On Calculating the Packing Efficiency for Embedding Hexagonal and Dodecagonal Sensors in a Circular Container

Hindawi Mathematical Problems in Engineering Volume 2019, Article ID 9624751, 16 pages https://doi.org/10.1155/2019/9624751

Research Article On Calculating the Packing Efficiency for Embedding Hexagonal and Dodecagonal Sensors in a Circular Container

Marina Prvan ,1 Julije ODegoviT ,1 and Arijana Burazin Mišura 2

Faculty of Electrical Engineering, Mechanical Engineering and Naval Architecture (FESB), University of Split, Split , Croatia University Department of Professional Studies, University of Split, Split , Croatia

Correspondence should be addressed to Marina Prvan; [email protected]

Received 19 March 2019; Accepted 8 June 2019; Published 10 July 2019

Academic Editor: Vyacheslav Kalashnikov

Copyright © 2019 Marina Prvan et al. Tis is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

In this paper, a problem of packing hexagonal and dodecagonal sensors in a circular container is considered. We concentrate on the sensor manufacturing application, where sensors need to be produced from a circular wafer with maximal silicon efciency (SE) and minimal number of sensor cuts. Also, a specifc application is considered when produced sensors need to cover the circular area of interest with the largest packing efciency (PE). Even though packing problems are common in many felds of research, not many authors concentrate on packing polygons of known dimensions into a circular shape to optimize a certain objective. We revisit this problem by using some well-known formulations concerning regular hexagons. We provide mathematical expressions to formulate the diference in efciency between regular and semiregular tessellations. It is well-known that semiregular tessellation will cause larger silicon waste, but it is important to formulate the ratio between the two, as it afects the sensor production cost. Te reason why we have replaced the “perfect” regular tessellation with semiregular one is the need to provide spacings at the sensor 2 vertices for placing mechanical apertures in the design of the new CMS detector. Archimedean {3, 12 } semiregular tessellation and its more fexible variants with irregular dodecagons can provide these triangular spacings but with larger number of sensor cuts. Hence, we construct an irregular convex hexagon that is semiregularly tessellating the targeted area. It enables the sensor to remain symmetric and hexagonal in shape, even though irregular, and produced with minimal number of cuts with respect to dodecagons. Efciency remains satisfactory, as we show that, by producing the proposed irregular hexagon sensors from the same wafer as a regular hexagon, we can obtain almost the same SE.

1. Introduction packing into a larger circle or a square is well addressed in the literature [9–13]. Packing is a common problem in many diferent felds, Packing problem can be mapped to the feld of sensor such as computer science, design, manufacturing, and engi- manufacturing [14–16]. In this case, packing is a problem of neering. It consists of embedding smaller shapes into a producing � sensors from a circular silicon wafer which is a larger one called container [1–4]. Basic requirements must be fxed-size container. Te goal is to cut out as many sensors as considered depending on the specifc application to fnd the possible from a given (in our case, circular) piece of silicon optimal packing efciency (PE). For example, inner shapes wafer material. Basic requirement is to minimize the silicon can be the same in area or mixed shapes can be considered [5]. waste, i.e., the unused part of a circular object. PE is equated Also, arranged components can be nonoverlapping or there this way with the efciency of the used silicon material or could be overlap allowed [6]. silicon efciency (SE). Tere are applications with diferent optimization goals, When producing a single sensor, it is known that SE is like minimizing the size of the container, or maximizing the maximal if hexagon shape is used with respect to triangle or number of arranged inner components [7, 8]. Te shape of square [17, 18]. Since hexagon is the closest to being circular the container and the packed items may vary from a fxed- with the largest area, it has been shown to provide minimized size circle, square, rectangle, triangle, etc. For example, circle silicon waste with respect to other geometrical shapes [19]. 2 Mathematical Problems in Engineering

Naturally, applying the vertex cuts and using the irregular sensors both decrease SE when � sensors are produced, because semiregular tessellation is expected to cause larger silicon waste than regular tessellation. Nevertheless, the goal is to analyse whether using an irregular convex hexagon is cost-efective. We discuss if it is as efcient as using a regular hexagon. We also compare its efciency with dodecagons, to see whether using the vertex cuts on tessellated regular hexagons is more efcient afer all in terms of packing and production cost. Te paper is organized as follows. First, related work is summarized in Section 2 and the theoretical background on mathematical tessellation is given in Section 3. Te used Figure 1: A hexagonal silicon sensor for CMS [30]. methodology of the research is explained in Section 4. Next, in Sections 5 and 6, packing regular and irregular polygons into a circular container is considered, and solutions to the defned problem statements are provided with the PE calcu- However, most papers in the literature concentrate only on lation formulas. An irregular convex hexagonal sensor shape producing rectangular or squared sensors. Namely, in the is proposed in Section 6. Results are provided in Section 7, by past, sensor dies were separated by the specifc laser cutting comparing the packings. We give a real-life example where procedure, applying only straight lines on the wafer in the a concrete SE is calculated for six-inch or eight-inch wafers manufacturing process [20, 21]. Te former is known as used. Section 8 concludes the paper together with the future “guillotine cutting” from one edge to the other [22]. Using the work and it is followed by the references used. single straight lines is allowed with the regular pattern, so this mechanism forced dies to have regular shapes such as squares 2. Background and rectangles. Nowadays, it is possible to use methods of forming a .. Packing Eﬃciency Calculation. Basic requirement when semiconductor die that does not require applying only the producing sensors is cost reduction by improving wafer single straight lines. Hence, dies having various nonrect- productivity. Tis is defned as the fraction of the used wafer angular shapes can be formed, increasing the wafer yield area to the total wafer area [31]. Szabo et al. [8] defned the [23, 24]. Polygon dies further maximize the wafer utilization density of a circle packing �(�, �),where� is a radius of the and multipolygon die packing may allow for a signifcant inner circle and � is size of the square container. Te PE of � reduction of the wafer silicon material from which the packed circles is given by the formula: polygon dies are produced [25]. Many applications require production and use of these ��2� � (�, �) = . (1) abovementioned polygonal sensors, especially hexagons [26, � �2 27]. One specifc application in our focus is the design upgrade of the new CMS detector. For this purpose, hexag- Tis means that packing density can be expressed as the ratio onal sensors are produced from six-inch or eight-inch wafer between total area covered by the packed items and the area sizes. With these produced sensors, the aim is to efciently of the usable container [8]. We adopt the former formula to cover a circular detector area of interest which is a circular approximate SE. container in the packing context [28, 29]. According to [1], the corners of the hexagons are fattened, providing a .. Related Work. Many studies are done on maximizing triangular gap wherever three hexagons meet. One regular wafer efciency, mostly concentrating on square or rectan- hexagonal sensor for this purpose is shown in Figure 1 gular sensors. A method for optimizing number of square [30]. It has cut out triangles or “vertex cuts” needed to cells that geometrically ft on a circular wafer and maximizing provide mechanical apertures when covering the detector wafer yield is developed in [32]. Ferris-Prabhu presented surface. in his work algebraic expressions that explicitly relate the With these vertex cuts, irregular dodecagon sensor shape number of produced cells to the wafer diameter and to the needs to be produced, where a regular dodecagon is obtained geometric parameters of the cells [33]. Some formulas are with maximal cut applied. SE is expected to be larger when developed estimating how many cells of a given area will a single sensor is produced, but with 12 straight cuts applied ft on a wafer and how much wafer area will be lost due to on the wafer in the manufacturing process. Te question that roundness of the wafer [34]. An iterative cutting procedure is posed in this study is can we construct a sensor whose was developed for maximizing number of rectangular cells geometrical shape can be as efcient in terms of minimal produced from six-inch and eight-inch wafers to maximize number of six cuts such as using a regular hexagon? Possibly, wafer utilization [35, 36]. for these irregular hexagonal sensors, no vertex cuts are Unlike in the feld of sensor manufacturing where the needed since they are provided by default in the tessellation region of interest is a circular silicon wafer or a container, the process. circular CMS detector represents region of interest that needs Mathematical Problems in Engineering 3

