text:
\section{Multitime hybrid differential game with \\curvilinear integral functional}

Let $t=(t^\alpha)\in \Omega_{0T}\subset \mathbb{R}^m_+$, $\alpha =1,...,m,$ be an evolution multi-parameter, called multitime. Consider an arbitrary $C^1$ curve $\Gamma_{0T}$ joining the diagonal opposite points $0=(0,\ldots,0)$ and $T=(T^1,\ldots,T^m)$ in the $m$-dimensional parallelepiped $\Omega_{0T}=[0,T]$ (multitime interval) in $\mathbb{R}^m_+$ endowed with the product order, a $C^2$ state vector $x:\Omega_{0T}\rightarrow \mathbb{R}^n, x(t)=(x^i(t)),$ $i=1,...,n$, a $C^1$ control vector $u(t)=(u_\alpha(t)):\Omega_{0T}\rightarrow U\subset \mathbb{R}^{qm},$ for the first equip of $m$ players (who wants to maximize), a $C^1$ control vector  $v(t)=(v_\alpha(t)):\Omega_{0T}\rightarrow V\subset \mathbb{R}^{qm},$ for the second equip of $m$ players (who wants to minimize), $u_\alpha(\cdot)=\Phi(\cdot,\eta_1(\cdot)), v_\alpha(\cdot)=\Psi(\cdot,\eta_2(\cdot)),$ a running cost $L_\alpha(t,x(t),u_\alpha(t),v_\alpha(t))dt^\alpha$ as a nonautonomous closed Lagrangian $1$-form (satisfies $D_\beta L_\alpha=D_\alpha L_\beta),$ a terminal cost $g(x(T))$ and the $C^1$ vector fields $X_\alpha=(X_\alpha^i)$ satisfying the complete integrability conditions (CIC)
 $D_{\beta}X_{\alpha}=D_{\alpha}X_{\beta}$ (m-flow type problem).

In our paper, a {\it multitime hybrid differential game} is given by a multitime dynamics,
as a PDE system controlled by two controllers (first equip, second equip)
and a target including a curvilinear integral functional. The approach we follow below is those in the paper
\cite{[2]}, but we must be more creative since our theory is multitemporal one (see also \cite{[8]}-\cite{[21]}).
More precisely, we introduce and analyze a multitime differential game whose Bolza payoff is the sum between a
path independent curvilinear integral (mechanical work) and a function of the final event (the terminal cost, penalty term),
and whose evolution PDE is an m-flow: {\it Find
$$\min_{v(\cdot)\in V}\max_{u(\cdot)\in U} J(u(\cdot),v(\cdot))=\int_{\Gamma_{0T}} L_\alpha (s,x(s),u_\alpha(s),v_\alpha(s))ds^\alpha+g(x(T)),$$
subject to the Cauchy problem
$$\frac{\partial x^i}{\partial s^\alpha}(s)=X^i_\alpha(s,x(s),u_\alpha(s),v_\alpha(s)),$$
$$x(0)=x_0, \,\,s\in \Omega_{0T}\subset \mathbb{R}_+^m, \,\,x\in \mathbb{R}^n.$$}

Let $D_{\alpha}$ be the total derivative operator and $[X_{\alpha},X_{\beta}]$ be the
bracket of vector fields. Suppose the piecewise complete integrability conditions (CIC)
$$ \left( \frac{\partial X_{\alpha}}{\partial u^a_\lambda}\delta^{\gamma}_{\beta} -  \frac{\partial X_{\beta}}{\partial u^a_\lambda}\delta^{\gamma}_{\alpha}\right)\frac{\partial u^a_\lambda}{\partial s^{\gamma}}+\left( \frac{\partial X_{\alpha}}{\partial v^b_\lambda}\delta^{\gamma}_{\beta} -  \frac{\partial X_{\beta}}{\partial v^b_\lambda}\delta^{\gamma}_{\alpha}\right)\frac{\partial v^b_\lambda}{\partial s^{\gamma}}=\left[ X_{\alpha},X_{\beta}\right] + \frac{\partial X_{\beta}}{\partial s^{\alpha}} - \frac{\partial X_{\alpha}}{\partial s^{\beta}},$$
where $a, b =1,...,q$, are satisfied throughout.

To simplify, suppose that the curve $\Gamma_{0T}$ is an increasing curve in the multitime interval $\Omega_{0T}$.
If we vary the starting multitime and the initial point, then we obtain a
larger family of similar multitime problems containing the functional
$$J_{x,t}(u(\cdot),v(\cdot))=\int_{\Gamma_{tT}} L_\alpha (s,x(s),u_\alpha(s),v_\alpha(s))ds^\alpha+g(x(T)),$$
and the evolution constraint
$$\frac{\partial x^i}{\partial s^\alpha}(s)=X^i_\alpha(s,x(s),u_\alpha(s),v_\alpha(s)),$$
$$x(t)=x, \,\,s\in \Omega_{tT}\subset \mathbb{R}_+^m,\,\, x\in \mathbb{R}^n.$$

We assume that each vector field
$X_\alpha:\Omega_{0T}\times \mathbb{R}^n\times U\times V\rightarrow \mathbb{R}^n$
is uniformly continuous, satisfying
$$\left\{\begin{array}{ll}
\Vert X_\alpha(t,x,u_\alpha,v_\alpha)\Vert\leqslant A_\alpha\\
\Vert X_\alpha(t,x,u_\alpha,v_\alpha) - X_\alpha(t,\hat{x},u_\alpha,v_\alpha)\Vert\leqslant A_\alpha\Vert x-\hat{x}\Vert,
\end{array}\right.$$
for some constant 1-form $A=(A_\alpha)$ and all $t\in \Omega_{0T}, \,\,x, \hat{x}\in \mathbb{R}^n, u\in U,v\in V.$

 Suppose the functions
$$g:\mathbb{R}^n\rightarrow \mathbb{R}, \quad L_\alpha:\Omega_{0T}\times \mathbb{R}^n\times U\times V\rightarrow \mathbb{R}$$
are uniformly continuous and satisfy the boundedness conditions
$$\left\{\begin{array}{ll}
\vert g(x)\vert\leqslant B\\
\vert g(x)-g(\hat{x})\vert\leqslant B\Vert x-\hat{x}\Vert,
\end{array}\right.$$

$$\left\{\begin{array}{ll}
\vert L_\alpha(t,x,u_\alpha,v_\alpha)\vert\leqslant C_\alpha\\
\vert L_\alpha(t,x,u_\alpha,v_\alpha)-L_\alpha(t,\hat{x},u_\alpha,v_\alpha)\vert\leqslant C_\alpha\Vert x-\hat{x}\Vert,
\end{array}\right.$$
for constant $1$-form $C=(C_\alpha)$ and all $t\in \Omega_{0T},\,\, x, \hat{x}\in \mathbb{R}^n,\,\, u\in U,\,\,v\in V.$

\begin{definition}
(i) The set $$\mathcal {U}(t)=\left\lbrace u_\alpha(\cdot):\mathbb{R}^m_+\rightarrow U \vert \ u_\alpha(\cdot) \mathrm{ \ is \ measurable \ and \ satisfies \ CIC}\right\rbrace $$ is called \textbf{the control set for the first equip of players}.
(ii) The set $$\mathcal {V}(t)=\left\lbrace v_\alpha(\cdot):\mathbb{R}^m_+\rightarrow V \vert \ v_\alpha(\cdot) \mathrm{ \ is \ measurable \ and \ satisfies \ CIC}\right\rbrace $$ is called \textbf{the control set for the second equip of players}.
\end{definition}

\begin{definition}
(i) A map $\Phi:\mathcal {V}(t)\rightarrow \mathcal {U}(t)$ is called \textbf{a strategy for the first equip of players}, if the equality $v(\tau)=\widehat{v}(\tau), t\leq \tau \leq s \leq T$ implies $\Phi[v](\tau)=\Phi[\widehat{v}](\tau).$
(ii) A map $\Psi:\mathcal {U}(t)\rightarrow \mathcal {V}(t)$ is called \textbf{a strategy for the second equip of players}, if the equality $u(\tau)=\widehat{u}(\tau), t\leq \tau \leq s \leq T$ implies $\Psi[u](\tau)=\Psi[\widehat{u}](\tau).$
\end{definition}

Let $ \mathcal{A}(t)$ be \textbf{\ the set of strategies for the first equip of players}
 and $\mathcal{B}(t)$ be \textbf{\ the set of strategies for the second equip of players}.

\begin{definition}
 (i) The function
$$m(t,x)=\min_{\Psi\in \mathcal{V}} \max_{u(\cdot)\in U} J_{t,x}( u(\cdot),\Psi[u](\cdot))$$ is called \textbf{the multitime lower value function}.
(ii) The function
$$M(t,x)=\max_{\Phi\in \mathcal{U}} \min_{v(\cdot)\in V} J_{t,x}(\Phi[v](\cdot),v(\cdot)) $$ is called \textbf{the multitime upper value function}.
\end{definition}

The multitime lower value function $m(t,x)$ and the multitime upper value function $M(t,x)$
are piecewise continuously differentiable (see below, the boundedness and continuity of the values functions).

\section{Properties of lower and upper values}

\begin{theorem}\textbf{(multitime dynamic programming optimality conditions)}
For each pair of strategies $(\Phi,\Psi),$ the lower and upper value functions
can be written respectively in the form
\begin{equation}\begin{split}
m(t,x)\ & =\min_{\Psi\in \mathcal{B}(t)} \max_{u_\alpha\in \mathcal{U}(t)}\bigg\{  \int_{\Gamma_{tt+h}} L_\alpha (s,x(s),u_\alpha(s),\Psi[u_\alpha](s))ds^\alpha \\& +m(t+h,x(t+h))\bigg\}
 \end{split}\end{equation}
and
\begin{equation}\begin{split}
M(t,x)\ &  =\max_{\Phi\in \mathcal{A}(t)}\min_{v_\alpha\in \mathcal{V}(t)}\bigg\{  \int_{\Gamma_{tt+h}} L_\alpha (s,x(s),\Phi [v_\alpha](s),v_\alpha(s))ds^\alpha \\ & +M(t+h,x(t+h))\bigg\},
\end{split}\end{equation}
for all $(t,x) \in \Omega_{tT}\times \mathrm{R}^n$ and all $h\in \Omega_{0T-t}.$
\end{theorem}

\begin{proof} First we recognize the Bellman principle
(we write the value of a decision problem at a certain point in multitime
in terms of the payoff from some initial choices and the value of the remaining
decision problem that results from those initial choices).

To confirm the first statement, we shall use the function
\begin{equation}\begin{split}
w(t,x)\ &=\min_{\Psi\in \mathcal{B}(t)} \max_{u_\alpha\in \mathcal{U}(t)}\bigg\{  \int_{\Gamma_{tt+h}} L_\alpha (s,x(s),u_\alpha(s),\Psi[u_\alpha](s))ds^\alpha \\ & +m(t+h,x(t+h))\bigg\}.
\end{split}\end{equation}

We will show that, for all $\varepsilon >0,$ the lower value function $m(t,x)$
will satisfies two inequalities, $m(t,x)\leq w(t,x)+2\varepsilon$ and
$m(t,x)\geq w(t,x)-3\varepsilon.$ Since $\varepsilon>0$ is arbitrary, it follows $m(t,x)=w(t,x).$

\begin{enumerate}[i)]

\item For $\varepsilon >0,$ there exists a strategy $\Upsilon\in \mathcal{B}(t)$ such that

\begin{equation}\begin{split}\label{eq.7}
w(t,x)\ &\geqslant \max_{u_\alpha\in \mathcal{U}(t)}\bigg\{  \int_{\Gamma_{tt+h}} L_\alpha (s,x(s),u_\alpha(s),\Upsilon[u_\alpha](s))ds^\alpha \\ & +m(t+h,x(t+h))\bigg\}-\varepsilon.
\end{split}\end{equation}

We shall use the state $x(\cdot)$ which solves the (PDE), with the initial condition $\overline{x}=x(t+h)$
(Cauchy problem) on the set $\Omega_{tT}\setminus \Omega_{tt+h},$
for each $\overline{x}\in \mathbb{R}^n$. We can write
\begin{equation}\begin{split}
m(t+h,\overline{x}) \ & =\min_{\Psi\in \mathcal{B}(t+h)} \max_{u_\alpha\in \mathcal{U}(t+h)}\bigg\{ \int_{\Gamma_{t+hT}} L_\alpha (s,x(s),u_\alpha(s),\Psi[u_\alpha](s))ds^\alpha  \\ & +g(x(T))\bigg\}.
\end{split}\end{equation}

