text:
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  \title{%servetheloop dump
  \large $\#$servetheloop dump (for the rLoop) }

\author{Ernest Yeung \href{mailto:ernestyalumni@gmail.com}{ernestyalumni@gmail.com}}
\date{27 mai 2017}
\keywords{Eddy current brakes}
\begin{document}

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\maketitle

\noindent gmail        : ernestyalumni \\
linkedin     : ernestyalumni \\
twitter      : ernestyalumni \\

\setcounter{tocdepth}{1}
\tableofcontents

\begin{multicols*}{2}

\begin{abstract}
servetheloop notes "dump" - I dump all my notes, including things that I tried but are wrong, here.

\end{abstract}

\part{Eddy Currents, Eddy Current Braking}

\section{Eddy Currents}

\emph{Keywords:} Eddy currents;

cf. Smythe (1968), Ch. X (his Ch. 10) \cite{Smyt1968}

Assume Maxwell's "displacement current" is negligible; this is ok if frequencies are such that wavelength $\lambda$ large compared to dimensions of apparatus $L$.  $\lambda \gg L$ or $\frac{c}{\nu} \gg L$.

I will write down the "vector calculus" formulation of electrodynamics, along side Maxwell's equations, or electrodynamics, over spacetime manifold $M$.  The latter formulation should specialize to the "vector calculus" formulation.

From
\[
\text{curl} \mathbf{E} = -\frac{d\mathbf{B}}{dt} \, (SI) \qquad \, \text{curl}\mathbf{E} = -\frac{1}{c} \frac{ \partial \mathbf{B}}{ \partial t} \, (cgs) \text{ or } \frac{ \partial B}{ \partial t} + \mathbf{d}E = 0
\]
Suppose $B = \text{curl} A$ or $B=\mathbf{d}A$ (EY 20170528: is this where the assumption above about $\lambda \gg L$ comes in?), then
\[
\begin{gathered}
	-\frac{\partial B}{ \partial t} = \mathbf{d}E \xrightarrow{ \int_S } \int_S \mathbf{d}E = \int_{ \partial S} E = -\int_S \frac{ \partial B}{ \partial t} \xrightarrow{ B = \mathbf{d}A } -\int_S \frac{ \partial }{ \partial t} \mathbf{d}A \xrightarrow{ \text{ flat space } } \int_S \mathbf{d}E = -\int_S \mathbf{d} \frac{ \partial A}{ \partial t}
\end{gathered}
\]
and so
\begin{equation}
	\mathbf{E} = \frac{-\partial \mathbf{A}}{ \partial t}
\end{equation}
up to gauge transformation, if $B = \mathbf{d}\mathbf{A} = \text{curl}\mathbf{A}$

Since this $\mathbf{E}$ field is formed in a conductor, Ohm's law applies.  Let's review Ohm's law.  Smythe (1968) refers to its 6.02 Ohm's Law - Resistivity section \cite{Smyt1968}.  Indeed, in a lab, the definition of resistance can be defined as this ratio:
\begin{equation}
R_{AB} := \frac{ -\int_A^B \mathbf{E} }{ I_{AB} } = \frac{ V_A- V_B}{ I_{AB} } = \frac{ \varepsilon_{AB} }{ I_{AB} }
\end{equation}
Moving along, the right way to think about resistivity $\rho$ is to consider conductivity.

Assume current density is linear to $\mathbf{E}$ field (as $\mathbf{E}$ field pushes charges along).  This linear response is reasonable.

Also, assume current density $\mathbf{J}$ is uniform over infinitesimal surface area $dA$ (i.e. surface $S$).

Define
\begin{equation}
\sigma \equiv \text{ \textbf{ conductivity } }
\end{equation}

Then the empirical relation/equation that underpins \emph{Ohm's law} is
\begin{equation}
\mathbf{J} = \sigma \mathbf{E}
\end{equation}
and define \emph{resistivity} from there:
\begin{equation}
\sigma := \frac{1}{\rho}
\end{equation}
where $\rho $ is the \emph{resistivity}.

Thus
\[
\begin{gathered}
	\sigma \int_A^B -\mathbf{E} = \sigma V_{AB} = \int_A^B \mathbf{J} = I_{AB} \frac{l}{A} \\
\frac{1}{\rho} R = \frac{l}{A} \text{ or } \boxed{ R = \rho \frac{l}{A} }
\end{gathered}
\]

From Maxwell's Equations,
\begin{equation}
	\mathbf{\delta}(B - 4\pi \mathbf{M} ) = \frac{4\pi }{c} J_{\text{free}}
\end{equation}

If $B=H+4\pi M = (1+4\pi \chi_m) H = \mu H$,
then
\begin{equation}
\mathbf{\delta}H = \mathbf{\delta} \frac{B}{\mu} = \frac{ 4\pi }{c} J_{\text{free} }\text{ or } \mathbf{\delta} B = \mu \frac{4\pi }{c} J_{\text{free}} \Longleftrightarrow \text{curl} B = \mu J_{\text{free}} \qquad \, (SI)
\end{equation}
and if $B=\mathbf{d}A$ and $\mathbf{d} \mathbf{\delta}A =0$.

%\subsubsection{Eddy Currents}

I build upon the physical setup proposed by Jackson (1998) \cite{Jack1998} in Section 5.18 "Quasi-Static Magnetic Fields in Conductors; Eddy Currents; Magnetic Diffusion."

For a system (with characteristic) length $L$, $L$ being small, \\
compared to electromagnetic wavelength associated with dominant time scale of problem $T$,
\[
\begin{gathered}
	f := \frac{1}{T} ; \quad \, \omega = 2\pi f ; \quad \, \omega \lambda = c \Longrightarrow \lambda = \frac{c}{ \omega } = \frac{c}{ 2\pi f } = \frac{Tc}{2\pi }  \\
\frac{L}{\lambda} = \frac{LTc}{2\pi } \gg 1
\end{gathered}
\]
From Maxwell's equations, in particular, Faraday's Law, and in its integral form (over 2-dim. \emph{closed} surface $S$),
\begin{equation}
\begin{gathered}
\mathbf{d}E + \frac{ \partial }{ \partial t } B = 0 \text{ or } -\mathbf{d}E =\frac{ \partial B}{ \partial t} \xrightarrow{ \int_S } \int_S \frac{ \partial B}{ \partial t} = -\int_S \mathbf{d}E = -\int_{\partial S} E
\end{gathered}
\end{equation}
So on $S$, changing magnetic flux $\int \frac{ \partial B}{ \partial t}$ results in $E$ field, circulating around boundary of $S$, $\partial S$.

We know that in a conductor, free conducting electrons get pushed around by $E$ fields, result in a current density $J$.

$J$ is related to $E$, \emph{empirically} (by Ohm's Law)
\[
J = \sigma E
\]
where $\sigma$ is the resistivity.

Then use the force law on this induced current $J$ from the $B$ field set up:
\[
F_{\text{net}} = \frac{1}{c} \int_S J\times B dA
\]
By working through the right-hand rule, $F_{\text{net}}$ the force on those currents induced in the conductor due to the $B$ that's there, is in the direction to help oppose changing (increasing or decreasing $\frac{\partial B}{ \partial t}$).

To find $B$, suppose $B=dA$, i.e. $B\in H^2_{\text{deRham}}(M)$, i.e. $B=\text{curl}A$.

For sure,
\[
\mathbf{\delta} (B-4\pi c \mathbf{M}) = 4\pi J \Longleftrightarrow \text{curl}(B-4\pi c \mathbf{M} ) = \text{curl} H = 4\pi J
\]
Be warned now that the relation $B=\mu H$ may not be valid on all domains of interest; $\mu$ could even be a tensor! (e.g. $B_{ij} = \mu^{kl}_{ij} H_{kl}$).  However, both Jackson (1998) \cite{Jack1998} in Sec. 5.18 Quasi-Static Magnetic Fields in Conductors; Eddy Currents; Magnetic Diffusion, pp. 219, and Smythe (1968), Ch. X (his Ch. 10), pp. 368 \cite{Smyt1968}, continues on \emph{as if} this relation is linear: $B=\mu H$.

Nevertheless, as we want to find $B$ by finding its "vector potential" $A$, we obtain a diffusion equation:
\begin{equation}\label{Eq:EddyCurrentsAdiffusion}
\begin{gathered}
	- \mathbf{\delta}B = \mathbf{*d*d}A = (-1) \mathbf{\delta d} A = (-1)( \Delta - \mathbf{d\delta} ) A \xrightarrow{ \mathbf{d\delta} A = 0 } - \Delta A = \\
	= 4\pi \mu J = 4\pi \mu \sigma E = 4\pi \mu \sigma \left( -\frac{ \partial A}{ \partial t} \right) \\
\Longrightarrow \boxed{ \Delta A = 4\pi \mu \sigma \frac{ \partial A}{ \partial t } }
\end{gathered}
\end{equation}
where in the first 2 steps (equalities), $- \mathbf{\delta}B = \mathbf{*d*d}A = (-1) \mathbf{\delta d} A$ it's interesting to note that the codifferential $\mathbf{\delta}$ for the 2 form $B$ had to be written out explicitly, and then the codifferential for the 1-form $A$ is \emph{different} from the $\mathbf{\delta}$ for $B$ by a(n important) factor of $(-1)$; where $\mathbf{d\delta} A=0$ must be satisfied by the form $A$ takes; and where, since $B=\mathbf{d}A$,
\begin{equation}
\begin{gathered}
	\mathbf{d}E + \frac{ \partial B}{ \partial t} = \mathbf{d} E + \frac{ \partial }{ \partial t} \mathbf{d} A = \mathbf{d} \left( E+ \frac{ \partial A}{ \partial t} \right) = 0 \Longrightarrow E = -\frac{ \partial A}{ \partial t} + \text{grad}\Phi \xrightarrow{ \Phi = \text{ constant } } E = -\frac{ \partial A}{ \partial t}
\end{gathered}
\end{equation}
whereas a choice of gauge for $E$ was chosen so that $\Phi=\text{constant}$ (and so a particular form for $E$ was chosen, amongst those in the \emph{same} equivalence class of $H^1_{\text{deRham}}(M)$.

To ensure that the differential geometry formulation is in agreement with the practical vector calculus formulation, compare Eq. \ref{Eq:EddyCurrentsAdiffusion} with Eq. (5.160) of Jackson (1998) \cite{Jack1998} and Eq. (10) in Sec. 10.00 of Smythe (1968) \cite{Smyt1968}.