 ∘ r  30 r /2 ∘ /2 r 15 r b ＞Ｃ va ＞Ｃ va b a a a a (a) (b) (c) (d)

Figure 2: Defned sensor geometrical shapes: (a) regular hexagon, (b) regular dodecagon, (c) irregular dodecagon, and (d) irregular convex hexagon. to be efciently covered. Also, in the feld of sensor networks, In this paper, we adopt the idea of using a regular grid sensor positions are crucial in the form of a hexagonal grid methodology to provide solutions of a packing problem. We to obtain better coverage efciency and to cover the sensing present mathematical constructions to calculate the packing area more efciently [37]. Te total number of sensors can density of embedding regular and irregular hexagons and be calculated using a known formula for centred hexagonal dodecagons in a circular container. Based on the literature numbers defning a ring of hexagons [38, 39]. Kim et al. [40] search, none of the papers deal with these specifc types apply a hexagon tessellation with an ideal cell size to deploy of cutting polygonal sensor shapes. Hence, we address that underwater WSN. Authors show that, for a circular region specifc type of packing problem when irregular hexagons of interest, one can calculate the number of hexagon rings and dodecagons are packed into circular container and we needed to fully cover the disk of a given radius. derive PE formulas. Tere are many research problems on packing and var- We defne four diferent sensor shapes produced from a ious shapes of packed objects and containers are studied. circular silicon wafer, that are presented in Figure 2. Irregular Using rectangular or square container approximation is dodecagon is gained from regular hexagon with small vertex provided in the literature, concentrating mostly on circu- cuts applied. Regular dodecagon is regular hexagon with lar packing problems. Litvinchev et al. [41] investigate the maximal vertex cuts applied. We adopt the irregular hexagon optimized packings of circles and “circular-like” objects in a with three symmetry axes in the triangular grid from [52]. rectangle to maximize the packed area and minimize waste. We extend the possible irregular hexagon types with respect Authors simplify the packing problem by using a regular to their size defned by the sides � and �. We tessellate those grid to model the inner positions of the packed items. Tis hexagons to cover a given area of interest and compare the formulation is further adopted by [42, 43] and applied for coverage efciencies. We prove that silicone waste would cutting and packing of circles and ellipses. Te goal is to pack be minimized when the ratio �/� is minimal. Increasing as many items as possible to increase the efciency and to the ratio is proportional to silicone waste. We also compare 2 decrease area of the container that needs to be produced. the tessellated irregular hexagons with {3, 12 } Archimedean Authors in [44] apply again a regular grid to a packing semiregular tessellation. We show that SE is negligible; these problem solution and derive a model for packing not only irregular hexagonal sensors can be produced with SE almost circles, but equal and unequal regular convex polygons such as in the regular semiregular tessellation. as hexagons in a rectangular region. Also, [45, 46] are We provide a study in two directions; frst, we analyze devoted to efcient solutions for packing convex objects in SE when � sensors are produced from a fxed-size circular the rectangular container. Diferent object shapes can be container (for each of the four sensor types). When �=1, adjusted to the given packing schemes and minimize the cost it represents the case when producing a single sensor. Next, by eliminating the unpacked areas. if all sensors are produced from the same wafer size, we compare SE when covering a circular surface that is fexible .. Contributions. Despite using a rectangular container in size. Generally, we propose new formulations to develop model as in the previous researches, it can be adjusted to a general model by using semiregular tessellation for the handle some other geometrical shape such as polygon or hexagon packing problem, giving the solution where the circle.Providedsolutionscansupportthesimplechangeof silicon waste is minimal. diferent container approximations [47]. Authors in [48, 49] provide a heuristic approach to maximize the number of unit 3. Tessellation Method items such as rectangles, squares, or circles inside a circle of known radius. Also, the polygon-shaped or circular container .. General Description. A tiling or a tessellation is covering is considered in [50, 51], with the aim to minimize the size of a fat surface using one or more tiles without any overlaps the polygon such that it contains a given set of fxed-size inner or gaps. Although the used tiles can have arbitrary shape, we circles and ellipses and to increase the fraction of the packed focus on regular polygons (hexagons and dodecagons). Con- container area. cerning the mathematical tiling context, edge is a boundary 4 Mathematical Problems in Engineering

