Introduction. Table of contents

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2 Introduction The advanced manufacturing sector is changing. New innovative and digital technologies are constantly being developed. Today s and tomorrow s employees need the right kind of mathematical skills to stay ahead of the curve and that s where NAMA's learning materials will help. Table of contents Part 1: Quantity... 3 L1: The number system... 4 L2: Relative size of numbers... 9 L3: Numerical sequences and patterns L4: Basic operations L5: Decimals, fractions and percentages L6: Mental calculations Part 2: Change and relationships L7: Types of change L8: Different representations of relationships L9: Approximations and estimations L10: Effective use of calculators Part 3: Space and shape L11: Units of measurement and scales L12: 2D-3D shapes and properties L13: Trigonometry and geometry L14: Spatial visualisation L15: Application of numerical skills to solve geometrical problems Part 4: Uncertainty and data L16: Probability L17: Introduction to types of data L18: Lists and tables L19: Graphs and charts L20: Measures of average L21: Comparison of different datasets

3 Recommended background The course is recommended for all who would like to develop their numerical skills, especially for: a) adult workers, currently employed or seeking employment in the Advanced Manufacturing sector, b) students participating in training courses or apprenticeships for Advanced Manufacturing jobs. There are no academic prerequisites for the course! 2

4 Part 1: Quantity 3

5 L1: THE NUMBER SYSTEM Theory The natural numbers are 0, 1, 2, 3, 4, 5, etc. There are infinitely many natural numbers. The set of natural numbers, {0, 1, 2, 3, 4, 5,...} is sometimes written N for short. Note: The sum of any two natural numbers is also a natural number (for example, = 2016), and the product of any two natural numbers is a natural number ( = 32000). This is not true for subtraction and division! The Integers The integers are the set of real numbers consisting of the natural numbers and their additive inverses: {..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5...} The set of integers is sometimes written Z for short. Note: The sum, product, and difference of any two integers is also an integer. But this is not true for division! The Rational Numbers The rational numbers are those numbers which can be expressed as a ratio between two integers. Examples: The fractions 1 2 and 33 5 are both rational numbers. All decimals which terminate are rational numbers. Decimals which have a repeating pattern after some point are also rational numbers: Examples: 3.42 can be written as 342, thus it is a rational number = 1, thus it is a rational number 12 Note: Given any two rational numbers, their sum, difference, product, and quotient is also a rational number (as long as we don't divide by 0). 4

6 The Irrational Numbers An irrational number is a number that cannot be written as a ratio (or fraction). If written in decimal form it never ends or repeats. Example: What number times itself equals 2? In other words, what are the solutions of the equation x 2 =2? The solutions are 2 and 2, with 2 being approximately equal to But = , which is close to 2. But you'll never hit exactly 2 by squaring a fraction (or a terminating decimal). The square root of 2 is an irrational number, meaning its decimal equivalent goes on forever, with no repeating pattern: 2 = Other examples of irrational numbers Other known irrational numbers are: the golden ratio, a number with great importance to biology and art: φ = = π (pi), the ratio of the circumference of a circle to its diameter: π = Euler s number: e = The Real Numbers The real numbers is the set of numbers containing all rational and irrational numbers. This set is sometimes written as R for short. The Complex Numbers The complex numbers are the set {a + bi a and b are real numbers}, where i is the imaginary unit (i 2 = 1). The complex numbers include the set of real numbers. This set is sometimes written as C for short. 5

7 Tasks Task 1 Is the sum a natural number? Task 2 Is (-3) (-6):9 an integer? Task 3 Is 1/3 an integer? Task 4 Is 3 an irrational number? Task 5 Is 3 3 a rational or an irrational number? Task 6 Is 3 + π a real number? Task 7 Are the solutions of the equation x 2 +x+1=0 real or complex numbers? 6

8 SOLUTIONS AND ANSWERS Task 1 Answer: Yes, it is. Task 2 Answer: Yes, it is. Task 3 Answer: Yes, it is. Task 4 Answer: Yes, it is. Task 5 Answer: It is a rational number, since it is equal to 3. Task 6 Answer: Yes, it is. Task 7 Answer: They are complex numbers. 7

9 Tasks 1-2 categorisation Quantity Mathematical subfields The number system Basic operations Formulae/data provided - Understanding basic numerical facts Performing simple operations Level 1 ALL Tasks 3-4 categorisation Quantity Mathematical subfields The number system Formulae/data provided - Understanding basic numerical facts Level 1 ALL Tasks 5-6 categorisation Quantity Mathematical subfields The number system Basic operations Formulae/data provided - Understanding basic numerical facts Performing simple operations Level 1 ALL Task 7 categorisation Quantity Mathematical subfield The number system Formulae/data provided - Solving equations Level 1 ALL 8

10 L2: RELATIVE SIZE OF NUMBERS Theory Relative size of numbers refers to the understanding of the relative magnitude of different types of numbers by comparing, ordering, and identifying equivalent forms of numbers within and across number formats using models, number lines, equality (=) and inequality symbols (,,, <, >) and explanations. The numbers can be: integers rational numbers common irrational numbers, including square roots numbers with whole-number or fractional bases and whole-number or fractional exponents absolute values numbers represented in scientific notation. Number lines The number lines can have a different format, according to the size of the numbers to be compared. Here are two examples: Note: The number line can be also used for addition, subtraction and multiplication! Very large and very small numbers Very large or very small numbers are usually expressed by scientific notation. A number in scientific notation is written as the product of a number (integer or decimal) and a power of 10. This number is always 1 or more and less than 10. 9

11 Example: There are approximately humans on earth. This number could be written in scientific notation as The number is equivalent to The number is equivalent to 10 9 or A number can be converted to scientific notation by increasing the power of ten by one for each place the decimal point is moved to the left. In the example above, the decimal point was moved 9 places to the left to form a number more than 1 and less than 10. Scientific notation numbers may be written in different forms. The number could also be written as 7e+9. The +9 indicates that the decimal point would be moved 9 places to the right to write the number in standard form. Two numbers written in scientific notation can be compared. The number with the larger power of 10 is greater than the number with the smaller power of 10. If the powers of ten are the same then the number with the larger factor is the larger number. Examples: is greater than is greater than Numbers written in standard form can be compared to numbers written in scientific notation by converting one number to the other format. Example: In order to compare and you need to firstly convert to or convert to

12 Tasks Task 1 Which number is 17 on the number line below? Task 2 By using the appropriate number line, put the following numbers in a descending order (from the largest to the smallest): 0.01, 0.009, 2 23, 2 25, Task 3 How will the result of the multiplication be demonstrated in a calculator? Task 4 The radius of a hydrogen atom is equal to m. Write that number in scientific notation. 11

13 SOLUTIONS AND ANSWERS Task 1 Answer: Task 2 Answer: 2 23 > 2 25 > 0,013 > 0,01 > 0,009 Task 3 Answer: 2 > 2 > > 0.01 > Task 4 Answer: 5e+32 12

14 Tasks 1-2 categorisation Quantity Mathematical subfield Relative size of numbers Formulae/data provided - Understanding basic numerical facts Level 1 ALL Task 3 categorisation s Quantity Change and relationships Mathematical subfields Relative size of numbers Effective use of calculators Formulae/data provided - Understanding basic numerical facts Level 1 ALL Task 4 categorisation Quantity Mathematical subfield Relative size of numbers Formulae/data provided - Understanding basic numerical facts Level 1 ALL 13

15 QUIZ 1. Which of the following numbers is the largest? a b c d Which of the following is equal to ? a b c d Which of the following numbers is between 2 and 3? a. 2/3 b. 3/2 c. 5/3 d. 5/2 4. Without making the operation, answer which of the following is true about the result of the multiplication a. The result is over 10 million. b. The result is over 1 million. c. None of the above is correct. 5. How many zeros does the number have? a. 31 b. 32 c. 33 d. 34 Answers: 1c, 2d, 3d, 4c, 5a 14

