text:
\section{Introduction}
Particles start from the points of a point process in $\mathbb{R}^d$ and perform
i.i.d.\ stochastic motions. Each particle carries a set (detection set) along
(typically, a ball of fixed radius). The goal is to find the time
until one of the particles detects a fixed set $K$.
This is a paper that should have been written long ago. Its first version,
\cite{KKP03} appeared in the proceedings of an obscure conference without many
proofs. After actually working out the Brownian case, we discovered
that Spitzer \cite{SPI} had done most of the job for $d=2$ or $3$.
A few years after that, we posed some open questions
\cite{K4B}, some of which were resolved
by Peres {\em et al.} \cite{PSSS}.
The purpose of this paper is twofold.
First, because we think that an overview of the topic is needed.
Second, because there are several interesting questions that can be asked.
This overview contains some new elements too. For example,
it contains a method for computing the exact distribution of the
detection time in odd dimensions and asymptotics of the
distribution in even dimensions.
We also compare what happens when we run the particles a Brownian motions
vs.\ linear motions with random speeds.
Roughly speaking, Brownian motions discover small objects
faster than linear motions.

The paper is organized as follows. First, we give an overview of
the general Boolean model. Then we pass on to the dynamic case.
We then pose the detection problem and provide some general formulas
involving capacity notions. The inertial Boolean model (particles move on
straight lines with constant speed) is presented next.
The Brownian Boolean model is then analyzed in detail.
We conclude with some open problems.
The first part of the paper, i.e., up to and including Section 4, is
largely an overview but with organized notation chosen so that the
model can be explained in much greater generality than the ``solvable'' one
(the Brownian Boolean model)
Section 5 is new (inertial Boolean model), but quite simple.
Section 6 deals with the Brownian Boolean model and explains the new
formulas regarding the distributions of detection times and their expectations,
separately in even and odd dimensions.

Throughout the paper we use the following notations:
$f(x) \sim g(x)$, as $x \to \infty$, means
$f(x)/g(x) \to 1$, as $x \to \infty$. Similarly, when $x$ tends to
another value. Also, $f(x)
\stackrel{\log}{\text{\Large $\sim$}}  g(x)$
means $\log f(x) \sim \log g(x)$.
The closed unit ball in $\mathbb{R}^d$ centered at the origin is denoted by $\mathbb B$.
The Euclidean norm of $x \in \mathbb{R}^d$ is denoted by $|x|$ and the $d$-dimensional
Lebesgue measure of a Borel set $A \subset \mathbb{R}^d$ is denoted as $\operatorname{vol}(A)$.
In particular, we let $\omega_d := \operatorname{vol}(\mathbb B)$, keeping in mind that
\[
\omega_d = \frac{\pi^{d/2}}{\Gamma(\frac{d}{2}+1)}, \quad d \in \mathbb{Z}_+,
\]
which gives
\[
\omega_{2n} = \frac{\pi^n}{n!}, \quad
\omega_{2n+1}= \frac{2^{n+1} \pi^n}{(2n+1)!!}, \quad
n \in \mathbb{Z}_+,
\]
where  $(2n+1)!! = 1 \cdot 3 \cdot 5 \cdots (2n+1)$.
When $A, B$ are sets (perhaps random) in $\mathbb{R}^d$, then $A\pm B$ is their Minkowski
sum (difference).
For example, $A$ could be a singleton and $B$ the trajectory of a stochastic
process. The modified Bessel function of the second kind and of order $\nu \in \mathbb{R}$
is denoted by $K_\nu$. It admits the integral representation
    \[
K_\nu(z) = \int_0^\infty \exp(-z\cosh t) \cosh(\nu t) \,\text{d} t,
\quad \operatorname{Re}(z) > 0.
\]
Alternative formulas for $K_\nu$ when $\nu$ is a half-integer
are in the appendix.
We remark also that the term ``Boolean model'' is not standard. Some people
refer to it as germ-grain model and allow the underlying point process to
be general. In this paper, we reserve the term ``Boolean model'' for the
case when the underlying point process is (homogeneous) Poisson
and use ``general Boolean model'' for the case where the point process is more
general.

\section{A little background on the general Boolean model}
The general Boolean model, also known as the germ-grain model
\cite[Ch.\ 4]{SW08},
 is one the basic objects of study of stochastic geometry.
Consider a random point process $\Phi$ on $\mathbb{R}^d$ (the ``{\em germs}'')
and a random compact set $G$ (a ``{\em grain}'').
Conditionally on $\Phi$, let $\{G_x, x \in \Phi\}$ be i.i.d.\
copies\footnote{We think of $\Phi$ both as a random discrete set,
and as a random point process. Thus $\{x\in \mathbb{R}^d: x \in \Phi\}$ stands for
the random set or for the support of the random measure. When $B$ is a Borel
subset of $\mathbb{R}^d$ we use $\Phi(B)$ to denote the value of the
random measure at $B$, i.e., the number of points of the random set
in $B$.} of $G$.
The {\em general Boolean model} is then defined
by\footnote{When $A, B \subset \mathbb{R}^d$, we let
$A+B := \{a+b:a\in A, b \in B\}$ (Minkowski addition).
Similarly, $A-B:=\{a-b:a\in A, b \in B\}$.
We let $x+B := \{x\}+B$.}
\begin{equation}
\label{booleanmodel}
\Xi:=\bigcup_{x \in \Phi} (x+G_x).
\end{equation}
(If $G$ is a deterministic set, then $\Xi$ can also be expressed as
$\Xi = \Phi+G$.)
Alternatively, if $\mathcal K$ is the collection of compact subsets of $\mathbb{R}^2$,
we may consider the point process
\begin{equation}
\label{mppN}
\mathcal N:= \{(x, G_x): x \in \Phi\}
\end{equation}
as a discrete random subset of the product space $\mathbb{R}^d \times \mathcal K$.
In fact, it is also a marked point process because to each $x$ there is
a unique $G_x \in \mathcal K$ such that $(x, G_x)$ is an element of the random
set $\mathcal N$. Putting an appropriate probability measure on the set of
marked point processes on $\mathbb{R}^d \times \mathcal K$ gives another
way of constructing a general Boolean model.

The {\em capacity functional} of the general Boolean model is defined as
\[
\operatorname{\sf T}_{\Xi}(K) := \mathbb P(K \cap \Xi \neq \varnothing),
\]
for $K$ a compact set.

The capacity functiona
 is a fundamental object in the theory of random
sets \cite{MOL10}.
If $X$ is a random locally compact subset of $\mathbb{R}^d$ then
\[
\operatorname{\sf T}_X(K) := \mathbb P(K \cap X \not = \varnothing)
\]
is defined on compact sets $K$.  It is submodular, i.e.,
$\operatorname{\sf T}_X(K_1 \cup K_2) \le \operatorname{\sf T}_X(K_1) + \operatorname{\sf T}_X(K_1) - \operatorname{\sf T}_X(K_1 \cap K_2)$
and upper semicontinuous, i.e., $\operatorname{\sf T}_X(K_n) \downarrow \operatorname{\sf T}_X(K)$
whenever $K_n$ is a decreasing sequence of compact sets with
$\bigcap_n K_n = K$.
An example of a capacity functional is the one given above. Another
example is obtained by considering a Brownian motion $\xi:=\{\xi(t),
t \ge 0\}$ in $\mathbb{R}^d$ started from some $\xi(0) \in \mathbb{R}^d$.
Letting
\[
\xi^{(t)} := \big\{\xi(s),~ 0 \le s \le t\big\}
\]
be the initial segment of $\xi$ up to time $t$, we have
$\operatorname{\sf T}_{\xi^{(t)}}(K) = \mathbb P(\xi^{(t)} \cap K \neq \varnothing)$.
If we set
\[
T_K := \inf\{t \ge 0: \xi(t) \in K\},
\]
we obtain
\[
\operatorname{\sf T}_{\xi^{(t)}}(K) = \mathbb P(T_K \le t).
\]
Thus, $t \mapsto \operatorname{\sf T}_{\xi^{(t)}}(K)$
is the distribution function of the random variable $T_K$.

Going back to the general Boolean model, we observe that
if (the law of) $\Phi$ is invariant under translations and ergodic, and
if $\operatorname{vol}$ denotes the Lebesgue measure on $\mathbb{R}^d$, then
$C_{\Xi}(\{0\})$  is the a.s.\ (and in $L^1$) limit of
$\operatorname{vol}(\Xi \cap [-h,h]^d)/h^d$, as $h \uparrow \infty$ and is
known as the {\em volume fraction} of $\Xi$.

An important particular case is when $\Phi$ is a {\em spatially homogeneous
Poisson process} with intensity $\lambda$. In this case, $\Xi$ is referred to as
a Boolean model\footnote{The terminology
is highly nonstandard. Some authors use the term ``Boolean model''
or ``germ-grain model'' for what we called ``general Boolean model''.
We decided to reserve the term ``Boolean model'' for
the case where $\Phi$ is a Poisson process.}
\cite{SKM08}.
The Boolean model has several computational advantages.
For example, owing to the thinning property of a Poisson process,
the set of points $x \in \Phi$ such that $x+G_x$ intersects a given
closed set $A$,
\[
\Phi_A:=\{x \in \Phi:~ A\cap (x+G_x) \neq \varnothing\},
\]
forms an inhomogeneous Poisson process in $\mathbb{R}^d$ with
intensity measure
\[
\mathbb E \Phi_A(\text{d} x) =
\lambda_A(x)\, \text{d} x = \lambda \, \mathbb P(A \cap (x+G) \neq \varnothing)\,\text{d} x.
\]
Indeed, conditionally on $\Phi$, the random
variables $\big\{ {\text{\Large $\mathfrak 1$}}(A \cap (x+G_x) \neq \varnothing),\, x \in \Phi\big\}$
are independent and ${\text{\Large $\mathfrak 1$}}(A \cap (x+G_x) \neq \varnothing)
\stackrel{\text{(d)}}{=} {\text{\Large $\mathfrak 1$}}(A \cap (x+G) \neq \varnothing)$, for all $x \in \Phi$.
Since $A\cap (x+G) \neq \varnothing$ is equivalent to
$(A-x) \cap G \neq \varnothing$, we hav
\[
\lambda_A(x) = \lambda \operatorname{\sf T}_G(A-x).
\]
Notice also that if $\Xi$ is a Boolean model then the marked
point process $\mathcal N$ introduced in \eqref{mppN} is a Poisson
process on $\mathbb{R}^d \times \mathcal K$ with intensity measure
$\lambda\, \text{d} x\, \mathbb P(G \in \text{d} g)$ on the space
$\mathbb{R}^d \times \mathcal K$.
Conversely, given a finite measure $\mu(\text{d} g)$ on $\mathcal K$, we can
construct a Poisson process $\mathcal N$ on
$\mathbb{R}^d \times \mathcal K$ with intensity measure $\text{d} x\, \mu(\text{d} g)$.
The projection of $\mathcal N$ on $\mathbb{R}^d$ is the set of germs and
the projection on $\mathcal K$ is the set of grains.

In applications of stochastic geometry, it is sometimes the case that the grains
depend on a parameter $t$ (time) and increase with $t$.
The general mobile Boolean
model has a time parameter $t$ which principally affects the locations
of the germs.

\section{The general mobile Boolean model}
\subsection{Definitions}
In its most general form, the general mobile Boolean model is defined
in terms of a germ point process $\Phi$ on $\mathbb{R}^d$ and of
a time-dependent grain process $G=\big\{ G(t), t \ge 0 \big\}$.
The latter is assumed to be a stochastic process with values in
$\mathcal K$ and continuous sample paths, when $\mathcal K$ is
equipped with the topology induced by the Hausdorff metric.
Conditionally on $\Phi$, let $\big\{ G_x, x\in \Phi \big\}$ be i.i.d.\ copies
of $G$ and define, for each $t \ge 0$, the general Boolean model
\[
\Xi(t) = \bigcup_{x \in \Phi} (x+G_x(t)).
\]
The {\em general mobile Boolean model} is the
{\em process} $\Xi=\big\{
\Xi(t), t \ge 0 \big\}$.
Physically, we think of particles located at points $x \in \Phi$ at time
$0$, each having a grain $G_x(0)$ ``around it''. At time $t$, point $x$
moves to a new position and, as a result, the grain around it becomes
$x+G_x(t)$.
Notice that the set
\[
\mathcal W(t) := \bigcup_{0 \le s \le t} \Xi(s)
\]
represents everything covered by the general Boolean model up to time $t$
and is itself too
a general Boolean model with grains the points of $\Phi$
and germs being i.i.d.\ copies of
\[
G^{(t)}:=\bigcup_{0 \le s \le t} G(s).
\]
Indeed,
\begin{equation}
\label{Wexpression}
\mathcal W(t) = \bigcup_{x \in \Phi} (x + \bigcup_{0 \le s \le t}
G_x(s)) = \bigcup_{x \in \Phi} (x + G_x^{(t)}).
\end{equation}
Note that $\mathcal W(t)$ is increasing  in $t$.

