Drapeability and Λ-Frames

Abstract

In recent years, two quite different tools have been employed to study global properties of arcs in the plane. The first is drapeability, which grew from ideas of J. R. Alexander in early 2000s defining an arc drapeable if it lies in the convex hull of a shorter convex arc. The second is $\mathrm{\Lambda }$-configuration, where an arc travels from one line to another and back. We investigate interrelations between these notions and in the process find drapeability criteria for open arcs, necessary and sufficient drapeability conditions for three-segment z-shaped arcs, and new bounds for the width of non-drapeable arcs.

1. Introduction

Drapeability and $\mathrm{\Lambda }$-configuration were independently introduced around 2005, and the connection between them investigated here is somewhat surprising. Ralph Alexander’s notion of a drapeable arc was developed and elaborated in the 2005 article [1] with intended applications in arc covering as we see in the quote from the article’s introduction: “When investigating covering problems, one quickly realizes that convex arcs are generally more easily dealt with than non-convex arcs… In this article we investigate some of the fundamental properties of drapeability in the hope of better understanding what it means for an arc not to be drapeable”. (Section 5.1 lists some results of [1]).

In arc covering one looks for a set that covers a family of arcs $\mathcal{C}$ (is a cover for $\mathcal{C}$), that is, the set contains a congruent copy of every arc in $\mathcal{C}.$ Such a set is called a worm cover if $\mathcal{C}$ is a family of unit arcs (worms). The shortest arc that does not fit inside a region is its escape path. Survey articles [2,3] describe a variety of worm covers and escape paths. Arc covering belongs to the field of discrete geometry while drapeability belongs to convex geometry. Here is an incomplete list of monographs in these fields [4,5,6,7,8,9].

Definition 1.

A drape $\delta$ is a convex arc lying in a box $a\times b$ whose endpoint segment, stand, is a. It is a drape for a set if its convex hull contains the set (Figure 1). An arc is drapeable if it has a drape of no greater length (Figure 2).

$\mathrm{\Lambda }$-configuration (see Section 5.2) of an arc between parallel lines traveling from one line to the other and back was defined in [14] that confirmed Besicovitch conjecture [15] on the escape path of the regular triangle. Similar configurations were used before [14], for example, in the 2000 theorem of Weinholtz [16] (Th. 3.1) a closed arc in ${\mathbb{R}}^n$ travels between parallel hyperplanes three times as it does in Proposition 8. Now $\mathrm{\Lambda }$-configurations are routinely employed in works establishing covering properties of convex sets, most notably in [12,13] confirming Wetzel’s conjecture on sector $\mathrm{\Pi }$ covering unit arcs. In relation to the problem of Moser [17,18] “Find the convex worm cover of least area”, currently, among all known worm covers, $\mathrm{\Pi }$ has the smallest area (see Section 4 in [3]).

An arc in the article is a parametrized curve $\gamma$ of length $|\gamma |$, which is closed if its endpoints coincide and open otherwise. The convex hull of $\gamma$ is denoted by $\mathcal{H}\left(\gamma \right)$, and $\hslash \left(\gamma \right)$ denotes the boundary curve of $\mathcal{H}\left(\gamma \right)$. Points of $\gamma$ are ordered in the sense of increasing parameter as follows: $\gamma \left(s\right)\prec \gamma \left(t\right)\mathrm{if}s<t.$

Definition 2.

A support angle of $\gamma$, formed by its two non-parallel support lines, has a Λ-triple, $\{{\gamma }_1,{\gamma }_2,{\gamma }_3\}$, of points of $\gamma$ if ${\gamma }_1\prec {\gamma }_2\prec {\gamma }_3$ and ${\gamma }_1,{\gamma }_3$ lie on one ray of the angle while ${\gamma }_2$ lies on the other. Support angles with $\mathrm{\Lambda }$-triples $\angle u,b$ and $\angle b,v$ of size $\alpha \in (0,{180}^{\circ })$ that share a ray, b, form a Λ-frame $\mathcal{F}\left(\alpha \right)$ bounded by base b and sides $u,v$ (Figure 3).

Definition 3.

Let $\gamma$ with a support line ℓ have $\left(\mathrm{proj}\gamma \right)$_ℓ = κη, and let segments $\kappa \xi$ and $\zeta \eta$ with $\gamma$’s points $\xi$ and $\eta$ be orthogonal to ℓ. The following arc

$${\delta }_{\ell }\left(\gamma \right)=\kappa \xi \cup {\hslash }_{\xi \zeta }\left(\gamma \right)\cup \zeta \eta .$$

(1)

is the smallest drape for $\gamma$ with the stand on a given line ℓ. For $\gamma$ with $\mathcal{F}\left(\alpha \right)$ it is called a base drape, if ℓ = b, a side drape if ℓ = u or ℓ = v, and a skew drape if ℓ $\not\supset b,u,v$ (Figure 3).

A frame $\mathcal{F}\left({90}^{\circ }\right)$, pictured in Figure 2b, appeared in [1] before the formal introduction of $\mathrm{\Lambda }$-configuration in [14]. Frames $\mathcal{F}\left({30}^{\circ }\right)$ appeared in [13] (Case 1) and [19]. Frames $\mathcal{F}\left(\alpha \right)$ of seemingly drapeable arcs were observed in Theorem 6 of [20] during the search for $\gamma$’s second support angle with a $\mathrm{\Lambda }$-triple when $\alpha /\pi$ is rational. They suggested that there could exist the widest interval $\mathcal{I}$ for $\alpha$ guaranteeing drapeability of every arc with $\mathcal{F}\left(\alpha \right),\alpha \in \mathcal{I}$. Indeed, such an interval exists and is $[{30}^{\circ },{180}^{\circ })$ for simple arcs. The main result of the article shows that $\mathcal{I}=[{30}^{\circ },{75}^{\circ }]$ (Section 2).

Given $\alpha$, a simple arc might not have a $\mathrm{\Lambda }$-frame $\mathcal{F}\left(\alpha \right)$; instead, we show, in Section 4, that every simple open arc $\gamma$ has a unique $\mathrm{\Lambda }$-frame $\mathcal{F}(\tau ,\sigma )$. The frame carries a fair amount of information on $\gamma$’s drapeability; for example, $\gamma$ is drapeable if $\tau ,\sigma \ge {30}^{\circ }$. For smaller $\tau ,\sigma$ drapeability of $\gamma$ often can be determined with methods of Section 3. Theorem 3 of this section obtains necessary and sufficient conditions for a 3-segment z-shaped arc, the z-arc, to be drapeable, that, in particular, allowed us to raise the lower bound for $\aleph =$ lub{widths of non-drapeable worms} from 4/15 to 0.31. They will also allow us to evaluate ℵ for z-worms (using a calculus software) to confirm or refute its conjectured value about 0.3286.

2. Draping Arcs with Frames $\mathcal{F}\left(\mathbf{\alpha }\right)$

Notation: 

${\gamma }_{P,Q,R,\dots }$ and ${\hslash }_{PQR\dots }$ are subarcs of $\gamma$ and $\hslash \left(\gamma \right)$ through $\gamma$’s points $P,Q,R,\dots$; $PQR\dots$ is a polygonal arc through points $P,Q,R,\dots$, though we allow $PQ$ denote both a line segment and its length; $\left[PQ\right]$ is the line extending $PQ$.

2.1. Lemma 1 and Non-Drapeable Arcs with $\mathcal{F}\left(\alpha \right),\alpha \notin [{30}^{\circ },{75}^{\circ }]$

In an angle $\angle u,b$ of size $\sigma \ge 0$ with points $L,E\in u$, $F\in b$, $\epsilon ={\left(\mathrm{proj}E\right)}_b$, let $LGE$ be the line from L to E reflected in b (Figure 4). Let $\beta =\angle FEL$, $\theta =\angle FLE$, and $\omega =\angle LF,b$. The following lemma tells us when the arcs $LFE$ and $LGE$ are longer than the arc $LE\epsilon$.

Lemma 1.

In Figure 4a,c, the arc $LFE$ is no shorter than the arc $LE\epsilon$ if and only if

$$\tan {\displaystyle \frac{\beta }{2}}+\tan {\displaystyle \frac{\theta }{2}}\ge {\displaystyle \frac{\sin (\beta -\sigma )}{\sin \beta }}.$$

(2)

In Figure 4b, the arc $LGE$ is no shorter than the arc $LE\epsilon$ if and only if

$$2\sin {\displaystyle \frac{\omega +\sigma }{2}}\ge cos{\displaystyle \frac{\omega -\sigma }{2}}.$$

(3)

Proof. 

With $h = \mathrm{dist}(F,u),$ $EF+FL-LE\ge E\epsilon$ is equivalent to

$$h{\displaystyle \frac{1}{\sin \beta }}+h{\displaystyle \frac{1}{\sin \theta }}-\left[h\cot \beta +h\cot \theta \right]\ge h{\displaystyle \frac{\sin (\beta -\sigma )}{\sin \beta }}\iff \left(2\right).$$

When $F=G$, one has $\beta =\omega +\sigma ,\theta =\omega -\sigma ,$ and $\omega ={\displaystyle \frac{\beta +\theta }{2}}$. Thus, (2) becomes

$${\displaystyle \frac{\sin \frac{\beta }{2}cos\frac{\theta }{2}+cos\frac{\beta }{2}\sin \frac{\theta }{2}}{cos\frac{\beta }{2}cos\frac{\theta }{2}}}\ge {\displaystyle \frac{\sin \frac{\beta +\theta }{2}}{2\sin \frac{\beta }{2}cos\frac{\beta }{2}}}\iff \left(3\right).$$

Note that (3) holds for $\sigma \ge {30}^{\circ }$, and so does (2), because $|LE\epsilon | \le |LGE| \le |LFE|.$ □

One can express (3) in several ways. In relation to $LGE$ in Figure 4b, let

$${\omega }_{\sigma }=arctan{\displaystyle \frac{4-5\sin \sigma }{3cos\sigma }},r={\displaystyle \frac{PE}{PL}},r_{\sigma }={\displaystyle \frac{2-4\sin \sigma }{2-\sin \sigma }},$$

(4)

Fact 1

(Proof in Section 5.3). The following are equivalent $|LGE|\ge |LE\epsilon |$ and

$$\tan {\displaystyle \frac{\omega }{2}}\ge {\displaystyle \frac{1-2\tan (\sigma /2)}{2-\tan (\sigma /2)}}\iff \left(3\right)\iff \omega \ge {\omega }_{\sigma }\iff r\ge r_{\sigma }.$$

Corollary 1.

