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Home News What is the Impact of 5g Radiation?
What is the Impact of 5g Radiation?
Manpreet Kaur
There is considerable situational anxiety across the world due to COVID-19, which is increasing by leaps and bounds due to the latest buzz about 5G radiation and its impact on health. Some believe 5G radiations is responsible for Coronavirus spreading across the globe, while others believe it causes cancer. Read on to know about the impact of 5G radiation on health.
What exactly is 5G radiation?
The term 5G means the fifth generation of mobile technology first widely deployed in 2019. It is the latest wireless cellular technology that employs millimeter waves in addition to microwaves used in older cellular technologies.
5G offers faster browsing, streaming, and downloading and much better connectivity. 5G is a natural evolution in our tech-driven world.
5G lies on the low-frequency end of the electromagnetic scale. Being less powerful than the visible light, 5G radiations are not strong enough to damage cells when compared to radiation at the higher frequency end of the spectrum like sun's rays and medical x-rays.
Is there any negative impact of 5G on health?
The latest report by Deloitte titled "Technology, Media, and Telecommunications Predictions 2021" debunks rumors related to the negative impact of 5g radiation on health. It is highly unlikely that 5G radiations from mobile will affect the health of people. Even users of any other generation of mobile phones as well as those living in close proximity of a mobile network and not using a cellphone are safe from health risks.
Some people believe 5G radiation causes cancer, while others fear it weakens the immune system, making one susceptible to COVID. The Deolitte report clearly states that 5G does generate radiations — but at safe levels. Also, none of it is radioactive radiation. So, there are no biological effects associated with 5G.
The report also stated that 5G is far more effective than older cellular mobile technologies. Besides higher speeds and low latency, 5G uses less power than 4G or 3G. Therefore, emits lesser power! Another best thing about 5G base stations is that they can be put into sleep mode at night when there are no active users. On the other hand, 4G networks transmit control signals even when there are no users in range.
According to a World Health Organization (WHO) report, based on research conducted to date no adverse health effects have been casually linked with exposure to wireless technologies. There are various studies performed across the entire radio spectrum. However, as of now, there are very few studies on 5G radiations.
Sure, radiofrequency fields heat tissues in the human body. But, considering the radiofrequency exposure levels from current technologies, they can cause a negligible rise in the human body temperature. Higher the frequency, lower the penetration into the body tissues. When radiofrequency increases, the energy adsorption becomes more confined to the skin and eye i.e surface of the body. As per WHO, overall exposure remains below international guidelines and no negative impact on public health is anticipated.
What are the international guidelines?
Most countries adhere to guidelines recommended by:
The International Commission on Non-Ionizing Radiation Protection (ICNIRP).
The Institute of Electrical and Electronics Engineers, through the International Committee on Electromagnetic Safety
Mobile phones operate in the range of the electromagnetic spectrum – from 100 kHz (0.1 MHz) to 300 GHz (3 lakh MHz), according to ICNIRP.
India follows ICNIRP guidelines and has set very strict standards on EMF radiations.
Airtel and Vodafone currently offer 4G in the 1800MHz and 2100MHz bands
Jio uses a lower band of 800 MHz.
The 2021 auction includes a band range of 2,500MHz.
The Department of Telecommunications (DoT) plans to set 5G in the 3300-3600 MHz frequency range.
All the above-mentioned frequency ranges including the frequencies under discussion for 5G are well within the permissible limit.
Are you still wondering is 5G bad for health?
Well, WHO is conducting the health risk assessment to figure out any biological effects associated with exposure to radiofrequency covering the entire radiofrequency range, including 5G and the report will be published by 2022.
WHO established the International Electromagnetic Fields (EMF) Project in the year 1996. This project examines adverse biological effects of exposure to electric and magnetic fields in the frequency range 0-300 GHz and suggests national authorities on EMF radiation protection.
Since the new 5G technology is deployed, the health organization can gather more public health data. This report will give us scientific evidence related to health risks or problems from 5G exposure.
So, friends, as of now there is nothing to worry about 5G. Wait till WHO releases its next report. Don't pay heed to "5G is bad for health" rumors anymore!
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SMS, Steve Saleen's new operation, wants to sell 2005-09 Mopar Hemi owners a supercharger that the company claims can bump underhood horsepower by as much as 45%. The SMS 296 supercharger has some unique features, including an internal intercooler and a low profile, letting the blower sit between the cylinder banks and fit entirely under the hood. Obviously, we can't verify any of the company's output claims, but if they'd send us a Challenger so-equipped, we'd be happy to amend this post with full test results. Steve? Anyone? Full release with specs after the jump.
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With over 30 years of racing experience Steve Saleen and the SMS design and engineering team have created the ultimate supercharger for Mopar enthusiasts. Now the aftermarket community can take advantage of the SMS 296 Supercharger™, which will be standard equipment on the highly anticipated 2009 SMS 570 Challenger™ and SMS 570X Challenger™, with power ratings from 500 to over 700 horsepower respectively!
The SMS 296 Supercharger™ is packed with innovative features created by SMS engineers to provide optimum boost in an exceptionally compact package.
At the top of the list is the Delta-Frame™ internal intercooler (patent pending), with dual thin-core intercoolers and single-pass coolant flow. The intercooler design provides for a considerable increase in surface area, which has the effect of minimizing pressure drop and optimizing thermal characteristics. The intelligent configuration also acts as a mechanical diverter for effective air transfer into the ports, while the angled manifold enables packaging of runners that are much longer than in conventional layouts.
The long intake runner optimizes airflow efficiency and provides the SMS 296 Supercharger™ with an extremely low profile. This arrangement means that it nestles comfortably within the V, and under the stock hood of all applications, as well as accommodating a bypass system for 3-way cruising off-boost.
At the heart of the supercharger are the asymmetric multi-lobe compressor screws, which are machined in-house by SMS on state-of-the-art CNC equipment. The screw concept is among the most efficient of all supercharger configurations. However, it is also one of the most challenging to manufacture.
In order to minimize parasitic losses, it is crucial to maintain incredibly tight tolerances between the profiles of the asymmetrical lobes. Running dry at speeds of up to 16,000 rpm, there is no room for error. In finishing the screw profiles, the SMS flexible CNC machining facility is working to exacting tolerances. The flexibility of the equipment also means that changes in the screw profiles can be accommodated for different applications.
The innovations on the SMS 296 Supercharger™ carry through to its exterior, with SMS Tailor Made™ custom finishes and powder coat color options to match most Mopar colors. Now enthusiasts can match the impact under the hood with the one at the rear wheels.
The complete SMS 296 Supercharger™ aftermarket assembly looks great and installs easily under the stock hood of all applications including; the Dodge Challenger SRT-8, Dodge Challenger R/T, Dodge Charger SRT-8, Dodge Charger R/T, Dodge Magnum SRT-8, Dodge Magnum RT, Chrysler 300C SRT-8, Chrysler 300C, Jeep Grand Cherokee SRT-8, and Select 5.7L Hemi Dodge Trucks.


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Whether you are a supplier, development partner, university, or research organization, we consider all opportunities to collaborate as a chance to combine our innovative forces to achieve a common goal.
Innovation is driven by exchanges: Let's drive innovation together!


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\section{\@startsection {section}{1}{\z@}{-3.5ex plus -1ex minus
    -.2ex}{2.3ex plus .2ex}{\large\bf}}
\def\subsection{\@startsection{subsection}{2}{\z@}{-3.25ex plus -1ex
    minus -.2ex}{1.5ex plus .2ex}{\normalsize\bf}}
\def\@fnsymbol#1{\ensuremath{\ifcase#1\or *\or 1\or 2\or 3\or 4\or
    5\or 6\or 7 \or 8\ or 9 \or 10\or 11 \else\@ctrerr\fi}}
\makeatother

\title{Geometric Permutations of Non-Overlapping Unit Balls Revisited%
  \thanks{This research was supported in part by NRF grant
    2011-0030044 (SRC-GAIA), and in part by NRF grant~2011-0016434,
    both funded by the government of Korea.}}

\author{Jae-Soon Ha%
  \thanks{KAIST, Korea, Email: jaesoonha@kaist.ac.kr,
    otfried@kaist.edu}
  \and
  Otfried Cheong\footnotemark[2]
  \and
  Xavier Goaoc%
  \thanks{Universit\'{e} Paris-Est Marne-la-Vall\'{e}e, France, Email:
    goaoc@univ-mlv.fr}
  \and
  Jungwoo Yang%
  \thanks{Aarhus University, Denmark, Email: jungwoo@madalgo.au.dk}
}

\begin{document}
\maketitle
\begin{abstract}
  Given four congruent balls $A, B, C, D$ in~${\mathbb{R}}^{d}$ that have
  disjoint interior and admit a line that intersects them in the
  order~$\p ABCD.$, we show that the distance between the centers of
  consecutive balls is smaller than the distance between the centers
  of~$A$ and~$D$.  This allows us to give a new short proof that $n$
  interior-disjoint congruent balls admit at most three geometric
  permutations, two if $n\ge 7$. We also make a conjecture that would
  imply that $n\geq 4$ such balls admit at most two geometric
  permutations, and show that if the conjecture is false, then there
  is a counter-example of a highly degenerate nature.
\end{abstract}

\section{Introduction}
A \emph{line transversal} to a family ${\mathfrak{F}}$ of pairwise disjoint convex
sets in~ ${\mathbb{R}}^d$ is a line that intersects every element of that
family. The study of line transversals, their properties, and
conditions for their existence started in the 1950s with the classic
work of Gr\"unbaum, Hadwiger, and Danzer; background about the
sizable literature on \emph{geometric transversal theory} can be
found in the classic survey of Danzer et al.~\cite{dgk-htr-63}, or the
more recent ones by Goodman et al.~\cite{gpw-gtt-93},
Eckhoff~\cite{e-hrctt-93}, Wenger~\cite{wenger1990geometric}, or
Holmsen~\cite{holmsen2008recent}.

An oriented line transversal~$\ell$ to a family~${\mathfrak{F}}$ induces a linear
order on~${\mathfrak{F}}$: Fig.~\ref{fig:intro1}(a) shows three oriented transversals
to a family of three congruent disks inducing the orders $A \prec C
\prec B$, $A \prec B \prec C$, and $B \prec A \prec C$.
\begin{figure}[h]
  \centerline{\includegraphics{intro1}}
  \caption{Orders and geometric permutations\label{fig:intro1}}
\end{figure}
For conciseness, we usually represent the order by the string listing
the elements, the three lines in Fig.~\ref{fig:intro1}(a) induce the
orders $\p ACB.$, $\p ABC.$, and $\p BAC.$. Natural questions in
geometric transversal theory are: Given a family of disjoint convex
objects, how many different orders can be realized by line
transversals?  How much can these orders differ? What becomes of these
questions if the objects have a more restricted shape, for instance if
they are balls or axis-aligned boxes?

