text:
\section{Introduction}

The development of integration on time scales is a recent
but already well studied subject. Available integration notions
on time scales include the Riemann delta integral \cite{Bh1,G:K:2002,G:K:2004},
the Riemann nabla integral \cite{G:K:2002}, the Riemann diamond-alpha integral
\cite{M:T:DSA}, the Lebesgue delta and nabla integrals \cite{Bh1,G:2003},
and the more general Henstock-Kurzweil delta and nabla integrals \cite{P:2006,Thomson}.
Other studies on time scales are dedicated to improper integrals \cite{B:G:2003}
and to multiple integration \cite{B:G:2005,B:G:2006}. Surprisingly enough, the
Stieltjes integral has not received attention in the literature of time scales.

In this paper we study the process of Stieltjes integration
on time scales, both in nabla and delta sense.
We trust that such integrals will find interesting applications in the study of dynamic equations on time scales, enabling to study more general situations than those treated before (\textrm{cf.} Section~\ref{sec:ConcFuture}).
Because delta and nabla theories are similar,
we avoid repetition by following the approach promoted by Bartosiewicz and Piotrowska \cite{BPi}: the box symbol $\Box$ is here used to represent the delta operator $\Delta$ as well as the nabla operator $\nabla$.

It is assumed that the reader is familiar with the time
scale calculus and the notations for delta and nabla differentiation.
For an introduction to time scales the reader is referred
to the book by Bohner and Peterson \cite{Bh}.
Throughout the paper $\mathbb{T}$ denotes a time scale.
Let $a,b\in\mathbb{T}$ and $a<b$. We distinguish $[a,b]$ as a real interval and we define $I=[a,b]_{\mathbb{T}}:=[a,b]\cap\mathbb{T}$.
In this sense, $[a,b]=[a,b]_{\mathbb{R}}$. Along the text
$I$ is a nonempty, closed, and bounded interval consisting
of points from a time scale $\mathbb{T}$.
Moreover, if $I=[a,b]_{\mathbb{T}}$, then we define
$I_{\Delta}:=[a,\rho(b)]_{\mathbb{T}}$ and $I_{\nabla}:=[\sigma(a), b]_{\mathbb{T}}$. By $I_{\Box}$ we denote one of them, where $\Box$
means either $\nabla$ or $\Delta$. Similarly, we use "$\Box$"
as  a common notation for the two kinds of derivatives on time scales: one can read $f^{\Box}$ either as $f^{\Delta}$
or as $f^{\nabla}$.

\section{The Riemann-Stieltjes $\Box$--integral}
\label{sec:RS}

Let $\mathbb{T}$ be a time scale, $a,b\in\mathbb{T}$, $a<b$,
and $I=[a,b]_{\mathbb{T}}$. A partition of $I$ is any finite ordered subset
\[
P=\{t_0, t_1,\ldots, t_n\}\subset [a,b]_{\mathbb{T}}, \quad
\text{where} \  a=t_0<t_1<\ldots<t_n=b\, .
\]
Let  $g$ be a real-valued increasing function on $I$. Each partition
$P=\{t_0, t_1,\ldots, t_n\}$ of $I$ decomposes $I$ into
subintervals $I_{\Box j}=[t_{j-1},t_j]_{\Box}$, $j=1,2,\ldots, n$, such
that $I_{\Box j}\cap I_{\Box k}=\emptyset$ for any $k\neq j$.
By $\Delta t_j=t_j-t_{j-1}$ we denote the length of the $j$th
subinterval in the partition $P$; by $\mathcal{P}(I)$ the
set of all partitions of $I$.

Let $P_m$, $P_n \in\mathcal{ P}(I)$. If $P_m\subset P_n$ we call $P_n$
a refinement of $P_m$. If $P_m$, $P_n$ are independently chosen, then
the partition $P_m\cup P_n$ is a common refinement of $P_m$ and
$P_n$.

Let us now consider a strictly increasing real-valued function $g$
on the interval $I$. Then, for the partition $P$ of $I$ we define
\[g(P)=\{g(a)=g(t_0), g(t_1), \ldots, g(t_{n-1}),g(t_n)=g(b)\}\subset g(I)\]
and $\Delta g_j=g(t_j)-g(t_{j-1})$. We note that $\Delta g_j$ is
positive and $\sum\limits_{j=1}^n\Delta g_j=g(b)-g(a)$. Moreover,
$g(P)$ is a partition of $[g(a),g(b)]_{\mathbb{R}}$. In what follows, for
the particular case $g(t)=t$ we obtain the Riemann sums for delta
integrals studied in \cite{Bh1}. We note that for a general $g$ the image $g(I)$ is not necessarily an interval in the classical sense, even for rd-continuous functions $g$, because our interval $I$ may contain scattered points. From now on let $g$ be always a strictly increasing real function on the considered interval $I=[a,b]_{\mathbb{T}}$.

\begin{lemma}
\label{lem:2}
Let $I=[a,b]_{\mathbb{T}}$ be a closed (bounded) interval in $\mathbb{T}$ and let $g$ be continuous on $I$. For every $\delta>0$
there is a partition $P_{\delta}=\{t_0,t_1,\ldots, t_n\}\in
\mathcal{P}(I)$ such that for each $j\in\{1,2,\ldots,n\}$ one has:
\[\Delta g_j=g(t_j)-g(t_{j-1})\leq \delta \quad \mbox{or} \quad
\Delta g_j>\delta \ \wedge \ \rho(t_j)=t_{j-1}.\]
\end{lemma}
\begin{proof}
Let $I=[a,b]_{\mathbb{T}}$ be closed (bounded) interval in $\mathbb{T}$. Firstly, let
us observe that if for $t\in I$ $g(t)\in [g(a),g(b)]_{\mathbb{R}}$, then
$a\leq t\leq b$ because $g$ is an increasing function on $I$. We
define two families of sets,
$B_j=(g(t_{j-1}),g(t_{j-1})+\delta]_{\mathbb{R}}\cap [g(a),g(b)]_{\mathbb{R}}$ and
$A_j=\{t\in I: g(t)\in B_j\}$. Inductively, we
construct the partition taking $t_0=a$ and
\[t_j=\left\{\begin{array}{lr} \sup A_j,  & \mbox{if} \ A_j\neq \emptyset \, ,\\
		\sigma(t_{j-1}), & \mbox{if} \ A_j= \emptyset \, .
\end{array}\right. \]
Then, we get $t_0<t_1<\ldots<t_n$, where $t_n=b$. It follows that
if $t_{j-1}<t_j$ and $t_j=\sigma(t_{j-1})$,
then $\rho(t_j)=t_{j-1}$ for any time scale $\mathbb{T}$.
\end{proof}

Let $f$ be a real-valued and bounded  function on the interval
$I$. Let us take a partition $P=\{t_0, t_1,\ldots, t_n\}$ of $I$.
Denote $I_{\Box j}=[t_{j-1},t_j]_{\Box}$, $j=1,2,\ldots,n$, and
\[m_{\Box j}=\inf\limits_{t\in I_{\Box j}} f(t), \quad  M_{\Box j}=\sup\limits_{t\in I_{\Box j}} f(t)
\, . \]
The \emph{upper Darboux--Stieltjes $\Box$--sum of $f$ with respect to the
partition $P$}, denoted by $U_{\Box}(P,f,g)$, is defined by
\[U_{\Box}(P, f, g)=\sum_{j=1}^nM_{\Box j}\Delta g_j \, ,\]
while the \emph{lower Darboux--Stieltjes $\Box$--sum of $f$ with respect to the
partition $P$}, denoted by $L_{\Box}(P,f,g)$, is defined by
\[L_{\Box}(P, f, g)=\sum_{j=1}^n m_{\Box j}\Delta g_j\,.\]

\begin{example}
\label{ex1}
Let $\mathbb{T}=\overline{q^{\mathbb{Z}}}$, $q>1$, $f(t)=t$,
$g(t)=t^2$, and $I=[0,1]_{\mathbb{T}}$. For the partition
$P=\{0,q^{-n+1}, \ldots, q^{-1},1\}$, where
$t_0=0<q^{-n+1}<\ldots<q^{j-n}<\ldots<q^{-1}<1=t_n$ with
$t_j=q^{j-n}$ for $j=1,\ldots, n$, we have $\Delta
g_j=t_j^2-t_{j-1}^2=q^{2(j-n-1)}(q^2-1)$  for $j= 2,\ldots,n$,
while $\Delta g_1=t_1^2-0=q^{2-2n}$.
Let us read $\Box=\Delta$. In this case
$M_{\Delta j}=\rho(t_j)$ and $m_{\Delta j}=t_{j-1}$. For our partition we have that
$I_{\Delta j}=[t_{j-1},\rho(t_j)]_{\mathbb{T}}=t_{j-1}$, $j= 2,\ldots,n$. Hence,
$m_{\Delta j}=M_{\Delta j}=t_{j-1}$ for $f(t)=t$, $j=2,\ldots,n$, and
$m_{\Delta 1}=0$, $M_{\Delta 1}=\rho(t_1)=q^{-n}$. The lower and upper
Darboux--Stieltjes $\Delta$--sums are, respectively,
\[L_{\Delta}(P,f,g)=\sum\limits_{j=2}^n t_{j-1}\Delta g_j
=\frac{q+1}{q^2+q+1}(1-q^{3(1-n)})\,  ,\]
\[U_{\Delta}(P,f,g)=q^{2-3n}+\sum\limits_{j=2}^n t_{j-1}\Delta g_j
=q^{2-3n}+L(P,f,g)=\frac{q+1+q^{2-3n}}{q^2+q+1}\, .\]
Consider now $\Box=\nabla$. In this case
$M_{\nabla j}=t_j$ and $m_{\nabla j}=\sigma(t_{j-1})$. For our partition we have that
$I_{\nabla j}=[\sigma(t_{j-1}),t_j]_{\mathbb{T}}=t_{j}$, $j=2,\ldots,n$. Hence,
$m_{\nabla j}=M_{\nabla j}=t_{j}$ for $f(t)=t$, $j=2,\ldots,n$, and
$m_{\nabla 1}=\sigma(0)=0$, $M_{\nabla 1}=t_1=q^{1-n}$. The lower and upper Darboux--Stieltjes
$\nabla$--sums are, respectively,
\[L_{\nabla}(P,f,g)=\sum\limits_{j=2}^n t_j \Delta g_j
=q\frac{q+1}{q^2+q+1}(1-q^{3(1-n)})\,  ,\]
\[U_{\nabla}(P,f,g)=q^{3-3n}+\sum\limits_{j=2}^n t_{j}\Delta g_j
=q^{3-3n}+L_{\nabla}(P,f,g)=q\frac{q+1+q^{2-3n}}{q^2+q+1}\, .\]
\end{example}

For computing the value of a Riemann integral on time scales one
uses the fact that $\Delta t_j\geq 0$. Since we assume that
$g$ is an increasing function, then $\Delta g_j\geq 0$. As we
shall see below, also some other properties of the Riemann-integral
(for function $g(t)=t$) are preserved for an arbitrary increasing
function $g$.

\begin{theorem}
\label{lem:1}
Suppose that $f$ is a bounded function on
$I=[a,b]_{\mathbb{T}}$, $a$, $b\in\mathbb{T}$.
Let $P\in \mathcal{P}(I)$ and $m\leq f(t)\leq M$ for all $t\in I$. Then,
\[m(g(b)-g(a))\leq L_{\Box}(P,f,g)\leq U_{\Box}(P,f,g)\leq M(g(b)-g(a))\]
and
\[L_{\Box}(P,f,g)\leq L_{\Box}(Q,f,g)\leq U_{\Box}(Q,f,g)\leq U_{\Box}(P,f,g)\]
for any refinement $Q$ of $P$.
\end{theorem}
\begin{proof}
For $\Box=\Delta$ the proof is an immediate consequence of \cite[Lemma~5.2]{Bh1}. For $\Box=\nabla$ the proof is done
in similar steps.
\end{proof}

\begin{definition}
\label{RSdef}
Let $I=[a,b]_{\mathbb{T}}$, where $a,b\in\mathbb{T}$.
The upper Darboux-Stieltjes $\Box$--integral from $a$ to $b$
with respect to function $g$ is defined by
\[\overline{\int_a^b}f(t)\Box g(t)=\inf_{P\in \mathcal{P}(I)} U_{\Box}(P,f,g)\, ;\]
the lower Darboux-Stieltjes $\Box$--integral from $a$ to
$b$ with respect to function $g$ is defined by
\[\underline{\int_a^b}f(t)\Box g(t)=\sup_{P\in \mathcal{P}(I)} L_{\Box}(P,f,g)\, .\]
If  $\overline{\int_a^b}f(t)\Box
g(t)=\underline{\int_a^b}f(t)\Box g(t)$, then we say that $f$ is
$\Box$--integrable with respect to $g$ on $I$, and the common
value of the integrals, denoted by $\int_a^bf(t)\Box
g(t)=\int_a^bf\Box g$, is called the Riemann-Stieltjes (or
simply Stieltjes) $\Box$--integral of $f$ with respect to $g$ on
$I$.