To summarize what's going on, I think one should at least understand in one's head how Maxwell's Equations apply, (and I will try to write in SI here)
\begin{equation}
\boxed{
\begin{gathered}
\int_S \frac{ \partial \mathbf{B}}{ \partial t} dA = -\oint \mathbf{E}\cdot d\mathbf{s} \Longrightarrow \mathbf{J}=\sigma \mathbf{E} \Longrightarrow \mathbf{F}_{\text{net}} = \int_S \mathbf{J} \times \mathbf{B} dA  \\
\text{ find } \mathbf{B} = ? \qquad \, \text{ using form } \mathbf{B} = \nabla \times \mathbf{A}, \\
\nabla^2 \mathbf{A} = \mu \sigma \frac{ \partial \mathbf{A} }{ \partial t} \qquad \, (SI)
\end{gathered}
}
\end{equation}
where, a change in magnetic flux over a surface $S$ over the conductor, $\int_S \frac{ \partial \mathbf{B}}{ \partial t}dA$ induces a circulation of $E$ field around $S$, $-\oint \mathbf{E} \cdot d\mathbf{s}$, and this $E$ field is pushing around \emph{free conducting charges} according to Ohm's law, $\mathbf{J} = \sigma \mathbf{E}$, with $\sigma$ being the conductivity of the conducting material, and this current density $\mathbf{J}$ is then acted upon by the prevailing $B$ field, according to the usual force law.  To find $\mathbf{B}$, one can find $\mathbf{A}$ and \emph{try} to find $\mathbf{A}$ analytically.

Keep in mind that for $\nabla^2 \mathbf{A} = \mu \sigma \frac{ \partial \mathbf{A}}{ \partial t}$, we had used, critically, the Maxwell equation $\mathbf{\nabla} \times \mathbf{H} = \mathbf{J}$, with $\mathbf{J}$ being the \emph{induced current of free conducting charges on the conductor}.  This $\mathbf{H}$ will contribute (through linear superposition) to the $\mathbf{B}$ that could already be there due to the permanent magnet.

What can we measure quantitatively?
\begin{itemize}
\item Can we measure $\mathbf{J}$ inside (on) the conductor?
\item Can we separate magnetization $\mathbf{M}$ of material from $\mathbf{B}$, to obtain the actual $\mathbf{B}$ (and then use linear superposition, $\mathbf{B}_{\text{total}} = \mathbf{B}_{\text{permanent magnet}} + \mathbf{B}_{\text{induced currents}}$?
\end{itemize}

Also, keep in mind the context that the conductor at hand is the long, almost semi-infinite rectangle of a conductor, aluminum sub-rail, specified by the SpaceX Hyperloop.  Force on that will cause an equal and opposite force on the pod, with its permanent magnets attached, and thus braking the pod.

\part{System Theory, including Avionics, Embedded Systems}

%\part{System Theory}

cf. Hermann (1974) \cite{Herm1974}.

For a given field $\mathbb{K} = \mathbb{R} \text{ or } \mathbb{C}$ (or some field, in general).

Let $U,X,Y$ be either vector spaces (or more generally, $R$-modules, with $R=\mathbb{K}$), or manifolds.   Let
\[
\begin{aligned}
	& U \equiv \text{ input space } \\
	& Y \equiv \text{ output space } \\
	& X \equiv \text{ state space } \\
\end{aligned}
\]

Consider maps $f,g$, s.t.
\begin{equation}
	\begin{gathered}
	\begin{aligned}
& f: X\times U \to X \\
	& g : x \times U \to Y \end{aligned} \\
\begin{aligned}
	& \frac{dx}{dt} = f(x,u) \\
	& y(t) = g(x(t),u(t))
\end{aligned}
\end{gathered}
\end{equation}

Ordered 5-tuple $\sigma = (U,X,Y,f,g)$ is a \textbf{system}.

$t \mapsto (x(t),u(t),y(t)) \in X\times U \times Y =: $ \textbf{trajectory} of system, they're solutions to $\frac{dx}{dt} = f(x,u)$.

\begin{definition}
	Input-output pair: $t\mapsto (u(t),y(t)) \in U\times Y$ if $\exists \, $ curve $t\mapsto x(t) \in X$ s.t.
\begin{equation}
t\mapsto (x(t),u(t),y(t))
\end{equation}
\end{definition}

Choose input curve $t\mapsto u(t)$ so to achieve some desired output $t\mapsto y(t)$.

If we have linear maps $f,g$ or at least "linearize" the system, then $f,g$ take forms
\begin{equation}
\begin{gathered}
\begin{aligned}
& f(x,u) = Ax+Bu \\
& g(x,u) = Cx+Du
\end{aligned}  \\
\begin{aligned}
	& A:X\to X \\
	& B: U\to X \\
	& C: X\to Y \\
	& D: U\to Y
\end{aligned}
\end{gathered}
\end{equation}

And so, for a (linear) system $(X,U,Y,A,B,C,D,)$, s.t.
\[
\begin{aligned}
	& \frac{dx}{dt} = Ax(t) + Bu(t) \\
	& y(t) = Cx(t) + Du(t)
\end{aligned}
\]
Discretize $t$ or replace differential with difference equations.  And so for  \\
$t\in $ additive semigroup $\mathbb{Z}^+$,
\begin{equation}
\begin{gathered}
\begin{aligned}
	& t \mapsto u(t) \in \mathbb{Z}^+ \to U \\
	& t \mapsto y(t) \int \mathbb{Z}^+ \to Y
\end{aligned} \\
	 \frac{du}{dt} \xrightarrow{ \text{ discretize } } \Delta u(t) = u(t+1) - u(t) \\
\begin{aligned}
	& \Delta u(t) = Ax(t) + Bu(t) \\
	& y(t) = Cx(t) + Du(t)
\end{aligned}
\end{gathered}
\end{equation}

Equivalence of systems with same $X,U,Y$.

Let $GL(X) = \lbrace g | g:X \to X \rbrace$.

Given $t \mapsto (x(t),u(t),y(t))$ s.t. it's a solution to
\begin{equation}
\begin{aligned}
	& \frac{dx}{dt} = Ax(t) + Bu(t) \\
	& y(t) = Cx(t) + Du(t)
\end{aligned} \qquad \, (3.2 of Hermann (1974))
\end{equation}
cf. Hermann (1974) \cite{Herm1974}.

Set
\begin{equation}
	x_1(t) = gx(t)
\end{equation}
Then
\[
\begin{gathered}
	\begin{gathered}
	\frac{dx_1(t)}{dt} = g\frac{dx}{dt} = g(Ax+Bu) = A_1x_1 + B_1u  \qquad \, \text{ and so }  \begin{aligned} & A_1 = gAg^{-1}  \\
& B_1= gB \end{aligned}
\end{gathered}
\begin{gathered}
y(t) = Cg^{-1}x_1 + Du = C_1x_1 + D_1 u \qquad \, \text{ with } \begin{aligned} & C_1 = Cg^{-1} \\
& D_1 = D \end{aligned}
\end{gathered}
\end{gathered}
\]
\[
\Longrightarrow \sigma_1 = (U,X,Y,A_1 ,B_1, C_1,D_1)
\]
$\sigma,\sigma_1$ automatically have the same input-output pairs.

\begin{definition}
\begin{equation}	\begin{aligned}
	& \sigma = (A,B,C,D) \\
	& \sigma_1 = (A_1,B_1,C_1,D_1) \end{aligned}
\end{equation}  2 linear systems with same state, input, output space $X,U,Y$,
$\sigma,\sigma_1$ \textbf{ algebraically equivalent  } iff $\exists \, g \in GL(V)$ s.t.
\begin{equation}
\begin{aligned}
& A_1 = gAg^{-1} \\
& B_1 = gB \\
& C_1 = Cg^{-1} \\
& D_1 = D
\end{aligned}
\end{equation}
\end{definition}
i.e. Let $\Sigma = \lbrace \sigma \rbrace =$ collection of linear systems $\sigma = (A,B,C,D)$, \\
i.e. $\Sigma = L(X,X) \times L(U,X) \times L(X,Y) \times L(U,Y)$.

Let $GL(X) = \lbrace g|g:X\to X, \, \forall \, g, \, \exists g^{-1} \rbrace$.

If $g\in GL(X)$ and
\[
\sigma = (A,B,C,D) \in \Sigma
\]
then
\[
g\sigma = (gAg^{-1}, gB,Cg, D)
\]
\begin{theorem}[3.2]
	equivalence classes in 1-to-1 correspondence with \textbf{orbit space } $GL(X) \backslash \Sigma$.
\end{theorem}

\subsubsection{cf. 4. Impulse response and Transfer Functions: Observatibility and Controllability of Hermann (1974) \cite{Herm1974}.  }

From Eq. (3.2) of Hermann (1974) \cite{Herm1974}, i.e. $\begin{aligned} & \quad \\
	& \dot{x} = Ax + Bu \\
	& y = Cx+Du \end{aligned}$, i.e. linear, time-invariant, "lumped parameter" systems.

"weighting parameter" $=$ "impulse response".

It's easy to solve
\[
\dot{x} = Ax + Bu
\]
with solution
\[
x(t) = \int_0^t \exp{ (A(t-s) )} Bu(s)ds + \exp{(At)} x(0)
\]
since
\[
y = \int_0^t Ce^{A(t-s)} Bu(s)ds + Ce^{At} x(0) + Du
\]
i.e. by variation of parameters method.

Consider
\[
\begin{gathered}
	\dot{x}^1 - Ax^1 = 0 \\
x^1 = \exp{(At)}
\end{gathered}
\]
If $x_p(t) = c_1(t)x^1(t) = c_1 \exp{ (At) }$,
\[
\begin{gathered}
\dot{x}_p = c_1e^{At} + c_1 Ae^{At} \\
\dot{x}_p = Ax_p - Bu = \dot{c}_1 + c_1 A - Ac_1 - Bu = 0  \\
\dot{c}_1 = Bu \text{ or } c_1 = \int_0^t ds Bu(s) \\
c_1(0) = x(0)   \qquad \, \text{ (EY : 20170602 I'm not sure if this is the right derivation) }
\end{gathered}
\]
\begin{definition}
	Function
\begin{equation}
	t\mapsto Ce^{At}B , \qquad \, R \to L(U,Y)
\end{equation}
is called \textbf{ impulse response } for system $\sigma = (A,B,C,D)$
\end{definition}

Its Laplace transform
\begin{equation}
	s=C(A\cdot s)^{-1}B
\end{equation}
is called \textbf{frequency response} of system.