16 L3: NUMERICAL SEQUENCES AND PATTERNS Theory A numerical sequence is a set of ordered values of a function, whose domain consists of the set of all natural numbers in ascending order of the numbers. The elements of a sequence are called the terms. The n-th term of a sequence is called the general term or variable of the sequence. The general term is denoted by a lower case letter with the subscript n., where n takes the values 1, 2, 3, A sequence can be finite or infinite. Examples: (2, 5, 9, 10) four elements sequence (1, 2, 3, 4, 5, 6,...) the infinite sequence of consecutive natural numbers. (2, 4, 6, 8, 10, 12, 14,...) the infinite sequence of positive even integers. (1, 1, 2, 2, 3, 3, 4, 4,...) alternating sequence of positive and negative numbers. (1, 1, 1, 1, 1, 1,...) the monotonically decreasing sequence of fractions (3, 9, 27, 81, 243,... ) the sequence of consecutive powers of 3. (-80, -77, -74, -71, -68, -65, ) - the monotonically increasing sequence. Examples: The sequence (1, 3, 5, 7, 9, ) contains the terms: a 1 = 1, a 2 = 3, a 3 = 5, etc. This is an arithmetical sequence; its general term is a n = 2n 1. The elements of an arithmetic progression: a, a+r, a+2r, are the terms of a numerical sequence with the general term a n = a+(n 1)r. The elements of a geometric progression: a, aq, aq 2, are the terms of a numerical sequence with the general term b n = aq n-1. Sequences whose elements are related to the previous elements in a straightforward way are often specified using recursion. To specify a sequence by recursion requires a rule to construct each consecutive element in terms of the ones before it. In addition, some initial elements must be specified so that new elements of the sequence can be specified by the rule. Example: The Fibonacci sequence can be defined using a recursive rule with two initial elements. The rule is that each element is the sum of the previous two elements, and the first two elements are 1 and 1: a 0 = 1, a 1 = 1, a n = a n 2 + a n 1 The consecutive elements of the sequence are called Fibonacci numbers: (1, 1, 2, 3, 5, 8, 13, 21, 34...) 15

17 Tasks Task 1 What is the general term of the sequence 2, 0, 2, 0,? Task 2 n 5 n The following sequence is given: a n = ( 1) for n = 1, 2, 3, Calculate the terms n 2 +3 a 2, a 3, a 6. Task 3 Calculate the fifth element of the following sequence: a { 1 = 3 a n+1 = a n + 5n Task 4 The image below describes a pattern for n = 1, 2, 3, 4 and 5. Can you find the general term of the sequence? Task 5 The table below shows the first terms of a sequence. Can you find the general term of the sequence? n a n

18 SOLUTIONS AND ANSWERS Task 1 Answer: a n = 1 ( 1) n. Task 2 Answer: a 2 = ( 1) = 3 7 a 3 = 1 6 a 6 = 1 39 Task 3 a 2 = a = = 8 a 3 = a = 18 a 4 = a = 33 a 5 = a = 53 Task 4 Answer: T n = n(n+1) 2 Task 5 Answer: a n = (n+1)(n+2) 17

19 Task 1 categorisation Quantity Mathematical subfield Numerical sequences and patterns Formulae/data provided - Performing simple operations Level 1 ALL Tasks 2-3 categorisation Quantity Mathematical subfields Numerical sequences and patterns Basic operations Formulae/data provided - Performing simple operations Level 1 ALL Tasks 4-5 categorisation Quantity Mathematical subfield Numerical sequences and patterns Formulae/data provided - Interpreting graphs and tables Level 2 ALL 18

20 QUIZ 1. The ninth term of the sequence 33, 37, 41, is equal to: a. 61 b. 63 c. 65 d. None of the above 2. Is it possible to have a sequence whose terms are decreasing? a. Yes b. No 3. The common ratio of the geometric sequence 3, 1, 1 3, 1 9, 1 27, is: a. 3 b. 1 c. 1/3 d. Other 4. Write the 40th term for the arithmetic sequence in which a 8 = 60 and a 12 = 48. a. -36 b. -32 c. -30 d A theatre has 24 seats in the first row. Each row has four more seats than the row before it. Which expression represents the number of seats in the nth row? a (n + 4) b n c (n + 1) d (n - 1) Answers: 1c, 2a, 3c, 4a, 5d 19

21 L4: BASIC OPERATIONS Theory Operations with natural numbers and integers The set of natural numbers is included in the set of integers, thus the rules for the operations on natural numbers apply also to the integers. Addition of Integers with the same sign Adding integers that have the same sign is done by adding the two numbers together and maintaining the sign. Examples: Both numbers are positive so we add the numbers together and the number remains positive: 3+7 = 10 Both addends are negative, so we add them together and maintain the negative sign: 3 + ( 2) = 5 Addition of Integers with Different Signs Adding integers with different signs is done by ignoring the signs at first and subtracting the smaller number from the larger. The final sum will maintain the sign of the larger addend. Example: : Since we are adding two numbers with different signs, ignore the signs and we are left with 13 and 8. The number 13 is larger than 8 so we subtract 8 from 13 which gives us 5. Of the two addends, 13 was the larger and was negative, so the final sum will be negative. Thus, our final sum is 5. Subtraction of integers When subtracting integers, we should change the sign of the subtrahend and then follow the rules of addition. Example: 7 ( 13) = = 6 20

22 Multiplication and division of integers Multiplying and dividing integers is similar to the multiplying and dividing of natural numbers. The only difference is that with integers, you must be aware of which sign to apply to the final answer. If the signs of the numbers you are multiplying or dividing are the same, then the answer will be positive. If the signs are different, then the answer will be negative. Examples: 3 ( 15) = 45 7 ( 4) = 28 Operations with rational numbers The rational numbers can be written as fractions or decimals. Thus, when you are asked to perform an operation between two or more fractions, you can firstly transform them into decimals and then perform the operations. Addition and subtraction of decimals Addition and subtraction of decimals is like adding and subtracting whole numbers. The only thing you should remember is to line up the place values correctly. The easiest way to do that is to line up the decimal points. Example: Multiplication and division of decimals When multiplying numbers with decimals, you should firstly multiply them as if they were whole numbers. Then, the placement of the number of decimal places in the result is equal to the sum of the number of decimal places of the numbers being multiplied. Example: When dividing a decimal number by a whole number you start by performing normal division and then put the decimal point in the same spot as the dividend (the number being divided). 21

23 Example: When dividing a decimal number by a decimal number you have to change the number you are dividing by to a whole number first, by shifting the decimal point of both numbers to the right. Example: In order to make the division 17.4:0.3 both numbers are multiplied by 10, thus we have the following division: 174:3=58. Addition and subtraction of fractions When adding or subtracting fractions, you must make sure that the fractions have the same denominator. If they do not, find the least common denominator (LCD) for the fractions and put each in its equivalent form. Then, simply add or subtract the numerators of the fractions. Example: = = 5 6 Multiplication and division of fractions When multiplying fractions, you simply multiply the numerators together and then multiply the denominators together. Then, if needed you can simplify the result. When you divide two fractions, you take the reciprocal of the second fraction, or bottom fraction, and multiply. 22

24 Example: 2 3 : 1 5 = = 10 3 Operations with real numbers The operations with real numbers follow the rules mentioned above. The result of such an operation depends on the context. Example: The sum = can be either left as it is or find its approximate value: 23

25 Tasks Task 1 Is the sum a natural number? Task 2 Is (-3) (-6):9 an integer? Task 3 Is 3 3 a rational or an irrational number? Task 4 Is 3 + π a real number? Task 5 n 5 n The following sequence is given: a n = ( 1) for n = 1, 2, 3, Calculate the terms n 2 +3 a 2, a 3, a 6. Task 6 Calculate the fifth element of the following sequence: a { 1 = 3 a n+1 = a n + 5n Task 7 The value of 1 stock of ABC Company costs 32 Euros and increased by 1.5%. What is the current value of 100 stocks? Task 8 The gross weight of a commodity is 150 kg and the tare weight is 3 kg. What percent of the gross weight is the net weight? Task 9 For a commodity of the value of 150 Euros you have to pay 183 Euros with VAT. Calculate what percent is the VAT tax. Task 10 The price of a stock increased from to Euros. Calculate by what percent was the price of stocks increased. 24

