u).$$
 As in the case $\Sigma$ is spacelike, a local parametrization of $\Sigma$ as well as $\langle \nabla f, N\rangle$ can be computed as follows.
\begin{lem}\label{tf}
\begin{enumerate}
\item If the rotation axis is spacelike
\begin{align*
	X(u,v)=(g(u)\cosh\theta+u\sinh\theta\cosh v,u\sinh v+a,g(u)\sinh\theta+u\cosh\theta\cosh v).
\end{align*}
\begin{align}\label {nts}
\langle \nabla f, N\rangle&= \frac1{\sqrt{[g'(u)]^2-1}}[(g(u)g'(u)+u)\sinh\theta\cosh\theta\cosh v + g(u)\cosh^2\theta+ug'(u)\sinh^2\theta\cosh^2 \nonumber\\
&\quad +(g(u)\sinh v+a)(g'(u)\sinh v)].
\end{align}
\item If the rotation axis is lightlike
\begin{align*
X(u,v)=\left(g(u)+\left[u-g(u)\right]\dfrac{v^2}{2}+a,-v[u-g(u)],u+\left[u-g(u)\right]\dfrac{v^2}{2}\right)\end{align*}
\begin{align}
\label{ntl}
\langle \nabla f, N\rangle&=\dfrac{1}{\sqrt{1-[g'(u)]^2}}\left[ ug'(u)(1+\sin^2v\sinh^2\theta)+[u+g(u)g'(u)]\dfrac{\sinh2\theta}{2}\sin v+g(u)\sinh^2\theta\right]\nonumber\\
&\quad+\dfrac{ag'(u)\cos v}{\sqrt{1-[g'(u)]^2}}.
\end{align}

\item If the rotation axis is timelike
\begin{align*
X(u,v)=(g(u)\cosh\theta\sin v + u\sinh\theta, g(u)\cos v+a, g(u)\sinh\theta\sin v+u\cosh\theta).
\end{align*}
\end{enumerate}
\begin{align} \label{ntt}
	\langle \nabla f, N\rangle&= \frac 1{\sqrt{[g'(u)]^2-1}}[(g(u)g'(u)+u)\sinh\theta\cosh\theta\sin v + g(u)\cosh^2\theta\sin^2v\nonumber\\
	&+g(u)\cos^2v+ug'(u)\sinh^2\theta+a\cos v].
\end{align}
\end{lem}

\begin{thm}\label{thm3}
A zero  $f$-mean curvature timelike surface of revolution in $\Bbb G^2\times\Bbb R$ is either a vertical plane containing the $z$-axis, the cylinder $x^2+y^2=1,$ or a timelike $f$-Catenoid.
\end{thm}

\begin{proof}
It is not hard to check that (\ref{ntl}) can not be constant, (\ref{nts}) is constant if and only if $\theta=0$ and $g'=0$ and (\ref{ntt}) is constant if and only if $\theta=a=0.$ That means if $l$ is lightlike, $\Sigma$ must be a vertical plane. By Lemma (\ref{lem1}) such a plane is of zero $f$-mean curvature if and only if  the plane contains the $z$-axis.
Now consider the case $l$ is timelike.  The condition $\theta=a=0$ means that $l$ is the $z$-axis.
If  $\gamma$ is a vertical line, then  $\Sigma$ is a circular cylinder. By Lemma (\ref{lem1}) a circular cylinder has non-zero $f$-mean curvature and only the cylinder $x^2+y^2=1$ is timelike $f$-minimal. Now consider the case $\gamma$ is not a vertical line. Locally we can parametrize $\Sigma$ as follows.
\begin{align}
X(u,v)=(u\sin v, u\cos v, g(u) ),\ \ 1-g'^2<0.
\end{align}

A direct computation shows that $\Sigma$ is timelike $f$-minimal if and only if $u$ is a solution of the following equation.
\begin{align}\label{tmse}
g''(u)u+g'(u)(1-g'^2(u))+u^2g'(u)(g'^2(u)-1)=0,\end{align}
Equation (\ref{tmse}) is equivalent to
\begin{align*}	g''(u)u=(1-u^2)g'(u)(g'^2(u)-1). \end{align*}
or

\begin{align}\label{tmse1}
	\dfrac{g''(u)}{g'(u)(g'^2(u)-1)}=\dfrac{1}{u}-u.\end{align}
	Intergrating both sides of (\ref{tmse1}), we obtain
\begin{align*}	\ln \dfrac{g'^2(u)-1}{g'^2(u)}=\ln u^2 -u^2 +C,\end{align*}
or
\begin{align*}	\dfrac{g'^2(u)-1}{g'^2(u)}=\dfrac{Cu^2}{e^{u^2}}
\end{align*}
Hence,
\begin{align*}
	g'(u)=\pm \sqrt{\dfrac{e^{u^2}}{e^{u^2}-Cu^2}}, \ \ e^{u^2}-Cu^2>0.
\end{align*}
Therefore,
\begin{align*}
g(u)=\pm\int_{u_0}^u \sqrt{\dfrac{e^{\tau^2}}{e^{\tau^2}-C\tau^2}}d\tau + B,
\end{align*}
where $e^{u^2}-Cu^2>0$  and $B$ is a constant.
Depending on the value of $C,$ the domain $D$ as well as the initial point $u_0$ are chosen as in the Subsection \ref{4.1}. The surface $\Sigma$ is a timelike $f$-catenoid.
\end{proof}

{\bf Acknowledgements.}  This research is funded by Vietnam National Foundation for Science and Technology Development (NAFOSTED) under grant number 101.04.2014.26.