(36 ) (44 ) (63 ) Figure 3: Tree regular tessellations (adjusted from [53]). which separates the two tiles, and vertex is an endpoint of an edge. We say that a tiling of the plane is an edge-to-edge tiling if it uses polygons only and all adjacent tiles share full edges. We say that a tiling of the plane is vertex-transitive if it uses only polygons and every vertex (where three or more tiles are connected) is the same as any other vertex, in terms of the number, kind, and sequence of the shapes surrounding it. Tiling is monohedral if all the tiles are of the same shape and size. Otherwise, it can be dihedral, trihedral, etc. Regular tiling is an edge-to-edge tiling by using congruent regular polygons. We could say it is a subtype of monohedral tiling. Uniform tiling is the tessellation of the plane by regular polygons with the restriction of being vertex-transitive. Tere are just three regular uniform tiling types given in Figure 3.

.. Regular Hexagon Tessellation

2 Proof. It is mathematically trivial to prove tessellation with Figure 4: Archimedean {3, 12 } semiregular tessellation (adjusted ∘ regular hexagons. Te sum of all angles in one vertex is 360 . from [54]). ∘ Te sum of the inner polygon angles is (� − 2) ∙ 180 .Asall ∘ inner angles are equal, it follows that �n =((n − 2) ∙ 180 )/n. Tere are � regular polygons in each vertex, and all the above single vertex is a meeting point for �1 regular polygons �1 and can be represented with the following equation (n ≥3): �2 regular polygons �2: ∘ ∘ ∘ (�1 − 2) ∙ 180 (�2 − 2) ∙ 180 ∘ (n −2) ∙ 180 ∘ � ∙ +� ∙ = 360 , (3) k ∙ = 360 . (2) 1 � 2 � n 1 2 where �1,�2,�1,�2 ∈�and �1,�2 ≥3(the smallest polygon �=2+4/(�−2) Relation (2) can be written as ,where � is a triangle). Tis can be also expressed as therearepossiblesolutionsonlyfor� ∈ {3,4,6}.Inthecase of regular hexagon where �=6,itgives�=3(number 1 1 1 1 �1 ∙( − )+�2 ∙( − )=1. (4) of hexagons in each vertex). Te other possible solutions 2 �1 2 �2 confrming tessellation are �=4and �=3. One possible solution for (4) is when �1 =3,�2 =12with �1 =1,�2 =2, and each vertex is a meeting point for .. Semiregular Tessellation. Archimedean or semiregular one equilateral triangle and two regular dodecagons. Tis is 2 tessellation uses more than one type of regular polygon with Archimedean {3, 12 } semiregular tessellation, as shown in the restriction of being vertex-transitive. Figure 4. Last possible solution allows usage of three diferent types of regular polygons in tessellation (it is easy to prove Proof. We show the proof of the simplest possibility which that it is not possible to use more than three). With similar allowsusageoftwodiferenttypesofregularpolygonsinthe argumentationasinprevioustwocases,weconcludethat tessellation. Te following equation represents the fact that there are eight possible semiregular tessellations [54]. Mathematical Problems in Engineering 5

4. Methodology .. Research Problem. Sensors are defned by four geometrical shapes, as shown in Figure 2. Te packing problem of embedding sensors in a circular container can be explained with the following problem formulations:

� R1 (i) Problem . Find the minimal circle radius such that R � 2 equal sensors of known size can be ft inside. R0 (ii) Problem . Find maximal size of a produced sensor, such that minimally � such sensors can be packed into a circular area of radius �. (iii) Problem . Find maximum number of sensors packed into a circular area with the known radius size. (iv) Problem . Derive PE (or SE) formulas.

.. Research Context. We use the following approach in solving the above problems. First, we start from a well-known Figure 5: Regular hexagon rings. regular hexagon tessellation and revisit the covering an area of interest without voids or gaps. We use formulas to fnd solutions for this general case when regular hexagons are (i) Producing � sensors and calculating SE: packed in the circular container. Next, since this “perfect” regular tessellation is not satisfactory in our case, not providing (1) �=1 spacings for mechanical apertures, we derive Archimedean 2 (a) Which one is the most efcient? {3, 12 } semiregular tessellation from the regular hexagon (b) Is producing an irregular hexagon cost- tessellation. Tis is done by cutting equilateral triangles at the efective? vertex which is common to three joined regular hexagons. (2) �>1 It is well known that PE is decreased when semiregular tessellation is applied with respect to the regular tessellation. (a) What is the SE ratio between regular and Te ratio is defned already for the problem of packing circles semiregular tessellation? into a squared container [9]. Even though it is expected to (b) Is producing � irregular hexagons cost- obtain lower PE when semiregular tessellation is used, we are efective? interested to calculate the ratio between the two to evaluate the diference of using regular and semiregular tessellation (ii) Packing sensors of known dimensions in the circular when packing sensors into a circle. area of interest and calculating PE which represents Because of the fexibility in choosing the size of the trian- coverage efciency: gular spacings with lowering the side of a regular dodecagon, we move from the regular dodecagons to the irregular (1) Flexible-size container in which we always pack ones (with 6 symmetry axes). We study the efciency when the same number of sensors packing them into a circular container. Irregular dodecagons (a) Which sensor shape requires the container can provide us the desired PE, but with larger number of to be the smallest? cuts when producing a single sensor (number of cuts is 12, (b) Which sensor shape reduces the unused compared to a regular hexagon that needs only 6 cuts in the packed space in the container and increases production). Hence, we examine the possibility of the sensor PE when area of interest is covered? to remain hexagonal in shape. We construct a sensor that is in (c) Which sensor shape enables that larger area a form of an irregular symmetric hexagon and we pack them of interest can be covered? in the circular area. To minimize the number of needed cuts of the sensor 5. Packing Regular Polygons into Circle and together with provided triangular spacings, we propose semiregular tessellation with irregular hexagons (with 3 .. Regular Hexagon Tessellation in the Circular Area. Let us symmetry axes) that is not edge-to edge. Our goal is to com- consider the regular hexagon tessellation with the coordinate pare our proposed packing with other shapes semiregularly system set like it is shown in Figure 5. Te circle of radius tessellating the plane, considering that triangles must be cut � centered in the coordinate system origin contains exactly out by default, so that we can have a fair comparison. Finally, one hexagon. We denote the central hexagon with �=0. we compare PE for each sensor type. Te nearest neighbors of the central hexagon create the frst “hexagon ring,” denoted by �=1. .. Research Questions. Te questions that need to be Number of hexagons in a ring is equal to 6. If we denote �ℎ answered with this study are formed by two aspects: the number of hexagons in � ring with ��,wecanseethat 6 Mathematical Problems in Engineering