26 Task 11 Find a number whose 20% is 1.5. Task 12 A worker got a salary rise of 120 Euros which is 15% of his salary. How much was he earning before the rise and how much is he earning now? Task 13 A client ordered steel cylinders from a company. The length of the metal plate, from which the cylinders are cut off, is 2000 mm, and its thickness is 100 mm. The length of cylinder is 240 mm and its width is equal to the thickness of the metal plate (ρ = 7,85 kg/dm 3 ). How many track containers are needed in order to transport the ordered materials to the client? The total mass of transport vehicle cannot be more than 40 t, the weight of an empty track container is around 20 t. 10 cylinders are packed in a box which weights 20 kg. Task 14 A company which produces steel cylinders transports the materials at first cost. How much will a client pay for cylinders and how much will the company earn from the client? The firm of the client is located in a distance of 200 km from the company. The cost of transport of 1 track container is 80 euros per hour. The cost of production of 1 cylinder is 53 euros. The profit margin comes to 10% for one cylinder. Task 15 One of the first things we have to do when producing an engineered item is to do the costing to ensure it is viable to make. The costs of the materials, the machining costs and the time to make it have to be considered. An aerospace customer requires a batch of machined shafts. A Computer Numerically Controlled (CNC) lathe is used to manufacture an aerospace component; the operating cost of the CNC machine in an engineering workshop is per hour the shaft takes 5 minutes to make on the machine. the material cost of each component is

27 1. How much does each component cost to manufacture taking into account the cost of material and machining time? 2. The customer requires a batch of 500, what is the price of the full batch? 3. When the customer gets the above quote they feel they should get a 15% discount. With the discount applied what would the customer be charged? Task 16 A Computer Numerically Controlled (CNC) lathe is used to manufacture an aerospace component: the shaft has length of 240 mm each component is cut from a 2000 mm long bar with an allowance of 5 mm per cut. In Factory I: the operating cost of the CNC machine in an engineering workshop is per hour the shaft takes 5 minutes to make on the machine the material cost of each component is 50 In Factory II: the operating cost of the CNC machine in an engineering workshop is per hour the shaft takes 6 minutes to make on the machine the material cost of each component is 49. A client wants to buy 1000 components. Which offer he should choose? Task 17 A Computer Numerically Controlled (CNC) lathe is used to manufacture an aerospace component and it is known that the shaft takes 5 minutes to make on the machine. 26

28 The effectiveness of the CNC operator decreases by 5% per hour from the moment of the beginning of work (the worker produces 5% less components than one hour before). Is it possible to produce 50 shafts during 8 hours of work? Task 18 A Computer Numerically Controlled (CNC) lathe is used to manufacture an aerospace component: the operating cost of the CNC machine in an engineering workshop is per 60 min. the shaft takes 5 minutes to make on the machine the material cost of each component is 50 the shaft has length of 24 cm each component is cut from a 2 m long bar with an allowance of 0.5 cm per cut. a) How many shafts will be produced during one hour by the machine? b) What is the production cost of one component by the machine? c) What is the total production cost of one component? d) What cost has to bear a client for 500 components? e) How much he will pay if he receives 15% of discount? f) How many shafts will the machine produce from one bar? g) How much scrap raw material will be left from each bar? h) What part of the whole is the scrap raw material left from each bar? 27

29 Task 19 A Computer Numerically Controlled (CNC) lathe is used to manufacture an aerospace component: the operating cost of the CNC machine in an engineering workshop is per 60 min. the shaft takes 5 minutes to make on the machine the material cost of each component is 50 the shaft has length of 24 cm each component is cut from a 2 m long bar with an allowance of 0.5 cm per cut. A client ordered shafts with the time limit of 10 days. Usually 2% of the ready products do not fulfil the quality norm. a) Calculate how many shafts the producer has to produce to meet the expectations of the client. b) Calculate how many minimum machines have to be motioned to realise the order without any delay if the company works in the system of 4 shifts (machines work continuously, 7 days/week, 24 hours) 28

30 SOLUTIONS AND ANSWERS Task 1 Answer: Yes, it is. Task 2 Answer: Yes, it is. Task 3 Answer: It is a rational number, since it is equal to 3. Task 4 Answer: Yes, it is. Task 5 Answer: a 2 = ( 1) = 3 7 a 3 = 1 6 a 6 = 1 39 Task 6 a 2 = a = = 8 a 3 = a = 18 a 4 = a = 33 a 5 = a = 53 Task 7 1.5% of 32 = = = Euros the price for 1 stock after increasing = 3248 Euros Task = 147 kg the net weight 29

31 = = = 98% Task = 33 Euros 33 = 11 = 22 = 22% - VAT tax Task = 1.12 Euros % 7.3% Task 11 Let x be the sought number 20% x = x = x = 7.5. Task 12 x the salary before the rise 15% x the rise 120 the rise 15% x = x = 120 x = 800. The worker is now earning = 920 Euros. Task 13 The weight of loading is = 20 t. 1 cylinder weights kg (why?). We will need 1000 boxes ( : 10). The total weight of 1 box with 10 cylinders is kg +20 kg = kg 30

32 : One track can be loaded with 118 boxes : We need 9 tracks. Task 14 The client will pay = 58.3 euros per 1 cylinder = euro per cylinders. The client will pay euros The average time for the distance of 200 km (if the average speed is 60 km/h) comes to 3h 20 min. The cost of the transport for 1 track is = euros for 8 tracks. 80 = = euros The company earns 5.3 euros on every cylinder, euros for cylinders = euros The company will earn euros. Task Cost of machining = 36/hour One shaft = 5 minutes The time to machine one = 60/5 = 12mins, 36/12 = 3.00 per shaft 2. The cost of the material = 50, therefore, total cost = = 53. If the customer requires 500, then 500 x 53 = 26, To apply a 15% discount = 26,500 x 15/100 = 3,975 26,500-3,975 = 22, 525 Task 16 Factory I: Time of production 5 min. The number of components produced in one hour 60min 5min = 12 The cost of production one shift by a machine = 3 The cost of material 50 The total production cost of one shaft = 53 31

33 The price of 1000 components = Factory II: Time of production 6 min. The number of components produced in one hour 60min 6min = 10 The production cost of one shaft by a machine = 3.4 The cost of material 50 The total production cost of one shaft = 52.4 The price of 1000 components = The offer of Factory II is preferable. Task 17 1 hour of work 60min 5min = 12 (12 shafts are produced) 2 hours of work 95% 12 = 11,4 (11 shafts are produced, the sum is 23) 3 hours of work 90% 11 = 9.9 (9 shafts are produced, the sum is 32) 4 hours of work 85% 9 = 7.65 (7 shafts are produced, the sum is 39) 5 hours of work 80% 7 = 5.6 (5 shafts are produced, the sum is 44) 6 hours of work 75% 5 = 3.75 (3 shafts are produced, the sum is 47) 7 hours of work 70% 3 = 2.1 (2 shafts are produced, the sum is 49) 8 hours of work 65% 2 = 1.3 (1 shaft is produced, the sum is 50) The worker is able to produce 50 shafts. Task 18 a) 5/60 of hour = 5 min. 60min /5 min = 12 items b) / 12 = 3 c) = 53 d) = e) 15 % = 15/ /100 = = f) 2 m = 200 cm 200 cm / 24 cm = items 32

34 Task 19 g) 8 items 24 cm =192 cm - the material used for the shafts 200 cm 192 cm = 8 cm - the rest of material together with 0.5 tolerance cm = 4 cm 8 cm 4 cm = 4 cm - the rest from one bar h) 40 mm /2000 mm = 1/50 = 0.02 of the whole material a) % = = the number of shafts to produce with the quality norm b) In one hour by one machine: 60min/5min = 12 items In 24 hours by one machine: = 288 items In 10 days by one machine: = 2880 The number of machines needed: 10200/2880 = machines have to be motioned. 33

35 Tasks 1-2 categorisation Quantity Mathematical subfields The number system Basic operations Formulae/data provided - Understanding basic numerical facts Performing simple operations Level 1 ALL Tasks 3-4 categorisation Quantity Mathematical subfields The number system Basic operations Formulae/data provided - Understanding basic numerical facts Performing simple operations Level 1 ALL Tasks 5-6 categorisation Quantity Mathematical subfields Numerical sequences and patterns Basic operations Formulae/data provided - Performing simple operations Level 1 ALL Tasks 7-12 categorisation Quantity Mathematical subfields Decimals, fractions, ratios Basic operations Formulae/data provided - Performing simple operations Level 1 ALL Task 13 categorisation s Space and shape, including measurements Quantity Mathematical subfields 2D-3D shapes and properties Basic operations Formulae/data provided Density of steel ρ = 7,85 kg/dm 3 volume of cylinder: V = πr 2 h Transformation of formula Switching between units Level 2 Machinist, buyer Task 14 categorisation s Space and shape, including measurements Quantity Mathematical subfields Formulae/data provided 2D-3D shapes and properties Basic operations Performing simple operations Costing a project Level 2 Machinist, Buyer 34