\subsection{The canonical form}
\label{cfsection}
The canonical way of introducing the distribution of the grain process
$\big\{ G(t), t \ge 0\big\}$
is by considering a specific point $h(G(0))$ of $G(0)$
as its ``center'' (where, formally, $h$ is a measurable function from
$\mathcal K$ into $\mathbb{R}^d$).
Then
\[
\xi(t) := h(G(t)), \quad t \ge 0,
\]
describes the motion of the center.
We may thus describe the law of $\big\{ G(t), t \ge 0\big\}$, in two steps.
First by specifying the law of $\big\{ \xi(t), t \ge 0 \big\}$ and then,
conditionally on $\big\{ \xi(t), t \ge 0 \big\}$, by specifying the law
of
\[
D(t):=G(t)-\xi(t),\quad t \ge 0.
\]
(In the simplest case, we may assume that $\big\{ D(t), t \ge 0\big\}$
is independent of $\big\{ \xi(t), t \ge 0 \}$.)
We can then write
\[
\Xi(t) = \bigcup_{x \in \Phi} (x+\xi_x(t) + D_x(t)),
\]
where $\xi_x(t) = h(G_x(t))$, $D_x(t)=G_x(t)-\xi_x(t)$.
The two representations for $\Xi(t)$ are absolutely equivalent. Notice,
however, that  if we let
\[
\Phi(t) :=\big \{x + \xi(t):~ x \in \Phi \big \}
\]
we can write, with a slight abuse of notation as regards the
indices of $D_x(t)$,
\[
\Xi(t) = \bigcup_{x \in \Phi(t)} (x + D_x(t)),
\]
and think of $\Xi(t)$ as a general Boolean model with
germs the points of $x\in \Phi(t)$ and grains $D_x(t)$.

In the simplest case, $\Phi$ is a homogeneous Poisson process
with intensity $\lambda$,
and $\{D(t), t \ge 0\}$
is independent of $\{\xi(t), t \ge 0\}$.
Then $\Phi(t)$ is again a homogeneous Poisson process
with intensity $\lambda$
Thus, if $\Xi(0)$ is a Boolean model and the trajectory of the center
is chosen independently of $D$, then,
for each $t>0$,
$\Xi(t)$ is also a Boolean model identical in distribution to $\Xi(0)$.

Note: Without further ado we shall, henceforth, define a mobile
Boolean model to be one for which the germ point process $\Phi$ is
homogeneous Poisson and the independence between $\xi(\cdot) = h(G(\cdot))$
and $D(\cdot)$ holds.
Dropping the adjective homogeneous gives an inhomogeneous mobile Boolean model.
An inhomogeneous mobile Boolean model is arguably a good model for
a sensor network. Sensing devices are initially located at the points $x$ of
the inhomogeneous Poisson process $\Phi$ and move according to independent
random motions $x+\xi_x(t)$ (assuming $\xi_x(0)=0$). The set $D_x(t)$
represents the part of space which can be sensed by the sensor at time $t$.
If the problem is to discover an unknown target, then inhomogeneity in $\Phi$
allows for the possibility of incorporating prior information about
the location of the target. Randomness in $D_x(t)$ may model the
different sensing abilities of the devices. And time-dependence
in $D_x(t)$ allows for modeling of loss  (or gain) of energy of the device.
Finally, randomness in the trajectories is natural too.

\section{The detection problem}
We now consider the problem of finding the distribution of
the first time that a general mobile Boolean model will detect a fixed
compact set $K$. The expressions \eqref{detGENERAL} and \eqref{detINHOMmbm}
below concern, respectively, a general mobile Boolean model and an inhomogeneous
mobile Boolean model.
The expressions \eqref{detHOMmbm}, and \eqref{detHOMmbmCANON}
concern both a homogeneous mobile Boolean model. The last
one relates the distribution of the detection time of $K$
by to the distribution of the first hitting time of a set
by the process $x+\xi(t)$ representing the
random motion of a sensing device initially located at the point $x$.

\subsection{Detection time for a general mobile Boolean model}
Let $K$ be a compact subset of $\mathbb{R}^d$ and let
\[
S_K := \inf\{t \ge 0:~ K \cap \mathcal W(t) \neq \varnothing\}.
\]
This is called the {\em detection time of the set $K$}.
We are interested in deriving information about the law of $K$.
Under natural assumptions on the law of $\Phi$ (e.g., if $\Phi$
is a Poisson process), and
if the grain $G$ has nonempty interior with positive probability,
then, due to compactness, the probability that $K$ is contained in
$\mathcal W(0)$ is positive.
Hence $\mathbb P(S_K=0)$ is typically (e.g., under the previous assumptions) positive.
By the monotonicity of $\mathcal W(t)$,
\begin{equation}
\label{detGENERAL}
\mathbb P(S_K \le t) = \mathbb P(K \cap \mathcal W(t) \neq \varnothing)
= \operatorname{\sf T}_{\mathcal W(t)}(K),
\end{equation}
and, by the expression \eqref{Wexpression} for $\mathcal W(t)$,
\begin{align*}
\operatorname{\sf T}_{\mathcal W(t)}(K)
&= \mathbb P\big(\bigcup_{x \in \Phi} K \cap (x+ G^{(t)}_x) \neq \varnothing\big)
\\
&= \mathbb P(\exists x \in \Phi  ~ K \cap (x+ G^{(t)}_x) \neq \varnothing).
\end{align*}
So, if we consider the point process
\begin{equation}
\label{thinnedpoisson}
\Phi_K:=\{x \in \Phi: ~ K \cap (x+ G^{(t)}_x) \neq \varnothing)\},
\end{equation}
we have
\[
\operatorname{\sf T}_{\mathcal W(t)}(K) = \mathbb P(\Phi_K \neq 0),
\]
which is the probability that  for some point $x$ of $\Phi$
the set $x+G_x^{(t)}$ intersects $K$.
This point process depends
on the general Boolean model in a rather complicated way.

\subsection{Detection time for a (possibly inhomogeneous) mobile
\label{dtimbm}
Boolean model}
Things become simple in the case of an inhomogeneous mobile Boolean model
where $\Phi$ is a Poisson process with intensity measure $\lambda(\text{d} x)$.
In this case, arguing as earlier, $\Phi_K$ is also a Poisson process
with intensity measure
\[
\mathbb P(G^{(t)} \cap(K-x) \neq \varnothing)\, \lambda(\text{d} x)
=\operatorname{\sf T}_{G^{(t)}}(K-x)\, \lambda(\text{d} x),
\]
and thus,
\begin{equation}
\label{detINHOMmbm}
\mathbb P(S_K \le t )= \operatorname{\sf T}_{\mathcal W(t)}(K) = 1-\exp
\bigg(-\int_{\mathbb{R}^d} \operatorname{\sf T}_{G^{(t)}}(K-x)\, \lambda(\text{d} x)\bigg).
\end{equation}
Since $(G^{(t)} + D) \cap (K-x) \neq \varnothing$
iff $x \in K-D-\xi(s)$ for some $s \le t$, we have,
by Fubini's theorem,
\begin{equation}
\label{byfubini}
\mathbb P(S_K > t)
=
\exp\bigg(-\mathbb E\, \lambda\bigg( \bigcup_{0 \le s \le t} [K-D-\xi(s)]\bigg)\bigg)
\end{equation}
If $\Phi$ is a spatially homogeneous Poisson process with
\[
\lambda(\text{d} x) = \lambda \cdot \text{d} x,
\]
then
\begin{equation}
\label{detHOMmbm}
\mathbb P(S_K > t) =
\exp\bigg(- \lambda\, \mathbb E\operatorname{vol}\bigg( \bigcup_{0 \le s \le t} [\xi(s) + D-K]
\bigg)\bigg).
\end{equation}

We point out that the sets involved in this union
are formed by translating the set $D-K =\{x-y: x\in D, y \in K\}$
by vectors $\xi(s)$.
Put it otherwise, a particle performing motion $\xi$ carries
a neighborhood with shape $D-K$. The set swept by
the particle up to time $t$ can be called the ($D-K$)-{\em sausage} of
$\xi$ up to time $t$. The term {\em Wiener sausage} is reserved for the
case when $\xi$ is a Brownian motion.

\subsection{Detection time for a homogeneous
mobile Boolean model in its canonical form}
Suppose now that
\[
G(t) = \xi(t) + D,
\]
where $D$ is a fixed deterministic compact set, for instance a closed ball,
and $\{\xi(t),\, t \ge 0\}$
a random process with continuous sample paths and $\xi(0)=0$;
see Section \ref{cfsection}.
Then
\[
G^{(t)} = \bigcup_{0 \le s \le t} (\xi(s) + D) = \xi^{(t)}+D,
\]
and
\[
G^{(t)} \cap (K-x) \neq \varnothing
\iff
(\xi^{(t)}+D) \cap (K-x) \neq \varnothing
\iff
(x+\xi^{(t)}) \cap(K-D) \neq \varnothing.
\]
Therefore, if we define the first hitting time
\[
T^x_B := \inf\{t \ge 0:\, x+ \xi(s) \in B\}
\]
of a closed set $B$ by the process $x+\xi(\cdot)$, we have
\[
\operatorname{\sf T}_{G^{(t)}}(K-x) = \mathbb P(T^x_{K-D} \le t).
\]
Using \eqref{detINHOMmbm} and \eqref{detHOMmbm} we  obtain
\begin{equation}
\label{detHOMmbmCANON}
\mathbb P(S_K > t) = \exp\bigg(-\lambda \int_{\mathbb{R}^d} \mathbb P(T^x_{K-D} \le t) \,\text{d} x\bigg).
\end{equation}
In particular, if $T^x_{K-D}$ has density $f^x_{K-D}(t)$ then
the hazard rate $h_K(t)$ of $S_K$, defined by
\[
\mathbb P(S_K \le t+\delta \mid S_K > t) = h_K(t)\delta+o(\delta),
\quad\text{ as } \delta \downarrow 0,
\]
exists and  is given by
\[
h_K(t) = \lambda \int_{\mathbb{R}^d} f^x_{K-D}(t)\, \text{d} x.
\]
Computing the distribution of $S_K$ exactly may be hard, but
asymptotics may be possible, via knowledge of the Laplace transform
of $T^x_{K-D}$ and Tauberian theorems.

\subsection{The isotropic case}
Suppose that the process $\{\xi(t),\, t \ge 0\}$, with $\xi(0)=0$, is
isotropic, i.e., that if $Q$ is a proper rotation of $\mathbb{R}^d$ then
$\{Q\xi(t),\, t \ge 0\}$ has the same law as $\{\xi(t),\, t \ge 0\}$.
Assuming further that
\begin{equation}
\label{Kset}
D =  r \mathbb B , \quad K = r_0 \mathbb B,
\end{equation}
where $\mathbb B = \{x\in\mathbb{R}^d:\, |x|\le 1\}$ is the unit ball
of radius 1 centered at the origin,
then, clearly, the integral in \eqref{detHOMmbmCANON} can be
simplified. Indeed, with $e_1=(1,0,\ldots,0)$,
\[
T^x_{K-D} = \inf\{t \ge 0:\, |x+\xi(t)| \le r+r_0\}
\stackrel{\text{(d)}}{=} \inf\{t \ge 0:\, \big|\,|x| e_1+\xi(t)\,\big| \le r+r_0\},
\]
so, letting
\begin{equation}
\label{Trrr}
T^{\rho}_{r+r_0} := \inf\{t \ge 0:\, |\rho e_1 +\xi(t)| \le r+r_0\}
\end{equation}
be the first hitting time of $r+r_0$ by the radial process
$|\rho e_1 +\xi(t)|$, we have
\begin{align*}
\int_{\mathbb{R}^d} \mathbb P(T^x_{K-D} \le t) \,\text{d} x
&= \sigma_{d-1} \int_0^\infty \mathbb P(T^{\rho}_{r+r_0} \le t)\, \rho^{d-1} \,\text{d}\rho
\\
&= \omega_d (r+r_0)^d + \sigma_{d-1}\int_{r+r_0}^\infty
\mathbb P(T^{\rho}_{r+r_0} \le t)\, \rho^{d-1} \,\text{d}\rho,
\end{align*}
where $\omega_d = \pi^{d/2}/\Gamma(1+d/2)$
is the $d$-dimensional Lebesgue measure of
the unit ball in $\mathbb{R}^d$ and $\sigma_{d-1} = d\omega_d$.
Hence
\begin{equation}
\label{SKrot}
\mathbb P(S_K > t)
= e^{-\lambda \omega_d (r+r_0)^d}
\exp \bigg(-\lambda\, \sigma_{d-1} \int_{r+r_0}^\infty
\rho^{d-1} \, \mathbb P(T^\rho_{r+r_0} \le t) \,\text{d} \rho \bigg).
\end{equation}
Although the integral has been reduced from a $d$-dimensional one to
$1$-dimensional, finding the distribution of \eqref{Trrr} is
still a $d$-dimensional problem.