�0 =1, �1 =6, �2 =12, �3 =18...�� =6�. Terefore, the total number of hexagons starting from the origin until the �ℎ � ring (included) will be

�� =�0 +�1 +�2 +⋅⋅⋅+�� =1+6+12+⋅⋅⋅+6�. (5) Afer summing the arithmetic sequence, we get the formula: R 2 1 �� =3� +3�+1. (6) � � With � we denote the radius of a circle containing ring R0 completely. Maximum distance from the hexagon vertex in �ℎ 2 2 2 R2 � ring from the origin is �� =(�/2) +(�� +2�∙��) , where the inner diameter �� of hexagon is calculated with the √ relation �� =� 3/2. Finally, we get the following:

√ 2 �� =�√�� =� 3� +3�+1. (7)

Solution (Problem ). Let us fnd the minimal value of radius � such that � equal nonoverlapping hexagons with known side value � can be packed inside. We will reformulate the � problem: for exact number of regular hexagons with side Figure 6: Regular dodecagon rings. �, how many rings does it take to cover them completely? If we know the number of needed rings �, then according to (7) � and (6), we also know �. Assumption √ 2 �� =�� ∙�ℎ,where�ℎ = (3 3/2)� is the area of a regular ��−1 ≤�<�� (8) hexagon. If we compare �� with a circle containing � rings area ��, results in we get the PE: 1 √12� − 3 � � ∙�ℎ 3√3 �=⌊ + ⌋. (9) � = � = ≈ 0.83. 2 6 � 2 2� (12) � �� Knowing the number of rings in the tessellation, according Tus, the ratio is constant and independent of � or � value. to (7), it is easy to determine radius �=��.

Solution (Problem ). Let us fnd the maximal size of a regular .. Archimedean Semiregular Tessellation. Using the same hexagon �, such that minimally � of them can be packed into logic with the notation like in the previous regular hexagon acirculararea� of radius �. tessellation case, we derive problem solutions on Archimede- {3, 122} By using the same assumption (8) as in Problem 1, we an semiregular tessellation. Te centre of the regular come to the same conclusion about the number of needed dodecagon’s circumscribed circle is matched with the sensor rings (9). Using (7), it follows that �=�/√3�2 +3�+1. centre (Figure 6). Dodecagons that are nearest neighbours to the central hexagon create a “dodecagon ring.” It includes few Solution (Problem ). How many regular hexagons of side equilateral triangles as well. Te total number of dodecagons �ℎ length � can be packed inside the circle with radius �? including the � ringisthesameasinthehexagonalcase(6). Again, we reformulate the problem: for a given circle of If �� is the radius of a circle containing ring � completely, � 2 2 2 radius , let us determine the maximum ring contained in then �� =(�/2) +(��+2�∙��) ,where�� is the inner diameter ∘ that circle. In this case, assumption of the dodecagons �� =(�/2)��75.Finally,weget � ≤�<� � �−1 � (10) √ 2 ∘ 2 �� = 1+��75 ∙ (2� + 1) . (13) 2 2 with notation �=(�/�) gives Solution (Problem ). Assumption in this semiregular tessel- 1 √12� − 3 lation case will be the same as with the regular tessellation �=⌊ + ⌋. (11) 2 6 (8) and due to (6), the conclusion about the number of rings needed is the same as (9): �=⌊1/2+√12� − 3/6⌋.Forgiven Using (6), the maximum number of packed sensors is �,using(15),wecalculate�=��. 2 �=��−1 =3� −3�−2. Solution (Problem ). Let us fnd maximal size of a regular Solution (Problem ). Let us derive the PE formula. Te area dodecagon, such that minimally � of them can be packed �ℎ of all hexagons from origin to the � ring �� (�� hexagons) is into a circular area � of radius �. Mathematical Problems in Engineering 7

Ｇ；ＲＲ ⇒ ；=＜

x b x b x a x a (a) (b)

Figure 7: (a) Irregular dodecagon, cut from regular hexagon. (b) For maximum x-cut value, dodecagon is regular.

Applying the same logic as in regular tessellation, using (9) we calculate � such that �� =�.Byusing(13),itfollows √ 2 ∘ 2 that �=2��/ 1+��75 ∙(2� + 1) .

Solution (Problem ). How many regular dodecagons with side length � can be packed inside the circle with radius �? Again, we use the same logic as in the regular hexagon 2 case. Assumption (10) with notation �=(2�/�) gives R1 1 R0 R �=⌊ (��75∘√�−1+1)⌋. 2 2 (14)

Terefore, the maximum number of packed sensors for a given ring � is �=��−1.