36 Task 15 categorisation Mathematical subfield Formulae/data provided Quantity Basic operations Decimals fractions and ratios Performing simple operations Using models of proportion Costing a project Level 1 Buyer, Estimator, Sales Engineer, Technical Support, Designer Tasks categorisation Quantity Mathematical subfield Basic operations Decimals fractions and ratios Formulae/data provided Performing simple operations Using models of proportion Costing a project Level 2 Buyer, Estimator, Sales Engineer, Technical Support, Designer 35

37 QUIZ 1. The result of the operation π + 3π is a: 2 2 a. Natural number b. Rational number c. Real number d. Cannot be determined 2. The result of the operation 3-a, where a is an integer, is always an integer. a. True b. False 3. Which of the following should be added to 1 3 in order to get as a result2 5? a. 1/2 b. 1/3 c. 1/5 d. None of the above 4. Which of the following operations gives the largest result? a b c d Which of the following divisions gives the smallest result? a. 2:π b. 3:π c. π:2 d. π:3 Answers: 1c, 2a, 3d, 4c, 5a 36

38 L5: DECIMALS, FRACTIONS AND PERCENTAGES Theory "Percent" is actually "per cent", meaning "out of a hundred". This meaning can be used, along with the fact that fractions indicate division, to convert between fractions, percentages, and decimals. Percentage to decimal You should just move the decimal point two places. Examples: 23% = % = % = Percentage and decimal to Fraction You should use the fact that "percent" means "out of a hundred", thus convert the percent to a decimal, and then to a fraction. Examples: 30% = 0.30 = 0.3 = % = = = 1 8 Decimal to Percent You should move the decimal point two places to the right. Examples: 0.29 = 29% 1.74 = 174% = 0.14% Fraction to Decimal and percent You should remember that fractions are division, thus all you have to do is that division. Then, if needed you can transform the decimal into a percent by following the process mentioned above. 37

39 Examples: = 3: 4 = 0.75 = 1: 3 = = 0. 3 Percentage calculations 1) Calculating percent of a number Example: 15% of 60 = 15% 60 = = 9 2) Calculate what percent of a number is a number Example: What percent of a number 20 is the number 5? Firstly, we can ask what fraction of the number 20 is number 5: 5 20 = 5 100% = 25% 20 3) By what percent is a number increased/decreased Example: The price of oranges at the marketplace is 1.50 Euros and 2 Euros per kg. a) By what percent are oranges cheaper at the marketplace than in a shop? = 0.50 the difference in the prices % = 25% b) By what percent are oranges more expensive in a shop than in a marketplace? % = 33. (3)% 1.50 Note that the percentage is higher in b) because we compare it with the smaller price. 4) Calculating a number when its percentage is given 38

40 Example: Find a number whose 25% is equal to 35. Let x be the sought number 25% x = 35 1 x = 35 4 x =

41 Tasks Task 1 The value of 1 stock of ABC Company costs 32 Euros and increased by 1.5%. What is the current value of 100 stocks? Task 2 The gross weight of a commodity is 150 kg and the tare weight is 3 kg. What percent of the gross weight is the net weight? Task 3 For a commodity of the value of 150 Euros you have to pay 183 Euros with VAT. Calculate what percent is the VAT tax. Task 4 The price of a stock increased from to Euros. Calculate by what percent was the price of stocks increased. Task 5 Find a number whose 20% is 1.5. Task 6 A worker got a salary rise of 120 Euros which is 15% of his salary. How much was he earning before the rise and how much is he earning now? Task 7 One of the first things we have to do when producing an engineered item is to do the costing to ensure it is viable to make. The costs of the materials, the machining costs and the time to make it have to be considered. An aerospace customer requires a batch of machined shafts. A Computer Numerically Controlled (CNC) lathe is used to manufacture an aerospace component; the operating cost of the CNC machine in an engineering workshop is per hour the shaft takes 5 minutes to make on the machine. the material cost of each component is

42 Task 8 1. How much does each component cost to manufacture taking into account the cost of material and machining time? 2. The customer requires a batch of 500, what is the price of the full batch? 3. When the customer gets the above quote they feel they should get a 15% discount. With the discount applied what would the customer be charged? A Computer Numerically Controlled (CNC) lathe is used to manufacture an aerospace component: the shaft has length of 240 mm each component is cut from a 2000 mm long bar with an allowance of 5 mm per cut. In Factory I: the operating cost of the CNC machine in an engineering workshop is per hour the shaft takes 5 minutes to make on the machine the material cost of each component is 50 In Factory II: the operating cost of the CNC machine in an engineering workshop is per hour the shaft takes 6 minutes to make on the machine the material cost of each component is 49. A client wants to buy 1000 components. Which offer he should choose? 41

43 Task 9 A Computer Numerically Controlled (CNC) lathe is used to manufacture an aerospace component and it is known that the shaft takes 5 minutes to make on the machine. The effectiveness of the CNC operator decreases by 5% per hour from the moment of the beginning of work (the worker produces 5% less components than one hour before). Is it possible to produce 50 shafts during 8 hours of work? Task 10 A Computer Numerically Controlled (CNC) lathe is used to manufacture an aerospace component: the operating cost of the CNC machine in an engineering workshop is per 60 min. the shaft takes 5 minutes to make on the machine the material cost of each component is 50 the shaft has length of 24 cm each component is cut from a 2 m long bar with an allowance of 0.5 cm per cut. a) How many shafts will be produced during one hour by the machine? b) What is the production cost of one component by the machine? c) What is the total production cost of one component? d) What cost has to bear a client for 500 components? e) How much he will pay if he receives 15% of discount? f) How many shafts will the machine produce from one bar? 42

44 g) How much scrap raw material will be left from each bar? h) What part of the whole is the scrap raw material left from each bar? Task 11 A Computer Numerically Controlled (CNC) lathe is used to manufacture an aerospace component: the operating cost of the CNC machine in an engineering workshop is per 60 min. the shaft takes 5 minutes to make on the machine the material cost of each component is 50 the shaft has length of 24 cm each component is cut from a 2 m long bar with an allowance of 0.5 cm per cut. A client ordered shafts with the time limit of 10 days. Usually 2% of the ready products do not fulfil the quality norm. a) Calculate how many shafts the producer has to produce to meet the expectations of the client. b) Calculate how many minimum machines have to be motioned to realise the order without any delay if the company works in the system of 4 shifts (machines work continuously, 7 days/week, 24 hours) 43

45 SOLUTIONS AND ANSWERS Task 1 1.5% of 32 = = = Euros the price for 1 stock after increasing = 3248 Euros Task = 147 kg the net weight = = = 98% Task = 33 Euros 33 = 11 = 22 = 22% - VAT tax Task = 1.12 Euros % 7.3% Task 5 Let x be the sought number 20% x = x = x =

46 Task 6 x the salary before the rise 15% x the rise 120 the rise 15% x = x = 120 x = 800. The worker is now earning = 920 Euros. Task 7 1. Cost of machining = 36/hour One shaft = 5 minutes The time to machine one = 60/5 = 12mins, 36/12 = 3.00 per shaft 2. The cost of the material = 50, therefore, total cost = = 53. If the customer requires 500, then 500 x 53 = 26, To apply a 15% discount = 26,500 x 15/100 = 3,975 26,500-3,975 = 22, 525 Task 8 Factory I: Time of production 5 min. The number of components produced in one hour 60min 5min = 12 The cost of production one shift by a machine = 3 The cost of material 50 The total production cost of one shaft = 53 The price of 1000 components = Factory II: Time of production 6 min. The number of components produced in one hour 60min 6min = 10 The production cost of one shaft by a machine = 3.4 The cost of material 50 The total production cost of one shaft = 52.4 The price of 1000 components =