\textbf{Remarks} A) \textbf{Resolvants}.  In functional analysis, operator-valued function $s\mapsto (A\cdot s)^{-1}$ is called resolvant of operator$A$.

\begin{theorem}[4.1 Hermann (1974) \cite{Herm1974}]
Let
\[
\begin{aligned}
&	\sigma = (U,X,Y,A,B,C,D) \\
&	\sigma_1 = (U,X_1,Y,A_1,B_1,C_1,D_1) \\
\end{aligned}
\]
Suppose $\sigma,\sigma_1$ equivalent, same input-output pairs.  Then
\begin{equation}
	C(A\cdot s)^{-1}B = C_1(A_1\cdot s)^{-1}B_1 \qquad \, s \in \mathbb{C}
\end{equation}
i.e. they have the same frequency responses.  Also $D=D_1$
\end{theorem}

\begin{proof}
Let $t \mapsto u(t)$ input curve, $x(0)$ initial state vector.

$t\mapsto y(t)$ output curve determined by $(t\mapsto u(t), x(0))$ in system $\sigma$.

Using Eq. (4.4) Hermann (1974) \cite{Herm1974},
\begin{equation}
\begin{gathered}
	y = \int_0^t Ce^{A(t-s) } Bu(s) ds + Ce^{At} x(0) + Du \\
y(t) = \int_0^t Ce^{A(t-s)} Bu(s) ds + Ce^{At } x(0) + Du(t)
\end{gathered}
\end{equation}
By hypotheses, $\sigma,\sigma_1$ equivalent, so  \\

state vector $x_1(0) \in X$, s.t.
\[
(t\mapsto u(t), x_1(0)) \text{ determines } t\mapsto y(t), \text{ i.e. }
\]
\[
y(t) = \int_0^t C_1 e^{A_1(t-s) } B_1u(s) ds + C_1 e^{A_1t } x_1(0) + D_1u(t)
\]

... 20170602 ...

\end{proof}

%\section{Embedded Systems}

\section{Finite State Machines}

Lee and Seshia \cite{LeSe2017}

\begin{definition}[memoryless, memoryless system]
Consider cont. time system $S: X\to Y$, where $X= A^{\mathbb{R}}$, $Y=B^{\mathbb{R}}$,
\[
S:(\mathbb{R}\to A) \to (\mathbb{R} \to B) \text{ i.e. } S:L(\mathbb{R},A) \to L(\mathbb{R},B)
\]
$A,B\in \textbf{Sets}$

$S$ \textbf{memoryless} if $\exists \, f :A\to B$ s.t. $\forall \, x$,
\begin{equation}
(S(x))(t) = f(x(t)) \qquad \, \forall \, t \in \mathbb{R}
\end{equation}
i.e. output $(S(x))(t)$ depends \emph{only} on $x(t)$
\end{definition}
cf. Sec. 2.3.2. Memoryless Systems of Lee and Seshia (2017) \cite{LeSe2017}

\begin{definition}[linear system; superposition]
system $S:X\to Y$ is \textbf{linear} if it obeys \textbf{superposition} property:
\begin{equation}
\begin{gathered}
	\forall \, x_1, x_2 \in X , \qquad \forall \, a,b, \in \mathbb{R} \\
S(ax_1 + bx_2) = aS(x_1) + bS(x_2)
\end{gathered}
\end{equation}
\end{definition}
cf. Sec. 2.3.3. Linearity and Time Invariance of Lee and Seshia (2017) \cite{LeSe2017}

\begin{definition}[delay]
\textbf{delay} $D_{\tau}:X\to Y$, $X,Y = C^1(\mathbb{R})$, so $D_{\tau}: C^1(\mathbb{R}) \to C^1(\mathbb{R})$
\begin{equation}
	\forall \, x \in X, \, \forall \, t \in \mathbb{R}, \, (D_{\tau}(x))(t) = x(t-\tau)
\end{equation}
\end{definition}
cf. Sec. 2.3.2. Memoryless Systems of Lee and Seshia (2017) \cite{LeSe2017}

\begin{definition}[time invariant]
	system $S:X\to Y$ \textbf{time-invariant} if
\begin{equation}
\begin{gathered}
\forall \, x \in X, \, \forall \, \tau \in \mathbb{R}, \\
	S(D_{\tau}(x)) = D_{\tau}(S(x))
\end{gathered}
\end{equation}
\end{definition}
Also, consider
\[
S(D_{\tau}(x))(t) = D_{\tau}(S(x))(t) = (S(x))(t-\tau)  \equiv y(t-\tau)
\]
e.g. Consider
\[
\dot{\theta}_y(t) = \frac{1}{I_{yy}} \int_{-\infty}^t\tau_y(t')dt'  \qquad \, \text{ so } \tau_y \mapsto \dot{\theta}_y
\]
\[
\begin{gathered}
	S(D_{\tau}\tau_y) = \frac{1}{ I_{yy}} \int_{-\infty}^t \tau_y(t'-\tau) dt' \\
D_{\tau}S(\tau_y)(t) = \frac{1}{ I_{yy}}\int_{-\infty}^{t-\tau} \tau_y(t')dt' = \frac{1}{I_{yy}}\int_{-\infty}^t \tau_y(u-\tau) du \equiv \frac{1}{ I_{yy}} \int_{-\infty}^t \tau_y(t'-\tau) dt' = \frac{1}{I_{yy}} \int_{-\infty}^t (D_{\tau}\tau_y)(t')dt' = S(D_{\tau}\tau_y)(t)  \\
\begin{aligned}
& t'=t-\tau \\
& t' + \tau = t\\
& u=t' + \tau \end{aligned}
\end{gathered}
\]

\begin{definition}[stable, bounded-input bounded-output stable (BIBO stable or just stable)]
system $S:X\to Y$, $X,Y = C^1(\mathbb{R})$ \textbf{stable} if $\forall \, w \in X$, s.t. $\exists \, A \in \mathbb{R}$ s.t. $|w(t) | \leq A$, \, $\forall \, t\in \mathbb{R}$,  \\
$S(w) = v$, $\exists \, B\in \mathbb{R}$ s.t. $|S(w)(t)| = |v(t)| \leq B$, \, $\forall \, t\in \mathbb{R}$.

\end{definition}

\subsubsection{Feedback Control: is it just an automorphism?}

Recall physics of the helicopter example:
\[
\begin{gathered}
	\dot{\theta}_y(t) = \dot{\theta}_y(0) + \frac{1}{I_{yy}} \int_0^t \tau_y(t')dt' \equiv (\text{int}(\dot{\theta}_y(0), I_{yy} ) )(\tau_y)(t), \text{ i.e. } \tau_y \xmapsto{ \text{int}(\dot{\theta}_y(0),I_{yy}) } \dot{\theta}_y
\end{gathered}
\]
Let $\psi$ be example of \emph{desired behavior}, input $\psi$ specifying desired angular velocity.

Introduce error measurement $e\equiv \text{err}$:
\[
\begin{gathered}
	\text{err}:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \\
\text{err}:\psi,\dot{\theta}_y \mapsto \psi- \dot{\theta}_y, \, \text{err}(\psi,\dot{\theta}_y)(t) = \text{err}(\psi(t),\dot{\theta}_y(t)) \equiv \text{err}(t) = \psi(t) - \dot{\theta}_y(t)
\end{gathered}
\]

\subsection{Discrete signals}

cf. 3.1 Discrete Systems of Lee and Seshia (2017) \cite{LeSe2017}

\begin{definition}[discrete signal]
\begin{equation}
\begin{aligned}
& e:\mathbb{R} \to \lbrace \text{ absent } \rbrace \cup X \equiv \lbrace \emptyset \rbrace \cup X \\
& e:t\mapsto e(t)
\end{aligned}
\end{equation}
\end{definition}

Let $T\subseteq \mathbb{R}$ s.t.
\[
T:= \lbrace t\in \mathbb{R} | e(t) \neq \text{ absent } \equiv \emptyset
\]
i.e. set of times when there's signal

If $\exists \, $ 1-to-1 function $f:T\to \mathbb{N}$ s.t. $f$ \textbf{order-preserving}, i.e. $\forall \, t_1, t_2 \in T$, where $t_1\leq t_2$, then $f(t_1) \leq f(t_2)$.

Example: Integration $I_i$:
\[
\begin{gathered}
	\begin{aligned}
	& I_i: \mathbb{R}^{\mathbb{R}_+} \to \mathbb{R}^{\mathbb{R}_+} \equiv I_i : (\mathbb{R}_+ \to \mathbb{R}) \to (\mathbb{R}_+ \to \mathbb{R}) \\
	& y= I_i(x)
\end{aligned}
\end{gathered}
\]
where $i\equiv $ initial value of integration and $x,y$ are cont.time signals, i.e. $x,y \in C^1(\mathbb{R})$.

e.g. If $i=0$, $\forall \, t \in \mathbb{R}_+$, $x(t) = 1$, then
\[
y(t) = i+ \int_0^t x(\tau) d\tau = t
\]

Example: Counter $C_i$:
\[
C_i: (\mathbb{R}_+ \to \text{Set})^P \to (\mathbb{R}_+ \to \lbrace \empty \rbrace \cup \mathbb{Z} )
\]
with $\text{Set } \in \textbf{Set}$ so $\text{Set}$ is a \emph{particular} set., \\
$P\in \textbf{Set}$, \\
e.g. $P\equiv $ set of input ports, e.g. $P=\lbrace \text{up}, \text{down} \rbrace$.

$C_i:(\mathbb{R}_+ \to \lbrace \text{ absent, present } \rbrace )^P \to (\mathbb{R}_+ \to \lbrace \text{ absent } \rbrace \cup \mathbb{Z})$

For particular time $t$,

suppose actor has input ports $P=\lbrace p_1, \dots p_N \rbrace$, \\
$\forall \, p \in P$, set $V_P \in \mathbf{Set}$ denotes values that maybe received on port $p$, when input is present $\equiv$ \textbf{type} of port $p$.