If an order can be realized by an oriented line, so can its reverse,
so the two are equivalent in this sense.  The equivalence classes,
that is, pairs of an order and its reverse, are called \emph{geometric
  permutations}.  Fig.~\ref{fig:intro1}(b) shows a set of five
congruent disks with the two geometric permutations~$\p ABCDE.$ and
$\p ACBDE.$, which could equally well be written as~$\p EDCBA.$ and
$\p EDBCA.$. In Fig.~\ref{fig:intro1}(b) the disks~$B$ and~$C$ touch
each other. We allow this, but a line transversal is not allowed to be
tangent to these disks in this common point.  Put differently, we can
remove the common points of contact from the objects to obtain a
family of disjoint convex objects with the same set of line
transversals.  It is convenient to allow such families, as
configurations are often easier to describe when objects touch.  We
will call a family of compact convex objects in~${\mathbb{R}}^{d}$ that may
touch, but whose interior is disjoint, a \emph{non-overlapping}
family.

The study of geometric permutations started in the 1980s with the work
by Katchalski et al.~\cite{katchalski1985geometric,
  katchalski1987geometric}. In the plane, $n$ convex objects admit at
most $2n-2$ geometric permutations and this bound is
tight~\cite{edelsbrunner1990maximum}.  One of the intriguing open
questions is the corresponding bound for three and higher dimensions:
$n$ convex objects in~${\mathbb{R}}^{d}$ can have $\Omega(n^{d-1})$ geometric
permutations~\cite{smorodinsky2000sharp}, but the best known upper
bound is only $O(n^{2d-3}\log n)$~\cite{Rubin2012}. For balls or similar
\emph{fat} objects, the lower bound of $\Omega(n^{d-1})$ is known to
be tight~\cite{smorodinsky2000sharp,katz2001tight}. Disjoint
\emph{congruent} balls, however, have only a constant number of
geometric permutations: In two dimensions, $n \geq 4$ congruent disks
have at most two geometric
permutations~\cite{smorodinsky2000sharp,asinowski2003geometric}.  In
dimension~$d \geq 3$, Cheong et al.~\cite{cheong2005geometric} proved
that $n$ non-overlapping congruent balls have at most three geometric
permutations, and at most two geometric permutations when~$n\geq 9$.

\bigskip

In this paper we revisit the problem of bounding the number of
geometric permutations of $n$~non-overlapping congruent balls
in~${\mathbb{R}}^d$.  Since we can arbitrarily choose the radius of the balls,
we will refer to them as \emph{unit balls}.  The earlier work of
Cheong et al.~\cite{cheong2005geometric} does not entirely settle the
question, as no construction of $n > 3$ non-overlapping unit balls is
known that admits more than two geometric permutations.  Furthermore,
the proof by Cheong et al.\ is quite technical and relies on delicate
geometric lemmas and tedious case analysis.

\bigskip

In the first part of this paper, we give a shorter and greatly
simplified proof that $n \geq 3$ non-overlapping unit balls have at
most three geometric permutations.  Unlike the previous
proof~\cite{cheong2005geometric}, it could be presented in its
entirety in an undergraduate course on transversal theory. Our main
theorem is the following:
\begin{theorem}
  \label{the:main}
  Let ${\mathfrak{F}}$ be a family of $n$ non-overlapping unit balls in ${\mathbb{R}}^d$.
  The number of geometric permutations of ${\mathfrak{F}}$ is at most three if $n
  \le 6$, and at most two if $n \ge 7$.
\end{theorem}

\noindent
Theorem~\ref{the:main} slightly improves the previous bound of Cheong
et al.~\cite{cheong2005geometric} by settling the question for $n=7$
and $8$. Our proof rests on the following lemma:

\begin{dilemma}
  If four non-overlapping unit balls $A$, $B$, $C$ and $D$ in ${\mathbb{R}}^d$
  have a line transversal with the order $ABCD$ then $|ad| >
  \max\{|ab|,|bc|,|cd|\}$.
\end{dilemma}
\noindent
(Here and throughout the paper we will use lower-case letters to
denote the centers of balls written with upper-case letters, so $a$,
$b$, $c$, and $d$ are the centers of $A$, $B$, $C$, and~$D$.)  The
lemma is not as obvious as it might appear: For three unit-balls, for
instance, the existence of a transversal with the order $\p ABC.$ does
not imply that $|ac| > |ab|$, as already evidenced by
Fig.~\ref{fig:intro1}(a).

We prove the distance lemma, in Section~\ref{sec:distance-lemma}, by
first modifying the given configuration into a canonical situation: We
shrink the balls, keeping them congruent, until we reach the smallest
radius for which they still have a transversal with the given order.
This idea has probably been used first by Klee~\cite{klee1954common}
and then by Hadwiger~\cite{hadwiger1957eibereiche}.  The resulting
canonical configuration~${\mathfrak{F}}$ has the property that the line
transversal~$\ell$ is \emph{pinned} (Lemma~\ref{lem:shrinking}): This
means that any arbitrarily small perturbation of~$\ell$ is no longer a
transversal of~${\mathfrak{F}}$.  In other words, $\ell$ is an isolated point in
the space of transversals of~${\mathfrak{F}}$.  The same method for deforming a
family of unit balls such that the line transversal becomes pinned has
been used by Cheong et al.~\cite{cheong2008helly}.  The correctness of
the method is there deduced from algebraic results by Megyesi and
Sottile~\cite{megyesi05} and by Borcea et al.~\cite{borcea2008line}.
This argument requires strict disjointness of the balls, and doesn't
meet our goal of a proof presentable to undergraduates.  We instead
observe that the fact we need is already implicit in a proof by
Holmsen et al.~\cite{holmsen2003helly}.  In Appendix~\ref{sec:holmsen}
we examine their proof to prove the correctness of the pinning method
for non-overlapping unit balls.

Before proving the distance lemma, we show, in
Section~\ref{sec:at-most-three}, that it readily simplifies various
steps of the proof of Cheong et al.~\cite{cheong2005geometric},
resulting in an elementary proof that the number of geometric
permutations of $n$~non-overlapping unit balls is at most three. On
the one hand, the distance lemma simplifies technical derivations.
For example, the fact that the geometric permutations $\p ABCD.$
and~$\p BADC.$ are \emph{incompatible} for non-overlapping unit balls,
that is, they cannot be realized at the same time by a family of four
balls, was given a delicate, five pages long,
proof~\cite[Section~4]{cheong2005geometric}; it follows immediately
from the distance lemma, since $\p ABCD.$ implies that $|ad| > |bc|$
and $\p BADC.$ implies that $|bc| > |ad|$, a contradiction. On the
other hand, using the distance lemma we can replace rather pedestrian
arguments by more conceptual analyses, for instance the mechanical
reduction from $n$ to $4$ balls~\cite[Section~2]{cheong2005geometric}
is done more concisely in Lemma~\ref{lem:ThreeFourGP}.

\bigskip

We conjecture that the geometric permutations $\p ABCD.$ and $\p
ACDB.$ are incompatible.  If proven, this would show that $n\ge 4$
non-overlapping unit balls have at most two geometric permutations,
thereby completely closing this question. In the second part of this
paper, we analyze the geometry of certain pinning configurations and
show that if our conjecture is false then it must admit
counter-examples of a highly contrived nature.

\section{At most three geometric permutations}
\label{sec:at-most-three}

We first use the distance lemma to reduce the problem from $n$~balls
to three or four balls (the same result was obtained by Cheong et
al.~\cite{cheong2005geometric} via a tedious case-analysis):
\begin{lemma}
  \label{lem:ThreeFourGP}
  If $n \geq 4$ non-overlapping unit balls in ${\mathbb{R}}^d$ have at least $k
  \in \{3,4\}$ geometric permutations, then $k$ of the balls have $k$
  distinct geometric permutations.
\end{lemma}
\begin{proof}
  Let ${\mathfrak{F}}$ be a family of $n \geq 4$ non-overlapping unit balls in
  ${\mathbb{R}}^d$. We call an element \emph{extreme} in a geometric permutation
  if it appears first or last in its order. We make two observations:
  \begin{enumerate}
  \item[(i)] Any two geometric permutations of ${\mathfrak{F}}$ have an extreme
    element in common. Indeed, if two geometric permutations $\sigma_1$
    and $\sigma_2$ of ${\mathfrak{F}}$ have disjoint sets of extreme elements
    $\{A_1,B_1\}$ and $\{A_2,B_2\}$ then applying the distance lemma to
    $\sigma_1$ yields $|a_1b_1|>|a_2b_2|$ and applying it to $\sigma_2$
    yields $|a_2b_2|>|a_1b_1|$, a contradiction.

  \item[(ii)] If two geometric permutations of ${\mathfrak{F}}$ share an extreme
    element $A$ then they differ on ${\mathfrak{F}} \setminus \{A\}$. Indeed, assume
    that the first geometric permutations writes $\p AB\ldots XY.$. If the
    second, which is distinct from the first, coincides with it on ${\mathfrak{F}}
    \setminus \{A\}$ then it must be $\p AYX\ldots B.$. The distance lemma
    then implies both that $|ay|>|ab|$ and that $|ab|>|ay|$, a
    contradiction.
  \end{enumerate}
  Assume that ${\mathfrak{F}}$ has three geometric permutations and let ${\mathfrak{G}}$ be a
  minimal subfamily of ${\mathfrak{F}}$ on which their restrictions~$\tau_1,
  \tau_2$, and $\tau_3$ are pairwise distinct.  There cannot be an
  extreme element common to all three~$\tau_i$ as observation~(ii)
  would contradict the minimality of~${\mathfrak{G}}$.  Hence, there exist three
  distinct elements $A,B,C \in {\mathfrak{G}}$ such that $A$ is extreme in
  $\tau_1$ and $\tau_2$, $B$ is extreme in $\tau_2$ and $\tau_3$ and
  $C$ is extreme in $\tau_1$ and $\tau_3$. Then the restrictions of
  $\tau_1$, $\tau_2$ and $\tau_3$ to $\{A,B,C\}$ are $\p ABC.$, $\p
  ACB.$ and $\p BAC.$, implying ${\mathfrak{G}} = \{A, B, C\}$.