Thus there exists a strategy $\Upsilon_{\overline{x}}\in \mathcal{B}(t+h)$ for which

\begin{equation}\begin{split}\label{eq.8}
 m(t+h,\overline{x})\ & \geqslant \max_{u_\alpha\in \mathcal{U}(t+h)}\bigg\{  \int_{\Gamma_{t+hT}} L_\alpha (s,x(s),u_\alpha(s),\Upsilon_{\overline{x}}[u_\alpha](s))ds^\alpha \\& + g(x(T))\bigg\}-\varepsilon.
\end{split}\end{equation}

Define a new strategy
$$\Psi\in\mathcal{B}(t), \Psi[u_\alpha](s)\equiv
\left\{\begin{array}{ll}
\Upsilon[u_\alpha](s) & s\in \Omega_{tt+h}\\
\Upsilon_{\overline{x}}[u_\alpha](s) & s\in \Omega_{tT}\setminus\Omega_{tt+h},
\end{array}\right.$$
for each control $u_\alpha\in \mathcal{U}(t).$ For any $u_\alpha\in \mathcal{U}(t)$,
replacing the inequality $\eqref{eq.8}$ in the inequality $\eqref{eq.7}$, we obtain
$$w(t,x)\geqslant   \int_{\Gamma_{tT}} L_\alpha (s,x(s),u_\alpha(s),\Psi[u_\alpha](s))ds^\alpha+g(x(T))-2\varepsilon.$$
Consequently
$$ \max_{u_\alpha\in \mathcal{U}(t)}\left\lbrace  \int_{\Gamma_{tT}} L_\alpha (s,x(s),u_\alpha(s),\Psi[u_\alpha](s))ds^\alpha+g(x(T))\right\rbrace\leq w(t,x)+2\varepsilon.$$

Hence
$$m(t,x)\leq w(t,x)+2\varepsilon.$$

\item On the other hand, there exists a strategy $\Psi\in\mathcal{B}(t)$ for which we can write the inequality

\begin{equation}\label{eq.9}
m(t,x)\geqslant \max_{u_\alpha\in \mathcal{U}(t)}\left\lbrace  \int_{\Gamma_{tT}} L_\alpha (s,x(s),u_\alpha(s),\Psi[u_\alpha](s))ds^\alpha+g(x(T))\right\rbrace-\varepsilon.
\end{equation}

By the definition of $w(t,x),$ we have
\begin{equation}\begin{split}
w(t,x) \ & \leqslant \max_{u_\alpha\in U(t)}\bigg\{  \int_{\Gamma_{tt+h}} L_\alpha (s,x(s),u_\alpha(s),\Psi[u_\alpha](s))ds^\alpha \\& +m(t+h,x(t+h))\bigg\}
\end{split}\end{equation}
and consequently there exists a control $u^1_\alpha\in \mathcal{U}(t)$ such that
\begin{equation}\begin{split}\label{eq.10}
w(t,x) \ & \leqslant   \int_{\Gamma_{tt+h}} L_\alpha (s,x(s),u^1_\alpha(s),\Psi[u^1_\alpha](s))ds^\alpha  \\&  +m(t+h,x(t+h))+\varepsilon.
\end{split}\end{equation}

Define a new control $${u_\alpha^\star}\in \mathcal{U}(t), {u_\alpha^\star}(s)\equiv
\left\{\begin{array}{ll}
u^1_\alpha(s) & s\in \Omega_{tt+h}\\
 u_\alpha(s) & s\in \Omega_{tT}\setminus\Omega_{tt+h},
\end{array}\right.$$
for each control $u_\alpha\in \mathcal{U}(t+h)$ and then define the strategy ${\Psi}^\star\in\mathcal{B}(t+h), \Psi^\star[u_\alpha](s)\equiv\Psi[{u_\alpha^\star}](s), s\in \Omega_{tT}\setminus\Omega_{tt+h}.$
We find the inequality
\begin{equation}\begin{split}
 \ & m(t+h,x(t+h)) \\ & \leq\max_{u_\alpha\in \mathcal{U}(t+h)}\left\lbrace \int_{\Gamma_{tt+h}} L_\alpha(s,x(s),u_\alpha(s),\Psi^\star[u_\alpha](s))ds^\alpha+g(x(T))\right\rbrace
\end{split}\end{equation}
and so there exists the control $u^2_\alpha\in \mathcal{U}(t+h)$ for which
\begin{equation}\begin{split}\label{eq.11}
 \ & m(t+h, x(t+h))\\&  \leq \int_{\Gamma_{tT}\setminus \Gamma_{tt+h}} L_\alpha(s,x(s),u^2_\alpha(s),\Psi^\star[u^2_\alpha](s))ds^\alpha+g(x(T)) +\varepsilon.
\end{split}\end{equation}

Define a new control
$$u_\alpha\in \mathcal{U}(t), u_\alpha(s)\equiv
\left\{\begin{array}{ll}
u^1_\alpha(s) & s\in \Omega_{tt+h}\\
u^2_\alpha(s) & s\in \Omega_{tT}\setminus\Omega_{tt+h}.
\end{array}\right.$$

Then the inequalities $\eqref{eq.10}$ and $\eqref{eq.11}$ yield
$$ w(t,x)\leq \int_{\Gamma_{tT}} L_\alpha(s,x(s),u_\alpha(s),\Psi[u_\alpha](s))ds^\alpha+g(x(T)) +2\varepsilon,$$
and so $\eqref{eq.9}$ implies the inequality
$$ w(t,x)\leq m(t,x)+3\varepsilon.$$

This inequality and  $m(t,x)\leq w(x,t)+2\varepsilon$ complete the proof.
\end{enumerate}
\end{proof}

\begin{theorem}\textbf{(boundedness and continuity of the values functions)}
The lower, upper value function $m(t,x)$, $M(t,x)$ satisfy the boundedness conditions
$$\vert m(t,x)\vert, \vert M(t,x)\vert\leq D$$
$$\vert m(t,x)-m(\hat{t},\hat{x})\vert, \vert M(t,x)-M(\hat{t},\hat{x})\vert\leq E\,\, \ell (\Gamma_{\hat{t}\,t})+ F\,\Vert x-\hat{x}\Vert,$$
for some constants $D, E, F$ and for all $t, \hat{t}\in \Omega_{0T}, x, \hat{x} \in \mathbb{R}^n.$
\end{theorem}

\begin{proof} We prove only the statements for upper value function $M(t,x).$

Since $\vert g(x)\vert\leqslant B, \vert L_\alpha(t,x,u_\alpha,v_\alpha)\vert\leqslant C_\alpha, \alpha=\overline{1,m}$, we find
\begin{equation}\begin{split}
\vert J_{t,x}(u(\cdot),v(\cdot))\vert \ & =\Big\vert \int_{\Gamma_{tT}} L_\alpha (s,x(s),u_\alpha(s),v_\alpha(s))ds^\alpha+g(x(T))\Big\vert \\& \leq \Big\vert \int_{\Gamma_{tT}} L_\alpha (s,x(s),u_\alpha(s),v_\alpha(s))ds^\alpha \Big\vert + \vert g(x(T))\vert \\& \leq \int_{\Gamma_{tT}} \Vert L_\alpha (s,x(s),u_\alpha(s),v_\alpha(s))\Vert \Vert ds^\alpha \Vert +\vert g(x(T))\vert \\& \leq \Vert C \Vert \int_{\Gamma_{tT}}  ds + B= \Vert C \Vert l(\Gamma_{tT}) +B\leq \Vert C \Vert l(\Gamma_{0T}) +B=D \\ & \Longrightarrow \vert M(t,x)\vert\leq D,
\end{split}\end{equation}
for all $u_\alpha(\cdot)\in \mathcal{U}(t),v_\alpha(\cdot)\in \mathcal{V}(t).$

Let $x_1,x_2\in \mathbb{R}^n,\,\, t_1, t_2 \in \Omega_{0T}.$
For $\varepsilon>0$ and the strategy $\Phi\in \mathcal{A}(t_1),$ we have

\begin{equation}\label{eq:1}
M(t_1,x_1)\leq \min_{v_\alpha \in \mathcal{V}(t_1)} J(\Phi[v_\alpha],v_\alpha)+\varepsilon.
\end{equation}

Define the control
$$\overline{v}_\alpha\in \mathcal{V}(t_1),\overline{v}_\alpha(s)\equiv
\left\{\begin{array}{ll}
{v}_\alpha^1(s) & s\in \Omega_{0t_2}\setminus\Omega_{0t_1}\\
{v}_\alpha(s) & s\in \Omega_{0T}\setminus\Omega_{0t_2},
\end{array}\right.$$
for any $v_\alpha \in \mathcal{V}(t_2)$ and some $v_\alpha^1 \in V$ and for each $v_\alpha \in \mathcal{V}(t_2), \underline{\Phi}\in \mathcal{A}(t_2)$ (the restriction of $\Phi$ over $\Omega_{0T}\setminus \Omega_{0t_1})$ by $\underline{\Phi}[v_\alpha]=\Phi[\overline{v}_\alpha], s\in \Omega_{0T}\setminus\Omega_{0t_2}.$

Choose the control $v_\alpha\in \mathcal{V}(t_2)$ so that
\begin{equation}\label{eq:2}
M(t_2,x_2)\geq  J(\underline{\Phi}[v_\alpha],v_\alpha)-\varepsilon.
\end{equation}

By the inequality $\eqref{eq:1},$ we have
\begin{equation}\label{eq:3}
M(t_1,x_1)\leq  J(\Phi[\overline{v}_\alpha],\overline{v}_\alpha)+\varepsilon.
\end{equation}

We know that the (unique, Lipschitz) solution $x(\cdot)$ of the Cauchy problem
$$\left\{\begin{array}{ll}
\frac{\partial x^i}{\partial s^\alpha}(s)=X^i_\alpha(s,x(s),u_\alpha(s),v_\alpha(s))\\
x(t)=x, \quad s\in \Omega_{tT}\subset \mathbb{R}_+^m, x\in \mathbb{R}^n, i=\overline{1,n}, \alpha =\overline{1,m},
\end{array}\right.$$
is the response to the controls $u_\alpha(\cdot), v_\alpha(\cdot)$ for $s\in \Omega_{0T}.$

We choose $x_1(\cdot)$ as solution of the Cauchy problem
$$\left\{\begin{array}{ll}
\frac{\partial x^i_1}{\partial s^\alpha}(s)=X^i_\alpha(s,x_1(s),\Phi[\overline{v}_\alpha](s),\overline{v}_\alpha(s))\\
x_1(t_1)=x_1, \quad s\in \Omega_{0T}\setminus \Omega_{0t_1}.
\end{array}\right.$$
Equivalently, $x_1(\cdot)$ is solution of integral equation
$$x_1(s)= x_1(t_1) + \int_{\Gamma_{t_1s}}X_\alpha(\sigma,x_1(\sigma),\Phi[\overline{v}_\alpha](\sigma),\overline{v}_\alpha(\sigma))d\sigma^\alpha.$$
Take $x_2(\cdot)$ as solution of the Cauchy problem
$$\left\{\begin{array}{ll}
\frac{\partial x^i_2}{\partial s^\alpha}(s)=X^i_\alpha(s,x_2(s),\underline{\Phi}[v_\alpha](s),v_\alpha(s))\\
x_2(t_2)=x_2, \quad s\in \Omega_{0T}\setminus \Omega_{0t_2}.
\end{array}\right.$$
Equivalently, $x_2(\cdot)$ is solution of integral equation
$$x_2(s)= x_2(t_2) + \int_{\Gamma_{t_2s}}X_\alpha(\sigma,x_2(\sigma),\underline{\Phi}[v_\alpha](\sigma),\overline{v}_\alpha(\sigma))d\sigma^\alpha.$$

It follows that
$$\Vert x_1(t_2)-x_1 \Vert = \Vert x_1(t_2)-x_1(t_1)\Vert \leq \Vert A\Vert \,\ell(\Gamma_{t_1t_2}).$$

 Since $v_\alpha=\overline{v}_\alpha$ and $\underline{\Phi}[v_\alpha]=\Phi[\overline{v}_\alpha],$
 for $s\in \Omega_{0T}\setminus\Omega_{0t_2}$, we find the estimation
\begin{equation}\begin{split}
\Vert x_1(s)-x_2(s)\Vert \ &  \leq \Vert  x_1(t_1) - x_2(t_2)\Vert + \Vert \int_{\Gamma_{t_1t_2}}\cdots \Vert\\ & \leq
\Vert A \Vert \ell(\Gamma_{t_1t_2})+ \Vert x_1-x_2\Vert
 ,\,\, \hbox{on}\,\, t_2\leq s\leq T.
\end{split}\end{equation}