redpajama_set_name: RedPajamaArXiv


text:
home and garden| antiques and collectibles
Paintings In Proust: A Visual Companion To ?in Search Of Lost Time?
byEric Karpeles
Paperback | September 19, 2017
A visually stunning and surprisingly accessible book that brings out subtle facets of Proust's masterpiece, as well as the artworks he cites. —Art in America
A la recherche du temps perdu by Marcel Proust is one of the most profoundly visual works in Western literature. Not only are there frequent references to specific works of art, but certain characters are also evoked by comparison to particular paintings. Bloch's appearance as a boy is likened to the portrait of Mehmet II by Gentile Bellini; Odette de Crécy strikes Swann by her resemblance to a figure in a Botticelli fresco. Even the lesser figure of a certain Mme. Blattin becomes the subject of Proustian mischief by being described as "exactly the portrait of Savonarola by Fra Bartolomeo." Eric Karpeles has identified and located the many paintings to which Proust makes reference and sets them alongside the relevant text from the novel; in other cases, where only a painter's name is mentioned to indicate a certain style or appearance, Karpeles has chosen a representative work to illustrate the impression that Proust sought to evoke.
With some 200 paintings beautifully reproduced in full color and texts drawn from the Moncrieff/Kilmartin/Enright translation, as well as concise commentaries on the evolving narrative, this book is an essential addition to the libraries of Proustians everywhere. The book also includes an authoritative introduction and a comprehensive index of artists and paintings mentioned in the novel.
Eric Karpeles is a painter as well as writer and translator. Author of Almost Nothing: The 20th-Century Art and Life of Józef Czapski, he is a fellow of the Czeslaw Milosz Institute at Claremont McKenna College.
Title:Paintings In Proust: A Visual Companion To ?in Search Of Lost Time?Format:PaperbackProduct dimensions:352 pages, 9.02 × 6.04 × 1.39 inShipping dimensions:9.02 × 6.04 × 1.39 inPublished:September 19, 2017Publisher:NortonLanguage:English
Books Home and Garden Antiques and Collectibles
Customer Reviews of Paintings In Proust: A Visual Companion To ?in Search Of Lost Time?
A visually stunning and surprisingly accessible book that brings out subtle facets of Proust's masterpiece, as well as the artworks he cites…Karpeles comes close to encapsulating Proust's vision…Paintings in Proust could serve effectively as either a concise preface or a meaningful afterword to the monumental novel. What's more, it can be appreciated entirely on its own.—Art in AmericaAn indispensable companion for readers of Proust…. Precisely what's been needed for about 90 years.—Rochester Post-Bulletin

redpajama_set_name: RedPajamaCommonCrawl


i see that you are a very creative person who has her hands in all kinds of crafts! happy pre birthday, friend. my husband did a similar thing to your husband so although my birthday is in may, we are going to a yarn factory in august as my birthday present - cash in hand!!! lol ... can we ever have enough stash? can you ever learn so much about quilting that you wont benefit from a great course?
the picture of you with your mother was simply charming.
there is nothing as draining as having an illness. i am so glad to hear that you enjoy relaxing in front of a game of cricket, needles in hand! i, too, can get lost in knitting and find it a true stress buster! i wish you very well in your fight!


text:
Will.i.am to Advise Branchless Banking App Atom
4/24/2017 by Billboard Staff
Jason LaVeris/FilmMagic
Will.i.am attends a benefit for onePULSE Foundation at NeueHouse Hollywood on Aug. 19, 2016 in Los Angeles.
Atom, the app-based bank, has tapped Black Eyed Peas artist will.i.am as its first strategic board advisor, providing the UK company with an outsider's perspective on technology and trends. The branchless bank launched a year ago as an iOS and Android app offering traditional banking services like savings accounts and mortgages, with more on the way.
"Our lives are faster than ever before, but the banking industry hasn't kept up," said the rapper-producer in a statement. "The scale of Atom's ambition to help people understand and manage their money better, and it's clever use of technology to give people an entire bank on their phones is awesome."
Will.i.am's other ventures outside music include being named "director of creative innovation" at Intel Corp. in 2011. Last year he partnered with Deutsche Telekom to launch a voice-activated smartwatch line.
The partnership between will.i.am and Atom was facilitated by ITB, a marketing agency that delivers such pairings for various brands.
Commented Anthony Thomson, Atom's founder and chairman: "Like us, will.i.am believes that technology can better and simplify our lives, which is at the heart of all that we are passionate about at Atom. As we develop the range of products and services on offer to customers, will.i.am's unique view on the world and knowledge of future trends will be a great asset for us."


This event is registered to The Savage Woods, LLC and the User ID for this account is 9702661333. The email address that has control of this account is Th∗∗∗ @ gmail.com. You must login with User ID 9702661333 to update this event.


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text:
namespace blink {

class CORE_EXPORT MediaFeatureOverrides {
 public:
  void SetOverride(const AtomicString& feature, const String& value_string);
  MediaQueryExpValue GetOverride(const AtomicString& feature) const {
    auto it = overrides_.find(feature);
    return it != overrides_.end() ? it->value : MediaQueryExpValue();
  }

 private:
  HashMap<AtomicString, MediaQueryExpValue> overrides_;
};