\section{The inertial Boolean model}
\label{inbm}
We take $\xi$ to be a linear stochastic motion. Let $v$ be
a random variable in $\mathbb{R}^d$ and let
the motion of a typical particle be
\[
\xi(t) := tv, \quad t \ge 0.
\]
The random set covered by time $t$ is
\[
\mathcal W(t) = \bigcup_{x \in \Phi} ( x+\{vs,\, 0 \le s \le t\} + K).
\]
Letting $\xi^{(t)} := \{\xi(s),\, 0 \le s \le t\}$,
and $G^{(t)} :=  \xi^{(t)} + D$, we have
\begin{multline}
\label{CapIne}
\operatorname{\sf T}_{G^{(t)}}(K-x)
= \mathbb P\big((\xi^{(t)} + D) \cap (K-x) \neq \varnothing\big)
\\
= \mathbb P\big(\{x+sv,\, 0 \le s \le t\}\,  \text{ intersects }K-D\big),
\end{multline}
which is laborious (but not impossible) to compute explicitly
(but we don't need the explicit formula).
Assume that the particles are initially placed at the points
of a Poisson process with intensity measure $\lambda(\text{d} x)$.
Then, from \eqref{byfubini}, the distribution of
the detection time of $K$ is
\[
\mathbb P(S_K > t) =
\exp\bigg(-\mathbb E\, \lambda\bigg( \bigcup_{0 \le s \le t} [K-D-sv]\bigg)\bigg).
\]
Take now $K= r \mathbb B$, $D=r_0 \mathbb B$, with $\mathbb B$ the unit ball in $\mathbb{R}^d$.
Then $K-D = (r+r_0) \mathbb B$. Let
\[
R:=r+r_0.
\]
Assuming further that $\lambda$ is isotropic
(invariant under rotations), so that
\[
\lambda(\text{d} x) = \text{d} \theta\, |x|^{d-1}\, \mu(\text{d} |x|),
\]
where $\text{d}\theta$ is the natural spherical measure on the boundary of the
unit ball and $\mu$ a measure on $\mathbb{R}_+$,
we have
\[
\lambda\bigg( \bigcup_{0 \le s \le t} (K-D-sv) \bigg)
= \frac{1}{2} \lambda(R \mathbb B) + \frac{1}{2} \lambda(R \mathbb B + t v e_1)
+ \lambda(C_{R,tv}),
\]
where $C_{R,tv}$ is a cylinder with height $tv$. Specifically,
$C_{R,tv}$ contains the $x \in \mathbb{R}^d$ such that
\begin{align*}
&x \cdot v \le t |v|^2,
\\
&|v|^2 x - (x \cdot v) v \in |v|^2 \in R \mathbb B.
\end{align*}
Even when the law of $v$ is assumed isotropic,
the integrals can be quite hard to compute exactly unless we
further assume invariance under translations for $\lambda$, i.e., take
now $\lambda$ to be a multiple of the Lebesgue measure:
\[
\lambda(\text{d} x) = \lambda \cdot \text{d} x.
\]
In this case, things are very simple:
\begin{equation}
\label{verysimple}
\lambda\bigg( \bigcup_{0 \le s \le t} (K-D-sv) \bigg)
= \lambda \cdot R^d \omega_d + \lambda \cdot R^{d-1} \sigma_{d-1}
t |v|.
\end{equation}
Therefore, if $K-D = R \mathbb B$, and if the original location of
particles is a homogeneous Poisson process with intensity $\lambda$
then, assuming that $\mathbb E|v|<\infty$,
\begin{equation}
\label{lawinertial}
\mathbb P(S_K > t) = \exp(-\lambda R^d \omega_d) \,
\exp(-\lambda R^{d-1} \sigma_{d-1} (\mathbb E |v|) t).
\end{equation}
This gives
\begin{equation}
\label{expectationinertial}
\mathbb E S_K = \frac{e^{-\lambda \omega_d R^d}}
{\lambda d \omega_d (\mathbb E|v|) R^{d-1}}
= \frac{1}{\lambda d \omega_d (\mathbb E|v|)} ~ \frac{1}{R^{d-1}}
- \frac{1}{d} R + o(R), \quad \text{as } R \downarrow 0.
\end{equation}
However, if $\mathbb E|v|=\infty$, then the Poisson process \eqref{thinnedpoisson}
has intensity measure proportional to \eqref{CapIne}
which integrates, with respect to the $d$-dimensional Lebesgue measure,
to the expectation of \eqref{verysimple} and this is infinity if $t >0$.
Therefore,
\[
\text{ if  $\mathbb E|v|=\infty$ then $S_K=0$, a.s.}
\]
A more elaborate problem is the computation of the law of
\[
\inf\{t \ge 0: r_0 \mathbb B \cap  \mathcal W(t) \neq \varnothing,~
a \not \in \mathcal W(t)\}
\]
where
\[
\mathcal W(t) = \bigcup_{x \in \Phi} ( x+\{vs,\, 0 \le s \le t\} + r \mathbb B),
\]
which is the  set covered up to time $t$ by the inertial Boolean model
when the particles carry balls of radii $r$ each.
In other words, the problem is that of finding information about the first time
that the particles will detect a fixed ball of radius $r_0$
before anyone of them goes close to some point $a$ (the enemy).
This problem will be addressed in the future.

\section{The Brownian Boolean model}

We now specialize further and take $\xi$ to be a standard
Brownian motion in $\mathbb{R}^d$. In other words, $\xi(0)=0$
and $\xi=(\xi_1, \ldots, \xi_d)$, where the $\xi_i$, $i=1, \ldots, d$
are i.i.d.\ standard Brownian motions in $\mathbb{R}$.
Let $D$ and $K$ be balls with radii $r$ and $r_0$, respectively,
as in \eqref{Kset}, and let $S$ be the detection time of $K$.
The distribution of $S$ depends on $r_0$ and $r$ through their sum, so we write
\[
R=r_0+r
\]
for convenience.
Let
\[
\varrho(t) :=|\rho e_1 +\xi(t)|.
\]
Then $\varrho$ is standard Bessel process
of dimension $d$ started at $\varrho(0)=\rho$,
denoted as BES$^d(\rho)$ by Revuz and Yor \cite{RY91}.
It is a strong Markov process (in fact, a Feller diffusion)
satisfying the It\^o equation
\[
\varrho(t)
= \varrho(0) + \beta(t) + \frac{d-1}{2} \int_0^t \frac{1}{\varrho(s)} \,\text{d} s,
\]
where $\beta$ is a standard Brownian motion in $\mathbb{R}$.
Letting $T^\rho_R$ be the first hitting time of the closed ball
of radius $R$ centered at the origin by the Bessel process started at $\rho$,
we have, from \eqref{SKrot},
\begin{equation}
\label{StailWiener}
\mathbb P(S>t) = \exp\bigg(-\lambda \sigma_{d-1} \int_0^\infty \rho^{d-1}\,
\mathbb P (T^\rho_R\le t)\, \text{d} \rho \bigg) =: e^{-\lambda V^R(t)}.
\end{equation}

The infinitesimal generator $\mathcal A$ of $\varrho$ is the radial
part of $\frac{1}{2} \Delta$, where $\Delta$ is the Laplacian on $\mathbb{R}^d$:
\[
\mathcal A f(\rho) =
\frac{1}{2} \frac{1}{\rho^{d-1}} \frac{\partial}{\partial \rho} \bigg(\rho^{d-1}
\frac{\partial f}{\partial \rho}\bigg)
= \frac{1}{2} f''(\rho) + \frac{d-1}{2\rho} f'(\rho),
\]
acting on $C^2$ functions.
Therefore, for $s>0$, the function
\[
L(\rho) := \mathbb E [e^{-s T^\rho_R} ]
\]
satisfies
\[
\mathcal A L = sLa
\]
i.e., the ODE
\begin{equation}
\label{BesselODE}
\frac{1}{2} L''(\rho) + \frac{d-1}{2\rho} L'(\rho) = sL(\rho),
\quad R < \rho < \infty,
\end{equation}
with boundary conditions
\begin{equation}
\label{bc}
L(R+)=1,
\quad
\lim_{\rho \to \infty} L(\rho)=0.
\end{equation}
Change variables using
\begin{equation}
\label{COV}
L(\rho) = \rho^b \widetilde L(a \rho),
\end{equation}
for appropriate constants $a >0, b \in \mathbb{R}$.
The ODE reduces to
\[
a^2 \rho^2 \widetilde L''(a\rho) + (d-1+2b) a\rho \widetilde L'(a\rho)
+(b^2-2b+bd -2s\rho^2) \widetilde L(a\rho)=0.
\]
Choosing
\[
b= \frac{d}{2}-1
\]
gives
\[
a^2 \rho^2 \widetilde L''(a\rho) +  a\rho \widetilde L'(a\rho)
-(b^2+2s\rho^2) \widetilde L(a\rho)=0.
\]
Letting
\begin{equation}
\label{a2s}
a=\sqrt{2s}
\end{equation}
(and letting $x:=a\rho$), we obtain the following ODE
\begin{equation}
\label{modbesode}
x^2 \widetilde L''(x) + x \widetilde L'(x) - (b^2+x^2) \widetilde L(x)=0.
\end{equation}
We recognize this as the modified Bessel ODE \cite[Sec.\ 3.7]{JACK}
the fundamental solutions of which are the modified
Bessel functions $I_{\pm b}$ and $K_{b}$.
The standard Bessel ODE differs from \eqref{modbesode}
by a change of sign in the last term. The fundamental solutions of the
standard Bessel ODE
are the Bessel functions of first and second kind $J_{\pm b}$, $N_b$,
whose series representations are easily obtained from the ODE;
see equations (3.82), (3.83) and (3.85) in \cite{JACK}.
The modified Bessel functions $I_{\pm b}$ and $K_{b}$
(of first and second kind, respectively)
are related to $I_{\pm b}$ and $K_{b}$ via
\begin{align*}
I_{\pm b}(x) &= i^{-b} J_{\pm b}(ix)
\\
K_b(x) &= \frac{\pi}{2} i^b
[ i J_b(ix)- N_b(ix)],
\end{align*}
and are real-valued, despite appearances; see (3.100), (3.101)
and (3.86) in \cite{JACK}.
Since $I_{\pm b}$ explodes as $x\to\infty$, we are left with
only one choice for \eqref{modbesode}:
\[
\widetilde L(x) = C K_{b}(x),
\]
where $C$ is a constant.
In terms of the original function, i.e., using the change of
variables \eqref{a2s}, \eqref{COV},
\[
L(\rho) = C \rho^{-b} K(\rho\sqrt{2s}).
\]
The boundary conditions \eqref{bc} determine $C$:
\[
C= R^{b}/K_{b}(R \sqrt{2s}).
\]
So the solution to the ODE \eqref{BesselODE}
with boundary conditions \eqref{bc}
is given by
\begin{equation}
\label{TLaplace}
L(\rho) = \mathbb E[e^{-sT^\rho_R} ]
= \frac{\rho^{-b} K_b(\rho\sqrt{2s})}
{R^{-b} K_b(R\sqrt{2s})}, \quad \rho \ge R.
\end{equation}

Compare now \eqref{StailWiener} with \eqref{detHOMmbm}.
Since $\bigcup_{0 \le s \le t} (\xi(s) + D-K) =
\bigcup_{0 \le s \le t} (\xi(s) + R \mathbb B)$, the quantity
in the  exponent
in \eqref{StailWiener} is the expected volume of the
Wiener sausage
\[
W^R(t) := \bigcup_{0 \le s \le t} (\xi(s) + R \mathbb B);
\]
see comments at the end of \mathbb S \ref{dtimbm}.
We have
\[
V^R(t):= \mathbb E \operatorname{vol}  W^R(t) = \int_{\mathbb{R}^d} \mathbb P(T^x_R \le t) \,\text{d} x
= \omega_d R^d +
\sigma_{d-1} \int_R^\infty \rho^{d-1} \mathbb P(T^\rho_R \le t)\,\text{d} \rho.
\]
Via \eqref{TLaplace}, we have an expression for
the Laplace transform of the the expected volume of
the Wiener sausage:
\begin{align}
\widehat V^R(s) := \int_0^\infty e^{-st} V^R(t) \,\text{d} t
&=
\frac{\omega_d R^d}{s}
+ \frac{\sigma_{d-1}}{s} \int_R^\infty \big(\mathbb E e^{-s T^\rho_R}\big)
\rho^{d-1} \,\text{d} \rho
\nonumber
\\
&= \frac{\omega_d R^d}{s}
+ \frac{\sigma_{d-1}}{s} \int_R^\infty
\frac{R^{\frac{d}{2}-1}}{\rho^{\frac{d}{2}-1}}
\frac{K_{\frac{d}{2}-1}(\rho\sqrt{2s})}{K_{\frac{d}{2}-1}(R\sqrt{2s})}
\rho^{d-1} \,\text{d} \rho
\nonumber
\\
&= \frac{\omega_d R^d}{s}
+ \frac{\sigma_{d-1} R^{\frac{d}{2}-1}}{sK_{\frac{d}{2}-1}(R\sqrt{2s})}
\int_R^\infty K_{\frac{d}{2}-1}(\rho\sqrt{2s}) \rho^{d/2} \,\text{d} \rho.
\label{VolLap}
\end{align}
We now use a couple of facts about the functions $K_b$; see \cite{AS}.
First, we have the recursion formula
\[
K_{b+1}(x) - K_{b-1}(x) = \frac{2b}{x} K_b(x).
\]
Second, we have the derivative
\[
K_b'(x) = \frac{b}{x} K_b(x) - K_{b+1}(x).
\]
Combining these we get
\[
\frac{\text{d} }{\text{d} x} \big( K_b(x) x^b \big) = - K_{b-1}(x) x^b,
\]
and so
\[
\int_x^\infty K_{\frac{d}{2}-1}(y) y^{\frac{d}{2}} \,\text{d} y
= K_{\frac{d}{2}}(x) x^{\frac{d}{2}},
\]
and, for $\lambda > 0$,
\[
\int_x^\infty K_{\frac{d}{2}-1}(\lambda y) y^{\frac{d}{2}} \,\text{d} y
=\frac{1}{\lambda} K_{\frac{d}{2}}(\lambda x) x^{\frac{d}{2}}.
\]
The last integral in \eqref{VolLap} evaluates to
\[
\frac{R^{\frac{d}{2}}}{\sqrt{2s}} K_{\frac{d}{2}}(R\sqrt{2s}),
\]
and so
\[
\widehat V^R_d(s) =  \frac{\omega_d R^d}{s}
+ \frac{\sigma_{d-1} R^{d-1}}{\sqrt{2s^3}}
\frac{K_{\frac{d}{2}}(R\sqrt{2s})}{K_{\frac{d}{2}-1}(R\sqrt{2s})},
\]
where we added a subscript $d$ to indicate dependence on the dimension.
We can save some space by observing that, due to Brownian scaling,
\begin{equation}
\label{Vscaling}
V^R_d(t) = R^d V^1_d(t/R^2),
\end{equation}
\[
\widehat V^R_d(s) = R^{d+2} \widehat V^1_d(R^2 s).
\]
and so it is only
\begin{equation}
\label{itisonly}
\widehat V^1_d(s) =  \frac{\omega_d }{s}
+ \frac{\sigma_{d-1} }{\sqrt{2s^3}}
\frac{K_{\frac{d}{2}}(\sqrt{2s})}{K_{\frac{d}{2}-1}(\sqrt{2s})}
\end{equation}
we should be looking for.
Since $K_b=K_{-b}$, the case $d=1$ is trivial:
\begin{equation}
\label{V11}
\widehat V^1_1(s) = \frac{2}{s} + \frac{2}{\sqrt{2s^3}}.
\end{equation}
Inverting this Laplace transform gives
\[
V_1^1(t) = 2 + \sqrt{\frac{8 t}{\pi}}.
\]
By the  scaling relation \eqref{Vscaling},
\[
V_1^R(t) = 2R + \sqrt{\frac{8 t}{\pi}},
\]
that is, the expected change of volume (=length) from its
initial value does not depend on $R$.