Thus the inequalities $\eqref{eq:2}$ and $\eqref{eq:3}$ imply
$$
 M(t_1,x_1)-M(t_2,x_2) \leq  J(\Phi[\overline{v}_\alpha],\overline{v}_\alpha])-J(\underline{\Phi}[v_\alpha],v_\alpha])+2\varepsilon
$$
$$\leq \Big\vert \int_{\Gamma_{t_1t_2}} L_\alpha (s,x_1(s),\Phi[\overline{v}_\alpha](s),\overline{v}_\alpha(s))ds^\alpha$$ $$+\int_{\Gamma_{t_2T}} (L_\alpha (s,x_1(s),\underline{\Phi}[v_\alpha](s),v_\alpha(s))  -L_\alpha (s,x_2(s),\underline{\Phi}[v_\alpha](s),v_\alpha(s)))ds^\alpha$$  $$+g(x_1(T))-g(x_2(T))+2\varepsilon\Big\vert
$$
$$\leq\int_{\Gamma_{t_1t_2}} \vert L_\alpha (s,x_1(s),\Phi[\overline{v}_\alpha](s),\overline{v}_\alpha(s))ds^\alpha\vert$$ $$+\int_{\Gamma_{t_2T}} \vert (L_\alpha (s,x_1(s),\underline{\Phi}[v_\alpha](s),v_\alpha(s))  -L_\alpha (s,x_2(s),\underline{\Phi}[v_\alpha](s),v_\alpha(s)))ds^\alpha\vert$$
$$+\vert g(x_1(T))-g(x_2(T))\vert +2\varepsilon$$
$$\leq \Vert C \Vert \ell(\Gamma_{t_1t_2}) +\Vert C \Vert \ell(\Gamma_{t_2T})\,(\Vert A \Vert \ell(\Gamma_{t_1t_2})+ \Vert x_1-x_2\Vert) +B\, \Vert x_1-x_2\Vert) +2\varepsilon$$
$$\leq \Vert C \Vert \ell(\Gamma_{t_1t_2}) +\Vert C \Vert \ell(\Gamma_{0T})\,(\Vert A \Vert \ell(\Gamma_{t_1t_2})+ \Vert x_1-x_2\Vert) +B\, \Vert x_1-x_2\Vert) +2\varepsilon.$$

Since $\varepsilon$ is arbitrary, we obtain the inequality
\begin{equation}\label{eq:7}
M(t_1,x_1)-M(t_2,x_2)\leq E\,\ell(\Gamma_{t_1t_2}) + F \,\Vert x_1-x_2\Vert.
\end{equation}

Let $\varepsilon>0$ and choose the strategy $\Phi\in \mathcal{A}(t_2)$ such that
\begin{equation}\label{eq:4}
M(t_2,x_2)\leq \min_{v_\alpha \in \mathcal{V}(t_2)} J(\Phi[v_\alpha],v_\alpha)+\varepsilon.
\end{equation}

For each control $v_\alpha \in \mathcal{V}(t_1)$ and $s\in \Omega_{0T}\setminus\Omega_{0t_2},$ define the control $\underline{v}_\alpha\in \mathcal{V}(t_2), \underline{v}_\alpha(s)=v_\alpha(s).$

For some $u^1_\alpha \in U,$ we define the strategy $\overline{\Phi}\in \mathcal{A}(t_1)$
(the restriction of $\Phi$ over $\Omega_{0T}\setminus\Omega_{0t_2}$) by
$$\overline{\Phi}[{v}_\alpha]=
\left\{\begin{array}{ll}
u^1_\alpha & s\in \Omega_{0t_2}\setminus\Omega_{0t_1}\\
\Phi[\underline{v}_\alpha] & s\in \Omega_{0T}\setminus\Omega_{0t_2}.
\end{array}\right.$$

Now choose a control $v_\alpha\in \mathcal{V}(t_1)$ so that
\begin{equation}\label{eq:5}
M(t_1,x_1)\geq  J(\overline{\Phi}[v_\alpha],v_\alpha)-\varepsilon.
\end{equation}

By the inequality $\eqref{eq:4},$ we have
\begin{equation}\label{eq:6}
M(t_2,x_2)\leq  J(\Phi[\underline{v}_\alpha],\underline{v}_\alpha)+\varepsilon.
\end{equation}

We choose $x_1(\cdot)$ as solution of the Cauchy problem (PDE system + initial condition)
$$\left\{\begin{array}{ll}
\frac{\partial x_1^i}{\partial s^\alpha}(s)=X^i_\alpha(s,x_1(s),\overline{\Phi}[v_\alpha],v_\alpha(s)), s\in \Omega_{0T}\setminus\Omega_{0t_1} \\
x_1(t_1)=x_1, \quad s\in \Omega_{0T}\setminus \Omega_{0t_1}
\end{array}\right.$$
and $x_2(\cdot)$ as solution of the Cauchy problem (PDE system + initial condition)
$$\left\{\begin{array}{ll}
\frac{\partial x_2^i}{\partial s^\alpha}(s)=X^i_\alpha(s,x_2(s),\Phi[\underline{v}_\alpha],\underline{v}_\alpha(s)), s\in \Omega_{0T}\setminus\Omega_{0t_2}\\
x_2(t_2)=x_2, \quad s\in \Omega_{0T}\setminus \Omega_{0t_2}.
\end{array}\right.$$
Using the associated integral equations, it follows that
$$\Vert x_1(t_2)-x_1 \Vert = \Vert x_1(t_2)-x_1(t_1)\Vert \leq \Vert A\Vert \,\ell(\Gamma_{t_1t_2}).$$
Also, for $s\in \Omega_{0T}\setminus\Omega_{0t_2}, v_\alpha=\underline{v}_\alpha$ and $\overline{\Phi}[v_\alpha]=\Phi[\underline{v}_\alpha],$ we find
\begin{equation}\begin{split}
\Vert x_1(s)-x_2(s)\Vert \ &  \leq \Vert  x_1(t_1) - x_2(t_2)\Vert + \Vert \int_{\Gamma_{t_1t_2}}\cdots \Vert\\ & \leq
\Vert A \Vert \ell(\Gamma_{t_1t_2})+ \Vert x_1-x_2\Vert,\,\, \hbox{on}\,\, t_2\leq s\leq T.
\end{split}\end{equation}

Thus, the relations $\eqref{eq:5}$ and $\eqref{eq:6}$ imply
$$
M(t_2,x_2)-M(t_1,x_1) =  J(\overline{\Phi}[v_\alpha],v_\alpha])-J(\Phi[\underline{v}_\alpha],\underline{v}_\alpha])+2\varepsilon
  $$
  $$
  =  - \int_{\Gamma_{t_1t_2}} L_\alpha (s,x_1(s),\overline{\Phi}[{v}_\alpha](s),{v}_\alpha(s))ds^\alpha
$$
 $$
+\int_{\Gamma_{t_2T}} (L_\alpha (s,x_1(s),{\Phi}[\underline{v}_\alpha](s),\underline{v}_\alpha(s)) - L_\alpha (s,x_2(s),{\Phi}[\underline{v}_\alpha](s),\underline{v}_\alpha(s)))ds^\alpha
 $$
 $$ +g(x_1(T))-g(x_2(T))+2\varepsilon $$
 $$\leq \Vert C \Vert \ell(\Gamma_{t_1t_2}) +\Vert C \Vert \ell(\Gamma_{0T})\,(\Vert A \Vert \ell(\Gamma_{t_1t_2})+ \Vert x_1-x_2\Vert) +B\, \Vert x_1-x_2\Vert) +2\varepsilon.$$

 Since $\varepsilon$ is arbitrary, we obtain the inequality
\begin{equation}\label{eq:8}
M(t_2,x_2)-M(t_1,x_1)\leq E\,\ell(\Gamma_{t_1t_2}) + F \,\Vert x_1-x_2\Vert.
\end{equation}

By $2.17$ and $2.22$, we proved the continuity of the lower and upper value functions.
\end{proof}

\section{Viscosity solutions of \\multitime (HJIU) PDEs}

\begin{theorem}\textbf{(PDEs for multitime upper value function, resp. multitime lower value function)}

The multitime upper value function $M(t,x)$ and the multitime lower value function $m(t,x)$
are the viscosity solutions of Hamilton-Jacobi-Isaacs-Udri\c ste (HJIU) PDEs:
\begin{itemize}
\item the multitime upper (HJIU) PDEs
$$\frac{\partial M}{\partial t^\alpha}(t,x)+\min_{v_\alpha\in \mathcal{V}} \max_{u_\alpha\in \mathcal{U}} \left\lbrace  \frac{\partial M}{\partial x^i}(t,x) X_\alpha^i(t,x,u_\alpha,v_\alpha)+L_\alpha(t,x,u_\alpha,v_\alpha)\right\rbrace =0,$$
with the terminal condition $M(T,x)=g(x),$
\item the multitime lower (HJIU) PDEs
$$\frac{\partial m}{\partial t^\alpha}(t,x)+\max_{u_\alpha \in \mathcal{U}} \min_{v_\alpha \in \mathcal{V}} \left\lbrace  \frac{\partial m}{\partial x^i}(t,x) X_\alpha^i(t,x,u_\alpha,v_\alpha)+L_\alpha(t,x,u_\alpha,v_\alpha)\right\rbrace =0,$$
with the terminal condition $m(T,x)=g(x).$
\end{itemize}
\end{theorem}

\begin{remark}
If we introduce the so-called upper and lower Hamiltonian $1$-forms defined respectively by
$$H^+_\alpha(t,x,p)=\min_{v_\alpha\in \mathcal{V}} \max_{u_\alpha \in \mathcal{U}}\lbrace p_i(t) X_\alpha^i(t,x,u_\alpha,v_\alpha)+L_\alpha(t,x,u_\alpha,v_\alpha)\rbrace,$$
$$H^-_\alpha(t,x,p)=\max_{u_\alpha\in \mathcal{U}} \min_{v_\alpha\in \mathcal{V}}\lbrace p_i(t) X_\alpha^i(t,x,u_\alpha,v_\alpha)+L_\alpha(t,x,u_\alpha,v_\alpha)\rbrace,$$
then the multitime (HJIU) PDE systems can be written in the form
$$\frac{\partial M}{\partial t^\alpha}(t,x)+H^+_{\alpha}\left( t,x,\frac{\partial M}{\partial x}(t,x)\right) =0$$ and
$$\frac{\partial m}{\partial t^\alpha}(t,x)+H^-_\alpha\left( t,x,\frac{\partial m}{\partial x}(t,x)\right) =0.$$
\end{remark}

The proof will be given in another paper.

\section{Representation formula of viscosity \\solutions for multitime (HJ) PDEs}

In this section, we want to obtain a representation formula for the viscosity solution
$M(t,x)$ of the multitime (HJ) PDEs system
\begin{equation}
 \frac{\partial M}{\partial t^\alpha}+H_\alpha\left( t,x,\frac{\partial M}{\partial x}(t,x)\right) =0, (t,x)\in \Omega_{0T}\times \mathbb{R}^n,\alpha=\overline{1,m},
\end{equation}
\begin{equation}
M(0,x)=g(x), x\in \mathbb{R}^n \,(\hbox{initial\, condition}),
\end{equation}
where the unique solution $M(t,x)$ satisfies the inequalities
\begin{equation}\label{eq:11}
\left\{\begin{array}{ll}
\vert M(t,x)\vert\leq D\\
\vert M(t,x)-M(\hat{t},\hat{x})\vert\leq E\,\,\ell(\Gamma_{t\hat t})+ F\,\,\Vert x-\hat{x}\Vert,
\end{array}\right.\end{equation}
for some constants $D, E, F$ (for $m=1,$ see also [4]).