  Assume now that ${\mathfrak{F}}$ has four geometric permutations and, again, let
  ${\mathfrak{G}}$ be a minimal subfamily of~${\mathfrak{F}}$ on which their restrictions
  $\tau_1, \tau_2, \tau_3$, and $\tau_4$ are pairwise distinct.  For $H
  \subseteq {\mathfrak{G}}$ let ${\tau_i}_{|H}$ denote the restriction of
  $\sigma_i$ to $H$. As we just argued, ${\mathfrak{G}}$ contains a triple $T =
  \{A,B,C\}$ such that $A$ is extreme in $\tau_1$ and $\tau_2$, $B$ is
  extreme in $\tau_2$ and $\tau_3$ and $C$ is extreme in $\tau_1$ and
  $\tau_3$; we further have
  \[
    {\tau_1}_{|T} = \p ABC., \quad {\tau_2}_{|T} = \p ACB., \quad {\tau_3}_{|T}
    = \p BAC..
  \]
  By observation~(i) the extreme elements of $\tau_4$ are among
  $\{A,B,C\}$, say $A$ and $C$. Since $\tau_1$ and $\tau_4$ have the
  same extreme elements but are different there must exist a pair
  $\{D,E\} \subset {\mathfrak{G}}$ such that the restrictions of $\tau_1$ and
  $\tau_4$ to $\{A,C,D,E\}$ are different. Assume that $B \notin
  \{D,E\}$ so that $Q=\{A,B,C,D,E\}$ has size five. We write
  ${\tau_1}_{|Q} = AX_1X_2X_3C$, ${\tau_2}_{|Q} = AY_1Y_2Y_3B$ and
  ${\tau_3}_{|Q} = BZ_1Z_2Z_3C$.  If $X_3=B$ then $|bc|<|ac|$ and we
  must set $Z_1=A$. This implies that $|ab|<|bc|$ and we must set
  $Y_1=C$. This implies that $|ac|<|ab|$ which contradicts the two
  previous inequalities. It must then be that $X_3 \neq B$. This
  implies that $|ab|<|ac|$ which forces $Y_3=C$. This implies
  $|bc|<|ab|$, which forces $Z_3=A$. This implies that $|ac|<|bc|$,
  again a contradiction with the two previous inequalities. As a
  consequence, $B \in \{D,E\}$ and $\tau_1, \ldots \tau_4$ are already
  distinct on the quadruple $\{A,C,D,E\}$.
\end{proof}

We can now easily prove that there cannot be more than three geometric
permutations.
\begin{theorem}
  \label{the:at-most-three}
  A family of non-overlapping unit balls in ${\mathbb{R}}^d$ has at most three
  geometric permutations.
\end{theorem}
\begin{proof}
  By Lemma~\ref{lem:ThreeFourGP} it suffices to prove the statement for
  families of size four. Let $A,B,C,D$ be four non-overlapping unit
  balls in ${\mathbb{R}}^d$ and assume that there is a line transversal in the
  order~$\p ABCD.$. The distance lemma implies that $|ad| >
  \max\{|ab|,|bc|,|cd|\}$ and no line can meet these balls in the
  order $\p ADCB.$ (which implies $|ab|>|ad|$), $\p BADC.$ (which
  implies $|bc|>|ad|$), $\p BDAC.$ (implying $|bc|>|ad|$), or $\p
  CBAD.$ (as this entails $|cd|>|ad|$). Of the twelve geometric
  permutations of four elements, this leaves the seven shown in
  Fig.~\ref{fig:incomp7gp} as candidates for the remaining geometric
  permutations of~${\mathfrak{F}}$.
  \begin{figure}[h]
    \centerline{\includegraphics{incomp7gp}}
    \caption{Proof of Theorem~\ref{the:at-most-three}.}
    \label{fig:incomp7gp}
  \end{figure}
  It is easy to verify that the geometric permutations connected by
  edges in Fig.~\ref{fig:incomp7gp} are also incompatible by the
  distance lemma.  The resulting graph has no independent set of size
  larger than two, and so ${\mathfrak{F}} = \{A, B, C, D\}$ has at most three
  geometric permutations.
\end{proof}

To prove the stronger statement of Theorem~\ref{the:main}, we need two
lemmas proven by Cheong et al.~\cite{cheong2005geometric} (their
proofs are short and self-contained).  For a directed line~$\ell$, we
write $\vec{\ell}$ for its direction vector, for points $p, q \in
{\mathbb{R}}^{d}$, we will write $\vc{pq}$ for the vector $q - p$ from~$p$
to~$q$.
\begin{lemma}[{\cite[Lemma~7]{cheong2005geometric}}]
  \label{lem:angle}
  Given three non-overlapping unit spheres $A$, $B$ and $C$ in
  ${\mathbb{R}}^{d}$, and a directed line~$\ell$ stabbing them in the order $\p
  ABC.$. Then $\angle(\vec{\ell}, \vc{ac}) < \pi/4$.
\end{lemma}
\begin{lemma}[{\cite[Lemma~6]{cheong2005geometric}}]
  \label{lem:cylinder}
  Let ${\mathcal{C}}$ be a cylinder of radius one and length less than
  $s\sqrt{2}$ in~${\mathbb{R}}^{d}$, for some $s \in {\mathbb{N}}$. Then ${\mathcal{C}}$ contains at
  most $2s$ points with pairwise distance at least two.
\end{lemma}
We analyze the intersection of two cylinders more carefully in the
following lemma:
\begin{lemma}
  \label{lem:cylinder6}
  Let ${\mathcal{C}}_1$ and~${\mathcal{C}}_2$ be cylinders of radius one and axes~$\sigma_1$
  and~$\sigma_2$ in~${\mathbb{R}}^{d}$.  If $\pi/4 < \angle(\vc{\sigma_1},
  \vc{\sigma_2}) \leq \pi/2$, then the intersection~${\mathcal{C}}_1 \cap {\mathcal{C}}_2$
  contains at most six points with pairwise distance at least two.
\end{lemma}
\begin{proof}
  We choose a coordinate system where $\sigma_1$ is the $x_1$-axis,
  and $\sigma_2$ is the line $(t\cos\theta, t\sin\theta, d, 0, \ldots,
  0)$, where $\theta = \angle(\vc{\sigma_1}, \vc{\sigma_2}) > \pi/4$
  and $d \geq 0$ is the distance between~$\sigma_1$ and~$\sigma_2$.
  The left side of Fig.~\ref{fig:cylinders} shows the projection on
  the $x_1 x_2$-plane.
  \begin{figure}[h]
    \centerline{\includegraphics{cylinders}}
    \caption{Projection of ${\mathcal{C}}_1$ and ${\mathcal{C}}_2$ on the $x_1 x_2$-plane.}
    \label{fig:cylinders}
  \end{figure}
  Consider the points $u = (1/\sin\theta, 0, 0, \dots, 0)$ and $v =
  (1/\sin\theta + \cot\theta, 1, 0, \dots, 0)$ marked in the figure.
  Since $\theta > \pi/4$, the distance between $u$ and~$-u$ is less
  than $2\sqrt{2}$, and so by Lemma~\ref{lem:cylinder} the section of
  ${\mathcal{C}}_1$ between $-u$ and~$u$ contains at most four points of pairwise
  distance at least two.  All remaining points in ${\mathcal{C}}_1 \cap {\mathcal{C}}_2$
  must project into the two symmetric shaded regions.  We will now
  show that these regions have diameter less than two and can
  therefore contain only one point each, proving the lemma.

  Let $p \in {\mathcal{C}}_1 \cap {\mathcal{C}}_2$ be a point with $u_1 < p_1 \leq v_1$ (we
  use indices for the coordinates in~${\mathbb{R}}^{d}$).  We will show that
  $|pr| < 1$, where $r = \big((u_1 + v_1)/2, \frac 12, \frac d2, 0,
  \dots, 0\big)$ (note that $r$ does not lie in the $x_1 x_2$-plane).
  Since $\theta > \pi/4$, we have $v_1 - u_1 = \cot\theta < 1$, and so
  $|p_1 - r_1| < 1/2$.  Let $\Pi$ denote the hyperplane $x_1 = u_1$,
  and let $p^\ast$ and $r^\ast$ be the orthogonal projection of $p$ and~$r$
  into~$\Pi$.  We observe that $p^\ast \in {\mathcal{C}}_1 \cap {\mathcal{C}}_2$. The
  intersection ${\mathcal{C}}_1 \cap \Pi$ is the unit-radius ball around the
  origin in~$\Pi$.  The intersection ${\mathcal{C}}_2 \cap \Pi$ is an ellipsoid
  with center~$q = (1/\sin\theta, 1/\cos\theta, d, 0, \dots, 0)$, see
  right hand side of Fig.~\ref{fig:cylinders}.  ${\mathcal{C}}_2 \cap \Pi$ contains exactly
  the points $x = (u_1, x_2, x_3, \dots, x_d) \in \Pi$ with
  \[
  (\cos\theta)^{2} (x_2-\frac{1}{\cos\theta})^{2} + (x_3 - d)^{2}
  + \sum_{i = 4}^{d} x_{i}^{2} \leq 1.
  \]
  Consider the ball~$D$ with center $s = (u_1, 1, d, 0, \dots, 0)$ and
  radius one.  For a point $x \in \Pi$ with $0 \leq x_2 \leq 1$, we
  have
  \[
  1 - x_2 \leq 1 - (\cos\theta)x_2 = (\cos
  \theta)(\frac{1}{\cos\theta} - x_2),
  \]
  and so $x \in {\mathcal{C}}_2$ implies $x \in D$.  It follows that $p^\ast \in
  {\mathcal{C}}_1 \cap D$. Since $|us| \geq 1$ and $r^\ast = (u + s)/2$ is the
  midpoint of the two centers we have $|p^\ast r^\ast| \leq \sqrt{3}/2$. It
  follows that $|pr|^{2} = |p_1 - r_1|^{2} + |p^\ast r^\ast|^{2} < 1/4 + 3/4
  = 1$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{the:main}]
  We proved the bound for $n \leq 6$ in
  Theorem~\ref{the:at-most-three}, so it remains to consider
  families~${\mathfrak{F}}$ of $n \geq 7$ balls.  We show that the geometric
  permutations $\p XYZU.$ and $\p XUYZ.$ are incompatible for~${\mathfrak{F}}$.
  Assume for a contradiction that $\ell$ is an oriented transversal
  inducing the order $\p XYZU.$, and $\ell'$ is an oriented
  transversal inducing the order $\p XUYZ.$.  By
  Lemma~\ref{lem:angle}, we have $\angle(\vec{\ell}, \vc{\ell'}) \leq
  \angle(\vec{\ell}, \vc{xz}) + \angle(\vc{\ell'}, \vc{xz}) <
  \pi/2$.  Since $\ell'$ meets $U$ before $Y$, we have
  $\angle(\vc{\ell'}, \vc{yu}) > \pi/2$, and by
  Lemma~\ref{lem:angle} again we have $\angle(\vec{\ell}, \vc{yu}) <
  \pi/4$, implying $\angle(\vec{\ell}, \vc{\ell'}) > \pi/4$.
  Consider now the cylinders~${\mathcal{C}}$ and~${\mathcal{C}}'$ of radius one with
  axes~$\ell$ and~$\ell'$.  Since $\ell$ and $\ell'$ are
  transversals for~${\mathfrak{F}}$, the centers of all balls in~${\mathfrak{F}}$
  are contained in ${\mathcal{C}} \cap {\mathcal{C}}'$. By Lemma~\ref{lem:cylinder6}, this
  implies $n \leq 6$, a contradiction.