47 The offer of Factory II is preferable. Task 9 1 hour of work 60min 5min = 12 (12 shafts are produced) 2 hours of work 95% 12 = 11,4 (11 shafts are produced, the sum is 23) 3 hours of work 90% 11 = 9.9 (9 shafts are produced, the sum is 32) 4 hours of work 85% 9 = 7.65 (7 shafts are produced, the sum is 39) 5 hours of work 80% 7 = 5.6 (5 shafts are produced, the sum is 44) 6 hours of work 75% 5 = 3.75 (3 shafts are produced, the sum is 47) 7 hours of work 70% 3 = 2.1 (2 shafts are produced, the sum is 49) 8 hours of work 65% 2 = 1.3 (1 shaft is produced, the sum is 50) The worker is able to produce 50 shafts. Task 10 a) 5/60 of hour = 5 min. 60min /5 min = 12 items b) / 12 = 3 c) = 53 d) = e) 15 % = 15/ /100 = = f) 2 m = 200 cm 200 cm / 24 cm = items g) 8 items 24 cm =192 cm - the material used for the shafts 200 cm 192 cm = 8 cm - the rest of material together with 0.5 tolerance cm = 4 cm 8 cm 4 cm = 4 cm - the rest from one bar h) 40 mm /2000 mm = 1/50 = 0.02 of the whole material Task 11 a) % = = the number of shafts to produce with the quality norm b) In one hour by one machine: 60min/5min = 12 items In 24 hours by one machine: = 288 items In 10 days by one machine: = 2880 The number of machines needed: 10200/2880 = machines have to be motioned. 46

48 Tasks 1-6 categorisation Quantity Mathematical subfields Decimals, fractions, ratios Basic operations Formulae/data provided - Performing simple operations Level 1 ALL Task 7 categorisation Quantity Mathematical subfield Basic operations Decimals fractions and ratios Formulae/data provided Performing simple operations Using models of proportion Costing a project Level 1 Buyer, Estimator, Sales Engineer, Technical Support, Designer Tasks 8-11 categorisation Quantity Mathematical subfield Basic operations Decimals fractions and ratios Formulae/data provided Performing simple operations Using models of proportion Costing a project Level 2 Buyer, Estimator, Sales Engineer, Technical Support, Designer 47

49 QUIZ 1. If the amount of 35 euros is increased by 100% it will become: a. 350 euros b. 70 euros c. 135 euros d. None of the above 2. The price of a stock increased from to Euros. By what percent was the price of stock increased? a. Approximately 27% b. Approximately 13% c. Approximately 3% d. None of the above 3. Is it true or false that 17.5% is the same with 0.175? a. True b. False 4. A commodity has lost half of its value the last months. In other words: a. It was decreased by 1/2% b. It was decreased by 25% c. It was decreased by 50% d. It was decreased by 100% 5. Which of the following is equal to 3/5? a. 60% b. 35% c. 3.5% d. None of the above Answers: 1b, 2a, 3a, 4c, 5a 48

50 L6: MENTAL CALCULATIONS Theory Mental calculations are arithmetical calculations by using only the brain, with no help from calculators, or written operations. Two key features of mental calculation are that numbers are treated as quantities rather than as digits and calculations that appear similar can be amenable to the use of different strategies depending on the numbers involved. Examples: = 14 can be used to calculate mentally other results: 7+ 6, 7 + 8, , , = 32 can be used to calculate 16 4 by doubling. 36 : 3 = 12 can be used to calculate 66 : 3 by partitioning 66 into and using the knowledge of 36 : 3 and 30 : 3 to reach the answer of is 91 (double 45 plus 1) (near doubles) is 56 plus 30 (86) plus 4 (90) plus 1, giving 91 (partitioning the smaller number and then bridging to 10) is 80 (40 plus 40) plus 15 (known fact: = 15), giving 95. Different strategies: 1) Counting forwards and backwards: Examples: count on 2 to 80 then 3 to count back in tens from 76 or count on in tens from count on in hundreds from count back in hundreds from 860 or count on in hundreds from ½ + ¾ count on in quarters count on in tenths 2) Reordering: Examples: = = = = = ) Partitioning 1 Using multiples of 10 and 100 Examples: = = = =

51 4) Partitioning 2 Bridging through multiples of 10 Examples: = = = ) Partitioning 3 Compensating Examples: = = = ½ + 1 ¾ = 3 ½ + 2 ¼ = ) Partitioning 4 Using near doubles Examples: is double 45 and add is double 250 and add 10 then add 20, or double 260 and add 10, or double 270 and subtract is double 300 and subtract 20 twice is double 2.5 and add 0.1 or double 2.6 and subtract is double 500 add 21 and then subtract 13 7) Multiplying and dividing by multiples of 10 Examples: = = : 10 = : 100 = = : 100 = ) Doubling and halving Examples: 24 5 = : = = = : 2 Half of 840 = 420 Quarter of 84 = Half of half of = 70 x = = % of 25 = 10% of = : 4 = (16 : 4) : 2 =

52 Tasks Task 1 1. Knowing that = 40, provide answers for 52 13, 51 12, Calculate by using different strategies 81 4, 81 61, 81 42, 81 78, Task 2 Calculate the following by using mental calculations: Task 3 Calculate the following by using mental calculations: Task 4 Calculate the following by using mental calculations: Task 5 Calculate the following by using mental calculations: : : Task 6 Find the half of 9 Find the half of 44 Find one third of 27 Find one tenth of 40 Find one fifth of 35 51

53 Find ½ of 46 Find ½ of 130 Find ½ of Find ½ of 730 Find ½ of Find 25% of 100 Find 80% of 100 cm Find 25% of

54 SOLUTIONS AND ANSWERS Task , 39, = = 77, = = 20, = = 40 1 = 39, = = = = 3, = = 20 2 = 18. Task = = ,7 4,5 = 5,7 4 0,5 Task = ,7 + 0,45 = 0,7 + 0,3 + 0,15 Task = = Task ,378 1, = 43 : = 21,5 100 = 2150 Task 6 Answers: 4,5, 22, 9, 4, 7, 23, 65, 5 + 0,70 = 5,70, = 365, ,50 + 0,15 = 41,65, 25, 80 cm, ½ 130 = 65 53

55 Tasks 1-6 categorisation Quantity Mathematical subfield Mental calculations Formulae/data provided Performing simple operations Level 1 ALL 54

56 Part 2: Change and relationships 55

57 L7: Types of change Theory Direct and inverse proportion Two variables x, y are proportional if a change in one is always accompanied by a change in the other, and if the changes are always related to a constant multiplier, which is called the coefficient of proportionality or proportionality constant, they are proportionally constant. If one variable is always the product of the other, the two are said to be directly proportional. In other words, x and y are directly proportional if the ratio y is x constant. If the product of the two variables is always equal to a constant, the two are said to be inversely proportional. In other words, x and y are inversely proportional if the product xy is constant. In order to express the statement "y is proportional to x we write an equation y = cx, for some real constant, c. In order to express the statement "y is inversely proportional to x" mathematically, we write the equation y = c/x. We can equivalently write "y is proportional to 1/x", which would be expressed by y = c/x. Examples: The circumference of a circle is directly proportional to its diameter, with the constant of proportionality equal to π. On a map drawn to scale, the distance between any two points on the map is directly proportional to the distance between the two locations that the points represent, with the constant of proportionality being the scale of the map. The number of tools you might buy and the amount you pay for them. If you buy twice as many tools as your friend, you pay twice as much. The connection between the cost and the number can be written as the equation Cost of tools = price per tool number of tools bought. This can also be written as y = kx, where k is the cost (the price per tool). If an object travels at a constant speed, then the distance travelled is directly proportional to the time spent traveling, with the speed being the constant of proportionality. The time taken for a journey is inversely proportional to the speed of travel; the time needed to build a wall is (approximately) inversely proportional to the number of workers building. Let s consider the rectangles of an area equal to 6. If the lengths of the rectangles sides are x and y, then x y =6, which means that = 6. The shorter the one side, the x 56

58 Proportion longer the second side. The lengths of the rectangles sides are inversely proportional. Proportion is an equality of two ratios, like in the following forms: a b = c d or a b = c d According to the main property of proportion: ad = bc, thus if three quantities in a proportion are given you can calculate the fourth by using the formulas: a = bc d, b = ad c, c = ad b, d = bc a. Example 25 = 3, 6 = 12 = Linear functions and equations are examples of proportions. A linear function has the form y = f(x) = ax + b A linear function has one independent variable and one dependent variable. The independent variable is x and the dependent variable is y. a is the coefficient of the independent variable. It is also known as the slope and gives the rate of change of the dependent variable. b is the constant term or the y intercept. It is the value of the dependent variable when x = 0. Combinations of linear equations Linear equations can be added together, multiplied or divided. Example: C(x) is a cost function and equals to: fixed cost + variable cost R(x) is a revenue function and equals to: selling price number of items sold P(x) is a profit function and equals to revenue less cost: P(x) = R(x) - C(x), where x = the number of items produced and sold Direct proportion to powers A value y can be directly proportional to x 2, x 3 and other powers of x. In these cases, an equation is formed with k, a constant multiplier (the constant of proportionality), at the start: y = kx n. 57