Also, we assume that $g:\mathbb{R}^n \rightarrow \mathbb{R}, H_\alpha:\Omega_{0T} \times \mathbb{R}^n\times \mathbb{R}^p\rightarrow \mathbb{R},$ satisfy the inequalities
$$\left\{\begin{array}{ll}
\vert g (x)\vert\leq B\\
\vert g(x)-g (\hat{x})\vert\leq B \Vert x-\hat{x}\Vert
\end{array}\right.$$
and
\begin{equation}\label{eq:8}
\left\{\begin{array}{ll}
\vert H_\alpha(t,x,0)\vert\leq K_\alpha\\
\vert H_\alpha (t,x,p)-H_\alpha(\hat{t},\hat{x},\hat{p})\vert\leq K_\alpha\,\, (\ell(\Gamma_{t\hat t})+\Vert x-\hat{x}\Vert +\Vert p-\hat{p}\Vert).
\end{array}\right. \end{equation}

\textbf{Max-min representation of a Lipschitz function as affine functions} (for $m=1,$ see also [2], [3]).

\begin{lemma}\label{l-2}

For each $\alpha$, let
\begin{equation}\label{eq:10}
\left\{\begin{array}{ll}
U=B(0,1)\subset \mathbb{R}^{n}\\
V=B(0,P)\subset \mathbb{R}^{n}\\
X_\alpha(u_\alpha)=K_\alpha u_\alpha,\, K_\alpha \in \mathbb{R}\\
L_\alpha(t,x,u_\alpha,v_\alpha)=H_\alpha (t,x,v_\alpha)-<K_\alpha u_\alpha,v_\alpha>.
\end{array}\right.
\end{equation}
Let $H_\alpha$ be a Lipschitz 1-form. For some constant $P>0$ and for each $t\in \Omega_{0T},\,\, x \in \mathbb{R}^n,$ we have
$$H_\alpha (t,x,{p})=\max_{v_\alpha \in V}\min_{u_\alpha \in U}\left\lbrace <X_\alpha(u_\alpha),{p}> + L_\alpha(t,x,u_\alpha,v_\alpha)\right\rbrace ,$$
if $\Vert {p}\Vert \leq P$.
\end{lemma}

\begin{proof}
In view of the assumption $H_\alpha (t,x,v_\alpha)-H_\alpha(t,x,{p})\leq K_\alpha \Vert {p}-v_\alpha\Vert,$
by the Cauchy-Schwarz formula, and by the condition $||u||\leq 1$,
we have for any $x\in \mathbb{R}^n,$
\begin{equation}\begin{split}
H_\alpha (t,x,{p})\ & =\max_{v_\alpha \in V} \left\lbrace H_\alpha(t,x,v_\alpha) - K_\alpha\Vert {p}-v_\alpha\Vert\right\rbrace \\ & =\max_{v_\alpha \in V}\min_{u_\alpha\in U}\left\lbrace H_\alpha(t,x,v_\alpha)\,+\, <K_\alpha u_\alpha,{p}-v_\alpha>\right\rbrace.
\end{split}\end{equation}
\end{proof}

\textbf{Max-min representation of a Lipschitz function as positive homogeneous functions} (for m=1, see also [2],[3]).

\begin{lemma}
Let $H_\alpha$ be a Lipschitz $1$-form which is homogeneous in $p,$ i.e.,
$$H_\alpha(t,x,\lambda p)=\lambda H_\alpha(t,x,p),\,\,\lambda \geq 0.$$
Then there exist compact sets $U, V\subset \mathbb{R}^{2n}$ and
vector fields
$$X_\alpha:[0,T]\times \mathbb{R}^n\times U\times V\rightarrow \mathbb{R}^n$$
satisfying
$$\Vert X_\alpha(x)-X_\alpha(\hat{x})\Vert\leqslant A_\alpha\Vert x-\hat{x}\Vert$$
and such that, for each $\alpha$,
$$H_\alpha (t,x,p)=\max_{v_\alpha \in V}\min_{u_\alpha \in U}\left\lbrace <X_\alpha(t,x,u_\alpha,v_\alpha),p> \right\rbrace ,$$
for all $t\in \Omega_{0T},x\in \mathbb{R}^n,p \in \mathbb{R}^n.$
\end{lemma}

\begin{proof}
Let $u_\alpha=(u^1_\alpha,u^2_\alpha),v_\alpha=(v^1_\alpha,v^2_\alpha)$ ($2n$-dimensional controls) and
\begin{equation}\label{eq:9}
\left\{\begin{array}{ll}
U=V=B(0,1)\times B(0,1)\subset \mathbb{R}^{2n}\\
L_\alpha(t,x,u^1_\alpha,v^1_\alpha)=H_\alpha (t,x,v_\alpha^1)-<K_\alpha u^1_\alpha,v_\alpha^1>\\
X_\alpha(t, x, u_\alpha, v_\alpha)=K_\alpha u^1_\alpha+ C_\alpha v^2_\alpha+ (L_\alpha(t,x,u^1_\alpha,v^1_\alpha)- C_\alpha)u^2_\alpha.\\
\end{array}\right.
\end{equation}
According to Lemma $\eqref{l-2}$ and the assumptions $\eqref{eq:9},$ if $\Vert\eta\Vert =1,$ we have
\begin{equation}\begin{split}
H_\alpha (t,x,\eta)\ &  =\max_{v^1_\alpha \in V^1}\min_{u^1_\alpha\in U^1}\left\lbrace <K_\alpha u^1_\alpha,\eta> + L_\alpha (t,x,u^1_\alpha, v^1_\alpha)\right\rbrace,
\end{split}\end{equation}
for $U^1=V^1=B(0,1)\in \mathbb{R}^n$.

For any $p\neq 0$, we can write
\begin{equation}\begin{split}
H_\alpha(t,x,p)\ & =\Vert p\Vert H_\alpha \left( t,x,\frac{p}{\Vert p\Vert}\right) \\ & =\max_{v^1_\alpha\in V^1}\min_{u^1_\alpha\in U^1}\left\lbrace <K_\alpha u^1_\alpha,p>+L_\alpha (t,x,u^1_\alpha,v^1_\alpha)\Vert p\Vert\right\rbrace .
\end{split}\end{equation}
Then, if we choose $C_\alpha>0$ such that $\vert L_\alpha\vert\leq C_\alpha,$ we find
\begin{equation}\begin{split}
H_\alpha (t,x,p)\ &  =\max_{v^1_\alpha \in V^1}\min_{u^1_\alpha\in U^1}\bigg\{ <K_\alpha u^1_\alpha,p> +C_\alpha\Vert p\Vert +( L_\alpha (t,x,u^1_\alpha, v^1_\alpha)-C_\alpha)\Vert p\Vert\bigg\}\\ & =\max_{v^1_\alpha \in V^1}\min_{u^1_\alpha\in U^1}\max_{v^2_\alpha \in V^1}\min_{u^2_\alpha\in U^1}\bigg\{ <K_\alpha u^1_\alpha,p> +<C_\alpha v^2_\alpha, p>\\ &  +( L_\alpha (t,x,u^1_\alpha, v^1_\alpha)-C_\alpha)< u^2_\alpha,p> \bigg\} \\ & =\max_{v_\alpha \in V}\min_{u_\alpha\in U}\bigg\{ <X_\alpha(t,x,u_\alpha, v_\alpha),p>\bigg\}.
\end{split}\end{equation}
Now, interchanging $\min_{u_\alpha^1\in U^1}$ and $\max_{v_\alpha^2\in V^1}$, the result in Lemma follows.
\end{proof}

We are now in a position to give the main result of this section.
\begin{theorem}
For each $t \in \Omega_{0T}$ and $x \in \mathbb{R}^n,$ the upper value function $M(t,x)$ verifies the equality
\begin{equation}\begin{split}
M(t,x)=\max_{\Phi\in \mathcal{U}(T-t)}\min_{v_\alpha\in V(T-t)}\bigg\{ \ & - \int_{\Gamma_{T-tT}} L_\alpha(T-s,x(s),\Phi[v_\alpha](s),v_\alpha(s))ds^\alpha \\ & +g(x(T))\bigg\} ,
\end{split}\end{equation}
where for each pair of controls $v_\alpha\in V(T-t)$,\,\, $u_\alpha=\Phi[v_\alpha]\in U(T-t),$
the state function $x(\cdot)$ solves the problem
\begin{equation}
\left\{\begin{array}{ll}
\frac{\partial x^i}{\partial s^\alpha}(s)=-F^i_\alpha u_\alpha(s), s\in \Omega_{0T}\setminus \Omega_{0T-t}\\
x(T-t)=x.
\end{array}\right.
\end{equation}
\end{theorem}

\begin{proof}
Let
$$H^1_\alpha(t,x,p)=\max_{v_\alpha\in V}\min_{u_\alpha\in U}\left\lbrace <X_\alpha(u_\alpha),p>+L_\alpha(t,x,u_\alpha,v_\alpha)\right\rbrace,$$
 $U=B(0,1)\subset \mathbb{R}^{pm}, V=B(0,P)\subset \mathbb{R}^{qm}$ and $X^i_\alpha, L_\alpha$
 Lipschitz functions with the assumptions $\eqref{eq:10}.$

Then $H_\alpha(t,x,p)=H^1_\alpha(t,x,p)$ provided $\vert p\vert\leq P.$
Since $M(t,x)$ satisfies $\eqref{eq:11},$ it follows that $M(t,x)$ is also the unique viscosity
solution of the multitime (HJ) PDEs system (for $m=1,$ see also [4])
\begin{equation}
 \frac{\partial M}{\partial t^\alpha}+H^1_\alpha\left( t,x,\frac{\partial M}{\partial x}(t,x)\right) =0, \,\,(t,x)\in \Omega_{0T}\times \mathbb{R}^n,\alpha=\overline{1,m},
\end{equation}
\begin{equation}
M(0,x)=g(x),\,\, x\in \mathbb{R}^n.
\end{equation}
If we take $M^1(t,x)=M(T-t,x),$ one observes that $M^1(t,x)$ is a viscosity solution of this system (for $m=1,$ see also [2])

\begin{equation}
 \frac{\partial M^1}{\partial t^\alpha}+H^+_\alpha\left( t,x,\frac{\partial M^1}{\partial x}(t,x)\right) =0,\,\, (t,x)\in \Omega_{0T}\times \mathbb{R}^n,\alpha=\overline{1,m},
\end{equation}
\begin{equation}
M^1(T,x)=g(x),\,\, x\in \mathbb{R}^n
\end{equation}
and
$$H^+_\alpha(t,x,p)=\max_{v_\alpha\in V}\min_{u_\alpha\in U}\left\lbrace -<X_\alpha(u_\alpha),p>+L_\alpha(T-t,x,u_\alpha,v_\alpha)\right\rbrace.$$
Using the above developments, we obtain
\begin{equation}\begin{split}
M^1(t,x)=M(t,x)=\max_{\Phi\in \mathcal{U}(t)}\min_{v_\alpha\in V(t)}\bigg\{ \ & - \int_{\Gamma_{tT}} L_\alpha(T-s,x(s),\Phi[v_\alpha](s),v_\alpha(s))ds^\alpha \\ & +g(xT))\bigg\} ,
\end{split}\end{equation}
where $x(\cdot)$ is the solution of the Cauchy problem
\begin{equation}
\left\{\begin{array}{ll}
\frac{\partial x^i}{\partial s^\alpha}(s)=-X^i_\alpha(u_\alpha(s))=-F^i_\alpha u_\alpha(s), s\in \Omega_{0T}\setminus \Omega_{0T-t}\\
x(t)=x,
\end{array}\right.
\end{equation}
for the control $u_\alpha(\cdot)=\Phi[v_\alpha].$
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Blackmagic outs sub-$2000 4K filmmaking camera
byLeigh :) Stark
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Some of the best technology for filmmaking appears on stills cameras, and folks like Sony and Panasonic are leading the charge. And yet on the horizon, an Aussie player could offer something unexpected.
We're always fans of a local story, and one of the technological stories few seem to be aware of in Australia is that of Blackmagic, an Australian company that started with video capture devices, developing solutions to help people capture from other sources and edit using their PCs.
The company still makes that sort of gear, but in the past few years, has been focused on pretty much all major aspects of what a filmmaker would set their eyes on, from colour grading to editing to streaming, and even to cameras themselves.
Since cameras tend to be the main way for films to start their journey from inception to reality, Blackmagic has been dabbling in the camera space, taking its unique understanding of the video capture hardware world and applying it to a camera you can make films with.
In the past few years, we've seen a few, including the big Blackmagic URSA and the more pocket friendly Micro Cinema and Pocket Cinema cameras, and these have delivered filmmaking at a fraction of the price normally charged for a proper big filmmaking camera.
This week, though, there's another model on the way, as Australia's Blackmagic makes a play for pockets one more time and focuses on the big 4K effort we're all looking at for TVs these days, coming in the Pocket Cinema Camera 4K.
Made to be small, this one is a mirrorless camera reliant on the Micro Four-Thirds sensor technology both Olympus and Panasonic uses, meaning the lenses will be compatible, while adaptors can be found for other brands, including that of Canon, Nikon, and Leica.
Under the hood is what makes this interesting, with the 4K sensor capturing video only in resolutions ranging from Full HD to Ultra HD (3840×2160), and even including the proper 4K film output of 4096×2160.
The sensor itself isn't just a stills sensor set to work mostly with video, as Blackmagic has developed this sensor to capture RAW video. Much like how RAW image support provides the ability to change light and exposure settings with individual images, RAW video does the same to video, with extended low-light support, 12-bit colour, and HDR imagery captured to video.
Blackmagic is capturing this data to either SD or CFast cards, but there's also the option to capture directly to USB-C devices, like Samsung's Type-C external drives, meaning the Blackmagic Pocket Cinema 4K can capture directly to a USB drive and then have its footage checked, edited, and shared from the one source.
"The new Blackmagic Pocket Cinema Camera 4K is exciting because it packs so much technology into an incredibly small design," said Grant Petty, CEO of Blackmagic Design.
"We feel it will go well beyond what customers expected. It's perfect for anyone that needs a truly professional ultra compact digital film camera," he said.
"This is not a consumer camera that's trying to do high end work, it's a true high end camera designed for high end work."
There's also a 5 inch touchscreen on the back, audio support over a mini-XLR jack with phantom power, and an HDMI port to test the video directly on a big 4K screen, while the camera itself can be charged over USB Type C, as well.
All up, it looks to be a fairly impressive little box, and it's one that won't actually come with as seriously high price tag as you might expect, netting a US price of $1295 which locally is translated to $1815 AUD.
Given what it could do, that's a very curious price, and could pose some interesting competition from the likes of Sony, Panasonic, Canon, and Nikon, as all offer 4K solutions for filmmaking. Granted, the photographic still capability isn't as high a megapixel as other cameras, sitting at 4K's 8 megapixel equivalent, but with an ISO range of up to 25600 and 13 stops of dynamic range, filmmakers are likely to be quite happy.
As for when you can expect to get your hands on this camera, you'll want to save that money for later this year, because the official word is "later this year", as in later this year we might see an Australian 4K camera to rival the big boys.
Leigh :) Stark
A technology journalist working out of Sydney, Australia, Leigh has written for publications including The Australian Financial Review, GadgetGuy, Popular Science, APC, PC & Tech Authority, as well as for radio and TV since 2007.
The Wrap - Australia's fastest technology roundup
The Wrap – What's hot at CES 2022
Sony brings Mini-LED, QD-OLED to big 4K, 8K TVs
LG ups power of OLED, QNED in 2022