For every $\alpha <{30}^{\circ }$, there exists a non-drapeable arc with $\mathcal{F}\left(\alpha \right)$.

Proof. 

Let $\sigma =\alpha$, and consider a symmetric arc $EGLRTV$ in Figure 5 (right) with $EGL$ being the arc of Lemma 1. Enlarged, $EGL$ is shown in Figure 5 (left) where $E\mu \perp b$, $\mu$ lies on $b^{″}\Vert b$ through L, $J={\left(\mathrm{proj}L\right)}_b$. We chose E so that (1) ${\displaystyle \frac{EP}{LP}}<r_{\sigma }$ as in (4) (no base drape as $|LE\epsilon |>|LGE|$ by Fact 1). (2) $E\mu +GJ>GL$ (no drape with stand on ℓ). (3) $E\eta >2GL$ where $E\eta \perp v$ (no side drape), which can be also achieved with long enough $LR$. □

Corollary 2.

For every $\alpha >{75}^{\circ }$, there exists a non-drapeable arc with $\mathcal{F}\left(\alpha \right)$.

Proof. 

Figure 6a: $▵LBS$ has $v=BL=BS=u$; $\angle v,u<{30}^{\circ }$; $P\in v$ with $\angle LPS<{30}^{\circ }$; $LGE$ is a line from L to $E\in v$ reflected in $PS$; $\epsilon ={\left(\mathrm{proj}E\right)}_{PS}$. We chose E to satisfy 3 conditions that make $\gamma =EGLS$ non-drapeable: (1) ${\displaystyle \frac{PE}{PL}}<r_{\sigma }$ as in (4). By Fact 1, $|SLE\epsilon |>|SLGE|=|\gamma |$. (2) $GS>LS\Rightarrow |SGEL|>|\gamma |$. (3) $GS>GL\Rightarrow |EGSL|>|\gamma |$.

Figure 6b: Segments $M^n{N}^n\Vert LS$ with endpoints on v and u approach $LS$. Arcs ${\gamma }^n=EGL{M}^n{N}^{n}S$ with $\mathcal{F}\left(\alpha \right)=▵LBS$ approach a non-drapeable $\gamma$. For large enough n, all ${\gamma }^n$ are non-drapeable by the Limit Theorem (Proposition 2). □

Lemma 2.

Let an arc γ have a support angle $\angle u,b$ of size σ with vertex P. Let $E\prec F\prec L$ be points of γ with $E,L\in v$, $F\in u$, and $PE<PL$. Let $\epsilon ={\left(\mathrm{proj}E\right)}_b$, $\kappa \eta ={\left(\mathrm{proj}\gamma \right)}_b$$(P\kappa <P\eta ),$ and a point $\xi \in {\gamma }_{A,F}$ be such that $\kappa \xi \perp b$ (Figure 7a,b). Suppose $|LE\epsilon | \le |LFE|$, then the end-arc ${\gamma }_{A,L}$ in Figure 7a,b has the following bound:

$$\kappa \xi +|{\hslash }_{\xi L}\left(\gamma \right)| \le |{\gamma }_{A,L}|.$$

(5)

Proof. 

If $P\epsilon \ge PF$ then $EL<FL$, $\kappa \xi + |{\hslash }_{\xi E}|<|{\gamma }_{A,F}|$, and $FL \le |{\gamma }_{F,L}|$ yield (5). If $P\epsilon <PF$, we use $\kappa \xi + |{\hslash }_{\xi L}\left(\gamma \right)|=\kappa \xi + |{\hslash }_{\xi E}| + EL$ to estimate length of the end-arc ${\gamma }_{A,L}$ as follows.

In Figure 7a, $\kappa \xi +|{\hslash }_{\xi E}|+EL \le |{\hslash }_{\xi E}|+\epsilon E+EL \le |{\hslash }_{\xi E}|+|EFL| \le |{\gamma }_{A,L}|.$

In Figure 7b, $\rho ={\left(\mathrm{proj}\xi \right)}_{\epsilon E}$ implies $\epsilon \rho =\kappa \xi ,\rho F<\xi F,$ and

$$\begin{array}{ccc}\kappa \xi +|{\hslash }_{\xi E}|+EL& =& |{\hslash }_{\xi E}|+|\epsilon EL|-E\rho \le |{\hslash }_{\xi E}|+|EFL|-E\rho \\ & \le & |{\hslash }_{\xi E}|+E\rho +|\rho FL|-E\rho \\ & =& |{\hslash }_{\xi E}|+|\rho FL| \le |{\hslash }_{\xi E}|+|\xi FL| \le |{\gamma }_{A,L}|.\end{array}$$

□

Corollary 3.

In Figure 7, if γ’s support angle $\angle u,b\ge {30}^{\circ }$, then (5) holds.

Proof. 

For $\gamma$ in Figure 7a,b the corollary is true by Lemmas 1 and 2.

Now consider $\gamma$ in Figure 7c. If $\xi \in {\gamma }_{F,L}$, then $\kappa \xi +|{\hslash }_{\xi L}\left(\gamma \right)| \le |{\gamma }_{F,L}| \le |{\gamma }_{A,L}|$. If $\xi \in {\gamma }_{A,F}$, there exists a support line $u^{\prime }$ of $\gamma$ touching ${\gamma }_{A,F}$ at E and ${\gamma }_{F,L}$ at $L^{\prime }.$ The $\mathrm{\Lambda }$-triple $\{E,F,L^{\prime }\}$ of $\gamma$ in a support angle $\angle u^{\prime },b>\angle u,b$, satisfies Lemma 2 with $L^{\prime }$ instead of L. Thus $\kappa \xi +|{\hslash }_{\xi L^{\prime }}\left(\gamma \right)| \le |{\gamma }_{A,L^{\prime }}|$ and we conclude that

$\kappa \xi + |{\hslash }_{\xi L}\left(\gamma \right)| = \kappa \xi + |{\hslash }_{\xi L^{\prime }}\left(\gamma \right)| + |{\hslash }_{L^{\prime }L}\left(\gamma \right)| \le |{\gamma }_{A,L^{\prime }}| + |{\gamma }_{L^{\prime },L}| = |{\gamma }_{A,L}|$.

□

2.2. Drapeable Arrangements of $\mathrm{\Lambda }$-Triples, Draping Simple Arcs

In a $\mathrm{\Lambda }$-frame $\mathcal{F}\left(\alpha \right)$ of Definition 2, there are five principal arrangements for points of $\mathrm{\Lambda }$-triples $\{{\gamma }_1,{\gamma }_2,{\gamma }_3\}\in \angle u,b$ and $\{{\overline{\gamma }}_1,{\overline{\gamma }}_2,{\overline{\gamma }}_3\}\in \angle b,v$. They are represented by so-called A-graphs A1, A2, A3, A4, and A5 in Figure 8, Figure 9 and Figure 10 below.

Example 1

(Figure 3b,c). Arcs with A-graphs A2 (Figure 9a,b) or A5 (Figure 10c) are clearly drapeable. []

Example 2

(Figure 11). Arcs with A-graphs A4, A5 (Figure 10) are drapeable by side drapes: (a) $\gamma =L\zeta RT\xi$; (b) $\gamma =\zeta RTY\xi$, where $R\xi \kappa <RY\xi$ by Lemma 1 in $\angle v,u$; (c) $\gamma =BDFL\xi R$, where $|\kappa \xi RD|<|R\xi LD|$ by Corollary 3 in $\angle v,u$. []

Example 3

(Figure 12b). The arc $\gamma$ in the figure has a $\mathrm{\Lambda }$-frame $\mathcal{F}\left(\alpha \right)$ with triples $\{F,L,T\}$ in $\angle u,b$ and $\{F,R,T\}$ in $\angle b,v$ forming A-graph A1. It also has $\hslash \left({\gamma }_{L,R}\right)={\hslash }_{LR}\left(\gamma \right)$, thus is clearly drapeable by a base drape (1). []

Theorem 1.

A simple γ with Λ-frame$\mathcal{F}\left(\alpha \right)$, $\alpha \in [{30}^{\circ },{180}^{\circ })$, is drapeable.

Proof. 

Simple open $\gamma$ with $\mathcal{F}\left(\alpha \right)$ must have 2 points on b. Thus its A-graph is A1 or A3 (Figure 8a and Figure 9c) with points $F\prec L\prec R\prec T$ where $F,T\in b$, $L\in u$, and $R\in v$. If $\alpha \ge {90}^{\circ }$ the arc $\gamma$ is visibly drapeable by the base drape (1) (Figure 3a). If $\alpha \in [{30}^{\circ },{90}^{\circ })$, $\gamma$ is drapeable by (1) after Corollary 3 is applied separately to end-arcs ${\gamma }_{A,L}$ and ${\gamma }_{R,Z}$ (Figure 12a). □

2.3. Draping Non-Simple Arc with A-Graphs A1

Definition 4.

Let $\alpha \in [{30}^{\circ },{75}^{\circ }]$. If $\gamma$ has a frame $\mathcal{F}\left(\alpha \right)$ with A-graph A1, it has points $F\prec L\prec R\prec T$ such that $F,T\in b$, $L\in u$, $R\in v$. If there are points $B\in {\hslash }_{LR}\left(\gamma \right)\cap {\gamma }_{A,F}$ and $Y\in {\hslash }_{LR}\left(\gamma \right)\cap {\gamma }_{T,Z}$, we choose $\omega \in {\gamma }_{L,R}\cap {\gamma }_{B,F}$ and $\chi \in {\gamma }_{L,R}\cap {\gamma }_{T,Y}$. When both B and Y exist, it is assumed that $\mathrm{dist}(B,b) \le \mathrm{dist}(Y,b)$, i.e., B is below Y.