The set of all functions that
are $\Box$--integrable with respect to $g$ in the Riemann-Stieltjes
(also Darboux-Stieltjes) sense will be denoted by
$\mathcal{R}_{\Box}(g,I)$.
\end{definition}

\begin{theorem}
\label{th.7}
If  $L_{\Box}(P,f,g)=U_{\Box}(P,f,g)$ for some $P \in \mathcal{P}(I)$,
then function $f$ is Riemann-Stieltjes $\Box$--integrable
and \[\int_a^bf\Box g=L_{\Box}(P,f,g)=U_{\Box}(P,f,g) \, .\]
\end{theorem}
\begin{proof}
Follows immediately from the definition of Riemann-Stieltjes
$\Box$--integral and Corollary~\ref{lem:1}.
\end{proof}

\begin{example}
Let $\mathbb{T}=\overline{q^{\mathbb{Z}}}$, $q>1$,
and let us continue Example~\ref{ex1}.
Consider functions $f(t)=t$ and $g(t)=t^2$ on the interval $I=[0,1]_{\mathbb{T}}$.
For the partition $P=\{0,q^{-n+1}, \ldots, q^{-1},1\}$, where
$t_0=0<q^{-n+1}<\ldots<q^{j-n}<\ldots<q^{-1}<1=t_n$ with $t_j=q^{j-n}$ for $j=1,\ldots, n$,
we have
\[\overline{\int_a^b}f(t)\Delta g(t)=\inf\limits_{P\in \mathcal{P}(I)} U_{\Delta}(P,f,g)
=\lim\limits_{n\rightarrow\infty}\frac{q+1+q^{2-3n}}{q^2+q+1}=\frac{q+1}{q^2+q+1}\]
and
\[\underline{\int_a^b}f(t)\Delta g(t)=\sup\limits_{P\in \mathcal{P}(I)} L_{\Delta}(P,f,g)
=\lim\limits_{n\rightarrow\infty}\frac{q+1}{q^2+q+1}(1-q^{3(1-n)})
=\overline{\int_a^b}f(t)\Delta g(t)\, .\]
Consequently, $\int_a^b f(t)\Delta g(t)=\frac{q+1}{q^2+q+1}$.
We also have that
\[\overline{\int_a^b}f(t)\nabla g(t)=\inf\limits_{P\in \mathcal{P}(I)} U_{\nabla}(P,f,g)=\frac{q^2+q}{q^2+q+1}\]
and
\[\underline{\int_a^b}f(t)\nabla g(t)=\sup\limits_{P\in \mathcal{P}(I)} L_{\nabla}(P,f,g)
=\lim\limits_{n\rightarrow\infty}\frac{q^2+q}{q^2+q+1}(1-q^{3(1-n)})
=\overline{\int_a^b}f(t)\nabla g(t)\, .\]
Hence, $\int_a^b f(t)\nabla g(t)=\frac{q^2+q}{q^2+q+1}=q\int_a^b f(t)\Delta g(t)$.
\end{example}

\begin{theorem}
Let $f$ be a bounded function on $I=[a,b]_{\mathbb{T}}$, $a$, $b\in \mathbb{T}$,
$m\leq f(t)\leq M$ for all $t\in I$,
and $g$ be a function defined and monotonically increasing on $I$. Then,
\[m(g(b)-g(a))\leq \underline{\int_a^b}f(t)\Box g(t)\leq
\overline{\int_a^b}f(t)\Box g(t)\leq M(g(b)-g(a))\, .\]
If $f\in \mathcal{R}_{\Box}(g,I)$, then
\[m(g(b)-g(a))\leq \int_a^bf(t)\Box g(t)\leq  M(g(b)-g(a))\, .\]
\end{theorem}
\begin{proof}
The definition of upper and lower Darboux-Stieltjes
$\Box$--integral implies that
$\underline{\int_a^b}f(t)\Box g(t) \leq
\overline{\int_a^b}f(t)\Box g(t)$. Thus,
\[\overline{\int_a^b}f(t)\Box g(t)=\inf_{P\in \mathcal{P}(I)}
U_{\Box}(P,f,g)\leq U_{\Box}(P,f,g) \leq \sum\limits_{j=1}^nM \Delta
 g_j=M(g(b)-g(a)) \, .\]
Similarly, $m(g(b)-g(a)) \leq \underline{\int_a^b}f(t)\Box g(t)$,
and the proof is done.
\end{proof}

\begin{theorem}(Integrability criterion)
\label{dor:criterion}
Let $f$ be a bounded function on $I=[a,b]_{\mathbb{T}},a,b\in \mathbb{T}$. Then,
$f\in\mathcal{R}_{\Box}(g,I)$ if and only if for every $\mathbb{\varepsilon}>0$
there exists a partition
$P\in \mathcal{P}(I)$ such that
\begin{equation}
\label{dor:criterion1}
U_{\Box}(P,f,g)-L_{\Box}(P,f,g)<\mathbb{\varepsilon}\, .
\end{equation}
\end{theorem}
\begin{proof}
Suppose that $f\in \mathcal{R}_{\Box}(g,I)$ and let $\mathbb{\varepsilon}>0$. Because
$$\int_a^b f\Box g=\inf\limits_{P\in
\mathcal{P}(I)}U_{\Box}(P,f,g)=\sup\limits_{P\in \mathcal{P}(I)} L_{\Box}(P,f,g)$$
there exist $P_1$ and $P_2\in \mathcal{P}(I)$ such that
\[\int_a^b f\Box g<U_{\Box}(P_1,f,g)< \int_a^b f\Box g+\frac{\mathbb{\varepsilon}}{2}\]
and
\[\int_a^b f\Box g-\frac{\mathbb{\varepsilon}}{2}<L_{\Box}(P_2,f,g)< \int_a^b f\Box g\, .\]
Hence, $U_{\Box}(P_1,f,g)-\int_a^b f(t)\Box g(t)<\frac{\mathbb{\varepsilon}}{2}$ and
$\int_a^b f(t)\Box g(t)-L_{\Box}(P_2,f,g)<\frac{\mathbb{\varepsilon}}{2}$. Let $P$ be
a common refinement of $P_1$ and $P_2$. Then, Theorem~\ref{lem:1}
implies that $U_{\Box}(P,f,g)-\int_a^b f(t)\Box g(t)<\frac{\mathbb{\varepsilon}}{2}$ and
$\int_a^b f(t)\Box g(t)-L_{\Box}(P,f,g)<\frac{\mathbb{\varepsilon}}{2}$. Thus,
$U_{\Box}(P,f,g)-L_{\Box}(P,f,g)<\mathbb{\varepsilon}$.

Conversely, let $f$ be a bounded function on $I$ and $g$ be an
increasing function on $I$. Suppose that for $\mathbb{\varepsilon}>0$ there exists
$P\in \mathcal{P}(I)$ such that $U_{\Box}(P,f,g)-L_{\Box}(P,f,g)<\mathbb{\varepsilon}$. The definition
of Riemann-Stieltjes $\Box$--integral and Theorem~\ref{lem:1}
imply that
\[L_{\Box}(P,f,g)\leq \underline{\int_a^b} f\Box g
\leq \overline{\int_a^b} f\Box g\leq U_{\Box}(P,f,g)\, .\]
It is obvious that $0\leq \overline{\int_a^b} f\Box
g-\underline{\int_a^b} f\Box g<\mathbb{\varepsilon}$. Since $\mathbb{\varepsilon}$ is
arbitrary, then $\overline{\int_a^b} f(t)\Box
g(t)$ $=\underline{\int_a^b} f\Box g$ and
$f\in\mathcal{R}_{\Box}(g,I)$.
\end{proof}

\begin{theorem}
Suppose that $P_{\delta}$ is a partition as stated in
Lemma~\ref{lem:2}. A bounded function $f$ on $I=[a,b]_{\mathbb{T}}$ is
integrable if and only if for each $\mathbb{\varepsilon}>0$ there exists  $\delta>0$
such that
\begin{equation*}
P_{\delta}\in \mathcal{P}(I)
\Rightarrow
U_{\Box}(P_{\delta},f,g)- L_{\Box}(P_{\delta},f,g)<\mathbb{\varepsilon} \, .
\end{equation*}
\end{theorem}
\begin{proof}
Part "$\Leftarrow$" follows immediately from the integrability
criterion (Theorem \ref{dor:criterion}).

For the proof of the part "$\Rightarrow$" suppose that $f \in
\mathcal{R}_{\Box}(g,I)$. Let $\varepsilon>0$. Then, there exist
partitions $\mathcal{P}_{\delta}^1$ and $\mathcal{P}_{\delta}^2$
such that
\begin{equation}
\label{1}
 U(\mathcal{P}_{\delta}^1,f,g)<\overline{\int_a^b} f\Box g+\frac{\varepsilon}{2}
\end{equation}
and
\begin{equation}
\label{2}
 \underline{\int_a^b}f\Box g-\frac{\varepsilon}{2}<L(\mathcal{P}_{\delta}^2,f,g).
\end{equation}
Let
$\mathcal{P}_{\delta}=\mathcal{P}_{\delta}^1\cup\mathcal{P}_{\delta}^2$.
Theorem~\ref{lem:1} and inequalities (\ref{1}) and (\ref{2}) imply that
\[\underline{\int_a^b}f\Box g-\frac{\varepsilon}{2}<L(\mathcal{P}_{\delta},f,g) \leq
  U(\mathcal{P}_{\delta},f,g)<\underline{\int_a^b}f\Box g+\frac{\varepsilon}{2}\]
Because $f \in \mathcal{R}_{\Box}(g,I)$, then
$\underline{\int_a^b}f\Box g=\underline{\int_a^b}f\Box g$. Hence,
$U(\mathcal{P}_{\delta},f,g)-L(\mathcal{P}_{\delta},f,g)<\varepsilon$.
\end{proof}

Lemma~\ref{lem:1} and the properties of the Riemann delta (nabla) integral imply the following:
\begin{theorem}
Let $I=[a,b]_{\mathbb{T}}$, $a$, $b\in\mathbb{T}$. Then, the condition $f\in
\mathcal{R}_{\Box}(g,I)$ is equivalent to each one of the following items:
\begin{enumerate}
  \item [i)]  $f$ is a monotonic function on $I$;

  \item [ii)] $f$ is a continuous function on $I$;

  \item [iii)] $f$ is regulated on $I$;

  \item [iv)] $f$ is bounded and has a finite number of discontinuity points on $I$.
\end{enumerate}
\end{theorem}

\begin{theorem}
\label{dor:3}
Let $f$ be bounded on $I=[a,b]_{\mathbb{T}}$, $a<b$. Then,
\begin{enumerate}
 \item[i)] If there exists an $\mathbb{\varepsilon}>0$ and a partition
 $P^*\in \mathcal{P}(I)$ such that the inequality (\ref{dor:criterion1})
 in Theorem~\ref{dor:criterion} is satisfied, then (\ref{dor:criterion1})
 is satisfied for every refinement $P$ of $P^*$.

 \item[ii)] If inequality (\ref{dor:criterion1})
 is satisfied for the partition $P$ given by
 $t_0=a<t_1<\ldots <t_n=b$, and $\xi_j$,
 $\tau_j\in I_{\Box j}=[t_{j-1},t_j]_{\Box}$ for $j=1, 2,\ldots, n$, then
  \begin{equation*}
   \sum_{j=1}^n\left|f(\xi_j)-f(\tau_j)\right|\Delta g_j<\mathbb{\varepsilon}\, .
  \end{equation*}

 \item[iii)] If $f\in \mathcal{R}_{\Box}(g,I)$ and inequality
 (\ref{dor:criterion1}) is satisfied for the partition $P$
 given by $t_0=a<t_1<\ldots <t_n=b$ and $\xi_j\in I_{\Box j}=[t_{j-1},t_j]_{\Box}$ for $j=1, 2, \ldots, n$, then
  \begin{equation*}
   \left|\sum_{j=1}^nf(\xi_j)\Delta g_j-\int_a^b f(t)\Box g(t)\right|<\mathbb{\varepsilon}\, .
  \end{equation*}
\end{enumerate}
\end{theorem}
\begin{proof}
Part i) follows from Theorem~\ref{lem:1}. Let us now prove part
ii). Let $f$ be a real-valued and bounded function on the interval
$I$. Let us take a partition $P=\{t_0, t_1,\ldots, t_n\}$ of $I$.
 Let $\xi_j$, $\tau_j\in
I_{\Box j}=[t_{j-1},t_j]_{\Box}$ for $j=1, 2,\ldots, n$. Then,
$M_{\Box j}-f(\xi_j)\geq 0$, $f(\tau_j)-m_{\Box j}\geq 0$, and
\[\sum_{j=1}^n\left|f(\xi_j)-f(\tau_j)\right|\Delta g_j
\leq \sum_{j=1}^n\left|\left(f(\xi_j)-f(\tau_j)\right)
+ \left(M_{\Box j}-f(\xi_j)\right)+\left(f(\tau_j)-m_{\Box j}\right)\right|\Delta g_j \, .\]
It follows that
\[\sum_{j=1}^n\left|f(\xi_j)-f(\tau_j)\right|\Delta g_j
\leq \sum_{j=1}^n\left|M_{\Box j}-m_{\Box j}\right|\Delta g_j=U_{\Box}(P,f,g)-L_{\Box}(P,f,g)< \mathbb{\varepsilon} \, .\]