redpajama_set_name: RedPajamaCommonCrawl


Vol. 65 April 15, 2012 First ACOs Appear with a Whimper, Not a Bang.
In the world of health care providers there has been much consternation, gnashing of teeth (maybe other body parts too), and exuberant, expert speculations about what an ACO was going to look like and by what means the ACO program was going to "revolutionize American medicine" as promised by Washington.
The first 27 ACO's are surprisingly underwhelming. None of them have been developed by the "big boys" with name recognition in the medical marketplace, unless you happen to live in Caldwell County, North Carolina, coastal Georgia, or Hackensack NJ . Half are small physician-led organizations. Less than half appear to involve hospitals. All totaled they cover less than 1% of Medicare patients. Only two of the twenty-seven are willing to take limited risk of losing money ("going over budget") in return for the potential of sharing in more savings ("providing care under budget").
Medicare beneficiaries (patients) will continue to be able to see ANY provider and unless the provider boasts about it or otherwise publicize it, the patient may not know that his or her physician is a member of an ACO. In my local community the physicians organization took out a full-page ad in Sunday's paper to "congratulate" themselves on being selected as an ACO. This particular physicians organization has had a contentious business relationship with the local hospitals for ten years, plans to cover only the minimum of 5,000 Medicare patients, and finds itself in competition with a recently spawned physician-hospital organization.
Another physicians organization formed just this year in a neighboring county plans to cover 6000 Medicare patients and does not involve its community hospital. The hospital itself is in a fierce, cost-cutting competition with a larger hospital up the road just 15 miles closer to the academic medical centers who are courting it for a merger.
Both situations illustrate how complicated the medical/political/economic environment is for these initial ACOs.
OR is it just one more step in our historically incremental evolution towards a system of universal health care?
Maybe the next 150 ACOs to be announced in July 2012 will give us more of a clue.
This entry was posted on Sunday, April 15th, 2012 at 11:37 PM and is filed under current events, government, health care reform, medical economics, politics.	You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
One Response to Vol. 65 April 15, 2012 First ACOs Appear with a Whimper, Not a Bang.


text:
Internet, Networking, & Security Home Networking
What Are WEP, WPA, and WPA2? Which Is Best?
WEP vs WPA vs WPA2: Why The Differences Matter
Melanie Uy
Former Lifewire writer Melanie Uy has 5+ years' experience writing about consumer-oriented technology and is an expert telecommuter.
The Wireless Connection
Installing & Upgrading
Wi-Fi & Wireless
The acronyms WEP, WPA, and WPA2 refer to different wireless encryption protocols that are intended to protect the information you send and receive over a wireless network. Choosing which protocol to use for your own network can be a bit confusing if you're not familiar with their differences.
Below is a look at the history of wireless encryption and a comparison of these protocols so that you can come to a solid conclusion about which you might want to use for your own home or business.
In short, WPA2 is what you want to use, if possible. Not all devices support it, but if yours does, it's important to adopt so that the most up-to-date security possible is available for all the devices that are currently connected and will eventually connect to your Wi-Fi network.
What They Mean and Which to Use
These wireless encryption protocols were created by the Wi-Fi Alliance, an association of hundreds of companies in the wireless network industry. The first protocol the group created was WEP (Wired Equivalent Privacy), introduced in the late 1990s.
WEP, however, had serious security weaknesses and has been superseded by WPA (Wi-Fi Protected Access). Despite being easily hacked, however, WEP connections are still in use and may be providing a false sense of security to the many people who are securing their networks with WEP.
Blend Images - John Lund/Brand X Pictures/Getty Images
The reason WEP is still used is likely either because the network administrator hasn't changed the default security on their wireless routers or because the device is too old and therefore doesn't support newer encryption methods like WPA.
Just as WPA replaced WEP, WPA2 has replaced WPA as the most current security protocol. WPA2 implements the latest security standards, including government-grade data encryption. Since 2006, all Wi-Fi certified products must use WPA2 security.
If you're looking for a new wireless card or device, make sure it's labeled as Wi-Fi CERTIFIED™ so that you know it complies with the latest security standard. For existing connections, make sure your wireless network is using the WPA2 protocol, particularly when transmitting confidential personal or business information.
Using WEP/WPA/WPA2 on a Router
During the initial setup, most modern wireless access points and routers let you select the security protocol to use. While this is, of course, a good thing, some people don't care to change it.
The problem with not changing the default security protocol used by the router is that it might be using WEP, which we now know isn't secure. Or, even worse, the router might be completely open with no encryption or password at all.
If you're setting up your own network, make sure to use WPA2 or, at the bare minimum, WPA.
How to Encrypt Your Wireless Network
Using WEP/WPA/WPA2 on the Client Side
When you try to establish a connection to a security-enabled wireless network for the first time — be it from your laptop, desktop, smartphone, etc. — you'll be prompted to enter the correct security key or passphrase to gain access to the network.
That password is the WEP/WPA/WPA2 code that the router administrator entered into the router when the network security was set up. If you're connecting to a business network, it's most likely provided by the network administrator.
Unlike from the admin's perspective where the router is set up to use a specific security protocol, you, as the user, can't change anything about the security method from the client side. When you enter the password for the Wi-Fi network, you're gaining access using whatever protocol has already been determined.
It's for this reason that you should take note of the network's security setting after connecting, to understand what's safe and not safe to transmit over that particular network.
How to Configure WPA Support in Microsoft Windows
What's the Best Encryption to Secure My Wi-Fi Network?
WEP (Wired Equivalent Privacy)
What Is Network Encryption?
What Is WPA3 Wi-Fi?
An Overview of Wireless Protected Access 2 (WPA2)
How to Set up Your Home Wi-Fi Network
What Is Wi-Fi Protected Access (WPA)?
WPA2 vs. WPA
What Is WPS and How Does It Work?
How to Stop People From Using Your Wi-Fi
What Is a Wireless Key?
The 9 Best VPN Routers of 2021
The 9 Best Secure Routers of 2021
How to Share a Wi-Fi Password From iPhone to Android


The aesthetics of vintage video games has inspired art everywhere, from chiptune and vaporwave music, to both animated and live-action feature films, and of course, modern video games.
Get your daily dose of nostalgia with Video Game Skies, a Tumblr collecting the most aesthetically pleasing skyboxes from retro video games.
The early days of video game design represented a time when creativity was put under extremely harsh restraints in the form of CPU memory. Games had to fit their entire playtimes into drives ranging as low as 8kb. Moreover, there were restrictions upon the number of sprites able to be displayed onscreen, and the number of audio tracks which could play at once.
The result was a series of art which thrived on simplicity, and retro video game soundtracks, sprites, and artwork are still celebrated to this day.
Check out a few of our favourites below, and find more Video Game Skies on Tumblr.


text:
Amanda Knox's rights violated in Italy murder probe, ECHR says
Thread starter JacksinPA
JacksinPA
https://www.bbc.com/news/world-europe-46985679
taly has been ordered to pay €18,400 (£16,000) to US citizen Amanda Knox, who spent years in prison for a murder of which she was later acquitted.
Ms Knox's long-running legal battle was closely followed by global media during years of retrials and appeals.
This poor young woman was the victim of a deranged prosecutor. I'm happy that she will now see some restitution.
PeteEU
Poor murderess...
Sent from my Honor 8X using Tapatalk
I am thinking she should cross Italy off of her future travel plans.
TurtleDude
warrior of the wetlands
longview said:
If I ever were to teach at a law school again, I would use this case as an example of how a government can totally Screw up substantive and procedural due process, among other things.
PeteEU said:
Except, of course, for the fact that she didn't murder anyone.
Tanngrisnir said:
many Euros who hate the USA use Amanda Knox as a surrogate for America.
Higgins86
Maybe, maybe not. Guarantee she was in the room when it went down though.
She did and got away with it.
Felis Leo
Moral clarity is needed
Perhaps she did, PeterEU. Perhaps she did not. I do not have any tribal skin in this fight. But I would not trust the verdict of the botched Italian justice system to come to anything approaching an appropriate legal standard by which to determine that guilt.
Nah. There's no credible evidence to back up that claim.
You're very much welcome to imagine otherwise.
TurtleDude said:
Eh, maybe, but if you saw the documentary on the case, that prosecutor was a full-on nutjob, thinking it was a satanic cult, etc...
yeah I did-I saw at least a coupled shows about it-wasn't there some dude that had a criminal background who had both opportunity and motive to kill the girl?
Yep, there was.
Felis Leo said:
You see this is my problem.. the insults by Americans against the Italian legal system. This is Italy, not the US so you cant transplant the very flawed US legal system on the Italian system. Did the police maybe make mistakes? Sure, not many murders in Italy relative to the very experienced American police, but what I make my verdict on is the publicly available information on how she acted after the murder. It screams guilty.
Toggle Almendro
One of the bright points in this case is, the Karmic backlash from what Italy did to Amanda and Raffaele resulted in Guede getting his LIFE sentence reduced, first to 30 years, and then to 16.
We're getting close to the point where Guede is going to be back out on the streets at night in an Italian college town. I'm expecting good things to happen.
"Good" in my opinion at least. The people of Italy might not like it so much when Karma comes calling.
Higgins86 said:
That is incorrect. The evidence is very clear that Amanda was nowhere near the place when Guede raped and killed Kercher.
Nope. The evidence shows that Guede did it all by himself.
Actually, she and Raffaele definitely had nothing to do with it. And the Italian officials knew that from the start.
Fishking
Good thing it was the ECHR that held them accountable and not the U.S. legal system. Question though...does the Italian legal system have a loophole in there somewhere that allows them to convict an innocent person because they are American?
You see this is my problem.. the insults by Americans against the Italian legal system.
It is perfectly appropriate to condemn backwards countries when they fail to adhere to civilized standards. And Italy's third world excuse for a justice system does not in any way meet civilized standards.
This is Italy, not the US so you cant transplant the very flawed US legal system on the Italian system.
True, but not in the way that you meant it. The Italian justice system will need to figure out simple concepts like evidence and proof before they can reasonably be compared to the justice systems in civilized countries.
Sure, not many murders in Italy relative to the very experienced American police, but what I make my verdict on is the publicly available information on how she acted after the murder. It screams guilty.
Setting aside the fact that, no it doesn't scream guilty at all, civilized countries base their verdicts on actual evidence.
Fishking said:
The Italian legal system simply ignores the law completely. They also ignore facts and evidence. That makes it easy for them to convict whoever they want.
Toggle Almendro said:
No it doesn't. She was in the house.
That is incorrect. She spent the night with her boyfriend Raffaele.
Had she been in the house, Guede would have raped and murdered her as well.
Hardly surprising coming from you. Now the ECHR must also be in cahoots with America.
If you believe Knox that would be true but there is no evidence that she wasn't in the house during that evening.