EY: At a reaction,
\[
\begin{gathered}
	y: P \to V_P \cup \lbrace \emptyset \rbrace \\
	y: p \mapsto y(p) \text{ or } \emptyset
\end{gathered}
\]

Assume absent at times $t$ where reaction doesn't occur.
e.g.
$Q = \lbrace \text{ count } \rbrace$

In summary, a \textbf{discrete signal} $x_i$, $\forall \, i = 1,2, \dots |P|$, $P\in \textbf{Sets}$, set of all input ports,  is
\begin{equation}
\boxed{
\begin{gathered}\begin{aligned}
& x_i: \mathbb{R} \to \lbrace \emptyset \rbrace \cup X_i \\
& x_i:t \mapsto x_i(t)
\end{aligned} X\in \textbf{Sets} \text{ or } X = \mathbb{R}
\end{gathered}
}
\end{equation}
Let
\begin{equation}
\boxed{ T_{x_i} = \lbrace t \in \mathbb{R} | x_i(t) \neq \emptyset \rbrace }
\end{equation}
be the set of times when there's signal.

Let
\begin{equation}
\boxed{ \begin{aligned}
& p_i : \mathbb{R} \to \mathbb{Z}_2 = \lbrace \text{ present , abset } \rbrace \\
& p_i: t\mapsto p_i(t)
\end{aligned} }
\end{equation}
$\forall \, i = 1,2, \dots |P | $ tell us if a signal is present or absent.

Similarly for the output $y_j$:

\section{Finite-State Machines}

cf. 3.3 Finite-State Machines of Lee and Seshia (2017) \cite{LeSe2017}

\textbf{Finite-state} machine; assume states $\equiv S \in \textbf{FiniteSets}$.

\subsection{Transitions; edges, directed edges }

cf. 3.3.1 Transitions of Lee and Seshia (2017) \cite{LeSe2017}

\textbf{transition} - directed edge.  $\forall \, $ transition, label with "guard/action", i.e. label with a "guard" and an "action"

\textbf{guard}- determines whether transition maybe taken on reaction \\
\textbf{action} - specfies \emph{what outputs are produced on each reaction} \\

guard is a \textbf{predicate} (boolean-valued) s.t. true when transition should be taken  \\
e.g. example 3.4., FSM model for garage counter \\
states $\equiv S = \lbrace 0 ,1, \dots M \rbrace$

\begin{tabular}{l l c}
	$\begin{aligned}
	\neg \\
	\sim \\
	!
	\end{aligned}$ & negation, not &  $\neg A$ true, iff $A$ false  \\ \hline
	$\begin{aligned}
	\wedge \\ \& \
	\end{aligned}$ & logical conjunction, and & if $A$ and $B$ both true, $T$, $A \wedge B$ true.  Else, it's false.    \\ \hline
	$\begin{aligned}
	& \vee \\ & + \\ & \parallel \end{aligned}$ & logical (inclusion) disjunction, or & if $A$ or $B$ (or both) true, $A \vee B$ false, if both false, $A\vee B$ false
	 \\ \hline
	$\begin{aligned}
	& \oplus \\
	& \veebar \end{aligned}$ & exclusive disjunction, xor & $A \oplus B$ true when either $A$ or $B$, but not both are true.
\end{tabular}

\begin{tabular}{c c c c c c}
\hline
$A$ & $B$ & $\neg A$ & $ A\wedge B$ & $A \vee B$ & $A \oplus B$ \\ \hline
T & T & F & T & T & F  \\
T & F & F & F & T & T \\
F & T & T & F & T & T \\
F & F & T & F & F & F
\end{tabular}

\textbf{action} (in reference to Lee, Seshia (2011) \cite{LeSe2017}), on a transition, action specifies resulting valuation on output ports when transition is taken.

"When assigning a value to an output port, we use notation name $:=$ value to distinguish assignment from a predicate, which would be written name $=$ value.

Let \\
ports $P_{\text{in}} = \lbrace p_1^{(\text{in})}, \dots p_{N_{\text{in}}}^{(\text{in})}  \rbrace$ \\
output ports $P_{\text{out}} = \lbrace p_1^{(\text{out})}, \dots p_{N_{\text{out}}}^{(\text{out})}  \rbrace$ \\

At a reaction,
\begin{equation}
\begin{gathered}
\begin{aligned}
& i: P_{\text{in}} \to V_{P_{\text{in}}} \cup \lbrace \emptyset \rbrace \\
& i: p \mapsto i(p)
\end{aligned} \\
\begin{aligned}
& o: P_{\text{out}} \to V_{P_{\text{out}}} \cup \lbrace \emptyset \rbrace \\
& o: p \mapsto o(p)
\end{aligned}
\end{gathered}
\end{equation}
Given
\[
i(p_1^{(\text{in})}, \dots p_{N_{\text{in}}}^{(\text{in}) } )  \in ( V_{P_{\text{in}}} \cup \lbrace \emptyset \rbrace )^{P_{\text{in}} } \times S
\]
where $S \equiv $ states of the system,  \\

\textbf{guard}, $g_{\text{guard}}$,
\begin{equation}
\begin{aligned}
& g_{\text{guard}} : ( V_{P_{\text{in}}} \cup \lbrace \emptyset \rbrace )^{P_{\text{in}} } \times S \to \mathbb{Z}_2 = \lbrace \text{true}, \text{false} \rbrace \\
& g_{\text{guard}}( p_1^{(\text{in})}, \dots p_{N_{\text{in}}}^{(\text{in})} ) \in \lbrace \text{true}, \text{false} \rbrace
\end{aligned}
\end{equation}

\textbf{action}  $a$,
\begin{equation}
\begin{aligned}
& a : g_{\text{guard}}, e \mapsto o( p_1^{(\text{out})}, \dots p_{N_{\text{out}}}^{(\text{out})} )
\end{aligned}
\end{equation}
where  $e\in E$, $E \equiv $ set of all possible transitions.

cf. Sec. 3.3.3. Update Functions of Lee, Seshia (2011)  \cite{LeSe2017}

(States, Inputs, Outputs, update, initial State) $\equiv  (S,I, O, u,s_0)$,

where $u: S\times I \to S \times O$, and update $\equiv $ transition function (synonymous)

Let sequence of reactions be indexed by $n \in \mathbb{N}$.

Then
\[
\begin{aligned}
& s:\mathbb{N} \to S \\
& s(0) = s_0
\end{aligned}
\]
$x:\mathbb{N} \to I$, $x\equiv $ input valuations \\
$y:\mathbb{N} \to O$, $y\equiv $ output valuation.

state machine dynamics
\[
(s(n+1),y(n)) = u(s(n),x(n))
\]

\[
\begin{gathered}
	\begin{aligned}
	& i: P \to \bigcup_{i=1}^{ P^{(\text{in})}_{N_{\text{in}} }} V_{ P_i^{(\text{in}) } } \bigcup \lbrace \emptyset \rbrace \\
	& i : p \mapsto i(p) \in V_P \cup \lbrace \emptyset \rbrace \equiv \text{ value of part $p$ }
\end{aligned} \\
i\in I \subset \text{Hom}(\textbf{Set}, \textbf{Set})
\end{gathered}
\]

\subsubsection{Graph $(V,E)$ viewpoint}

Let $V=S \equiv $ states if $S\in \textbf{FiniteSet}$ (finite number of states), then for convenience, we can label, 1-to-1, $\forall \, s \in S$ with $\alpha = 0,1,\dots |S|$.

$\forall \, s_{\alpha} S$, \\
$\exists \, \lbrace p_i^{(\alpha)} \rbrace_{i=1}^{ P^{(\alpha)} }$ input ports, $\lbrace ( x_i^{(\alpha)}, p_i^{(\alpha)} \rbrace | x_i^{(\alpha)} \in (\mathbb{R} \to X_i^{(\alpha)} ) , p_i^{(\alpha)} \in \lbrace \text{ present } , \emptyset \rbrace \rbrace_{i=1}^{ |P^{(\alpha)}| }$    \\
$\exists \, \lbrace q_j^{(\alpha)} \rbrace_{j=1}^{ Q^{(\alpha)} }$ output ports, $\lbrace ( y_j^{(\alpha)}, q_j^{(\alpha)} \rbrace | y_j^{(\alpha)} \in Y_j^{(\alpha)}  , q_j^{(\alpha)} \in \lbrace \text{ present } , \emptyset \rbrace \rbrace_{j=1}^{ |Q^{(\alpha)}| }$

Consider $e\equiv (s_{\alpha},s_{\beta}) \in E$, $s_{\alpha}$ the initial vertex (state), $s_{\beta}$ terminal vertex (state).

$\forall \, (s_{\alpha}, s_{\beta})$, associate a guard $g: (x_1^{(\alpha)} \dots x_{P^{(\alpha)}}^{(\alpha)} ) \mapsto \lbrace T,F \rbrace$ and $y^{(\alpha)} = (y_1^{(\alpha)} \dots y_{Q^{(\alpha)}}^{(\alpha)}$

If $\forall \, (x_1^{( \alpha)} \dots x_{P^{(\alpha)}}^{(\alpha)})$, $\exists \, $ at most 1 $(s_{\alpha}, s_{\beta} )$, system is deterministic.

\subsection{Nondeterministic Finite-State Machine (FSM); Formal Model}

cf. 3.5.1. Formal Model of Lee and Seshia (2017) \cite{LeSe2017}

\textbf{nondeterministic FSM}
\begin{equation}
	(\text{States, Inputs, Outputs, possibleUpdates, initialStates}) \equiv (S,I,O,u,I_0)  \equiv (S,X,Y,u,I_0)
\end{equation}
with update relation (or \textbf{transition relation}) $u$
\[
u:S\times I \to 2^{S\times O}
\]
where powerset of $S\times O$, $2^{S\times O}$, is the set of possible (next state, output valuation) pairs, \\
$I_0 = $ set of initial states.

Let $\Sigma = \lbrace \text{ apart, together } \rbrace = \lbrace \text{ apa, tog } \rbrace$

\[
I_0 \ni \lbrace \begin{gathered}
x_1(0) := x_{1,0} \\
x_2(0) := x_{2,0} \\
\dot{x}_1(0) = \dot{x}_2(0) = 0 \end{gathered}
\]
\[
(\text{apa},\text{tog}) , g= \text{ if } x_1(t) = x_2(t)
\]
action $\equiv$ set action, use momentum conservation: $m_1 \dot{x}_1 + m_2 \dot{x}_2 = (m_1+m_2) \dot{x} $ or
\[
\dot{x}(t) = \frac{m_1 \dot{x}_1(t) + m_2\dot{x}_2(t) }{ m_1 + m_2 }
\]

Let $s \equiv F_{\text{stick}} \in \mathbb{R}_+$, represent the stickiness of the 2 masses.
\[
F_2-F_1 = -k_2 (x(t) - p_2) + k_1(x(t) - p_1) > F_{\text{stick}} \text{ or } -(k_2- k_1) x(t) + k_2 p_2 - k_1 p_1 > F_{\text{stick}}
\]

Example 4.7 of Lee and Seshia (2017) \cite{LeSe2017}

Consider a bouncing ball.