Example 4

(Figure 13a). Suppose ${\gamma }_{\omega ,\chi }\cap {\hslash }_{LR}=\varnothing$. Points $\omega$, $\chi$ can be identified by parameters $\omega =\gamma \left(t_1\right)=\gamma \left(t_2\right)$, $\chi =\gamma \left(s_1\right)=\gamma \left(s_2\right)$ and order relation:

$$A\prec \gamma \left(t_1\right)<F<L<\gamma \left(t_2\right)<\gamma \left(s_1\right)<R\prec T\prec \gamma \left(s_2\right).$$

There exists a support line ℓ of $\gamma$ touching subarcs ${\gamma }_{A,\gamma \left(t_1\right)}$ and ${\gamma }_{\gamma \left(s_2\right),Z}$. Because we have $|FT| < |{\gamma }_{F,\gamma \left(t_2\right),\gamma \left(s_1\right),T}|$, the skew drape (1) ${\delta }_{\ell }\left(\gamma \right)=\kappa \xi \cup {\hslash }_{\xi T}\cup TF\cup {\hslash }_{F\zeta }\cup \zeta \eta$ drapes $\gamma$. []

Example 5

(Figure 13b). Let a support line ℓ at Y touch ${\gamma }_{L,R}$. If ∠ℓ, $b\ge {30}^{\circ }$, then $|\eta FT|<|FLT| \le |{\gamma }_{F,L,\chi ,T}|$ by Lemma 1. Thus, ${\delta }_{\ell }\left(\gamma \right)$ in (1) drapes $\gamma$. []

The arc ${\delta }_{\ell }$ might not drape $\gamma$ if ∠ℓ, $b<{30}^{\circ }$ and $\alpha \in ({60}^{\circ },{75}^{\circ }]$ (Figure 14b) but the base drape ${\delta }_b$ does, we show in Section 2.5, after the longest Section 2.4 confirms that for $\alpha \in [{30}^{\circ },{75}^{\circ }]$ an example of a non-dreapeable arc similar to the arc in Corollary 2 cannot be built.

2.4. Arcs LRSY, LYRI, FLRTY, and BFLRTY

Definition 5.

Let $FLRTY$ be an arc through points of $\gamma$ in Definition 4 with $F, T\in b, L\in u,$$R\in v,$ and Y above $LR$. Let $YSR$ be the line from Y to R reflected in b, and let $\beta =\angle LY,b$, $\sigma =\angle YR,b$, $I={\left(\mathrm{proj}R\right)}_b$ (Figure 15a).

In Example 5 we learned that the arc $FLRTY$ has a skew drape if $\beta \ge {30}^{\circ }$. By studying arcs $LRSY$, $LYRI$ we show that it has a base drape when $\beta <{30}^{\circ },\alpha \in [{60}^{\circ },{75}^{\circ }]$ and regardless of $\beta$ when $\alpha \in [{30}^{\circ },{60}^{\circ }]$.

Fact 2.

Let $LY \le LR$ in Figure 15. Then $|LYRI|<|LRSY|$.

Proof. (a) $\mathbf{\sigma }\ge {\mathbf{30}}^{\circ }$. Here $LY+|YRI|\stackrel{\mathit{Lemma}1}{\le }LY+|RSY|\le LR+|RSY|$.

(b) $\mathbf{\sigma }<{\mathbf{30}}^{\circ }$. Let $\theta$ denote the angle of reflection of the line $YSR.$ Then $\theta +\sigma =\angle YRS$.

(1) $\theta +\sigma \ge 90$, we use $\angle YRS=\theta +\sigma$ in $▵YRS$ to have $YR<YS.$

(2) $\theta +\sigma <90$, let a point $S^{\prime }\in b$ be such that $Y{S}^{\prime }=YS$, and let $X=Y{S}^{\prime }\cap LR$. Now $\angle S^{\prime }YR={180}^{\circ }-(\theta +\sigma )$ is obtuse; thus, $YR<XR$ and we conclude that

$LY + YR < YX + LX + XR = YX + LR < Y{S}^{\prime } + LR = YS + LR.$□

Example 6.

Arcs $LYR$ with $LY\le LR$: (a) $\beta \le {30}^{\circ }$ and either $\alpha \le {60}^{\circ }$ or R is below L;

(b) $\angle \beta +\sigma \le {90}^{\circ }$, for instance, $\alpha \le {45}^{\circ }$ or $\left\{Y\mathrm{is}\mathrm{below}L\mathrm{or}R\right\}$. []

Lemma 3.

Let $LY>LR$ and $\beta =\angle \left[LY\right],b<{30}^{\circ }$ in Figure 16. Then $|LYRI|\le |LRSY|$.

Proof. 

Because of Fact 2 and Example 6, where $LY\le LR$, we must have R above L and $\beta +\sigma >{90}^{\circ }$, which with $\beta <{30}^{\circ }$ implies that $\alpha ,\sigma >{60}^{\circ }$. We prove two cases. In the first case Y lies on v, in the second case Y lies in the interior of the frame.

Case 1. $Y=V\in v$ (Figure 16a). In this case, $\sigma =\alpha$; thus, $\beta +\alpha >{90}^{\circ }$. Let $RUV$ be the line from R to V reflected in b and $N\in v$ be such that $LN\Vert b$. Then $LN>LV>LR$ and $LN-LR<\mathrm{dist}(N,RI)<IQ$. A claim, $|RUV|-|VRI|>|IQ|$, establishes the lemma because

$$|RUV|-|VRI|>IQ>LN-LR>LV-LR\Rightarrow |LRUV|>|LVRI|.$$

Let $VQ=1$ and $r=RQ$ vary from 0 to 1. Then $IQ=rcos\alpha$ and

$$y^2={\left(Y{R}^{\prime }\right)}^2=1+r^2-2rcos\left(2\alpha \right)={(1-r)}^2+4r{\sin }^2\alpha$$

make the inequality $|RUV|-|VRI|>IQ$ equivalent to

$$\begin{array}{c}{(1-r)}^2+4r{\sin }^2\alpha >{(1-r+r\sin \alpha +rcos\alpha )}^2\\ \iff 4r{\sin }^2\alpha >2(1-r)r(\sin \alpha +cos\alpha )+r^2{(\sin \alpha +cos\alpha )}^2\\ \iff f(r,\alpha )=4{\sin }^2\alpha -2(1-r)(\sin \alpha +cos\alpha )-r{(\sin \alpha +cos\alpha )}^2>0.\end{array}$$

This is true because $f(0,\alpha )=4{\sin }^2\alpha -2(\sin \alpha +cos\alpha )>0$ on $[{60}^{\circ },{75}^{\circ }]$, while $f_r^{\prime }(r,\alpha )=2(\sin \alpha +cos\alpha )-{(\sin \alpha +cos\alpha )}^2>0$ for all $\alpha$.

Case 2. $Y\notin v$ (Figure 16b). We extend $LY$ to $V\in v$ and draw the arc $LRUV$ of Case 1. When straightened arcs $YSR$ and $VUR$ meet at $R^{\prime }$, angles $\angle VY{R}^{\prime }>\angle VYR=\beta +\sigma >{90}^{\circ }$ make $VR>RY$ and $R^{\prime }V>R^{\prime }Y$. Thus, if $X\in R^{\prime }V$ is such that $R^{\prime }X=R^{\prime }Y$, we have $\angle YXV>{90}^{\circ }$; thus, $YV>VX$, and

$$\begin{array}{c}VX=R^{\prime }V-R^{\prime }Y=|RUV|-|RSY|=|LRUV|-|LRSY|.\\ |LVRI|-|LYRI|=YV+VR-RY>YV>VX=|LRUV|-|LRSY|.\end{array}$$

The last inequality is $|LRSY|-|LYRI|>|LRUV|-|LVRI|$ which is positive by Case 1. □

Lemma 4

(A geometric property of isosceles triangles). Let $PZQ$ be an isosceles triangle of base $PQ$ and base angle $\alpha \in [{45}^{\circ },{60}^{\circ }]$. Let $ZUR$ be a line from $R\in ZQ$ to Z reflected in $PQ$. Then $|PZQ|<|PRUZ|$ (that is, arc $PZQ$ drapes every arc $PRUZ$ with $R\in ZQ$ and $U\in PQ$).

Proof. 

Let $R^{\prime }$ be reflection of R in $PQ$. The following are agreement and notation in relation to Figure 17a: $\mathit{PZ}=\mathbf{1},\mathit{s}=\mathit{PR},\mathit{y}=\mathit{ZU}+\mathit{UR}={\mathit{ZR}}^{\prime },\mathit{r}=\mathit{RQ},\mathit{t}=\mathbf{4}{\mathbf{cos}}^{\mathbf{2}}\mathit{\alpha }-\mathbf{1}$.

Observe that $t\in [0,1]$ and $r=\sqrt{t}$ when $s=1$. Also $y>1$ because $\alpha \ge {45}^{\circ }$. Using the law of cosine in $▵ZQ{R}^{\prime }$ for y and $▵PQR$ for s, we show that $y+s\ge 2:$

$$\begin{array}{ccc}\hfill y^2& =& 1+r^2-2r(2{cos}^2\alpha -1)=1+r^2-rt+r.\hfill \\ \hfill s^2& =& r^2+4{cos}^2\alpha -4r{cos}^2\alpha =r^2+t+1-r(t+1)\hfill \\ \hfill y^2+s^2& =& 2+r^2+r^2-2rt+t^2-t^2+t\ge 2.\hfill \end{array}$$

(6)

$$\begin{array}{ccc}\hfill y^2{s}^2& =& [1+r^2-rt+r][1+r^2-rt-r+t]\hfill \\ & =& 1+{(r^2-rt)}^2+(r^2-rt)(1+t)+t\hfill \\ & \ge & 1+{(r^2-rt)}^2+\underset{r}{min}(r^2-rt)(1+t)+t\hfill \\ & =& 1+{(r^2-rt)}^2-(t^2/4)(1+t)+t\ge 1.\hfill \end{array}$$

(7)

Now (7), (6) yield $ys={\left(y^2{s}^2\right)}^{1/2}\ge 1$ and $y+s={(y^2+s^2+2ys)}^{1/2}\ge 2$. □

Lemma 5.