Finally, let us prove part iii). Let the inequality
(\ref{dor:criterion1}) be satisfied for the partition $P$ given by $t_0=a<t_1<\ldots <t_n=b$. Let $\xi_j\in I_{\Box j}=[t_{j-1},t_j]_{\Box}$ for
$j=1, 2, \ldots, n$. Then, $\sum_{j=1}^n f(\xi_j)\Delta g_j\leq
U_{\Box}(P,f,g)$ and $L_{\Box}(P,f,g)\leq \int_a^b f(t)\Box g(t)$. Hence,
\[ \left|\sum_{j=1}^nf(\xi_j)\Delta g_j-\int_a^b f(t)\Box g(t)\right|
\leq |U_{\Box}(P,f,g)-L_{\Box}(P,f,g)|<\mathbb{\varepsilon}\, . \]
\end{proof}
Theorem~\ref{integal_box} gives a comparison between the two kinds of Riemann--Stieljes integrals on time scales and
the Riemann--Stieljes integral on the real interval. In the particular case $g(t) = t$ one gets the result as stated in \cite{BPi}.
\begin{theorem}
\label{integal_box}
Let $a,b\in\mathbb{T}$ and $g:[a,b]\rightarrow\mathbb{R}$ be a strictly increasing function on $[a,b]$. Let $f:[a,b]\rightarrow\mathbb{R}$. Denote by $\left.f\right|_{\mathbb{T}}$ and $\left.g\right|_{\mathbb{T}}$ the restrictions of functions $f$ and $g$ to the time scale $\mathbb{T}$. Then,
\begin{enumerate}
  \item [i)] $\int_a^b \left.f\right|_{\mathbb{T}}(t)\Delta \left.g\right|_{\mathbb{T}}(t)\leq \int_a^b f(t)dg(t)\leq \int_a^b \left.f\right|_{\mathbb{T}}(t)\nabla \left.g\right|_{\mathbb{T}}(t)$
      if $f$ is strictly increasing on $[a,b]$;

  \item [ii)] $\int_a^b \left.f\right|_{\mathbb{T}}(t)\nabla \left.g\right|_{\mathbb{T}}(t)\leq \int_a^b f(t)dg(t)\leq
\int_a^b \left.f\right|_{\mathbb{T}}(t)\Delta \left.g\right|_{\mathbb{T}}(t)$
  if $f$ is strictly decreasing on $[a,b]$.
\end{enumerate}
\end{theorem}
\begin{proof}
We prove here the item i). The proof of ii) is similar. If $f$ is strictly increasing on $[a,b]$, then $\left.f\right|_{\mathbb{T}}$
is regulated on
$[a,b]_{\mathbb{T}}$ and all integrals exist for an increasing function $g$. Let $P\in \mathcal{P}\left([a,b]_{\mathbb{T}}\right)$ and $P=\{t_0,\ldots,
t_n\}$, where $a=t_0<t_1<\ldots<t_n=b$. Then, $\int_a^bf(t)dg(t)\leq
\sum_{j=1}^n f(t_{j})\Delta g_j= U_{\nabla}(P,f,g)$. Taking the infimum of the right-hand side over all partitions from
$\mathcal{P}\left([a,b]_{\mathbb{T}}\right)$  we get
$\int_a^bf(t)dg(t)\leq\overline{\int_a^b}\left.f\right|_{\mathbb{T}}(t)\nabla
\left.g\right|_{\mathbb{T}}(t)$.
Similarly, $\int_a^b f(t)dg(t)\geq \sum_{j=1}^n f(t_{j-1})\Delta g_j= L_{\Delta}(P,f,g)$. Taking now the supremum of the right-hand side we get that $\int_a^b f(t)dg(t)\geq\underline{\int_a^b}\left.f\right|_{\mathbb{T}}(t)\Delta
\left.g\right|_{\mathbb{T}}(t)$. For $\left.f\right|_{\mathbb{T}}\in
R_{\Box}(\left.g\right|_{\mathbb{T}},[a,b]_{\mathbb{T}})$, then also $\int_a^bf(t)dg(t)\leq
\int_a^b \left.f\right|_{\mathbb{T}}(t)\nabla \left.g\right|_{\mathbb{T}}(t)$ and $\int_a^b f(t)dg(t)\geq
\int_a^b \left.f\right|_{\mathbb{T}}(t)\Delta \left.g\right|_{\mathbb{T}}(t)$.
\end{proof}

\begin{corollary}
Let $a,b\in\mathbb{T}$, $I=[a,b]_{\mathbb{T}}$, $g: I \rightarrow\mathbb{R}$ be
a strictly increasing function, and $f: I \rightarrow\mathbb{R}$. Then,
\begin{enumerate}
  \item [i)] $\int_a^b f(t)\Delta g(t)\leq  \int_a^b f(t)\nabla g(t)$
if $f$ is strictly increasing on $I$;

  \item [ii)] $\int_a^b f(t)\nabla g(t)\leq  \int_a^b f(t)\Delta g(t)$ if  $f$
is strictly decreasing on $I$.
\end{enumerate}
\end{corollary}

\section{Algebraic properties of the Riemann-Stieltjes $\Box$--integral}

In this section we prove some algebraic properties of the
Riemann-Stieltjes integral on time scales. The properties are valid for an arbitrary time scale $\mathbb{T}$
with at least two points.
We define $\int_a^a f(t)\Box g(t)=0$ and $\int_a^b f(t)\Box g(t)=-\int_b^a f(t)\Box g(t)$ for $a>b$.
\begin{theorem}
Let $I=[a,b]_{\mathbb{T}}$, $a,b\in\mathbb{T}$. Every constant function
$f:\mathbb{T}\rightarrow\mathbb{R},$ $f(t) \equiv c$,
is Stieltjes $\Box$--integrable with
respect to $g$ on $I$ and
 \[\int_a^bc\Box g(t)=c\left(g(b)-g(a)\right).\]
\end{theorem}
\begin{proof}
Let $P\in \mathcal{P}(I)$ and $P=\{t_0,\ldots,t_n\}$. Then,
\begin{equation*}
\begin{split}
 L_{\Box}(P,f,g) &=U_{\Box}P,f,g)=c\sum\limits_{j=1}^n\Delta g_j\\
 &= g(t_1)-g(a)+g(t_2)-g(t_1)+\cdots+g(b)-g(t_{n-1})\\
 &=g(b)-g(a).
\end{split}
\end{equation*}
Hence, $\underline{\int_a^b} f\Box g=\overline{\int_a^b} f\Box
g=c\left(g(b)-g(a)\right)$.
\end{proof}

\begin{theorem}
Let $t\in\mathbb{T}$ and  $f:\mathbb{T}\rightarrow\mathbb{R}.$ Then, $f$ is Riemann-Stieltjes
$\Delta$--integrable with respect to $g$ from $t$ to $\sigma(t)$ and
\begin{equation}
\label{eq:2}
 \int_t^{\sigma(t)}f(\tau) \Delta g(\tau)
 =f(t)\left(g^{\sigma}(t)-g(t)\right)\, ,
\end{equation}
where $g^{\sigma} = g \circ \sigma$.
Moreover, if $g$ is $\Delta$--differentiable at $t$, then
\begin{equation}
\int_t^{\sigma(t)}f(\tau) \Delta g(\tau)
=\mu(t)f(t)g^{\Delta}(t)\, . \label{eq:3}
\end{equation}
\end{theorem}
\begin{proof}
For $\sigma(t)=t$ both equations (\ref{eq:2}) and (\ref{eq:3})
hold. When $\sigma(t)>t$, only one partition for
$I=[t,\sigma(t)]_{\mathbb{T}}$ is possible. In that case
$P=\{t,\sigma(t)\}$ and we have one
set $I_{\Delta 1}=[t,\rho(\sigma(t))]_{\mathbb{T}}=\{t\}$. Hence, $m_{\Delta 1}=M_{\Delta 1}=f(t)$, and
\[\underline{\int_t^{\sigma(t)}} f\Delta g
=\overline{\int_t^{\sigma(t)}} f\Delta g
=U_{\Delta}(P,f,g)=L_{\Delta}(P,f,g)=f(t)(g(\sigma(t))-g(t)) \, .\]
As $\sigma(t)>t$, then
also $\mu(t)\neq 0$ and $g^{\sigma}(t)-g(t)=\mu(t)g^{\Delta}(t)$.
\end{proof}

\begin{theorem}
Let $t\in\mathbb{T}$ and  $f:\mathbb{T}\rightarrow\mathbb{R}.$ Then, $f$ is Riemann-Stieltjes
$\nabla$--integrable with respect to $g$ from $\rho(t)$ to $t$ and
\begin{equation}
\label{eq:25}
 \int_{\rho(t)}^{t}f(\tau) \nabla g(\tau)
 =f(t)\left(g(t)-g^{\rho}(t)\right)\, ,
\end{equation}
where $g^{\rho} = g \circ \rho$.
Moreover, if $g$ is $\nabla$--differentiable at $t$, then
\begin{equation}
\int_{\rho(t)}^{t}f(\tau) \nabla g(\tau)
=\nu(t)f(t)g^{\nabla}(t)\, . \label{eq:36}
\end{equation}
\end{theorem}
\begin{proof}
For $\rho(t)=t$ both equations (\ref{eq:25}) and (\ref{eq:36})
hold. When $t>\rho(t)$, only one partition for
$I=[\rho(t), t]_{\mathbb{T}}$ is possible. In that case
$P=\{\rho(t), t\}$ and we have one
set $I_{\nabla 1}=[\sigma(\rho(t)),t]_{\mathbb{T}}=\{t\}$. Hence, $m_{\nabla 1}=M_{\nabla 1}=f(t)$, and
\[\underline{\int_{\rho(t)}^{t}} f\nabla g
=\overline{\int_{\rho(t)}^{t}} f\nabla g
=U_{\nabla}(P,f,g)=L_{\nabla}(P,f,g)=f(t)(g(t)-g(\rho(t))) \, .\]
As $\rho(t)<t$, then
also $\nu(t)> 0$ and $g(t)-g^{\rho}(t)=\nu(t)g^{\nabla}(t)$.
\end{proof}

\begin{corollary}
Let $a,b\in\mathbb{T}$ and $a<b$.
\begin{enumerate}
\item [i)] If $\mathbb{T}=\mathbb{R}$, then a bounded function $f$ on $[a,b]_{\mathbb{T}}$ is
Riemann-Stieltjes $\Box$--integrable with respect to the
increasing function $g$ from $a$ to $b$ if and only if $f$ is
Riemann-Stieltjes integrable on $I$ in the classical sense.
Moreover, then  $\int_a^bf(t)\Box g(t)=\int_a^bf(t)dg(t)$,
where the integral on the right hand side is the classical
Riemann-Stieltjes integral (see, \textrm{e.g.}, \cite[Chapter~4]{B:garden}).

\item [ii)] If  $\mathbb{T}=\mathbb{Z}$, then each function $f:\mathbb{Z}\rightarrow\mathbb{R}$ is
Riemann-Stieltjes $\Box$--integrable from $a$ to $b$ with respect to an
arbitrarily increasing function $g:\mathbb{Z}\rightarrow\mathbb{R}$. Moreover,
$\int_a^bf(t)\Delta g(t)=\sum_{t=a}^{b-1}f(t)(g(t+1)-g(t))$,  $\int_a^bf(t)\nabla g(t)=\sum_{t=a+1}^{b}f(t)(g(t)-g(t-1))$.
\end{enumerate}
\end{corollary}
\begin{proof}
i) Notice that for $\mathbb{T}=\mathbb{R}$ Definition~\ref{RSdef} coincides
 with the classical definition of the Riemann-Stieltjes integral.
 Moreover, since for $\mathbb{T}=\mathbb{R}$ the $\Box$--differential coincides
 with the standard differential, it follows that
 \[\int_a^bf(t)\Box g(t)=\int_a^bf(t)g'(t)dt=\int_a^bf(t)dg(t) \, .\]

ii) Let $I=[a,b]_{\mathbb{Z}}$, $a<b$, $a$, $b\in\mathbb{Z}$. Consider the partition
  $P_{\delta} \in \mathcal{P}(I)$ given by $a=t_0<t_1<\ldots<t_n=b=a+n$,
  where $t_j=a+j$ for $j=0,\ldots, n$.
  Notice that $[t_{j-1},\rho(t_j)]_{\mathbb{T}}=\{t_{j-1}\}$ for each $j=1,\ldots, n$
  and $\Delta g(t_j)=g(t_j)-g(t_j-1)$. A direct calculation shows that
  $L_{\Delta}(P_{\delta}, f, g)=U_{\Delta}(P_{\delta},f,g)
  =\sum_{j=1}^nf(t_{j-1})\Delta g(t_j)
  =\sum_{t=a}^{b-1}f(t)(g(t+1)-g(t))$.
  Hence, $U_{\Delta}(P_{\delta},f,g)=L_{\Delta}(P_{\delta},f,g)$
  and Theorem~\ref{th.7} imply the desired formula for the $\Delta$--case. Similarly proof holds for the $\nabla$--case.
\end{proof}

\begin{theorem}(Linearity)
\label{lin}
Let functions $f_1$ and $f_2$ be Riemann-Stieltjes
$\Box$--integrable on the interval $[a,b]_{\mathbb{T}}$ with respect to  $g$, and $c$ be a constant. Then,
\begin{enumerate}
 \item [i)] $cf_1 \in \mathcal{R}_{\Box}(g,I)$
 and $\int_a^bcf_1 \Box g = c \int_a^b f_1\Box g$;