Solution (Problem ). Let us derive the PE formula. Te area 2 ∘ of a regular dodecagon is �� = 3� ��75 . Consequently, the �ℎ area of dodecagons from origin to the � ring is �� =�� ∙��. If we compare �� with area of circle containing k rings ��,we get the PE: Figure 8: Irregular dodecagon rings. � � ∙�� 36� � = � = � . � 2 ∘ (15) � �� [3��15 +16�� −4] �ℎ 2 2 dodecagon vertex in � ring from origin is �� =(�/2) +(V� + 2 Terefore, the ratio depends only on � value. 2� ∙ V�) .UsingthefactthatV� = �/(2��(�/2)),itfollowsthat

6. Packing Irregular Polygons into Circle 2 �√ 2� + 1 �� = 1+( ) . (16) .. Irregular Convex Dodecagon. If we take regular hexagon 2 �� (�/2) and starting from each vertex, we move along the side for the same distance �, like it is shown in Figure 7, and fnally we connect ending points, and we get irregular dodecagon. Solution (Problem ). Te number of dodecagons in each Regular dodecagon is just a special case where �=�which is ring and total number of dodecagons �� are the same as 2 true for the cut value �=�/(√3+2). in Archimedean {3, 12 } semiregular tessellation. Hence, by Let us now defne the irregular dodecagon rings (Fig- using the same assumption, the result about number of ure 8), whose number of contained sensors is the same as needed rings � is given by (9). For that � we calculate in all three previous cases. Maximum distance from irregular appropriate radius �=�� using (16). 8 Mathematical Problems in Engineering

20 20

18 18 b = 2

16 16

14 14 b = 2 12 12

10 10

8 8

6 6 a = 4

4 4 b = 1 b = 1 2 a = 2 a = 3 2 a = 3

0 0 20181614121086420 20181614121086420 (a) (b)

Figure 9: (a) Irregular hexagon �(1),(b)irregularhexagon�(2).

Triangular spacing is equilateral triangle with side � Te hexagonal shape below is adjusted from [40]. We 2 whose area is defned as �=�√3/4. defne hexagon size by two parameters; sides � and �.Te ratio �/� is fxed and it is defned by small triangle side ratio Solution (Problem ). Let us fnd the maximal size of a as shown in Figure 9. We extend the class of hexagons by regular dodecagon, such that minimally � of them can be increasing the ratio �/�, as shown in Figure 9. As we have packed into a circular area � of radius �. Applying the already stated before, we construct irregular hexagon which same logic as in regular tessellation, using relation (9), we decreases the number of cuts in the single sensor production calculate � such that �� =�. From (13), it follows that �= and enables the fexibility of the triangular spacings. We 2 defne irregular hexagon types denoted with �(Δ),where 2��/√1 + ((2� + 1)/��(�/2)) . Δ=|�−�|. �(1) Solution (Problem ). Assumption (10) is used. With notation We refer to irregular hexagon from Figure 9(a) as 2 �(2) �=(2�/�) it gives and from Figure 9(b) as , while hexagons in Figure 10 are referred to as �(4) and �(6), respectively. From the relation 1 � ∘ �=⌊ (�� √�−1+1)⌋. � + � = 120 (refer to Figure 2(d)), it follows that 2 2 (17) � √3 �� = . (20) By using (6), the maximum number of packed sensors is 2 1+2(�/�) �=��−1. Terefore, it is enough to know the ratio �/� (or �/�)to Solution (Problem ). To calculate the PE formula, frst determine the inner hexagon angles. For example, if �/� = ∘ ∘ we determine the area of an irregular dodecagon: �� = 7/5,thenitfollowsthat�≈71 and �≈49. 2 3� (sin �+sin �), where circumference circle radius � is Triangular spacing is equilateral triangle with side � that calculated by �=�/(2sin(�/2)). Te area of all dodecagons is defned as �=Δ/2. It follows that the triangle area is ��ℎ �� contained from origin to ring � is Δ2√3 2 2 �� = . (21) �� =�� ∙��=(3� +3�+1)∙3� (sin �+sin �) . (18) 16 ∘ Using (16) and because �+�=60, PE can be expressed as Terefore, based on the size of the triangular spacing, higher Δ provides larger triangle area that gives larger waste in the 2 ∘ � 3(3� +3�+1)∙cos (� − 30 ) sensor production, but it provides larger area for mechanical � = . 2 2 2 (19) apertures when � sensors cover a circular area of interest. �� (sin (�/2) + cos (�/2)(2� + 1) ) For �>�, the area of this irregular hexagon can be calculated by using the formula: .. Irregular Convex Hexagon. Let us consider an irregular � 3√3 2 ∘ convex hexagon with three symmetry axes in the triangular ��ℎ = � sin (150 −�), (22) coordinate system (Figures 9(a) and 9(b)). 2 Mathematical Problems in Engineering 9

20 20 18 18 16 16 14 14 12 12 10 10 8 8 6 6 b =2 4 4 b =1 2 a=6 2 a=5 0 0 20181614121086420 20181614121086420 (a) 20 20 18 b=2 18 16 16 14 14 a=8 12 12 10 10 8 8 6 6 4 4 b=1 a=7 2 2 0 0 20181614121086420 20181614121086420 (b)

Figure 10: Irregular hexagon. (a) �(4);(b)�(6).

where circumscribed circle radius � is calculated as �= Parameters V�, V� are shown in Figure 2(d). Finally, it �/(2 sin(�/2)). follows that If we compare areas of the regular and irregular hexagon with the same circumscribed circle radius, we get � � 3 √3 2 � = √1+[�� (1 + �) + �] (24) �ℎ 1 � 2 2 2 2 = ∘ . (23) ��ℎ sin (150 −�) or using �/� ratio Consequently,forthecasewheretheratio�/� = 7/5,it follows that �ℎ/��ℎ ≈ 1/0.98.WecanconcludethatSE � 1 � 2 � = √1+ ((3� + 1) + (3� + 2)) . (25) would be almost the same when one regular or irregular � 2 3 � hexagon �(1) is produced from a circular wafer of a fxed radius. Solution (Problem ). For the irregular hexagon rings, the We apply the same logic in describing irregular hexagon number of contained hexagons in each ring is the same as in rings(Figure11),aswithregularhexagonsused.Tetotal ��ℎ the regular hexagon case. Tus, by using the same logic from number of hexagons starting from the origin until the ring (8), we come to the same number of needed rings (9). As for is again given by (6). With �� we denote the circle radius that agiven�, we can calculate the appropriate radius �=�� by contains ring � completely, since maximum distance from �ℎ 2 2 using (32). hexagon vertex in � ring from origin is �� =(�/2) +(V� + 2 �∙V�) ,whereV� +V� = V� (it is easy to show that �� described Solution (Problem ). Let us determine the maximal size of using (shorter) side � has the higher value than the one with an irregular hexagon, such that minimally � of them can be � side). packed into a circular area � of radius �. Te same logic is 10 Mathematical Problems in Engineering