Let $\alpha \in [{45}^{\circ },{60}^{\circ }]$ and $LY>LR$ in Figure 17b and Figure 18. Then $|LYRI|<|LRSY|$.

Proof. Case 0: $Y=Z,S=U$ (Figure 17b). Let $L^{\prime }\in PR$ be such that $L^{\prime }R=LR$. Then $PL>P{L}^{\prime }$, which with Lemma 4 yields the desired inequality:

$$|LZRI|<|LZQ|=|PZQ|-|PL|<|PRUZ|-|P{L}^{\prime }|=|RUZ|+|L^{\prime }R|=|LRUZ|.$$

Case 1: $Y\in v$ (Figure 18a). Let $LRUZ$ be the arc of Case 0. Straightened arcs $RSY$ and $RUZ$ meet at $R^{\prime }$. We mark $X\in UZ$ such that $R^{\prime }X=R^{\prime }Y$ and observe that $\angle ZYL=\beta +\sigma >{90}^{\circ }$ yields $LZ>LY$, which with $ZY>ZX=|RUZ|-|RSY|$ gives us

$$\begin{array}{c}|LZRI|-|LYRI|=LZ+ZY-LY>LY+ZY-LY>ZX=|LRUZ|-|LRSY|\\ \Rightarrow |LRSY|-|LYRI|>|LRUZ|-|LZRI|>0\mathrm{by}\mathrm{Case}0.\end{array}$$

Case 2: $Y\notin v$ (Figure 18b). We use the argument of Case 2 in Lemma 3. □

Corollary 4.

Let $|LYRI|\le |LRSY|$. Then $|LYRI|\le |LRTY|$ if $I\notin ST$ (Figure 19a) and $|LYRT|<|LRTY|$ if $I\in ST$ (Figure 19b).

Proof. 

By the law of light $|RSY|\le |RTY|$ for any $T\in b$. Combined with the assumption this yields the first inequality for arcs in Figure 19a. In particular,

$$LY+YR+RI<LR+RI+IY<LR+RI+TY.$$

We add $RT$ to each side to have $|LYRT|<|LRTY|$ for arcs in Figure 19b. □

Let $\mathrm{\Psi }\left(\gamma \right)=BFLRTY$ be an arc through points of $\gamma$ in Definition 4. If ${\left(\mathrm{proj}LR\right)}_b$$=JI$, then ${\left(\mathrm{proj}\mathrm{\Psi }\right)}_b$ can be $FT,JT,JI,FI$. The following theorem is true when $\alpha \in [{30}^{\circ },{60}^{\circ }]$ and $\alpha \in [{60}^{\circ },{75}^{\circ }],\beta <{30}^{\circ }$, that is, when $\gamma$ satisfies Fact 2, Example 6, Lemmas 3 and 5.

Theorem 2.

Let $|LYRI|\le |LRSY|$. Then Ψ has a base drape Δ equals $FLBYRT$ in Figure 20a, $JLBYRT$ in Figure 20b, and $JLBYRI$ if $FT\subset JI$ or $FLBYRI$ if $JT\subset FI$.

Proof. 

Let $N={\left(\mathrm{proj}B\right)}_b$ and $X=BN\cap LY$. Since B is below Y in Definition 4, the angle $NBY$ is obtuse making $|BY|<|XY|$ and, consequently,

$$\begin{array}{c}|LBY|<|LXB|+|XY|=|BX|+|LY|<|BF|+|LY|,\\ |LBYRT|<|BF|+|LY|+|YRT|=|BF|+|LYRT|\stackrel{\mathit{Corollary}4}{<}|BF|+|LRTY|.\end{array}$$

The last inequality together with $JL\le FL$ yield $|FLBYRT|<|\mathrm{\Psi }|$ in Figure 20a and $|JLBYRT|<|\mathrm{\Psi }|$ in Figure 20b. □

Remark 1.

In Figure 20c, the arc $\widehat{\mathrm{\Gamma }}\subseteq \mathcal{H}(\Delta )$ is drapeable by $\Delta$ if $\mathrm{\Psi }\left(\widehat{\mathrm{\Gamma }}\right)$ satisfies Theorem 2.

2.5. Base Drape for Arcs in Definition 4 Satisfying Theorem 2

Consider $\gamma$ in a $\mathrm{\Lambda }$-frame $\mathcal{F}\left(\alpha \right)$ of Definition 4 with $\mathrm{\Psi }\left(\gamma \right)$ as in Theorem 2. The algorithms of Propositions 3 and 4 can produce its simple approximation, $\pi$, with $\gamma$’s finitely many points fixed (some could be renamed due to re-parametrizations) here, the endpoints $A,Z$, vertices $F,L,R,T$ of $\mathrm{\Lambda }$-triples of $\gamma$, and points $B,Y$ when exist. If neither B nor Y exists, $\gamma$ is drapeable (see Figure 12b). In $\pi$, points $\omega \in {\gamma }_{L,R}\cap {\gamma }_{B,F}$ and $\chi \in {\gamma }_{L,R}\cap {\gamma }_{T,Y}$ are represented by $\omega ,\overline{\omega }$ and $\chi ,\overline{\chi }$ (Figure 21). Observe that $\pi$ might not inherit $\mathrm{\Lambda }$-triples or remain inside the $\mathrm{\Lambda }$-frame of $\gamma$. Subarcs of $\pi$ that fall outside $\mathcal{F}\left(\alpha \right)$ are substituted with segments of $\mathcal{F}\left(\alpha \right)$, e.g., $CDE$ is replaced by $CE$ in Figure 21b.

Example 7

(Only Y exists). Let $\gamma$ in Definition 4 have $\mathrm{\Psi }\left(\gamma \right)=FLRTY$ satisfying Theorem 2. Let $\pi$ be $\gamma$’s simple approximation pictured in Figure 22a. We remove open subarcs ${\pi }_{A,\overline{\omega }},{\pi }_{F,\omega },{\pi }_{Y,Z}$ preserving ${\hslash }_{LR}\left(\pi \right)$ and connect ${\pi }_{\omega ,\chi }$ to ${\pi }_{\overline{\omega },L}$ and ${\pi }_{R,\overline{\chi }}$ using bridges $\overline{\omega }\omega$ and $\chi \overline{\chi }$ (Figure 22b). Proposition 3 guarantees that the length of the bridges is less than $\epsilon$. Re-parametrization of the modified $\pi$ with added bridges gives us a non-simple arc $\mathrm{\Gamma }$ that satisfies Definition 4 with $\mathrm{\Psi }\left(\mathrm{\Gamma }\right)=\mathrm{\Psi }\left(\gamma \right)=FLRTY$ draped by $\Delta =JLYRI$ by Theorem 2:

$$\mathrm{\Gamma }={\pi }_{F,\overline{\omega }}\cup \overline{\omega }\omega \cup {\pi }_{\omega ,\chi }\cup \chi \overline{\chi }\cup {\pi }_{\overline{\chi },R,T,\chi }\cup \chi \overline{\chi }\cup {\pi }_{\overline{\chi },Y}.$$

Let points $p\in JL,q\in LY,s\in YR,r\in RI$ and $\tau \in {\hslash }_{LY},\sigma \in {\hslash }_{YR}$ be such that open segments $Jp,qY,rI,sR$ and $\tau Y,\sigma R$ have no points of $\mathrm{\Gamma }$. We construct an arc $\widehat{\mathrm{\Gamma }}\subseteq \mathcal{H}(\Delta )$ consisting of subarcs of $\mathrm{\Gamma }$ and $\Delta$ with properties mentioned in Remark 1 (Figure 23):

$$\widehat{\mathrm{\Gamma }}={\mathrm{\Gamma }}_{F,p}\cup pLq\cup {\mathrm{\Gamma }}_{q,R}\cup Rr\cup {\mathrm{\Gamma }}_{r,T,s}\cup sY.$$

Let $\kappa \eta ={\left(\mathrm{proj}\mathrm{\Gamma }\right)}_b$ with $\kappa \xi ,\eta \zeta \perp b$ be so that $\kappa \xi \cap \mathrm{\Gamma }=\left\{\xi \right\}$, $\eta \zeta \cap \mathrm{\Gamma }=\left\{\zeta \right\}$. We show that the base drape $\delta =\kappa \xi \cup {\hslash }_{\xi \zeta }\left(\mathrm{\Gamma }\right)\cup \zeta \eta$ in Figure 22b drapes $\mathrm{\Gamma }$ as follows.

$$\begin{array}{ccccccc}|{\delta }_{\kappa \tau }|& =& \kappa \xi +|{\hslash }_{\xi \tau }|& <& Jp+|{\mathrm{\Gamma }}_{p,\xi }|+|{\hslash }_{\xi \tau }|& +& |pLq|-|pLq|,\\ |{\delta }_{\tau \sigma }|& =& \tau Y+|{\hslash }_{Y\sigma }|& <& qY+|{\mathrm{\Gamma }}_{\tau ,q}|+|{\hslash }_{Y\sigma }|& +& Ys-Ys,\\ |{\delta }_{\sigma \zeta }|& =& \sigma R+|{\hslash }_{R\zeta }|& <& sR+|{\mathrm{\Gamma }}_{\sigma ,s}|+|{\hslash }_{R\zeta }|& +& Rr-Rr,\\ |{\delta }_{\zeta \eta }|& & & <& rI+|{\mathrm{\Gamma }}_{\zeta ,r}|.& \end{array}$$

Adding the lines and using $|{\mathrm{\Gamma }}_{p,\xi }|+|{\hslash }_{\xi \tau }|+|{\mathrm{\Gamma }}_{\tau ,q}|<|{\mathrm{\Gamma }}_{p,q}|$ we obtain the first bound for

$$|\delta |<|\Delta |+|{\mathrm{\Gamma }}_{p,q}|+|{\hslash }_{Y\sigma }|+|{\mathrm{\Gamma }}_{\sigma ,s}|+|{\hslash }_{R\zeta }|+|{\mathrm{\Gamma }}_{\zeta ,r}|-|pLq|-Ys-Rr.$$

Then we use bounds $|{\hslash }_{Y\sigma }|+|{\mathrm{\Gamma }}_{\sigma ,s}|<|{\mathrm{\Gamma }}_{Y,s}|,$ $|{\hslash }_{R\zeta }|+|{\mathrm{\Gamma }}_{\zeta ,r}|<|{\mathrm{\Gamma }}_{R,r}|,$ and $|\Delta |<|\widehat{\mathrm{\Gamma }}|$ to have

$$|\delta |<|\widehat{\mathrm{\Gamma }}|+|{\mathrm{\Gamma }}_{p,q}|+|{\mathrm{\Gamma }}_{Y,s}|+|{\mathrm{\Gamma }}_{R,r}|-|pLq|-Ys-Rr=|\mathrm{\Gamma }|.\left[\right]$$

3. Drapeability of z-Arcs

A z-arc is a three-segment non-convex simple arc $\gamma =ABCD$ with $AD\cap BC\ne \varnothing$. Its shape is defined by four angles; here, $\alpha =\angle BAC$, $\omega =\angle BCA$, $\theta =\angle CBD$, $\beta =\angle CDB$. We find drapeability conditions for a z-arc in Figure 24a in terms of these angles. In [1] such conditions were found for symmetric z-arcs in Figure 25a in terms of $AB/|\gamma |$ and $AC/|\gamma |$.