 \item [ii)] $f_1+f_2 \in \mathcal{R}_{\Box}(g,I)$
 and $\int_a^b\left(f_1+f_2\right)\Box g
 =\int_a^bf_1\Box g+\int_a^bf_2\Box g$.
\end{enumerate}
\end{theorem}
\begin{proof}
Let $I_j=(t_{j-1},t_j]_{\mathbb{T}}$, and
$p\in\mathcal{P}(I)$ be any partition of the interval
$I$. Denote $I_j=[t_{j-1},t_j]_{\Box}$ and
 \begin{eqnarray*}
  m_j^{f_1}=\inf\limits_{t\in I_{\Box  j}}f_1(t) \, , \quad
  \quad
  M_j^{f_1}=\sup\limits_{t\in I_{\Box  j}}f_1(t) \, ,\\
  m_j^{f_2}=\inf\limits_{t\in I_{\Box  j}}f_2(t)\, , \quad
  \quad
  M_j^{f_2}=\sup\limits_{t\in I_{\Box  j}}f_2(t) \, .
 \end{eqnarray*}
 \emph{i)} Let us notice that for $c=0$ function
 $cf_1 \in \mathcal{R}_{\Box}(g,I)$. Suppose that $c \neq 0$ and denote
 \[\mathfrak{m}_j=\inf\limits_{t\in I_{\Box  j}}(cf_1)(t)\quad
 \text{and}\quad
 \mathfrak{M}_j=\sup\limits_{t\in I_{\Box  j}}(cf_1)(t) \, .\]
 Then,
 \[\mathfrak{m}_j=cm_j^{f_1}\quad
 \text{and}\quad
 \mathfrak{M}_j=cM_j^{f_1}\quad
 \text{for}\;c>0;\]
 \[\mathfrak{m}_j=cM_j^{f_1}\quad
 \text{and}\quad
 \mathfrak{M}_j=cm_j^{f_1}\;\;\;\text{for}\;c<0\,.\]
 This implies that
 \[
 L_{\Box}(P,cf_1,g) =\left\{\begin{array}{ll}
					 cL_{\Box}(P,f_1,g), & \hbox{for\;$c>0$} \\
					 cU_{\Box}(P,f_1,g), & \hbox{for\;$c<0$}
\end{array}\right.\]
and
\[U_{\Box}(P,cf_1,g)=\left\{\begin{array}{ll}
  cU_{\Box}(P,f_1,g), & \hbox{for\;$c>0$} \\
  cL_{\Box}(P,f_1,g), & \hbox{for\;$c<0$.}
\end{array}\right.\]
Thus,
\begin{equation}
\label{war}
  U_{\Box}(P,cf_1,g)-L_{\Box}(P,cf_1,g)=|c|(U_{\Box}(P,f_1,g)-L_{\Box}(P,f_1,g))\,.
\end{equation}
Because $f_1\in\mathcal{R}_{\Box}(g,I)$, then for any $\varepsilon>0$ there
exists a partition $P$ such that
$U_{\Box}(P,cf_1,g)-L_{\Box}(P,cf_1,g)<\frac{\varepsilon}{|c|}$.
Together with (\ref{war}) this leads to
$U_{\Box}(P,cf_1,g)-L_{\Box}(P,cf_1,g)<\varepsilon$. Hence,
$cf_1 \in \mathcal{R}_{\Box}(g,I)$. Moreover,
 \[\underline{\int_a^b}cf_1 \Box g
 =\sup_{P\in \mathcal{P}(I)} L_{\Box}(P,cf_1,g) = c\sup _{P\in \mathcal{P}(I)}L_{\Box}(P,f_1,g)
 = c\underline{\int_a^b}f_1\Box g\]
 and
 \[\overline{\int_a^b}cf_1\Box g=\inf_{P\in \mathcal{P}(I)} U_{\Box}(P,cf_1,g) = c\inf_{P\in \mathcal{P}(I)} U_{\Box}(P,f_1,g)=
   c\overline{\int_a^b}f_1\Box g\,. \]
 Since $f_1 \in \mathcal{R}_{\Box}(g,I)$, then
 $\overline{\int_a^b}f_1\Box g=\underline{\int_a^b}f_1\Box g$,
 which proves the intended conclusion. \\

 \emph{ii)} Let
  \[\mathfrak{m}_j=\inf\limits_{t\in I_{\Box  j}}(f_1+f_2)(t)\quad
  \text{and}\quad
  \mathfrak{M}_j=\sup\limits_{t\in I_{\Box  j}} (f_1+f_2)(t) \, .\]
 Then, $\mathfrak{m}_j \geq m_j^{f_1}+m_j^{f_2}$
 and $\mathfrak{M}_j \leq M_j^{f_1}+M_j^{f_2}$.
 This implies
\begin{equation*}
\begin{gathered}
L_{\Box}(P,f_1,g)+L_{\Box}(P,f_2,g) \leq L_{\Box}(P,f_1+f_2,g)\\
U_{\Box}(P,f_1,g)-U_{\Box}(P,f_2,g) \leq U_{\Box}(P,f_1+f_2,g)\,.
\end{gathered}
\end{equation*}
 The integrability criterion implies the existence of partitions
 $P_1\in\mathcal{P}(I)$ and $P_2\in\mathcal{P}(I)$ such that the
 inequalities
 \begin{equation}
 \label{3.4}
  U_{\Box}(P_1,f,g)-L_{\Box}(P_1,f,g)<\frac{\varepsilon}{2}\quad
  \text{and}\quad
  U_{\Box}(P_2,f,g)-L_{\Box}(P_2,f,g)<\frac{\varepsilon}{2}
 \end{equation}
 also hold on their common refinement $P_{\varepsilon}=P_1\cup
 P_2$. From (\ref{3.4}) follows that
 $U_{\Box}(P_{\varepsilon},f_1+f_2,g)-L_{\Box}(P_{\varepsilon},f_1+f_2,g)<\varepsilon$.
 Hence, $f_1+f_2 \in \mathcal{R}_{\Box}(g,I)$. Additionally,
 \begin{multline*}
  \underline{\int_a^b}(f_1+f_2)\Box g=\sup_{P\in \mathcal{P}(I)} L_{\Box}(P,f_1+f_2,g) \\
   \leq \sup_{P\in \mathcal{P}(I)} L_{\Box}(P,f_1,g)+\sup_{P\in \mathcal{P}(I)} L_{\Box}(P,f_2,g)=
   \underline{\int_a^b}f_1\Box g+\underline{\int_a^b}f_2\Box g
 \end{multline*}
 and
 \begin{multline*}
  \overline{\int_a^b}(f_1+f_2)\Box g=\inf_{P\in \mathcal{P}(I)} U_{\Box}(P,f_1+f_2,g) \\
  \geq \inf_{P\in \mathcal{P}(I)} U_{\Box}(P,f_1,g)+\inf_{P\in \mathcal{P}(I)} U_{\Box}(P,f_2,g)=
   \overline{\int_a^b}f_1\Box g+\overline{\int_a^b}f_2\Box g \, ,
 \end{multline*}
 so that
 \[\int_a^b(f_1+f_2)\Box g \geq\overline{\int_a^b}f_1\Box g+\overline{\int_a^b}f_2\Box g \ \ \  \mbox{and} \ \ \
 \int_a^b(f_1+f_2)\Box g \geq
   \underline{\int_a^b}f_1\Box g+\underline{\int_a^b}f_2\Box g \, .\]
 Because $f_1 \in \mathcal{R}_{\Box}(g,I)$ and $f_2 \in \mathcal{R}_{\Box}(g,I)$,
 then $\overline{\int_a^b}f_1\Box g=\underline{\int_a^b}f_1\Box g$ and
 $\underline{\int_a^b}f_2\Box g=\overline{\int_a^b}f_2\Box g$.
\end{proof}

Similarly to the proof of Theorem~\ref{lin} item i),
one can show the following:

\begin{theorem}
Let $f\in \mathcal{R}_{\Box}(g_1,I)$ and $f\in \mathcal{R}_{\Box}(g_2, I)$,
where $g_1$ and $g_2$ are increasing functions on $[a,b]_{\mathbb{T}}$. Then,
$f \in \mathcal{R}_{\Box}(g_1+g_2,I)$ and $\int_a^bf\Box\left(g_1+ g_2\right)
= \int_a^bf\Box g_1+\int_a^bf\Box g_2$.
\end{theorem}

\begin{theorem}
Let $a$, $b$, $c\in\mathbb{T}$ with $a<b<c$. If $f$
is bounded on $[a,c]_{\mathbb{T}}$
and $g$ is monotonically increasing on $[a,c]_{\mathbb{T}}$, then
\begin{equation*}
\int_a^cf\Box g
= \int_a^bf\Box g+ \int_b^cf\Box g\, .
\end{equation*}
\end{theorem}
\begin{proof}
 There exist partitions $P_1 \in \mathcal{P}([a,b]_{\mathbb{T}})$ and $P_2 \in
 \mathcal{P}([b,c]_{\mathbb{T}})$ such that for $\varepsilon>0$
 \[U_{\Box}(P_1,f,g) - L_{\Box}(P_1,f,g)<\frac{\varepsilon}{2}\quad
 \text{and}\quad
 U_{\Box}(P_2,f,g) -
 L_{\Box}(P_2,f,g)<\frac{\varepsilon}{2} \, .\]
Then, there is a partition $P \in \mathcal{P}([a,c]_{\mathbb{T}})$ such that
 $U_{\Box}(P,f,g)=U_{\Box}(P_1,f,g)+U_{\Box}(P_2,f,g)$ and
 $L_{\Box}(P,f,g)=L_{\Box}(P_1,f,g)+L_{\Box}(P_2,f,g)$ so that
 $U_{\Box}(P,f,g)-L_{\Box}(P,f,g)<\varepsilon$. Hence, $f$ is $\Box$--integrable
 with respect to $g$ and
 \begin{multline*}
 \int_a^c f \Box g \leq U_{\Box}(P_1,f,g)+U_{\Box}(P_2,f,g) \\
 \leq  L_{\Box}(P_1,f,g)+L_{\Box}(P_2,f,g)+\varepsilon
 \leq \int_a^b f \Box g +\int_b^c f \Box g +\varepsilon \, .
 \end{multline*}
Similarly,
 \[\int_a^c f \Box g \geq \int_a^b f \Box g +\int_b^c f \Box g -\varepsilon \, .\]
\end{proof}

\begin{theorem}(Integration by substitution)
Let $\tilde{\mathbb{T}}$ be a time scale.
Assume that $\varphi:\tilde{\mathbb{T}} \mbox{$\rightarrow$} \mathbb{R}$ is a strictly increasing
continuous function and $\mathbb{T}=\varphi(\tilde{\mathbb{T}})$ is a time scale.
Moreover, $\varphi$ maps an interval $[A, B]_{\tilde{\mathbb{T}}}$ onto $[a,b]_{\mathbb{T}}$.
Let $g$ be monotonically increasing on $[a, b]_{\mathbb{T}}$ and
$f\in \mathcal{R}_{\Box}(g,[a,b]_{\mathbb{T}})$. Then,
$f\circ\varphi\in \mathcal{R}_{\Box}(g\circ\varphi, [A,B]_{\tilde{\mathbb{T}}})$ and
\begin{equation*}
\int_a^b f(t) \Box g(t)=\int_A^B f(\varphi(s))\Box g(\varphi(s)).
\end{equation*}
\end{theorem}
\begin{proof}
Since $\varphi$ is a strictly increasing continuous function,
there is a one-to-one correspondence between the partition
 $Q=\{s_0, s_1,\ldots, s_n\}\subset [A, B]_{\tilde{\mathbb{T}}}$,
 where $A=s_0<s_1< \ldots <s_n=B$,
 and the partition
 $P=\{t_0, t_1, \ldots, t_n\}\subset [a,b]_{\mathbb{T}}$,
 where $a=t_0=\varphi(s_0)<t_1=\varphi(s_1)< \ldots <t_n=\varphi(s_n)=b$.
 Since $f([t_{j-1},t_j]_{\mathbb{T}})=f([\varphi(s_{j-1},\varphi(t_j)]_{\tilde{\mathbb{T}}})$
 for each $j$, then $L_{\Box}(P,f,g)=L_{\Box}(Q,f\circ\varphi, g\circ\varphi)$
 and $U_{\Box}(P,f,g)=U_{\Box}(Q,f\circ\varphi, g\circ \varphi)$.
 The result follows from Theorem~\ref{dor:criterion}.
\end{proof}

\section{From the Riemann-Stieltjes to Riemann $\Box$--integral}

Theorem~\ref{dor:transition} below establishes a relation between
the Riemann-Stieltjes $\Box$--integral and the Riemann $\Box$--integral for a function $g$ being $\Box$--differentiable on the interval $I$ of integration.
We begin by noting that one can easily reformulate to nabla version, with the interval opened from the left side,
the delta mean value theorem \cite[Theorem~1.14]{Bh1}:

\begin{theorem}
\label{thm:Th1.14:nabla}
Let $f$ be a continuous function on $[a, b]_{\mathbb{T}}$
that is $\nabla$ -- differentiable on $(a, b]$. Then there exist $\xi$, $\tau\in (a,b]$ such that
\[f^{\nabla}(\xi)\leq \frac{f(b)-f(a)}{b-a}\leq f^{\nabla}(\tau)\, .\]
\end{theorem}