7. Results and Discussion PE is calculated with using every sensor geometrical shape, approximating SE in the sensor production cost and PE which represents coverage efciency when packing sensors of known dimensions in the circular area of interest. In the former context, we separate the following cases: R1 (a) A single sensor (�=1)isproducedbyafxed-size circular silicon wafer given by radius �. R 0 (b) � sensors are produced (�>1)byafxed-size circular silicon wafer given by radius �.

R2 (c) Sensors are produced from the fxed wafer size and they are packed into a circular container whose size is fexible. It is defned with the condition that always afxednumberofsensors� is contained inside (� sensor rings).

.. Single Sensor Packing in Circle (�=1). When a single sensor is produced, one must consider the useful wafer area availableinthemanufacturingprocess,asshowninFigure12. Figure 11: Irregular hexagon rings. SE is the largest for dodecagon sensor shape (≈96%) since it is the closest to the circle, while for regular hexagon SE ≈83%. When the proposed irregular hexagon �(Δ) is used, used as in solving previous Problem 2, which gives the needed efciency is presented in Figure 13 which is in concordance number of rings � such that �� =�. with the area ratios presented in Figure 14. Naturally, based In the previous discussion, we have narrowed the choice on relation (20), when �/� approaches the value 1, �(Δ) ∘ on irregular hexagons that have two diferent sides (�, �)that approaches the SE of a regular hexagon as � approaches 60 alternate forming a symmetric geometrical shape. Irregular angle (Figure 15). Naturally, �(1) is the closest to the regular hexagons with diferent side values can have the same area. hexagon, so it obtains the highest SE (where the minimal So, to determine the required type of irregular hexagon, we value is ≈80%). Hence, silicon waste is negligible when must know at least one of sides (�, �)or�/� ratio. Terefore, using this type of irregular hexagonal shape with respect to by using (20) and (24) we get the needed values. a regular hexagon. However, it is shown that using other irregular hexagons is also efcient. We can see in Figure 13 Solution (Problem ). Again, assumption is formulated with �� > 55 Δ≤5 2 that %for , which means that more than half of the inequation (10). By using notation �=(2�/�) it follows the wafer is utilized. that For a single sensor in the selected class of irregular 1 � hexagons (or for a specifc Δ), it is possible to calculate needed �=⌊ (√3 (�−1) +2+ )⌋ . 3 (1+�/�) � (26) criteria to satisfy the required SE. It follows that √ ∘ √ �0 3 3 sin (150 −�) 3 3 sin (5�/6 − �) Solution (Problem ). Let us derive the PE formula. = = . (29) � 2� 2� Te area of an irregular hexagon � is ��ℎ = 0 2 ∘ (3√3/2)� sin(150 −�). Te area of all hexagons from origin We can see from the above formula (29) that SE depends on �ℎ to � ring �� (�� hexagons in total) can be expressed as the angle � and this formula is true for every Δ.Hence,we can see for the specifc Δ which ratios �/� are satisfactory. For � =� ∙��ℎ � � example, if required SE is 75%, the above formula gives the � < 1.4821 � < 84.92∘ �(3) √ (27) angle or . Tis means that, in ,the 2 3 3 2 ∘ �≥3 =(3� +3�+1)∙ � sin (150 −�). required SE is achieved for . 2 Since regular dodecagon is the closest to the circular

If we compare the area of one circle containing � rings ��, shape (Figure 12), it is reasonable to conclude that, when with area �� we get the PE: comparing a single sensor production in the regular and irregular dodecagons used (with increased ratios �/� > 1), � � SE will be decreasing. �� >1 3√3� (30∘ +�) (28) .. Comparison of Hexagon Packings ( ) = � sin . 2 2� (1 + [�� (�/2)(1+(3/2) �) +(√3/2) �] ) sin2 (�/2) ... Regular and Semiregular Tessellation. When � sensors are produced from a fxed-size wafer, we use mathemati- Tus, the ratio depends on � and � value. cal tessellation mechanism with cantered-centric approach, Mathematical Problems in Engineering 11

SE=82.699% SE=95.71% SE=?

Figure 12: Straight lines applied in single sensor production.

0.85 110 105 0.8 100 95 0.75 90 85 0.7 80 75 0.65 ALPHA ANGLE 70

PACKING EFFICIENCY PACKING 0.6 65 60 0.55 12345678910 12345678910 SIDE B SIDE B H(1) H(4) H(1) H(4) H(2) H(5) H(2) H(5) H(3) H(6) H(3) Figure 15: Irregular hexagon �(Δ) angle �. Figure 13: SE for irregular hexagon �(Δ), Δ = 1, 2, 3, 4, 5.