If any of the angles is obtuse, $\gamma$ is drapeable (see, e.g., Figure 24b). We assume that $\sigma =\omega -\theta \ge 0$. Thus, $H=\mathrm{dist}(B,AC)\ge h=\mathrm{dist}(C,BD).$ Let $D\eta =\mathrm{dist}(D,[AC\left]\right)$ and $A\mu =\mathrm{dist}(A,[BD\left]\right)$. We have

$$\begin{array}{cc}\mu ACD\mathrm{drapes}\gamma \mathrm{if}AC\le BC;& \\ ABD\eta \mathrm{drapes}\gamma \mathrm{if}BD\le BC,& \mathrm{or}BD\le CD,\mathrm{or}(\mathrm{Lemma})\sigma \ge {30}^{\circ }.\end{array}$$

(8)

We now assume that $\gamma$ satisfies (9) below which, in particular, every non-drapeable $\gamma$ does:

$$AC>BC,BD>BC,BD>CD,\alpha ,\beta ,\theta ,\omega <{90}^{\circ },\sigma <{30}^{\circ }.$$

(9)

3.1. Symmetric z-Arc

A symmetric z-arc $\gamma$ has $\omega =\theta$, $\alpha =\beta$ (Figure 25a). Its drapeability in [1] (Th. 4) is given by a cubic $9a^3-9a^{2}s+3a{s}^2+5s^3-15a^2+6as-3s^2+7a-s-1\le 0$ in $a={\displaystyle \frac{AB}{|\gamma |}}$, $s={\displaystyle \frac{AC}{|\gamma |}}$. It appears that angles defining $\gamma$’s shape yield simpler conditions. Let $D\iota =\mathrm{dist}(D,[AB\left]\right)$ and $A\upsilon =\mathrm{dist}(A,[CD\left]\right)$ (Figure 25b).

Corollary 5.

A symmetric z-arc ($\omega =\theta$, $\alpha =\beta$) is drapeable if and only if

$$\tan {\displaystyle \frac{\beta }{2}}+\tan {\displaystyle \frac{\theta }{2}}\ge 1.$$

(10)

Proof. 

The arc $ABD\eta$ drapes $\gamma$ if $BCD\ge BD\eta$. We apply Lemma 1 to $BCD$ with $BCD=LFE$, $BD\eta =LE\epsilon$, and $\sigma =0$ (Figure 4c).

If $ABD\eta$ does not drape $\gamma$, neither does $\upsilon ABD$, because $AC=BD$ and $BD>CD$ by (9) make $A\upsilon =\mathrm{dist}(A,[BC\left]\right)>D\eta$ (Figure 25b). □

Note that, if $\theta =\beta$, i.e., the arc has all angles equal (Figure 2c), then (10) is

$$2\tan {\displaystyle \frac{\beta }{2}}\ge 1\iff \tan \beta \ge {\displaystyle \frac{4}{3}}\iff \mathrm{dist}(BD,AC)\ge {\displaystyle \frac{4}{15}}|\gamma |.$$

(11)

The last inequality is equivalent to $AC\le {\displaystyle \frac{2}{5}}|\gamma |\iff AD\le {\displaystyle \frac{\sqrt{97}}{15}}|\gamma |$ in [1] (p. 644).

3.2. Z-Arc ABCD with AC and BD Parallel

Angles $\alpha ,\beta ,\omega =\theta$ define the shape of a z-arc with $AC\Vert BD$ (Figure 25).

Corollary 6.

A z-arc with $\theta =\omega$$(\alpha \le \beta )$ in Figure 25 is drapeable if and only if either $ABD\eta$ drapes γ; that is, (10) is true, or $ACD\iota$ drapes γ; that is,

$$\tan {\displaystyle \frac{\alpha }{2}}+\tan {\displaystyle \frac{\theta }{2}}\ge \sin \alpha (\cot \beta +\cot \theta ).$$

(12)

Proof. 

If $ABD\eta$ does not drape $\gamma$, neither do $\mu ACD$ because $AC\ge BD$ and $\upsilon ABD$, because $BD>CD$ by (9) implies $A\upsilon >h.$ However, $ACD\iota$ might drape $\gamma$ if $D\iota$ is small enough, precisely, if $D\iota \le AB+BC-AC$, that is,

${\displaystyle \frac{h}{\sin \alpha }+\frac{h}{\sin \theta }-h(\cot \alpha +\cot \theta )\ge h(\cot \beta +\cot \theta )\sin \alpha \iff \left(12\right).}$□

3.3. General z-Arc

Along with angles $\alpha ,\omega ,\beta ,\theta$ defining the shape of the z-arc in Figure 26a, we employ angle $\sigma =\omega -\theta$. Observe that $BD=h(\cot \theta +\cot \beta )$, $AC=H(\cot \alpha +\cot \omega )$, and the altitudes of the z-arc’s vertices have the following relations.

$$H=h{\displaystyle \frac{\sin \omega }{\sin \theta }},D\eta ={\displaystyle \frac{h\sin (\beta -\sigma )}{\sin \beta }},A\mu ={\displaystyle \frac{H\sin (\alpha +\sigma )}{\sin \alpha }}.$$

(13)

Theorem 3.

A z-arc is drapeable if and only if one of the following holds

$$\tan {\displaystyle \frac{\beta }{2}}+\tan {\displaystyle \frac{\theta }{2}}\ge {\displaystyle \frac{\sin (\beta -\sigma )}{\sin \beta }}min\left\{1,{\displaystyle \frac{(\cot \alpha +\cot \omega )\sin \omega \sin \beta }{\sin \theta }}\right\}.$$

(14)

$$\tan {\displaystyle \frac{\alpha }{2}}+\tan {\displaystyle \frac{\omega }{2}}\ge {\displaystyle \frac{\sin (\alpha +\sigma )}{\sin \alpha }}min\left\{1,{\displaystyle \frac{(\cot \beta +\cot \theta )\sin \theta \sin \alpha }{\sin \omega }}\right\}$$

(15)

Proof. 

Either the arc $ABD\eta$ or the arc $\upsilon ABD$ drapes $\gamma$ if

$$BC+CD-BD\ge min\{D\eta ,A\upsilon \}=D\eta min\left\{1,AC/CD\right\},\mathrm{that}\mathrm{is}$$

$${\displaystyle \frac{h}{\sin \theta }}+{\displaystyle \frac{h}{\sin \beta }}-h\cot \theta -h\cot \beta \ge D\eta min\left\{1,{\displaystyle \frac{AC}{CD}}\right\}\stackrel{\left(13\right)}{⟺}\left(14\right).$$

Similarly, $\mu ACD$ or $ACD\iota$ drapes $\gamma$ if $AB+BC-AC\ge min\{A\mu ,D\iota \}$, i.e.,

$${\displaystyle \frac{H}{\sin \alpha }}+{\displaystyle \frac{H}{\sin \omega }}-H\cot \alpha -H\cot \omega \ge A\mu min\{1,{\displaystyle \frac{BD}{AB}}\}\stackrel{\left(13\right)}{⟺}\left(15\right).$$

If neither $ABD\eta$, $\upsilon ABD$ nor $\mu ACD$, $ACD\iota$ drape $\gamma$, no other drape does. For example, consider a drape $\rho ABD\chi$ in Figure 26b, whose stand $\rho \chi$ lies on a line through C. We show that $A\rho +D\chi >D\eta .$ Let C be the center of symmetry in segments $D{D}^{\prime },\chi {\chi }^{\prime },\eta {\eta }^{\prime }$. Let a point ${\chi }^{″}\in D^{\prime }{\eta }^{\prime }$ be such that $D^{\prime }{\chi }^{″}=D^{\prime }{\chi }^{\prime }$ and segment $D^{\prime }E$ be the extension of $D^{\prime }{\chi }^{\prime }$ to $AC$. Now, ${\chi }^{″}{\eta }^{\prime }<{\chi }^{\prime }E<\rho A$; hence, $\rho A+D^{\prime }{\chi }^{\prime }>D^{\prime }{\eta }^{\prime }$, that is, $A\rho +D\chi >D\eta$. Thus $|\rho ABD\chi |>|ABD\eta |$. But the arc $ABD\eta$ is longer than $\gamma$ because it does not drape $\gamma$. □