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text:
About Florian
Radu Paladi remembered in film
Florian's first solo CD, Following the River: Music along the Danube (Acousence records ©2017, ACO-CD 13317), includes a piece called Rondo a capriccio by Radu Paladi. Paladi (1927-2013) was a Romanian pianist, composer and conductor who also had a long career teaching at the Universitatea Națională de Artă Teatrală și Cinematografică 'Ion Luca Caragiale' (national university of film and theatrical art) in Bucharest. Florian writes,
Radu Paladi trained as a pianist under Florica Musicescu, who was also the teacher of Dinu Lipatti and Radu Lupu, among many others. This solid technical grounding made him an excellent composer for the piano. He went on to become a prolific composer, particularly distinguished for choral, chamber and film music. His style has been described as resistant to avant-garde language: instead he aimed at a more direct, simple and pure musical language. This is precisely how his sparkly Rondo a capriccio flows: beguilingly pure and unfussy. Here Paladi's virtuosic writing for the piano is an interesting blend of Mendelssohn and Debussy, with added hints of Bartók and Prokofiev. Most importantly, the integration of Romanian folk material, typical of his entire output, feels refreshingly natural here. The music is very descriptive, and unfolds in an almost cinematic way.
Florian had the privilege of knowing Radu Paladi and as a tribute to him wanted not only to record one of his piano pieces, but also to honour his work in film and his fascination with the folk tradition. This short film blends shots of Florian at home in his practice room with images from rural Romania which capture the mood of the piece and its sense of a place vividly remembered. The film is not itself great art, but is made by way of thanks to a man whose music is little known outside Romania, but deserves wider recognition and more recordings. To see the film at its best, you can watch it on Vimeo by following the link button below, but you can also see it in one of Florian's Facebook posts, where you can let Florian know what you think and share it with others who might enjoy it. Do you like it? Would you like to see more? Let Florian know and he will get working! The short film is also available on Florian's YouTube channel.
Categories News, Recording, Reflections Tags Inspiration, Recording
Florian's first Edinburgh Festival Fringe
'Following the River' receives award
Finchcocks: A light at the end of the tunnel
My response to Covid-19
© 2023 Florian mitrea|Whistling Dog


Abstract: This paper starts from the premise that, although augmented reality (AR) proposes a new perceptual experience situated at the limit between the actual and the virtual, when one sees it in the larger context of contemporary art practice one can observe its strong connections with previous artistic forms. Installation art and virtual reality (VR) are the two important reference points in the genealogy of AR art, their specific ways of using space having a particular relevance in AR|s own spatiality. In this sense, I see AR mainly as a spatial paradigm situated at the confluence between installation-type physical space and the digital spatiality of VR. I would suggest that AR creates a new species of space, an |augmented space| in which physical reality and digital images coexist. The present essay focuses on this intermediary zone, discussing the problems of interactivity, virtuality, illusionism and representation following two seminal artworks in the field of AR.
Keywords: augmented space; augmented reality; real space; digital space; installation art; virtual reality; representation; technology; interactivity; virtuality; illusionism; digital spatiality; spatial paradigm.