  We now assume that ${\mathfrak{F}}$ has three geometric permutations. By
  Lemma~\ref{lem:ThreeFourGP} there is a subset ${\mathfrak{G}} = \{A, B, C, D\}$
  of four balls such that ${\mathfrak{G}}$ already has three geometric
  permutations. We can assume $\p ABCD.$ is one of them.  The
  incompatible pair $(\p XYZU., \p XUYZ.)$ implies that $\p ACDB.$,
  $\p ADBC.$, $\p CABD.$, and $\p BCAD.$ cannot exist.  Of the
  geometric permutations shown in Fig.~\ref{fig:incomp7gp}, this only
  leaves $\p ACBD.$, $\p ABDC.$, and $\p BACD.$. Since we already know
  $\p ABDC.$ and $\p BACD.$ to be incompatible by the distance lemma
  (see Fig.~\ref{fig:incomp7gp}), the last pair must therefore include
  $\p ACBD.$.  But this permutation is incompatible with the other two
  because they form pairs of the form~$(\p XYZU., \p XUYZ.)$.
\end{proof}

\section{Proof of the distance lemma}
\label{sec:distance-lemma}

We say that a family ${\mathfrak{F}}$ \emph{pins} a line $\ell$ or that $\ell$ is
pinned by ${\mathfrak{F}}$ if $\ell$ is a line transversal to ${\mathfrak{F}}$ and any
arbitrarily small perturbation of $\ell$ is not a line transversal to
${\mathfrak{F}}$. \footnote{Equivalently, a line is pinned by ${\mathfrak{F}}$ if it is an
  isolated point in the space of line transversals to ${\mathfrak{F}}$ endowed
  with the natural topology on the space of lines, for instance as
  given by the Grassmann-Pl\"ucker coordinates.} It is often
convenient to deform a family of balls and lines into a configuration
where the lines are pinned.  The following lemma describes such a
deformation.
\begin{lemma}
  \label{lem:shrinking}
  Let ${\mathfrak{F}}(t)=\{B_1(t),B_2(t), \ldots, B_n(t)\}$ be a parameterized
  family of non-overlapping balls of radius~$t \in [0,1]$ in ${\mathbb{R}}^3$,
  with the property that $B_{i}(s) \subset B_{i}(t)$ for any $1 \leq i
  \leq n$ and $0 \leq s < t \leq 1$. If ${\mathfrak{F}}(1)$ has a line transversal
  in the order $B_1(1)B_2(1)\ldots B_n(1)$ then there exists $t^\ast \in
  [0,1]$ such that ${\mathfrak{F}}(t^\ast)$ has a pinned line transversal in the
  order $B_1(t^\ast)B_2(t^\ast)\ldots B_n(t^\ast)$.
\end{lemma}

\noindent
The proof of Lemma~\ref{lem:shrinking} is already implicit in Holmsen
et al.~\cite{holmsen2003helly}.  For completeness, we revisit their
proof in Appendix~\ref{sec:holmsen} and make the necessary
adjustments.

The following lemma allows us to reduce the dimension in which we have
to prove our statements.
\begin{lemma}
  \label{lem:dim-red}
  Let ${\mathfrak{F}}$ be a family of non-overlapping unit balls in ${\mathbb{R}}^d$ with a
  line transversal~$\ell$, and let $\mathfrak{S}$ be an affine subspace
  containing the centers of all balls in~${\mathfrak{F}}$.  The orthogonal
  projection~$\ell'$ of $\ell$ into the subspace~$\mathfrak{S}$ is a line
  transversal to ${\mathfrak{F}}$ realizing the same geometric permutation.
\end{lemma}
\begin{proof}
  If $p$ is a point in a ball $B \in {\mathfrak{F}}$ and $p'$ is the projection of
  $p$ into~$\mathfrak{S}$, then $|bp'| \leq |bp|$, and so $p' \in B$. The lemma
  follows.
\end{proof}

We will use the following folklore characterization of triples of
balls pinning a line (we include a proof for completeness).
\begin{lemma}
  \label{lem:tri-tang}
  A set $\{A,B,C\}$ of three non-overlapping unit balls in ${\mathbb{R}}^3$ pins
  a line $\ell$ if and only if they are tangent to $\ell$, their
  centers $a,b,c$ are coplanar with $\ell$, and in that plane $\ell$
  separates the center of the middle ball (in the order of tangency)
  from the other two centers.
\end{lemma}
\begin{proof}
  If $\{A,B,C\}$ satisfies the condition then the ball $B$ is
  separated from $A$ and $C$ by the plane $\Pi$ perpendicular in
  $\ell$ to the plane of centers. Any line transversal to $\{A,B,C\}$
  in the same order as $\ell$ must be contained in $\Pi$, and thus
  $\ell$ is pinned.

  Conversely, assume that $\{A,B,C\}$ pins $\ell$.  If $\ell$ is not
  contained in the plane of centers, then rotating $\ell$ toward its
  orthogonal projection into that plane decreases the distances to all
  centers as in Lemma~\ref{lem:dim-red}.  Since any intermediate line
  in this rotation is therefore a line transversal, $\ell$ is not
  pinned. The line $\ell$ is thus contained in the plane of centers, and
  the necessity of the separation condition is easily checked.
\end{proof}

\begin{proof}[Proof of distance lemma.]
  We first argue that the statement follows from the case~$d=3$.
  Indeed, let $\mathfrak{S}$ denote a $3$-dimensional space containing the four
  balls' centers (if they are not coplanar, then $\mathfrak{S}$ is uniquely
  defined). The space $\mathfrak{S}$ intersects the four non-overlapping
  $d$-dimensional unit balls in four non-overlapping $3$-dimensional
  unit balls with the same centers. Let $\ell'$ denote the orthogonal
  projection of $\ell$ into $\mathfrak{S}$. By Lemma~\ref{lem:dim-red}, $\ell'$ is
  a line transversal to the four $3$-dimensional balls with the
  ``same'' geometric permutation as~$\ell$.  It therefore suffices to
  prove the claim for the $3$-dimensional balls and~$\ell'$.

  We now assume that we are in the case~$d=3$.  We shrink the balls
  uniformly around their center. By Lemma~\ref{lem:shrinking} we will
  reach a configuration with transversal $\ell$ that is pinned by four
  non-overlapping unit balls~$\{A,B,C,D\}$. If the four centers are
  collinear then the statement is clear, so we assume otherwise.
  \begin{figure}[ht]
    \centerline{\includegraphics{FigLemmaDistanceABCD}}
    \caption{Notation for the proof of the distance lemma (here all
      four balls are shown tangent to the pinned line).}
    \label{fig:LemmaDistanceABCD}
  \end{figure}

  We will use the following notation (refer to
  Fig.~\ref{fig:LemmaDistanceABCD}). Let $\ell^{\perp}$ denote a plane
  orthogonal to~$\ell$. For $x \in \{a,b,c,d\}$, let $x'$ denote the
  orthogonal projection of $x$ onto $\ell$ (that is, the point closest
  to $x$ on~$\ell$), and let $x^\ast$ denote the orthogonal projection
  of $x$ onto~$\ell^{\perp}$.  We set $d_1=|a'b'|$, $d_2=|b'c'|$,
  $d_3=|c'd'|$, and $\Delta = 2(d_1d_2 +d_1d_3 + d_2d_3)$.
  We have:
  \[
  |ad|^2 = (d_1 + d_2 + d_3)^2 + |a^\ast d^\ast|^{2}, \quad
  |ab|^2 = d_{1}^{\,2}+|a^\ast b^\ast|^2, \quad
  |bc|^2 = d_{2}^{\,2}+|b^\ast c^\ast|^2, \quad
  |cd|^2 = d_{3}^{\,2}+|c^\ast d^\ast|^2.
  \]
  Since $\ell$ is a line transversal with order~$\p ABCD.$, we have
  $d_{1}, d_{2}, d_3 > 0$ and therefore $\Delta > 0$.

  Our goal is to prove that
  \begin{align}
    \label{i:c1}
    |ad|^2 - |ab|^2 & =
    \Delta + d_2^{\,2}+d_3^{\,2} +|a^\ast d^\ast|^{2}-|a^\ast b^\ast|^2 > 0 \\
    \label{i:c2}
    |ad|^2 - |bc|^2 & =
    \Delta + d_1^{\,2}+d_3^{\,2} +|a^\ast d^\ast|^{2}-|b^\ast c^\ast|^2 > 0 \\
    \label{i:c3}
    |ad|^2 - |cd|^2 & =
    \Delta + d_1^{\,2}+d_2^{\,2} +|a^\ast d^\ast|^{2}-|c^\ast d^\ast|^2 > 0
  \end{align}

  Assume first that three of the balls already pin $\ell$.  There are
  essentially two cases:
  \begin{itemize}
  \item If the three balls are $\{A,B,C\}$ then, by
    Lemma~\ref{lem:tri-tang}, $a^\ast = c^\ast$ and $|b^\ast c^\ast|^2 = |a^\ast
    b^\ast|^2 =4$. The fact that $a^\ast = c^\ast$ immediately implies
    Inequality~\eqref{i:c3}. Since $C$ and~$D$ are non-overlapping we
    have
    \[
    d_{3}^{\,2} + |{a^\ast}{d^\ast}|^2 = d_{3}^{\,2} + |{c^\ast}{d^\ast}|^2 =
    |cd|^2 \ge 4,
    \]
    which implies Inequalities~\eqref{i:c1} and~\eqref{i:c2}. The case
    where the three balls are $\{B,C,D\}$ is symmetric.