}  // namespace blink

#endif  // THIRD_PARTY_BLINK_RENDERER_CORE_CSS_MEDIA_FEATURE_OVERRIDES_H_

redpajama_set_name: RedPajamaGithub


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text:
Joining the System
We've realized gradually over the course of the past 1.5 years or so that we have another alter forming — this makes 9 of us in total, at least that we're aware of — so I thought we'd talk a bit about what it's like when a new alter forms!
For most of us (me, Steve, Sawyer, Fox, Bucky) we were here as parts or fragments before becoming fully-fledged alters. I was part of Maddie until the situation with her abusive girlfriend at the time was too much for her to handle alone, and at that point I broke off into an alter. Likewise, Fox was a part of me until a trigger of his pushed him to splinter off. Mal, Maddie, and Percy are the exceptions — as far as we can tell, they skipped the fragment stage and showed up as alters right away. I can't speak for other systems' experiences with alter formation, but this has consistently been ours.
When an alter begins to form in our system they don't have a name or voice yet, and we can't see them the way we can clearly see each other. (Side note: this is also what it was like when we were first diagnosed and hadn't yet nailed down intrasystem communication.) As time goes on and the new kid starts settling in, we generally get a name and gender, and it's not too long after that until we can see and hear them.
At this early stage, when I've been fronting and then the developing alter (I'll call her Nine for now) pops in, I can feel what she's feeling. I can tell those feelings aren't coming from me, but that's as much as I can gather. Nine has come out just a handful of times over the past year, and almost always for just a minute or two.
She made an appearance yesterday while Steve and I were in class, and she was immediately on the verge of a panic attack. I was aware of the physiological symptoms — heart racing, shortness of breath, tears — but didn't feel any anxiety myself. This is how it's felt every time she's shown up. Yesterday was the first time I was able to get a sense of anything about her, and even at that all I could glean was a sense of pronoun. We don't have an age, name, face, voice–anything. It was the longest she's ever been around, too; she was co-con with us for about 10 minutes. I can't feel her presence right now, but I generally can't feel Maddie's either. I know they're here, they're just much, much quieter than everyone else.
We'll include more updates about Nine as we learn more about her; stay tuned, and feel free to reach out with any questions!
Hollywood and DID