text:
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Judge's dismissal of atheist's harassment claim against Muslim makes waves
By Dugald McConnell and Brian Todd, CNN
(CNN) - A protester who ridiculed the Muslim prophet Mohammed claims he was assaulted by a Muslim who was offended by the stunt, but a judge has sympathized with the alleged perpetrator, in a case that has drawn national attention.
Self-proclaimed atheist Ernie Perce marched in a Halloween parade in Mechanicsburg, Pennsylvania last October, dressed in a costume mocking Mohammed.
In a YouTube video he posted, Perce can be seen wearing a long fake beard, a white turban and green face paint, calling out provocative phrases like: "I am the prophet Mohammed! Zombie from the dead!" Perce and someone else in a zombie-themed pope costume are carrying a banner that reads "The Parading Atheists of Central Pennsylvania / Ghoulish – Godless – God-Awful."
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Then a man who is not seen on the video can be heard saying, "Take it down." Amid sounds of a scuffle, Perce can be heard saying "Hey, he's attacking me!"
Perce told CNN affiliate WHTM that the man "grabbed me, choked me from the back, and spun me around, to try to get my sign off that was wrapped around my neck."
Based on Perce's complaint, a Muslim named Talaag Elbayomy was charged with harassment. But on December 6, District Judge Mark Martin dismissed the case, saying it was one person's word against another's, and that there was no other evidence or eyewitness testimony to prove that Elbayomy had harassed or touched the alleged victim.
The judge also scolded Perce, saying he'd been needlessly provocative on an issue sensitive with Muslims.
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"You have that right, but you're way outside your bounds of First Amendment rights," Martin said, according to a recording Perce made of the court hearing. "I think our forefathers intended that we use the First Amendment so that we can speak our mind, not to piss off other people and other cultures, which is what you did."
The judge went on to point out that in many Muslim countries, ridiculing Mohammed could warrant the death penalty under Islamic law.
Critics say Martin's lecture shows he used Muslim cultural grounds to excuse a deplorable assault, and failed to defend an atheist's First Amendment rights.
"That's greatly disturbing to people that believe in free speech," said George Washington University law professor Jonathan Turley. "You can say things that are hurtful to others. We hope that you don't, but you most certainly can be protected. People like Thomas Paine spent his entire life ticking off people across the colonies."
Former terrorism prosecutor Andrew C. McCarthy, writing on the blog of National Review, accused the judge of allowing the Muslim suspect to invoke a "Sharia defense – what he claimed was his obligation to strike out against any insult against the prophet Mohammed."
And Perce said of Judge Martin, "He let a man who is Muslim, because of his preference of his culture and his way of life, walk free, from an attack."
The judge, in a phone interview with CNN, defended his ruling.
"The commonwealth didn't present enough evidence to show me that this person is guilty beyond a reasonable doubt," Martin said. "That's why I dismissed the case. Nothing as nefarious as what everyone's thinking, that I'm a Muslim or I'm biased. I'm actually a Lutheran."
Martin added that he has served three tours of duty, totaling more than two years, in Iraq and Afghanistan, where he learned more about Muslim culture.
"It just amazes me that people think that I'm biased towards Islam," he added. "I got sniped at once, I got ambushed once, I got attacked by a mob once... I've served close to 27 years in the military - and have gone overseas - exactly to preserve that right [freedom of speech.]"
But Martin also repeated his criticism of the atheist protester. "With rights come responsibilities. The more people abuse our rights, the more likely that we're going to lose them," he said. " We need to start policing up our own actions, using common sense, in how we deal with others."
Attorney R. Mark Thomas, who represented the Muslim suspect, blamed Perce for the Halloween altercation. "The so-called victim was the antagonist," he told WHTM. "I think this was a good dressing down by the judge."
A blog post by the group American Atheists disagrees. "That a Muslim immigrant can assault a United States citizen," it says, "in defense of his religious beliefs and walk away a free man, while the victim is chastised and insulted... is a horrible abrogation."
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Filed under: Atheism • Halloween • Islam • TV-The Situation Room
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AverageJoe76
Ok, I'm a little conflicted with this one. Not sure of the case details (evidence provided/ witness testimony) to conclude that is was a good vs. bad decision. We should never condone violence as a response to free speech. BUT........ I don't support free speech that is intentionally harmful ( and trust me.... saying that is wierd for me because I LOVE free speech, and would not want to be hampered in ANY way). I believe the atheiest knew how much of an outrage that is to Muslims. Basically pushing a button to provoke a hot-button response. Was the Muslim correct to attack? Definitely not. This is actually a story of; "just because I can, does it mean I should...?" I'm not in bed with any religion, but I won't go out of my way to urinate on thier beliefs.
Average Joe 76, your statements "I don't support free speech that is intentionally harmful" and "I LOVE free speech and would not want [it] to be hampered in any way" are in conflict. Please decide which you support, as you cannot support both.
Phineas.... I started by stating I was conflicted by this case. I do support free speech. I just don't like the kind that is intentionally/ purposely harmful. Words DO hurt. Everyone knows that. One side of me wants people to speak their minds freely, the other half wants that free expression to go through a "common-sense" filter in the individual's head. It's painfully apparrent I cannot have it both ways. What would happen if the KKK put on a parade through South Side Chicago wearing black face shouting, "get out darkies" or something? I got a feeling it'll be a problem. Is it right they get attacked? Depends on who you ask; the moral one or the lawful one. Sometimes they don't mix, although we'd like them to.
BucketDrop
Huh, no other witnesses saw this event? The guy wearing the Christian costume didn't see it? The people on the streets didn't see it? Really? Something is fishy with this verdict.
BabbleOn
I believe that it's your cologn that you're smelling. 😉
Nonimus
I'm confused, the article states:
"Martin said. 'That's why I dismissed the case. Nothing as nefarious as what everyone's thinking, that I'm a Muslim or I'm biased. I'm actually a Lutheran.'"
But the AA site linked to in the article says:
"Hence forth comes the accusations of sensationalist journalism. I guess Judge Martin shouldn't have said he was one, then. I quote,
'I'm a Muslim, I find it offensive.'
Those are his words, as recorded and transcribed."
Armyguy1209
Bits of the audio were deleted by the atheist. Martin really said "IF I'm a muslim, I find it offensive."
I don't think it was edited, but I'm not sure the AA site heard the judge correctly, I'd like to see the court transcript.
I was only listening to the youtube version, so I may not have the best copy, but what I heard was "[unintelligble] I'm a Muslim, I'd be offended."
I know for a fact 100% it was edited. This atheist decided to turn everything against the judge because he didn't rule in his favor. It's very easy to tamper with audio. The recorder belonged to the atheist. He recorded it, and he turned everything against Martin. Martin is being criticized so much over everything, and really, he had the balls to make the correct ruling in court. The atheist is now going around like a big baby spreading lies and everything. And no, it's not "I believe" it was edited. it's "I know" it was edited. The judge is actually a Lutheran. He got a chance this morning to tell it how it is.
@Armyguy1209,
Don't be an as.s, please. I'm trying to get facts not wild accusations.
@Army1209,
p.s. I also agree about the "I'm a Muslim" phrase on the audio, it was garbled, but I didn't sound, to me, like the text version on the AA site.
DrewNumberTwo
"what kind of person... would do something so offensive and then be supposedly shocked at the result?" Rosa Parks?
"Finally, try putting yourself in the position of the Muslim guy. Maybe his emotions and beliefs are different from ours, but I find it fairly easy to imagine how stressed out he was. If what he was doing was trying to rip off the guy's costume, not beat on him, it becomes even more understandable." Would it have been okay to rip Rosa Parks' clothes off, as long as her sitting in the wrong seat stressed you out enough? People who are offended don't get special rights to do whatever they want.
I agree with the judge. I am an atheist, but believe that the Jesus was a great moral guide. I also firmly believe in the first amendment rights and our duty to fight for them. I was also in the service in the late 1970's. The judge was correct in his assessment. The jerk was purposely antagonistic. He was not protesting; he was searching for a fight. He was not making any statement other than what an idiot he is. He is lucky the judge didn't nail him for inviting a riot, though would have been a small riot, or disturbing the piece. If there was no evidence to prove this guy did it, he ruled correctly. So the jerk got chewed out. Lets wait for the waaambulance. Dude, put on your big boy pants and get on with life. I doubt that he is a true atheist. That is just an easy excuse for justifying his behavior. Shame on him.
Please make rules to tell people what to and what not to wear for halloween. will you?
Wear and do what you want. I don't care. Just don't come whinning to me for consequences suffered for doing something stupid.
Rickey, like so many others, you're asking to have it both ways. Either you support the right of freedom of expression, or you agree that offensive speech (particularly speech YOU find offensive) should be curtailed. Those two positions are mutually exclusive.
Islam is based in murder and oppression screw them
obviously you havent read the Koran. i am christian and have read the Koran out of curiosity. please bring facts to the table before spewing lies
re: want2believe
Why should he be dismissed?! The outcome of the case had nothing to do with this man's 1st amendment rights.
"District Judge Mark Martin dismissed the case, saying it was one person's word against another's, and that there was no other evidence or eyewitness testimony to prove that Elbayomy had harassed or touched the alleged victim."
"The commonwealth didn't present enough evidence to show me that this person is guilty beyond a reasonable doubt"
Did you miss that or just choose to ignore it?
I read the article. The judge, by his own description of events, lied. He concocted the verdict to cover his bases and sound legal.
Sounds like you will ignore the truth if the curtain is made opaque.
What the judge failed to remember:
What instigated the attack on the Twin Towers and the Pentagon?
And what drives today's 24/7 mosque/imam-planned acts of terror and horror?
The koran, Mohammed's book of death for all infidels and Muslim domination of the world by any means. Muslims must clean up this book removing said passages admitting that they are based on the Gabriel myth and therefore obviously the hallucinations and/or lies of Mohammed.
Until then, no male Muslim can be trusted anytime or anywhere..................................
Suuuure...and there is nothing violent in the bible.
that is the religion of piece?!
So your on the whole – 'IAll Muslims are terrorist' bandwagon" huh? The Koran does not teach terrorism. plain and simple. i am christian and have read the koran out of curiosity. i have many muslim friends also (i suppose that makes me some kind of traitor huh?). Every religion has their 'wackos'. Look at Jim Jones. The people that attacked the US on 9/11 werent true Muslims. they were terrorist that used religion as an excuse to do what they did .
o	On the koranic passages and world domination:
"Mohammed could not have known the size of the world, but several passages in the Koran show that he envisioned Islam dominating all of it, however large it might be: "He it is who sent his messenger . . . that he may cause it [Islam] to prevail over all religions´(Koran 9:33, M.M. Ali; see also 48:28 and 61:9). M.M. Ali designates these three passages as "the prophecy of the ultimate triumph of Islam in the whole world."
Mohammed's successors, the caliphs, quoted passages like these to inspire Muslim armies as they advanced out of Arabia, imposing Islam by the sword upon a peacefully unsuspecting Middle East and North Africa, as I described in the previous chapter.
Islamic armies, imbued with what Mohammed claimed was divine authorization, imposed Islam by force over vast areas, all the while extorting wealth from subjugated Jews and Christians to fund their ongoing conquests. As I noted, major defeats at Tours, France, in A.D. 732, and again at Vienna, Austria, in A.D. 1683, halted Islam's attempt to take all of Europe by force. Gradually Islamic forces were forced to retreat from Europe, except for part of the Balkans. But Islam has again set its sights on a conquest of Europe and of European civilization, wherever the latter has spread to North and South America and other regions. Muslim strategists ask their followers, Why do we find in these modern times that Allah has entrusted most of the world's oil wealth primarily to Muslim nations?
Their answer: Allah foresaw Islam's need for funds to finance a final politico-religious victory over what Islam perceives as its ultimate enemy: Christianized Euro-American civilization. So, Islam follows Nazism, fascism and communism as the world's latest hostile takeover aspirant.
Nazis, fascists and communists failed. Does Islam have a better chance at success? I believe it will flounder if we awaken to its threat in time; yet, if there is not adequate planned resistance, Islam does have a better chance of succeeding. Communism's world takeover attempt was guaranteed to fail because its economic policy was naively contrary to human nature. Advocating the rubric What is mine is thine, and what is thine is mine, communism failed to see that human nature will not keep those two balanced propositions in equilibrium. Like
a female black widow spider consuming her mate, the latter part of the formula makes a meal of the former, leading to the collapse of any system based upon that formula.
In contrast, political systems do well if they can persuade people to adhere to What's mine is mine and What's thine is thine maxims.
Only if a strong religious incentive is added does such an idealistic formula have any long-term chance. Even then success will be spotty. But communism (and Nazism, for that matter) excluded religion. And that mistake was the final nail eventually clamping a lid on communism's coffin. Communism, on a historical scale, perished while still in its childhood.
Islam is not repeating communism's mistake. Mating political cunning and incredible wealth with religious zeal, Islam does have a chance to succeed and will succeed unless major parts of the Western world unite to take appropriate countermeasures. But many Western leaders, unable to believe that a mere religion could possible be a serious political threat, keep proclaiming themselves as Islam-friendly, reasoning that all religions are good-aren't they?
A Muslim strategist in Beverly Hills, California, declared several years ago, as quoted by a friend of mine: "Now that the struggle between Western democracies and international communism is winding down, it is time for the real and final struggle to begin, and we are going to win!"
Actually, it is the Old Testament that lists the god-approved atrocities:
•Exodus 32: 3,000 Israelites killed by Moses for worshipping the golden calf.
•Numbers 31: After killing all men, boys and married women among the Midianites, 32,000 virgins remain as booty for the Israelites. (If unmarried girls are a quarter of the population, then 96,000 people were killed.)
•Joshua: ◦Joshua 8: 12,000 men and women, all the people of Ai, killed.
◦Joshua 10: Joshua completely destroys Gibeon ("larger than Ai"), Makeddah, Libnah, Lachish, Eglon, Hebron, Debir. "He left no survivors."
◦Joshua 11: Hazor destroyed. [Paul Johnson, A History of the Jews (1987), estimates the population of Hazor at ?> 50,000]
◦TOTAL: if Ai is average, 12,000 x 9 = 108,000 killed.
•Judges 1: 10,000 Canaanites k. at Battle of Bezek. Jerusalem and Zephath destroyed.
•Judges 3: ca. 10,000 Moabites k. at Jordan River.
•Judges 8: 120,000 Midianite soldiers k. by Gideon
•Judges 20: Benjamin attacked by other tribes. 25,000 killed.
•1 Samuel 4: 4,000 Isrealites killed at 1st Battle of Ebenezer/Aphek. 30,000 Isr. k. at 2nd battle.
•David: ◦2 Samuel 8: 22,000 Arameans of Damascus and 18,000 Edomites killed in 2 battles.
◦2 Samuel 10: 40,000 Aramean footsoldiers and 7,000 charioteers killed at Helam.
◦2 Samuel 18: 20,000 Israelites under Absalom killed at Ephraim.
•1 Kings 20: 100,000 Arameans killed by Israelites at Battle of Aphek. Another 27,000 killed by collapsing wall.
•2 Chron 13: Judah beat Israel and inflicted 500,000 casualties.
•2 Chron 25: Amaziah, king of Judah, k. 10,000 from Seir in battle and executed 10,000 POWs. Discharged Judean soldiers pillaged and killed 3,000.
•2 Chron 28: Pekah, king of Israel, slew 120,000 Judeans
•TOTAL: That comes to about 1,283,000 mass killings specifically enumerated in the Old Testament/Torah.
The New Testament has only one major atrocity, that of god committing filicide assuming you believe in this Christian mumbo jumbo. Said atrocity should be enough to vitiate all of Christianity.
snowdogg
"The judge also scolded Perce, saying he'd been needlessly provocative on an issue sensitive with Muslims."
Oh boo, hoo!
Judge Martin should be removed from the bench.
Troll Dave
Jesus and Mohammad are gay lovers everyone knows that!
I think the judge got it right. people siter the 1st ammendment and take it too far. sorry but this amounted to hate speech. i'm a christina but it seems today that everyone thinks its free game to attack, bash muslims. they think "its ok because the muslim religion is a terrorist religion'. this is bull! we all believe in the same God (sorry atheists). this kid was looking for trouble and publicity. it would have been the same if this kid would have walked around Harlem, NY or Compton, Ca. in a Klan outfit shouting 'I hate N*****s" . You think he's not looking for an @$$ whooping?
hey christina-how u doin?
"we all believe in the same God (sorry atheists)" I think every other religion in the world that isn't judaism, Christianity (including Mormonism), and Islam, would like to refute that comment (don't worry, it's only a few thousand denominations of faith with billions of believers).
Everything else you said was just patently wrong. Words don't equal an invitation for violence unless the expressly say so, just beause someone was offended doesn't make it okay. And the victim din't wander into a mosque wearing a zombie Muhammad costume, he was in a prade of other atheists, don't like atheists, you probably shouldn;t be watching.
True – i stand corrected. i was indeed talking about christian, islam, and judaism believing in the same god. sorry in being so general. See i can admit my mistakes 🙂
And guess what?! How many Clam marches in black or Jewish areas have been UPHELD as protected speech, NOT provocative?!! This is just one judge's choice to strike back against atheism. After all, the judge is a Lutheran.
ok so its a give and take. christians cant pray in public, have nativty displayed in public, no 10 commandments in public, and an atheist gets punched in the face. lol joking. in reality i dont believe in violence but this kid was just looking for 'you tube' fame. he could care less about atheist rites'.
Audio of judge's rant where he defends Islam and discusses Shariah law as though it has anything to do with someone's first amendment rights. He goes on to say he is Muslim and is offended by what the victim was doing.
http://www.youtube.com/watch?v=Bf11F3y9LOE&feature=youtu.be
If some bully like this Perce idiot went up to you and started saying nasty things about your wife with her standing there, and then you punched his lights out, should he be able to hide behind the first amendment? Same thing, only people want to get all stirred up because it involved Islam. The judge did exactly the right thing.
That moniker suits you well ;(
1- Yes he would, as long as when you ignored him or tried to walk away, or told him to leave you alone first he stopped.
2- He didn't run up to a Muslim man and start laughing at him, he was walking as part of a group when the man came after him.
If the husband assaulted him for expressing his freedom of speech rights, then yes.
If you punch somebody because of something they say, you belong in prison, end of story. You never have the right to put your hands on somebody over words.
So you are advocating physical violence in return for words of insult? Do you allow your second grader to punch the other second grader's lights out because he was called a poo-head? Really? If so, please learn how to not be this way.
Thanks PGA!
The other two are prime examples of forgetting what's right while attempting to follow the letter of the law. It's one of the things wrong with our country right now. We're being run by lawyers who write the laws in English that only another attorney can understand or interpret. Freedom of speech was intended to keep those that run the country (potential tyrants) from shutting us up. It was never intended to give us the freedom to insult each other without any retribution. Why anyone can't understand this is a mystery.
Actually, yes. It is legal to speak, even when what you have to say is nothing anyone else would like to hear. However, it is not legal to assault another person, no matter what they say to offend you. That's the law. That the atheist was deliberately provoking Muslims is irrelevant.
Do we realize that a small incident has been turned into a major incident.
The judge dismissed the case. He, then, gave his opinion which is his right as much it is our right to our opinions. Let's leave it at that and move on to more important issues facing this country.
Ray From New Orleans
It is not the judges right or duty to express his personal opinions or beliefs while sitting on the bench ,especially while giving a verdict.Judges are there to make sure due process and a fair and impartial hearing or trial takes place for all those involved.this particular judge's rant on the muslim world based on his views from serving in the military had absolutely no bearing on the case and should not have even been brought up much less used to chastise the victim with. I would suggest that this judge overstepped his bounds and trampled all over the first amendment rights of the victim.
theinfidel2
Judge Martin failed to see the irony in his own words: "With rights come responsibilities. The more people abuse our rights, the more likely that we're going to lose them," he was referring to the atheist but it is actually more applicable to the Muslim guy – HE is the one who abused our rights and HE should realize it is HIS responsibility to abide by our laws and NOT ATTACK someone whose views are offensive to him. Also, one of the rights lost – lost cuz the judge himself helped it to be lost – was the right to free speech even in cases when such speech is insulting. This judge should be disrobed simply on the basis of his own stupidity.
We need an amendment to the first amendment that clearly rewords "freedom of religion" to state "freedom of religion, and from religion"!
Wow, it is amazing how many people here see that Perce said he was attacked and automatically assume it was true. Did nobody else read the article all the way through? The judge dismissed the charges because the plaintiff couldn't supple sufficient evidence to prove there was an attack... for those of you not smart enough to understand, that falls under the phrase "innocent until proven guilty". Perce couldn't prove the attack happened.
Of course there was no evidence. The judge threw it all out – the video tape and the testimony of the arresting officer. The judge had no intention of ever hearing the case and simply wanted to make an example of an upity heathen.
Shane, true; but nothing you have said addresses what some of us see as the more critical point in all this–did the judge trample Pearce's 1st amendment protections when he dressed him down in open court?
Phineas – I would say no, Perce was purposefully trying to get a rise out of others, and was doing so by desecrating what a religion holds dear. Why are so many people upset because someone got scolded by a judge for being unprovokedly antogonistic? What happened to having respect for others and the beliefs of others? If you disagree, fine, don't go around trying to belittle them or ridicule them. In my opinion, Perce got what he deserved.
wilypagan
Actually it is completely untrue that there was no evidence of physical contact. I have listened to the audio recording of the proceeding which is available on the internet (although Judge Martin tried to prevent its release by threatening the victim with contempt of court). The victim testified as to the physical contact and the officer who interviewed the perpetrator immediately after the incident testified that the perpetrator had admitted the physical contact.
This judicial official bears careful watching. The time he spent overseas appears to have warped his understanding of American values. Who does he think he is to decide that the victim's expression was not protected under the First Amendment? There was no question of incitement here. The victim was walking in a public place and was approached by the perpetrator, and no one else.
Muslims, including Judge Martin (listen to the tape of the proceeding), need to learn that old saying we learned as kids "Sticks and stones can break my bones, but words can never hurt me". Until Judge Martin re-learns American values, he has no place in this great nation, much less on the bench.
The judge didn't need to at all go on about Shariah law, but even then he still wasn't trampeling any 1st amendment rights.
Shane, read very carefully what you have written: "...if you disagree, fine. Don't go around trying to belittle them or ridicule them..." That is precisely what the judge did in open court to the alleged victim. When an agent of the state–in this case, a judge–tells someone his protected actions were wrong, it has what is commonly referred to as having a "chilling" effect on speech, expression, etc. And that is a trampling of rights, rights that very same agent of the state is sworn to uphold.
Caveman
I will say this. In no way shape or form does this have to do with religion or beliefs. This boils down to "assault". The second that muslim put his hands on that man, for whatever reason "OTHER THAN SELF DEFENSE" it bacame felonious assault, and shame on the judge for allowing it. Secondly, is that to say that if we attend a muslim parade and physically attack someone for their rants, that the person who attacked wouldn't get charged with assault. C'mon... You and I both know that this has nothing to do with religion and has everything to do with the act itself, which was felonious in this country, regardless of the ridicule of a religious figure. They burn the American flag all the time in other countries, so what.. do we beat them down for it? .. No.
This judge should not be a judge for he has issue making good judgement. This is USA, if we can make fun of president, why can't we make fun of Muslim or atheist? Even we do not like to be making fun of, violence is not the solution and be dismissed.
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Social innovations to watch out for | September 2021
Social media never stops evolving, and each new month welcomes brand-new updates across all of the beloved social media platforms. For your brand or clients, you need to know about them… Here are our favourite innovations for September 2021
1. TikTik launches 'Creative Solutions': a guiide to building effective Tick Tok campaigns
As the rise of TikTok continues, having recently overtaken Facebook as the most downloaded social media app in the world, many more marketers are now looking to utilise this platform to help boost awareness of their brand.
To support the influx of marketers to the platform, TikTok has recently launched a 'Creative Solutions' guide to help businesses find success by providing a breakdown of best practices as well as tips and tricks to support the utilisation of the app.
2. Twitter launches eCommerce pilot in US that allows users to shop from their profile
Twitter has recently launched a new "Shop Module' that will offer brands the opportunity to showcase their products to users directly from their business profile. In this module, users are now able to scroll through a carousel format of product images and tap through on a product they're interested in purchasing, linking to the brand's website within the Twitter platform itself.
Whilst Twitter is in early explorations of this new feature with it only available in beta to a handful of brands in the US, we hope to see it being implemented internationally soon for all ecommerce brands.
3. Instagram introduces Reel Ads
After successfully testing Reel ads in select countries, Instagram has officially launched this feature to all users. The new ad placement aims to help businesses reach wider audiences by providing a new place for people to discover inspiring new content from brands and creators. According to Instagram, "Reels [are] the best place to reach people who don't follow you and a growing global stage where brands and creators can be discovered by anyone." Alongside this exciting new ad space, reels have also been extended from 30 to 60 seconds, providing more time to showcase your messaging.
4. Snapchat launches 'Snapchat Trends'
Snapchat has recently launched a new marketing tool, much like Google Trends, but with a focus on Snapchat data to provide marketers valuable insight into the platform's trends.
The new feature "highlights the most popular keywords shared with the Snapchat community via public Stories and My Stories that have been viewed by a large group, in addition to a searchable database of terms. With these insights, Snap is able to provide a new level of visibility into organic chatter on Snapchat, looking at what our community is talking about and showcasing popular Snaps that exemplify each trending topic." Not only will the new tool provide valuable insight into the Snapchat generation, it will provide businesses an opportunity to learn more about the Snapchat community and the key topics they care about.
What this new tool will include:
Product Market Research – using data trends to evaluate overall market fit for new products.
Copywriting – as new lingo emerges from Snapchat users, their data trends will assist in writing contextually relevant copy for your audiences.
Competitor Research – helping you to understand the market that your brand and products operate in.
5. Linkedin adds new native video meeting feature
After a year of facilitating third-party video providers, Linkedin has now rolled out its own native video option to provide further ways for users to connect within the platform.
This new feature aims to make it even easier for users to connect remotely and to enhance discussions further. Following a slow implementation of visual tools such as stories and live-streaming to compete with other platforms, this feature will provide users with more accessible ways of communicating.
Interested in how one (or more!) of these social innovations could help your brand reach more people on social media? Bring Digital Cake into the mix and send us a message
Interested in how one (or more!) of these social media innovations could help your brand reach more people on social media? Bring Digital Cake into the mix and send us a message.