1 Tis means that maximal number of sensors that can be 0.95 produced is always limited by the number of rings �,aswe 0.9 want sensors to be tessellated inside the circular area. Also, it �=� ̸ � 0.85 means that in the case where �, even if we need only sensors, more sensors need to be produced. Tus, we calculate 0.8

AH/AIH the sensor dimensions, such that � rings of them can be ft 0.75 inside a fxed-radius and calculate the used PE (SE). 0.7 As we can see in Table 1, PE is constant for regular tessella- 0.65 tion (83%), while Archimedean semiregular tessellation with 12345678910 regular dodecagon has �� ≈ 77%. Since regular dodecagon SIDE B is larger for �=0than a regular hexagon, its area becomes H(1) H(4) smaller compared to a regular hexagon when larger number H(2) H(5) of rings is packed in the same circular area. Tis causes H(3) H(6) lower PE with Archimedean semiregular tessellation used. Tis is expected, because silicon waste in the semiregularly Figure 14: Comparison of regular and irregular hexagon area. tessellated case is obtained by cutting triangles at the sensor vertices. Let us compare several semiregular tessellations with meaning that central sensor centre corresponds to the wafer irregular dodecagons used, where smaller triangles are cut centre. We form a requirement that than in the case of Archimedean tessellation. We can see in Figure 16 that the larger the ratio is between sides of an irregular dodecagon, the more we approach to the efciency �≤�. � (30) of the regular tessellation. Tis is a natural conclusion 12 Mathematical Problems in Engineering

Table 1: Regular Hexagon and Regular Dodecagon for 8” wafer (R=100 mm).

Regular hexagon Regular dodecagon kSk a [mm] Ah SE a [mm] Ad SE 01 100 25980.8. 51.8 30004.01 . 1 7 37.8 3711.5 . 17.8 3544.9 . 2 19 22.94 1367.4 . 10.7 1282.7 . 337 16.4 702.2 . 7.7 655.3 . 461 12.8 425.9 . 5.95 396.7 . 591 10.5 285.5 . 4.9 265.6 . 6127 8.9 204.6 . 4.12 190.2 . 7169 7.7 153.7 . 3.6 142.9 . 8217 6.8 119.7 . 3.15 111.2 . 9271 6.07 95.9 . 2.8 89.1 . 10 331 5.5 78.5 . 2.6 72.9 .

0.83 It is already shown that SE is decreasing with the increased 0.82 Δ (Figure 13), and we confrm this again in Figure 17 graphs. Δ 0.81 We see that, based on relation (20), as is larger, the ratio �/� approaches zero, getting �(Δ) further away from the SE 0.8 of a regular hexagon. Naturally, �(1) is the most efcient. We 0.79 can obtain larger SE when increasing the ratio �/� for the 0.78 individual irregular hexagon architecture with Δ>1. � PACKING EFFICIENCY PACKING 0.77 SE of the produced irregular hexagons is larger when the number of sensor rings �≥4, than in the case of 0.76 2 12345678910 Archimedean {3, 12 } semiregular tessellation where �� ≈ NUMBER OF RINGS K 77%. Tis is the case

a/b=1 a/b=2.5 (i) for every �/� ratio in �(1) a/b=1.5 a/b=3 (ii) �/� ≥ 0.5 in �(2), �(3), �(4). a/b=2 �<4 Figure 16: SE for semiregular tessellations with dodecagons (8” For lower number of sensor rings ,irregularhexagon wafer, R=100 mm). would be almost as efcient as Archimedean semiregular tessellation for any Δ. However, if one needs to obtain larger SE, it can be accomplished with �/� selected accordingly, such �/� that it is closer to a regular hexagon. from the fact that when the side ratio in the irregular � dodecagon is larger, smaller triangles are cut at the vertices, We can conclude that one can produce irregular causing the smaller amount of silicon waste. Tis is because hexagons with packing almost as efcient or even more the cut triangle side is �, so smaller size means smaller area efcient than in the case when using Archimedean semireg- and larger ratio �/�. ular tessellation. We can further increase the efciency by Maximal waste is obtained for the Archimedean semireg- using some other more efcient semiregular tessellation with ular tessellation, since maximal triangles are cut out from the irregular dodecagons, when smaller triangles are cut out at wafer, causing the largest silicon waste. the vertices, but the number of cuts when producing a single dodecagonal sensor is double. According to (21), we can see that the size of the trian- � ... Irregular Hexagon. When producing sensors that are gular spacing depends on Δ, where these two variables are irregular hexagons, we can see that, with the increasing num- proportional. When �(Δ) is increased, the triangular spacing � ber of rings packed in the circular wafer, SE is increasing, is increasing as well, causing larger amount of waste when while this was not the case with the irregular dodecagon. Te N irregular sensors are produced from the circular wafer. approximated value is based on the following limit functions, Tis decreases the SE but requires minimal number of cuts for irregular hexagon and irregular dodecagon, respectively: when a single sensor is produced. Also, it increases triangular spacings for mechanical apertures where PE in covering a � 3√3 (30∘ +�) � = sin circular area remains satisfactory. lim 2 ∘ (31) ��→∞ �� 2� cos (30 −�/2) ∘ ∘ .. Packing Fixed-Size Sensors into a Circular Container. Tis �� 9 cos (� − 30 ) 9 cos (30 −�) lim = = . (32) section will obtain the same PE results as the efciency in ��→∞ � 4� 2 (�/2) 4� ��2 (30∘ −�/2) � cos the previous one expressed in a form of SE. Namely, when Mathematical Problems in Engineering 13

H(1) H(2) 0.82 0.82 0.8 0.8 0.78 0.78 0.76 0.74 0.76 0.72 0.74 0.7 PACKING EFFICIENCY PACKING PACKING EFFICIENCY PACKING 0.72 0.68 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 NUMBER OF SENSOR RINGS (K) NUMBER OF SENSOR RINGS (K)

b=1, a=2 b=3, a=4 b=1, a=3 b=3, a=5 b=2, a=3 b=4, a=5 b=2, a=4 b=4, a=6

H(3) H(4)

0.82 0.82 0.8 0.78 0.77 0.76 0.74 0.72 0.72 0.7 0.68 0.67 0.66 0.64 PACKING EFFICIENCY PACKING PACKING EFFICIENCY PACKING 0.62 0.62 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 12345678910111213141516 NUMBER OF SENSOR RINGS (K) NUMBER OF SENSOR RINGS (K)

b=1, a=4 b=3, a=6 b=1, a=5 b=3, a=7 b=2, a=5 b=4, a=7 b=2, a=6 b=4, a=8

Figure 17: SE for irregular hexagon �(Δ), Δ=1,2,3,4.