  \item If the three balls are $\{A,B,D\}$ then, by
    Lemma~\ref{lem:tri-tang}, $a^\ast = d^\ast$ and $|a^\ast b^\ast|^2 =
    |b^\ast d^\ast|^2 =4$. Since $B$, $C$ and~$D$ are non-overlapping, we
    have
    \[
    d_2^{\,2} = |bc|^2 - |{b^\ast}{c^\ast}|^2 \ge 4- |{b^\ast}{c^\ast}|^2
    \quad \hbox{and} \quad d_3^{\,2} = |cd|^2 - |{c^\ast}{d^\ast}|^2 \ge
    4- |{c^\ast}{d^\ast}|^2 = 4 - |{c^\ast}{a^\ast}|^2.
    \]
    Using $|a^\ast d^\ast|^{2} = 0$ and $|a^\ast b^\ast|^{2} = 4$, we can bound
    \begin{align*}
      \Delta + d_2^{\,2}+d_3^{\,2} +|a^\ast d^\ast|^{2}-|a^\ast b^\ast|^2 & \geq
      \Delta + 4-|{b^\ast}{c^\ast}|^2 + 4- |{c^\ast}{a^\ast}|^{2} - 4 =
      \Delta + 4-|{b^\ast}{c^\ast}|^2 - |{c^\ast}{a^\ast}|^{2} \\
      \Delta + d_1^{\,2}+d_3^{\,2} +|a^\ast d^\ast|^{2}-|b^\ast c^\ast|^2 & \geq
      \Delta + d_1^{\,2}+4- |{c^\ast}{a^\ast}|^2 -|b^\ast c^\ast|^2 \\
      \Delta + d_1^{\,2}+d_2^{\,2} +|a^\ast d^\ast|^{2}-|c^\ast d^\ast|^2 & \geq
      \Delta + d_1^{\,2}+ 4- |{b^\ast}{c^\ast}|^2 -|c^\ast a^\ast|^2
    \end{align*}
    Since $c^\ast$ lies in the disk of diameter~$a^\ast b^\ast$, the
    triangle $a^\ast b^\ast c^\ast$ is right or obtuse, and
    \[
    |a^\ast c^\ast|^2+|c^\ast b^\ast|^2 \le |a^\ast b^\ast|^2 = 4.
    \]
    This implies Inequalities~\eqref{i:c1}--\eqref{i:c3}. The case
    where the three balls are $\{A,C,D\}$ is symmetric.
  \end{itemize}

  \medskip

  It remains to handle the case where no three balls in $\{A,B,C,D\}$
  pin $\ell$.  This implies that $\ell$ is tangent to all four balls,
  and the points $a^\ast, b^\ast, c^\ast, d^\ast$ lie on the unit circle
  around $\ell^\ast=\ell \cap \ell^{\perp}$ in~$\ell^{\perp}$.  We let
  $\alpha$ be the angle made by $a^\ast$ and $b^\ast$ at $\ell ^\ast$,
  $\beta$ the angle made by $b^\ast$ and $c^\ast$ at $\ell ^\ast$, and
  $\gamma$ the angle made by $c^\ast$ and $d^\ast$ at $\ell ^\ast$, see
  Fig.~\ref{fig:LemmaDistanceABCD}.  All the angles are measured
  counterclockwise (even if they are larger than $\pi$) so that the
  angle made by $a^\ast$ and $d^\ast$ is the same as $\alpha+\beta+\gamma$
  modulo $2\pi$.

  We define a function~$g$ by $g(\phi) = \sqrt{2 + 2\cos\phi}$ for
  $\phi \in {\mathbb{R}}$. We claim that for any three angles $x, y, z$ we have
  \begin{equation}
    \label{i:g}
    g(x + y + z) \le g(x) + g(y) + g(z),
  \end{equation}
  and that the inequality is strict unless two of $x,y,z$ are equal to
  $\pi$ modulo $2\pi$. Indeed, let $f(\phi)=\sqrt{2 - 2\cos\phi} =
  g(\pi-\phi)$ and observe that $f(\phi)$ is the distance between two
  points on the unit circle that make an angle of $\phi$ at the center
  of the unit circle.  The triangle inequality immediately implies
  that for any two angles $\phi$ and $\theta$ we have $f(\phi+\theta)
  \leq f(\phi) + f(\theta)$, where equality holds only if $\phi$ or
  $\theta$ is equal to~$0$ modulo~$2\pi$. Thus, for any three angles
  $x, y, z$ we have
  \begin{align*}
  f\big(\pi - (x + y + z)\big) & = f\big(3\pi - (x + y + z)\big) \\
  & = f\big((\pi - x) + (\pi - y) + (\pi - z)\big)
  \leq f(\pi - x) + f(\pi - y) + f(\pi -  z),
  \end{align*}
  and Inequality~\eqref{i:g} follows, with equality only if two of
  $x,y,z$ are equal to $\pi$ modulo~$2\pi$.

  Consider the angles~$\alpha$, $\beta$, and~$\gamma$.  If $\alpha =
  \beta = \pi$, then by Lemma~\ref{lem:tri-tang}, $\{A, B, C\}$
  already pin~$\ell$, a contradiction.  If $\beta = \gamma = \pi$,
  then $\{B, C, D\}$ already pin~$\ell$, again a contradiction.
  We thus have
  \begin{align}
    \label{eq:g}
    g(\alpha + \beta + \gamma) & \leq g(\alpha) + g(\beta) + g(\gamma),
  \end{align}
  and the inequality is strict unless $\alpha = \gamma = \pi$. In this
  case\footnote{Theorem~\ref{the:min4pinning} actually implies that in
    this case $\ell$ is not pinned at all, the argument here keeps the
    proof self-contained.} $|a^\ast b^\ast| = |c^\ast d^\ast| =
  2$, and therefore $|ab| > 2$ and~$|cd| > 2$.

  We observe that
  \begin{align*}
    |ab|^2 & = d_1^{\,2} + 2 - 2\cos\alpha = d_{1}^{2} + 4 -
    g(\alpha)^{2},\\
    |bc|^2 & = d_2^{\,2} + 2 - 2\cos\beta = d_{2}^{2} + 4 -
    g(\beta)^{2},\\
    |cd|^2 & = d_3^{\,2} + 2 - 2\cos\gamma = d_{3}^{2} + 4 -
    g(\gamma)^{2}.
  \end{align*}
  Since the balls are non-overlapping, this implies $d_{1} \geq
  g(\alpha)$, $d_{2} \geq g(\beta)$, and $d_{3} \geq g(\gamma)$.  In
  particular, when Inequality~(\ref{eq:g}) is not strict, then $d_{1}
  > g(\alpha)$ and $d_{3} > g(\gamma)$.

  Recall that $\Delta = 2(d_{1}d_{2} + d_{1}d_{3} + d_{2}d_{3})$ and
  set $\Delta' = 2\big(g(\alpha)g(\beta) + g(\alpha)g(\gamma) +
  g(\beta)g(\gamma)\big) \leq \Delta$.
  We can write
  \begin{align*}
    |ad|^2 & = (d_{1} + d_{2} + d_{3})^{2} + 4 - g(\alpha + \beta + \gamma)^{2} \\
    & = d_{1}^{\,2} + d_{2}^{\,2} + d_{3}^{\,2} +
    \Delta + 4 - g(\alpha+\beta+\gamma)^{2} \\
    & \geq d_{1}^{\,2} + d_{2}^{\,2} + d_{3}^{\,2} +
    \Delta' + 4 - g(\alpha+\beta+\gamma)^{2}.
  \end{align*}
  We can now complete the proof.
  \begin{align}
    |ad|^2 & \geq d_{1}^{\,2} + d_{2}^{\,2} + d_{3}^{\,2} +
    \Delta' + 4 - g(\alpha+\beta+\gamma)^{2} \nonumber\\
    & \geq d_{1}^{\,2} + g(\beta)^{2} + g(\gamma)^{2} +
    \Delta' + 4 - g(\alpha+\beta+\gamma)^{2}
    \label{eq:g-2} \\
    & = \big(g(\alpha) + g(\beta) + g(\gamma)\big)^{2}
    - g(\alpha+\beta+\gamma)^{2}  + d_{1}^{\,2} + 4 - g(\alpha)^{2} \nonumber\\
    & \geq d_{1}^{\,2} + 4 - g(\alpha)^{2} = |ab|^2
    \label{eq:g-3}
  \end{align}
  If Inequality~(\ref{eq:g-3}) is not strict, then
  Inequality~(\ref{eq:g-2}) is strict, and so $|ad| > |ab|$.
  \begin{align}
    |ad|^2 & \geq d_{1}^{\,2} + d_{2}^{\,2} + d_{3}^{\,2} +
    \Delta' + 4 - g(\alpha+\beta+\gamma)^{2} \nonumber\\
    & \geq g(\alpha)^{2} + d_{2}^{\,2} + g(\gamma)^{2} +
    \Delta' + 4 - g(\alpha+\beta+\gamma)^{2} \label{eq:g-4} \\
    & = \big(g(\alpha) + g(\beta) + g(\gamma)\big)^{2} -
    g(\alpha+\beta+\gamma)^{2}  + d_{2}^{\,2} + 4 - g(\beta)^{2}
    \nonumber \\
    & \geq d_{2}^{\,2} + 4 - g(\beta)^{2} = |bc|^2
    \label{eq:g-5}
  \end{align}
  Again, if Inequality~(\ref{eq:g-5}) is not strict, then
  Inequality~(\ref{eq:g-4}) is, and we have $|ad| > |bc|$.
  By symmetry, we also obtain $|ad|>|cd|$.
\end{proof}

\section{Conjectures on four unit balls and two lines}
\label{sec:conjecture}

We conjecture that the geometric permutations $\p ABCD.$ and $\p
ACDB.$ are incompatible for non-overlapping unit balls:
\begin{conjecture}
  \label{conj:abcd-acdb}
  There is no set of four non-overlapping unit balls in~${\mathbb{R}}^3$
  admitting the geometric permutations $\p ABCD.$ and $\p ACDB.$.
\end{conjecture}

\noindent
As we have seen in the proof of Theorem~\ref{the:main} in
Section~\ref{sec:at-most-three}, Conjecture~\ref{conj:abcd-acdb} would
imply that a family of at least four non-overlapping unit balls
in~${\mathbb{R}}^{d}$ has at most two geometric permutations, settling our
question entirely. In this section, we study what a counter-example to
our conjecture would look like.