3.4. The Width of Non-Drapeable z-Worms

The width of a set is the smallest distance between the set’s parallel support lines. The broadworm’s construction (see [3], p. 2), implies that the isosceles ${60}^{\circ }$ v-worm of width $\sqrt{3}/4$ is the thickest v-worm. Thus, ref. [21] argues, the width of any z-worm is smaller than that. Intuitively, the width of a non-drapeable z-worm must be even smaller. For a family of unit arcs F we define

$$\aleph \left(F\right)=\mathrm{lub}\{\mathrm{widths}\mathrm{of}\mathrm{non}-\mathrm{drapeable}\gamma \in F\}=min\{r:\gamma \in F\mathrm{is}\mathrm{drapeable}\mathrm{if}{\mathrm{width}}_{\gamma }\ge r\}$$

and called it ℵ when F is the family of all unit arcs. Only $\aleph \left(F_S\right)=4/15$ is computed for the family $F_S$ of symmetric z-worms $\alpha =\beta =\omega =\theta$ in Figure 2c. We compute it for the family $F_{sym}$ symmetric z-worms $\alpha =\beta$; $\omega =\theta$ (Figure 25a) and outline computations for all z-worms. A preliminary study shows that the width of a non-drapeable z-arc is H or w (Figure 27a) and only H if the arc is symmetric. Observe that slight reductions of angles in equality of (10) convert drapeable z-arcs to non-drapeable. Hence,

$$\aleph \left(F_{\mathrm{sym}}\right)=max\left\{H={\left({\displaystyle \frac{2}{\sin \beta }}+{\displaystyle \frac{1}{\sin \theta }}\right)}^{-1},\tan {\displaystyle \frac{\beta }{2}}+\tan {\displaystyle \frac{\theta }{2}}=1\right\}\approx 0.27.$$

In Figure 27b, a sequence of non-drapeable z-worms $A{B}^{n}C{D}^n$, whose first terms have widths $H>0.31$, approaches a unit v-arc $ABC$ with base angles $\angle BCA={30}^{\circ }$ and $\angle BAC={\alpha }_{\infty }$ defined by ${\displaystyle \tan \frac{{\alpha }_{\infty }}{2}+\tan \frac{\pi }{12}=\frac{\sin ({\alpha }_{\infty }+\pi /6)}{\sin {\alpha }_{\infty }}}$ (which yields ${\alpha }_{\infty }\approx 73.5^{\circ }$).

Conjecture 1.

${\displaystyle \aleph (\mathrm{z}-\mathrm{worms})={\mathrm{width}}_{ABC}=\frac{\sin {\alpha }_{\infty }}{2\sin {\alpha }_{\infty }+1}\approx 0.3286}$.

Conjecture 1 can be resolved using software when ℵ(z-worms) is computed as $max\left(H\right)={\left({\displaystyle \frac{1}{\sin \beta }}+{\displaystyle \frac{1}{\sin \theta }}+{\displaystyle \frac{\sin \theta }{\sin \omega \sin \beta }}\right)}^{-1}$ constrained by equalities in (14) and (15).

4. The $\mathrm{\Lambda }$-Frame $\mathcal{F}(\mathbf{\tau },\mathbf{\sigma })$ of a Simple Open Arc

Proposition 1

([20], Th. 4). Every simple open arc γ in the plane has a unique pair of parallel support lines b and $b^{″}$ with a Λ-triple $\{F,M,T\}$.

Remark 2.

We assume that γ is not a line segment; the plane containing γ is vertical with left/right directions defined by an observer in front of the plane; and b is below $b^{″}$. The two types of frame defined in Definition 6 below are determined by γ’s points $F,T\in b$ and $M,\mathrm{\Omega }\in b^{″}$ that span the largest intervals on the lines. Parametrically listed $F,M,T,\mathrm{\Omega }$ are vertices of either a convex C-graph or a Z-graph (Figure 28). Finally, γ is placed so that ${\gamma }_{A,F}\cap b^{″}=\varnothing$ (Figure 29).

4.1. Definition of the Frame and Proof of its Existence

The frame is build around a simple open $\gamma$ that is in $\mathrm{\Lambda }$-configuration of Proposition 1. We employ clockwise/counterclockwise rolls of a line about the convex hull $\mathcal{H}\left(\gamma \right)$; see (1.2) [20]. At every position this line coincides with a support line w of $\gamma$. The boundary points of the compact set $\gamma \cap w$ are pivots of the rolls.

Fact 3. (i) Every γ in Remark 2 has a support angle $\angle \tilde{u},b$ with a Λ-triple (Figure 29).

(ii)Suppose an obtuse support angle $\angle \tilde{u},b$(or $\angle b,\tilde{v}$)has a Λ-triple. Then the support angle $\angle u,b={90}^{\circ }$ ($\angle b,v={90}^{\circ }$, Figure 30b) also has a Λ-triple.

Proof. (i) Since $\gamma$ is not a segment, it can be placed so that ${\gamma }_{A,F}\cap b^{″}=\varnothing$ (Figure 29).

(ii) Let $\tilde{L}$ and L be pivots of a line rolling about $\mathcal{H}\left(\gamma \right)$ clockwise when it coincides with $\gamma$’s support lines $\tilde{u}$ and $u\perp b$ respectively (Figure 29 (right)). If the pivot $\tilde{L}$ were in ${\gamma }_{A,F}$, the entire set $\gamma \cap \tilde{u}$ of simple $\gamma$ would be in ${\gamma }_{A,F}$. This would contradict to the assumption that $\angle \tilde{u},b$ has a $\mathrm{\Lambda }$-triple. Thus, $F⪯\tilde{L}⪯L⪯M$ making $\{F,L,T\}$ a $\mathrm{\Lambda }$-triple of $\angle u,b$. □

Definition 6

(Figure 30 and Figure 31). Frame $\mathcal{F}(\tau ,\sigma ),\tau ,\sigma \in (0,{90}^{\circ }]$ is formed by the largest support angles with Λ-triples $\angle u,b$ and $\angle b,v$ of sizes τ and σ.

Frame $\mathcal{F}(\tau ,0),\tau \in (0,{90}^{\circ }]$ is assigned to an arc γ that in Λ-configuration between b and $b^{″}$ has Z-graph $FMT\mathrm{\Omega }$ (Remark 2) as follows.

Let $\angle u^{″},b^{″}$ of size ${\tau }^{″}$ be the largest support angle with a Λ-triple not exceeding ${90}^{\circ }$. If ${\tau }^{″}<\tau$, we rotate the plane with γ by ${180}^{\circ }$ and then relabel lines $b\leftrightarrow b^{″}$, $u\leftrightarrow u^{″}$, angles $\tau \leftrightarrow {\tau }^{″}$, and points $F\leftrightarrow \mathrm{\Omega },M\leftrightarrow T.$ The Λ-frame $\mathcal{F}(\tau ,0)$ is a semi-strip bounded by $u,b,b^{″}$ (Figure 31).

Theorem 4.

Every simple open arc γ that is not a line segment has a unique Λ-frame $\mathcal{F}(\tau ,\sigma )$ described in Definition 6.

Proof. 

Finding $\angle u,b$. As $\gamma$ is placed according to Remark 2, there exists a unique support line $\tilde{u}$ touching ${\gamma }_{A,F}$ and ${\gamma }_{F,M}$. It can be reached by rolling a line about $\mathcal{H}\left(\gamma \right)$ clockwise from its position at b, pivoting on points $E\in {\gamma }_{A,F}$, until it touches a point $\tilde{L}\in {\gamma }_{F,M}$. Then $\tilde{u}=\left[E\tilde{L}\right]$ cannot be b, because $\gamma$ is not a segment, nor $b^{″}$, because $b^{″}\cap {\gamma }_{A,F}=\varnothing$. The angle $\angle \tilde{u},b$ has a $\mathrm{\Lambda }$-triple $\{F,\tilde{L},T\}$ and, if not obtuse, is the desired $\angle u,b$ with $L=\tilde{L}$. It has another $\mathrm{\Lambda }$-triple $\{E,F,L\}$ (Figure 30a, Figure 31a and Figure 32b) unless $EL=FM$ (Figure 30b and Figure 31b). If $\angle \tilde{u},b$ is obtuse, $\angle u,b={90}^{\circ }$ (Fact 3, Figure 29 (right) and Figure 32a).

Finding $\angle b,v$. (i) $b^{″}\cap {\gamma }_{T,Z}=\varnothing$, i.e., $\gamma$ has C-graph $FM\mathrm{\Omega }T$. We show that $0<\sigma \le {90}^{\circ }$. There exists a unique support line $\tilde{v}$ touching ${\gamma }_{T,Z}$ and ${\gamma }_{M,T}$ that can be reached by rolling a line about $\mathcal{H}\left(\gamma \right)$ counterclockwise and pivoting on points $X\in {\gamma }_{T,Z}$ from its position at b until it touches a point $\tilde{R}$ in ${\gamma }_{M,T}$ (Figure 30b). Then $\tilde{v}=\left[X\tilde{R}\right]$ and is not b because $\gamma$ is not a segment, nor $b^{″}$, because $b^{″}\cap {\gamma }_{T,Z}=\varnothing$. The angle $\angle b,\tilde{v}$ has a $\mathrm{\Lambda }$-triple $\{F,\tilde{R},T\}$ and, if not obtuse, is our $\angle b,v$ with $R=\tilde{R}$. It has another $\mathrm{\Lambda }$-triple $\{R,T,X\}$ (Figure 32) unless $RX=MT$. If $\angle b,\tilde{v}$ is obtuse, $\angle b,v={90}^{\circ }$ (Fact 3, Figure 30b).

(ii) $b^{″}\cap {\gamma }_{T,Z}\ne \varnothing$, i.e., $\gamma$ has Z-graph $FMT\mathrm{\Omega }$. The angle $\angle b,v$ of size 0 is formed by b and $b^{″}$ with $\mathrm{\Lambda }$-triples $\{F,M,T\}$ and $\{M,T,\mathrm{\Omega }\}$ (Figure 31). □

4.2. Draping Simple Open Arcs $\gamma$ Using $\mathrm{\Lambda }$-Frames

It was mentioned that, given $\alpha >0$, an arc $\gamma$ might not have a frame $\mathcal{F}\left(\alpha \right).$ If it has, its $\mathrm{\Lambda }$-frame $\mathcal{F}(\tau ,\sigma )$ has $\tau ,\sigma \ge \alpha$ (Figure 33), unless $\alpha >{90}^{\circ }$; in that case, $\tau =\sigma ={90}^{\circ }$. The argument of Theorem 1, where Corollary 3 was applied separately to the end-arcs ${\gamma }_{A,L}$ and ${\gamma }_{R,Z}$, shows that $\gamma$ is drapeable when $\tau ,\sigma \ge {30}^{\circ }$. Note that in the frame $\mathcal{F}(\tau ,\sigma )$ the case pictured in Figure 7c does not occur. To establish drapeability of $\gamma$ when $\tau$ or $\sigma$ is less than ${30}^{\circ }$, we use Lemmas 1 and 2 and the arguments similar to those for z-arcs in Section 3.