\subsection{Odd dimensions}
\label{odddimensions}
Consider now the case where
\[
d=2n+1, \quad n =0,1,\ldots
\]
We will produce an algorithm for computing $\widehat V^1_{2n+1}(s)$ recursively,
and carry out its first few steps.
The modified Bessel functions of half-integer order have a simple
form:
\[
K_{n+\frac{1}{2}}(x) = \sqrt{\frac{\pi}{2}} \frac{e^{-x}}{\sqrt{x}}
y_{n}(1/x), \quad n=0,1,\ldots,
\]
where $y_n(x)$ is a polynomial of degree $n$ with integer coefficients:
\[
y_n(x) := \sum_{k=0}^n \frac{(n+k)!}{(n-k)! k!} \bigg(\frac{x}{2}\bigg)^k,
\quad n=0,1,\ldots,
\]
known as the Bessel polynomial of degree $n$; see \cite[formula (3)]{KF}.
Note, in particular, that
\[
y_n(x) = (2n-1)!! x^n + (2n-1)!! x^{n-1} + \cdots + \frac{n(n+1)}{2} x + 1,
\]
i.e., the coefficients of the two highest powers are equal to the double factorial
\[
(2n-1)!! = \frac{(2n)!}{n! 2^n} = (2n-1)(2n-3)(2n-5) \cdots 3 \cdot 1.
\]
See Appendix \ref{modbes} for a table of the first few Bessel polynomials
and their corresponding Bessel functions.
Consequently,
\begin{equation}
\label{Vhat}
\widehat V^1_{2n+1}(s) = \frac{\omega_{2n+1}}{s}
+\frac{\omega_{2n+1}}{s}~ \frac{2n+1}{\sqrt{2s}}~
\frac{y_n(1/\sqrt{2s})}{y_{n-1}(1/\sqrt{2s})},
\quad n=0,1,\ldots,
\end{equation}
where
\[
y_{-1}(x) :=1,
\]
as follows by comparison to \eqref{V11}. We thus have
\begin{align*}
\frac{1}{\omega_3} \widehat V^1_3(s)
&= \frac{1}{s} +\frac{3}{s} ~\frac{\sqrt{2s}+1}{2s}
\\
\frac{1}{\omega_5} \widehat V^1_5(s)
&= \frac{1}{s} + \frac{5}{s}~ \frac{(2s)^{3/2} + 6s + 3 \sqrt{2s}}
{(2s)^{3/2}~[\sqrt{2s}+1]}
\\
\frac{1}{\omega_7} \widehat V^1_7(s)
&= \frac{1}{s} + \frac{7}{s}~
\frac{(2s)^{3/2} + 12 s + 15 \sqrt{2s} + 15}
{\sqrt{2s}~ [(2s)^{3/2} + 6s + 3 \sqrt{2s}]}
\\
\frac{1}{\omega_9} \widehat V^1_9(s)
&= \frac{1}{s} + \frac{9}{s}~
\frac{(2s)^{5/2} + 40 s^2 + 45 (2s)^{3/2} +210 s + 105\sqrt{2s}}
{(2s)^{3/2}~ [(2s)^{3/2} + 12 s + 15 \sqrt{2s} + 15]}.
\end{align*}
These Laplace transforms can, in principle, be inverted by
using partial fraction expansion and the fact that (see
Erd\'elyi {\em et al.} \cite[Ch.\ 7, p.\ 233]{ERD})
\[
\frac{1}{\sqrt{s}+\beta} =
\int_0^\infty e^{-st}
\bigg[\frac{1}{\sqrt{\pi t}} - \beta e^{\beta^2 t} \operatorname{erfc}(\beta \sqrt{t})\bigg]
\,\text{d} t,
\]
where
\[
\operatorname{erfc}(t):=1-\operatorname{erf}(t),\quad
\operatorname{erf}(t) := \frac{2}{\sqrt{\pi}} \int_t^\infty e^{-u^2} \,\text{d} u.
\]
For example, writing
\[
\frac{1}{\omega_3} \widehat V^1_3(s)
= \frac{1}{s} + \frac{3}{\sqrt{2s^3}} + \frac{3}{2s^2},
\]
we obtain
\[
\frac{1}{\omega_3}V^1_3(t) = 1 + \frac{6}{\sqrt{\pi}}\sqrt{t} +\frac{3}{2} t.
\]
Expanding $\widehat V^1_5(s)$, we obtain
\[
\frac{1}{\omega_5} \widehat V^1_5(s)
=\frac{1}{s} + \frac{10}{2s}~
\bigg[\frac{1}{\sqrt{2s}+1} + \frac{3}{\sqrt{2s}(\sqrt{2s}+1)}
+ \frac{3}{2s(\sqrt{2s}+1)} \bigg].
\]
Since
\begin{align*}
\frac{1}{s(\sqrt{s}+1)} &= \int_0^\infty e^{-st} [1-e^t \operatorname{erfc}(\sqrt{t})] \,\text{d} t
\\
\frac{1}{s\sqrt{s}(\sqrt{s}+1)} = \frac{1}{s\sqrt{s}} - \frac{1}{s(\sqrt{s}+1)}
&=
\int_0^\infty e^{-st}
\bigg[2\sqrt{\frac{t}{\pi}} -1 + e^t \operatorname{erfc}(\sqrt{t}) \bigg] \,\text{d} t
\\
\frac{1}{s^2(\sqrt{s}+1)} &=
\int_0^\infty e^{-st} \bigg[
1+t - 2\sqrt{\frac{t}{\pi}} - e^t \operatorname{erfc}(\sqrt{t})
\bigg] \,\text{d} t,
\end{align*}
letting $g_1(t), g_2(t), g_3(t)$ be the functions in the square brackets
of the last three lines,
we have
\begin{align*}
\frac{1}{\omega_5} V^1_5(t)
= 1 + 10
\bigg[\frac{1}{2} g_1(t/2) + \frac{3}{2} g_2(t/2) + \frac{3}{2} g_3(t/2)\bigg]
&=
6 -5 e^{t/2} \operatorname{erfc}(\sqrt{t/2})+ \frac{15}{2} t.
\end{align*}
Using the scaling relation \eqref{Vscaling}, we can now
obtain $V^R_d(t)$, for $d=1,3,5$ and, therefore,
the distribution of the detection time via \eqref{StailWiener}.
\begin{align}
d=1:&  \qquad \mathbb P(S> t) = \exp\big(-2\lambda R -4\lambda\sqrt{t/\pi}\big).
\label{d1}
\\
d=3: & \qquad
\mathbb P(S>t) = \exp\big(-\frac{4\pi\lambda}{3}R^3
-8\sqrt{\pi t}\lambda R^2 - 2\pi \lambda t R\big).
\label{d2}
\\
d=5: & \qquad
\mathbb P(S>t) = \exp\big(-\frac{16 \pi^2}{5}R^2
+ \frac{8R^5}{3} e^{t/2R^2} \operatorname{erfc}(R^{-1} \sqrt{t/2})
-4\pi^2 R^3 t\big)
\label{d3}
\end{align}
Notice that the exponent is not a polynomial in $R$,
as is apparent for the $d=5$ case.\footnote{Answering a question posed
by G\"unter Last to me a few years ago.}

A more efficient way of doing the above is by means of the recursion formula
\[
y_n(x) = (2n-1) x y_{n-1}(x) + y_{n-2}(x), \quad n=1,2,\ldots,
\]
with initial conditions $y_{-1}(x)=y_0(x)=1$. See \cite[\mathbb S 7]{KF}.
Let
\[
H_n(s) := \frac{\sqrt{2s}}{2n+1} \bigg(s \frac{\widehat V^1_{2n+1}(s)}{\omega_{2n+1}}
-1 \bigg).
\]
From \eqref{Vhat} we have
\[
H_n(s)
= \frac{y_n(1/\sqrt{2s})}{y_{n-1}(1/\sqrt{2s})},
\]
and so, from the recursion formula for Bessel polynomials,
\[
H_n(s) = \frac{2n-1}{\sqrt{2s}} + \frac{1}{H_{n-1}(s)},
\quad n=1,2,\ldots,
\]
where $H_0(s)=1$.
This gives a recursive way for computing $\widehat V^1_{2n+1}(s)$.

Let
\[
a_n(s) := \frac{2n-1}{\sqrt{2s}}.
\]
A ``closed'' formula can also be obtained:
\begin{align}
\label{contfrac}
\frac{\widehat V^1_{2n+1}(s)}{\omega_{2n+1}}
&= \frac{1}{s} +  \frac{a_{n+1}(s)}{s}
\,
\left\{
a_n(s) + \cfrac{1}{a_{n-1}(s)
	+ \cfrac{1}{a_{n-2}(s) + \cfrac{1}
{\ddots +\frac{1}{a_1(s)+1} }}}
\right\}
\\
\label{contfrac2}
&= \frac{1}{s} + \frac{2n+1}{2s^2}\,
\left\{
(2n-1) + \cfrac{\sqrt{2s}}{(2n-3)
        + \cfrac{\sqrt{2s}}{(2n-5) + \cfrac{\sqrt{2s}}
{\ddots +\frac{\sqrt{2s}}{1+\sqrt{2s}} }}}
\right\}
\end{align}

\subsection{Large time asymptotics in all dimensions}
\label{largetime}
Expression \eqref{itisonly} allows us to find logarithmic asymptotics
for $\mathbb P(S > t)$, as $t \to \infty$, in any dimension $d$. We
repeat the expression here:
\[
\frac{1}{\omega_d}\widehat V^1_d(s) =  \frac{1}{s}
+ \frac{d}{\sqrt{2s^3}}
\frac{K_{\frac{d}{2}}(\sqrt{2s})}{K_{\frac{d}{2}-1}(\sqrt{2s})} =:
\frac{1}{s} + \widehat g_d(s).
\]
The following asymptotics are known for Bessel functions \cite{AS}. For
any $b > 0$, as $z \to 0$,
\[
K_b(z) \sim 2^{b-1} \Gamma(b) z^{-b}.
\]
Therefore, for $d \ge 3$, we have
\[
\widehat g_d(s) \sim \frac{d(d-2)}{2s^2}, \quad s \to 0.
\]
Hence
\[
g_d(t) \sim \frac{d(d-2)}{2} t, \quad t \to \infty.
\]
By the scaling equation \eqref{Vscaling} and expression
\eqref{StailWiener}, we obtain
\begin{align*}
\mathbb P(S>t) &= \exp(-\lambda \omega_d R^d -\lambda \omega_d R^d g_d(t/R^2)).
\\
&\stackrel{\log}{\text{\Large $\sim$}}
\exp\bigg(-\lambda \omega_d \frac{d(d-2)}{2} R^{d-2} t\bigg),
\quad d \ge 3.
\end{align*}
The case $d=2$ has to be treated differently as it requires the
behavior of $K_0$ near zero which is different:
\[
K_0(z) \sim \log(1/z) , \quad z \to 0.
\]
Hence
\[
\widehat g_2(s) = \frac{2}{\sqrt{2s^3}}
\frac{K_1(\sqrt{2s})}{K_0(\sqrt{2s})}
\sim
\frac{2}{\sqrt{2s^3}}
\frac{1}{ \sqrt{2s }\log \frac{1}{\sqrt{2s}}}
=\frac{2}{s^2 \log(1/s)}
\]
Notice that
\[
\widehat g_2(s) = s^{-2} \ell(1/s),
\]
where the function
$\ell(z) = 2/\log(z)$ is slowly varying at infinity, viz.,
$\ell(\kappa z)/\ell(z) \to 1$, as $z \to \infty$, for all $\kappa >0$.
Combining Karamata's Tauberian theorem with the monotone
density theorem ($g_2(t)$ is increasing function of $t$)
we conclude that
\[
g_2(t) \sim  t \ell(t) = \frac{2t}{\log t},\quad \text{  as $t \to \infty$.}
\]
Arguing as before, this means that, as $t \to \infty$,
\begin{align*}
\mathbb P(S>t) &= \exp(-\lambda \omega_2 R^2 -\lambda \omega_2 R^2 g_2(t/R^2))
\\
&\stackrel{\log}{\text{\Large $\sim$}}
\exp\bigg(-\frac{2\pi\lambda t}{\log t}\bigg).
\end{align*}
This agrees with the result of \cite[Theorem 2]{SPI}.