\bigskip

We first employ the shrinking technique to obtain a configuration
where both line transversals are pinned.

\begin{lemma}
  \label{lem:shrink-conj}
  If Conjecture~\ref{conj:abcd-acdb} is false then there exist four
  non-overlapping unit balls in~${\mathbb{R}}^3$ that pin two lines realizing
  the geometric permutations $\p ABCD.$ and $\p ACDB.$.
\end{lemma}
\begin{proof}
  Consider four non-overlapping unit balls~${\mathfrak{F}} = \{A, B, C, D\}$
  in~${\mathbb{R}}^{3}$ that admit line transversals with orders~$\p ABCD.$
  and~$\p ACDB.$.  We uniformly shrink the four balls about their
  centers.  By Lemma~\ref{lem:shrinking}, we will reach a radius~$t_1
  > 0$ where the transversal~$\sigma_1$ for one of the two orders is
  pinned, while a transversal for the other order still exists.  For
  each ball $X \in {\mathfrak{F}}$, we pick a point $x_0 \in X \cap \sigma_1$.  We
  continue shrinking the balls, but now we shrink $X$ with homothety
  center~$x_{0}$.  By Lemma~\ref{lem:shrinking}, we will reach a
  radius~$t_2 > 0$ where the transversal~$\sigma_2$ for the second
  order is pinned.  Since the points~$x_0$ lie in the shrunken balls,
  $\sigma_1$ is still a transversal, and since the balls of
  radius~$t_2$ lie inside the balls of radius~$t_1$, $\sigma_1$ is
  still pinned.  By scaling the balls back to unit radius, we obtain
  the configuration announced by the lemma.
\end{proof}

\subsection{Configurations with two pinned transversals}

In this section we restrict the geometry of configurations as in
Lemma~\ref{lem:shrink-conj}. We start with some geometric
preliminaries. Throughout this section we will be dealing with
families of at most four balls.

\begin{lemma}
  \label{lem:xyz-angle}
  If three non-overlapping unit balls $X, Y$ and $Z$ in ${\mathbb{R}}^{3}$ admit
  a line transversal with order~$\p XYZ.$, then the angles
  $\angle(yxz)$ and $\angle(xzy)$ are acute.
\end{lemma}
\begin{proof}
  Using Lemma~\ref{lem:shrinking}, we shrink the balls uniformly around
  their centers until a line transversal $\ell$ with order~$\p XYZ.$
  is pinned. By Lemma~\ref{lem:tri-tang}, $\ell$ is parallel to~$xz$
  and $y$ lies inbetween $x$ and $y$ in the projection on~$\ell$,
  implying the claim.
\end{proof}

\begin{lemma}
  \label{lem:xyz,xzy}
  If three non-overlapping unit balls $X, Y$ and $Z$ in ${\mathbb{R}}^{3}$ admit
  two line transversals with the orders $\p XYZ.$ and $\p XZY.$, then
  the triangle $\triangle xyz$ is acute and $|yz|<2\sqrt{2}$.
\end{lemma}
\begin{proof}
  Lemma~\ref{lem:xyz-angle} implies that $\triangle xyz$ is acute. To
  prove the last statement, we shrink the balls uniformly around their
  centers.  By Lemma~\ref{lem:shrinking}, there exists $t > 0$ such that
  without loss of generality the following holds: $X$, $Y$, $Z$ are
  now balls with radius~$t \leq 1$ pinning a line transversal~$\ell$
  with order~$\p XYZ.$, and the balls have a second line
  transversal~$\ell'$ with order~$\p XZY.$.  By
  Lemma~\ref{lem:tri-tang}, $\ell$ lies in the plane containing~$x, y,
  z$, and separates $y$ from $x$ and~$z$. Let $x', y', z'$ be the
  projection of $x, y, z$ onto~$\ell$, they appear in this order
  along~$\ell$. If $|yz| \geq 2\sqrt{2}$, then $|y'z'| \geq 2$, and
  there is a plane orthogonal to~$\ell$ that separates $Z$ on one side
  from $X$ and~$Y$ on the other side. But that contradicts the
  existence of~$\ell'$.
\end{proof}

We can now state our restriction on a possible counter-example of
Conjecture~\ref{conj:abcd-acdb}.
\begin{theorem}
  \label{the:double-pinning}
  If four non-overlapping unit balls $\{A, B, C, D\}$ in ${\mathbb{R}}^3$ pin
  two lines with the geometric permutations $\p ABCD.$ and~$\p ACDB.$,
  then these lines are not pinned by a proper subset of the balls.
\end{theorem}
\begin{proof}
  Consider four non-overlapping unit balls $\{A,B,C,D\}$ in ${\mathbb{R}}^3$
  that pin two lines $\sigma_1$ and $\sigma_2$ realizing,
  respectively, the geometric permutations $\p ABCD.$ and~$\p ACDB.$.
  Let us first remark that
  \begin{equation}\label{eq:abs1}
    \pi/4 < \angle(\vc{ab}, \vc{\sigma_1}) < \pi/2
  \end{equation}
  Indeed, since $\sigma_1$ meets $A$ before $B$, we have
  $\angle(\vc{ab}, \vc{\sigma_1}) < \pi/2$. Moreover, if
  $\angle(\vc{ab}, \vc{\sigma_1}) \leq \pi/4$, since
  Lemma~\ref{lem:angle} yields that $\angle(\vc{\sigma_1}, \vc{bd})<
  \pi/4$, we would have $\angle(\vc{ab},\vc{bd}) \leq
  \angle(\vc{ab},\vc{\sigma_1}) + \angle(\vc{\sigma_1}, \vc{bd}) <
  \pi/2$, a contradiction with $\triangle abd$ being acute by
  Lemma~\ref{lem:xyz,xzy}. Let us also remark that
  \begin{equation}\label{eq:ab}
    |ab| < 2\sqrt{2}
  \end{equation}
  as otherwise any line meeting~$A$ before~$B$ would make an angle less
  than~$\pi/4$ with~$\vc{ab}$, contradicting
  Equation~\eqref{eq:abs1}.

  \begin{figure}[ht]
    \centerline{\includegraphics{double-pinning}}
    \caption{Proof of Theorem~\ref{the:double-pinning}}
    \label{fig:double-pinning}
  \end{figure}

  \medskip

  Let us assume for a contradiction that~$\sigma_2$ is pinned by three
  of the four balls. We first remark that~$\sigma_2$ can only be
  pinned by $A$, $D$, $B$ (refer to Fig.~\ref{fig:double-pinning}):
  %
  \begin{denseitems}
  \item If $\sigma_2$ is pinned by $C$, $D$, and~$B$, then by
    Lemma~\ref{lem:tri-tang}, $\sigma_2$ lies in the plane of $bcd$
    and is parallel to~$cb$.  Since $\sigma_2$ meets~$A$ before~$C$,
    the bisecting plane of $C$ and~$B$ separates $B$ from $A$ and $C$,
    a contradiction to the existence of~$\sigma_1$.

  \item If $\sigma_2$ is pinned by $A$, $C$, $D$, then it is the only
    line meeting the three balls in this order.  But then $\sigma_2 =
    \sigma_1$, a contradiction.

  \item If $\sigma_2$ is pinned by $A$, $C$, $B$, then by
    Lemma~\ref{lem:tri-tang} it lies in the plane of $acb$ and is
    parallel to~$ab$. Lemma~\ref{lem:xyz,xzy} implies that $|bc| <
    2\sqrt{2}$ and so $\angle(\vc{cb}, \vc{\sigma_2}) > \pi/4$.
    Lemma~\ref{lem:angle} yields, however, that $\angle(\vc{cb},
    \vc{\sigma_2}) < \pi/4$, a contradiction.
  \end{denseitems}
  %
  So assume that $\sigma_2$ is pinned by $A$, $D$, and $B$. By
  Lemma~\ref{lem:tri-tang}, $\sigma_2$ lies in the plane of $adb$ and
  is parallel to~$\vc{ab}$. Thus, $\angle{(\vc{ab},\vc{cb})} =
  \angle{(\vc{\sigma_2},\vc{cb})} < \pi/4$ by Lemma~\ref{lem:angle}.
  It follows that $c$ is contained in $\mathfrak{C}(b,\vc{ba},\pi/4)$, where
  $\mathfrak{C}(u, \vec{v}, \alpha)$ denotes the cone of all points~$p$ such
  that $\angle(\vc{up}, \vec{v}) \leq \alpha$.  Since $\sigma_1$
  intersects $C$ after $A$ and~$B$, the center~$c$ must lie in the
  cone with apex~$a$ spanned by the ball of radius~$2$ and center~$b$.
  With $\gamma = \arcsin(2/ab)$, this cone is $\mathfrak{C}(a, \vc{ab},
  \gamma)$. By Equation~\eqref{eq:ab}, $2 \leq |ab|<2\sqrt{2}$, and so
  $\pi/4 < \gamma \leq \pi/2$.  Let $p$ be a point in the plane
  containing $a$, $b$ and $c$ and such that $\angle(bap) = \gamma$ and
  $\angle(abp) = \pi/4$ (see Fig.~\ref{fig:ADB}). Notice that $c$ lies
  in the triangle~$\triangle abp$. We claim that any point in this
  triangle is at distance less than $2$ from $a$ or $b$. In particular,
  there is no way to place $c$ so as to make $A$, $B$, $C$
  non-overlapping unit balls and $\sigma_2$ cannot be pinned by $A$,
  $D$, and $B$, or, more generally, by three of the balls.

  \begin{figure}[ht]
    \centerline{\includegraphics{FigADB}}
    \caption{$\triangle abp$ for three values of~$\gamma$.}
    \label{fig:ADB}
  \end{figure}

  It remains to prove the claim on~$\triangle abp$. Since $\gamma >
  \pi/4$ we have $|bp| > 2$.  Let $q$ be the point on the segment $bp$
  with~$|bq| = 2$.  Since the segment~$ap$ touches the circle of
  radius~$2$ around~$b$, we have $\angle(aqp) > \pi/2$, and so $|aq| <
  |ap|$. We claim that $|ap| < 2$. Indeed, the law of sines gives
  $|ab|/\sin(3\pi/4-\gamma) = |ap|/\sin(\pi/4)$, and so
  \[
  |ap| = \frac{\sin(\pi/4)}{\sin(\pi/4 + \gamma)} |ab|
  = \frac{\frac 12 \sqrt{2}}{\sin{(\pi/4+\gamma)}}
  \frac{2}{\sin{\gamma}}
  = \frac{\sqrt{2}}{\sin(\pi/4+\gamma)\sin \gamma}
  = \frac{2}{\sin \gamma(\sin \gamma + \cos \gamma)}
  \]
  Define $f(x) = \sin x(\sin x + \cos x) = \sin^{2}x + \frac 12 \sin
  2x$.  Since $f'(x) = \sin 2x + \cos 2x$, the function~$f$ is
  (strictly) increasing from $x = 0$ to $x = 3\pi/8$, and (strictly)
  decreasing from $x = 3\pi/8$ to $x = 7\pi/8$. Since $f(\pi/4) =
  f(\pi/2) = 1$, it follows that $f(x) > 1$ for $\pi/4 < x < \pi/2$;
  this proves our claim that $|ap| < 2$. Now, if a point~$u \in \triangle
  apb$ lies to the left of the vertical line through~$q$, then it is
  at distance less than~$2$ from~$a$, if $u$ lies to the right of~$q$
  then it has distance less than~$2$ from~$b$.

  \medskip

  It follows that $\sigma_2$ cannot be pinned by any three of the
  balls, and is tangent to all four. We now assume, for a
  contradiction, that $\sigma_1$ is pinned by three of the balls.
  Again, we easily dismiss three of the cases:
  %
  \begin{denseitems}
  \item If $\sigma_1$ is pinned by $B$, $C$, and~$D$, then by
    Lemma~\ref{lem:tri-tang} $\sigma_1$ lies in the plane of $bcd$ and
    is parallel to~$bd$.  Since $\sigma_1$ meets~$A$ before~$B$, the
    bisecting plane of $B$ and~$D$ separates $D$ from $A$ and $B$, a
    contradiction to the existence of~$\sigma_2$.

  \item If $\sigma_1$ is pinned by $A$, $C$, $D$, then $\sigma_2 =
    \sigma_1$, a contradiction.