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Look Good on a Personal Profile Analysis
Use a point system for interviews
The disadvantages of competency-based interviews
Convert Izod to Charpy
The Advantages of MANOVA Over ANOVA
HOMEPAGE MONEY
The Pro & Cons of Aptitude Tests in Interviews
Written by: Priti Ramjee
John Rowley/Digital Vision/Getty Images
Many employers rely on aptitude tests with interviews in their hiring process to select the best candidate. Aptitude tests assess IQ, verbal reasoning, logic, mathematical skill and an applicant's personality type.
The outcome of an aptitude test demonstrates how an applicant performs in different tasks and reacts to different situations. It identifies strengths and weaknesses and the appropriateness for the position of interest. While an aptitude test with an interview is designed to select the right candidate, it has benefits and concerns.
Advantages to the Interviewer
With an aptitude test and the interview itself, the interviewer does not need to use any other method to screen applicants, saving valuable time. The tests are difficult to cheat on and, therefore, interviewers consider them reliable. If there is more than one open position for the same job, the test can be cost-effectively reused. An aptitude test in an interview accurately helps an interviewer determine the right applicant for the job.
With an aptitude test and the interview itself, the interviewer does not need to use any other method to screen applicants, saving valuable time.
Disadvantages to the Company
There are disadvantages to aptitude tests as part of the interview process for the company. If a company is structuring interviews for various positions, it can be costly to provide aptitude tests because the tests are designed for each job. The test is based on a person's experience and, therefore, cannot be used for applicants who do not have work experience. While the tests are generally accurate, the interviewer is aware that candidates can react to the stress of the test, which could affect its reliability.
There are disadvantages to aptitude tests as part of the interview process for the company.
If a company is structuring interviews for various positions, it can be costly to provide aptitude tests because the tests are designed for each job.
Less Pressure for the Applicant
An applicant could feel less pressure with an aptitude test if he is made aware of it in advance to give him a chance to prepare. Tests are available online for the applicant to become familiar with the types of questions and methods of answers. The verbal tests allow the applicant to practice reading carefully to find the obvious answer. An aptitude test is not about finishing all of the questions but getting as many right as possible. As a small but important detail, the applicant has the advantage of bringing a calculator that he is familiar with.
An applicant could feel less pressure with an aptitude test if he is made aware of it in advance to give him a chance to prepare.
The verbal tests allow the applicant to practice reading carefully to find the obvious answer.
Technique Can be Developed
With aptitude tests becoming more common as part of an interview, you have the advantage of practicing to increase your skill. Number and word puzzles help to develop comprehension and mathematical skills. You can develop your exam technique of speed and accuracy by calculating the number of questions and the allotted time for the test to gauge how much time to allocate per question.
Advantages & Disadvantages of Traditional Assessment→
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Career Tests Guide: Online Career Aptitude Tests
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Express: Increase Your Speed and Accuracy by taking Job Aptitude Test
University of Kent: Numerical Reasoning Test
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Donkey Kong Country: Tropical Freeze Sells a Mere 36,000 Units in Japan
1 POSTED: 22 Feb 2014 09:05
Donkey Kong Country: Tropical Freeze sold through only 35% of it's first-week shipment Japan.
Donkey Kong Country: Tropical Freeze may have only just released in the US, but Japanese gamers have been enjoying it for over a week now. Media Create, the Japanese company that gathers and analyzes data from the games industry, has now reported that that week has been a somewhat... disappointing one for DK's return, as the game only managed to sell through 35% of its initial shipment, at around 36,000 copies.
To put that into perspective for you, 2010's Donkey Kong Country Returns for the Wii sold 163,000 units in its first week while the 3DS port sold 104,000 units.
That's quite a downward trend, and Media Create has a few theories as to why this is the case. First up, the Wii U's considerably smaller install base is partly to blame, but it also speculates that people aren't being enticed by the incremental upgrades of Tropical Freeze, which is, as our own Jonathan Bolding put it, "the same thing [as Donkey Kong Country Returns] updated to 2014".
However, Media Create would like to caution people that heavy snowfall, in the Tohoku and Kanto regions of Japan, some of the heaviest it has seen in decades, may have also been a factor in the game's poor performance.
Media Create does point out that more Donkey Kong Country: Tropical Freeze sales will probably come over a longer period of time due to its evergreen nature.
source: Nintendo Everything
Foolery
Heh. Snowfall preventing people from picking up a copy of Tropical Freeze. That's funny.
Anyway, the smaller install base makes sense. That, and I don't think Nintendo IPs like Donkey Kong have the influence or appeal they once did.
CriticalMiss
I think that is the first time I have heard of the weather being used as an excuse for poor game sales. Also aren't WiiU games download only or am I misinformed? If they are digital, or at least available digitally, heavy snow doesn't seem like it would have this much of an impact unless it knocked out service for those regions.
CriticalMiss:
The Wii U indeed still uses physical disks in addition to digital downloads.
HOWEVER, my area (Tochigi) as well as the neighboring prefectures of Gunma and Fukushima were hit with quite severe power failures following the storm, so quite a lot of people WERE actually without power.
But, I do believe Media Create was referring to physical copies being unable to be transported to individual shops due to roads being blocked. My local store, for example, did not have the game in stock until yesterday.
It looks like a really fun platformer. I'd buy it for PC.
Super Not Cosmo
Adam Jensen:
It does indeed! Hell, I too would probably give it a go if it made it's way to the PS3.
novem
I attached a two terabyte hard drive to my WiiU and download most of the games I buy for the system. Anytime I see numbers like this I assume a fair number of other people do the same since these reports almost never include digital sales.
Shadow-Phoenix
Cool lack of sales and there being snowfall in Japan apparently doesn't contribute to bad sales, it obviously contributes to Nintendo going PC only!, is what I'm mainly expecting this thread to devolve into.
I was snowed in for a couple of days last week so I guess that's a pretty good reason to not rush out and grab a new DK games...a new DK game with an arctic motif...I think TF would be doing a bit better if DKCR wasn't re-released on the 3DS to be honest...I can just see a scenario in which people who bought both are burned out on the gameplay.
Aiddon
10 POSTED: 22 Feb 2014 11:49
I remember the LAST time people went on about a Nintendo title underperforming with Super Mario 3D World. Turned out they jumped the gun and the game had legs which was something that has become lost in recent years. Let's face it, you can't judge Nintendo games by first week sales. They just don't operate on the same logic other games subscribe to
Trishbot
Aiddon:
True, but I'd say the main reason is still simply because the Wii U has not taken off. Even Nintendo admits that. Several great Nintendo games on a system that just isn't selling results in fewer sales.
A more accurate picture is this: The attach rate of console owners to game buyers.
12 million people owned a Wii when DK Returns hit in Japan. 163,310 bought it at launch. That's an attach rate of only 1.3%.
1.75 million people currently own a Wii U in Japan. Around 36,000 copies sold. That's an attach rate of 2.1%.
That's a HIGHER attach rate than the Wii version had. Nearly double, in fact. A higher percentage of people with Wii U's bought Donkey Kong: Tropical Freeze than the percent of Wii owners that bought DK Returns.
So, you know, perspective.
MrMan999
Is it really wise to start calling Doom and Gloom already? I mean Nintendo games have a history of being steady sellers over the course of a console cycle. And Donkey Kong Country has always been more of a western aimed series anyway.
CelestDaer
heavy snowfall, in the Tohoku and Kanto regions of Japan, some of the heaviest it has seen in decades, may have also been a factor in the game's poor performance
So, irony, then? They're blaming irony on poor sales?
I bought it today since I burned the shit out of my feet and couldn't go to the convention I was going to attend. It's good, but like Super Mario 3D world I don't really feel it's worth $60.
Full Metal Bolshevik
Weaver:
Any game worth 60$ nowadays? Cause everyone says that about almost every game. It seems the only games worth that much are games with a lot of emphasis in online, because it makes the game way longer.
Full Metal Bolshevik:
That's a fair point. Your game will be longer than an average FPS campaign.
I think the problem with is, I usually buy games < $10 after they've been out for awhile. Nintendo games never go cheaper and I'm just trying to desperately pretend I didn't waste a bunch of money on a WiiU :P
Clive Howlitzer
If a game doesn't immediately sell 50,000,000 units, it is a huge failure. Clearly.
AzrealMaximillion
MrMan999:
While this is true I'd take into consideration that the most units that anything sells, whether it's a movie, album or video game, is on the first week.
I agree with you to a point. With Mario and Legend of Zelda your statement is true. It hasn't been true however for not only the majority of WiiU exclusives (Zombie U, Wonderful 101), but other 1st party titles. DK hasn't been consistently popular for a while now. Donkey Kong Returns sold really well don't get me wrong, but Tropical freeze having a third of the first week sales as its predecessor is something that can't be ignored.
EHKOS
Well, with the Wii U sales being what they are, it's hard not to point a finger in that direction. I would love to play the game, but I haven't bought a Wii U yet, and I might not ever. But sticking it to a single console that has it's own sales troubles might be a good reason why this is happening. That and DKCR got some mixed reviews, at least in my circles. I liked it, but I can understand why fans of the SNES series wouldn't, and there were core bits that I didn't care for. It's a nice and colorful game, and I want to see it and Nintendo recover.
Clive Howlitzer:
There's a difference between failing to achieve CoD's megasales figures and failing to meet an indie game sales figures.
But as Trishbot stated, the game had a much higher attach rate compared to it's previous installment.
So its less the game failing and more the platform itself failing to find any traction.
Arkayn
It is a reason (the game is) for more people to finally get a Wii U.
People want the games to buy a system so here is another great software titles to justify the Wii U. It needing to be on another platform is not a justification any more than saying "I don't want a PS3 as I have a PC ALREADY so Sony should put Gran Turismo on PC to suit me".
I really agree with the Call Of Duty comment above. Not everything has to sell CoD or GTA numbers to be a success. GTA V sold the most of any title in 2013 but they spent 100million making it and another 100million marketing it (that is the actual figure btw). DK did not get that spent on it by far.
Though 36,000 units, jeepers. People need to start buying the freaking Wii U rather than sitting on the fence. Heck I bought a PS4 with far less justification.
Hairless Mammoth
Time for my bigoted Metroid fan comment of the day: "Well, it's time Retro made a sequel to the Metroid Prime trilogy." But, in all seriousness, the Wii U doesn't have the install base of the Wii, and the snowstorms in Japan would affect sales and shipping of all goods. Even North American cities use to snow are getting hit hard this year.
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No, just No.


