48 24.674% respectively, also the Auxetic design has lower RSD than the original design approximately about 15%. Conclusions The results showed that the Reuleaux triangle honeycomb structure has the lowest RSD at the top and front views. This can explain that the Reuleuax triangle honeycomb structure has a side as the curve, which can spread the force and affect lower stress to the endpoint of a curve than the re-entrant honeycomb structure which has the endpoint of the side as a line, and the Re-entrant structure class has every unit cell are connected directly, so that can spread the stress more than another class. Then, we concluded that the Reuleaux triangle honeycomb structure is the best force distribution. Another result showed that the new helmet design has a lower RSD than the original helmet design around 15%. Because the Auxetic structure applied to the helmet has properties are force distribution, strength, and absorption energy. Then, it can decrease the chance of the head getting force attacked on the spot, but the force attack will spread outside the helmet. So, we concluded that the new helmet design can decrease the chance of injury from accidents by around 15%. Acknowledgments We would like to acknowledge and give my warmest thanks to our advisors, Dr. Saknarin Channark and Mr. Jirakoon Erbim. Their guidance and advice carried me through all the stages of writing our project. Would also like to thank my project members for their brilliant comments and suggestions, thanks to you. Finally, would like to thank our school, for gift us the opportunity to do the project. References [1] Meena, K., & Singamneni, S. (2019). A new auxetic structure with significantly reduced stress concentration effects. [2] Influence of auxetic structure parameters on dynamic impact energy absorption. viewed 18 August 2022, <https://hal.archives-ouvertes.fr/hal035830 95> [3] Kelkar, P. U., Kim, H. S., Cho, K.-H., Kwak, J. Y., Kang, C.-Y., & Song, H.-C. (2020). Cellular Auxetic Structures for Mechanical Metamaterials: A Review. [4] Daniel Acuna, Francisco Gutiérrez, Rodrigo Silva, Humberto Palza, Alvaro S. Nunez & Gustavo Düring. viewed 25 October 2022, <https://www.nature.com/articles/s42005- 022-00876-5> [5] VARUN VACHHAR. (2018). Reuleaux Polygons. viewed 22 October 2022, <https://varun.ca/reuleaux-polygons/> [6] Emily VanSchmus. Four-Leaf Clover Facts to Know for St. Patrick's Day. viewed 26 October2022, <https://www.bhg.com/holidays/st-patricksday/traditions/fun-facts-about-four-leafclovers/> [7] Gary Meisner. Is the Nautilus shell spiral a golden spiral? Viewed 26 October 2022, <https://www.goldennumber.net/nautilusspiral-golden-ratio/> [8] Mohammad Nasim, Md.Jahid Hasan, &Ugo Galvanetto (2022). Impact behavior of energy absorbing helmet liners with PA12 lattice structures: A computational study. [9] Harris, I. M., Lee, K. C., Deeks, J. J., Moore, D. J., Moiemen, N. S., & Dretzke, J. (2020). Pressure-garment therapy for preventing hypertrophic scarring after burn injury. Cochrane Database of Systematic Reviews. RAY (2022). Calculating the Centroid of Compound Shapes Using the Method of Geometric Decomposition. viewed 3 October 2022, <https://owlcation.com/stem/How-to-SolveCentroids-of-Compound-Shapes> [10]
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 49 Patitta Dansikaew1 and Supanicha Srisopa1 Advisors: Wararat Sri-od2 Special Advisor: Attaphum Tubtim3 1,2Princess Chulabhorn Science High School Phitsanulok, Makham Sung, Mueang, Phitsanulok, 65000, Thailand Abstract The purpose of this mathematics project at the high school level is to explore the creation of Pa Daeng weaving patterns using the GSP mathematical software. The goal is to not only understand the pattern of Pa Daeng weaving cloth, but also to develop visualization and problem-solving skills, and to integrate mathematics and technology in the learning process. This project aims to promote multiple intelligences, including linguistic intelligence, logical intelligence, dimensional relations and art, which can be applied to everyday life. The study found that the Geometer's Sketchpad (GSP) program was effective in creating the desired pattern of Pa Daeng weaving fabric, and a satisfaction study conducted among the consumer group showed that most of the participants found the pattern beautiful and well-ordered. The use of this program can help increase the value of Pa Daeng weaving fabrics and provide a source of income for the people in the community. A study of Geometric Transformations in Conjuction with the GSP Program for the design of Pa Daeng Weaving Patterns. First, The summary of the study found that creating patterns on fabric using a combination of transformations and the GSP (Geometer's Sketchpad) software makes the pattern creation process easier. There is no need to create multiple patterns; instead, the use of translations allows for greater convenience. The GSP program enables the creation of multiple patterns by incorporating mathematical principles into the design process. In the consumer evaluation, it was found that the majority of consumers rated the patterns as aesthetically pleasing and well-organized. Keywords: The Geometer's Sketchpad program (GSP), Pa Daeng weaving fabric. Introduction Mathematics is a subject that deals with reasoning, thinking processes, and problem solving, and plays a significant role in the development of human thinking. It helps improve creativity, reasoning, systematic thinking, and enables one to analyze problems or situations thoroughly and carefully. Mathematics also aids in forecasting, planning, making decisions, solving problems, and its usage in daily life is crucial. According to the Office of Academic and Educational Standards (2008), mathematics is an important tool in science, technology, and other sciences. However, teaching and learning in mathematics still faces challenges, as can be seen from research reports on problems and solutions for teaching and learning management that affects its development. The quality of learners at the basic education level has been found to be affected by teaching and learning activities that are not well-informed or not connected to real-life experiences. The mathematics project is a way for learners to accumulate knowledge and create their own works, as stated by Ladda (2011) and Phattadon (2014). Phattadon's research studied the mathematics learning outcomes of
50 Mathayomsuksa 5 students who were taught mathematics through projects, and found that their learning outcomes were higher after the project than before. Local wisdom is a body of knowledge that is accumulated through experience, development, systematic and continuous education. Encouraging children and youth to inherit wisdom from the knowledgeable community will help preserve and continue Thai arts and crafts. Taphan Hin District in Phichit province is an area with a rich source of local knowledge learning resources. The study of the mathematics project integrated with local wisdom aims to understand the learning outcomes, mathematical process skills, and satisfaction with learning mathematics, as well as guide improvements to make teaching mathematics more effective. Pa Daeng weaving cloth is a local wisdom and the native cloth of Ban Pa Daeng, which is woven using a loom. It is known for its exquisite beauty, unique patterns, durability, and color fastness. The price of Pa Daeng woven cloth is also lower than other hand-woven fabrics, and it can be cut into dresses for various important occasions, with patterns passed down from ancestors such as vine, geometric, floral, and animal patterns, and even special characters as required by customers. The organizing committee aimed to bring together the knowledge of geometry and the GSP program to design new patterns for Pa Daeng weaving cloth, adding value and generating income for the people in the community. The study was conducted to evaluate the efficiency of the project and assess the satisfaction of the consumer group, with most results indicating that the new patterns were beautiful and well-ordered. Materials and Methods Step 1: Preparing the Fabric Design - In this step, the first step towards creating the Pa Daeng weaving pattern using the GSP program is to familiarize oneself with the concepts of geometric transformations, reflections, rotations, iterations, and parallel shifts. This knowledge, combined with experimenting with the GSP program, will help in preparing for the design of the fabric pattern. Step 2: Fabric Design Process In this step, the fabric patterns are designed using the Geometer's Sketchpad program by KrabiGsp. There are two design processes described: the S-pattern design process and the design process of the ears of rice pattern. The S-pattern design process involves the following steps: 1. Open the Geometer's Sketchpad program by KrabiGsp 2. Draw a square ABCD 3. Move the square parallel along the vectors WX and YZ to get the desired letter S shape. 4. Hide dots and lines as needed and keep scrolling parallel until a reasonable distance is reached. 5. Add simple basic patterns to the design.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 51 2.2 The design process of the ears of rice pattern involves the following steps: 1. Open the Geometer's Sketchpad program by KrabiGsp 2. Design a rice basket pattern. 3. Design a tree pattern. 4. Design a turbine pattern. 5. Combine all three patterns to complete the design. Step 3: Consumer Satisfaction Study In this step, the organizers collect data from 20 sample groups of consumers to study their satisfaction with the Pa Daeng weaving patterns designed using knowledge of geometric transformations and the GSP program. The data is collected through a questionnaire and the results are analyzed. Satisfacation/ Subject Very good fair improve beauty 12 4 4 orderliness 15 4 1 * Data based on 20 individual Results and Discussion The first part of the study focused on utilizing geometric transformations in conjunction with the Geometer's Sketchpad (GSP) program. The aim was to familiarize with the tools in the GSP program and apply them to the creation of patterns through parallel shifts, rotations, and reflections. Part 2: Designing Pa Daeng Weaving Patterns Using Geometric Transformations and GSP Program The study of using the GSP program in conjunction with geometric transformations has shown that by incorporating rotations, reflections, and parallel shifts, unique and desired fabric patterns can be created. This is demonstrated through the use of parallel scrolling in the GSP program to produce images. Part 3: A study of consumer satisfaction toward Pa Daeng weaving patterns designed using the knowledge of conversion. Conclusions The results of the study showed that using the GSP program in combination with conversion made it easier to create fabric patterns. The program allows for the application of mathematical principles, and with the use of parallax scrolling, it is more convenient to produce a variety of patterns. In the evaluation of consumer satisfaction, the majority of participants rated the Pa Daeng weaving patterns as attractive and well-designed.