\subsection{Expectations}
For $d=1$, using $\int_0^\infty \exp(-\sqrt{t}) \,\text{d} t=2$, we can compute
the expectation of $S$ explicitly by integrating
\eqref{d1}:
\[
\mathbb E S := \frac{\pi}{8} ~ \frac{e^{-2\lambda R}}{\lambda^2}.
\]

For $d=2$, we have no explicit expression, but the earlier asymptotic
expression for large $t$ tells us that, as $R \to 0$, $\mathbb E S$
converges to a constant. This is a manifestation of the fact
that Brownian motion is neighborhood recurrent in 2 dimensions.

For $d=3$, we can use the integral
$\int_0^\infty \exp(-\sqrt{t}-t) \,\text{d} t=
1+ \frac{\pi e^{1/4}}{2}(\operatorname{erf}(1/2)-1)$ to integrate \eqref{d3}:
\[
\mathbb E S =
\frac{e^{-\frac{4}{3} \pi \lambda R^3}}{ 2(\pi\lambda R)^{3/2}}
\bigg[
\sqrt{\pi\lambda R} + 2\sqrt{2}\pi\lambda R^2 e^{8\lambda R^3}
\big( \operatorname{erf}( 2\sqrt{2\lambda} R^3/2 )-1 \big)
\bigg]
\]

For higher dimensions, we can translate the previous asymptotics
for $t \to \infty$ into asymptotics for $R\downarrow 0$ and obtain that
\begin{equation}
\label{expectationbrownian}
\mathbb E S \sim \frac{c_d}{R^{d-2}}, \text{ as } R \downarrow 0,
\end{equation}
where $c_d$ is a constant depending on $d$ only. This
estimate holds for all $d \ge 2$.
Comparing \eqref{expectationbrownian} with \eqref{expectationinertial}
we find that in any dimension $d\ge 2$, it is better to
make particles (sensors) perform Brownian motions
rather than random straight lines with finite mean velocity
if the goal is to detect a small object.\footnote{This answers
a question posed by Venkat Anantharam a few years ago, who, jokingly,
commented that an engineer would never have sensors perform
Brownian motions.}

\section{Concluding remarks and open problems}
In this paper, we reviewed the mobile Boolean model, focusing,
in particular, in the inertial and Brownian cases.
For the inertial case, we have an explicit expression
\eqref{lawinertial}
for the distribution of the detection
time of in any dimension.
For the Brownian case, we have explicit expressions
\eqref{d1}, \eqref{d2}, \eqref{d3} in dimensions $d=1$, $3$ and $5$,
an algorithm for computing an explicit expression when $d$ is odd
(Section \ref{odddimensions})
and logarithmic asymptotics when $d$ is even (Section \ref{largetime}).
We worked with a target set $K$ which is a ball. This enabled us to reduce
the a multi-dimensional problem to one dimension.
We also gave formulas or estimates for expectations.

For general compact set $K$,
and dimension $d=3$,
Spitzer's \cite{SPI} paper gives asymptotic estimates for
the expected volume of a $K$-Wiener sausage,
in terms of the Newtonian capacity $\operatorname{cap}(K)$ of $K$,
using entirely probabilistic methods:
\[
\mathbb E \operatorname{vol}(\xi^{(t)} + K) =
\operatorname{vol}(K) + \operatorname{cap}(K)\,t
+ 4(2\pi)^{-3/2} \operatorname{cap}(K)^2\, \sqrt{t}
+o(\sqrt{t}), \quad d=3.
\]
Translating this into a detection time probability estimate,
and using the scaling relation, we have
\[
P(S_{R K} > t)
= \exp\big(
\operatorname{vol}(K)\,R^3 +  \operatorname{cap}(K)\,R^2 t
+  4(2\pi)^{-3/2} \operatorname{cap}(K)^2\, R \sqrt{t}
+o(R \sqrt{t})
\big).
\]

We next pose some open problems.

\paragraph{Open problem 1.}
We did not touch at all the coverage problem, i.e., the law of
the random variable $\inf\{t \ge 0:~ K \subset \mathcal W(t)\}$.
For the Brownian Boolean model, and when the diameter of $R$ tends to infinity,
the problem has been solved in \cite{PSSS}. What are the corresponding
asymptotics for the inertial cases?

\paragraph{Open problem 2.}
Let $K_1$ and $K_2$ be two sets (e.g., balls of radii $r_1, r_2$),
and let $S_{K_1}, S_{K_2}$ be their detection times.
Find the probability $P(S_{K_1} < S_{K_2})$.

\paragraph{Open problem 3.}
For the inertial Boolean model, find the distribution of the
detection time of a ball before a fixed point is hit.
(See remarks at the end of Section \ref{inbm}.)

\paragraph{Open problem 4.}
Investigate further the algorithm of Section \ref{odddimensions}
and, in particular, the ``closed'' formula \eqref{contfrac}-\eqref{contfrac2}.

\paragraph{Open problem 5.}
Let there be an independent space-time
Poisson process $\Psi$ in $\mathbb{R}^d \times \mathbb{R}_+$ with fixed intensity.
Interpret its points as ``customers''. The Brownian Boolean model
is a space-time serving mechanism clearing points whenever it meets them.
Find necessary and sufficient stability conditions.
This problem is related to a number of recent stability problems in
queueing theory where the spatial dimension is just as important as
the time dimension \cite{SF,BFL,RFK}.

redpajama_set_name: RedPajamaArXiv


text:
The starving at the West's gate
I didn't hear our (not so)youth(ful) worker's annual Christian Aid spiel at church yesterday as I was doing my turn in the creche for fretful babies (full marks for my forward planning there). I would have liked to have heard the sermon given by Edward Lucas at the Christian Aid Week service in Canterbury Cathedral yesterday, and which forms the basis of this article in today's Times. I might send a copy to our YW for her attention, hem!
The reason for Third World poverty was already obvious centuries ago. Just ask Amos.
CHRISTIAN AID WEEK is rightly a time for warmheartedness. But that is no excuse for softheadedness. It is sloppy thinking, for example, to believe poverty in one place is caused by wealth in another. To share the wealth of the rich world evenly among the poor would temporarily dent poverty, not end it. Redistribution invariably destroys wealth in one place; it rarely creates it in another. The redistributed money would mostly go on short-term consumption or be stolen by corrupt officials. The root cause of poverty, above all injustice, would remain.
That still eludes much of the anti-poverty lobby. But the prophet Amos, writing three millennia ago, clearly spotted the link between poverty and injustice. For I know how many are your transgressions, and how great are your sins — you who afflict the righteous, who take a bribe, and push aside the needy in the gate.
When Amos mentioned those who "push aside the needy in the gate" he was referring chiefly to law courts but also the route to markets in towns and cities, where the poor and powerless were at the mercy of corrupt gatekeepers. Third World small businesses on the way to market suffer similarly from corrupt bureaucrats and policemen today.
Technology and capitalism now have made cheap and accurate weighing scales widely affordable, ending one of the most common ways in which the poor are cheated. But the grossly unjust taxes imposed by fiddling with money, through inflation and non-convertibility, continue. The rich and powerful can use hard currencies and foreign bank accounts; for the weak and poor, the lack of a safe way to save is yet another burden.
Since Amos's day it has become pretty clear how wealth is created. There is no example of a country where trade, competition and the rule of law have not brought prosperity. Bad government, the favouring of elites, protectionism and monopoly all entrench poverty. Modern prosperity is the result of specific institutions and habits. But like water flowing downhill, wealth trickles away unless it is well husbanded.
Polite Christian society does not celebrate the wondrous wealth-creating processes of global capitalism. It winces at them. Worries about inequality (ultimately a secondary question to poverty) and, worse, a distaste for wealth, eclipse the extraordinary way in which the embrace of capitalism and global trade in India and China have lifted hundreds of millions of people from poverty in the past two decades.
This anti-capitalist attitude is as absurd as a Christian distaste for the laws of physics. It also leads to a very damaging conflation of private generosity with public policy. The overwhelming lesson of five decades of Third World aid is that, paid from taxation, it takes money from poor people in rich countries and gives it to rich people in poor ones.
And talking of the poor there is a report issued today, commissioned by the Department of the Deputy Prime Minister (yes him) which reveals that Muslims are the UK's "Most deprived minority" At time of writing the Times only has a brief paragraph on it as breaking news. The BBC, of course, carries the report as a "Top story"
The report revealed Muslims were more likely than any other faith group to be jobless and living in poor conditions.
It said half of Muslims aged over 25 were unemployed, and one in three lived in the most deprived areas of England.
Interestingly the Times comment, brief as it is, reports this particular fact as "Half of Muslims aged over 25 are not in the formal labour market," an apparently slight, but telling difference.
University researchers in Birmingham, Derby, Oxford, and Warwick also found they had poorer levels of education.
"Taking the Muslim population as a whole, they face some of the most acute conditions of multiple deprivation," the report said.
John Prescott's former department, the Office of the Deputy Prime Minister (ODPM), commissioned the academics to review data on the Hindu, Sikh and Muslim communities.
So how are the Hindu and Sikh communities fixed? I bet they are working hard, getting on and doing well. An asset not a burden. And what are the poor hard up Muslims doing in the BBC's picture? Rolling out a prayer mat.
Don't they realise, Laborare est Orare. Time they were told.

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Why is radioactive decay dependent on amount of substance available? Radioactive decay is an attribute of unstable nucleus. When we represent it in equation, we don't involve any macroscopic attribute of substance.
But still, rate of radioactive decay is proportional to amount of substance available (a macroscopic attribute). Why?
What's really the relation with others when it comes to bleeding out its unstability?

A: The core reason is that the chance for a given atom to decay in the next $n$ seconds is always the same.
You can get a more intuitive feel for this by considering a simple game with dice.

*

*Start with some initial population of ordinary dice (I'll assume we're using the basic six-sided (cubical) ones, but you table top RPG fiends can use polyhedral ones if you want).

*On each round roll all the surviving dice. Collect, count and remove any that show a 1; these are the ones that decayed and they are removed from the surviving population (they do not participate in future rounds). Record the number that decayed and the number that remain.

*Continue with step #2 until none survive (or you get bored).

*Repeat the whole of 1--3 several time to get some sense of the variation.