In Corollary~\ref{box_mean} we write together
\cite[Theorem~1.14]{Bh1} and Theorem~\ref{thm:Th1.14:nabla}
with our "$\Box$"--notation:

\begin{corollary}
\label{box_mean}
Let $f$ be a continuous function on $[a, b]_{\mathbb{T}}$
that is $\Box$ -- differentiable on $[a, b]_{\Box}$. Then there exist $\xi$, $\tau\in[a, b]_{\Box}$ such that
\[f^{\Box}(\xi)\leq \frac{f(b)-f(a)}{b-a}\leq f^{\Box}(\tau)\, .\]
\end{corollary}

\begin{theorem}
\label{dor:transition}
Let $I=[a,b]_{\mathbb{T}}, a,b\in\mathbb{T}$. Suppose that $g$ is an increasing
function such that $g^{\Box}$ is continuous on $(a,b)_{\mathbb{T}}$
and $f$ is a real bounded function on $I$.
Then, $f\in\mathcal{R}_{\Box}(g, I)$ if and only if
$fg^{\Box}\in \mathcal{R}_{\Box}(g,I)$. Moreover,
\begin{equation*}
 \int_a^bf(t)\Box g(t)=\int_a^bf(t)g^{\Box}(t)\Box t \, .
\end{equation*}
\end{theorem}
\begin{proof}
 Let $\mathbb{\varepsilon}>0$. Since $g^{\Box}$ is continuous, then $g^{\Box}\in \mathcal{R}_{\Box}(t,I)$,
\textrm{i.e.}, $g^{\Box}$ is Riemann--Stieltjes $\Box$--integrable on $I$. Hence, there
 exists a partition $P=\{a=t_0, t_1,\ldots, t_n=b\}\in
 \mathcal{P}(I)$ such that $U_{\Box}(P,g^{\Delta},t)- L_{\Box}(P,g^{\Delta},t)<\frac{\mathbb{\varepsilon}}{M}$,
 where $M=\sup_{t\in I}|f(t)|$.
From the $\Box$--version of the mean value theorem on time scales (see Corollary~\ref{box_mean}),
 for each $j=1,\ldots,n$ there are
 $\xi_j, \tau_j\in I_j=[t_{j-1},t_j]_{\Box}$ such that
 \[g^{\Box}(\tau_j)\Delta t_j\leq \Delta g_j\leq  g^{\Box}(\xi_j)\Delta t_j \ .\]
 From Theorem~\ref{dor:3} part ii), for any $s_j\in I_j$, $j=1, \ldots, n$,
 \[\sum_{j=1}^n \left|g^{\Box}(\xi_j)-g^{\Box}(s_j)\right|\Delta t_j<\mathbb{\varepsilon}\, .\]
 Then, $\sum_{j=1}^n f(s_j)\Delta g_j\leq \sum_{j=1}^n f(s_j)g^{\Box}(\xi_j)\Delta t_j$ and
 \[\left|\sum_{j=1}^n f(s_j)\Delta g_j -\sum_{j=1}^n f(s_j)g^{\Box}(s_j)\Delta t_j\right|\leq
 \left|\sum_{j=1}^n f(s_j)g^{\Box}(\xi_j)\Delta t_j-\sum_{j=1}^n f(s_j)g^{\Box}(s_j)\Delta t_j\right| \, .\]
 Hence,
 \begin{equation}
 \label{dor1}
  \left|\sum_{j=1}^n f(s_j)\Delta g_j -\sum_{j=1}^n f(s_j)g^{\Box}(s_j)\Delta t_j\right|
  \leq M\left|\sum_{j=1}^n \left(g^{\Box}(\xi_j)-g^{\Box}(s_j)\right) \Delta t_j\right|<\mathbb{\varepsilon}
 \end{equation}
 for any $s_j\in I_j$. Thus, $\sum_{j=1}^n f(s_j)\Delta g_j\leq U_{\Box}(P, fg^{\Box},t)+\mathbb{\varepsilon}$
 and $U_{\Box}(P,f,g)\leq U_{\Box}(P,fg^{\Box},t)+\mathbb{\varepsilon}$. Inequality (\ref{dor1}) implies that
 $U_{\Box}(P,fg^{\Box},t)\leq U_{\Box}(P,f,g)+\mathbb{\varepsilon}$. Thus,
 \[\left|U_{\Box}(P,f,g)-U_{\Box}(P,fg^{\Box},t)\right|\leq \mathbb{\varepsilon} \,.\]  Moreover,
 \[\left|\overline{\int_a^b}f(t)\Box g(t)-\overline{\int_a^b}f(t)
 g^{\Box}(t)\Box t \right|\leq \mathbb{\varepsilon}\, .\]
 Since $\mathbb{\varepsilon}$ is arbitrary, we conclude that
 \[\overline{\int_a^b}f(t)\Box g(t)=\overline{\int_a^b}f(t) g^{\Box}(t)\Box t.\]
 In a similar way one prove that
 $\underline{\int_a^b}f(t)\Box g(t)
 =\underline{\int_a^b}f(t) g^{\Box}(t)\Box t$.
 \end{proof}

\begin{theorem}(Delta integration by parts)
Let $I=[a,b]_{\mathbb{T}}, a,b\in\mathbb{T}$. Suppose that $g$ is an increasing
function such that $g^{\Delta}$ is continuous on $(a,b)_{\mathbb{T}}$
and $f$ is a real bounded function on $I$. Then,
\begin{equation*}
\int_a^bf\Delta g=\left[fg\right]_a^b-\int_a^b g^{\sigma}\Delta f \, ,
\end{equation*}
 where, as usual, $g^{\sigma}$ means $g \circ \sigma$.
\end{theorem}
\begin{proof}
 Theorem~\ref{dor:transition} and integration by parts for
 the Riemann $\Delta$--integral on time scales (see \cite{Bh1}) imply
 that $\int_a^bf(t)\Delta g(t)=\int_a^bf(t)g^{\Delta}(t)\Delta t$
 and
$\int_a^bg^{\sigma}f^{\Delta}\Delta t +\int_a^bf\Delta t=[fg]_a^b$.
 Hence, $\int_a^bf\Delta g=\left[fg\right]_a^b-\int_a^b g^{\sigma}\Delta f$.
\end{proof}

\begin{theorem}(Nabla integration by parts)
Let $I=[a,b]_{\mathbb{T}}, a,b\in\mathbb{T}$. Suppose that $g$ is an increasing
function such that $g^{\nabla}$ is continuous on $(a,b)_{\mathbb{T}}$
and $f$ is a real bounded function on $I$. Then,
\begin{equation*}
\int_a^bf\nabla g=\left[fg\right]_a^b-\int_a^b g^{\rho}\nabla f \, ,
\end{equation*}
 where  $g^{\rho}=g \circ \rho$.
\end{theorem}
\begin{proof}
 Theorem~\ref{dor:transition} and integration by parts for
 the Riemann $\nabla$--integral on time scales  implies
 that $\int_a^bf(t)\nabla g(t)=\int_a^bf(t)g^{\nabla}(t)\nabla t$
 and
$\int_a^bg^{\rho}f^{\nabla}\nabla t +\int_a^bf\nabla t=[fg]_a^b$.
 Hence, $\int_a^bf\nabla g=\left[fg\right]_a^b-\int_a^b g^{\rho}\nabla f$.
\end{proof}

\section{Conclusion and Future Perspectives}
\label{sec:ConcFuture}

This article is about the concept of
Riemann-Stieltjes delta integration on time scales.
The results of the paper may be used, \textrm{e.g.},
to generalize the $\mathbb{T} = \mathbb{R}$ inequalities
proved in \cite{D:2004,D:2002}
to a general time scale $\mathbb{T}$. Then,
as the particular case $g(t)=t$, one would
obtain the previous inequalities
proved for the Riemann integral
on time scales \cite{R:R:D:JIA,R:D:JMS:2009}.

Another interesting line of research is to
investigate the possibility of extending all
previous notions of integration on time scales
by putting together our present results with
the Henstock-Kurzweil integrals introduced
by Peterson and Thompson in \cite{P:2006,Thomson}.
Such Henstock-Kurzweil-Stieltjes integrals on time scales
are under study and will be addressed elsewhere.

\begin{acknowledgement}
The first author was supported by Bia\l ystok Technical University through W/WI/7/07 grant; the second and third authors by the Portuguese Foundation for Science and Technology
(Paw\l uszewicz through the program ``1 000 Doutorados
para Institui\c{c}\~{o}es Cient\'{\i}ficas Portuguesas'';
Torres through the R\&D unit CEOC of the University
of Aveiro, cofinanced by the European Community fund
FEDER/POCI 2010).
\end{acknowledgement}