Inelastic collision with ground.

\[
\Sigma = \lbrace \text{ free } \rbrace
\]
$(\text{free},\text{free}) $ if $y(t) = 0 $, $\dot{y}(t) = -a\dot{y}(t)$, constant $a$, $0<a<1$.

Cont. state varialbes of free mode (free state)
\[
\begin{gathered}
	s(t) \equiv X(t)  = \left[ \begin{matrix} y(t) \\
\dot{y}(t) \end{matrix} \right] \text{ so } \dot{X}(t) = \left[ \begin{matrix} \dot{y}(t) \\ \ddot{y}(t) \end{matrix} \right] = \left[ \begin{matrix} \dot{y}(t) \\ -g \end{matrix} \right] = \left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] X + \left[ \begin{matrix} 0 \\ -g \end{matrix} \right]
\end{gathered}
\]
\[
\begin{aligned}
	& \ddot{y}(t) = -g \\
	& \dot{y}(t) = -gt \\
	& y(t) = \frac{-1}{2} gt^2 + h_0
\end{aligned}
\]

\part{Screws, Bolts, Fasteners (through Shigley's Mechanical Engineering Design)}

cf. Budynas and Nisbett (2014) \cite{BuNi2014}.

\section{Screws, Fasteners, Design of Nonpermanent Joints}

cf. Ch. 8 "Screws, Fasterners, and the Design of Nonpermanent Joints", Budynas and Nisbett (2014) \cite{BuNi2014}.

cf. Section 8-1, "Thread Standards and Definitions", pp. 410, Budynas and Nisbett (2014) \cite{BuNi2014}.

pitch $\equiv $ distance between adjacent thread forms, measured parallel to thread axis.

$N :=$ number of thread forms per unit length (per inch U.S.); pitch $=1/N$

$d\equiv $ major diameter $:=$ largest diameter of screw thread.

$d_r \equiv $ minor (or root) diameter $=$ smallest diameter of screw thread.

$l \equiv $ lead $=$ distance nut moves parallel to screw axis when nut given $1$ turn for \\
single thread $=l=p$ ($p$ is \emph{pitch}) \\
multiple-threaded $- 2 $ or more threads out beside each other (imagine 2 or more strings wound side by side around a pencil).  \\
double-threaded $\to l = 2p$  \\
triple-threaded $\to l=3p$

Imagine square-threaded power screw, single thread, unrolled or developed, for exactly single turn.

$d_m:= $ mean thread diameter circle.  $\lambda := $ lead angle of thread.

Consider sum of all forces \textbf{on} the single screw thread.

To raise load, force $P_R$ acts to the right \textbf{on} single screw thread.   \\
$\mathbf{P}_R = (P_R,0)$ ; \, $P_R >0$ \\
$\mathbf{N}= N (-\sin{\lambda}, \cos{\lambda} )$  \\
friction force $\mathbf{fN} \, \text{ (Shigley) } \equiv - \mu N(\cos{\lambda}, \sin{\lambda})$ \\
$\mathbf{F} =$ sum of all axial forces acting upon normal thread area $=(0,-F)$.

\begin{equation}
	\mathbf{P}_R + \mathbf{N} + \mathbf{fN} + \mathbf{F} = (P_R-N\sin{\lambda} , - \mu N \cos{\lambda}, N\cos{\lambda} - \mu N\sin{\lambda}, -F)
\end{equation}
Surely $(\mathbf{P}_R + \mathbf{N} + \mathbf{fN} + \mathbf{F})_x =0$, so that $P_R = N( \sin{\lambda} + \mu \cos{\lambda})$.

We can imagine $m_{\text{thread}} a_{\text{thread}} = N (\cos{\lambda} - \mu \sin{\lambda}) -F = \frac{P_R}{ \sin{\lambda} + \mu \cos{\lambda} }(\cos{\lambda} - \mu\sin{\lambda} ) -F \neq 0 $

Lowering load, $P_L$ acts to the left \textbf{on} simple screw thread.

$\mathbf{P}_L = (-P_L ,0)$; $\mathbf{fN} \equiv \mu N(\cos{\lambda}, \sin{\lambda})$.

\begin{equation}
\sum_{\text{ on single screw thread} } \mathbf{F}_i = \mathbf{P}_L + \mathbf{N} + \mathbf{fN} + \mathbf{F} = (-P_L - N\sin{\lambda} + \mu N \cos{\lambda}, N\cos{\lambda} + \mu N \sin{\lambda} - F)
\end{equation}
Surely $(\mathbf{P}_L + \mathbf{N} + \mathbf{fN} + \mathbf{F})_x =0$ so that $P_L = N(\mu \cos{\lambda} - \sin{\lambda} )$.

We can imagine
\[
m_{\text{thread}} a_{\text{thread},y} = F_{\text{net},y} = N(\cos{\lambda} + \mu \sin{\lambda}) - F = \frac{ P_L}{ \mu \cos{\lambda} - \sin{\lambda} } (\cos{\lambda} +\mu \sin{\lambda} ) - F
\]

At equilibrium of forces,

For raising load,
\[
\begin{gathered}
	\mathbf{\tau} = \mathbf{r} \times \mathbf{P} = \mathbf{r} \times \mathbf{P}_R = \mathbf{r} \times \left( F \left( \frac{ \sin{ \lambda } + \mu \cos{\lambda} }{ \cos{ \lambda} - \mu \sin{\lambda}  } \right) \right) = \mathbf{r} \times \left( F \left( \frac{ \tan{ \lambda } + \mu  }{ 1 - \mu \tan{\lambda}  } \right) \right) = \\
= \mathbf{r} \times F \left( \frac{ \frac{ l}{ \pi d_m } + \mu }{ 1 - \mu \frac{l}{\mu d_m } } \right) = \mathbf{r} \times F \left( \frac{ l + \mu d_m \pi }{ \mu d_m - \mu l } \right)
\end{gathered}
\]
For lowering load,
\[
\begin{gathered}
	\mathbf{\tau} = \mathbf{r} \times \mathbf{P} = \mathbf{r} \times \mathbf{P}_L = \mathbf{r} \times \left( F \left( \frac{  \mu \cos{ \lambda } -  \sin{\lambda} }{ \cos{ \lambda} + \mu \sin{\lambda}  } \right) \right) = \mathbf{r} \times \left( F \left(   \frac{ \mu -  \tan{ \lambda }   }{ 1 + \mu \tan{\lambda}  } \right) \right) = \\
= \mathbf{r} \times F \left( \frac{ \mu -  \frac{ l}{ \pi d_m }  }{ 1 + \mu \frac{l}{\mu d_m } } \right) = \mathbf{r} \times F \left( \frac{ \pi \mu d_m - l }{ \pi d_m + \mu l } \right)
\end{gathered}
\]

Load lowers itself, by causing screw to spin without external effort.
\begin{equation}
	\mathbf{\tau}_L  = \mathbf{r} \times F \left( \frac{ \pi d_m \mu - l }{ \pi d_m + \mu l } \right)  \leq 0 \text{ or } \mu \leq \tan{\lambda}
\end{equation}
If $\tau_L >0$, screw is \textbf{self-locking}; in this case, $\mu > \tan{\lambda}$

Efficiency expression (for power screws)
\begin{equation}
\begin{gathered}
	\tau_0(\mu = 0)  = \frac{d_m}{2} F\left( \frac{l }{ \pi d_m} \right)  = \frac{Fl}{2\pi }
\end{gathered}
\end{equation}
Since thread friction has been eliminated, torque required only to raise the load.

Then define efficiency $\epsilon_{\text{load}}$ as
\begin{equation}
\begin{gathered}
	\epsilon_{\text{load}} = \frac{ \tau_0 }{ \tau_R} = \frac{Fl}{2\pi \tau_R} = \frac{Fl}{ 2\pi  \frac{d_m}{2} F \left( \frac{ l + \mu d_m \pi }{ \pi d_m - \mu l } \right) } = \frac{l}{ \pi d_m} \left( \frac{ \pi d_m - \mu l }{ l + \mu d_m \pi } \right) = \frac{1 - \frac{ \mu l }{ \pi d_m } }{ 1 + \frac{ \mu d_m \pi }{ l }  }
\end{gathered}
\end{equation}

\section{Ball Screws; ball screw}

cf. \href{https://www.thk.com/sites/default/files/documents/uk_pdf/product/general/a/ee_A15.pdf}{THK Ball Screw, General Catalog}

THK provided an excellent overview of the equations/relations (and thus the physics) behind ball screws (THK, \href{https://www.thk.com/sites/default/files/documents/uk_pdf/product/general/a/ee_A15.pdf}{THK Ball Screw, General Catalog}).

I will include THK's notation and my notation, side by side.

Let \\
$\beta \equiv \lambda \equiv$ lead angle  \\
$d_p \equiv $ Ball center-to-center diameter \\
$\text{Ph} \equiv l \equiv $ feed screw lead (feed screw is that long piece, the screw; lead is distance but moves in 1 turn) (mm) \\
\begin{equation}
	\tan{\beta} = \frac{l}{\pi d_p}
\end{equation}

\subsection{Driving torque required to gain thrust}

Let
$T \equiv \tau := $ driving torque ($N \cdot mm$) \\
$Fa \equiv f_a := $ frictional resistance on guide surface ($N$) \\
$Fa= \mu \times mg \equiv \mu Mg$ \\
$\mu := $ frictional coefficient of the guide surface  \\
$g := $ gravitational acceleration ($9.8 \, m/s^2$)  \\
$gm \equiv M =$ mass of trasnferred object (kg).  \\
$\eta_1 :=$ positive efficiency of feed screw (see Fig. 1 on A-682 (rotational to linear), from Fig. 1, $0.003 \leq \mu \leq 0.02$
\begin{equation}
\tau =  \frac{ f_a l }{ 2\pi \eta_1 }
\end{equation}
$\eta_1$ efficiency is either rotational to linear energy, or force.  I believe it's a ratio between magnitude of forces.

cf. \href{http://web.mit.edu/2.75/fundamentals/FUNdaMENTALs\%20Book\%20pdf/FUNdaMENTALs\%20Topic\%206.PDF}{FUNdaMENTALS of Design, Topic 6, Power Transmission Elements II, by Alexander Slocum, 2008}.

cf. \href{https://www.cs.cmu.edu/~rapidproto/mechanisms/chpt2.html}{Introduction to Mechanisms; 2 Mechanisms and Simple Machines}.