sensor size is fxed, and the area of the container is adjusted (83%). However, triangular spacings are smaller and again accordingly, this is the same as when we have a fxed region we need to apply maximal number of cuts in a single sensor of interest and we adjust the sensor size. Te reason is that we production. always use the cantered tessellation approach defned by the Concluded from Figure 17, irregular hexagonal sensor number of sensor rings �. shape will reduce the unused packed space in the container When sensors are produced from the fxed-size wafers and increases PE when area of interest is covered. PE will and we pack them in the circular container, we can discuss be larger than Archimedean semiregular tessellation, with the size of the container that is required. Namely, the size smaller triangular spacings, but these are provided naturally of the container is fexible because we want to pack sensors by the packing process. Tere are no cuts needed at the of fxed dimensions with the centred tessellation mechanism regular hexagon vertices since a single irregular hexagon and the restriction that all packed sensors are whole, there are sensor can be produced with minimal number of six cuts in no partial sensors. Te size of a container is always defned total. such that a fxed and tessellated number of sensors with the We can compare the size of the container that we need valid restriction (30) can be ft inside. depending on how many sensor rings do we pack inside. When the size of the container is defned by the number of Sensor is produced from a fxed-size wafer and example for sensor rings that are packed inside, PE is the smallest for regu- 6” sensors is given in Figure 18(b). Since wafer radius is the lar dodecagon or Archimedean semiregular tessellation. Tis same for all sensor shapes, regular dodecagon will require the means that leaving largest triangles unused in the sensor ver- largest container (because we assume that all sensors must be tices causes the smallest coverage of the circular surface. On whole with no partial sensors allowed). Te reason for this is the other hand, we get the largest spacing area for mechanical that regular dodecagon is the largest in area and the closest apertures, but individual sensors are produced with maximal to a circular shape. Hence, regular dodecagon shape enables number of cuts. Using larger irregular dodecagons (like in that the largest area of interest can be covered. the previous section) can reduce the unused packed space Irregular dodecagons packed in the container cause the because triangular spacings are smaller, which increases PE fact that its size can be smaller, and closer to the size of towards the most efcient regular hexagonal tessellation used the container when regular hexagons are used. In this case, 14 Mathematical Problems in Engineering

1600 1600 1400 1400 1200 1200 1000 1000 800 800 600 600 400 400 200 200 0 0 12345678910 12345678910 NUMBER OF SENSOR RINGS K RADIUS OF THE CIRCULAR CONTAINER [MM] THE CIRCULAR CONTAINER OF RADIUS

RADIUS OF THE CIRCULAR CONTAINER [MM] THE CIRCULAR CONTAINER OF RADIUS NUMBER OF SENSOR RINGS K

H(3) a=1 b=4 H(1) a=1 b=2 regular hexagon irregular dodecagon a/b=2 H(2) a=1 b=3 regular hexagon irregular dodecagon a/b=1.5 regular dodecagon (a) (b)

Figure 18: Size of the fexible container used for packing (6” wafer, R=75 mm) compared to a regular hexagon. (a) Irregular hexagons, (b) irregular dodecagons.

the smallest container can be used for packing since regular increases the waste as expected. However, we can produce hexagons have the largest PE. On the other hand, since the larger number of smaller sensors from a fxed-size circular area of an irregular hexagon is smaller than the area of a waferinthecentredtessellatedmanner.TeSEofaregular regular hexagon, smaller container is needed for the same hexagon remains constant, and we derived the ratio between number of sensors that are packed inside (Figure 18(a)). Te ��.ℎ�� ≈83%and��.�� ≈77%, when � sensors are drawback can be that the same number of smaller sensors produced. Naturally, SE is increased with using dodecagonal can cover smaller area of interest. Tis means that we place sensor shapes, as waste of the silicon material is smaller. the same number of smaller sensors in the container knowing Te number of cuts needed in producing a single sensor that its size will be smaller, covering the smaller area with the is increased in double when dodecagon sensor shape is chosen sensor types. used compared with hexagon. Terefore, in this paper, we answered the question if we can use some other more efcient sensor shape that keeps the �� almost as if a regular hexagon 8. Conclusions was used, while the number of straight cuts when sensors are produced is minimal. With the proposed irregular hexagons, Regular tessellation enables area to be covered with no voids providing triangular space is enabled by default with semireg- or gaps, ensuring the constant �� ≈ 83%. Also, the same SE ular tessellation and applying the vertex cuts on the sensors is is obtained when a single sensor is produced in a form of a avoided. Te number of cuts needed in the sensor production regular hexagon. In this “perfect” tessellation, one important is reduced because sensor remains hexagonal, even though drawback is that there are no spacings for mechanical irregular in shape. Choosing a specifc group of irregular apertures and mounting, when � such produced sensors are hexagons �(Δ) enables us to decide on the size of the required packed inside the circular area. Tis is needed in a specifc triangular spacing with the �� that remains satisfactory. �� of application such as the design of the new CMS detector [1]. producing a single irregular hexagonal sensor is satisfactory Te spacings can be obtained with truncated tips at the ver- for Δ = 1, 2, 3 with �� > 65%, with the largest efciency tices of the hexagonal sensors. When using these sensors with obtained for the smallest Δ=1. Also, producing � such cut out maximal triangles at the vertices, we semiregularly sensors is cost-efective, with the SE larger than Archimedean 2 tessellate the area of interest causing Archimedean {3, 12 } semiregular tessellation. Te efciency is growing with larger semiregular tessellation. Tis way, sensor becomes a regular number of sensor rings produced, which is not the case in dodecagon. Also, using irregular dodecagonal sensor shape dodecagonal sensors used. When more sensors are produced, can provide us with some additional fexibility in adjusting SE is getting closer to the desired 83% of a regular hexagon, thesizeofthetriangularspacings,whereloweringthespacing which means that we can produce these sensors with almost area causes the enhanced PE of the area. thesameSE. 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