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Home > Energy & Commodities
World's biggest battery race gets new contender in Australia
Tue, Jan 12, 2021 - 12:29 PM
[SYDNEY] Another big battery project has been announced in Australia, as energy storage developers rush to profit from the nation's abundant sunshine and wind resources.
Origin Energy Ltd. issued an expression of interest to firms to supply and install a 700-megawatt facility near the Eraring coal plant in New South Wales, with the first phase planned to start by the end of next year, the Sydney-based company said on Tuesday. That comes as utilities around the world race to install warehouse-sized batteries to back up wind and solar generation, with the current 300-megawatt record holder starting in California last month.
While coal is still the biggest power source in Australia, companies and states are seeking to address an expected power crunch as the nation's ageing plants are retired. Batteries can also be a very profitable investment, with the three biggest in South Australia earning as much as a third of their initial capital costs when the state was temporarily separated from the rest of the nation's electricity grid last February, according to BloombergNEF.
"The deployment of this battery at Eraring will support Origin's orderly transition away from coal-fired generation by 2032" and help New South Wales meet its energy policy objectives, said Greg Javis, Origin executive general manager for energy supply and operations. "We have an important role to play in positioning Origin's electricity generation portfolio to support Australia's rapid transition to renewables." Massive batteries are becoming a staple of power grids as a larger share of generation capacity tilts to intermittent sources such as solar and wind. Battery operators can purchase that clean energy during excess periods and re-sell it when the sun isn't shining or wind isn't blowing.
Other companies also have big plans for energy storage, with France's Neoen SA submitting a scoping study for a 500-megawatt battery in New South Wales to regulators in December, after in November saying it would partner with Tesla Inc to install a 300-megawatt facility in Victoria. Even larger systems than Origin's have been announced for California, China and Australia.
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text:
Science Discussion & News
General & Off Topic Discussion
Real World News
SpaceX StarLink satellite internet [UPDATES]
By DocM, February 24, 2015 in Science Discussion & News
Draggendrop 5,747
OS: Windows/VMware
Phone: Dixi cups and string
Just the prior approved constellation was staggering...and now a new VLEO plus frequency additions/re-alignment...puts this on a level that no one will be able to compare to when the systems are completed.
The consumer terminal will be the "interesting aspect"...but I believe that they can pull this off..in stages. I would think the trick will be to have variants of the consumer terminal..starting at basic and upping the ante as the constellations mature and new features introduced to consumers.The consumers will not just be for the "net". A whole range of different consumers will eventually be in the mix, which will also be in competition with the regulars such as iridium and SES. The new customers of the "future" may lead to..aircraft internet and data/position tracking, marine communications and tracking, land traffic, military communications, continental "feed" transmission for broadband and generic consumer consumption for all broadband uses. This will start out small but be a "monster constellation" in a decade.
Edited October 26, 2018 by Draggendrop
Unobscured Vision 2,681
Resident Nerd 7th Class
Location: /home/
OS: Windows / Mint
What I think they'll do .. and this is just my thinking ... they'll launch the first batches with the already-approved frequency ranges (as they can build those first), then intersperse the constellation(s) with the later batches with the upgrades to the electronics that can accommodate the other frequencies as-approved down the line. Nothing saying the later batches can't travel into their respective positions either. Difficult and tricky, yes; not desirable either, but do-able.
With the number of birds being discussed and the bandwidth requirements (and likely consumer demand) they'll need to co-locate multiple birds to service each node and learn how to safely (read: at minimal risk) add to the nodes as needed.
As I said, this is just my thinking.
There is a way around this. SpaceX already has the 4425 constellation approved. They have applied and received a limited band allotment to add to it. They also waived another higher band for now. The problem is that the FCC is unsure how to designate that waived band. Boeing just put a request for a similar band as well but will go nowhere for now. The waived bands in question will be eventually released and the use will be determined as intended.
If I were running it, I would proceed as intended with the 4425 constellation. The birds for it should already be equipped with the 2 new bands. Now if one were to also put the equipment for the "waived bands, but leave it turned off and software configurable, when the time comes for allotment...just turn it on. The wived bands are so close to the upper 2 V bands that they could be incorporated but software limited in use for now.
The 7518 will be a new constellation and can be designed with everything from the start, including the software configurable bands...Ground use can follow the same strategy.
The cost is not going to be much higher, they are already in the bands range.
Summary..constellation #1 has same as before with new allocation, but software restricted for present licence. The second constellation as designed with same software restriction until bands released. If the next year and half required for design finalization/production, it gives them 4 years to put up approx 525 for #1 and approx 900 for #2 each year for 4 years and use BFR/BFS for the final 3 years to put the other 50% of both constellations up. Just depends on sat size and max number in a F9B5 fairing.
Beittil 584
A piece on SpaceX and StarLink popped up on Reuters today!
https://www.reuters.com/article/us-spacex-starlink-insight/musk-shakes-up-spacex-in-race-to-make-satellite-launch-window-sources-idUSKCN1N50FC
Seems like Elon wasn't satisfied with the speed at which things were going eh
Within hours of landing, Musk had fired at least seven members of the program's senior management team at the Redmond, Washington, office, the culmination of disagreements over the pace at which the team was developing and testing its Starlink satellites, according to the two SpaceX employees with direct knowledge of the situation.
Known for pushing aggressive deadlines, Musk quickly brought in new managers from SpaceX headquarters in California to replace a number of the managers he fired. Their mandate: Launch SpaceX's first batch of U.S.-made satellites by the middle of next year, the sources said.
DocM 16,649
Replaced by Musk for not meeting deadlines, replaced with people from within SpaceX. We know he doesn't have much tolerance for this, and often takes over projects himself. With the Department of Defense as a potential anchor customer...
Apparently those fired came from environments like Microsoft and Google which have a slower development culture.
Statement to Ars
https://arstechnica.com/information-technology/2018/10/unhappy-elon-musk-went-on-firing-spree-over-slow-satellite-broadband-progress/
The SpaceX Redmond office is an essential part of the company's efforts to build a next-generation satellite network that can link the world with reliable and affordable broadband service, reaching those who have never been connected before. Given the success of our recent Starlink demonstration satellites, we have incorporated lessons learned and re-organized to allow for the next design iteration to be flown in short order. This is a very similar approach of rapid iteration in design and testing which led to the success of Falcon 1, Falcon 9, Falcon Heavy and Dragon.
http://nrg.cs.ucl.ac.uk/mjh/starlink-draft.pdf
SpaceX has filed plans with the US Federal Communications Committee (FCC) to build a constellation known as Starlink, consisting of 4,425 low Earth orbit communication satellites.
It will use phased array antennas for up and downlinks and laser communication between satellites to provide low-latency high bandwidth coverage over most of the world. To understand the latency propertes of such a network, we built a simulator based on public details from the FCC filings. We evaluate possible ways to use the laser links to provide a network, and look at the problem of routing on this network. We provide a preliminary evaluation of how well such a network can provide low-latency communications, and examine its multipath properties. We conclude that a network built in this manner can provide lower latency communications than any possible terrestrial optical fiber network for communications over distances greater than about 3000 km.
We already knew this a year ago. Better than predicted functionality, impossible throughput realized, SpaceX will be printing their own money from the results.
Maybe they should be asking us. Heh.
It does please me! Premature failures at the higher altitude would have stayed up there a 1000+ years being debris creators. 550km would be very responsible.
https://twitter.com/IridiumBoss/status/1060701864696647681
FCC filing for altitude mod...and a few extra's...
Application for Fixed Satellite Service by Space Exploration Holdings, LLC
https://fcc.report/IBFS/SAT-MOD-20181108-00083
The sub links are at the above link...such as technical...pdf downloads available.
TECHNICAL INFORMATION TO SUPPLEMENT SCHEDULE S
83 page pdf
https://fcc.report/IBFS/SAT-MOD-20181108-00083/1569860.pdf
In coming down to 550km the StarLink LEO segment's latency drops to 15 milliseconds 😵
Xenon 7,878
Location: USA and France
OS: Windows 10 Pro & Android
Phone: Yes
The one thing about this besides it being very cool is I hope they have a way to de-orbit those sats when they end of life. That would be a lot of useless hardware up there.
30 minutes ago, Xenon said:
At similar altitudes there is still enough atmosphere to cause drag. This is why the ISS needs regular re-boosting to a higher altitude by visiting spacecraft.
As such, a damaged or dead LEO StarLink satellite at 550km will passively de-orbit in 1-3 years, 5 tops. No action needed. That's per the FCC filings.
A normally retired StarLink LEO satellite (4,400+ in number) will use its plasma engine to deorbit even faster.
The even lower VLEO StarLink satellites (7,518 of them) will need even more frequent plasma engine use to prevent even faster passive de-orbiting. A self-disposing orbit.
No longer necessary. The orbital decay ensures that each satellite is gone within 5 years.
Haha, @DocM beat me to the punch by seconds.
This is friggin' cool, though. Latency dropping to ~15ms at the altitude the constellation will operate at. An already fantastic selling point becomes more optimal. Financial and Business sectors will be extremely pleased with this development like Mr. Desch is.
Even in terms of a standard downlink budget, more power, better spot beam control and helps user terminal constraints. The debris mitigation is very important to me...I like....😊
I thought that it may be of help to clear up a few items that a few news outlets are having difficulty with.
Over the next little while, if feeling okay, I will explain general goodies from what I have data mined so far.
First is latency....it is important...for some applications.... irrelevant for others. An example of importance would be live chat or stock market transactions where sooner is better and/or less noticeable. Where it is not so important is in buffered applications such as general file transfer and watching a netflix movie.
Latency will depend on altitude of the satellite in question as well as other considerations of your "satellite link budget". The budget is what we use to calculate power losses due to variables that affect the frequency in use, the tighter the budget, the more likely of interruption. An example is satellite TV during a rain storm...those great looking colored squares, because the budget only allots a general loss for certain atmospherics, this is required to enable the system to be affordable for the consumer.
The above video is for one altitude and does not contain the latest data from FCC filing in respect to both constellations. The video is good enough to portray activities in general.
To get up to speed, SpaceX filed and received a permit for a 4425 constellation, Ka and Ku band in 6 planes.
As of November 6th, 2018, a request for modification has been entered as well as an application for another constellation. This modification will remove the 1600 satellites in the 1150 km orbit and put 1584 satellites at 550 km instead which makes the 4425 sat constellation now at 4409 satellites.
The modification proposal also asks to include V band communications in the 4409 constellation...this is important for the future. The next constellation will be 7518 satellites, in V band only...we can talk about that.....much later.
This is the modification of orbit...
This is the full constellation...eventually.....
This adds up to 4409 as per FCC filing. This will also hopefully be authorized with V band...so it plays well with the next constellation.
Apogee and perigee will be maintained to within 30 km and inclinations to 0.5 degree.
from the filing...
Within the 1584 satellites, the first deployment will start with 1st generation satellites with a simpler design to start the constellation sooner and then iterations will commence quickly, starting in 2019. Sound familiar. The first batch will use Ku then will have Ku and Ka introduced. Ground stations will start with dish and move to an array.
teaser...final plan implementation of both constellations is V band to consumer terminal....you read that right, will cover later. Right now, it's get it up fast, iterate as they go and move quick...average sat life is 5 years which will get the older ones out fast for the "new toys". They will need BFR within a few years.
The V band integration will follow...
Steerable pattern controlled by constellation ground station.
FREQUENCY BANDS REQUESTED (for the above Ku and Ka)
10.7-12.7 GHz
12.75-13.25 GHz
13.85-14.5 GHz
I won't bother with power densities until a later time.
I will cover debris mitigation shortly....The higher orbits can use centuries to de-orbit if left unattended. SpaceX has a comprehensive plan that I will go over later....
Made a boo-boo in the above post...
(SpaceX filed and received a permit for a 4425 constellation, Ka and Ku band in 6 planes) should have been 5 altitude orbits. Having too much fun with 3 monitors full of data and I got ahead of myself with planes. 4409 satellites in 5 altitudes which contain 75 planes total and the lower V band frequencies have been approved for the 4409 constellation...SpaceX removed a few extra higher V band frequencies via waiver but I am pretty sure it's temporary because the bands are not even allotted at this time. SpaceX will be the ones pushing and defining the overall use of these upper bands.
I will have to save the debris mitigation topic for the next post. I felt that the following was important enough to post first. I apologize for the wall of text but hopefully someone will read a sentence or two.
From FCC filing, attachment A, November 8, 2018
Sat Insertion
For its proposed operations at 550 km, SpaceX expects the insertion altitude for the modified 1,584 satellites generally to be 300-350 km and raise by thrusters. I would hazard to guess that initial launches on F9B5 will be from LC39A and LC40 with instantaneous windows. Since this is LEO, I have a hunch that the "fairing" will be a "volume restriction" for the number of satellites on a dispenser. Not a lot of information out in the wild but estimates that I have seen lean around 25 to 40 satellites...we will see later when more data available.
GPS particulars
A reddit user dug up a Space Act Agreement between NASA and SpaceX which in itself is not unusual...until one notices JPL/Caltech's RTGx software and reference to Linux use on a Xilinx Ultrascale+ quad A53's with dual frequency GPS receiver, signed 17th of August 2018 with a completion date of 3 months from signing, which is approximately next week. This is a contract for evaluation, recommendation and modifications for optimal use of the RTGx system, fastest real time GPS system with WAAS, presently available. WAAS is a Wide Area Augmentation System, LAAS is local. These systems add correction factors to drastically improve accuracy. Quick examples for GPS...average system much like cell phones, around a meter. Survey/Scientific systems are in the cm range and military systems are kept quiet but it is real time and very accurate. This is why a quality survey/science receiver with augmentation can cost upwards of $50,000...your cellphone must keep costs down.
I was going to cover this in detail for the GPS3 launch coming up but will throw a bit here. Ephemeris (Almanac) data is formatted and stored in all stations, sats and terminals. It contains all data for the satellites geospatial particulars such as position, altitude, velocity, unique identifiers, corrections, sat health and future movement. This is updated...the rate of which depends on end user capabilities. This is very important, in this case use, for user terminal beam geometry, inter-satellite communications and for routing algorithms.
Speed of light.
The altitude that the first batch will be working in is 550 km. This altitude still has very minor atmospheric constituents but for RF communications, we can call it a vacuum for short distance use. One things that gets lost in news articles is the context of velocity and bandwidth. There are several methods of modulation and depth of modulation to a carrier. This example will be a very simplistic approach to get the concept.
If I have a carrier frequency and use 10% of it for data, this is an easy multiple to do in your head. 1 Mhz carrier allows 100 Khz BW, 100 Mhz carrier allows 10 Mhz BW and a 10 Ghz carrier allows 1 Ghz BW.
Next is velocity. In a pure vacuum medium, RF communication is at the speed of light. It does not matter what the frequency is, they are all the same velocity. Laser communications is just at a higher frequency in the radio magnetic spectrum. One can now see that the higher in the spectrum that we operate, the same level of modulation results in a larger BW to contain data. The issue is when we are not operating in a pure vacuum. Materials/environments represent a plethora of characteristics which slow signal propagation or outright absorb/reflect it. There are trade offs with every frequency band we use and the transmission medium we employ. A vacuum, fiber optic, waveguide, planar strip, plasma, dielectric or plain metal conductor all represent trade offs for a particular frequency range. An example is quality fiber will have a transmission velocity at half of "c". A piece of copper wire can act like an inductance or capacitance at various frequencies. A mixture of topology can cause a phase error which must be corrected on the system. An example is "intelligence" being ready for the next process in a mixed circuit but must be buffered/delayed until another piece of "intelligence" in a different substraight arrives.
Summary...vacuum transmission is fast. Satellite components, terrestrial components, atmospherics and processes/algorithms are the issue. This is why I believe software processes and routing for this complex system will take many iterations. Google's involvement, possibly a bit of "ai", will be instrumental for fast iteration.
This is just my opinion, you folks may have a different take but...history repeats itself.
A few of us have followed the progression of technology. There has been excitement, achievements and the usual detractors.
First it can't be done.
Then when getting close, focus of complaints narrow.
When accomplished, focus shifts to false economies for other attributes while some get their butts kicked.
There are still many critics, some tied to other loyalties, political arena of mega military contractors and location of "pork industries", lack of STEM knowledge, lack of interest or just plain hatred of Elon Musk.
It will never go away.
This is the adventure of renewable energy and responsible transportation.
When it gets close...focus shifts to poor comparisons of quality, range and convenience.
When it is on its way, attack the company on safety, rebates, short selling, cash flow or as usual, pick on Elon.
Tesla has helped lead the change world wide. EV's are here to stay, diesels are beginning to be removed from production, combustion engines are "souring" with the awareness of "climate change". Solar energy and battery walls have proved their worth. Other countries that have been on-board before are increasing the pace to renewables. Europe, China and Southeast Asia have and continue amazing work towards lessening reliance on fossil fuel.