52 Acknowledgments This project has been successfully completed and I would like to express my gratitude to all those who have helped and contributed to its completion. Special thanks to Ajarn Wararat Sri Aod, my professor who has advised and guided me throughout the project by providing knowledge, support, and feedback. His suggestions, guidelines and advice have been invaluable in helping me to complete this project. I would also like to thank Ajarn Attaphum Tubtim, my co-educator, for his support in writing the English article and abstract. Additionally, I would like to extend my gratitude to the entire project team, even those not mentioned by name, for their help and support. I hope that this project report will be useful for anyone who is interested in further research. References Donnaya Suvetwethin. (2018). Ban Pa Daeng Weaving Group. Retrieved on January 8, 2023 from the website https://www.thaihealth.or.th/Ban Pa Daeng Weaving Group/. Play and learn with Teacher Tam (2021). Rice pattern design Using GSP program. Retrieved 8 January 2023 from the website https://youtu.be/A7DR1kU9rT8. NuPenDekDee. (2019). Apichat Petchploy. (2014). pattern from zoom and reproduction in GSP. Retrieved January 8, 2023, from https://youtu.be/zJPqIcJ6xMs. SongTV Thailand. (2016). Making Trees in GSP. Retrieved January 8, 2023, from https://youtu.be/m27_WfXQWL8. Patthadon Wananam. (2016). A Study of Mathematics and Local Wisdom Projects and Elevating Learning Achievement. Retrieved on January 8, 2023, from the website https://so05.tcithaijo.org/index.php/rmutsbhs/article/view/97831. Lect. Dr. Usarat Ratanakanumn. (2014). IC 361 Science of Fibers and Fabrics. Retrieved on January 8, 2023 from www.science.mju.ac.th/chemistry/downl oad/u_ratanakamnuan/IC%20361%20 Fiber and Fabric Science 1-57.pdf.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 53 Wattima Pintong1 and Piyapat Pethin1 Adisor : Pacharee Chaipeth2 1,2Princess Chulabhorn Science High School Trang, Bangrak, Mueang, Trang, 92000, Thailand Abstract This project aim to the number of polygons in an equilateral triangle n units aims to study methods for counting polygons that can occur in an equilateral triangle n units by using the principles of sequence and series form. in order to study and prove the simple form of the total number of formed images The case studies are divided as follows: 1. The number of triangles is divided into 2 cases as follows: ( )( ) 1 2 1 2 3 1 8 n n n + + − ; odd n I + ( )( )( ) 1 2 1 2 8 n n n + + ; even n I + 2. The number of rhombus is divided into 2 cases as follows: ( )( )( ) 1 1 2 3 1 8 n n n + + − ; odd n I + ( )( )( ) 1 2 1 2 8 n n n − + ; even n I + 3. The number of parallelogram is divided into 2 cases as follows: number of polygons = ( )( )( ) 1 2 3 1 1 8 n n n − − + ; odd n I + The number of polygons in an equilateral triangle n units number of polygons = ( )( )( ) 1 2 2 2 8 n n n − + ; even n I + 4. The number of trapezoid= ( ) 1 3 2 4 15 29 24 2 n n n − + − 5. The number of equilateral hexagon is divided into 3 cases as follows: n mod 0 3 ( ) : ( )( ) 1 2 3 8 n n − n mod 1 3 ( ) : ( )( ) 1 2 2 1 8 n n + − n mod 2 3 ( ) : ( )( ) 1 2 2 1 8 n n − + Keywords: polygons , n units Introduction An equilateral triangle is composed of a unique feature that includes all the same edges and all angles of the same size. Allow manufacturers to choose to use equilateral triangles to create projects. The content of the demonstration project is about taking any n unit sized side triangles. Our project will study five geometric structures, namely equilateral triangles. Diamonds, parallelograms, trapezoids, and equilateral hexagons. Materials and Methods Experimental process 1. Always select any n squares as the minimum value (1 unit) first. 2. Draw a triangle divided into sub triangles to set an example that displays the counting method from 1 to 10 units. For certain formats, it may be necessary to draw additional values to facilitate counting and accurately observe the number format and counting method. 3. Always calculate the required number of squares starting from a shape with a side length of 1 unit. 4. Observation mode counting method.
54 5. Observe the relationship between counting and the number of photos to obtain a simple photo as an example. 6. Model the total number of images. 7. Verifying serial accuracy through mathematical induction Part 1 The number of Equilateral triangles 1.1 Sub equilateral triangles 1 unit From this process and method, we can seen that number And then we will proof this by mathematical Induction 2 1 (2 1) i n i i n = = − = () 2 1 (2 1) i n i i n = = − = Base step : (1) get 1 1 (2 1) i i i = = − thus 1 = 1 2 ∴ (1) is true. Inductive step : () is 1 (2 1) i k i i = = − = + + + + − 1 3 5 ... (2 1) k Suppose that () is true. () : 2 1 3 5 ... (2 1) + + + + − = k k ( + 1) : 1 3 5 ... (2 1) 1 3 5 ... (2 1) (2 1) + + + + + = + + + + − + + k k k 2 = + + k k (2 1) 2 = + ( 1) k ∴ ( + 1) is true. Simplify P(n) is true from that P(k) is true and made P(k + 1) is also true. 1 . 2 equilateral triangle, 2 units or more (bottom down) From the preliminary count, the following counting patterns can be observed: From the nature of the calculation and validation as well. Part 2 The number of Rhombuses 2.1 sub rhombus 1 unit Then we get number 1 1 3 i n i i = − = = ( )( ) ( )( ) 3 1 1 1 2 3 1 2 n n n n = − − + = − ( ) 1 1 1 2 1 1 ( ) 1 ( 1)( 1 1) 2 2 ( 1 1) 2 (2 ) 2 i n i n i i i i n n n n n n n n n n = = − = = = + = + + − − + = + + − = = 2(1) 1 1 = − =
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 55 2.2 Common rhombus Ex. 5 units Ex. 6 units It can be noted that the parity of n affects the counting method. from the counting methods mentioned above. will lead to a simple image formatting It will find the subseries that occur before checking. According to the parity of n, it can be concluded that Common rhombus is ( )( )( ) 1 3 1 2 1 8 = − − − n n n ; odd n I + ( )( )( ) 1 2 2 5 8 = − − n n n ; even n I + Part 3 The number of Parallelograms And base side effect for counting. Sub parallelogram = − − (n n n )( 1 2 )( ) 3.1 Common parallelogram Inside triangle 9 units we can seen 2 × 3,4,5,6 3 × 4,5,6 4 × 5 Notice how the addition of the number of figures is specific to the parity of n so that the sum of all figures is a nested series. by before splitting the case. The characteristics of the sum will be written as follows: number ( )( ) ( )( ) 0 1 1 3 2 1 2 n k k i k k k i i i = = = = = + By is member of series. when odd n I +
56 that we get k n = − 1,3,5,..., 4 so k m = − 2 1 ; 1, 2,3,... m = ; 5,7,9,... n = the last word 2 1 4 n k l n = − = − l =1, 2,3,... ∴ 3 2 n l − = when even n I + get k n = − 2, 4,6,..., 4 so k p = 2 ; 1, 2,3,... p = ; 6,8,10,... n = the last word k q n = = − 2 4 ∴ 4 2 n q − = When calculating the values in each case and checking them through mathematical induction, the conclusion is drawn Number when odd n I + ( )( )( ) 1 2 3 1 4 1 8 = − − − + n n n n and when even n I + ( )( )( ) 1 2 4 2 8 = − − n n n Part 4 The number of Trapezoids 4.1 Sub-Trapezoid (Inverted base) From observing how to count in a pattern and then calculate and check that. number in this case ( )( )( ) 1 1 1 2 = − + n n n 4.2 Sub-Trapezoid (Supined base) . number in this case ( )( )( ) 1 2 1 2 = − − n n n 4.3 Common-Trapezoid (Inverted base) Different forms can be generated based on the cardinality, in which case the number of births is the same as in the case of 4.2 4.4 Common-Trapezoid (Supined base) For any triangle of n units, the calculation mode is the same. In the next line, the number will gradually decrease and will gradually decrease to There is only one photo in a row, and the same counting feature was also found in larger photos. Due to the fact that a triangle has three sides, it can be calculated based on three identical cardinal numbers. From the calculations and examination, it can be seen that: number ( )( )( ) 1 432 2 = − − − n n n Part 5 The number of Equilateral hexagon for 6 for 7 for 8 for 9 From the appearance of the hexagons, it shows that the 7 and 8 units triangles have the same hexagon pattern as the 6 units triangle, but in the 9 units triangle there is one larger hexagon, so the hexagonal pattern in any n units triangle will correspond to the remainder from dividing n by 3. case 1 : n 0 mod3 ( )