59 Quadratic functions The quadratic function has the form: f(x) = y = a x 2 + bx + c where a, b, and c are numerical constants and a is not equal to zero. Exponential and logarithmic proportionality A variable y is exponentially proportional to a variable x, if y is directly proportional to the exponential function of x, that is if there are non-zero constants k and a such that y = ka x A variable y is logarithmically proportional to a variable x, if y is directly proportional to the logarithm of x, that is if there are non-zero constants k and a such that y = k log a (x) Exponential change is a change that does not progress at a steady rate, like linear change, but occurs at an increasing rate. Example: A single bacteria cell might divide into two cells once every 20 minutes (this is actually how fast e coli multiplies). This is known as its doubling rate. So, after 20 minutes, we have 2 e coli cells. After 40 minutes, each of those has divided into two, and we have four e coli cells. After an hour, they have divided again, and we have eight e coli cells. In another hour, we have 64 cells. And so on. We are not just adding e coli cells to the mix, we are multiplying them, and so the number of cells increases at an increasing rate. Example: Name few more cases of exponential growth. When solving exponential problems, two situations often occur: Time is given; you are expected to find the amount at that given time. This usually just involves evaluating the amount function. The amount is given; you are expected to determine at what time this amount occurs. This usually involves solving an exponential equation, which means logarithms will be needed. A logarithm is an exponent which indicates to what power a base must be raised to produce a given number: y = a x x = log a y exponential form logarithmic form 58

60 Examples: x is the logarithm of y to the base a (a > 0, a 1, y > 0) log a y is the power to which we have to raise a to get y x = log x = log 16 4 x = log 2 8 x = log 5 25 This means the logarithm of to the base 10 (called common logarithm). It is the exponent to which 10 must be raised to get We know that 10 4 = Therefore, x = 4. This means the logarithm of 4 to the base 16. It is the exponent to which 16 must be raised to get 4. We know that 16 = = 4. Therefore x = 1 2. This means the logarithm of 8 to the base 2. It is the exponent to which 2 must be raised to get 8. We know that 2 3 = 8. Therefore, x = 3. This means the logarithm of 25 to the base 5. It is the exponent to which 5 must be raised to get 25. We know that 5 2 = 25. Therefore, x = 2. Example: The table shows the sound level intensity (in W/cm 2 ) of various sources. The hearing threshold level for the human ear is around W/cm 2. The sound level intensity which is painful is around 10-4 W/cm 2. Source of sound Sound intensity I [W/cm 2 ] Jet plane 10-2 Pain level 10-4 Rock concert 10-5 Pneumatic drill 10-6 Conversation Whispering Hearing threshold level You can measure the sound intensity level by comparing it with the intensity of another sound. In practice we use the measure called noise level. It compares the intensity of any sound with the hearing threshold level I 0 = W/cm 2 : For example, for a pneumatic drill we have: L = 10 log I I 0 db (decibel) L = 10 log = 10 log 1010 = = 100 db 59

61 Trigonometric change The trigonometric functions are functions of angles, since they relate the angles of a right triangle to the lengths of its sides. The most known trigonometric functions are the sine, cosine, and tangent y = sinx, Domain: (, ), Range: [ 1, 1] y = cosx, Domain: (, ), Range: [ 1, 1] y = tanx, Domain: All reals numbers except odd multiples of π 2 Range: (, ) Graphing a sine, cosine and tangent functions 60

62 Tasks Task 1 We know that y is directly proportional to x. We also know that when x = 15 then y = 5. Find the constant of proportionality and the value of x when y = 10. Task 2 The daily costs of a company producing cats fodder are described by the formula y = 2x + 800, where 800 is a constant daily cost, x the number of cans with fodder produced in 1 day and 2 PLN is the cost of production of one can. The incomes of the company can be described by the formula y = 4x, where x is a number of sold cans per day and 4 PLN is the price of 1 can. Assuming that the number of produced and sold cans per day is the same: a) for what rate of production per day the company s income is equal to 0 (incomes are equal to production costs)? b) for what rate of production per day the company can achieve a daily profit of 400 PLN and for what rate 1400 PLN? Task 3 Let us consider rectangles with the sides of x and y, the area equal to 10 and the perimeter no more than 40. Write a formula describing the relation of y as a function of x. What is the domain of the function? Task 4 How much more time will a journey of the distance of 150 km take at a speed 45 km/h than at a speed of 90 km/h? Task 5 The relation between the average speed v and a time t needed to cover the distance of 60 km is described by the formula: t = 60. Fill the following table and draw a graph. v v [km/h] t [h] 61

63 Task 6 A cardboard model of a car is part of an outdoor display. Its height is 0.5 m. The actual car is 4.5 m long, 1.5 m high and 1.8 m wide. Find the length and the width of the model, if its dimensions are proportionate to the real car. Task 7 A company receives 63 for each unit of output sold. It has a variable cost of 32 per item and a fixed cost of What is its profit if it sells: (a) 40 items? (b) 120 items? (c) 350 items? Task 8 Calculate: Task 9 a) log 3 81 b) log c) log d) log 9 3 e) log By using the table of values of common logarithms shown calculate: a. log 2 b. log 1.41 c. log 1.75 d. log

64 Task 10 Source of sound Sound intensity I [W/cm 2 ] Jet plane 10-2 Pain level 10-4 Rock concert 10-5 Pneumatic drill 10-6 Conversation Whispering Hearing threshold level ) Using the formula L = 10 log I concert and conversation. I 0 db calculate the noise level of a jet plane, rock 2) How many times is the sound intensity level increased if the noise level is increased Task 11 by: a) 10 db b) 20 db c) 40 db In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 30 o. The drill should reach hole 2 at the latest after 0.8 s. What is the minimum average movement speed in m/min? 63

65 Task 12 In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 45 o. The drill should reach hole 2 at the latest after 0.8 s. What is the minimum average movement speed in m/min? Task 13 In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 30 o. The drill reaches hole 2 in 0.8 s and the drilling of a hole lasts 0.5 s. The process is repeated for a number of holes. How many holes can be bored within 2 min? What is the distance between first and the last hole? Task 14 In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 30 o. The drill reaches hole 2 in 0.8 s. and the drilling of a hole lasts 0.5 s. The process is repeated for a number of holes for 2 min. Given that the distance of the centre of hole 1 is 2 cm from both edges of the block, calculate the minimum dimensions of the block in order for the drilling process to be completed. Task 15 A numerically controlled machine tool moves with a constant speed of 6.5 m/min. The drilling of a hole lasts 0.5 s. 64

66 Find the shortest routes that the drill should follow in order to complete the patterns shown in the images below. Task 16 Materials have different properties, one of them is hardness FORCE kg F Ball Indenter D Width of indentation d 2F HB = πd(d (D 2 d 2 ) A Brinell Hardness test is carried out to determine the hardness of a material. This can be especially useful when looking at castings to check they are conforming to the desired specification. Particularly, the Brinell hardness test measures the permanent width of indentation produced by a carbide indenter. A load is applied to a test specimen, for a given length of time. The material thickness and time the load is applied have to be constant to produce ideal or fair conditions. The drawing shows the Brinell method of hardness testing. HB is the number used to specify the hardness. Calculate the impression diameter d, for a test that has a load F of 3000kg, a Ball Diameter of 10mm and a specimen that has specified HB of 130 Task 17 Calculate HB, when: F = 3000kg D 10 mmm d = 1 mm Task 18 For a certain material HB is 150. Let F 1 = 1000kg, F 2 = 3000kg, F 3 = 5000kg. Assume that D = 10 mmm. Examine d at different pressures F. 65

67 Task 19 2F HB = πd(d (D 2 d 2 ) a. If force F will be double, how HB will change? b. Is the following statement correct or wrong? If D doubles then HB halves. Task 20 Investigate how d varies, while D is changing, but HB and force F are fixed (HB = 100 and F = 3000kg). Make calculation by using D 1 = 10 mm, D 2 = 15 mm, D 3 = 20 mm. Is this change regular? Comment on your results. Task 21 Assuming that R 1 = 20 Ω, R 2 = 20 Ω, R 1 and R 2 are connected in parallel and I = 2 A, how much is V? Task 22 An engineer measures the current flowing in a 24V DC circuit with a multimeter and discovers it is 1.2A. The resistor R1 is 100 Ω but R2 is not known, its value is required in order to verify the circuit. Calculate the value of R 2. Task 23 The current system (see the figure) has dropped to a minimum and is 200mA. Resistor R 1 is still 100 Ω. Investigate what resistor should be plugged parallelly, in order to maintain the voltage of 24V. 66