  \item If $\sigma_1$ is pinned by $A$, $B$, $D$, then it lies in the
    plane of $abd$ and is parallel to~$ad$.  Since $|bd| < 2\sqrt{2}$
    by Lemma~\ref{lem:xyz,xzy}, we have $\angle(\vc{bd},
    \vc{\sigma_1}) > \pi/4$, contradicting Lemma~\ref{lem:angle}.
  \end{denseitems}
  \begin{figure}[ht]
    \centerline{\includegraphics{FigABC}}
    \caption{When $\sigma_1$ is pinned by $A$, $B$, $C$}
    \label{fig:ABC}
  \end{figure}
  So $\sigma_1$ must be pinned by $A$, $B$, and~$C$ and lie in the
  plane spanned by~$abc$, see Fig.~\ref{fig:ABC}. Since $\sigma_1$ is
  a transversal, the center~$d$ of ball~$D$ must lie inside the
  cylinder~${\mathcal{C}}$ of radius~one with axis~$\sigma_1$.  Since $\sigma_2$
  meets $A$, $D$, and $B$ in this order, $d$ must also lie in the
  cylinder of radius $2$ with axis~$ab$. Since $\angle(abd) < \pi/2$
  by Lemma~\ref{lem:xyz,xzy}, $d$ lies above the plane orthogonal
  to~$ab$ through~$b$, and since $\sigma_1$ meets $D$ after $C$, $d$
  lies to the right of the plane orthogonal to $\sigma_1$ through~$c$.
  In the projection on the $abc$-plane, this restricts $d$ to the
  shaded area in Fig.~\ref{fig:ABC}.  Let~$p$ be the rightmost point
  of this feasible region for~$d$, that is, the point in the
  $abc$-plane such that $|bp| = 2$ and $\angle(abp) = \pi/2$. For any
  point $u$, let $\sigma^{\perp}_1(u)$ be the plane orthogonal to $\sigma_1$
  passing through~$u$. The center~$d$ lies in the cylinder~${\mathcal{C}}$
  between the planes~$\sigma^{\perp}_1(c)$ and~$\sigma^{\perp}_1(p)$.

  We will now show that any point~$d$ in this cylinder has distance
  less than two from $b$ or~$c$, and so $D$ cannot be non-overlapping
  with $B$ and~$C$, a contradiction.  Clearly it suffices to show this
  for the disk~$S$ of radius one around~$\sigma_1$ in the
  plane~$\sigma^{\perp}_1(p)$.  Let $q$ be a point on the boundary of~$S$
  with $|bq|=2$ (see right side of Fig.~\ref{fig:ABC}). It suffices to
  show that $|cq| < 2$. Let $\alpha = \angle(\vc{ab},
  \vc{\sigma_1})$.  By Equation~\eqref{eq:abs1},
  $\pi/4<\alpha<\pi/2$.  Since
  $|qb^\ast|=|pb^\ast|=2\sin{(\pi/2-\alpha)}=2\cos{\alpha}$ and $\angle(b^\ast
  \!qc^\ast)=\pi/2$, $|qc^\ast|^2=4-4\cos^2{\alpha}= 4\sin^2{\alpha}$.  We
  have $|cc^\ast| =|ac^\ast| -|ac|= |ab|\cos{\alpha}+|bb^{^\ast}|-|ac|$, and
  with $|ab|=2/\sin{\alpha}$, $|bb^\ast| = 2\cos(\pi/2-\alpha) =
  2\sin\alpha$, and $|ac|\geq 2$ this implies $|cc^\ast| \leq
  2\cot{\alpha} + 2\sin{\alpha} - 2$.  Thus $|cq|^2 = |qc^\ast|^2 +
  |cc^\ast|^2 \leq 4 \sin^{2}{\alpha} + (2\cot{\alpha} + 2\sin{\alpha}
  -2)^2$.

  We have
  \begin{align}
    \sin^{2}{x} + \big(\cot {x} + \sin{x} - 1 \big)^2 - 1
    & =\big(\cot{x}+\sin{x}-1\big)^2-\cos^{2} x \nonumber\\
    & = \big(\cot{x}+\sin{x}+\cos{x}-1 \big)
    \big(\cot{x}+\sin{x}-\cos{x}-1 \big) \nonumber \\
    & = \big(\cot{x}+\sin{x}+\cos{x}-1 \big)
    (\cot x - 1)(1 - \sin x).
    \label{eq:abc1}
  \end{align}
  On the interval $\pi/4 <x< \pi/2$, we have $\sqrt{2}/2 < \sin x <
  1$, $\sqrt{2}/2 > \cos x > 0$, and $1 > \cot x > 0$, implying $\cot
  x -1 < 0$ and $1 - \sin x > 0$.  Furthermore
  \[
  \sin{x}+\cos{x}
  =\sqrt{2}\sin{(\frac{\pi}{4}+x)} > 1 \qquad \text{for}\; \pi/4 < x < \pi/2,
  \]
  and so the first term in Equation~(\ref{eq:abc1}) is positive. It
  follows that $\sin^{2} x + (\cot x + \sin x - 1)^{2} < 1$.  This
  implies $|cq|^2 < 4$, and we arrived at the contradiction for this
  final case.
\end{proof}

\subsection{Minimal pinnings by four balls}

By Lemma~\ref{lem:shrink-conj} and Theorem~\ref{the:double-pinning},
if Conjecture~\ref{conj:abcd-acdb} is false then there exist four
non-overlapping unit balls with two transversals with the orders~$\p
ABCD.$ and~$\p ACDB.$ that are both pinned by the four balls but no
three of them. We now analyze the geometry of such \emph{minimal
  pinnings} by four balls (Theorem~\ref{the:min4pinning}) and derive a
statement equivalent to Conjecture~\ref{conj:abcd-acdb} but more
restrictive (Conjecture~\ref{conj:double-pinning}).
\begin{figure}[ht]
  \centerline{\includegraphics{FigMiniFourPinning}}
  \caption{An alternating hyperboloidal configuration\label{fig:althyp}}
\end{figure}

If $X$ is a ball tangent to a line $\ell$, the
\emph{ridge}~$r_\ell(X)$ of $X$ with respect to $\ell$ is the line
tangent to $X$ and perpendicular to $\ell$ in $X \cap \ell$. We say
that four or more lines are in \emph{hyperboloidal configuration} if
they are all contained in the same family of rulings of a hyperbolic
paraboloid or a hyperboloid of one sheet (see~\cite{hcv} for a
classical discussion of such configurations). An \emph{alternating
  hyperboloidal configuration} is a pair $({\mathfrak{F}},\ell)$ where ${\mathfrak{F}}$ is a
family of four unit balls balls and $\ell$ is a line tangent to every
member of ${\mathfrak{F}}$ satisfying the two following conditions:
\begin{denseitems}
\item[(i)] the ridges $\{r_\ell(X) \mid X \in {\mathfrak{F}}\}$ are in
  hyperboloidal configuration, witnessed by a hyperbolic
  paraboloid~$\mathcal{H}$,
\item[(ii)] the normals to $\mathcal{H}$ at its tangency point with
  the balls of ${\mathfrak{F}}$, directed towards the center of that ball and
  ordered along $\ell$, point to alternating sides of $\mathcal{H}$.
\end{denseitems}
Since in an alternating hyperboloidal configuration the ridges are all
perpendicular to $\ell$, and therefore parallel to a common plane, the
quadric they span can only be a hyperbolic paraboloid. Condition~(i)
then forces the line~$\ell$ to intersect~$\mathcal{H}$ in at least
four points and therefore $\ell \subset \mathcal{H}$. This in turn
implies that every ball $X \in {\mathfrak{F}}$ is tangent to $\mathcal{H}$ in $X
\cap \ell$ as the tangent plane to both $X$ and $\mathcal{H}$ in $X
\cap \ell$ contains the two lines (but $X$ does not need to intersect
$\mathcal{H}$ in a single point).  That and the fact that a hyperbolic
paraboloid~ separates~${\mathbb{R}}^3$ into two connected components makes
condition~(ii) well-defined.

\begin{theorem}
  \label{the:min4pinning}
  If a family ${\mathfrak{F}}$ of $4$ non-overlapping unit balls in ${\mathbb{R}}^3$
  minimally pins a line $\ell$ then $({\mathfrak{F}},\ell)$ is an alternating
  hyperboloidal configuration.
\end{theorem}
\begin{proof}
  Let $\sigma$ be a line minimally pinned by $\p ABCD.$ in that order.
  We assume that $\sigma$ coincides with the $z$-axis. The center of
  ball $X$ is denoted by $x$ and its contact point with $\sigma$ is
  denoted by $x'$. We parameterize the space of lines by ${\mathbb{R}}^4$ using
  the coordinates of the intersections with the planes $z=0$ and
  $z=1$: the parameters $(u_1,u_2,u_3,u_4)$ corresponds to the line
  through $(u_1,u_2,0)$ and $(u_3,u_4,1)$. The lines lying in a plane
  with constant $z$ are not represented, but this will not be an
  issue. The point $(0,0,0,0)$ corresponds to~$\sigma$.

  To every ball $X \in \{A,B,C,D\}$ we associate the \emph{screen}
  $S_\sigma(X)$ that is the intersection of the closed halfspace
  bounded by the tangent plane to $X$ in $x'$ that contains $X$, and
  the plane perpendicular to $\sigma$ in $x'$. The screen
  $S_\sigma(X)$ is a halfplane that lies in a plane perpendicular to
  $\sigma$ and is bounded, in that plane, by the ridge~$r_\sigma(X)$.
  Now, the line transversals to $S_\sigma(X)$ form, in our ${\mathbb{R}}^4$, a
  halfspace $H(X)$ bounded by a hyperplane through the
  origin~\cite[pp.~4--5]{aronov2011lines}. We let $n(X)$ denote the outer
  normal of $H(X)$ and observe that the boundary of $H(X)$ is the set
  of lines intersecting the ridge $r_\sigma(X)$.