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Pasta e Ceci ( Pasta and Chickpeas) is my favourite Italian comfort food. Economical to make and extremely nourishing, it is a little wet like a soup, but substantial like a meal. Akin to Pasta e Fagioli ( Pasta and beans), I begin to make this dish for lunch once Autumn turns the next dark corner.
There are many versions around, but they are all noted for their simplicity. Don't feel the need to doctor it or add things. Classic Italian dishes taste so good because of restraint.
3 large tomatoes, peeled and chopped, or the equivalent from a can, plus a little juice.
Soak chick peas overnight and cook the following day until done, or use the quick soak method. Drain when cooked.
Add the olive oil to a heavy based soup pot. Fry the soffrito ( the rosemary, garlic and chilli) gently for a minute, then add the tomatoes. Cook for a little, add a splash of water. Meanwhile cook the pasta in salted water, drain, and retain some of the cooking water.
Below- a few soffrito ingredients) Add the cooked chickpeas to the hot tomato mixture, add salt and pepper to taste and 1/2 cup of pasta water. Heat, then add the cooked pasta. Serve in large, warmed shallow pasta bowls with a drizzle of your best oil and some grated parmigiana cheese. Red wine and bread ? Yes please.


Among its many claims to fame, Philadelphia's prominent role in the "Rocky" series has led many film lovers to associate the city with the world of boxing. The city's reputation as a haven for the sport has reached new heights that—even if you don't fancy yourself a budding Rocky Balboa—you can get in on for fun and fitness!
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A Mentor establishes a relationship with another person that makes available and holds accountable for certain specific information, skills, networks and their application throughout his or her spiritual journey. The end result of this relationship is an apprenticeship that points the Mentee to Christ and spiritual practices that promote Christ-likeness. Training required. Potential Spiritual Gifts: Encouragement, Mercy. Potential Abilities: Social, Investigative. Potential Demonstrated Passions: Repairing, Serving, Defending. Potential Felt Passions: Mentoring, Adults of various life stages, Lonely Individuals, New Christians.
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All Upcoming Adria Arjona Movies and TV Shows
Adria Arjona has this weird custom of uttering 'F*** it, man' before they say action while shooting. According to her belief, the ritualistic cursing gives her the power to perform. Even though finished her high school in Miami and acting course at Lee Strasberg Theater & Film Institute, New York, Adria Arjona has Latic America in and around her persona. Being the daughter of renowned Latin American musician, Ricardo Arjona, she is half Puerto Rican and half Guatemalan As the legendary Guatemalan singer and songwriter, Arjona has sold over 20 million records and is the winner of international awards.
Adria has an unusual childhood on the road with her father. Travelling, seeing other cultures and knowing other people close has become her life since then. In that day, she used to have breakfast in Guatemala and dinner Mexico. In almost all her interviews Adria remembers those day and the fathers who gave her wings to fly. After the Lee Strasberg Theater & Film Institute days, she appeared in a CBS drama 'Person of Interest.' Two years later, she picking up Spanish movies and small roles in the movie series 'Unforgettable.' With the series 'True detective' catapulted her into new heights. Her character alongside Colin Farrell and Taylor Kitsch, as a naive woman trapped in a cobweb of lies, became a huge hit.
Adria never looked back from there. She played popular characters in 'Narcos,' 'Emerald City,' 'Pacific Rim: Uprising,' and 'Life of the Party.' She is also famous for he open commented on typecasting of Hispanic actors in Hollywood. So, you must be wondering what is the next Adria Arjona movie/tv show? Well, the good news is that we have a clear answer to that question. Here's the list of upcoming films/shows of Adria Arjona scheduled to release in 2019 and 2020. We can only hope that the latest Adria Arjona movies/shows would be as good as some of her recent ones.
1. Triple Frontier (2019)
'Triple Frontier' features five former American special agents who come out of retirement to pull off a heist. The movie features Ben Affleck, Charlie Hunnam, Oscar Isaac, Pedro Pascal, and Garrett Hedlund in the lead roles. Adria Arjona plays Yovanna, a humble woman who tries to find a better place and life for her and her brother. She speaks her native tongue, Spanish, in the movie, with Yovanna is constructed from her Hispanic roots. Set in the backdrop of the dangerous border zone area between Paraguay, Argentina and Brazil, 'Triple Frontier' is directed by J.C. Chandor. The movie hits the U.S. theaters on 13 March 2019 after a long wait.
2. Good Omens (TV Mini-Series, 2019)
Based on the acclaimed novel by Terry Pratchett and Neil Gaiman, 'Good Omens' is a six-part TV Mini-series for Amazon and BBC Two. The series blends humor, horror, fantasy, and drama to narrate the plot unfolds in apocalyptic future. Neil Gaiman of 'American Gods' wrote the script for the series and plays the role of the Showrunner.
As the end approaches, humanity is on the verge of extinction and an angel named Aziraphale, played by Michael Sheen, and a demon named Crowley, played by David Tennant worry about the mortals. They love earth with all her virtues and vices as both of them have lived among human beings since the starting of time. Adria plays the practical occultist named 'Anathema Device,' who is the descendant of the ancient witch Agnes Nutter in the series. Good Omens will have its world premiere in over 200 countries in 2019.
3. Underground (2019)
Michael Bay's upcoming action thriller '6 Underground' has Ryan Reynolds and Adria Arjona in the lead roles. The movie is based on an idea Reynolds' 'Deadpool' buddies, Rhett Reese and Paul Wernick. Even though official details of the plot are under wrap, various reports suggest that the movie follows Reynold's character, who is the leader of an elite squad that wages war on all the evil forces on earth, with the core team, comprises of four men and two women. Other cast members include Mélanie Laurent, Corey Hawkins, Manuel Garcia-Rulfo, Ben Hardy, Lior Raz, Payman Maadi, and Dave Franco. '6 Underground' will be premiered on Netflix soon.
4. Morbius (2020)
'Morbius' features the tragic story of Dr. Michael Morbius, played by Jared Leto, a scientist who turns himself into a vampire in an attempt to cure a rare blood disease he is suffering from. Arjona plays Martine Bancroft, Dr. Michael Morbius' fiancée. Even though Martine desperately tries to find a cure for Morbius' trauma, she fails to stop him transforming into a living vampire. Daniel Espinosa directs the movie from a script by Burk Sharpless and Matt Sazama. 'Morbius' will hit the theaters on July 31, 2020.
Read More: Upcoming Charlie Hunnman Movies
5. Wasp Network (TBA)
Olivier Assayas's ambitious and secretive project, 'Wasp Network' is based on the book of Brazilian journalist Fernando Morais titled 'The Last Soldiers of the Cold War. The intense drama tells the story of 5 Cuban citizens incarcerated in Miami during the early 1990s. Wasp network is the nickname given to Cuban spies who infiltrated anti-Castro groups operating from the U.S. coast of Florida. Arjona joins the cast list alongside Penélope Cruz, Pedro Pascal, Gael Garcia Bernal, Edgar Ramirez, and Wagner Moura.
Read More: Upcoming Garrett Hedlund Movies