*Graph both the (average) number that decay and the (average) number that survive as a function of the number of rounds. (For more insight, do this both on linear paper and on semi-log paper.)
You'll notice that the odd for any particular die to fail and be removed on any given round is always 1/6, which is equivalent to the constant odds for a unstable nucleus to decay in some give time frame.
You can also simulate this kind of experiment with a code like:
// Rolling dice radioactive decay demo simulator.
//
// During each round any die that rolls a 1 is removed from the
// population, and we plot the average number of dice surviving as a
// function of the number of rolls.
//
// Relies on the ROOT framework. http://root.cern.ch/
//
// Compile with
//
//    g++ -O3 $(root-config --cflags --ldflags --libs) decay.cc -o decay
//
// then run with
//
//     ./decay [<initial population> [<number of rolls> [<number of trials>]]]
//
// and find the output in "decay.png"
#include <iostream>

#include <TProfile.h>
#include <TCanvas.h>
#include <TRandom3.h>
#include <TF1.h>

int main(int argc, char*argv[]){

  // running parameters and default values
  int population = 1000;
  int timeticks = 25;
  int trials = 20;

  // trivial argument handling
  switch (argc) {
  default: /* FALL-THOUGH */
  case 4:  /* FALL-THOUGH */
    trials = std::atoi(argv[3]);
  case 3:  /* FALL-THOUGH */
    timeticks = std::atoi(argv[2]);
  case 2:  /* FALL-THOUGH */
    population = std::atoi(argv[1]);
  case 1:  /* FALL-THOUGH */
  case 0:
    break;
  }

  std::cout << "Doing " << trials
        << " trials of " << population
        << " dice over " << timeticks << " rolls."
        << std::endl;

  // Setup a profile histogram of the number of dice surviving at a
  // particular time over some number of trials
  TProfile*hS = new TProfile("hS","Surviving population",
                 timeticks+1,-0.5,timeticks+0.5);
  hS->GetXaxis()->SetTitle("Rolls");
  hS->GetYaxis()->SetTitle("Surviving dice");

  TProfile*hD = new TProfile("hD","Number decaying",
                 timeticks+1,-0.5,timeticks+0.5);
  hD->GetXaxis()->SetTitle("Rolls");
  hD->GetYaxis()->SetTitle("Decaying dice");

  // A PRNG
  TRandom*r=new TRandom3(0);

  for (int p=0; p<trials; ++p) { // Several passes to establish a profile
    if (p%5 == 0) std::cout << "   Trial " << p+1 << "..." <<std::endl;
    int count = population;
    int t=0;
    for (t=0; t<timeticks; t++) { // several time buckets to observe decay
      hS->Fill(t,count);
      int decay = 0;
      for (int i=count; i; --i) { // Try each surviving die
    if ( r->Integer(6) == 0) {
      --count; // discard this die if it rolls 1
      ++decay;
    }
      }
      hD->Fill(t,decay);
    }
    hS->Fill(t,count);
  }

  // Fit the data
  hS->Fit("expo","","",0.5,timeticks+0.5);
  hS->GetFunction("expo")->SetLineColor(kRed);
  hS->GetFunction("expo")->SetLineStyle(kDotted);

  hD->Fit("expo","","",0.5,timeticks+0.5);
  hD->GetFunction("expo")->SetLineColor(kBlue);
  hD->GetFunction("expo")->SetLineStyle(kDashDotted);

  // Output results
  TCanvas*c1 = new TCanvas("c1","c1",600,900);
  c1->Divide(1,2);
  c1->cd(1);
  hS->DrawCopy("");
  hD->DrawCopy("SAME");
  c1->cd(2);
  gPad->SetLogy();
  hS->Draw("");
  hD->Draw("SAME");
  c1->Print("decay.png");
}

Using the default values (a great many dice for many trials of many rounds) you get something like

or with a number of dice, rounds and repeats that wouldn't bore a human to tears you get something like

The fits are exponentials to show the functional dependence.

A: The chance for a fixed nucleus to decay doesn't depend on the number of nuclei. In a fixed amount of time all the nuclei have a certain chance to decay. Increasing the number of nuclei will increase the number of nuclei that decay, but that's really just what you'd expect.
It's like rolling lots dice, the number of dice showing a certain digit will be proportional to the number of dice you roll. Continuing this analogy, say you remove a die if it shows 1. This correspond to a nucleus decaying. Then at first many dice will be removed, but as the number of dice grow smaller so will the number that are removed.

redpajama_set_name: RedPajamaStackExchange


As avid Paneristi are aware, today's powerhouse Panerai watch brand had its humble beginnings in a retail watch store in Florence, founded by the Panerai family in 1860. As inspiration for the latest version of its Radiomir 1940 3 Days Acciaio – 47mm, the company reached back to that historical early era, in which the family-owned Orologerie Suizzera (precursor to the modern Officine Panerai) sold not only wristwatches and pocketwatches but also table clocks and pendulum clocks. It is a particular example of this last item that provided the template for the Art Deco dials of the two new models.
The dial of a pendulum clock displayed on the first floor of Panerai's Florentine shop on the Piazza San Giovanni is re-created in two distinct dial versions, both in 47-mm cases made of polished stainless steel — ivory (Ref. PAM00791) and black (PAM00790). Both retro-look dials are notable for their large, Art Deco hour numerals; peripheral railway minute track track and additional interior ring; and lacquered, spear-shaped hour and minute hands, a style used for the first time on a Panerai timepiece.
The cushion-shaped Radiomir 1940 case, now a mainstay of Panerai's collection, is exceptionally corrosion-resistant and resists water pressures down to 100 meters — a feature calling to mind the Radiomir's watch's origins as a military timepiece for the Royal Italian Navy's elite combat divers. Inside the case and on display through the clear sapphire caseback is the in-house-made movement, the manual-winding P.3000, which boasts a three-day power reserve held in twp mainspring barrels. Mechanical details in full view include the three brushed-finish bridges protecting the movement's interior parts, and a large 3-Hz balance held firmly by a bridge with twin supports. Caliber P.3000 is also equipped with a quick-adjustment function that allows the user to move the hour hand independently forward or backward without affecting the minute hand or the continuous running of the watch.
Panerai's P.3000 caliber is visible through the caseback.
The new, vintage-clock-inspired Radiomir 1940 3 Days Acciaio watches have historically inspired, period-appropriate Plexiglas crystals, slightly domed, over their dials. They are mounted on a sturdy, natural leather strap with beige contrast stitching and an impressed "OP" (Officine Panerai) logo. The price for both special editions is $9,200. What do you think of Panerai's latest ticking tributes to its pre-military Florentine history? Let us know in the comments below.
At around 40mm it would be hard to say no. At 47mm… No way!
Fantastic look with Panerai quality could've been a winner. But I guess Italy didn't get the memo… 47mm is so yesterday!!
47mm is too big but the style is nice. Hope Richemont will find a way to restore the confidence of the buyers on his brands. IWC and Panerai lost a lot of ground due to the numerous mistakes and lack of proper care and attention to quality and customers satisfaction.
That is the most beautiful dial. I would enjoy seeing an unlimited production with a smaller case size.


Since 2003, Gonzaga has played one of its December games on the other side of the mountains for an event that was dubbed the Battle in Seattle. It was a smashing success in that inaugural year with the Zags muscling out an overtime win against then third-ranked Missouri.
But it became a little less of a battle the past couple of years with opponents like South Alabama and Cal Poly not exactly drawing massive crowds to Key Arena. It seemed that last year, even Coach Mark Few was losing some interest in the annual trip to the Emerald City.
The Gonzaga athletic department made some big strides today, however, when they announced that the 2015 opponent would be Tennessee. That game is slated for December 19.
True, the Bruce Pearl years are long behind the Volunteers and the team went just 16-16 last year, but there is some mojo in Knoxville that makes this a good score. Also, just having a big, major conference team with name recognition is a coup for the event.
Right now, former Texas head coach Rick Barnes is in the process of rebuilding Tennessee and has made some quick work doing just that, it appears. Here are some of their off-season moves.
And in case you're having some trouble remembering where exactly Zag basketball is at after that trip to the Elite Eight and loss to eventual NCAA champion Duke: Yes, Pangos and Bell are gone, but Wiltjer, Sabonis, Karnowski, Draginis and all those young point guards are back. A way-too-early poll has them at No. 21 in the nation.


text:
League Standings

<form method ="post" action="/Standings/data" name="sortLeague">
    Sort By:
    <select name="sortReq1">
        <option value="league" id="league-option">League</option>
        <option value="conference" id="conference-option">Conference</option>
        <option value="division" id="division-option">Division</option>
    </select>
    <select name="sortReq2">
        <option value="city" id="city-option">City</option>
        <option value="team" id="team-option">Team</option>
        <option value="standing" id="standing-option">Standing</option>
    </select>

    <input type="submit" value="Go">
</form>

{league}

     {conference}

        NFL Team

        W
        L
        T
        Net Points

    {teams}

        {name}
        {wins}
        {losses}
        {ties}
        {netpoints}

    {/teams}

{/league}

redpajama_set_name: RedPajamaGithub


The Irish Horseracing Regulatory Board Veterinary Officer examined Ah Littleluck, trained by Thomas Gibney, at the request of the Stewards and reported the animal to be blowing hard post race.
The Stewards enquired into an incident after the turn into the straight involving Tony The Gent, ridden by C.T. Keane, placed fourth, Perfect Soldier, ridden by G.P. Halpin, placed third and Prosecution (GB), ridden by T.P. Madden, placed second, where it appeared that Perfect Soldier collided with Tony The Gent causing the latter to check. Evidence was heard from the riders concerned. Having viewed the recording of the race and considered the evidence, the Stewards were satisfied that any interference caused was accidental in nature and took no further action.
C.P. Hoban, rider of Giorni Felice trained by Edward Lynam, reported to the Clerk of Scales that his mount hung left during this race.
W.J. Lee, rider of Patuano trained by W. McCreery, reported to the Clerk of Scales that his mount did not stride out fluently at any stage of this race.
Under the provisions of Rule 90 (iii), the Stewards allowed Plinth, trained by Joseph Patrick O'Brien, to run on receipt of the trainer's written assurance that the horse was correctly vaccinated in accordance with Rule 91.
Under the provisions of Rule 90 (iii), the Stewards allowed Project Bluebook (FR), trained by John Quinn (in GB), to run on receipt of the trainer's written assurance that the horse was correctly vaccinated in accordance with Rule 91.
The Stewards fined Gordon Elliott, trainer, €65 for failing to fit blinkers to Cartwright (GB), prior to entering the parade ring.
D.N. Russell, rider of The Game Changer trained by Gordon Elliott, reported to the Clerk of Scales that his mount was pulled up after his saddle had slipped.
Under the provisions of Rule 90 (iii), the Stewards allowed Guido Reni, trained by Karl Thornton, to run on receipt of the trainer's written assurance that the horse was correctly vaccinated in accordance with Rule 91.
Under the provisions of Rule 90 (iii), the Stewards allowed Mash Potato, trained by Noel C. Kelly, to run on receipt of the trainer's written assurance that the horse was correctly vaccinated in accordance with Rule 91.
The Irish Horseracing Regulatory Board Veterinary Officer examined Townshend (GER), trained by W.P. Mullins, at the request of the Stewards and reported the animal to be post race normal.
The Stewards met to consider reports from Mr. Frank Clune, IHRB Security Officer and Dr. Lynn Hillyer, IHRB Senior Veterinary Officer concerning the manner in which William Coleman O'Brien, trainer, had failed to cooperate when requested on arrival to present the passport for his charge Antiqua Son. Mr. Clune stated that he requested the passport for Antigua Son from Mr O'Brien when the animal was entering the stable yard. He further stated that Mr O'Brien questioned his role in requesting passports as he was not a member of the veterinary team and acted in an aggressive manner. Dr. Hillyer stated that she introduced herself to the trainer but she had also encountered an aggressive attitude which was both unnecessary and unacceptable. Mr O'Brien denied that he had acted in an aggressive manner but accepted during interrogation that he was unaware of the changes to Rule 90 (i) (Identification and Vaccinations Checks). He apologised for his failure to be familiar with the Rules pertaining to the presentation of passports at the racecourse. Having considered the reports and all the evidence, the Stewards were satisfied that William Coleman O'Brien was in breach of Rule 272 in that his behaviour to Officials was unacceptable and furthermore found him to be in breach of Rule 147 (i) by failing to familiarise himself with the change to Rule 90 (i). Having taken his previous record into consideration, the Stewards fined William Coleman O'Brien €500.
Under the provisions of Rule 90 (iii), the Stewards allowed Moondust Magic, trained by Seamus Fahey, to run on receipt of the trainer's written assurance that the horse was correctly vaccinated in accordance with Rule 91.