68 Task 24 A third resistor is connected to the system, displayed at the Fig. 1, (see. Fig.2). The power source is fixed at 24V. How is the current flowing (I)? R 2 R 1 R 3 Fig. 1 Fig.2 Task 25 a. Investigate how parallel connections of successive resistors of the same resistance (expressed in Ω) affect R T. Consider n = 5, n = 10. Formulate a conclusion. b. Investigate how parallel connections of successive resistors of the same resistance (expressed in Ω) affect I (current flowing). Formulate a conclusion. Task 26 How does V depend from R, if I is constant? Create a table that demonstrates that dependence. Take, that I = 2.4 A. Task 27 Examine the relationship between I and R when V is constant and equal to 24V. Create a table that demonstrates that dependence. Take, that the smallest resistance R = 5 Ω. Draw a graph of this dependence. Formulate a conclusion. Task 28 You have to make the resistance R T being equal to 20 Ω, by combining parallel connections of other resistors. You can choose from the following types of resistors: R 1 = 100 Ω, R 2 = 25 Ω, R 3 = 50 Ω, R 4 = 35 Ω. Which will you choose? Has the task only one solution? You can freely choose the number of resistors. Task 29 You have to make the resistance R T being equal to 20 Ω, by combining parallel connections of other resistors. You can choose from the following types of resistors: R 1 = 100 Ω, R 2 = 25 Ω, R 3 = 50 Ω, R 4 = 35 Ω. The prices for resistors are the following: R 1 : 5 euros (50 pieces), R 2 : 6 euros (50 pieces), R 3 : 7 euros (50 pieces) and R 4 : 6.5 euros (50 pieces). Which combination is the most cost-effective? 67

69 Task 30 An electric motor is a device that converts electrical energy into mechanical energy. An induction motor consists of a fixed stator and a rotor that has electro-magnets attached to it (see diagram below). When an AC current is passed through the windings in the stator, it produces a rotational magnetic field and so the rotor turns within the stator. Motors vary widely and are chosen for their application this can include the required constant speed or torque requirements. An induction motor needs to be chosen for its speed. The synchronous speed (ns) of an induction motor is based on the supply frequency and the number of poles in the motor winding and can be expressed as: ns = 120 f/n where: ns = revolutions per minute (rpm) f = frequency [cycles per sec] in Hertz (Hz) n= number of motor poles The actual speed of the motor is less than its synchronous speed; this is because the rotor will never catch up to the stator's rotating magnetic field. This difference between actual speed and synchronous speed is called slip. The actual speed of the motor (na) = 1425 rpm. 1. Calculate the synchronous speed of a motor with a frequency of 50Hz and 4 poles. 2. If slip is a ratio of actual speed to the synchronous speed, calculate the % slip (s) when s = (ns na/ ns) and (na) = 1425 rpm Task 31 An electric motor is a device that converts electrical energy into mechanical energy. An induction motor consists of a fixed stator and a rotor that has electro-magnets attached to it (see diagram below). 68

70 When an AC current is passed through the windings in the stator, it produces a rotational magnetic field and so the rotor turns within the stator. Motors vary widely and are chosen for their application this can include the required constant speed or torque requirements. An induction motor needs to be chosen for its speed. The synchronous speed (ns) of an induction motor is based on the supply frequency and the number of poles in the motor winding and can be expressed as: ns = 120 f/n where: ns = revolutions per minute (rpm) f = frequency [cycles per sec] in Hertz (Hz) n= number of motor poles The actual speed of the motor is less than its synchronous speed; this is because the rotor will never catch up to the stator's rotating magnetic field. This difference between actual speed and synchronous speed is called slip. The torque, which is directly proportional to slip is given by the formula: Task 32 T = Power 9550 speed Torque is measured in Nm, Power is measured in kw and speed is measured in rpm. A motor with an efficiency of 95% will require an input power (P) of 7.8kW to obtain an output power of 7.5W. Calculate the Torque. A mechatronic while designing a pneumatic-hydraulic system has to use a pomp of a maximum pressure of 6 bars (P e1 ). He/she has to calculate the hydraulic pressure (P e2 ) and the liquid volume (V c ) considering the stroke given in the pneumatic-hydraulic switch presented in the picture. 1) What is the hydraulic pressure (P e2 ), if one does not consider the friction loss? 2) What is the hydraulic pressure given that the efficiency coefficient is 85%? 3) What is the liquid volume (V c ) provided by the pressure switch given that the stroke is 50 mm? 69

71 Task 33 A grinding wheel with an external diameter d = 240 mm works with the permissible peripheral speed of 32 m/s. What is the maximum rotation speed that the drive engine can work with? Task 34 An operator is supposed to make a 220 mm long (L w ) conical reamer whose length of the conical part is L = 130 mm and respective diameters are D = 34mm, d=30mm. What is the tailstock shift V R necessary to make the conical part? Task 35 The table below contains some specific cutting force indicators. An operator/mechanic is supposed to choose the type of the cutting machine to make a series of details on the basis of the accepted technological data, that is to calculate the thickness of the cut layer h and read from the table the value of kc. Then he/she is supposed to calculate the cutting force Fc and the cutting power Pc. Particularly, a rod made from the steel (16 Mn Cr 5) is cut with the cutting depth a = 5 mm, feed f = 0.32 mm, cutting edge angle x = 75 and cutting speed vc = 160 m/ min. Calculate the following: Thickness of the cut layer h (mm) Specific cutting force kc (N / mm) Cutting force Fc (N) Cutting power Pc (KW) Section of the cut layer A (mm²) 70

72 Task 36 While designing a detail a designer made a part of it that is meant to be subjected to measurement control by an operator with the use of a gage block. The length of the gage block is L1 = 100 mm and thermal expansion coefficient is α = / C. What is the extension of the gage block given that its temperature increases due to the hand warmth from 20 C to 25 C. Task 37 The permissible load capacity of an electric lift is 4000 kg. The engine of the particular electric lift uses the power of 8.4 kw. The engine and the mechanism of the lift have the total efficiency of 82%. What load can be raised by the lift at the height of 4 m within 20 seconds? Task 38 A locksmith is supposed to prepare the details for further cutting. His task is to cut the material with a cutting blade. The contact pressure is F = 800 N. A mechanic has to calculate the cutting forces of a one-sided cutting blade in order to properly cut the material. What are the cutting forces F 1 and F 2 of the cutting blade? 71

73 Task 39 A locksmith is supposed to prepare and sharpen a mechanic saw blade s teeth, since when the saw is working there is an excessive movement resistance. Calculate the cutting force F c and feed cutting force F f of a single saw blade s tooth if a contact pressure is F = 3500 N. Task 40 A stock whose load makes up a force of 4000 N is stored on a 6 m long slack ramp of the height h = 1.2 m. What F force is necessary to prevent the load F G making up a force of 4000 N from rolling down? (friction is not taken into consideration) In the image shown: Haltenkraft F: Stabilising force F 72

74 SOLUTIONS AND ANSWERS Task 1 Since y is proportional to x it will be y = kx Also when x = 15 then y = 5 To find the value of k we substitute the values y = 5 and x = 15 into y = kx 5 = k 15, so k = 5 = To find the value of x when y = 10 we substitute y = 10 and k = 1/3 into y = kx 10 = (1/3) x So x = 30 when y = 10. Task 2 Hint: Draw the two diagrams shown in the image below a) 2x +800 = 4x x = 400 b) 4x - (2x+800) = 400 2x = 1200 x = 600 4x - (2x+800) = x = 2200 x =

75 Task 3 Let x, y the lengths of sides of a rectangle x y = 10, which means that y = 10 x. In order to find the domain of the function we solve the inequality: 2x + 2y 40 x + y 20 x + 10 x 20 x + 10 x 20 0 x 2 20x+10 x 0 x(x 2 20x + 10) 0 x = 0, x = , x = x (0, ]. Task 4 Twice more time: With the speed 45 km/h time of journey is 200 min. With the speed 90 km/h time of journey is 100 min. Task 5 Answer: v [km/h] t [h]