Taking a look at \emph{2 Mechanisms and Simple Machines}, Sec. \emph{2.1.1 Screw Jack}, for Introduction to Mechanisms by Yi Zhang, with Susan Finger and Stephannie Behrens, for \href{https://www.cs.cmu.edu/~rapidproto/home.html}{39-245} Rapid Design through Virtual and Physical Prototyping at Carnegie Mellon University, let \\
$\mathbf{W} :=$ heavy weight to raise $= W(-\mathbf{e}_z)$, \\
$\mathbf{F}:=$ much smaller force applied at handle, \\
$p \equiv l :=$ pitch of screw (distance advanced in 1 complete turn) $=l$, then \\
neglecting friction, \\
force multipled by distance through which it moves in one complete turn $=$ weight lifted times distance through which it's lifted in same time,

and so by energy conservation
\[
F(2\pi R) = Wp \equiv Wl \text{ or } F = \frac{Wp }{2\pi R}
\]

\subsection{Kinematics}

\subsubsection{Helix Geometry/Mathematics}

Consider a circular helix of radius $a$, slope $\frac{b}{a}$ (or pitch $2\pi b$), described by parametrization
\[
\begin{aligned}
	& x(t) = a \cos{(t)} \in \mathbb{R} \\
		& y(t) = a \sin{(t)} \in \mathbb{R} \\
	& z(t) = b(t) \in \mathbb{R}
\end{aligned}
\]
Clearly the radius is $a$ as $x^2 + y^2 = a^2 = r^2$, and the rotation about the (symmetric) axis of rotation (in $z$ direction) is $\theta \equiv \theta(t) = 0$, in \emph{radians}.

For rotation $\theta$, we travel in the $z$-direction by $z(t) =b\theta$.  So for 1 full rotation (revolution), $\theta = 2\pi$, we have in the $z$-axis by $z(t=2\pi) =2\pi b$.

In engineering (in general), in particular for screws (lead screws or ball screws), we have the \emph{lead} $l$: \\
$l:= \text{thread lead, i.e. linear distance either screw or nut moves, i.e. (linear) translation length (for 1 full rotation (revolution)) }$.  Clearly
\begin{equation}
l = 2\pi b
\end{equation}

So given input torque $\tau$, efficiency for rotational motion to linear motion (i.e. linear output$/$ rotational input), $\eta_1$,
\begin{equation}
\begin{gathered}
\eta_1 \tau \theta = Fz \text{ or } \eta_1\tau \theta = F(b\theta) = F \left( \frac{l}{2\pi \theta } \right) \text{ or } \boxed{ F = \frac{2\pi \eta \tau }{l } }
\end{gathered}
\end{equation}

Other sources I looked at:
Vahid-Araghi and Golnaraghi (2010) \cite{VaGo2010}

\section{State Machine; Finite State Machine and (finite) graphs; graphs}

cf. Stallings (1983) \cite{Stal1983}

\subsection{Paths}

cf. pp. 553, Section 2. Paths in Stallings (1983) \cite{Stal1983}; pp. 14, Subsection 2.1 Graphs in Serre (1980) \cite{Serr1980}
\begin{definition}[path]
path $p$ in $\Gamma$, of length $n=|p|$, with initial vertex $u$ and terminal vertex $v$, $n$-tuple of edges of $\Gamma$, $p=e_1 e_2 \dots e_n$, s.t. $\forall \, i = 1\dots n-1$, then $\tau(e_i) = i(e_{i+1})$ s.t. $u=i(e_1)$, $v=\tau(e_n)$

\end{definition}

For $n=0$, given any vertex $v$, $\exists \, !$ unique path $\Lambda_v$, $| \Lambda_v| =0$, s.t. $u=v$, i.e. \\
standard arc of length $n$, $\Lambda_n$, described as interval $[0,n]$ subdivided at integer points, path $p$ is map of graphs $p:A_n \to \Gamma$ s.t. $p(0)=u$, $p(n)=v$.

\part{Avionics; Embedded Systems; Embedded C}

\textbf{Introduction to Embedded Systems: Interfacing to the Freescale 9S12} from Valvano was meant to be an introduction (Valvano (2010) \cite{Valv2010}).  Valvano in there suggests \textbf{Embedded Microcomputer Systems} (Valvano (2011) \cite{Valv2011}) for an advanced treatment.

\end{multicols*}

\begin{thebibliography}{9}

\bibitem{Hugh2000}
Scott B. Hughes.  \textbf{Magnetic braking: Finding the effective length over which the eddy currents form}.
\href{https://drive.google.com/file/d/0Bwo3W0v5P04LX29XT2NFeVY0a1E/view}{Magnetic braking: Finding the effective length over which the eddy currents form}

\bibitem{Smyt1968}
William R. Smythe, \textbf{Static and Dynamic Electricity}.  3rd ed. (McGraw-Hill, New York, 1968).

\bibitem{Jack1998}
J.D. Jackson.  \textbf{Classical Electrodynamics} Third Edition.  Wiley.  1998.   ISBN-13: 978-0471309321

\bibitem{Herm1974}
Robert Hermann.  \textbf{Linear Systems Theory \& Introductory Algebraic Geometry (Interdisciplinary Mathematics Series)}.  Math Science Pr (June 1974)
ISBN-13: 978-0915692071

\bibitem{BuNi2014}
Richard Budynas, Keith Nisbett.  \textbf{Shigley's Mechanical Engineering Design} (McGraw-Hill Series in Mechanical Engineering) 10th Edition.  McGraw-Hill Series in Mechanical Engineering.  ISBN-13: 978-0073398204

EY : 20170606 -  I was only able to obtain a copy of the 9th edition to use.  If you'd like to help us keep up to date with the 10th edition or to help the rLoop in general, please donate to the \href{rLoop}{www.rLoop.org}!

\bibitem{VaGo2010}
Orang Vahid-Araghi, Farid Golnaraghi.  \textbf{Friction-Induced Vibration in Lead-Screw Drives}.  Springer.  2010.  e-ISBN 978-1-4419-1752-2

\bibitem{Stal1983}
John R. Stallings.  "Topology of finite graphs."  \textbf{Invent. math. 71}, 551-565 (1983)  \textbf{Inventiones mathematicae}.  Springer-Verlag 1983.

\bibitem{Serr1980}

Trees (Springer Monographs in Mathematics) 1st ed. 1980. Corr. 2nd printing 2002 Edition
by Jean-Pierre Serre (Author), J. Stilwell (Translator)

\bibitem{Valv2010}
Jonathan W. Valvano.  \textbf{Introduction to Embedded Systems: Interfacing to the Freescale 9S12}.  1st. Ed. 2010.  Cengage Learning.  USA.

\bibitem{Valv2011}
Jonathan W. Valvano.  \textbf{Embedded Microcomputer Systems: Real Time Interfacing}.  3rd. Ed. 2011.  Cengage Learning.  USA.

\bibitem{LeSe2017}
%Edward Ashford Lee and Sanjit Arunkumar Seshia.  \textbf{Introduction to Embedded Systems} \textbf{
Edward A. Lee and Sanjit A. Seshia, \textbf{\href{http://leeseshia.org/index.html}{Introduction to Embedded Systems}}, A Cyber-Physical Systems Approach, Second Edition, MIT Press, ISBN 978-0-262-53381-2, 2017.

\end{thebibliography}

\end{document}

redpajama_set_name: RedPajamaGithub


Whilst i was a way the gaming news did not stop so this leads me to do a catch up post.
THQ may still be taken by the Clearlake Capital Group according to an announcement by the publisher now the doors are open for other companies to bid for THQ as a whole or is assets. This offer for $60 million is the first in hopefully many bid for this company.
All gamers pretty much mess them selves at all the gaming awesomeness that happens at E3 and to make the wait worse "Major Nelson" posted a countdown on his blog.
E3 does come around every year at the same time but with chatter around the circuit of Microsoft new console if this counter counts down to anything other then the announcement of a next gen console lots of people are going to be disappointed and annoyed.
Finally Bioshock Infinite is really going to happen and is going to be released on March 26, there's no denying that when this game comes out it will be awesome.
Now Feast your eyes on gaming glory.
reports of the mythical "steam box" a HDTV optimizes PC from Valve was announced by valve and the mini PC maker Xi3 reviled a device code named "Piston" at CES in Las Vegas mot much is none on the device other then it's small and can fit in your hand and is being developed between two companies specifically to work with Steam's Big Picture Mode.
"Today marks the beginning of a new era for Xi3. This new development stage product will allow users to take full advantage of their large high-definition TV displays for an amazing computer game experience.
another post from Polygon has a few more details on the device.
no release date for the device but the device the piston is based off the X7A mini- PC which has a $999 price tag so we may have to fork out quite a bit of cash.


U.S. FrontLine | CK Marketing Solutions Inc. U.S. FrontLine – CK Marketing Solutions Inc.
USFL is the most widely circulated free Japanese paper in the United States. In addition to a complete renewal of the website, we reinvented their logo and introduced social media tools to expand USFL's online presence. The website was created with usability as the priority, while updating the look and feel without changing USFL's brand image and perception.