text:
Journal of Disaster and Emergency Research
COVID-19 Pandemic: The Mental Health Impact on Elders
Document Type : Letter to The Editor
Mostafa Roshanravan 1
Shandiz Moslehi 2
1 Department of Health in Disasters and Emergencies, School of Health Management and Information Sciences, Iran University of Medical Sciences, Tehran, Iran
2 1.Health Management and Economics Research Center, Health Management Research Institute, Iran University of Medical Sciences, Tehran, Iran 2.Health Management and Economics Research Center, Department of Health in disasters and emergencies, Iran University of Medical Sciences, Tehran, Iran
preventive measures
Pandemics can have significant psychological effects such as anxiety, insomnia and depression (1). The novel coronavirus (COVID-19) and factors such as the lack of accurate information about the virus, the risk of virus transmission to individuals and family members, and the lack of a definitive drug to treat it, increase the psychological effects of the elderly (2). COVID-19 pandemic is known as an adult problem and therefore encourages older people to separate from younger people, thus creating an age difference in understanding the consequences of COVID-19 (1).
A recent study reported that at lateadulthood social isolation has a more considerable risk of anxiety and depression (3). Also due to ongoing quarantines cannot reach their outpatient clinics in order to obtain their monthly prescriptions, these results in perception of lack of treatment and exacerbation of pre-existing psychiatric symptoms (4). Increased suicidal ideations and attempts consequent to stress, on the background of the already existing suicidality risk in the elderly, is an added concern (5).
Elderly individuals may be more exposed to these complications because they are vulnerable and sensitive (1). There are several reasons why the elderly constitutes an especially vulnerable people in disaster, for example: advanced age itself is a pre-disposing factor to physical and mental health issues. Older people generally have only a close circle of friends and family with whom they routinely interact. There is the problem of access to medicines and health facilities, as the elderly may have physical problems that make this difficult. The flow of information about epidemics may be a source of anxiety and stress in the elderly (2). Also the elderly who are not good financial situation have more mental health problems (6).
According to the stated cases to improve the mental health conditions of the elderly, the tips should be considered. Including:
Interventions, such as using online mental health services and online social support (7).
Telephonic contacts with friends, family, health caregivers, and voluntary organizations while isolation period during the COVID-19 pandemic, can give the elderly hope (7).
Using telemedicine services which can
help to take care of the elderly during their seclusion (7).
Access to general and psychiatric medical care, healthy contact with family, access to relevant and updated information, and respecting their personal space are important components of mental health care in the elderly (8).
It is helpful to change the lifestyle of the elderly and perform daily activities. Activities such as music, painting, and literature enrich their daily lives (9).
Exercise can be helpful for controlling major depressive disorder (MDD) and other psychiatric disorders and chronic diseases. Many physiological mechanisms involved in the endocrine axis and immune function can be improved with regular exercise, reducing both the severity of mental disorders and the possible severity of COVID-19 infection (9).
Other family members should take some time to talk to older members of the family and get involved in some of their daily routines (10).
Stores offer special hours for the elderly, such as the early hours of the day.
Government and non-governmental organizations should cooperate in delivering the equipment and medicines needed by the elderly.
Giving financial aid will help to create peace of mind and a sense of financial security in the elderly (2).
Allocating a hot-line for the elderly to call if necessary.
Giving the right information through the media can be helpful in reducing their
anxiety (11).
Health policy makers have an important role to play in overcoming these barriers, and their cooperation will provide high-quality and timely psychological services to the elderly (4).
In short, the health of the elderly needs more attention, especially when public health emergencies occur. In such cases, the society
and professionals must take the necessary interventions to reduce the psychological impact in the elderly (12).
Authors' contribution
All authors contributed to this project and article equally. All authors read and approved the final manuscript.
Bergman YS, Cohen-Fridel S, Shrira A, Bodner E, Palgi Y. COVID-19 health worries and anxiety symptoms among older adults: the moderating role of ageism. International psychogeriatrics. 2020;32(11):1371-5.
Philip J, Cherian V. Impact of COVID-19 on mental health of the elderly. Int J Community Med Public Health. 2020;7:2435.
Santini ZI, Jose PE, Cornwell EY, Koyanagi A, Nielsen L, Hinrichsen C, et al. Social disconnectedness, perceived isolation, and symptoms of depression and anxiety among older Americans (NSHAP): a longitudinal mediation analysis. The Lancet Public Health. 2020;5(1):e62-e70.
Yang Y, Li W, Zhang Q, Zhang L, Cheung T, Xiang Y-T. Mental health services for older adults in China during the COVID-19 outbreak. The Lancet Psychiatry. 2020;7(4):e19.
Armitage R, Nellums LB. COVID-19 and the consequences of isolating the elderly. The Lancet Public Health. 2020;5(5):e256.
Chandola T, Kumari M, Booker CL, Benzeval M. The mental health impact of COVID-19 and lockdown-related stressors among adults in the UK. Psychological medicine. 2020:1-10.
Shuja KH, Aqeel M, Khan EA, Abbas J. Letter to highlight the effects of isolation on elderly during COVID‐19 outbreak. International Journal of Geriatric Psychiatry. 2020;35(12): 1477-8.
Banerjee D. The impact of Covid‐19 pandemic on elderly mental health. International journal of geriatric psychiatry. 2020.
Grolli RE, Mingoti MED, Bertollo AG, Luzardo AR, Quevedo J, Réus GZ, et al. Impact of COVID-19 in the mental health in elderly: psychological and biological updates. Molecular Neurobiology. 2021;58(5):1905-16.
Javed B, Sarwer A, Soto EB, Mashwani ZuR. The coronavirus (COVID‐19) pandemic's impact on mental health. The International journal of health planning and management. 2020;35(5):
993-6.
Ozamiz-Etxebarria N, Dosil-Santamaria M, Picaza-Gorrochategui M, Idoiaga-Mondragon N. Stress, anxiety, and depression levels in the initial stage of the COVID-19 outbreak in a population sample in the northern Spain. Cadernos de saude publica. 2020;36.
Meng H, Xu Y, Dai J, Zhang Y, Liu B, Yang H. Analyze the psychological impact of COVID-19 among the elderly population in China and make corresponding suggestions. Psychiatry research. 2020;289:112983.
Roshanravan, M., Moslehi, S. (2021). COVID-19 Pandemic: The Mental Health Impact on Elders. Journal of Disaster and Emergency Research, 4(2), 66-68.
Mostafa Roshanravan; Shandiz Moslehi. "COVID-19 Pandemic: The Mental Health Impact on Elders". Journal of Disaster and Emergency Research, 4, 2, 2021, 66-68.
Roshanravan, M., Moslehi, S. (2021). 'COVID-19 Pandemic: The Mental Health Impact on Elders', Journal of Disaster and Emergency Research, 4(2), pp. 66-68.
Roshanravan, M., Moslehi, S. COVID-19 Pandemic: The Mental Health Impact on Elders. Journal of Disaster and Emergency Research, 2021; 4(2): 66-68.
Creative Commons Attribution 4.0 International License.


Need help getting organized for your big day?
With all the varieties of items to choose from, it's no wonder why I loose track of what I'm intentionally browsing for. Shopping online can become a stressful and confusing experience.
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Are You Taking Christmas Photos This Year?


Kareena Kapoor Khan: If someone will leave you awestruck with her no-makeup look, it has to be Taimur Ali Khan's mother, Kareena Kapoor Khan. Regardless her age, Kareena is just glowing as the... Makeup is a great way to emphasize and highlight things you like about your face. Without it, you can still do that. Choosing colors that �work� with your skin tone and eye color is a good first step.
They're used to piling on the foundations, powders and everything in between, but when the lights go out and the cameras go off what do these stars look like w how to make a whale cake topper It makes anyone's skin look flawless," he says. Here's how to get in on the miracle. Follow these Here's how to get in on the miracle. Follow these make-up steps !
There are probably many times you�d love to go out without makeup and avoid the hassle, but get nervous of how your skin looks without it. Makeup free and gorgeous skin first starts with a good cleansing routine. how to make money without a job at 17 Makeup is a great way to emphasize and highlight things you like about your face. Without it, you can still do that. Choosing colors that �work� with your skin tone and eye color is a good first step.
Makeup is a great way to emphasize and highlight things you like about your face. Without it, you can still do that. Choosing colors that �work� with your skin tone and eye color is a good first step.
How to Fake Flawless Skin-Without Caked On Makeup! January 29, 2015 by Melissa 26 Comments. Disclosure: This post may contain affiliate links. I receive a small commission at no cost to you when you make a purchase using my link.


text:
Posted on June 20, 2016 May 30, 2017
Shelby County's online-only real estate tax sale begins Wednesday
MEMPHIS, Tenn. (June 20,2016) By Linda A. Moore of The Commercial Appeal
Shelby County will holds its third online-only real estate tax sale Wednesday through Friday with an expected 400 properties available for purchase.
The sale begins at 8 a.m. Wednesday and is scheduled to end at 8 p.m. Friday unless bidding on properties continues.
The vacant lots, commercial buildings and houses will be randomly grouped to open and close at different times, allowing investors an opportunity to bid throughout the sale.
The opening bids represent the delinquent city and county taxes, accrued interest and penalties, court costs, attorney fees, service and title costs. Also, owners of a property have until the close of business the day before the sale to pay the amount owed and remove their property from the sale.
Anyone interested in participating can register at www.CivicSource.com.
The sale is conducted four times a year and is a joint effort of the chancery court clerk and master and the county trustee's office.
The county held it's first online-only tax sale in January, followed by a second in April.
SHELBY COUNTY TO HOST THE THIRD ONLINE TAX SALE JUNE 22-24
Sending more properties to tax sales fights blight; puts pressure on tax-delinquent owners to pay
MEMPHIS, Tenn. (June 16, 2016) – In the third of four innovative online tax sale auctions, Shelby County Government will offer over 400 residential, commercial and vacant properties at www.CivicSource.com, beginning Wednesday, June 22 at 8 a.m. through Friday, June 24 at 8 p.m..
The tax sale is conducted four times each year and is a joint effort of the Chancery Court Clerk & Master and the Shelby County Trustee's Office. For the third auction, the 400 available properties are randomly grouped into four batches, with each batch having a different closing time. This will allow investors to redeploy unused capital throughout the sale.
As in the past, the opening bid for a property includes delinquent city and county taxes, accrued interest and penalty charges, court costs, attorney fees, as well as service and title costs. Owners of the property may still pay the amount of taxes governing the sale up until the close of business prior to the day of the sale.
"Nearly 50 properties were purchased in the last tax sale which was held in April. For the next to last sale for the year, we hope the pool of prospective bidders broadens to aid in delinquent collections and our fight against blight," said Shelby County Trustee David C. Lenoir. "Everyone pays more because of the blight in our communities. Blighted properties represent money lost in unpaid taxes and decreased property values."
Interested parties can sign up for the tax sale at http://www.civicsource.com. Register now for a free account and familiarize yourself with several investor friendly features such as custom watch-lists, sliding close and proxy biding.
Additional questions can be directed to the Shelby County Trustee, at (901) 222-0200, or the Chancery Court Clerk & Master's office at (901) 222-3900.
Real Estate Road Show Returns June 16
Memphis Daily News
MEMPHIS, TN (June 12, 2016) A primer on Shelby County property tax sales is June 16 at the Memphis Bioworks Foundation auditorium, 20 S. Dudley St., from 2:30 p.m. to 4 p.m., starting with a 2 p.m. sign-in.
The Real Estate Road Show session comes in advance of the three-day online tax sale June 22-24, which is expected to include about 400 properties – residential, commercial and vacant land.
Real estate information company Chandler Reports, which is owned byThe Daily News Publishing Co., is a sponsor of the road show.
The road show is an explanation of how the sales work, with representatives from Chancery Court and the Shelby County Trustee's office present to walk those interested through the steps and the conditions for buying property.
Opening bids include delinquent city and county taxes, accrued interest and penalty charges, court costs, attorney fees, and service and title costs.
The briefing will include an explanation of recent changes in the redemption period for properties.
The online sales began in January through CivicSource.com, and those attending the road show are encouraged to bring laptops or other mobile devices to test the new software.
Register for the road show at reroadshowmemphis.eventbrite.com.
– Bill Dries
D.A.'s office leading effort to sell tax delinquent properties
L'Observateur
Published 12:10 am Wednesday, June 8, 2016
EDGARD — There have been 11 adjudicated properties in St. John the Baptist Parish sold through a partnership between the St. John the Baptist Parish District Attorney's Office and auction house CivicSource. District Attorney Bridget A. Dinvaut said three more properties are being auctioned this month. Dinvaut said she is impressed with CivicSource for numerous reasons, one being CivicSource gives new property owners title insurance.


text:
Q: Visualize large dimension clusters in R using k-means I have a dataset with several rows and 5 dimensions (all numeric). After normalization, I applied the k-means algorithm in order to clusterize the data.
clus2_k3<-kmeans(clus2,centers=3)

After this step I would like to visualize the result, but as it has more than 3 dimensions it is not possible to use 2D or 3D plot.
Is there any command or algorithm to plot it, or if not, an alternative way to reduce the number of dimensions without losing the information from the substracted ones?

A: It's fairly standard to do pairwise plots, e.g.

This is 4 dimensional data. The same approach still works for 5 dimensions, too.
See the R command pairs from your R manual for details.

A: Okay, it is completely unreadable as a comment...
require(ggplot2)
data("iris")

pca_res <- prcomp(as.matrix(iris[, 1:4]), center = TRUE, scale. = TRUE)
plot_data <- cbind(as.data.frame(pca_res$x[, 1:2]), labels = iris[, 5])

ggplot(plot_data, aes(x = PC1, y = PC2, colour = labels)) +
  geom_point()

Edit: You may try different combinations of center and scale. params, as e.g. this set looks a bit better with both set to FALSE:

Edit:
To see the "loss of information" mentioned in the comment one may use the summary() function:
summary(pca_res)

# Importance of components:
#                           PC1    PC2     PC3     PC4
# Standard deviation     1.7084 0.9560 0.38309 0.14393
# Proportion of Variance 0.7296 0.2285 0.03669 0.00518
# Cumulative Proportion  0.7296 0.9581 0.99482 1.00000

Here PC1 and PC2 stand for 0.96 of cumulative prop. of variance, which means 96% of "information" is stored in those two components.