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No Sunday service. Morning timetables are found at the top of this page, afternoon timetables at the bottom.
All local routes (routes 3-6) use a "Hail and Ride" system. That means you don't have to wait at a formal bus stop. You can wave down the bus anywhere along the route where it is safe and legal for the bus to stop. Please make yourself visible to the bus driver.
Route 5 note: the Kawai St South is only used at off-peak times.
These times are approximate. Please be at the stop five minutes before the stated time.
Routes & Timetables Route 1: Nelson > Bishopdale > Stoke > Richmond - Weekdays Route 1: Nelson > Bishopdale > Stoke > Richmond - Weekends Route 1: Richmond > Stoke > Bishopdale > Nelson - Weekdays Route 1: Richmond > Stoke > Bishopdale > Nelson - Weekends Route 2: Nelson > Tahunanui > Stoke > Richmond - Weekdays Route 2: Nelson > Tahunanui > Stoke > Richmond - Weekends Route 2: Richmond > Stoke > Tahunanui > Nelson - Weekdays Route 2: Richmond > Stoke > Tahunanui > Nelson - Weekends Route 3: Nelson > The Wood > Atawhai >The Wood > Nelson Weekdays Route 3: Nelson > The Wood > Atawhai > The Wood > Nelson - Saturday Route 4: Nelson > NMIT > The Brook > NMIT > Nelson - Weekdays Route 4: Nelson > NMIT > The Brook > NMIT > Nelson - Saturday Route 5: Nelson > Victory > Hospital > Victory > Nelson - Weekdays Route 5: Nelson > Victory > Hospital > Victory > Nelson - Saturday Route 6: Nelson > Washington Valley > Tahunanui > Nelson -Weekdays Route 7: Stoke Loop Route 8: Richmond Loop East > Champion Road - Weekdays Route 8: Richmond Loop East > Champion Road - Saturday Route 8: Richmond Loop West > Three Brothers Corner - Weekdays Route 8: Richmond Loop West > Three Brothers Corner - Saturday
Nelson > Victory > Hospital > Victory > Nelson
Nelson Terminus - Wakatu Square 7:15:00 am 7:45:00 am 8:15:00 am 9:00:00 am 10:10:00 am 11:00:00 am
St Vincent St - Countdown 7:17:00 am 7:47:00 am 8:17:00 am 9:02:00 am 10:12:00 am 11:02:00 am
St Vincent St - Victory Sq 7:20:00 am 7:50:00 am 8:20:00 am 9:05:00 am 10:15:00 am 11:05:00 am
Abraham Heights 7:25:00 am 7:55:00 am 8:25:00 am 9:10:00 am 10:20:00 am 11:10:00 am
5 Emano St 7:28:00 am 7:58:00 am 8:28:00 am 9:13:00 am 10:23:00 am 11:13:00 am
Totara St 7:30:00 am 8:00:00 am 8:30:00 am 9:15:00 am 10:25:00 am 11:15:00 am
207 Kawai St South 9:20:00 am 11:20:00 am
Nelson Hospital 7:34:00 am 8:04:00 am 8:34:00 am 9:25:00 am 10:35:00 am 11:25:00 am
Vanguard St - Totara St 12:00:00 pm 12:00:00 pm 12:00:00 pm 9:26:00 am 10:36:00 am 11:26:00 am
Vanguard St - Victory Sq 7:36:00 am 8:06:00 am 8:36:00 am 9:27:00 am 10:37:00 am 11:27:00 am
Nelson 7:42:00 am 8:12:00 am 8:42:00 am 9:33:00 am 10:37:00 am 11:33:00 am
Weekdays p.m
Nelson Terminus - Wakatu Square 12:05:00 pm 1:00:00 pm 2:00:00 pm 3:00:00 pm 4:00:00 pm 4:30:00 pm 5:00:00 pm 5:30:00 pm 6:00:00 pm
St Vincent St - Countdown 12:07:00 pm 1:02:00 pm 2:02:00 pm 3:02:00 pm 4:02:00 pm 4:32:00 pm 5:02:00 pm 5:32:00 pm 6:02:00 pm
St Vincent St - Victory Sq 12:10:00 pm 1:05:00 pm 2:05:00 pm 3:05:00 pm 4:05:00 pm 4:35:00 pm 5:05:00 pm 5:35:00 pm 6:05:00 pm
Abraham Heights 12:15:00 pm 1:10:00 pm 2:10:00 pm 3:10:00 pm 4:10:00 pm 4:40:00 pm 5:10:00 pm 5:40:00 pm 6:10:00 pm
5 Emano St 12:18:00 pm 1:13:00 pm 2:13:00 pm 3:13:00 pm 4:13:00 pm 4:43:00 pm 5:13:00 pm 5:43:00 pm 6:13:00 pm
Totara St 12:20:00 pm 1:15:00 pm 2:15:00 pm 3:15:00 pm 4:15:00 pm 4:45:00 pm 5:15:00 pm 5:45:00 pm 6:15:00 pm
207 Kawai St South 1:20:00 pm
Nelson Hospital 12:24:00 pm 1:25:00 pm 2:19:00 pm 3:19:00 pm 4:19:00 pm 4:49:00 pm 5:19:00 pm 5:49:00 pm 6:19:00 pm
Vanguard St - Totara St 12:26:00 pm 1:26:00 pm 2:20:00 pm 3:20:00 pm 4:20:00 pm 4:50:00 pm 5:20:00 pm 5:50:00 pm 6:20:00 pm
Vanguard St - Victory Sq 12:29:00 pm 1:27:00 pm 2:21:00 pm 3:21:00 pm 4:21:00 pm 4:51:00 pm 5:21:00 pm 5:51:00 pm 6:21:00 pm
Nelson 12:32:00 pm 1:33:00 pm 2:27:00 pm 3:27:00 pm 4:27:00 pm 4:57:00 pm 5:27:00 pm 5:57:00 pm 6:27:00 pm
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Court shoots down Yakama tobacco suit
Company must pay into settlement escrow fund
Published: April 8, 2013, 5:00pm
YAKIMA — A tobacco company owned by a Yakama Nation tribal member must pay into an escrow account established under a 1998 settlement with big tobacco companies, a federal judge has ruled.
The 1998 settlement requires big tobacco companies to pay money to 46 states each year to offset public health costs from their products. Smaller companies are required to pay into an escrow account each year, but that money can be returned to each company after 25 years if no health claims are made, under the settlement.
King Mountain Tobacco claimed it should be exempt from paying into the account under the Yakama Nation's 1855 treaty with the federal government. U.S. District Court Judge Lonny Suko ruled against the company on Friday, granting Washington state's motion to dismiss the case.
King Mountain, owned by Yakama tribal member Delbert Wheeler, grows tobacco and manufactures cigarettes and roll-your-own tobacco on the tribe's reservation south of Yakima in south-central Washington. That tobacco is shipped to Tennessee, where it is threshed, and to North Carolina to be blended with tobacco there.
In 2009, about 3 percent of the tobacco used in King Mountain's cigarettes was grown on the reservation, rising to nearly 38 percent in 2011.
"Based on the finding above that the finished cigarettes and roll-your-own tobacco are not directly derived from trust land, King Mountain can prove no set of facts in support of the claim that Washington's escrow statutes are in conflict with the Treaty or federal law," Suko wrote.
In a statement Monday, Wheeler said the ruling was a disappointment, but not a surprise.
"Our Treaty guarantees us the exclusive use and benefit of our land. It also guarantees us the right to travel and trade off our reservation, without government conditions and restrictions that apply to tribes that don't have our Treaty," he said.
King Mountain has been registered as a tobacco manufacturer and has paid into the escrow each year since 2007 — more than $5 per carton sold to nontribal members, both on and off the reservation.
"We're very pleased," said David Hankins, senior counsel for the Washington state Attorney General's Office.
The strange case of the casino, the Senate leader and the defense bill
1972 ushered in post-termination era of change for U.S. tribes
Northwest January 9, 2022
Politics November 18, 2021