Kaylon Layne - Page 2 of 10 - Messy, imperfectly beautiful motherhood.
FREE Mother's Day Printable Your Mother will LOVE!!!
I've decided to start bringing my thrift finds to you. Get excited! You won't want to miss these thrift finds. You can join my new thrift finds group HERE.


text:
Foreign direct investment to Pakistan slows down to an eight-month low
Written by Hamza Zakir Aug 18 · 1 min read>
The net foreign direct investment (FDI) being directed to various sectors of Pakistan's economy, including energy, telecommunications and banking, has slowed down to an eight-month low. As of July 2021 – the first month of the current fiscal year 2021-22 – the figure stands at $89.9 million.
As reported by The Express Tribune, this drop in FDI can be attributed in part to a temporary lull in the second phase of the China-Pakistan Economic Corridor (CPEC) and Belt Road Initiative that was observed this month.
The gross inflows of FDI amounted to $176.3 million in July 2021 which was partially higher compared to $168.7 million recorded in the same month last year. However, the gross outflows more than doubled to $86.4 million in the month under review compared to $40 million in the corresponding month last year. According to the State Bank of Pakistan's (SBP) data that was mentioned in the report, the net inflows dropped to an eight-month low at $89.9 million.
In terms of FDI specifics in July 2021, Singapore has been the single largest investor in Pakistan over the course of the month. It invested $16.4 million in the month compared to net $2.3 million in the same month last year, according to the SBP data.
United States, meanwhile, emerged as the second largest investor with net $14.9 million in the month under review compared to net $11.5 million in the corresponding month last year.
Hong Kong invested $12.3 million in the month compared to $4.7 million in the same month last year. China, which was the leading investor with a net investment of $44.1 million in July 2020, poured $6.6 million in July 2021 after a significant outflow of $15.2 million in the month.
Speaking sector-wise, the power sector has amassed the single largest investment in July 2021 ($41.9 million), followed by financial business ($27 million), and the oil and gas exploration sector ($22.4 million).
Foreign Direct Investment foreign investment pakistan Pakistan pakistan FDI Pakistan news
Written by Hamza Zakir
Platonist. Humanist. Unusually edgy sometimes. Profile
Up Next: The world's largest chipmaker has overtaken Tencent to become Asia's most valuable company

redpajama_set_name: RedPajamaCommonCrawl


Essentially the idea is that based on some seed data, some complicated sums are done to give a location.
People get to that location for a meetup.
A map tool is available which does the sums for you. You set the date, click your area and it gives you a location.
Due to problems with the seed data (US stock market) and time zones a new rule has been introduced today for people east of 30 degrees west. This is taken care of automatically by the map tool. There are several pieces of code for implementing this - though most have yet to be updated to reflect the 30W rule.
The idea is that the seed data is processed using an algorithm called md5. This algorithm produces a 'hash' of the data. it is difficult to find alternate data which produces the same hash. A small change in the data produces a big change in the hash.
The idea of a hash is a way of producing a 'fingerprint' of a file. I.e. I could send you a file, but how would you know it hadn't been tampered with? Well, I could phone you, you could recognise me and I could read you the hash of that file (which you can then generate and check).
A hash can also be used as a zero knowledge proof. I.e. I wanted to prove to you that I had discovered some fact. I might not want you to know the fact (yet). For example, I might know the first line of the 'Times' editorial for next saturday. I could generate a hash of that line and give it to you - when the paper is published that information can be checked.
In this case, the md5 algorithm is used to give a reasonable pseudo-randomisation of one number into another. It's just a bit of fun.
I've not gone to a geohash event myself - but I like the concept.


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A closely-spaced motorcade of double-size Van Hool buses now trundles up and down Telegraph Avenue at all hours. I regularly observe the middle and Berkeley end of Bus Line No. 1 doing business on Telegraph and occasionally getting paint at Kelley-Moore at Telegraph and 42nd. Morning, noon, and night, I see an average range of six to 16 passengers occupying these cavernous vehicles. Telegraph Avenue in Oakland is in poor condition already. The greatest damage to city streets is done by buses, according to the paving engineer hired by City of Berkeley that I spoke to when North Shattuck Avenue was last repaved. And we all see how little budget there seems to be for street maintenance and repair in Berkeley and Oakland. Considering how much fuel is being consumed to deploy so many heavy buses to move so few passengers, BRT deserves a good deal more before-the-fact public disclosure and scrutiny than it has received.
Listening to public input is not an AC Transit forté. The public agency is determinedly deaf to shouts of dismay from citizens who don't want Telegraph to become any harder to negotiate than it already is. As much so as when both the riding public and bus drivers strongly criticized the elimination of bus lines in Oakland, and the use of more Van Hool buses. Top management "stayed the course" by moving ahead with reduction of services to areas that had no other transit service, and what has been called a sweetheart contract with the Belgian bus manufacturer. Today, as it touts its Telegraph Avenue project, AC Transit is removing bus stops elsewhere.
Fran Haselsteiner wants us to know that BRT has been "fully vetted" by the public. But the many voices raised in opposition to this plan today suggest that "fully vetted" means something less than "exhibited to all in full public view." Much like other schemes that affect hundreds of thousands of people—the aborted land grab at the Ashby BART and the stalled land grab at North Shattuck Avenue come to mind—BRT has suddenly emerged full-blown after the tritest gesture to public process. Now Fran urges us to sample the bliss from assurances by architects of this scheme that 9,300 new riders "may" accrue from its implementation. This bit of prefabricated spin—quintessetial pie-in-the-sky—misses the mark among level-headed Berkeley and Oakland citizens who oppose BRT.
Here's what bugs me about BRT: 1) It is grossly wasteful and destructive to our infrastructure, as described above. 2) AC Transit does not own Telegraph Avenue and has no right to push everyone who uses the street into one lane. Neither does Berkeley or Oakland. Telegraph was constructed and is maintained (!) with citizens' money; the citizens own it. 2) Depriving the citizenry of full use of their street flies in the face of democratic principle and process. I choose to ride my bicycle for 80 percent of my transit needs. As a property manager I drive my car on Telegraph to obtain paint and supplies in Oakland. I don't need AC Transit or Fran to "discourage" me by making the experience even more unpleasant and time-consuming (and thereby more gas-consuming) than it already is.
If AC Transit is after more riders, let's start by admitting that pre-BRT capacity is more than adequate for the foreseeable future. How about asking for state and federal transit money to offer low-cost or free bus service instead of clogging up our main intercity artery? There is already BART between the two cities. Restore bus service to some neighborhoods. That will increase ridership for sure, not 'maybe'.
BRT looks like a high-visibility bid for state and federal funds. The more insistently its spokespeople attempt to refute the obvious disadvantages of this plan, the more I wonder how much money is involved and where it is intended to go.
Glen Kohler lives in Berkeley's South Campus area.


You can learn the basics of scriptwriting and video production, including how to write and produce a script on issues you and your friends may face (teen pregnancy, violence, abuse, etc.). The final script for production will be decided by the participants. Once the scripts are written, you choose how you want to be involved in the video production process— video crew (behind the camera) or the cast (in front of the camera). You will then produce Public Service Announcements (PSAs) that can be viewed on Rochester Channel 15 and Time Warner Channel 18.
In our 12-15 week program, you will have the opportunity to express your feelings, thoughts and life experiences through music. Professional instructors with real-world music production experience will teach you how to record, mix, edit and produce digital music. You will learn the principles of sound recording and music production. Key concepts include songwriting and musical arrangement, in addition to learning how to create/modify sounds and program drum beats and patterns. This is also a way for you to gain a real understanding of the music industry with workshops on contracts, copyrights and radio.
Our Theatre/Drama Program is designed to educate and train teens who may be interested in the field of theater. You will also gain an understanding of the theater industry, have the opportunity to see a professional production and tour a professional theatre. There will also be opportunities to rehearse a script and perform a show that will address the issues of teen pregnancy, STI, HIV, abuse, etc.
Our Photography Program is designed to teach entry level photography development and provide an introduction to graphic arts and design. Professional cameras are used in this group. In Control collaborates with Icon Photography to oversee educational workshops, field trips, and all onsite photography at local events, fairs, festivals and at all In Control events.
The Color Guard is an auxiliary group that works in the use of dance, movement as well as the use of flags, rifles and other performance equipment to create and tell a story. Classical music is usually used as a sound backdrop. The Drum Line is our newest division of the Mighty Liberators and consists of 25 drummers. These programs will increase your confidence and give you leadership skills.
Our Dance Program is designed to give you an opportunity to express yourself through dance. The program involves teaching the art of dance and choreography, while building self-esteem.


text:
Le Monument des Fusillés de Tavannes commémore le massacre de 15 résistants français, un belge et un inconnu. Ces hommes courageux furent tués à l'aube du .

Site
Le monument des fusillés de Tavannes est situé sur la Route nationale 913a, dans la Meuse, en France. Il dépend du village de Vaux-devant-Damloup (55400) ou se trouve aussi le Fort de Vaux. Pour voir le monument, il faut descendre à l'entrée du ravin du Bois de la Laufée. Ce monument est élevé devant le charnier où furent retrouvés, en 22 décembre 1944 près de l'entrée nord du tunnel de Tavannes, les corps mutilés de 16 patriotes résistants, massacrés par la Gestapo. Parmi ces corps, se trouvaient quinze Français, dont un inconnu, et un Belge. Une plaque porte les noms de ces hommes courageux qui tombèrent en héros, face aux Allemands.

Découverte du drame
Nous citons :

Le 22 décembre 1944, M. Scherrer, chef de canton à la SNCF, circulant sur le chemin qui, partant de la sortie du tunnel de Tavannes, aboutit à la route du fort de Vaux au point le plus creux, eut son attention attirée par une forte odeur de chair putréfiée qui semblait provenir d'une petite sablière en bordure du chemin. En grattant le sol, il vit un bras décharné et se rendit bientôt compte que plusieurs corps étaient là. Il prévint la gendarmerie. Les autorités firent le nécessaire et on découvrit seize cadavres masculins portant chacun la trace d'une balle dans la tête. Ils furent transportés à Verdun et déposés dans un local du service des Sépultures Militaires en vue de leur identification. Il devint vite évident qu'il s'agissait d'un crime de la Gestapo qui, à l'approche des libérateurs avait emmené précipitamment ce groupe d'hommes enfermés à la prison de Verdun et les avait lâchement assassinés, sans jugement, dans le coin reculé d'une forêt.

Les victimes
Treize corps ont été identifiés rapidement. Il s'agissait d'habitants de la région dont deux prêtres.

En octobre 1945, une mission belge devait reconnaître en l'un des trois corps restés inconnus François De Kinder, héros de la Résistance, beau-frère d'Hubert Pierlot, Premier Ministre de Belgique. Ce résistant fut arrêté à Paris par la Gestapo et fut conduit à Fresnes puis à Verdun où il arriva au début août. Il fut emmené avec les autres victimes.

Un deuxième inconnu fut identifié, le  est resté anonyme. Il sera inhumé à l'entrée du cimetière national du Faubourg-Pavé.

Reconstitution du drame

Le , près de l'entrée nord du tunnel, on découvre une fosse commune contenant les corps de 16 hommes exécutés par la Gestapo lors de leur retraite vers l'Allemagne. On n'a pu identifier à ce jour les coupables de ce crime. Les Allemands abandonnent Verdun le . Dans leur fuite, ils feront couler le sang français.

Le responsable de la Gestapo en France, Helmut Knochen a pourtant couvert ces assassinats multiples et été condamné à mort puis gracié, sa peine commuée en 20 ans de prison.