The issue is the mega fossil fuel empire. They have pull and will cause frustration till the end. Dealerships tied to political strings, States tied to oil interests and delaying solar deployment (Florida ring a bell). Auto manufacturers now need to retool and reduce combustion production.
This will not go away...Tesla is fighting against big oil, auto manufacturers, parts suppliers for combustion, energy giants and of course politicians and lobbyists. Last but not least are the usual pundits...climate change deniers, the STEM iliterate and of course the "I hate Elon at all costs" group.
The adventure for universal internet access begins. This system, when complete will have capabilities far beyond broadband internet and may eventually encompass pretty much anything handled by "everyone's" satellite network..let that sink in.
Right now we are at the can't be done phase with some quietly showing that it actually could be done. Constellation squabbles have already started, a lot with misleading information from competitors evidenced by the FCC filings.
There will be several iterations, satellites will take at least 6 years for the full initial constellation and a few more for the second constellation and full BFR/BFS use.
In the mean time...the usual pundits will talk about poor initial results which will eventually tail off.
Starlink will have a lot of "enemies" such as broadband companies, local providers, satellite operators that know the capabilities as Starlink progresses.
This will involve big business...world wide with the politics, as the system grows.
This will not go away...and of course...insert the usual "Elon haters" for another venue of attack.
This will get ugly and the technological enthusiasts will have to put up with the "bad press" for quite some time to come.
Patience will be valuable during this period and the "bad press" and "pundit" involvement will exceed that of TESLA.
That's just my opinion though.
On the lighter side....we are going to see a few more companies "edging in"
Deliver your content simply anywhere! To any device, anytime! Eutelsat CIRRUS is the reliable, high quality answer to reach multiple screens. Watch the demo and find out more at #AfricaCom
https://twitter.com/Eutelsat_SA/status/1061513865014308866
We will eventually see SES,Orbcom, Intelsat, Iridium,and a lot more plying their niche's like IOT, TV , music, film cloud service access , inventory tracking....bottom line is their profitability.
This will increase as the constellation develops.
Updated analysis
This looks better with the 550 km initial orbital change. I believe that the routing will obviously be determined by software, but in my opinion, he may have made a few assumptions that I would have refrained from. This will be one constellation, multiple use with inter-satellite communications as well as ground support.
The level of inter- satellite interaction will be determined as the system evolves, as implied in the various filings. I have a hunch but prefer not to change feet at the moment. There is a lot of critical data missing, which is very intentional and very understandable. Even the size of the initial payloads are a bit ambiguous even when we have an idea of thruster size, fuel tank mass, the mass of the 4 gyro's...etc...but I have collected several images that I will post later.
Found this of interest, but I will stop posting these as several companies are ramping up the "customer" acquisition attempt, with what they have at the time. Some subtle, some not so, but it was expected to start some time, might as well be now. Particularly when Elon has stated a quicker introduction and iteration of the constellation system.
On 14 November at 13:20, @INTELSAT's @jpgillet will participate in a discussion at #AfricaCom around how hybrid networks are the key to solving the digital divide and #satellite technology is essential to building out Africa's network infrastructure and meeting broadband demands!
https://twitter.com/INTELSAT/status/1062473346435170304
Bit boring...skip if you hate walls of text...
Back to where I left off before being sidetracked...
From the table we can see that the 1584 satellites will be in the 550 km orbital altitude. The number of planes is 24. Use the equator as reference and 360 degrees divided by 12 planes (2 planes opposed) means a spacing between plane sets of 30 degrees. A visual idea is one plane set crossing at 30 degrees and the plane set beside it will be crossing at 60 degrees. All plane sets equally spaced by 30 degrees and each plane set will have an inclination of 53 degrees east and 180-53 degrees west.
Each plane has 66 satellites. Earth radius is 6371 km, then add 550 km for a total from earth core of 6921 km. Circumference is 2Piradius which gives us 43485.926 km. this circumference divided by 66 satellites will give a distance of approximately 660 miles for and aft in a plane.
Note...One may ask why opposed planes...A series of single inclined plane at that altitude and the beam patterns from the FCC filing will result in an approximately 30% loss of geography coverage...I did a rough calculation/simulation.
Few calculations for fun...
Orbital velocity of a satellite is roughly v=SQRT(G*Mcentral/R)
G is 6.673 x 10-11 N•m2/kg2
R was 43485.92551 km before rounding
Mcentral is Mearth which is 5.98 x 10^24 kg
550 km orbit is 7.5932 km/sec
5727 seconds to orbit, or approx 95.4 minutes orbital period.
(ISS is approx 92.68 minutes at 7.66 km/sec for approx 400 km altitude, inclination of 51.6 degrees. The ISS moves west each pass by 23 degrees because the earth rotates at 0.25 degree per minute. The SpaceX constellations various planes will move approximately the same)
From data mining an orbital plane at 1110 km and 53.8 degrees inclination,I found an orbital period of 6420 seconds or 107 minutes.
1110 km plane is 107 minute orbital period
550 km plane is 95.4 minute orbital period
ISS at 400 km is approximately 92.7 minute orbital period. ISS varies from 330 km to 435 km presently. Lower limit may need to be modified later.
VLEO 7518 sat constellation at approx 340 km will be around 91.6 seconds orbital period
A quick visualization is putting the ISS tracker on your browser and imagine the ISS as a constellation satellite with a satellite for and aft by of this at 660 miles....then visualize the other planes criss crossing and the sets spaced across the equator by 30 degrees.
At the equator, each degree of latitude is 111 km, 30 degrees is approximately 3330 kilometers and narrows as longitude moves to the poles. This seems large but with an inclination of 53 degrees, approximately 2/3 of the planet, centered on the equator, will have coverage.
The above is for the initial 1584 satellite constellation layer. There are 4 more altitude bands to fill for the full 4409 satellite constellation. We have not mentioned the 7518 yet. Both constellations, when complete will provide a very robust world broadband network. Non fibre networks do not stand a chance against this when implemented.
Debris mitigation
The upper 4 altitude bands are the ones that require proper mitigation since an unattended satellite can be up there for centuries. The 550 km 1584 satellite batch and the full VLEO of 7518 satellites will be easy.
From the FCC filing November 8, 2018 Attachment A
Rapid, Passive Disposal
SpaceX intends to perform an active disposal of all of its satellites at the end of their life, in which the satellites first drop to a perigee of approximately 300 km while maintaining an apogee at approximately 550 km.
For the new lower shell of satellites, this "active" phase of the deorbit sequence will take a few weeks for each vehicle, after which several weeks to months of "passive" disposal follow, with the exact time depending on solar activity. Even this phase is not fully passive – to minimize the risk of debris even further, SpaceX satellites will continue to perform conjunction avoidance until the high atmospheric torques from low altitudes cause the vehicle to be uncontrollable.
At all times during this descent, including the period during which they will traverse the orbital altitude of the ISS and other NASA assets, the spacecraft will retain sufficient fuel to perform maneuvers. After all propellant is consumed, the spacecraft will be reoriented to maximize the vehicle's total cross-sectional area. Finally, the spacecraft will begin to passivate, drawing batteries down to a safe level and powering down.
While SpaceX expects its satellites to perform nominally and deorbit actively as described above, in the unlikely event a vehicle is unable to finish its planned disposal maneuver, the denser atmospheric conditions at 550 km provide fully passive redundancy to SpaceX's active disposal procedures.
The natural orbital decay of a satellite at 1,150 km requires hundreds of years to enter the Earth's atmosphere, but the lower satellites at an altitude of 550 km will take less than five years to do so, even considering worst-case assumptions.
Due to the very lightweight design of the new spacecraft, SpaceX achieves a very high area-to-mass ratio on its vehicles. Combined with the natural atmospheric drag environment at 550 km, this high ratio ensures rapid decay even in the absence of the nominally planned disposal sequence. Thus, even assuming an extreme worst-case scenario – i.e., the spacecraft fails while in the operational orbit (circular at 550 km), has no attitude control, and solar activity is at a minimum – the longest decay time is still only approximately 4.5-5 years.
In reality, this confluence of worst-case assumptions is unlikely to be realized for a number of reasons, not the least of which is that overall solar activity is ramping up into the next decade, meaning a more realistic worst-case decay time of 1 to 3 years.
But even assuming the unlikely 5-year decay period, SpaceX satellites will reach demise well within the prevailing 25-year deorbit standard. In fact, SpaceX will exceed new stricter parameters NASA recently determined for safe operation of large constellations28 by achieving a 100% success rate of post-mission disposal within 5 years even assuming worst-case conditions, directly addressing one concern previously identified by the Commission.
Nonetheless, SpaceX's nominal disposal plan that it anticipates for nearly every spacecraft will result in a lifetime of less than six months after SpaceX initiates disposal, an advantage of operating at the lower altitude proposed in the modification.
// note...This is why I really like this idea...launch the first series of prototype payloads to this altitude and deal with the issues at hand...before moving up to the higher orbits where a dead sat stays. This method enables GNC telemetry observations for a larger sample size which will help to develop a robust payload to operate in a non
// forgiving location. Debris in numbers at that altitude can at some point, render the altitude useless.
Atmospheric Demise
//(shorten this up)
The spacecraft's small mass and predominantly aluminum construction maximize the likelihood of atmospheric demise on re-entry. To verify this, SpaceX utilized NASA's Debris Assessment Software ("DAS").The SpaceX design remains below the necessary demise threshold, even using worst-case satellite configurations.
Although SpaceX made efforts to avoid the use of components resistant to disintegration, some scenarios were unavoidable at this time. DAS analysis indicates that three unique components may have a chance of reaching the Earth's surface with sufficient energy to result in human casualty.
These components are listed in the table below.
altitude demise requirements
Comparing these results against those from the Original Applications yields the following observations. First, the total DCA of the communications components has decreased slightly, from 2.79 m2 to 2.34 m2. Second, by far the biggest change is the addition of steel reaction wheels, which were not a potential source of human casualty risk in the prior analysis. SpaceX plans to deploy two versions of its initial satellites with slightly different configurations and each will only carry a subset of the components identified above.
Done for the day...may have made a few typo's will check again when more time...
FCC approves StarLink's 7,518 satellite VLEO segment, final approval for the full 12,000+ satellite system.
Washington Post...
Elon Musk's SpaceX wins FCC approval to put 7,000 Starlink Internet satellites into orbit
Federal regulators are allowing entrepreneur Elon Musk to use an expanded range of wireless airwaves for his plan to deliver cheap, high-speed Internet access - from space.
The decision Thursday by the Federal Communications Commission paves the way for SpaceX to build its full network of about 12,000 satellites intended to blanket the earth in wireless Internet access. Proponents say next-generation satellite Internet technology could help developing countries and rural areas connect to economic opportunities currently out of reach for them because they lack competitive Internet access.
"I'm excited to see what these services might promise and what these proposed constellations have to offer," said FCC Chairman Ajit Pai. "Our approach to these applications reflects this commission's fundamental approach to encourage the private sector to invest and to innovate and allow market forces to deliver value to American consumers."
Edited November 15, 2018 by DocM
FCC approves SpaceX, Telesat, LeoSat and Kepler internet constellations
WASHINGTON — The U.S. Federal Communications Commission voted Nov. 15 to approve four proposed satellite constellations, authorizing nearly 8,000 small telecom satellites to serve U.S. entities from low Earth orbit.
The new regulatory approvals set the stage for two companies, SpaceX of Hawthorne, California, and Telesat of Ottawa, Canada, to expand constellations already approved last year with more satellites in the rarely used V-band spectrum.
Canadian startup Kepler Communications and LeoSat, a company licensed from the Netherlands, also received approvals, Kepler for 140 Ku-band satellites and LeoSat for 78 Ka-band satellites.
The FCC's approval triggers a regulatory deadline whereby the companies must place at least half their constellations in orbit within six years, and the full systems in nine. At the end of nine years, the FCC freezes approval at the number of satellites in use.
SpaceX, Telesat and Kepler all asked for leniency with the six-year rule, but were rebuffed by the FCC.
FCC Commissioner Jessica Rosenworcel said the constellations that advanced today are evidence of a major shift in space activities, since the number of approved satellites nearly equals the 8,126 objects that have been launched into space since Sputnik in 1957.
"Do the math: it adds up to a next generation space race," she said. "New commercial models, players and technologies are all coming together and rapidly multiplying the range of satellite services. With these new services come all kinds of opportunities."
FCC Commissioner Brendan Carr cautioned that how successful the new constellations will be is not up to the FCC.
"We let these four companies move forward, and allow the market decide their success," he said. "For my part, I'm excited to see what services these companies will offer, and I'm glad we are clearing the way for more choices and more connections for Americans regardless of where they live."
Commissioners Rosenworcel and Michael O'Rielly cautioned that more work remains to be done on orbital debris concerns given the large number of satellites planned to launch within the next decade, but said approving the four constellations was important to do now regardless.
"While there are still issues to be explored, including communications with [Earth stations in motion] and orbital debris, and policy calls that we may not have gotten quite right, such as how we handle in-line interference, the commission continues to take necessary steps to allow investment and future
https://spacenews.com/fcc-approves-spacex-telesat-leosat-and-kepler-internet-constellations/
The starting pistol...er...shotgun...has been fired.
This starts the "pressure cooker" for SpaceX and subsidiaries.
The Starlink lower orbit modification filing has been accepted by the FCC.
FCC filing acceptance
This starts the 30 day discussion period..or should I say whining period for some...
This should pass...it is the better option for debris mitigation.
SpaceX raising $500 million to help build satellite broadband network
SpaceX is raising $500 million from investors to help build its worldwide satellite broadband network, The Wall Street Journal reported today.
The company run by Elon Musk has agreed on financing terms with existing shareholders and new investor Baillie Gifford & Co., who will pay $186 per share for new stock, valuing the company at $30.5 billion, according to Journal sources. SpaceX hasn't received the money yet but could announce the deal by the end of December, the Journal reported.
The funding round would pay for initial costs but not the entire project, which the Journal report said could cost as much as $10 billion. We contacted SpaceX about the funding today but the company declined to comment.
SpaceX last month received Federal Communications Commission approval to deploy up to 11,943 broadband satellites for its planned "Starlink" service.
That includes a previously approved first round of 4,425 satellites, which are expected to orbit at altitudes of 1,110km to 1,325km, a fraction of the altitude of traditional broadband satellites. The second batch of 7,518 satellites is set to operate at even lower altitudes of 335km to 346km. SpaceX has said it will provide gigabit speeds with latencies comparable to cable and fiber systems.
SpaceX's initial FCC approval from March of this year requires it to launch 50 percent of the first 4,425 satellites by March 2024 and all of them by March 2027. SpaceX's approval for the second batch of 7,518 satellite similarly requires it to deploy half within six years and the remaining satellites within nine years. However, SpaceX may seek a waiver that would lower the number of satellites it has to deploy in the first six years.
The slow progress led Musk to fire some of the project's senior managers. Still, the schedule hasn't changed much since November 2016, when SpaceX's FCC license application mentioned 2019 as a possible time for launching satellites into orbit. A Reuters report in October this year said SpaceX is planning to launch satellites in mid-2019 and offer broadband service in 2020.
more at the link...
https://arstechnica.com/information-technology/2018/12/spacex-raising-500-million-to-help-build-satellite-broadband-network/
pay walled, but here is the link for verification...
Elon Musk's SpaceX Is Raising $500 Million in Funding
Elon Musk's rocket company, Space Exploration Technologies Corp., is set to raise $500 million at a $30.5 billion valuation, in a bid to help get its internet-service business off the ground, according to people familiar with the fundraising......
https://www.wsj.com/articles/elon-musks-spacex-is-raising-500-million-in-funding-11545142054
This is basically keeping venture funding with an internal prior investment firm...this is a good move since they understand the frame work of the project.
reference for launch timeline...
Musk shakes up SpaceX in race to make satellite launch window: sources
The management shakeup and the launch timeline, previously unreported, illustrate how quickly Musk wants to bring online SpaceX's Starlink program, which is competing with OneWeb and Canada's Telesat to be first to market with a new satellite-based Internet service.
SpaceX spokeswoman Eva Behrend told Reuters the Redmond office remains an essential part of the company's efforts to build a next-generation satellite network.
"Given the success of our recent Starlink demonstration satellites, we have incorporated lessons learned and re-organized to allow for the next design iteration to be flown in short order," Behrend said.
She noted the strategy was similar to the rapid iteration in design and testing which led to the success of its rockets.
more importantly...
JUNE 2019 LAUNCH GOAL
SpaceX has said it would launch its satellites in phases through 2024. It goal of having Internet service available in 2020 is "pretty much on target" with an initial satellite launch by mid-2019, one of the sources said.
OneWeb aims for a first launch between December and February 2019, while Telesat was targeting 2022 for broadband services.
This funding round and a few more to follow, have one purpose...to place an initial constellation in orbit...as quickly as possible...to generate income which will pay for the rest of the constellation, an estimated $10B venture.
The initial launch campaign will be tentatively starting June 2019...This means that F9 and possibly a few FH's later on, will be the work horses for the initial 2 to 3 years.
This initial constellation is expected to begin limited operations in 2020 ...this removes any newer vessel from helping F9 with the launch at this time.
With other competitors on their tail and 2024 not that far away for the FCC mandated constellation numbers....they need to move now.
Starship and the super heavy booster will not be ready in time for this initial deployment phase.
Getting service started in 2020 means 800 birds launched.
The next two years should be fun...