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 57 Considering that there are always n-2 sub hexagons in the bottom row, reducing one in each row for larger hexagons. There is a hexagonal number n-4 in the bottom row, and then n-6, n-8, and 1 are reduced. Which always forms one. As the number of arithmetic sequences increases: i k = + − 1 3 1 ( ) ;k I + So that ; i k n = − − 1, 4,7,...,3 2,..., 2 Consider in the last row: i n n = − 2 We seen that 3 2 2 r n − = − ; n k r = 3 n r = Number ( )( ) 0 1 1 2 n i i i i i i = = = + From the calculations and examination, it can be seen that: number ( )( ) 1 2 3 18 = − n n case 2 : n 1 mod3 ( ) Similarly, considering the largest hexagon, it will only form two. 0 i = 2 . As the number of arithmetic sequences increases: i k = + − 2 3 1 ( ) ;k I + i k n = − − 2,5,8,...,3 1,..., 2 Consider in the last row: i n n = − 2 We seen that 3 2 2 z n − = − ; n k z = 1 3 n z − = Number ( )( ) 0 1 1 2 n i i i i i i = = = + From the calculations and examination, it can be seen that: Number ( )( ) 1 2 2 1 18 = + − n n case 3 n 2 mod3 ( ) Similarly, considering the largest hexagon, it will only form three. i 0 = 3 . As the number of arithmetic sequences increases: i k = + − 3 3 1 ( ) ;k I + i k n = − 3,6,9,...,3 ,..., 2 Consider in the last row: i n n = − 2 We seen that 3 2 y n = − ; n k y = 2 3 n y − = Number ( )( ) 0 1 1 2 n i i i i i i = = = + From the calculations and examination, it can be seen that: Number ( )( ) 1 2 2 1 18 = − + n n Results and Discussion After studying all subformats in each case, the creator adds them up and formats them again. To lead to the final result that show it in conclusions part.
58 Conclusions 1. The number of triangles is divided into 2 cases as follows: ( )( ) 1 2 1 2 3 1 8 n n n + + − ; odd n I + ( )( )( ) 1 2 1 2 8 n n n + + ; even n I + 2. The number of rhombus is divided into 2 cases as follows: ( )( )( ) 1 1 2 3 1 8 n n n + + − ; odd n I + ( )( )( ) 1 2 1 2 8 n n n − + ; even n I + 3. The number of parallelogram is divided into 2 cases as follows: ( )( )( ) 1 2 3 1 1 8 n n n − − + ; odd n I + ( )( )( ) 1 2 2 2 8 n n n − + even n I + 4. The number of trapezoid = ( ) 1 3 2 4 15 29 24 2 n n n − + − 5. The number of equilateral hexagon is divided into 3 cases as follows: n mod 0 3 ( ) : ( )( ) 1 2 3 8 n n − n mod 1 3 ( ) : ( )( ) 1 2 2 1 8 n n + − n mod 2 3 ( ) : ( )( ) 1 2 2 1 8 n n − + Acknowledgments The triangle project in the n unit equilateral triangle is completed very well. Thank you to the Mathematics Department for providing you with a place to learn. Thank you to Ms. Pajaree Chaipetch, the project consultant.Thank you to all the project organizers who worked together to think and complete the project. Finally, thank you to Churabhorn Rajavidyalai Science School. Trang provides a great opportunity for all students to complete the project so that they can apply the knowledge in class. References [1] Chulatutor. What are sequences and sequence abstracts. Retrieved on January 8, 2022, fromhttps://www.chulatutor.com/blog [2] wikipedia. (2564). Mathematical induction. Retrieved on December 25, 2021, from https://th.wikipedia.org/wiki/%E0%B8%81 %E0%
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 59 Thanapol Yaba1* Russalee Leemanan1 Natthawara Lerdariyapongkul1* Kamtorn Chailuek2 1Princess Chulabhorn Science High School Satun 2Faculty of Science, Prince of Songkla University *Email: [email protected] Abstract This mathematical project on parallel parking trajectory equation was caused by a notice of people’s parallel parking problems, which needs to take into account the parking angle and space. In addition, parallel parking is one of the driving tests for obtaining a personal driver's license, where test takers need to park their car properly and consider various factors affecting their parking. The project team were therefore interested in the function equation, making it possible to perform a parallel parking through only one attempt as well as to determine the total parking space through such function equation graph. According to the results of this project, the parallel parking trajectory equation and the total parking space can be determined by. () = ℎ + 0.5 1 + 4 (ℎ+0.5) where ℎ is parked car width Keywords: Footpath, Parking, Logistic function The Parallel Parking Trajectory Equation Introduction Currently, parallel parking requires a lot of driver skills and expertise as well as must take into account the car size, car parking angle, and external factors causing problems in parking, such as total parking space and driver experience. In addition, it is one of the driving tests for obtaining a personal driver's license, where test takers need to park their car properly and accurately and consider various factors affecting their parking. From the above-mentioned problem, the team saw the importance of parallel parking trajectory as a problem of most people today. Therefore, this project was conducted to examine the parallel parking trajectory equation to make it possible to perform a parallel parking through only one attempt and an advance of technology invented creating a new technology of automatic parking and also determine the total parking space by using knowledge of mathematics to solve the problem, which can bring maximum benefits to both the drivers and the general public. Materials and Methods 1. According to the study through highangle images and parallel parking principles, it was found that the car should be parked at an angle of 45 degrees with the horizontal line by referring to the footpath edge. The parking trajectory was similar to the logistic curve or logistic function graph FiParked car width Distance between 2 cars 0 Driver’s car width Figure 1 Parallel parking trajectory equation graph using the logistic function equation with graph description
60 with the following equation: () = 1 + −(−0) Where 0 is midpoint of logistic function graph; is car width; and is logistic growth rate or curve slope. Considering the curve slope with a value of 1, was substituted with 0. In addition, any could be substituted on the graph with a slope of 1 Because tan 45° = 1, the derivative of the function with respect to is calculated using and as constants. ( 1+−) = − (1+−) 2 = Graph slope Substituting x=0 from considering the slope of the curve with a value of 1 as in Figure 2 −(0) (1 + −(0)) 2 = 1 4 = 1 = 4 When considering the distance from the axis = 0 to = in Figure 1 , it is discovered that it is equal to the total of the width of the parked car and the distance between the two cars, which is approximately 0.5 m. = ℎ + 0.5 where ℎ is parked car width. Substituting0 = 0 and = 4 ℎ+0.5 ; () = ℎ + 0.5 1 + − 4 (ℎ+0.5) Because in Thailand, parking compared to the footpath, the car will reverse from the left side by substituting = − the parallel parking trajectory equation could be determined that; () = ℎ + 0.5 1 + 4 (ℎ+0.5) where ℎ is parked car width. The parallel parking trajectory equation was used to create a graph using the GeoGebra program by substituting ℎ with 2 to observe the condition of the graph. 2. According to the study through high-angle images and parallel parking principles, it was found that the car should be parked at an angle of 45 degrees with the horizontal line by referring to the footpath edge. The parking trajectory was also similar to the inverse tangent function graph with the following general equation: () = tan−1 The inverse tangent function graph was shifted up for a distance of 2 above the x-axis so that it could be compared with the logistic function equation graph. The equation was as follows: () = (tan−1 + 2 ) Figure 2 The logistic function equation graph with the slope of the curve 1