Example 8 (Figure 34) Consider $\gamma$ with $\mathrm{\Lambda }$-frame $\mathcal{F}(\tau ,0),\tau <{30}^{\circ }$ and $\gamma$’s projections on the sides of the frame: $\kappa \eta ={\left(\mathrm{proj}\gamma \right)}_b$, $\mu \nu ={\left(\mathrm{proj}\gamma \right)}_{b^{″}}$, $\iota \phi ={\left(\mathrm{proj}\gamma \right)}_u$. We identify points of $\gamma$ lying on its support lines: $\xi \in {\gamma }_{A,E}\cap \kappa \mu$, $\varphi \in {\gamma }_{A,E}\cap \phi \iota$ and $\zeta \in {\gamma }_{\mathrm{\Omega },Z}\cap \eta \nu$, $\rho \in {\gamma }_{\mathrm{\Omega },Z}\cap \phi \iota$. Let $\overline{\gamma }=\xi EFLMTX\mathrm{\Omega }\zeta$ and $\widehat{\gamma }=\varphi \xi EFLMTX\mathrm{\Omega }\rho \zeta Y$ be polygonal approximations of $\gamma .$ Note that $\gamma$ and $\widehat{\gamma }$ share projections on $b,u,$ and $b^{″}.$ Let $\overline{\iota }\overline{\phi }={\left(\mathrm{proj}\overline{\gamma }\right)}_u.$

We check whether $\gamma$ is draped by drapes with stands on $b,u,$ or $b^{″}.$

${\mathit{\delta }}_{\mathbf{b}}$: If $|LFE|\ge LE+\mathrm{dist}(E,b)$ then by Corollary 3, $|{\gamma }_{A,L}|\ge |\kappa \xi \cup {\hslash }_{\xi L}\left(\gamma \right)|$. Thus,

$$|{\gamma }_{A,M}|=|{\gamma }_{A,L}|+|{\gamma }_{L,M}|\ge |\kappa \xi \cup {\hslash }_{\xi M}\left(\gamma \right)|.$$

(16)

If $|MT\mathrm{\Omega }|>M\mathrm{\Omega }+\zeta \eta$, which is surely true if $|MT\mathrm{\Omega }|>M\mathrm{\Omega }+\mathrm{dist}(b,b^{″}),$ then

$$|{\gamma }_{M,\zeta }|\ge M\mathrm{\Omega }+\zeta \eta +|{\hslash }_{\mathrm{\Omega }\zeta }\left(\gamma \right)|=|{\hslash }_{M\zeta }\left(\gamma \right)|+\zeta \eta .$$

(17)

Combined, (16) and (17) imply that ${\delta }_b=\kappa \xi \cup \hslash \xi \zeta \cup \zeta \eta$ drapes $\gamma$. If ${\delta }_b$ does not drape $\gamma$, we check whether ${\delta }_{b^{″}}$ drapes $\gamma$.

${\mathit{\delta }}_{{\mathbf{b}}^{″}}$: Recall that $\angle u^{″},b^{″}\ge \angle u,b$ in the frame $\mathcal{F}(\tau ,0)$. If

$$|X\mathrm{\Omega }Y|\ge XY+\mathrm{dist}(Y,b^{″}),$$

(18)

which is true, in particular, by Lemma 1, if $\angle u^{″},b^{″}\ge {30}^{\circ }$ (that seems to be the case in Figure 34) then, by Corollary 3,

$$|{\gamma }_{X,Z}|\ge |\nu \zeta \cup {\hslash }_{\zeta X}\left(\gamma \right)|.$$

(19)

If also $\tan {\displaystyle \frac{\angle MFT}{2}}+\tan {\displaystyle \frac{\angle MTF}{2}}\ge 1$, then

$$|TMF|>TF+\mathrm{dist}(b,b^{″})>TF+\xi \mu .$$

(20)

Together, inequalities (18)–(20) yield

$$\begin{array}{ccc}|\gamma |=|{\gamma }_{Z,V,T}|+|{\gamma }_{T,M,F}|+|{\gamma }_{F,A}|& >& \nu \zeta +|{\hslash }_{\zeta T}\left(\gamma \right)|+TF+\xi \mu +|{\hslash }_{F\xi }\left(\gamma \right)|\\ & =& \nu \zeta +|{\hslash }_{\zeta \xi }\left(\gamma \right)|+\xi \mu =|{\delta }_{b^{″}}|.\end{array}$$

If ${\delta }_b$ and ${\delta }_{b^{″}}$ do not drape $\gamma$, we could similarly check whether ${\delta }_u\left(\gamma \right)$ does. But for this we would rather check if ${\overline{\delta }}_u$ drapes $\overline{\gamma }$. It can be shown that, when one of the drapes ${\overline{\delta }}_b=\kappa \xi ELM\mathrm{\Omega }\zeta \eta ,{\overline{\delta }}_u=\overline{\iota }\rho \zeta XTF\xi \overline{\phi },{\overline{\delta }}_{b^{″}}=\nu \zeta XTF\xi \mu$ drapes $\overline{\gamma }$, then drapes with stands on the same line drape $\widehat{\gamma }$ and $\gamma$. []

Comments on applications of $\mathrm{\Lambda }$-frames.

(1) The shortest drape for $\gamma$ is the drape ${\delta }_{\ell }$ in (1) where ℓ is a support line of $\gamma$. It seems that placing an open arc $\gamma$ in its frame $\mathcal{F}(\tau ,\sigma )$ reduces the search for this line ℓ to a finite number of lines.

(2) If $\gamma$ is not simple The Limit Theorem (Proposition 2) and Propositions 3, 4 allow us to drape simple approximations of $\gamma$ in their $\mathrm{\Lambda }$-frames.

(3) Example 8 shows that with frames $\mathcal{F}(\tau ,\sigma )$ we can test the second width conjecture on the largest width of non-drapeable unit arcs, which intuitively should be smaller than the largest width of unit arcs $b_0$.

Conjecture 2.

$\aleph (\mathrm{unit}\mathrm{arcs})=\aleph (\mathrm{unit}\mathrm{z}-\mathrm{arcs})$.

5. Review of Known Results: Proof of Fact 1: Summary

5.1. “Drapeability” [1]

Recall that In [1] an arc $\gamma$ is drapeable if it lies in the convex hull of a drape $\delta$ of Definition 1 such that $|\delta |\le |\gamma |$. It is clear that drapeability is preserved under isogonal affine transformations, thus often is studied for unit arcs (worms).

Topological properties and criteria for drapeability.

Closed arcs and arcs $\gamma$ with $\hslash \left(\gamma \right)\supset PQ\ge A{Z}_{\gamma }$ are drapeable [1], Theorem 1.

Wrapped arcs (Figure 2a), arcs that are intersected at least twice by every line crossing $AZ$, and unilateral arcs $\gamma$ (Figure 2b), arcs whose open endpoint segments $AZ$ lie the unbounded component of ${\mathbb{R}}^2\setminus \gamma$, are drapeable [1], Theorems 10, 16, 17.

Conjecture 3

(Wetzel [1], p. 651). The length of any wrapped γ satisfies $|\gamma |\ge 3AZ$.

Proposition 2

(Limit Theorem [1], Th. 8). If drapeable arcs ${\gamma }_n\rightrightarrows \gamma$ in the Hausdorff distance and $|{\gamma }_n|\to |\gamma |$ as $n\to \infty$, then γ is drapeable.

The following three propositions were designed to manage topological arguments in curve covering (see [13,14]) and drapeability like in Corollary 2, Example 7 here.

Proposition 3

(Simple Approximation Theorem [1], Th. 14). Given open γ and $\epsilon >0$, there is a simple polygonal path π through γ’s endpoints $A,Z$ with $|\gamma |-\epsilon \le |\pi |\le |\gamma |$ and the Hausdorff distance from γ less than ε.

The argument, whose flexibility allowed us to build $\pi$ through 6 points in Section 2.5, requires moving vertices of a polygonal arc to form a nearby simple arc:

Proposition 4

([1], Lemma 13). Let $\omega =V_0{V}_1{V}_2\dots V_n$ be a polygonal arc and let $\epsilon >0$. If each of the vertices of a polygonal path ${\omega }^{\prime }=V_0^{\prime }{V}_1^{\prime }{V}_2^{\prime }\dots V_n^{\prime }$ lies within ${\displaystyle \frac{\epsilon }{3n}}$ of the corresponding vertex of ω, then $||{\omega }^{\prime }|-|\omega ||<\epsilon$ and the Hausdorff distance between the traces of ${\omega }^{\prime }$ and ω is less than ε.

Proposition 5

(Corollary to Proposition 3). If a convex set covers simple polygonal arcs of length L, it covers all arcs of length L.

Metric criteria of drapeability for unit arcs $\gamma$, see [1] (pp. 644–647).

A unit arc $\gamma$ is drapeable if either of the 2 bounds is satisfied:

$A{Z}_{\gamma }\le {\displaystyle \frac{\pi -2}{\pi +2}}$ [1] (Th. 7) or ${\mathrm{diam}}_{\gamma }\le {\displaystyle \frac{\sqrt{3}-1}{2}}$ [1] (p. 646), [22] (Th. of Pál).

The smallest width of drapeable unit arcs exists by the Limit Theorem. It lies between $min\{{\mathrm{width}}_{\gamma }:\gamma \mathrm{in}\mathrm{Figure}\mathrm{c}\}=4/15$ and the width of the broadworm $b_0$.