A: You can use plot function only but you need to specify particular column which K-means throws, it will throws result as list with name cluster.
Try this code:
plot(clus2_k3$cluster)


text:
How MP3 Files Work
by Marshall Brain
Distributing Original Music
If you are an artist who is recording music at home or in a small studio, you can use MP3 files and the Web to distribute your music to a larger audience. The first step is to create a song, either on a cassette tape, minidisc or CD. If it's on a CD, you can use the ripper and encoder tools described in the previous section to create an MP3 file. If it's on a cassette or other source, you can connect the output of the audio source to the line-in or microphone jack of your sound card and record the music digitally on your computer. Then you can encode that file to create the MP3.
Once you have an MP3 file in hand, you have two distribution options:
You can go to an MP3-distribution site and let them distribute your music. The advantage of this approach is that large MP3-distribution sites gets millions of visitors every month, so your potential audience is very large.
You can create your own Web site for your music or band and promote the site yourself. This gives you more control and individuality, but requires you to get the word out on your own. See How Web Pages Work for details on creating and hosting your own Web site.
Some musicians distribute their music through a blog. Jonathan Coulton, known for his comedic folk songs, uses a blog to keep his fans informed of what's going on in his life. They can also find links to purchase his music or listen to several of his songs for free. Coulton's success with this untraditional approach may inspire other musicians to follow suit.
One good option is to make your MP3 files available on a large Web site and then link to the download area from your band's Web site. This lets you get the best of both worlds, and you can take advantage of the larger site's servers for those big MP3 files.
For more information on the MP3 format, MP3 sites and related topics, check out the links below.
Do you have a license for that band?
Putting your music out on the Web can be a great promotional tool, but it might also require you to give away your hard work for free. If you want to maintain control over your work, you might want to look into a Creative Commons license. These licenses are free and allow you to maintain some control of your work while making it available to other people.
Can I run my MP3 player on methanol?
How MP3 Players Work
MP3 Player Quiz
How iPods Work
How iTunes Works
How Podcasting Works
How CD Burners Work
How CDs Work
How File Compression Works
How File Sharing Works
How Sound Cards Work
MPEG.ORG
Decker, Logan. "Do Higher MP3 Bit Rates Pay Off?" Maximum PC. April 19, 2007. (Aug. 15, 2008) http://www.maximumpc.com/article/do_higher_mp3_bit_rates_pay_off
Digital Audio Systems. "MP3 Encoding." (Aug. 15, 2008) http://www.digital-audio.net/technical_enc.shtml
Fraunhofer IIS. "MP3: MPEG Audio Layer III." (Aug. 15, 2008) http://www.iis.fraunhofer.de/EN/bf/amm/projects/mp3/index.jsp
Levine, Robert. "The Death of High Fidelity." Rolling Stone Magazine. Dec. 27, 2007. (Aug. 14, 2008) http://www.rollingstone.com/news/story/17777619/the_death_of_high_fidelity/print
Loukides, Mike. "High Bit Rate MP3s: Are they Necessary?" O'Reilly Media. March 1, 2000. (Aug. 14, 2008) http://www.oreillynet.com/pub/a/oreilly/web/news/high_bit_rate_0300.html
MP3-Converter.com. "Bit Rates and Sound Quality." (Aug. 15, 2008) http://www.mp3-converter.com/bitrates.htm
MPEG.org. (Aug. 15, 2008) http://www.mpeg.org/MPEG/mpeg-pointers-and-resources/
Google Goes Incognito With Vanishing Gmail
5 Factors That Affect Cloud-based Data Upload and Retrieval
How Data Centers Work


The widest varieties of laces are included in the collection of the most reputed Lace Wholesalers in Kolkata. Accessorize your garments with pretty laces.
Nothing can make a woman feel more feminine than wearing a lacy attire. Lace fabrics can readily make a style statement and this is why it is always considered as the most graceful fabric of every season. Lacy dresses give a beautiful look and at the same time, it can make a woman look classy and sensuous. Wearing lacy attire is the ideal way to express every wearer's personal style. Varieties of laces are included in the assortment of the best Lace Wholesalers in Kolkata besides the other varieties of these pretty accessories.
There are generally two types of handmade laces. The first variety is named as the needle laces that is made using needle and thread and by creating variations on the buttonhole stitch. This is the most time-consuming and the most flexible lace making technique. The second variety is termed as the bobbin lace that is made by plaiting and twirling different threads. Bobbins and pillow are used to make the bobbin lace. Chantilly lace is a category of bobbin lace and is popularly known as the bridal lace. Moreover, these handmade laces are designed beautifully and differently, each created in a distinct way using different tools.
Crochet, tatting, and knitting are the most common lace varieties that are made using a single thread method and knotted lace, needle lace, and embroidery lace is made using multiple thread procedure. Laces replicated or made using machines are known as the machine-made laces. Cutwork, knotted lace, tape lace, tatted lace, crocheted lace, knitted lace, etc. are made differently in the textile industries. Threads are removed from the woven background for making white-work lace or cutwork lace. The left over threads are either filled with intricate embroidery works or wrapped neatly whereas tape laces are made by using handmade textile strips or using a machine for creating the designs.
Knitting technique is used to make big or tubular items. Long needles are used in this case. The popular Shetland lace is a type of knitted lace. Tatted lace is made using an oval-shaped shuttle with shuttling needle or a pointed end. Crochet laces use steel hook approximately of 6 inches long. Pineapple crochet, apple crochet, and filet crochet are all the different varieties of crocheted lace. The embroidered lace is made on machine-made textile using a plastic or wooden frame and needle.
Different types of machines are needed to manufacture laces. For example, a stocking frame, hand embroidery machine, barmen machine, and bobbinet machine. The bobbinet machine was invented first and then came the leaver and the pusher machine. The leaver machine is one of the most efficient machines for making the patterned laces. The hand embroidery machine makes multiple copies of any embroidered lace.
Lingerie accessorized with laces or fully made with lace fabrics attracts hundreds of modern women. Moreover, lacy fabrics that were once used only for the intimate wears have now conquered the entire fashion world. Whether it is a layered top or a stylish dress, lacy fabrics looks elegant in every case. Lacy border on even the simplest outfit can make the attire look urbane and beautiful. Several reputed fashion designers have created ample of endearing lace creations via uncountable ways and have been able to instill lace into the contemporary clothing styles. Many Fabric Importers in Kolkata have exposed the love and demand for lace fabrics through their huge collection.
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text:
Pioneers of Television
Season 4 Episode 2 | 53m 41s
From George Clooney on "ER" to Richard Chamberlain on "Dr. Kildare," television's long love affair with doctors and nurses shows no signs of letting up. Noah Wyle, Anthony Edwards, Gloria Reuben and Eriq LaSalle open up about the secrets of "ER"; Howie Mandel, Ed Begley Jr. and Christina Pickles revisit "St. Elsewhere."
Acting Funny
A look at the backstage techniques of America's favorite comedic actors.
S4 Ep4 | 53m 11s
Breaking Barriers
Trace the story of people of color on American television.
Review television's long love affair with doctors and nurses.
Standup to Sitcom
See how America's top standup comics made the transition to the sitcom format.
Doctors and Nurses Preview
Explore television's long love affair with doctors and nurses from ER to Dr. Kildare.


text:
MOT takes ownership of Atlanta building
June 15, 2007 — 10.35pm
Macquarie Office Trust has taken full ownership of the Promenade II office building in Atlanta.
Purchasing the remaining 50 per cent stake in the class-A office building will cost the fund $US122 million ($A146.24 million.)
"This acquisition allows the trust to redeploy the proceeds of a previous successful asset sale into a higher growth opportunity through a section 1031 transaction," chief executive Adrian Taylor said.
"Midtown is a location of choice for many of Atlanta's most successful businesses.
"This preference is expected to intensify with the concentration of existing and planned high-end hotel and residential amenities, as well as further development of the Midtown Mile retail corridor," he said.
The purchase will be partly funded by the trust's sale in December of One Liberty Square, Boston, which returned $US50.4 million ($A60.41 million), a net gain of $US13.5 million ($A16.18 million).
The 38-storey Promenade II officer tower was built in 1990 and is 98 per cent leased.
Macquarie Office Trust (MOT) purchased the final stake from private equity firm, The Blackstone Group.
MOT is in discussion with Blackstone about buying them out of a further four remaining joint venture investments.
While the discussion are confidential, MOT said it both parties had agreed to market their interests in 10 and 30 South Wacker Drive, Chicago.


Composite Materials and Sandwich Structures A Primer - Defense ...sandwich construction have resulted in lighter structural designs with superior performance. ... Advanced composite materials used in structural applications are obtained by reinforcing a matrix material with continuous fibers having high strength and stiffness properties. The selection of a composite material for any application will involve selection of reinforcing fiber and matrix, and their fractional volume in the resulting ..... The mains disadvantages of honeycomb construction are-.
Composite Materials for Aircraft Structures - Montana State Universitycomposite Materials for Aircraft Structures. Dr. Douglas S. Cairns,. Lysle A. Wood Distinguished Professor. Department of Mechanical and Industrial Engineering. Montana State University. ME 463 composites,. Fall 2009 2009 .... SummaryAdvantages and disadvantages of composite Materials. Advantages. disadvantages. Weight reduction. (approximately 20-50%). Some higher recurring costs. Higher nonrecurring costs. Corrosion resistance. Fatigue resistance.
Composite construction - SlideShare21 Jan 2016 ... AL-NAHRAINUNIVERSITY COLLEGEof ENGINEERING CIVIL ENGINEERINGDEPARTMENT ADVANCESTRUCTURALDESIGN composite construction Prepared By: Hashim Dheyaa.
Composites And Its Classification, Advantages, Disadvantages ...composites are defined as the combination of two or more chemically different constituents which combines macroscopically to produce a useful material. Natural composite Materials Are: Wood, Granite and Bone. Artificial composite Materials Are: Plywood, concrete, fibrous composites, fiber glass, plastic, alloys metals along with the impurities etc.,. composite Material Usage: Sports. In Sports they are used for the Bicycles, Golfing, Boats, badminton, Tennis and Hockey.
Nonmetallic Aircraft Materials Part Two (Advantages ...Bonded construction eliminates joints and fasteners. Easily repairable. The disadvantages of composites include: Inspection methods difficult to conduct, especially delamination detection (Advancements in technology will eventually correct this problem.) Lack of long term design database, relatively new technology methods. Cost. Very expensive processing equipment. Lack of standardized system of methodology. Great variety of materials, processes, and techniques.


text:
Q: Cannot Resolve Symbol "Scalatest" I am trying to use scalatest, but Intellij cannot recognize:
import org.scalatest._

Here is my build.sbt file, located in the same directory as my scalatest.jar file.
scalaVersion := "2.11.2"

libraryDependencies += "org.scalatest" % "scalatest_2.11" % "2.2.4" % "test"

Thanks

A: So you have by convention two source folders:
src/main/scala/...
src/test/scala/...

The first is shown blue, the second green in IntelliJ IDEA. The library dependencies in sbt are associated with either of these, so
"org.foo" % "bar_2.11" % "1.2.3"

Is a main dependency, available to main sources (and also test, because test depends on main). And
"org.foo" % "bar_2.11" % "1.2.3" % "test"

Is a test dependency, only available to test sources. The idea is that these are libraries which are not required for your product, but only to run the unit tests.

In your example, Scala-Test is only available to test sources, so trying to import it from main sources will fail.


text:
Q: JS Declare and assign multi variable inLine es6? it possible to declare and assign multi variable in same line with es6 syntax ?
just example here,
suppose i need to give value "50" to mX,mY,mXDZ,mYDZ.
How i can perform this ?
let [mX,mY,mXDZ,mYDZ] = 50;
let mX,mY,mXDZ,mYDZ = 50; // !just why not!

I am mainly looking for readability.

A: No, that's not exactly possible. Destructuring needs a thing with multiple values1 to be assigned to multiple targets, not a single value. You can however cheat:
function* repeat(x) { while(true) yield x; }

let [mX, mY, mXDZ, mYDZ] = repeat(50);

Apart from that, in a multiple declaration every variable needs its own initialiser:
let mX = 50, mY = 50, mXDZ = 50, mYDZ = 50;

for which you can of course also use one shared variable:
let val = 50, mX = val, mY = val, mXDZ = val, mYDZ = val;

1: You can also destructure the same property multiple times (let {val:mX, val:mY, val:mXDZ, val:mYDZ} = {val:50};), but it still requires an object.

A: You can declare all the variables and then in a separate statement initialize them:
let mX,mY,mXDZ,mYDZ;
mX = mY = mXDZ = mYDZ = 50;

In a variable declaration statement (let, var, or const), the syntax is such that each individual variable can have an initialization expression. There are no provisions for multiple initializations from one expression.
With destructuring, you could do this:
let [mX, mY, mXDZ, mYDZ] = [50, 50, 50, 50];

but that doesn't seem "better" in any way to me, at least for readability. It seems error-prone.
It's very important to note that
let mX = mY = mXDZ = mYDZ = 50;

is definitely not correct.  That statement is parsed as
let mX = (mY = mXDZ = mYDZ = 50);

That will create one local variable ("mX") and three implicit global variables.  In "strict" mode it would be an error; in non-strict mode it's also an error but it's harder to debug.


Like clockwork, ESPN has rolled right into the 2015 NFL Draft following the conclusion of 2014's. Toddy McShay has already put together a way-too-early look at how next year's draft could shake out. It may be a bit premature, but the Worldwide Leader has to start popping them out now if it wants to reach the quota of 986 mock drafts over the next 12 months.
McShay uses Aaron Schatz's projected win totals to lay out the draft order. To little surprise, he has Oakland, Cleveland and Jacksonville selecting near the top again. He also includes draft-eligible non-seniors in his mock.
If there is one thing to learn from this mock draft, it's that Florida State is loaded (which is probably a bit of an understatement). McShay has six FSU players being selected in the top 26 picks.
Without further ado, here is 2015 NFL Mock Draft No. 1 of 986.