text:
Updated: Driver Found Guilty in Draper Crash Case - WBOC-TV 16, Delmarvas News Leader, FOX 21 -
Updated: Driver Found Guilty in Draper Crash Case
Monday, September 10, 2018 1:37 PM EDT
By Madeleine Overturf
moverturf@wboc.com
By Kye Parsons
kparsons@wboc.com
GEORGETOWN, Del.- Following two days of testimony, a Sussex County Superior Court jury convicted the driver charged in last year's bicycle crash that led to the death of longtime WBOC owner Thomas Draper.
Shawn Armstrong was found guilty of operation of a vehicle causing the death of another person and inattentive driving in connection with the Sept. 7, 2017 incident that occurred in Slaughter Beach outside of Milford.
Armstrong could face up to 30 months behind bars when sentenced on Oct. 26, according to the Delaware Department of Justice. Armstrong did not speak to WBOC after the verdict, but his attorney, Michael Abram, expressed disbelief as he left the courthouse.
"I don't know what evidence the jury saw," he said. "I'm confused."
Abram told WBOC he assumed they would appeal the verdict. Members of Draper's family, meanwhile, say the verdict brought some sense of relief.
"The reality is the damage is done for us. There is no bringing my father back," said Draper's son, Hank Draper. "I don't personally have any ill will or vengeful feelings about Shawn Armstrong, but he is going to have to live with this for the rest of his life."
"I am just glad it was proved my dad didn't erratically pull out in front of him as if he does not know how to ride a bicycle," Draper concluded. Draper told WBOC he plans to speak at the sentencing in October.
Armstrong's trial got underway Monday in Sussex County Superior Court with jury selection, opening statements and four witnesses called to the stand.
The trial resumed Tuesday morning with the defense stating its case.
The defense called to the stand Brian Jackson, one of Armstrong's employees. Jackson, who works with Armstrong for EH Custom Homes, was building a home addition in Slaughter Beach the day of the accident.
Jackson testified that on his way to the project site, he passed a man riding a bicycle and wearing a green vest. He said that at around 7:30 a.m. he saw the man biking on the shoulder by the fire house. Jackson said he infers the man was Mr. Draper because he saw the same biker a number of times during the six to seven months he worked on the project in Slaughter Beach, but never saw the man after the accident.
Jackson testified the biker was always in the shoulder when he saw him. He also said on the day of the accident, the sun was bright but there were no other distractions.
During cross examination, the prosecution poked holes at Jackson's motive and memory, pointing out that Jackson never spoke to police, and he could not recall the exact weather, bike or helmet color that day. The prosecution also said Jackson's description of the biker's location was roughly three-quarter of a mile before the crash site, and 20-30 minutes ahead of Armstrong's timeline, so the timing would not make sense. The prosecution also pressed Jackson that he can't be sure the man he saw was Mr. Draper.
Defendant Testimony
After a brief re-direct by the defense, Armstrong took the stand. He described his job as a supervisor for EH Custom Homes and the work site in Slaughter Beach. Again, Armstrong said he saw Mr. Draper biking on the shoulder. Armstrong testified that when he first noticed Mr. Draper, he was a couple hundred yards behind him. He also testified he was not distracted-- he said he was not eating breakfast or playing with the radio before the crash. During Monday's courtroom action, speed, impairment, and cellphone use were essentially ruled out as contributing factors in the crash, as per a Delaware State Police investigation.
Armstrong testified that Mr. Draper veered off a foot or two into the roadway before the accident.
"There was nothing I could've done. It just happened so quick," Armstrong told the court. Armstrong also described how he tried to call 911 and due to a phone error, he could not hear 911 at first and began to panic. He said he was then next to Mr. Draper with his hand on his back "to let him know I was there." Armstrong described trying to keep Mr. Draper still and then holding his head when the state trooper arrived.
Next, the defense played the 911 call from the scene. In the call, Armstrong was clearly emotional and upset and told the dispatcher, "I just hit a guy on the side of the road...He was in the shoulder of the road...I don't know if something moved (audio a bit unclear) him, but he swerved out into my lane."
The 911 dispatcher told Armstrong to keep Mr. Draper still and let Mr. Draper know him help was on the way, after asking his condition (Armstrong described it as heavy breathing).
Armstrong said, "I can't believe this," and described trying to text an employee to come to the scene. Armstrong mentioned multiple times that Mr. Draper was trying to get up, with Armstrong responding, "Help is on the way, please stay down." Draper can be heard groaning in pain on the call.
Armstrong once again expressed disbelief, saying, "I don't know why (this happened)." Armstrong stayed on the phone until the trooper arrived, and the dispatcher told him he can hang up.
Next the defense played Armstrong's interview that day with Cpl. Kenneth Argo, the crash investigator. Once again, Armstrong attested that Mr. Draper swerved across the line into the roadway at the last second. Armstrong again said he doesn't know why this happened, this time saying, "I don't think he knew I was there. I had no reaction time."
Armstrong said he may have been going about 50-55 miles per hour, just at or above the speed limit. Armstrong said he stopped as soon as he hit Mr. Draper and that "for some reason he came out in front of the car."
Armstrong stressed that he maintained in his lane and did not move over. He said he was listening to the Elvis Duran radio morning show and it had just switched to a song when the crash occurred. He said the air conditioner was on, windows up, but the music was not loud. When referring to Mr. Draper's biking, Armstrong said, "He wasn't wobbling" and "he wasn't near the grass."
Armstrong further stated that his vehicle had automatic lights on, but not the headlights. Armstrong told Cpl. Argo that he didn't see Draper look back, but he didn't look unsecure so he just maintained his position in the lane.
"I wish I didn't," he then said.
Armstrong also testified that had seen Mr. Draper before on the road. He said he always saw Mr. Draper on the shoulder, and never biking with anybody. He described that anytime he drove past a bike, he maintained his lane, thinking the less corrections there are, the less chances there are for accidents.
During cross examination, co-prosecutor Nichole Gannett questioned Armstrong's passing habits, or lack thereof. She pointed out that Armstrong knows people can fall off bikes, so why wouldn't he pass. Gannett said that Armstrong would give a car with people inside it more room than a bicyclist.
During cross examination, Armstrong admitted "I suppose" he was legally allowed to pass, but affirmed "he [Mr. Draper] looked steady on his bike." , Armstrong also testified he never had a problem the previous times he passed Draper, so he had no indication he should change lanes.
Cpl. Argo Recalled to the Stand
Once again, Argo was called to the stand. Defense attorney Mike Abram asked if there is a duty for the driver to go to the opposite lane if the bike was in the shoulder? Argo says there is not. Argo also said Armstrong was cooperative that day. Abram pressed Argo on the fact that Curt Esposito (Mr. Draper's longtime cycling partner) was interviewed months after the accident, and the crash reconstruction map was printed five days after the accident. During cross examination, (in front of the prosecution) Argo said he came to the conclusion Mr. Draper was in the roadway.
"I believe he was in the roadway the entire time based on the condition of the shoulder," Argo testified under oath.
Esposito Returns to the Stand
Curt Esposito made a brief return to the stand, and spoke more about biking in Slaughter Beach. He said a bicyclist could hear a car or truck coming a ways back, and Mr. Draper always heard it first. Esposito also said, "Nobody could go in the shoulder," because of its condition.
After a brief break, the jury was read its instructions. The judge reviewed the charges and what the jury needs to be convinced of beyond a reasonable doubt to find Armstrong guilty. He described the definition of the charges: Operation of a vehicle causing the death of another person and inattentive driving.
The jury must be unanimous in its verdict.
Closing Arguments
The state began its closing arguments just before noon. Co-prosecutor Nichole Gannett stated "the state doesn't have to prove to you why this happened [...] No matter what the reason, the defendant did not maintain a proper lookout."
Gannett pointed out that Armstrong said Mr. Draper wasn't near the grass, and pictures show there was grass on the shoulder, meaning Mr. Draper was in the roadway. The prosecutor also reaffirmed Mr. Draper's biking habits, and said where Esposito testified Mr. Draper biked in regards to the shoulder was exactly where he was struck.
Gannett also pointed out that multiple times Armstrong said Mr. Draper was steady but asked could he be holding steady on a shoulder with grass growing through it?
Gannett also said Mr. Draper would have heard the car coming, as he was not wearing headphones and Esposito testified about Mr. Draper's hearing. Gannett also reaffirmed Jackson's (Armstrong's employee) spotty memory.
Gannett admitted how Armstrong has maintained his story, but if he had maintained a proper lookout, he would have noticed Mr. Draper was in the roadway. She again questioned "what reasonable person" would give a car more of a lookout than a bike, referring to Armstrong's lack of passing habits.
At noon the defense began its closing arguments. Abram pointed out that the gouge marks (the reason why the Delaware State Police investigation concluded Mr. Draper was in the roadway) was 1.7 feet from the line, but Esposito testified Mr. Draper was only ever 8-10 inches from the line. He also pointed out Esposito was not there at the time of the accident, saying "every day is different. Things change," referring to Draper's bike habits.
Abram also pointed out that there are no pictures of the shoulder ahead of the accident scene, so there is no way to show the conditions that could have allowed Mr. Draper to bike there. He even mentioned that Esposito said the two had road bikes that could go on some gravel, which is how Abram described the shoulder. Abram also said Argo was a crash specialist, "not a bike riding specialist," so how could he know how Mr. Draper was biking?
Abram also pointed out that Armstrong called 911 right away, and one minute into the call he said, "He pulled out in front of me." Abram asked how could Armstrong concoct a story in that short of a time frame? He also pointed out that his client admitted he may have been speeding (albeit slightly) to Argo.
"If his story was fabricated, why would he implicate himself in another way?" Abram asked the jury. He told jurors they had a hard job, and again reminded them that sympathy cannot come into play. He also told the jury "this is not a 50-50 game," meaning they must be sure beyond a reasonable doubt, and he says the state did not do that with the evidence the jury has.
Then the state got a chance to make its rebuttal, the final piece. Prosecutor Kevin Gardner began with a story about his nephew loving Oreos, but when the Oreos went missing and his nephew blamed his brother. Gardner said if his nephew had crumbs around his mouth, that is evidence that his nephew ate the Oreos. He used the analogy saying the reconstruction map showed Armstrong hit Mr. Draper in the roadway, and Armstrong blamed Mr. Draper like Gardner's nephew blamed his brother.
Gardner also said the skid marks were parallel to the shoulder, showing Mr. Draper was in the roadway, and that Jackson is Armstrong's employee, so why would he testify against his boss?
Gardner asked why Jackson never told his story to police before Tuesday. Gardner again referred to the quality of the shoulder and Armstrong admitting he saw Mr. Draper from a distance.
"We have no doubt he saw Mr. Draper, but that split second of not paying attention, what did that cost?" Gardner asked the jury.
He finished his rebuttal referring to the Oreo analogy again and ended the trial how he began: the devil is in the details.


text:
import { bindActionCreators } from 'redux';
import { connect } from 'react-redux';
import NewCategoryContainer from './Container';
import * as NewCategoryActions from './action';

function mapDispatchToProps(dispatch) {
  return bindActionCreators(NewCategoryActions, dispatch);
}

export default connect(null, mapDispatchToProps)(NewCategoryContainer);

redpajama_set_name: RedPajamaGithub