  Now let $I=H(A) \cap H(B) \cap H(C) \cap H(D)$. A necessary
  condition for the balls $\{A,B,C,D\}$ to pin $\sigma$ is that $I$
  has empty interior~\cite[Lemma~9]{cheong2012lower}. This implies
  that the family of normals $\{n(A), n(B), n(C), n(D)\}$ is linearly
  dependant (since, in ${\mathbb{R}}^4$, four halfspaces with linearly
  independent normals intersect with non-empty interior). It could be
  that two, three or all four vectors are \emph{minimally} linearly
  dependent. Geometric interpretations of these situations were given
  in~\cite[Lemma~15]{aronov2011lines}:

  \begin{denseitems}
  \item If two normals are dependent then the two corresponding ridges
    are equal. This cannot happen for non-overlapping balls.
  \item If three normals are dependent, then it must be that the three
    corresponding ridges are either coplanar with or concurrent on
    $\sigma$. Concurrency is again ruled out for non-overlapping
    balls, and we rule out coplanarity in the next paragraph.
  \item If no three normals are dependent then the four ridges
    are in hyperboloidal configuration (the other case with concurrent
    ridges can again not occur in our situation).
  \end{denseitems}

  Let us observe that the case where three normals are linearly
  dependent cannot correspond to a minimal pinning of $\sigma$ by the
  four balls. As mentioned, the three corresponding ridges must lie,
  together with $\sigma$, in some plane $\Pi$. Let us denote them
  $r_1$, $r_2$ and $r_3$ in the order in which $\sigma$ meets them,
  and let $X_i$ be the ball corresponding to~$r_i$. Since a triple of
  balls does not suffice to pin~$\sigma$, $\Pi$ does not separate
  $X_2$ from $X_1$ and~$X_3$; by symmetry we can assume that either
  all three balls are on the same side of $\Pi$ or $\Pi$ separates
  $X_1$ and $X_2$ from $X_3$.  Since $I$ has empty interior, the
  fourth ridge also lies in the plane~$\Pi$. Then, either three of the
  balls pin~$\sigma$ or all four do not, a contradiction.

  We must therefore be in the situation where the four ridges are in
  hyperboloidal configuration. Observe that since any three normals
  are linearly independent, the intersection $I$ must be exactly a
  line in~${\mathbb{R}}^{4}$.  That line $I \subset {\mathbb{R}}^4$ corresponds to the set
  of lines intersecting all four ridges. Let $Q$ denote the quadric
  formed by the union (in ${\mathbb{R}}^3$) of these line transversals to the
  four ridges. Let us orient $Q$, that is choose an outward normal
  (defined continuously over all of $Q$).  There are two connected
  components in ${\mathbb{R}}^3\setminus Q$, which we call \emph{sides}
  of~$Q$. As we move a point $p$ along~$\sigma$, the outward normal
  of~$Q$ in~$p$ keeps pointing into the same side but rotates
  continuously around $\sigma$; as $p$ ranges over all of $\sigma$,
  that normal turns by a total angle of~$\pi$.  For each ball $X \in
  \{A,B,C,D\}$, $S(X)\setminus r(X)$ is contained either in the
  positive side or in the negative side of~$Q$.

  Now consider the orthogonal projection of the four screens on the
  plane $z=0$. The circular order in which the projections of the
  ridges appear matches the order in which $\sigma$ meets the screens;
  indeed, the projection of ridge $r(X)$ is simply the trace in $z=0$
  of the plane tangent to~$Q$ in $x'$, and we observed that the
  tangent plane turns continuously, and by a total angle of~$\pi$, as
  the contact point ranges over all of~$\sigma$.  Moreover, (the
  relative interiors of) any two consecutive screens are contained in
  opposite sides of~$Q$ as otherwise we can perturb the plane $z=0$ so
  that the projections of the four screens intersect with non-empty
  interior, and $I$ cannot have empty interior. Altogether, this
  proves that a quadruple of balls minimally pinning a line must
  form an alternating hyperboloidal configuration.
\end{proof}

Theorem~\ref{the:min4pinning}, Lemma~\ref{lem:shrink-conj}, and
Theorem~\ref{the:double-pinning} imply that
Conjecture~\ref{conj:abcd-acdb} can be reformulated in the following
form:
\begin{conjecture}
  \label{conj:double-pinning}
  There is no set ${\mathfrak{F}}$ of four non-overlapping unit balls in~${\mathbb{R}}^3$
  with two line transversals $\sigma_1$, $\sigma_2$ that realize the
  geometric permutations $\p ABCD.$ and $\p ACDB.$ and such that
  $({\mathfrak{F}},\sigma_1)$ and $({\mathfrak{F}},\sigma_2)$ are alternating hyperboloidal
  configurations.
\end{conjecture}

\section{Concluding remarks}

Conjectures~\ref{conj:abcd-acdb} and~\ref{conj:double-pinning} can be
expressed by asking whether a system of low-degree polynomial
equations and inequalities in a small number of variables has a
solution (see Appendix~\ref{sec:semialgebraic}).  In principle, such
systems can be solved by computer algebra software based on Gr\"obner
basis computation such as the \textsc{raglib} maple library.
Inequalities are harder to handle than equalities by these solvers, so
this is one reason why Theorem~\ref{the:min4pinning},
Lemma~\ref{lem:shrink-conj}, and Theorem~\ref{the:double-pinning} are
interesting: Conjecture~\ref{conj:double-pinning} replaces most of the
inequalities in Conjecture~\ref{conj:abcd-acdb} by equalities.

Our attempts using algebraic solver software were inconclusive.  These
questions may constitute interesting challenges for the computer
algebra community.

\medskip

We speculate that Theorem~\ref{the:min4pinning} can be turned into an
equivalence, at least when the balls are disjoint. One approach could
be to remark that if $({\mathfrak{F}},\sigma)$ is an alternating hyperboloidal
configuration where $\sigma$ realizes the geometric permutation $\p
ABCD.$, then the direction $\vec{u}$ of $\sigma$ is on the boundary of
the so-called \emph{cone of directions} of transversals to the triples
$\p ABC.$ and $\p BCD.$ (using ideas such as
eg.~\cite[Proposition~3]{borcea2008line}). These cones are strictly
convex in $\vec{u}$, are mutually tangent in $\vec{u}$ (their common
supporting great circle is the set of directions of transversals to
the four ridges) and the alternating property of the configuration
$({\mathfrak{F}},\sigma)$ ensures that this tangency is \emph{external}.  Spelling
out this outline requires non-trivial technical developments, all the
more if one cares for the setting of non-overlapping balls, and is not
needed for our main result of Section~\ref{sec:conjecture}; we thus
leave it to the interested reader to check the validity of this
approach.

\section*{Acknowledgments.}

The experiments with the computer algebra software leading to the
polynomial systems given in Appendix~\ref{sec:semialgebraic} were
conducted in collaboration with Guillaume Moroz, to whom the authors
are very grateful.

redpajama_set_name: RedPajamaArXiv


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text:
KLEMENT GOTTWALDadmin2019-04-29T08:12:23+00:00
KLEMENT GOTTWALD
Czechoslovak communist president Klement Gottwald was born on 23 November 1896 in a small Moravian town Vyškov.
He joined the Social Democratic Party in 1912 and became a charter member of the Communist Party of Czechoslovakia in 1921. He edited party's newspapers in Bratislava until 1926. In 1925 he became a member of the Communist Party of Czechoslovakia's Central Committee and became the Secretary General of the Party in 1927. He was elected to the National Assembly two years later.
During the Second World War he spent seven years in exile in Moscow.
After the War he returned to Czechoslovakia and became the Party Chairman in 1946 and led the party to 38% votes in the free elections. After the Communist coup in Czechoslovakia in 1948 Gottwald became the most powerful man in the country and after the Edvard Beneš's resignation from presidency on 2 June 1948 he became the president of Czechoslovakia on 14 June 1948.
During Gottwald's presidency the farms were collectivised and country's industry was nationalised. During his leadership series of purges were also held including the purge with party's General Secretary Rudolf Slánský and the Foreign Minister Vlado Clementis who were later executed.
Gottwald died from a heart attack on 14 March 1953, five days after Stalin's funeral.

redpajama_set_name: RedPajamaCommonCrawl


Captain Howe's rig for close-reefing topsails, differs from the common rig, by having double topsail yards. The lower topsail yard is trussed to the lower cap, and instead of slings, is supported from below by a crane upon the forward rim of the top.
The yard now is entirely suspended to the cap. The lower topsail, therefore, is the size of the close-reefed sail of the ordinary rig, and sets entirely by the sheets. The upper topsail sets upon the part of the topmast above the cap, and has its foot laced to a jackstay upon the fore-cant of the yard below, so that no wind can escape between the "two topsails". This arrangement of the yards has many advantages. Labour and time are saved in reefing; a ship can be reduced to close-reefed topsails at any time, by lowering the upper topsails, which will then lie becalmed before the lower topsails, remaining perfectly quiet in the roughest weather, and can be furled or not.
In squally weather, then, this rig is invaluable, for whole topsails can be carried to the last moment, and instantly reduced to close-reefed topsails with certainly of action, without the necessity of a man leaving the deck. Its economy in the wear of canvass must also be very great, for the sails are of manageable size, and have neither bunt-lines, reef-tackles, or clue-lines to chafe them.
Any ship with the ordinary rig can adopt the new, by a yard to the cap and cutting the topsails in two; and, if thought proper, enlarge the breadth of the head, so as to spread more canvass on the same length of yards, as there is no room required for reefing outside the brace bands: the reef tackles and blocks will make the braces for the yard at the cap.
It may be observed that in the ordinary rig the trestle-trees are never relieved from the continual heavy pressure of the weight above, until the topsail-yard is on the cap, and they are frequently found defective from this cause alone. In Howe's rig much relief is given by the half-sail and light upper yard at the topmast head; and when both the two topsails are set, if the topsail haulyards are let go, the weight of the upper topsail is no longer resting upon the trestle-trees, as is the case in the old rig.
Again, should a ship lose her three topmasts on a lee shore, blowing hard, by cutting under the lower topsails and courses, which she would have no chance to do under the old rig, especially in cold weather with a Lascar crew, or shorthanded, as many vessels are now sometimes obliged to leave the Colonial ports to sail round Cape Horn.
A ship with this rig is more seaworthy, because she may always be considered as under close-reefed topsails, and may be worked with fewer men than a vessel of the same size having the old rig. It looks rather clumsy in port, and this, we believe, is the principal objection urged against it by those who do not comprehend its advantages at sea. Ships, however, are rigged for service at sea, and not for show in port; that, therefore, which is the most serviceable is certainly the best.
To Determine the Size of a Topsail on Howe's Rig.
1. The hoist of upper-topsail. -- Allow 2 inches for every 3 feet in hoist or measure taken from the spider-hoop, down to the centre of the lower topsail yard (see p. 6). -- 2. The hoist of lower-topsail. -- Subtract 1 ft., for drift and sheeting home, from the distance the lower yard is below the top of the cap. -- 3. Head of upper and lower-topsails. -- Subtract 2 ft. from the hounded lengths of the tow topsail yards. -- 4. Foot of upper and lower-topsails. -- Subtract 3 ft. from the hounded length of the yard at the cap, and the lower-yard.
From Robert Kipping: Elementary Treatise on Sails and Sailmaking with Draughting, and the Centre of Effort of the Sail . . . Together with an Appendix, Comprising the Inventions of Self-Reefing Topsails of Captain Cunningham &etc. and Captain Howe; and Also a Cutting Board or Table, by Mr. Pittard. 5th ed., enl. and greatly improved.
C. Wilson, London, 1860-61 (5th). pp 168-169 [?].
Weale's Rudimentary Series, No. 149. First published in 1847.
Transcribed by by Lars Bruzelius.
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