Notes et références

Bibliographie

Liens externes
 Souvenir Belge lu le 15/04/2020
 Reportage Youtube : le Monument aux fusillés lu le 15/04/2020
 Le Monde en guerre : le Monument aux fusillés lu le 15/04/2020

Références

Monument dans la Meuse
Mémorial de la Seconde Guerre mondiale en France
Mémoire de la Résistance française

redpajama_set_name: RedPajamaWikipedia


text:
Email Login | Register
TLP Founder Creates New Non-Profit Organization for Cystic Fibrosis
June 23, 2019 May 10, 2018 by Cips
As many of you know, our founder Danitsia was born with Cystic Fibrosis. At the age of 5 her family was hit with a disastrous event in where she lost both her brothers, Junior was 21 and Rey was 12 at the time, to a fatal car accident in Venezuela. The death of her brother Rey had always been a looming day in the life of her family because Rey, like Dani, was born with Cystic Fibrosis though he was misdiagnosed most of his life and did not achieve the diagnosis until age 7. By then the damage to his lungs had been significant and other organs had been affected. Once the tissue in your lungs has been scarred and damaged beyond function, it cannot be reversed. Rey was on the verge of multiple organ failure and was put on a double lung transplant list, a liver transplant list, and a kidney transplant list. The odds of Rey surviving past the age of 18 were slim to none.
The car accident had been a blessing in many ways because Rey would have suffered a slow, painful, and agonizing death to multiple organ failure due to Cystic Fibrosis. Rey was killed on impact. However, Junior was more than likely a carrier of the gene but did not suffer from CF at all. He was healthy, smart, and compassionate and helped his family care for his brother and sister. No one expected death to touch him. It was a grave loss all around and the feelings associated with both brothers' passing were difficult, conflicting, and confusing. The family really had very little support to offer a grieving mother, a desperate older sister, a father that pushed his own feelings down, and a little young girl that was very confused by it all and dealing with the fact that she had the same disease that would have ultimately killed her brother Rey.
It is because of this very fact we have created a new foundation called the Appreciate Every Breath Foundation. Our efforts with this foundation is to solicit the donation of time by counseling and mental health professionals to members of the foundation, who are patients suffering from Cystic Fibrosis. The foundation will also be extending benefits to caregivers, family, and friends of those affected by Cystic Fibrosis. We are looking to build a network of qualified mental health professionals that are willing to donate one hour per week to the cause for one year. Once we have established a wide list of counselors that register with us, we will begin opening the doors to patients and their families for private counseling sessions.
All patients, family members, and friends will be allowed access to a private Discord Server where they can interact and help each other. Our counselors will lurk about in that server to offer counsel and help as needed during their availability as some counselors will want to do more than just one hour. We will also hold group sessions and provide monthly support groups for patients themselves, spouses/partners of patients, as well as family members and friends of patients with Cystic Fibrosis.
Once a year our foundation will also participate in the annual "Adding Tomorrows" initiative with the Cystic Fibrosis Foundation to raise funds for Cystic Fibrosis research in an effort to bring us one step closer to a cure — a cure that will help thousands of affected families throughout the world. It is our sincere hope that all our friends and family help us launch this effort. If you or anyone you know would like to participate in a Great Strides Walk, Xtreme Hike, Cycle for Life or any other "Adding Tomorrows" fundraiser or if you are or someone you know is a mental health professional that may wish to donate your services, please let us know. Email us directly at donate@aebfoundation.org.
Jake "Cips" Knoll, COO
The Last Prophecy Gaming, Inc.
Categories Community News Tags Charity, Community, Cystic Fibrosis, Foundation
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Public Remarks: 2011 Camper on a cleared lot with a shed on site. Short distance to most amenities. Amenities include pool, dock fishing pier, crabbing pier, marina, pavilion, clubhouse, mini-golf, playground, 24hr security, and so much more. Seller also owns lot 10 & 11 which are also for sale. Great opportunity to own one, two or three lots.


Do you have essential bottles everywhere in your house? Are there bottles next to your diffuser, bottles in your bathroom, bottles in your purse, and bottles in your kitchen?
Do you have a central location where you put all your essential oil so you can organize it to see what you have?
You want to use it! And in order to use your oil you've got to keep it close to where you use it.
You work hard to maintain a surplus of what you need! And for good reason! But because you have extra you need to be organized and intentional about what you use. That essential oil that you're using, while it does a lot of good, costs money!
My name is Jennifer Osuch. I'm the author of the Preparedness Planner Series and I'm also one of the lead teachers at Self Reliant School–where my husband and I teach preserving, cooking, gardening and all around natural living skills.
I've been using natural remedies, including essential oils, for years.
Have you ever made a blend you love and then forgot the recipe? I have! And it's frustrating!
Aroma therapy has really helped me sleep and relieve anxiety. There was one blend that really helped me sleep, but the oil company who mixed it no longer carried it. Because it was a natural product and all the ingredients were listed. I was able to make my own blend and come really close to the blend I was used to–the one that helped me fall asleep.
Then one spring my kids and I were spring cleaning, and one of my son accidentally threw away my recipe. It was so frustrating because I didn't have one place to store my blend recipes.
Determined to be more organized I created The Preparedness Planner Essential Oil Edition so not only could I keep track of my blend but I could keep track of all my oils. Knowing exactly what I have on hand allows me to take advantage of sales at my favorite oil company, and get the oils I need to replenish. The planner also allows me to keep a journal of oils that work on for certain situations.
Making your own essential oil blends helps you save money and allows you to use oils that you love in combinations that are tailored to fit you.
Preparedness Planner Essential Oil Edition is for those that have been using essential oils for year or are just starting to explore the world of natural remedies.
All the forms you'll need to keep track of ALL your essential oils.
The Essential Oil Edition allows you to not only keep track of your stockpile but also allows you to keep track of your oil rotation.
Easily track when bottles run out and when you add to your stockpile.
Keep track of your homemade oil blends and journal about your experience with essential oils for easy reference later.
Complete with step-by-step directions and a brief overview of essential oil.
Details about why essential oils are important.
Extensive list of essential oils on the market today.
Keep track of your inventory and helps you see what oils you have in your stockpile or need to purchase.
Quickly reference which oil worked on which family member for what condition with the essential oil journal.
Blank recipe card pages for you to write your favorite homemade oil blends.
I've also included a few of my favorite homemade blends to get you started.
Balance sheet that allows you to track your stockpile and usage from year to year.
PDF format allowing you to print as many or as few pages as fits your families needs.
Three FREE Gifts For You….
I wanted to include a few extra special things too, because I want you to have all the tools you need to organize your essential oils, so I thought you might like to have all the forms included in the planner in a downloadable spreadsheet with all of the formulas and calculators done for you.
As a bonus to you we've included all the forms in the planner in spreadsheet format so you can keep track of your essential oils on your computer, tablet or phone.
Each of the mini measuring spoons are named dash, pinch, smidgen and nip for easy reference and hold the following amount of oil.
We're so sure you're going to love our Preparedness Planner Essential Oil Edition that we're offering a 30-Day money back guarantee.
If for any reason you decide that keeping track of your essential oils is not for you then we'll refund 100% of your investment. No questions asked!
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CNBC running "pump and dump" Bitcoin scam to fleece viewers, warns Peter Schiff
The economic prophet who correctly predicted the 2008 housing collapse and ensuing "Great Recession" has said during a recent podcast that financial news network CNBC continues to promote the viability of Bitcoin despite the fact that the cryptocurrency is headed for collapse.
Appearing on the Quoth the Raven podcast, Peter Schiff said the network hasn't had him back as a guest in about a year following a "bitcoin versus gold debate" he had with another guest, Brian Kelly of BK Capital Management, in which he disparaged the cryptocurrency.
Schiff said that the network pledged to have him back but that it hasn't happened. Meanwhile, Kelly, who took the side of bitcoin, is a frequent guest.
The prophetic financial advisor accused CNBC of running an "organized pump and dump" scam by pushing bitcoin hard while ignoring warning signs that cryptocurrencies are not just risky but potentially financially destabilizing as well.
"I think they know I'm negative on bitcoin and they don't want any negative perspectives," he said on the podcast. "They're there to constantly try to convince their viewers to buy these coins so that the people that come on the show can get out of theirs. They're trying to create demand."
He noted that in the current economy, bitcoin "is deflating," and "if you want to get out, somebody has to come in and provide the buying. And so that's what they're (CNBC) doing."
Schiff opined that the network no longer wanted to book him "not because I'm bearish on the economy but because I'm bearish on bitcoin."
"I mean, they can't even spot that bubble? That's how bad CNBC is at spotting bubbles," he noted, adding that the network "did not cover bitcoin at all" when it was far cheaper to buy.
Bitcoin hypocrisy at CNBC
The QTR host noted that Kelly, who has been at the network for years, never mentioned bitcoin when the price was low. But then one day when he was hyping bitcoin after its alleged value increased dramatically, he said Kelly was "going down a checklist of ways you can buy bitcoin, and conveniently…third on the list" was his own firm.
"So, this guy, who has never mentioned [bitcoin] before, is now presenting it to viewers, like, 'not only go out and buy it but buy it through me, like, I have my own crypto asset management company,'" the QTR host said. (Related: Young cryptocurrency morons think they will own the world without effort… welcome to Millennial delusions gone digital.)
Schiff, who served as an economic advisor to then-Rep. Ron Paul, R-Texas, who ran for president in 2008 as GOP candidate, also discussed how Keynesian and Austrian economic theories differ, as well as the politics surrounding POTUS Donald Trump's economics, the recent volatility in tech stocks and how inflation affects the middle class, as well as other subjects.
But he also addressed how wrong the regular and financial media, as well as 'mainstream' economists, were in 2008, an accusation he's also had to deal with recently because of his bearish outlook.
"It's a total double standard because it shows you their way of thinking," he said. "If you look at all of these experts that were completely wrong now that we're 10 years from the financial crisis…by 2007, the bubble had burst…even after it was so completely obvious. I was predicting it. They didn't figure it out until everything had imploded."
He also recalled how, during television appearances in 2008, he would say the country was in a recession, but he got pushback from other economists who had concluded there was "no recession in sight."
As for bitcoin, Schiff criticized Kelly for hyping bitcoin at $20,000 and $30,000 to viewers but then announcing that he himself had sold his holdings during a rally.
"And now after the fact you're disclosing that you sold into that rally?" Schiff said, noting the hypocrisy.
Read more about bitcoin's coming demise at BitcoinCrash.news.
ZeroHedge.com
Tagged Under: bearish, bias, bitcoin, Brian Kelly, Bubble, CNBC, Collapse, cryptocurrencies, devalue, economic policy, economics, finance, financial news, Peter Schiff, propaganda, U.S. dollar, worthless
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