On Sunday we can expect sunny weather. Morning temperatures will be around freezing point and will reach a maximum of 20 °C during the day. Low temperatures at night will be around 0°C. Southwesterly winds are weak.
On Monday we can expect sunny weather. Morning temperatures will be around freezing point and will reach a maximum of 17 °C during the day. Low temperatures at night will be around 2°C. Southwesterly winds are weak.


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Photography ©
Voscar, The Maine Photographer.
Maine Crafts
Maine Lodging
Around Maine
Maine's Best in Whitewater Rafting
It doesn't take long after you've gotten into a raft to realize this is an exciting fun filled way to spend a day, or more if you opt for an extended trip. The first rapid, Rock Garden, on the Kennebec is just the start of a heart stopping several mile series of rapids that intensifies in its strength and power until you drop over Magic Falls. But that is far from the end of the trip. No, there is still more white water to come along with calm sections where you can get out of the raft and frolic in the river. Better yet, that is just one of the three rivers Rafting Outfitters based in the Jackman Region and Millinocket - Baxter State Park area run.
The last great log drive in Maine, on the Kennebec River in 1975, ended one era and began another. Prior to 1975, there was little recreational use of Maine's wilderness rivers, except for the occasional angler. After that however, a few adventurers began taking passengers down the Kennebec in wooden boats call dories, and the phenomenally popular sport of whitewater rafting began in Maine.
How times have changed. From a few hardy people who thought it was fun to run the rivers, to today, when there is a concentration of whitewater rafting companies in the Jackman-Forks and Millinocket - Baxter State Park areas that run thousands of people during the course of a season. Daily trips are scheduled for the Kennebec, Dead and Penobscot Rivers. And, Maine is the only state in the northeast that can guarantee white water for an extended season - from late April to mid-October, due to dam controlled release.
The 12 mile Kennebec River trip begins on Indian Pond and roars for several miles of up to class IV rapids through the spectacular Upper Kennebec Gorge. The last part of the trip includes smaller rapids and good spots for a swim.
The Dead River provides the longest stretch of continuous whitewater in the East. The 16 mile trip begins at Grand Falls and runs through class IV and V rapids to the confluence of the Kennebec River in The Forks.
There are a limited number of big water releases on the Dead. Sport book up early for this trip. So call the outfitter ahead of time to reserve space. There are several smaller raft releases later in the year.
September is a fantastic time to take a Dead River run. Not only is the water fun, but the foliage is out of this world.
The first two miles of the Penobscot River Trip descend Class V rapids from McKay Station through the breathtaking Ripogenus Gorge ending with a run through the Cribworks. There is little time to practice the moves your guide will instruct you in before hitting Exterminator and the Staircase. The remaining 12 miles of Class III and IV rapids stretches from Big Eddy to Pocwockamus Falls. Although everyone talks about the upper part of the river, there are surprises around every bend. This is an action packed trip from start to finish. Moose and bald eagles are commonly seen feeding on the river with an occasional black bear making an appearance.
Anyone who enjoys the outdoors and is looking for adventure will be delighted by whitewater rafting. People of all ages and athletic abilities, with no previous experience raft the rivers. Minimum suggested age is 12 on the upper Kennebec and eight on the lower section; 15 on the Upper Penobscot and 10 on the lower, and 15 is the minimum age on the Dead River.
Everything you need is provided by the outfitter. Transportation, life jackets (PFDs), helmets and paddles, a pre-trip orientation and safety meeting, and a hearty, open-grill lunch are all included.
SOME TRIP SUGGESTIONS
Kennebec River -- One of Maine's most popular adventures. A full day of awesome whitewater and tranquil beauty invigorates and inspires you. Begin by plunging into the gorge with it's towering canyon walls. The upper section is all action with one standing wave after another. Professional guides will navigate you through infamous rapids like 'Big Mama', 'Whitewash,' and 'Magic Falls.' Most outfitters provide a hearty riverside lunch, after which you will be invited to kick back, float, and relax, or maybe try your hand running the rapids of the lower section in an inflatable kayak. The Kennebec offers class III-V whitewater all season long thanks to a daily average dam release of 5,000 cubic feet per second. (cfs) Spring offers the biggest water and lowest rates. Summer is the most popular season offering warm water and big waves. (Book early as the State of Maine limits the number of passengers, to preserve this wild resource from over use). Fall brings the return of lower rates and brilliant colors. So no matter when you go, this 14 mile river trip offers great rafting for first timers and seasoned paddlers alike. Most outfitters have a minimum suggested age requirement for the gorge of 13 years. Younger participants and those interested in class III whitewater can board the boats below the gorge in time to join the others for lunch.
Dead River -- Rafting the Dead is a unique privilege thanks to scheduled dam releases only selected dates. This exhilarating class III-V run is just plain fun, with16 miles of almost nonstop action including memorable rapids like 'Haydens Landing' and 'Poplar Falls.' Most outfitters feed you a snack beforehand and a hearty lunch at the end of the journey. Again, most outfitters have a minimum suggested age of 15 years for The Dead River.
The Kennebec and Penobscot Rivers offer fun and exciting class 4 releases every day from mid-April through mid-October. In addition, there are special release dates scheduled on the Kennebec and Dead Rivers for 2014 are below:
Dead River Release Dates
Saturday, May 3, 2014 - 7000+ CFS (Cubic Feet Per Second)
Saturday, May 10, 2014 - 7000+ CFS (Cubic Feet Per Second)
Sunday, May 25, 2014 (Memorial Day Weekend) - 5500 CFS (Cubic Feet Per Second)
Saturday, May 31, 2014 - 5500 CFS (Cubic Feet Per Second)
Saturday, June 7, 2014 - 3500 CFS (Cubic Feet Per Second)
Sunday, June 8, 2014 - 2400 CFS (Cubic Feet Per Second)
Saturday, July 19, 2014 - 2400 CFS (Cubic Feet Per Second)
Saturday, August 9, 2014 - 2400 CFS (Cubic Feet Per Second)
Sunday, August 31, 2014 (Labor Day Weekend) - 5500 CFS (Cubic Feet Per Second)
Saturday, September 13, 2014 - 3500 CFS (Cubic Feet Per Second)
Saturday, October 4, 2014 - 6000 CFS (Cubic Feet Per Second)
Kennebec Turbine Test High Water
Saturday, June 14, 2014 - 8000 CFS (Cubic Feet Per Second)
Monday, July 7, 2014 - 8000 CFS (Cubic Feet Per Second)
Saturday, September 6, 2014 - 8000 CFS (Cubic Feet Per Second)
Supporting Whitewater Rafting Businesses
Bishop's Motel - Located in Jackman, Maine
Bishop's Country Inn Motel is becoming a very popular business choice, offering wireless internet and large in-room tables to work from. All rooms include a full working phone system (cell phone use in this area is very limited), refrigerators, air conditioning, microwaves, coffee makers, irons with ironing boards and remote control color flat screen TV. We try to surpass all your expectations with all the comforts of home as well as knowledgeable, friendly service. Relax in a spacious room that is spotlessly clean. We are centrally located in the town of Jackman within walking distance to three local restuarants.
Cabins on Moosehead Lake - Moosehead Hills Cabins - Located in Greenville, Maine
You have a choice of two beautiful lakes when you visit us at Moosehead Hills Cabins. Our log cabins at our home base are nestled on 50 acres of woods and nature trails with waterfront on Moosehead Lake. We also have a new log cabin with its own private dock and waterfront on idyllic Wilson Pond, a 7-mile long lake in Greenville. Each of our log cabins has either two or three bedrooms, a kitchen, a fireplace, and a full bath or whirlpool tub. Enjoy our beautiful sunsets, and the views of the lake and mountains from the deck of your log cabin.
North Country Rivers - Located in Bingham, Maine
North Country Rivers offers Deluxe four season Riverside Log Cabin Rentals in Bingham, Maine. Fully furnished with a Full Kitchen, Bathroom, Private Bedrooms, Living Room, Dining Area, Cable TV, Heat and AC. Recreational opportunities include Whitewater Rafting, Kayaking, Moose Safari's, ATV Trails, Hiking, Biking, Fishing, Hunting, Snowmobiling and much more. All credit cards accepted, call 1-800-348-8871 today to start planning your Maine Outdoor Vacation!
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