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 61 As wishing the graph to have a height equal to the car width () and a slope () of 45 degrees, the equation was as follows: () = (tan−1 + 2 ) Mathematical Proof lim→∞ () = lim→∞ (tan−1 + 2 ) = ( lim→∞ tan−1 + lim→∞ 2 ) = ( 2 + 2 ) = () = Then, the function derivative could be determined as follows: (tan−1 + 2 ) = ( 2 2 + 1) Substituting with 0 ′ (0) = Because the car angle should be of 45 degrees to the horizontal line, the curve slope was 1. = 1 = the parallel parking trajectory equation could be determined as follows: () = (tan−1 + 2 ) When considering the distance from the axis = 0 to = in Figure 1, it is discovered that it is equal to the total of the width of the parked car and the distance between the two cars, which is approximately 0.5 m. = ℎ + 0.5 where ℎ is parked car width. Therefore, the parallel parking trajectory equation could be determined using the inverse tangent function equation as follows: Figure 3 Parallel parking trajectory equation graph using the logistic function equation when substituting ℎ = 2 Figure 4 Parallel parking trajectory equation graph using the inverse tangent function equation when shifted up for a distance of 2 above the x-axis
62 () = ℎ + 0.5 (tan−1 (ℎ + 0.5) + 2 ) In this case, the function can be adapted for parking relative to the sidewalk when the car reverses from the left by substituting = − Therefore, the parallel parking trajectory equation could be determined as follows: () = ℎ + 0.5 (−tan−1 (ℎ + 0.5) + 2 ) where ℎ is parked car width. The parallel parking trajectory equation was used to create a graph using the GeoGebra program by substituting ℎ = 2 to observe the condition of the graph. Results and Discussion 1. From the study on parallel parking trajectory through high-angle images and parallel parking principles, it was found that the car should be parked at an angle of 45 degrees to the horizontal line to make it possible to perform a parallel parking through only one attempt.. Considering the trajectory line from the beginning until the end of the parking, the logistic function equation graph and the inverse tangent function equation graph could be compared as follows: According to the comparison of the equation of the parking trajectory compared to the footpath using the logistic function equation and the inverse tangent function equation shown in Figure 6, the logistic function equation will reach the L value faster and more consistently than the inverse tangent function equation graph. Furthermore, when evaluating the inverse tangentfunction equation graph, it takes longer to get from the sidewalk to the sidewalk than it does to form a 45 degree angle with the horizontal. Therefore, the parallel parking trajectory equation could be determined as follows: () = ℎ + 0.5 1 + 4 (ℎ+0.5) where ℎ is parked car width. 2 .From the study of the principle of parking compared to the footpath, it was found that the parking distance should be at least 1 . 5 times the width of the driver's car. When substituting the width of different types of car into the equation of the parking trajectory compared to the footpath using the logistic From Table 1 shows the relationship between parking distance to vehicle length. when substituting the width of the car length of the car from type Makes and models of different types of cars were then used to graph the equation of the parking trajectory compared to the footpath using the logistic function equation. Then calculate the length from the moment the car starts to park function equation Then study the correlation between the parking distance and the length of different types of passenger cars. The results are as follows. an angle of 45 degrees with the horizontal until the end of the parking to find the minimum parking distance that the car can park on the sidewalk at once. In addition, the corrleation between the parking distance to the length of the driver’s car in various type: sedan, pickup, van and truck can be summarized as 1.89, 1.59, 1.51 and 1.10 respectively. Figure 5 Parallel parking trajectory equation graph using the inverse tangent function equation when substituting ℎ = 2
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 63 Conclusions 1. From the study on parallel parking characteristics and principles, it was found that they were similar to the logistic function equation and the inverse tangent function equation. Then, the graph characteristics of these two equations were compared, and it was found that. The logistic function equation curve enters the value faster and more stable than the inverse tangent function equation curve and can shorten the time it takes for the car to make a 45 degree angle to the horizontal. which will get the equation of the trajectory of parking compared to the footpath as follows; () = ℎ + 0.5 1 + 4 (ℎ+0.5) where ℎ is parked car width. 2. When the length from the moment the car begins to park at an angle of 45 degrees with the horizontal until the end of the parking is measured on the graph, it is discovered that the driver's car length is inversely proportional to the correlation between the parking distance and the driver’s car length when compared to various types of parked car. Acknowledgments This mathematical project on parallel parking trajectory equation was successfully completed with great supports and advices from Kamtorn Chailuek, Nurainee Chumsalaeh, who kindly examined correctness and errors and gave valuable and useful comments and suggestions in the preparation of this project. We would hereby like to say thanks for their great contributions. References [1] Howard, A. and Stephen, D. Calculus, 8 th ed.; USA. John Wiley & Sons, Inc., 2005. [2] Department of Land Transport. (2022). Parallel parking. (online). Available on (19 September 2022). The logistic function. (2 0 2 2 ) . The logistic function. (online). Available on (19 September 2022) Type Brand/model Parked car width (m) Parking space (m) correlation between the parking distance and the driver’s car length Sedan correlation between the parking distance and the driver’s car length Pickup correlation between the parking distance and the driver’s car length Van correlation between the parking distance and the driver’s car length Truck Sedan HONDA city S 1.748 8.2 1.82 1.55 1.58 1.05 Toyota vios entry 1.700 8.1 1.80 1.53 1.56 1.04 Nissan Almera 1.740 8.2 1.82 1.55 1.58 1.05 Pickup Isuzu D-Max X-Series Hi-Lander Double Cab 1.870 8.5 1.89 1.60 1.63 1.09 Toyota Hilux Revo Double Cab 1.855 8.4 1.87 1.58 1.62 1.08 Nissan Navara Double Cab 4WD VL 7AT 1.850 8.4 1.87 1.58 1.62 1.08 Van Toyota Hiace 1.950 8.6 1.91 1.62 1.65 1.10 Nissan Urvan 1.880 8.4 1.87 1.58 1.62 1.08 Hyundai H-1 2022 1.920 8.5 1.89 1.60 1.63 1.09 Truck HINO FG8JJ1A 240 2.435 8.8 1.96 1.66 1.69 1.13 HINO FL8JN1A 2.460 8.9 1.98 1.68 1.71 1.14 Isuzu FTR34QXXXU 2.435 8.8 1.96 1.66 1.69 1.13 Average correlation between the parking distance and the driver’s car length when compared to various types of parked car 1.89 1.59 1.51 1.10 Table 1 correlation between the parking distance and the driver’s car length when compared to various types of parked car.