The shortest drape for non-drapeable unit arcs has length at most $\sqrt{2}$, because every unit arc fits in some rectangle ℓ × w with ℓ + 2w $\le \sqrt{2}$ by [1] (p. 647). Note that the smallest rectangular worm cover ${(1-b_0^2)}^{1/2}\times b_0$ in [10] has ℓ + 2w > 1.77, see [2] (p. 649).

5.2. Λ-Configurations

Definition 7.

Two support lines (planes) of arc γ meet γ in Λ-configuration if there exists a Λ-triple ${\gamma }_1\prec {\gamma }_2\prec {\gamma }_3$ points of γ such that ${\gamma }_1$ and ${\gamma }_3$ lie on one line (plane) and ${\gamma }_2$ lies on the other. A simply connected region bounded by parallel support lines (planes) with a Λ-triple is called Λ-strip (Λ-slab).

Proposition 6

([14] Th. 5.1, [20] Corollary 1). Every planar arc has a Λ-strip.

Called the Λ property of arc, Proposition 6 has two space versions. One is

Proposition 7

([20] Th. 3). Given a support plane $\overline{p}$ of an arc γ, there exists a Λ-slab of γ whose boundary planes are normal to $\overline{p}$.

The other, inspired by Theorem of Weinholtz [16] (Th. 3.1) for closed arcs, was obtained to improve bounds for the ratio length/width of true space arcs.

Proposition 8

(Ghomi, [16] Corollary 3.2). For any rectifiable arc $\gamma :[a,b]\to {\mathbb{R}}^n$ there exist four points ${\gamma }_1\prec {\gamma }_2\prec {\gamma }_3\prec {\gamma }_4$ of γ and a pair of parallel support hyperplanes such that ${\gamma }_1,{\gamma }_3$ lie on one plane, while ${\gamma }_2,{\gamma }_4$ lie on the other.

Proposition 8 gives at least one $\mathrm{\Lambda }$-slab for a space arc, but it does not give a $\mathrm{\Lambda }$-strip for a planar arc as the arc’s two hyperplanes coincide. Proposition 7 gives a $\mathrm{\Lambda }$-slab with the boundary parallel to u for every direction $u\in S^2$. That could be just enough $\mathrm{\Lambda }$-slabs to prove (b) or (a) or both in the following conjecture.

Conjecture 4.(a) In Proposition 8 there exist a pair of parallel support hyperplanes such that ${\gamma }_1,{\gamma }_4$ lie on one plane, while ${\gamma }_2,{\gamma }_3$ lie on the other.

(b) For a non-planar arc in ${\mathbb{R}}^3$, Proposition 7 implies Proposition 8.

Proposition 9.

(a)An arc that does not touch vertices of any of its support angles of size α has at least one pair of such angles with Λ-triples (Th. 6) [20].

(b)For simple open arc this pair of support angles is unique, and angles’ vertices lie in the Λ-strip of Proposition 1, separated by γ in it (Figure 35a) (Th. 5) [20].

The second statement of Proposition 9 was used to setup Wichiramala’s brilliant proof by 4 reflections in [13]. Uniqueness of the pair of angles in this statement is not necessary in his proof. An inspection of about a dozen figures like Figure 35b suggests the following:

Conjecture 5.

An arc γ that does not touch vertices of any of its support angles of size α has at least one pair of such angles with Λ-triples whose vertices lie and are separated by γ in a Λ-strip.

5.3. Proof of Fact 1

In Figure 36 we see an angle $\angle u,b$ of size $\sigma$ with vertex at $P,$ a line $LGE$ connecting points L and E on u via reflection in b, and $\epsilon ={\left(\mathrm{proj}E\right)}_b$. We let $\omega$ be the reflection angle of line $LGE$, and we assume that $PL=1$ and $r=PE$. Then ${\displaystyle \tan \omega =\frac{(1+r)}{(1-r)}\tan \sigma }$.

In (4) we defined ${\displaystyle {\omega }_{\sigma }=arctan\frac{4-5\sin \sigma }{3cos\sigma }}$ and ${\displaystyle r_{\sigma }=\frac{2-4\sin \sigma }{2-\sin \sigma }}$. We now prove the following.

Fact 1. The four inequalities below are equivalent and each is equivalent to $|LE\epsilon |\le |LGE|$.

$$\tan {\displaystyle \frac{\mathbf{\omega }}{\mathbf{2}}}\ge {\displaystyle \frac{\mathbf{1}-\mathbf{2}\tan \frac{\mathbf{\sigma }}{\mathbf{2}}}{\mathbf{2}-\tan \frac{\mathbf{\sigma }}{\mathbf{2}}}}\stackrel{\mathit{C}\mathbf{1}}{⟺}\left(3\right)\stackrel{\mathit{C}\mathbf{2}}{⟺}\mathbf{\omega }\ge {\mathbf{\omega }}_{\mathbf{\sigma }}\stackrel{\mathit{C}\mathbf{3}}{⟺}\mathbf{r}\ge {\mathbf{r}}_{\mathbf{\sigma }}.$$

Proof. 

Recall that Lemma 1 implies $\left(3\right)\iff |LE\epsilon |\le |LGE|$. Now

$$\begin{array}{cccccc}C1:& \tan {\displaystyle \frac{\omega }{2}}& \ge & {\displaystyle \frac{1-2\tan (\sigma /2)}{2-\tan (\sigma /2)}}& \iff & \\ & 2\tan {\displaystyle \frac{\omega }{2}}+2\tan {\displaystyle \frac{\sigma }{2}}& \ge & 1+\tan {\displaystyle \frac{\omega }{2}}\tan {\displaystyle \frac{\sigma }{2}}& \iff & \\ \left(3\right)& 2\sin {\displaystyle \frac{\omega +\sigma }{2}}& \ge & cos{\displaystyle \frac{\omega -\sigma }{2}}& ♢& \\ C2:& 4{\sin }^2{\displaystyle \frac{\omega +\sigma }{2}}& \ge & {cos}^2{\displaystyle \frac{\omega -\sigma }{2}}& \iff & \\ & 4-4cos(\omega +\sigma )& \ge & 1+cos(\omega -\sigma )& \iff & \\ & 3+3\sin \omega \sin \sigma & \ge & 5cos\omega cos\sigma & \iff & \\ & 3/cos\omega cos\sigma & \ge & 5-3\tan \omega \tan \sigma & \end{array}$$

with $x=\tan \omega ,y=\tan \sigma ,$ the last inequality squared is

$$\begin{array}{cccccc} & 9(1+x^2)(1+y^2)& \ge & 25-30xy+9x^2{y}^2& \iff & \\ & 9x^2+30xy+9y^2-16& \ge & 0& \iff & \\ & {(3x+5y)}^2-16(1+y^2)& \ge & 0& \iff & \\ & 3x+5y-4\sqrt{1+y^2}& \ge & 0& \iff & \\ & {\displaystyle 3\tan \omega +5\tan \sigma -4/cos\sigma }& \ge & 0& \iff & \\ & 3\tan \omega & \ge & {\displaystyle (4-5\sin \sigma )/cos\sigma }& ♢& \\ C3:& {\displaystyle \left(\frac{1+r}{1-r}\right)\frac{\sin \sigma }{cos\sigma }}& \ge & {\displaystyle \frac{4-5\sin \sigma }{3cos\sigma }}& \iff & \\ & r& \ge & {\displaystyle \frac{2-4\sin \sigma }{2-\sin \sigma }}& =& {\displaystyle r_{\sigma }.}\end{array}$$

□

5.4. Summary of Properties of a Non-Drapeable Arc

Properties established in [1]:

1. A non-drapeable arc crosses its open endpoint segment, and the set of lines meeting the segment exactly once has positive geometric measure.

2. A non-drapeable arc has an endpoint segment longer than $0.222L$.

3. A non-drapeable arc has diameter larger than $0.366L$.

4. A non-drapeable symmetric z-arc $\gamma =ABCD$ in Figure 25a satisfies a cubic inequality $9a^3-9a^{2}s+3a{s}^2+5s^3-15a^2+6as-3s^2+7a-s-1>0$, where $a={\displaystyle \frac{AB}{|\gamma |}}$, $s={\displaystyle \frac{AC}{|\gamma |}}$.

5. A non-drapeable symmetric z-arc with equal sides has width less than ${\displaystyle \frac{4}{15}}L$.

Properties established here:

6. A non-drapeable arc with a $\mathrm{\Lambda }$-frame $\mathcal{F}\left(\alpha \right)$ has $\alpha \notin [{30}^{\circ },{75}^{\circ }].$

7. The $\mathrm{\Lambda }$-frame $\mathcal{F}(\tau ,\sigma )$ of a non-drapeable simple arc (uniquely assigned to every simple open arc) has $\tau$ or $\sigma$ smaller than ${30}^{\circ }$.

8. The necessary and sufficient conditions for a z-arc to be drapeable in terms of angles $\alpha ,\beta ,\omega ,\theta$ defining the arc’s shape are given by inequalities (14), (15), A drapeable z-arc satisfies one of the inequalities taken with “≥”.

9. A non-drapeable z-arc satisfies both inequalities (14) and (15) with “<”. In particular,

10. A symmetric z-arc in Figure 25a is non-drapeable if and only if $\tan {\displaystyle \frac{\beta }{2}}+\tan {\displaystyle \frac{\theta }{2}}<1.$

11. A non-drapeable symmetric unit z-arc must have width less than $0.27$.

12. A non-drapeable unit arc, in particular, unit z-arc can have width larger than $0.31$.

13. The least upper bound for the width of a non-drapeable unit z-arc $ABCD$ can be calculated using $AB+BC+AC=1$ and (14), (15) with “=”.

Questions related to worm covers:

Question 1.

Can a non-convex escape path of a convex set be drapeable?

Question 2.

What is the shape of a convex worm cover of width $b_0$ that has the least area?

Question 3.

Must the smallest convex worm cover of a given shape have a proper convex subset of a smaller width that covers non-drapeable unit arcs?

Question 4.

Would the smallest convex cover for non-drapeable unit arcs be smaller than the smallest convex cover for all unit arcs?