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Bulbophyllum lasioglossum är en orkidéart som beskrevs av Robert Allen Rolfe och Oakes Ames. Bulbophyllum lasioglossum ingår i släktet Bulbophyllum och familjen orkidéer. Inga underarter finns listade i Catalogue of Life.

Källor

Orkidéer
lasioglossum

redpajama_set_name: RedPajamaWikipedia


Having never owned a car, the bicycle plays an important role in Anik See's life. In her book, A Fork in the Road – Tales of Food and Travel, she writes about her cycling and travel adventures through Asia, the Middle East and Latin America. Long distance cycle touring has also taken her to Sri Lanka, India, and Scotland. She currently resides in bike-friendly Amsterdam and you can learn more at her website aniksee.com. Anik See welcome to The Pursuit Zone.


Completed in 1998, the Anacostia Community Services Center is a 740 square meter, single-story building located on a one-half acre site in the center of the Anacostia section of Washington, D.C. The facility contains offices, classrooms, conference rooms and a 120 square meter auditorium oriented to serve the community's needs. The Center provides a multitude of services to individuals and families, including supplies, housing, emergency services, youth development, basic computer skills, health and nutrition education, and a meeting facility available to community and neighborhood organizations. The site provides 23 parking spaces for employees and visitors and a loading area for receiving and distributing food and clothing to the members of the community.
The project work included the razing and preparation of the construction site and tying into city utilities, including the installation of a sand filtration system, new storm water drain system, and electrical, water, and gas connections. Also installed in the new building were an HVAC system, office facilitation, kitchen, multipurpose meeting room, parking, and building security that included both a perimeter fence and brick walls.