64 Natthapong Aonkaew1 , Wimonsiri Khempomma1 and Warisara Keskesorn1 Advisors: Natnarin Boontima2 and Surasuk Boontima2 Special Advisor: Kittiphong Lalun3 1,2Loeianukulwittaya School, Mueang Loei, Loei, 42000, Thailand 3Department of Agricultural Engineering, Khon Kaen University, Khon Kaen, 40002, Thailand Abstract The aim of this project were to prepare and developed low cost charcoal solid fuel stick by used the linear programming techniques. The properties of charcoal solid fuel stick correspond with the constrains, weight total within 100 g per piece, ignite easily within 3 minutes, without smoke, heating rate within 5C per minute, provide heating continuous within 80C for more than 30 minutes and moisture absorption within 15% at open ambient condition. The decision variables were weight of tamarind charcoal (x), mixture wood charcoal ( y) and volume of 40% natural rubber latex (Z). Investigate optimal dosage and decision variables coefficient by linear programming techniques. Formulation of objective function was presented as: = x + + Where is the lowest costs. , and are decision variables coefficient or unit cost of x, y and z, respectively. The optimal dosage of x and y were found to be 65.0 g and 15.8 g, respectively. Value of Z was found to be 48 cm3 only. Because, if volume less than or more than can’t compress the product to stick. Decision variables coefficient of objective function , and equal to 0.012, 0.008 and 0.018, respectively. Therefore, production process charcoal solid fuel stick lowest cost is 1.7704 Baht/piece Keywords: Linear programming ,objective function and charcoal solid fuel stick The Linear Programming for Produce Charcoal Solid Fuel Stick Introduction Lack of clean energy access and use lie at the heart of many pressing issues currently facing people around the planet. Many families, particularly in low income countries, rely on solid, polluting and dangerous fuels such as wood, dung, charcoal and kerosene to meet their energy needs for cooking, heating. According to new analysis from the WHO Global Household Energy Database, around 3.1 billion people living in low and middle-income countries rely on polluting fuels and technologies for cooking. The use of these fuels in inefficient stoves results in the emission of harmful pollutants which contributes to household air pollution (HAP). Overall, 74% of households in Asia and more than 60% in Thailand report use of solid fuels, mostly in the form of biomass. The most common fuel used for cooking and heating is wood, followed by other solid biomass fuels, such as charcoal, agricultural residues and solid alcohol fuel [1]. Charcoal is a solid fuel used for heating and cooking that is created through the process of carbonization, which is a process where complex carbon substances such as wood or other biomass are broken down through a slow heating process into carbon and other chemical compounds [2]. Charcoal is charred wood, which has lost all moisture and most volatile contents, charcoal making is a traditional industry across the world charcoal is an energy dense fuel that can easily. In the charcoal production process, many small pieces of charcoal scraps are left with not usable which has produce are charcoal solid fuel stick [3]. Therefore, the aim of this study were to develop and fabricate lowest cost charcoal solid fuel stick, which as high-energy fuel briquettes consisting of various waste mix ratios. The raw material consisting of tamarind charcoal scraps, mixture wood charcoal scraps and natural rubber latex. Investigate optimal dosage and determination of relation coefficient for decision variables by linear programming techniques. Linear programming techniques will always produce an optimal solution to an linear programming problem. Nonbinding constraints are not associated with the feasible solution space; i.e., they are redundant and can be eliminated from the matrix. Nonzero slack or surplus is associated with a binding constraint. This
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 65 approach has proven useful in guiding quantitative decisions across different fields such as business planning, industrial engineering, and to some extent, in the social and physical sciences [4]. Materials and Methods Formulation of linear programming problem The problems for this project are prepare and develop charcoal solid fuel stick shown the costs with least. The decision variables consist of three raw material, tamarind charcoal scraps, mixture wood charcoal scraps and natural rubber latex. The constraints of product, final product has weight within 100 g, can be ignite easily within 3 minutes, without smoke, heating rate within 5C per minute, provide heating continuous, at 30 minutes after ignited shown as temperature within 80C and moisture absorption within 15% at open ambient condition. The formulation of linear programming model, three components are: 1. The decision variables x is weight of tamarind charcoal scraps (g) y is weight of mixture wood charcoal scraps (g) z is volume of 40% natural rubber latex (cm3 ) 2. The Objective Function = x + + Where is the lowest cost. , and are decision variables coefficient or unit cost of x, y and z, respectively 3. The constrains Non-negativity constrains: x 0 y 0 z 0 Final product weight constrains (A) : 1 + 2 + 3 100 Where 1, 2 and 3 are variables coefficient. Ignite constrains (B): 1 + 2 + 3 3 Where 1, 2 and 3 are variables coefficient. Heating rate constrains (C): 1 + 2 + 3 5 Where 1, 2 and 3 are variables coefficient. Provide Heating continuous constrains (D): Temperature: 1 + 2 + 3 80 Where 1, 2 and 3 are variables coefficient. Moisture absorption constrains (E): 1 + 2 + 3 15 Where 1, 2 and 3 are variables coefficient. Preparation of charcoal solid fuel stick Charcoal solid fuel stick was preparation by mixed of raw materials with vary ratio of decision variables several formula. Preparation process as shown in figure 1. Fig. 1. The preparation process of low cost charcoal solid fuel stick
66 Testing properties of products The combustion efficiency and thermal properties tested as shown in figure 2. Fig. 2. Samples testing for investigation properties of products. The results were test constrains value at several weight ratio of decision variables, final products weight constrains (A) and ignite constrains (B) shown in table 1. Heating rate constrains (C), provide heating continuous constrains (D) and moisture absorption constrains (E) shown in table 2, respectively. Table 1: The result were tested final products weight constrains and Ignite constrains values of samples Ratio No. Ratio of x : y : Z (g : g : cm3 ) The constrains values A (g) B (minute) 1 60.0 : 20.8 : 48.0 96.35 3.11 2 62.5 : 18.3 : 48.0 99.46 2.26 3 65.0 : 15.8 :48.0 98.33 2.48 4 67.5 : 13.3 : 48.0 99.19 2.27 5 70.0 : 10.8 : 48.0 99.44 2.49 6 72.5 : 8.3 : 48.0 99.61 2.45 7 75.0 : 5.8 : 48.0 98.57 3.32 8 77.5 : 3.3 : 48.0 98.93 3.21 Table 2: The result were tested of heating rate constrains, provide heating continuous constrains, and moisture absorption constrains values of samples Ratio No. The constrains values C (C/minute) D (C) E (%) 1 4.59 72.60 25.48 2 5.61 75.05 18.33 3 5.83 83.35 13.45 4 5.05 80.45 14.77 5 5.49 81.05 13.64 6 5.74 80.15 13.66 7 5.88 86.65 14.71 8 6.44 86.35 12.55 The experimental for investigate values of constrain which as vary ratios of decision variables consist weight of tamarind charcoal scraps, weight of mixture wood charcoal scraps and volume of 40% natural rubber latex. From the results has 4 ratios was shown properties complete with as specified of constrains, ratio number 3, 4, 5 and 6, respectively. Results and Discussion The forecasting equation of linear programming The unit cost of decision variables survey from local market in Loei province, Thailand. Tamarind charcoal scraps price 36 Baht per 3,000 g or 0.012 Baht/g, mixture wood charcoal scraps price 24 Baht per 3,000 g or 0.008 Baht/g. 40% of natural rubber latex for production process, optimal dosage was found to be 48 cm3 only. Because, if volume less than or more than can’t compress the product to stick. Therefore, value of 40% natural rubber latex is constant, 48 cm3 for every ratio. 40% natural rubber latex price 18 Baht per 1,000 cm3 or 0.018 Baht/cm3 . From the objective function, D1 = 0.012, D2 = 0.012 and D3 = 0.018, respectively. The equation of linear programming from objective function was present as: = 0.012x + 0.008 + 0.018Z This equation to be used as a forecasting equation for evaluation the lowest costs () of charcoal solid fuel stick production process. When as x, y and Z in equation above was substitution with optimal dosages from table 1, such as ratio number 3 were x = 65.0 g, y = 15.8 g and Z = 48.0 cm3 to obtain: Pmin = 0.012(65.0) + 0.008(15.8) + 0.018(48.0) = 0.78 + 0.1264 + 0.864 = 1.7704 The lowest costs of charcoal solid fuel stick production process present in table 3.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 67 Table 3: The optimal dosages for produce charcoal solid fuel stick Ratio No. Optimal dosage of decision variables Pmin (Baht) x (g) y (g) Z (cm3 ) 3 65.0 15.8 48.0 1.7704 4 67.5 13.3 48.0 1.7804 5 70.0 10.8 48.0 1.7904 6 72.5 8.3 48.0 1.8004 From table 3 show optimal dosages of decision variable in production process charcoal solid fuel stick. But, when substituting the values in the equation of linear programming was found to be 1.7704 Baht/piece is lowest cost according to the specified condition. Linear programming is the most important for solve relationship variables for the best of values. Because, linear programming is a mathematical modeling technique involves maximizing or minimizing a linear function while taking into account various constrain.. Conclusions Linear programming, also known as linear optimization, is a method for achieving the best possible outcome in a mathematical model where the requirements are defined by linear relationships. Preparation process lowest cost charcoal solid fuel stick, optimal dosage of decision variables x, y and Z were found to be 65.0 g, 15.8 g and 48 cm3 , respectively. The decision variables coefficient of objective function , and equal to 0.012, 0.008 and 0.018, respectively. Therefore, production process charcoal solid fuel stick lowest cost is 1.7704 Baht/piece according to the specified condition. The constraints of product, final product has weight within 100 g, can be ignite easily within 3 minutes, without smoke, heating rate within 5C per minute, provide heating continuous, at 30 minutes after ignited shown as temperature within 80C and moisture absorption within 15% at open ambient condition. Acknowledgments The authors would like to thank the Department of Agricultural Engineering, Faculty of Engineering, Khon Kaen University for providing research facilities. The financial support from Loeianukulwittaya school, Office of the Basic Education Commission, Education ministry is gratefully acknowledged. References [1] Gil, A. Challenges on Waste-to-Energy for the Valorization of Industrial Wastes: Electricity, Heat and Cold, Bioliquids and Biofuels. Environ. Nanotechnol. Monit. Manag. 2022. [2] O. Sławomir, “Analysis of usability of potato pulp as solid fuel,” Fuel Processing Technology, vol. 94, no. 1, pp. 183–192, 2012 [3] J. K. Odusote and H. O. Muraina, “Mechanical and combustion characteristics of oil palm biomass fuel briquette,” Journal of Engineering Technology, vol. 8, no. 1, pp. 14-29, 2017. [4] Rager, J. M. F. Urban energy system design from the heat perspective using mathematical programming including thermal storage, n 6731 (Ph.D. thesis). École Polytechnique Fédérale de Lausanne, Lausanne, Switzerland. 2015.
68 Yu Sakamoto1 1 Nara Women’s University Secondary School Abstract In an earlier study, I resarched the process of folding the origami paper to express that the length of the fold is an irrational number, such as the square root of 2 or 3.This led to my further interest in origami folds, wherein the length of the origami fold was an irrational number other than the square root. The present study shows the process of folding an origami paper so that the length of the foldachieved by folding the paper is of the size of pi. First, we express the length of the circumference of a regular n-sided polygon inscribed in a circle with a diameter of 1 as the length of the origami fold. Subsequently, by performing the same operation for a sufficiently large n, pi is expressed in the length of the origami folds. Keywords: origami , Pi , irrational numbers The Pi in Origami Introduction In this study, the Huzita-Hatori axioms were used wherein each axiom consists of the following seven axioms. (1) Given two lines L1 and L2, to fold a line placing L1 onto L2. (2) Given two points P1 and P2, to fold a line placing P1 onto P2. (3) Given two points P1 and P2, to fold a line passing through both P1 and P2. (4) Given one point P and one line L, to fold a line passing through P and perpendicular to L. (5) Given two point P1 and P2 and one line L, to fold a line placing P1 onto L and passing through P2. (6) Given two point P1 and P2 and two lines L1 and L2, to fold a line placing P1 onto L1 and placing P2 onto L2. (7) Given one point P and two lines L1 and L2, to fold a line placing P onto L1 and perpendicular to L2. Materials and Methods The law of cosine was used to express the length of the circumference of a regular n-sided polygon inscribed in a circle of diameter 1 as the equation.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 69 Results and Discussion In this study, an approximate value of pi was expressed in the length of the origami folds. The result is owing to the fact that pi is a transcendental number. The adove images were shown = = 1 2 , ∠ = 360 ; Given Applying the low of cosines to 2 = 2 + 2 − 2 cos ∠ = 1 2 ∙ 1 2 + 1 2 ∙ 1 2 − 2 ∙ 1 2 ∙ 1 2 ∙ cos 360 = 1 2 ( 1 2 + 1 2 − 2 ∙ 1 2 ∙ cos 360 ) = 1 2 (1 − cos 360 ) = √ 1 2 (1 − cos 360 ) ∴ = √ 1 2 (1 − cos 360 ) Conditions for folding the paper are as follows: = 2^, not using the tool, square has a side of 4. The following image shows the process of folding the length of Pi using an origami paper. The green line expresses the length of 1 2 (1 − cos 360° ) The red line expresses the length of √ 1 2 (1 − cos 360° ) On multiply this by and we obtain . Conclusions This study shows that the approximate value of pi was expressed in the length of the origami folds. In the future research, we plan to study the process of folding the length of the Napier numbers, which are transcendental numbers. References [4] Koshiro Hatori, “Origami versus StraightEdge-and-Compass”K’sOrigami,2015, https://origami.ousaan.com/index.html, (2023-08-28) [5] 三谷 純.「折り紙の中の数学」.『数 学通信』.26 巻, 3 号,p51-60
70 Sanada Yuya1 and Tanigawa Haruki1* Advisors: Ishida Akio2 1,2National Institute of Technology (KOSE N), Kumamoto College, 2659-2, Suya, Koshi Kumamoto, 861-1102, Japan 1*e-mail: [email protected] c.jp 1 e-mail: [email protected] Abstract Multidimensional arrays, which we learn in programming classes, are also called tensors, and are one of the important data structures. In this research, we will develop a tool that leads to support for understanding the handling of this tensor and the calculation method. Prior research on this development dealt with several 3-d puzzles, called “ball-counting puzzles” and “coloring puzzles”. In that research, these puzzles were represented by integer-valued tensors and solved using a tensor calculation. The calculation is called the matrix unfolding, an operation that converts a tensor into a matrix. Keywords: 3-D puzzle, tensor calculation, multidimensional array, matrix unfolding, math and information Solving 3-d Puzzles Using Tensor Calculation Introduction These days, multidimensional arrays, also known as tensors, are one of the important data structures studied in programming classes. However, there are problems such as a lack of exercises and difficulty in understanding of operations. In previous research, it has been developed learning materials using 3D puzzles to assist students in understanding the concepts and operations of multidimensional data and to solve the lack of practice tasks. In this research, we created learning materials using 3D puzzles with LEGO blocks. We consider that this will allow learners to experience the process of solving 3D puzzles with their own hands and help them to understand. In addition, it is difficult for us to understand what the elements inside the 3D puzzle are, but by preparing real objects, students can visually understand the elements inside the 3D puzzle, which they have been able to understand sensitively through their imagination, making it easier for them to solve the puzzle. Therefore, we think that this is appropriate as an easy-to-understand teaching material for beginning learners. It is expected that the amount of exercise contents for learning support about tensors will become sufficient by continuing this research because this research can support understanding of various tensor knowledge by changing the type of 3D puzzles. Method 1 Algorithm for 1-mode matrix unfolding of 3- order tensor Input 3-order tensor A of size × × Output 1-mode matrix unfoldiong A(1) Operate1 Put the elements of in row i and column {( − 1) + } of matrix A(1) Operate2 Cut from left to right and arrange each resulting matrix horizontally.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 71 Algorithm for 2-mode matrix unfolding of 3- order tensor Input 3-order tensor A of size × × Output 2-mode matrix unfolding A(2) Operate1 Put the elements of in row j and column {( − 1) + } of matrix A(2) Operate2 Cut from front to rear and turn over like a moving the top edge to the right edge and arrange each resulting matrix. Algorithm for 3-mode matrix unfolding of 3-order tensor Input 3-order tensor A of size × × Output 3-mode matrix unfoldiong A(1) Operate1 Put the elements of in row k and column {( − 1) + } of matrix A(3) Oprate2 Cut from top to bottom and turn over like a moving the top edge to the bottom edge and arrange each resulting matrix. Algorithm for folding of 3-order tensor Input 3-mode matrix unfolding A Output 3-order tensor A of size × × Operate Put the subscripts of each element of the tensor into the formula for the n-mode matrix unfolding, and put the data in that position into each element. Result 1 We visualize the matrix unfolding and folding with LEGO bricks. Matrix unfolding is performed using the Operate1 and Operate2 shown in the Method, respectively. The following is a representation of a 3-order tensor using LEGO bricks. Figure 2 Front view Figure 4 2-mode matrix unfolding of 3-order tensor Figure 5 3-mode matrix unfolding of 3-order tensor Figure 1 The example of 3- order tensor Figure 3 1-mode matrix unfolding of 3-order tensor
72 Method 2 Algorithm for 1-mode matrix unfolding of 4-order tensor Input 4-order tensor A of size × × × Output 1-mode matrix unfolding A(1) Operate Put the elements of in row i and column{( − 1) + ( − 1) + } of matrix A(1) Algorithm for 2-mode matrix unfolding of 4-order tensor Input 4-order tensor A of size × × × Output 2-mode matrix unfolding A(2) Operate Put the elements of in row j and column{( − 1) + ( − 1) + } of matrix A(2) Algorithm for 3-mode matrix unfolding of 4-order tensor Input 4-order tensor A of size × × × Output 3-mode matrix unfolding A(3) Operate Put the elements of in row k and column{( − 1) + ( − 1) + } of matrix A(3) Algorithm for 4-mode matrix unfolding of 4-order tensor Input 4-order tensor A of size × × × Output 4-mode matrix unfolding A(4) Operate Put the elements of in row j and column{( − 1) + ( − 1) + } of matrix A(4) Result 2 We visualize the matrix unfolding and folding with LEGO bricks. Matrix unfolding is performed using the Operate1 and Operate2 shown in the Method, respectively. The following is a representation of a 3-order tensor The following is a representation of a 4-order tensor using LEGO bricks. Figure 6 The example of 4-order tensor Figure 7 1-mode matrix unfolding of 4-order tensor Figure 8 2-mode matrix unfolding of 4-order Discussion LEGO bricks can be easily increased in tensor dimension by arranging them in a row, and the color of the blocks can be used to easily determine the elements. The matrix unfolding of a tensor becomes more complex as the dimension increases, and there are as many types of matrixes unfolding as the number of dimensions (1,2,3... n-mode) The matrix unfolding using the LEGO bricks model is very useful for understanding the motion, because we can see the process of how the tensors are cut and arranged in the matrix unfolding process.