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Editorial Special issue: Links in the air transport chain: Where and when does optimisation begin?
Eddy van de Voorde University of Antwerp
https://doi.org/10.18757/ejtir.2008.8.2.3334
The air transport sector has recently enjoyed a period of strong economic growth and boom, marked by the entry of new carriers, expanding business, and considerable profit-making. Clearly, though, this has not always been the case. What is more, in the first half of 2008, we have witnessed a new wave of mergers and takeovers, as well as bankruptcies and market exits, which would seem to suggest that the boom is over, for a while at least. As in the maritime sector, thought and action in the airline business have evolved rapidly in recent years, with respect to both passenger and freight transport. Increasingly, players are approaching the industry from the perspective of air transport chains. Prospective customers are no longer selecting airports and airlines on the basis of their individual merits, but because they belong to an air transport chain that meets their preferences maximally and corresponds to their willingness to pay. Hence, the success of those airports and airlines depends crucially on whether or not they belong to a successful air logistics chain. This process coincides with a degree of specialisation, air freight being a typical example. Whereas air freight used to be regarded as a 'side product' of passenger transport, there are now a number of carriers focusing exclusively on freight. Likewise, there are now airports (albeit smaller ones) who consider freight transport as their core business. This growing significance of full-freighter services has been occasioned by a combination of factors, including insufficient freight capacity and stricter safety regulations on passenger planes, a trend towards scale enlargement, and substantial imbalances between incoming and outgoing flows. It is important that research should stay apace with this rapid evolution in the airline industry. In the present issue of the European Journal of Transport Infrastructure Research, we pick up on the now widely accepted transport chain perspective. The six contributions focus on different links in the air transport chain, but without losing sight of interdependencies and the possibility of integration.
Voorde, E. van de. (2008). Editorial Special issue: Links in the air transport chain: Where and when does optimisation begin?. European Journal of Transport and Infrastructure Research, 8(2). https://doi.org/10.18757/ejtir.2008.8.2.3334
Hilde Meersman, Eddy van de Voorde, Thierry Vanelslander, The Air Transport Sector after 2010: A Modified Market and Ownership Structure , European Journal of Transport and Infrastructure Research: Vol. 8 No. 2 (2008)
Eddy van de Voorde, Book review Christian von Hirschhausen, Torsten Beckers and Kay Mitusch (eds.) Trends in Infrastructure Regulation and Financing. International Experience and Case Studies from Germany. Cheltenham, Edward Elgar , European Journal of Transport and Infrastructure Research: Vol. 6 No. 2 (2006)


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Cell Phone Radiation, RF 101
New studies claim cell phone radiation causes male infertility, kills sperm cells and even causes erectile dysfunction (ED)
Updated on January 18th, 2017 at 09:09 am
According to RF Safe, cell phone radiation has now been linked to erectile dysfunction (ED), lower sperm count and male infertility in the Environmental Health Trust's latest newsletter. RF Safe provides men several proven ways for reduce excessive cell phone radiation exposure.
As news of the Environmental Health Trust's latest March 28th, 2014 newsletter links cell phone radiation to male reproductive problems. The report reviews the scientific literature to date and shows that ten studies have found significant changes in sperm exposed to cell phone radiation.
These studies show men who carried their phones in a pocket or on the belt were more likely to have lower sperm counts and/or more inactive or less mobile sperm.
One study was conducted at the Medical University of Graz, in Austria; the other at the Queens University in Canada. Men in the studies who reported cell phone use showed lower levels of the luteinizing hormone (LH), an important reproductive hormone that is secreted by the pituitary gland in the brain, and higher levels of circulating testosterone.
Preliminary findings for the research was published in the US National Library of Medicine National Institutes of Health on Apr 26, 2013. http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3921848/
Men talking on the phone for more than an hour a day had 17% fewer highly motile sperm than men who talked less than 15 minutes a day.
Men carrying a phone in a hip pocket or on a belt had 11% fewer mobile sperm than men who kept their phone elsewhere on their body.
Men carrying a cell phone on a belt and using it intensively during a 5-day test period had a 19% drop in highly motile sperm from previous levels.
In the most striking findings, the studies prove that health effects happen at levels of exposure far below the current SAR safety guidelines imposed by the FCC.
Dr. Conrado Avendano, heading a team of researchers in Argentina, reported that they studied semen samples from 29 healthy, fertile men and measured sperm motility after a 4-hour period of exposure to WiFi radiation.
Sperm in the control group was maintained at the same temperature for the same amount of time, without exposure to WiFi.
WiFi Group: 25% of the sperm stopped swimming.
Control Group: only 14% stopped.
WiFi Group: 9% of the sperm showed DNA fragmentation or irreversible damage in the genetic code.
Control Group: only 3% showed these signs.
"There are precautions that can be taken. Most important might be using the cellphone's speaker option and texting more often. Also avoid carrying a phone against the body at all times without microwave shielding between the device and users. These precautions can help", says Coates.
The simplest solution is to start using EMF protection products that provide protection from cell phone radiation. Dont settle for just any products claiming to provide protection. Shop carefully. Do the research, as RF Safe has pointed out many products marketed to reduce radiation can actually increase exposure.
Changing cell phone habits: It is common for men who talk on a cell phone using a Bluetooth device or other headset, that the phone is kept in a pants pocket or clipped to a holster. This exposes reproductive organs to cell phone radiation which lowers sperm count and/or poorer sperm quality in men who use their phones this way than in those who do not.
RF (Radio Frequency) Radiation passes right through most all non conductive materials unimpeded until being either reflected or its electromagnetic energy is absorbed. RF Safe smart phone flip cover cases are designed with special silver plated shielding materials inside Peel n Shield™ flip covers that deflect radiation by lining the entire front cover. By keeping this super thin performance radiation shielding between the device and user, the users body absorbs much less microwave frequency radiation radiating from the cell phone when using simple safety accessories.
To stay up-to-date in the concerns facing radiation from cell phones and the best cell phone radiation protection from the #1 RF Safety company with over 15 yrs of phone RF shielding experience. Share www.rfsafe.com
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2019 has finally arrived, and this year, we're taking a proactive, "new year, all you" approach to our lives. Rather than waiting for things to just happen to us, we're going after what we want — at work, in our relationships, and within ourselves.
Inspired by Paramount's What Men Want, a movie all about taking control of your life and breaking down barriers, we're giving you that extra push you need to approach all of 2019's challenges and successes with the same boss attitude, by giving you a mantra to live by, according to your sign.
Let this mantra serve as your guiding light this year — whether it's something you repeat every morning, write on your mirror, or keep stored in your phone's Notes app for those times you need it most. It'll be your first line of defence when tackling that 2 a.m. "u up" text from your ex or your boss' bad attitude about that project you slaved over. This year, you're becoming the person who's big enough, brave enough, and loving enough to wield the power positive change requires. Now go ahead and get what you want.