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 73 Even though the formulas may look similar at first glance, the actual matrix unfolding shows completely different ways of cutting and arranging the tensors, and it is fun to easily find the regularity of each mode, making it suitable for beginning tensor students who should try it first. Conclusions We surveyed the LEGO blocks to see how effective they are in explaining higher-order tensors. We asked the participants to look at the matrix unfolding of the 3x3x3 tensor and guess the front view of the original 3x3x3 tensor. The results of the questionnaire were as follows. The results of the questionnaire showed that the percentage of correct answers to the problems solved without LEGO blocks was 90%, while the percentage of correct answers to the problems solved with LEGO blocks was 100%. When asked whether LEGO blocks were useful in understanding matrix unfolding, 90 % of the respondents answered that LEGO blocks were useful. In the process of solving the puzzles without LEGO blocks, it was necessary to give hints in the process of solving, and it was found that it took more than three times as much time as the method using LEGO blocks. Many of the participants commented that "it was easy to see the process of moving the blocks" and "it was easy to understand what they were doing by actually moving the blocks by hand. The results of the above questionnaire also indicate that LEGO blocks are suitable teaching materials for supporting the understanding of higher-order tensors. In addition, by changing the colors of the LEGO blocks and the way they are assembled, we can expect that LEGO blocks can be used as teaching materials to support understanding of operations on tensors other than matrix unfolding. Acknowledgments First, we would like to thank the TJ-SSF management team and the teachers at Kumamoto National College of Technology for giving us the opportunity to participate in TJ-SSF. We would also like to thank Dr. Naoki Yamamoto and Dr. Akio Ishida, who belongs to Kumamoto National College of Technology, for their advice and various ideas for this research. References [1] Naoki Yamamoto, Akio Ishida, Kazuki Ogitsuka, Nobuhiro Oishi, and Jun Murakami, "Development of Online Learning Material for Data Science Programming Using 3D Puzzle," International Journal of Information and Education Technology vol. 11, no. 4, pp. 154-163, 2021. [2] L.D.Lathauwer, B.Moor, and J. Vandewalle, “A multilinear singular value decomposition,” SIAM Journal on Matrix Analysis and Applications, vol. 21, no. 4, pp. 1253-1278, 2000 [3] Naoki Yamamoto, Akio Ishida, Nobuhiro Oishi, and Jun Murakami, “A method to solve 3-d puzzles by using multidimensional data decomposition technique”, SHOTOH SUGAKU vol.83, 201
74 Mitsunaga Manami1 and Murazaki Kurea1 Advisors: Hashimoto Junya2 1,2Kumamoto National Institute of Technology, Department of Architecture and Civil Engineering, 2627 Hirayamashin-Machi, Yatsushiro, Kumamoto Abstract We will conduct research and analysis to determine what paths would provide a more efficient movement for housework efficiently in the house. We will conduct our research using the following process. 1.We research the housing environments in Japan. Here, we will use the drawings of houses created in the "Basic Drawing II" class in the first grade at the KOSEN. 2.Based on those floor plans and elevations, we analyze the movement that allow housework to be performed in the shortest path with the least amount of labor. The movement for housework includes the movement from the laundry to the clothes-drying area, the movement from the kitchen to the dining room, the movement for cleaning, and so on. 3.We will analyze the results of Process 2 and investigate the characteristics of efficient housework movements. 4.Based on these results, we propose a house design that allows efficient housework. Conversely, we will also present examples of housing environments that are good for people to live in, even if the movements are long. We will discuss why this is so. We have found a movement for efficient housework. Flow lines for housework efficiently Furthermore, we were able to propose a house that included the line. Keywords: movement, housework, efficiently, architectural design, house design Introduction Many people value the time they spend at home relaxing with their families and enjoying their hobbies. To create such time, we believe that we can reduce the time and effort we spend on housework. So, we thought maybe they could make time, by our house design. We propose a house design that allow for efficient housework. Methods People-Flow Analysis (1) Record the time and place of daily housework. We researched two families. (2) Prepare some houses designs. We used 8 plans, drawn for class assignment. (3) Calculate the length of movement when doing the chores described in (1) above. (4) Identify the characteristics of the house with the shortest total length of movement. Proposing a house design (5) Based on (4), consider the layout of the main rooms in the house with the shortest housework movement. (6) Design the house based on (5). Results and Discussion People-Flow Analysis 1.Organize the data We focus on Table1 that shows the observations of housework movement. Table 1: Shows the observation of housework movements Placa number A Kitchen D Living Room B Laundry Room E Dining Room C Drying Spease F Bathroom
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 75 Sample1 6:15 making breakfast A 6:50 making a lunch box A 8:00 doing the dishes A 17:00 filling the bath F 17:02 cooking rice A 17:30 making dinner A 18:00 running the laundry machine B 18:30 making dinner A 19:01 setting the table E 19:30 clearing the table E 19:31 doing the dishes A 20:00 taking the out of the laundry machine B 20:01 running the laundry machine B 20:02 folding the laundry C 20:10 putting away the laundry B 20:11 putting away the laundry A 20:12 hanging the laundry C 21:00 taking the out of the laundry machine B 21:01 hanging the laundry C 22:00 cooking rice A Sample2 7:00 making breakfast A 7:12 running the laundry machine B 7:13 making breakfast A 7:20 hanging the laundry on hangers B 7:22 hanging the laundry C 7:52 hanging the laundry on hangers B 7:58 hanging the laundry C 7:58 hanging the laundry on hangers B 7:59 hanging the laundry C 16:30 taking in the laundry C 16:31 taking in the laundry D 18:15 making dinner A 19:00 setting the table E 20:00 doing the dishes A 20:05 filling the bath F 21:00 folding the laundry D 21:16 putting away the laundry B 22:50 running the laundry machine B The sections that were moved most frequently were the Laundry Room-Drying Room, Kitchen-Laundry Room, and KitchenDining Room. If the housework movements as shown in Table 1 in the eight houses designs we have prepared, we output the total length in housework movement. Table2: Shows the total length in housework movement of sample plans 2.Considerations of data Based on the results of the two tables, we consider what we discovered about Characteristics of efficient housework movements. 1 The places where housework done need to be closer to each other. 2 In both patterns, there was a lot of movement between the Laundry room and the Drying room. We assume that other houses have a lot of that movement. So, the length between the sample Sample1 (mm) 1 2 3 4 AF 8707 11934 23275 16152.5 AB 7735 10232 21455 13195 AE 2275 5823 5130 5460 BC 9870 9824 10495 13195 AC 8307 6566 27150 4095 Total 91389 108814 206960 130585 5 6 7 8 AF 11819 6028 9862 6424 AB 9311 4617 9255 3980 AE 1593 4020 2953 4051 BC 10227 10120 10248 11658 AC 10073 10278 4523 9669 Total 105584 84863 93185 87202 sample Sample2 (mm) 1 2 3 4 AB 7735 10232 21455 13195 AE 2275 5823 5130 5460 AF 8705 11934 23275 16152 BC 9870 9824 10495 13195 BD 7250 5347 3145 15015 CD 8307 4243 5975 10010 Total 91497 103162 149000 144462 5 6 7 8 AB 9311 4617 9255 3980 AE 1593 4020 2953 4051 AF 11819 6028 9862 6424 BC 10227 10120 10248 11658 BD 7214 8068 6053 7836 CD 7214 2960 4195 3090 Total 98274 79427 94773 84024
76 Laundry room and the Drying room should be shortened. 3 If the places to do housework are clustered on the same floor, the housework movement is short. 4 The total length is longer because of the longer Laundry Room to Drying Room distance in all samples. 5 The position of the room door affects the length. 3. Floor Plan for efficient housework The plan we designed is shown below. Figure 1. the layout of the main rooms in the house with the shortest housework movement The sections that were moved most frequently were the Laundry Room-Drying Room, Kitchen-Laundry Room, and KitchenDining Room. Therefore, these sections were designed to be next to each other. The doors were positioned so that they lined up on the straight line. Table3: Shows the total length in housework movement of our plan The total length of our plan was shorter than 8 samples. Compared to the average of eight samples. Sample1 -69.857m Sample2 -79.129m The actual movements during the observation are marked with red lines. Figures 3 and 4 show that most of the movement is between the laundry room and the drying space, then between the laundry room and the kitchen. Therefore, the drying space and kitchen were placed near the laundry room. Figure2. Shows movements in the actual house at the time of observation (Sample1) Figure3. Shows movements in the actual house at the time of observation (Sample2) Proposing a house design Figure4 is floor plan of house with the layout of Drawing1. Design conditions are “a family of four (father, mother, two children) “and “site area is 143 square meters (General site Our plan (mm) in Japan)”. Sample1 Sample2 AF 4985 AB 1954 AB 1954 AE 1895 AE 1895 AF 4985 BC 2297 BC 2297 AC 3965 BD 9693 CD 2202 Total 34443 Total 35720
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 77 First floor ground plan Second floor ground plan Figure4. house design with the layout of Drawing1 We designed a wooden house, which is the mainstream of detached houses in Japan. This structure, called the wooden framework construction method, supports the building by means of pillars and beams. First, pillars are erected on a concrete foundation. Next, beams are combined with the pillars to form the framework, and after the roof is put on, walls and other structures are attached. Conclusions We now understand the characteristics of a house with a short housework movement. Just by changing the movements, we can do housework efficiently. For example, through this research, we felt that it is better to be close to water. After recording my housework, we began to think about housework in my daily life. Although I referred to many designs in this study, we cannot be stated unconditionally that only houses with short housework movements are superior. A better housework movement depends on the rhythm of the residents’ lives. We would like to design a house that is close to the people who live in it by using what We have learned in this research. Acknowledgments In acknowledgement you may thank all the people/organizations who provided their assistance to you in forms of advice, suggestions, and any others.
78 Charunsak Kaeothaisong 1 and Sasikan Boonkrong1 Advisors: Itsara Wichitpan2 and Wasana Boonthawee2 1,2Khumuangwittayakhom School , Buriram, Thailand Abstract This project aims to 1. find a formula for finding the number of subsets that have the specified elements quickly and correctly and 2. study satisfaction with doing exercises Subsets with Combinations. The results of the project found that the formula for finding the number of subsets with elements as specified in the problem is: Giver p is the total number of elements in the set and q is the number of elements in the subset to be found. The formula for finding the number of subsets that have q elements is: ( ) p,q p p! C = = q p - q !q! . And satisfaction with formula for finding the number of subsets that have the specified elements after doing exercises about “Subset with Combinations” which evaluates the results from the average score of the entire questionnaire. At a very satisfied level. Subset with Combinations Introduction From studying the subject of subsets. It has only stated the method for finding all subsets, for example n(A) = 8, the total number of subsets of A is 28 = 256 subsets. But not have finding the number of subsets that have the specified elements quickly and correctly, for example how many subsets of A if A have 6 elements? So we thought There should be a way to quickly and accurately find the number of subsets that have the elements we want. We have tried to apply with the principles of Combinations. Let's apply which can produce accurate results. Objective 1. find a formula for finding the number of subsets that have the specified elements quickly and correctly 2. study satisfaction with doing exercises Subsets With Combinations. Process 1. Define the problem, which is to find the number of elements of a subset that has the specified number of elements defined problem. 2. Study problems related to sets, specifically focusing on finding subsets. From books and the internet, such as Example1 Let A = {1,2,3,4,5} Find the number of all subsets of A that have 0,1,2,3,4,5 elements. Solution All subsets of A consist of , {1} , {2},{3},{4},{5},{1,2},{1,3},{1,4},{1,5},{2,3}, {2,4},{2,5},{3,4},{3,5},{4,5},{1,2,3}, {1,2,4}, {1,2,5},{1,3,4},{1,3,5},{1,4,5},{2,3,4},{2,3,5}, {2,4,5},{3,4,5},{1,2,3,4}, {1,2,3,5}, {1,2,4,5}, {1 , 3 , 4 , 5 }, {2 , 3 , 4 , 5 }, {1 , 2 , 3 , 4 , 5 }. So . there are 32 total subsets of A. - Subset of A with 0 element is . So subset of A with 0 element has 1 subset. - Subset of A with 1 elements is {1}, {2}, {3},{4},{5} . So subset of A with 1 element has
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 79 5 subset. - Subset of A with 2 elements is {1,2},{1,3},{1,4},{1,5},{2,3},{2,4}, {2,5}, {3,4},{3,5},{4,5}. So subset of A with 2 elements has 10 subsets. - Subset of A with 3 elements is {1,2,3}, {1,2,4},{1,2,5},{1,3,4},{1,3,5},{1,4,5},{2,3,4}, {2,3,5},{2,4,5},{3,4,5}. So subset of A with 3 elements has 10 subsets. - Subset of A with 4 elements is {1,2,3,4}, {1,2,3,5},{1,2,4,5},{1,3,4,5},{2,3,4,5} . So subset of A with 4 elements has 5 subsets. - Subset of A with 5 elements is {1,2,3,4,5}. So subset of A with 5 elements has 1 subsets. Example2 Find m×n when m is the number of subsets of set A with 3 elements and n is the number of subsets of set B with 4 elements. Given set A = {1,2,3,4,5} , B = {a,b,c,d,e,f} Solution - Subset of A with 3 elements is {1,2,3}, {1,2,4},{1,2,5},{1,3,4},{1,3,5},{1,4,5}, {2,3,4},{2,3,5},{2,4,5},{3,4,5}. So subset of A with 3 elements has 10 subset m = 10. - Subset of B with 4 elements is {a,b,c,d}, {a,b,c,e},{a,b,c,f},{a,b,d,e}, {a,b,d,f}, {a,b,e,f},{a,c,d,e},{a,c,d,f},{a,c,e,f}, {a,d,e,f},{b,c,d,e},{b,c,d,f},{b,c,e,f},{b,d,e,f}, {c,d,e,f}. So subset of A with 4 elements has 15 subset n = 15. So m×n = 10×15 = 150 3. Notice the answers from up examples. and try out with the principles of Combinations. If r objects are to be combined from n objects, i.e. selected in any order, then the number of ways in which this can be done is: ( ) n,r n n! C = = r n - r !r! Therefore, from Example 1, get 1. Subset of A with 0 element is 5,0 5! 5! 1 (5 0)!0! 5! C = = = − subset 2. Subset of A with 1 element is 5,1 5! 5! 5 (5 1)!1! 4! C = = = − subsets 3. Subset of A with 2 element is 5,2 5! 5! 10 (5 2)!2! 3!2! C = = = − subsets 4. Subset of A with 3 element is 5,3 5! 5! 10 (5 3)!3! 2!3! C = = = − subsets 5. Subset of A with 4 element is 5,4 5! 5! 5 (5 4)!4! 4! C = = = − subsets 6. Subset of A with 5 element is 5,5 5! 5! 1 (5 5)!5! 5! C = = = − subsets from Example 2, get 1. Subset of A with 3 element is 5,3 5! 5! 10 (5 3)!3! 2!3! C = = = − subsets m = 10 2. Subset of A with 4 element is 6,4 6! 6! 15 (6 4)!4! 2!4! C = = = − subsets n = 15 So m×n = 10×15 = 150 4. The formula for finding the number of subsets with elements as specified in the problem is: 5. Take the formula that has been summarized and test it with various related problems to see the results and find any problems that may arise occur. Given : p is the total number of elements in the set q is the number of elements in the subset to be found. The formula for finding the number of subsets that have q elements is: ( ) p,q p p! C = = q p - q !q!
80 6. Make the exercises. " Subset with Combinations" to try out with friends in the same grade. And evaluate satisfaction with the formula for finding the number of subsets with the specified elements. Results and Discussion Results 1. The formula for finding the number of subsets with elements as specified in the problem is: Given : p is the total number of elements in the set q is the number of elements in the subset to be found. The formula for finding the number of subsets that have q elements is: ( ) p,q p p! C = = q p - q !q! 2. Satisfaction with formula for finding the number of subsets that have the specified elements after doing exercises about “Subset with Combinations” which evaluates the results from the average score of the entire questionnaire. At a very satisfied level. Discussion 1. Finding the formula for finding the number of subsets with elements as specified in the problem. Use observation of results from many samples and summarize into a formula. 2. Satisfaction with formula at a very satisfied level. Because formula for finding the number of subsets with elements as specified in the problem can help students to find answers more quickly. Including easy to learn and understand formulas. Acknowledgments In acknowledgement you may thank all the people/ organizations who provided their assistance to you in forms of advice, suggestions, and any others. References [1] Prakong Karnasutta. (1995). Statistics for behavioral science research. Bangkok: Chulalongkorn University. [2] IPST. (2016). Foudation mathematics textbook. Volume 1, Mathayomsuksa 4-6, Secondary School Curriculum, B.E. 2008. 10th printing. Bangkok: Teachers' Council of Ladprao Printing House. [3] Samai Laowanich. (1978). Mathematics M. 4-5-6. Bangkok: Charoendee Publishing.
Thailand – Japan Student Science Fair 2023 “Seeding Innovations through Fostering Thailand – Japan Youth Friendship” 79 Contributors for TJ-SSF 2023 Ministry of Education Embassy of Japan in Thailand Office of The Basic Education Commission Ministry of Education, Culture, Sports, Science and Technology Khon Kaen University Japan International Cooperation Agency Loei Rajabhat University Chulabhorn Royal Academy
80 Contact Persons of TJ-SSF 2023 Name Tel. Language Mr. Songkran Buttawong 095-194-1926 TH Miss Wiriya Tasee 087-224-2291 TH/ENG Mrs. Chanunphat Khotewong 091-706-7815 TH Mr. Satawat Tudmala 095-865-2380 TH Miss Chayaporn Montha 080-746-3378 TH/ENG Miss Parita Chamontree 095-659-0484 JP