76 Task 6 x the length of the car cardboard model 1.5 = x 1.5x = x = 2.25 : 1.5 x =1.50. y the width of the car cardboard model y = 0.6. The length of the car cardboard model is 1.5 m and the width is 0.6 m. Task 7 R(x) = 63x C(x) = x P(x) = 63x - ( x) = 31x Task 8 (a) For x = 40: P(40) = = -340 which is a loss (b) For x = 120: P(120) = = 2140 (c) For x = 350: P(350) = = 9270 Answers: 75

77 a) 4, b) - ½, c) 3/2, d) ½, e) 4. Task 9 Answers: a b c d Task 10 1) Jet plane: 10 log = = 140 Rock concert: 10 log = = 110 Conversation: 10 log 10 6 = 10 6 = 60 2) a) 10 times, b) 100 times, c) 1000 times. Task 11 Answer: v = 6.3 m/min Task 12 Answer: m/min Task 13 Answer: 92 holes Distance: m Task 14 Answer: Min. dimensions of the block are: a = 3.85 m, b = 6.67 m. Task 16 Firstly, transpose the equation so that d is the subject d = D² - (D 2F/(HB x πd)² Once this is completed insert the numbers (be careful with the units!) D = 10mm, F = 3000kgF, HB =

78 d = 10² [10 (2x 3000/130 x π x 10)²] d = 100 [10 ( )² d= d = 27,221 = 5.217mm Task 19 Answer: a. HB will be half b. wrong Task 21 1/R T = 1/ R 1 + 1/ R 2, so 1/R T = 1/20 + 1/20 = 1/10, thus R T = 10 Ω Because V = I R, then V = 2A 10 Ω = 20 V Task 22 V = I R R = V/I = 24V/1.2A =20 Ω, using Ohm s Law: 1/R T = 1/ R 1 + 1/ R 2 R 2 = 25 Ω. Task 24 1/R T = 1/ R 1 + 1/ R 2 + 1/ R 3, which give us R T = 10. Because V = I R, then I = 24V/10Ω = 2.4 A Task 27 Answer: We can use the relation: V= IR, Then I = V/R R I /3 1 1/2 24/100 Task ns = 120 f/n = 120 x 50Hz/4 = 1500 rpm 2. s = (ns na/ ns) = ( )/1500 x 100 = 75/1500 x 100 = 5% slip [Note: x100 to get the %] 77

79 Task p e1 A 1 = p e2 A 2 => p e2 = pe1 A1 A2 2. p e = p e2 η = 384 bars 0.85 = bar 3. V = s A 2 = 5 cm 4.9 cm 2 = 24.5 cm 3 Task 33 Answer: n max = 2546 rotations / min Task 34 V R = Task mm 30 mm mm 220 mm 3.38 mm A = a x f = 5x 0.32 = 1.6 mm 2 = 384 bar Fc = A x kc = 1.6 mm 2 x 1990n/mm 2 = 3184 N h = f x sinx = 0.31 mm P = Fc x vc = cutting power = 3184N x 160 m/ 60 s 8491W = 8.49 kw Task 36 ΔL = L x α x Δt Δt = 25 C - 20 C = 5 C ΔL = 100 mm x / C x 5 C = mm = 8 µm Task 37 P 1 = 8.4 kw = 8400 W = 8400 N x m / s P 2 = η x P 1 = 6888 W = 6888 N x m / s P 2 = W / t = FG x h / t FG = P2 x t / h = N FG = m x g ; m = FG / g = 3511 kg Task 38 tan β = F F1 F 1 = F tan 30 = 800 N = 1386 N 78

80 sin β = F F2 F 2 = F sin 30 = 800 N 0.5 = 1600 N Task 39 Task 40 sin α = Fc F F x s = F G x h cos α = Ff F F c = F sin 80 = 3500 N = 3447 N F f = F cos 80 = 3500 N = 608 N F = F G x h / s = 4000 N x 1.2 m / 6 m = 800 N 79

81 Task 1 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided - Performing simple operations Level 1 ALL Task 2 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided - Performing simple operations Solving equations Costing a project Using graphs and diagrams Level 2 ALL Task 3 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided Area of rectangle: E = x y Performing simple operations Solving inequations Level 2 ALL Task 4 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided - Performing simple operations Level 1 ALL Task 5 categorisation Change and relationships Mathematical subfields Types of change Formulae/data provided - Performing simple operations Using graphs and diagrams Level 1 ALL Task 6 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided - Performing simple operations Using proportion Level 1 ALL Task 7 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided - Performing simple operations Using simple functions Level 1 80

82 ALL Task 8 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided - Performing simple operations Solving equations Level 1 ALL Task 9 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided - Using tables Level 1 ALL Task 10 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided Table given Performing simple operations Using tables Transforming formulas Level 2 ALL Tasks categorisation s Space and shape, including measurements Change and relationships Mathematical subfields 2D-3D shapes and properties Trigonometry Formulae/data provided Trigonometric table, v=s/t Transformation of formula Switching between units Calculating distance by using trigonometric functions Level 1 Machinist, CNC machinist Task 14 categorisation s Space and shape, including measurements Change and relationships Mathematical subfields 2D-3D shapes and properties Trigonometry Formulae/data provided Trigonometric table, v=s/t Transformation of formula Switching between units Calculating distance by using trigonometric functions Level 2 Machinist, CNC machinist Task 15 categorisation s Space and shape, including measurements Change and relationships Mathematical subfields 2D-3D shapes and properties Trigonometry Formulae/data provided Trigonometric table, v=s/t Transformation of formula Switching between units 81

83 Trial and error Level 3 Machinist, CNC machinist Task 16 categorisation Change and relationships Mathematical subfield Types of change 2F Formulae/data provided HB = where πd(d (D 2 d 2 ) D = Ball Diameter in mm d = impression diameter in mm F = Load in kgf HB = Brinell Hardness No. Performing simple operations Using formulas Level 2 Metallurgist, Designer, Test Engineer, Quality Assessor Tasks categorisation Change and relationships Mathematical subfield Types of change 2F Formulae/data provided HB = where πd(d (D 2 d 2 ) D = Ball Diameter in mm d = impression diameter in mm F = Load in kgf HB = Brinell Hardness No. Performing simple operations Using formulas Level 1 Metallurgist, Designer, Test Engineer, Quality Assessor Task 19 categorisation Change and relationships Mathematical subfield Types of change 2F Formulae/data provided HB = where πd(d (D 2 d 2 ) D = Ball Diameter in mm d = impression diameter in mm F = Load in kgf HB = Brinell Hardness No. Using formulas Level 2 Metallurgist, Designer, Test Engineer, Quality Assessor Task 20 categorisation Change and relationships Mathematical subfield Types of change 2F Formulae/data provided HB = where πd(d (D 2 d 2 ) D = Ball Diameter in mm d = impression diameter in mm F = Load in kgf HB = Brinell Hardness No. Using formulas Level 3 Metallurgist, Designer, Test Engineer, Quality Assessor Tasks categorisation Change and relationships Mathematical subfields Types of change Formulae/data provided Ohm s Law: Vs = I T x R T 82

84 1/R T = 1/ R 1 + 1/ R 2 Formula for resistors in parallel Performing simple operations Using formulas Level 1 Electronics Technician PCB Designer Tasks categorisation Change and relationships Mathematical subfields Types of change Different representations of relationships Formulae/data provided Ohm s Law: Vs = I T x R T 1/R T = 1/ R 1 + 1/ R 2 Formula for resistors in parallel Using formulas Identifying regularities Level 3 Electronics Technician PCB Designer Task 29 categorisation Change and relationships Mathematical subfields Types of change Different representations of relationships Formulae/data provided Ohm s Law: Vs = I T x R T 1/R T = 1/ R 1 + 1/ R 2 Formula for resistors in parallel Using formulas Identifying regularities Costing a project Level 3 Electronics Technician PCB Designer Task 30 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided ns = 120 f/n, where: ns = revolutions per minute (rpm) f = frequency [cycles per sec] in Hertz (Hz) n= number of motor poles Performing simple operations Transforming formulas Level 1 Electrician, Maintenance Task 31 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided ns = 120 f/n, where: ns = revolutions per minute (rpm) f = frequency [cycles per sec] in Hertz (Hz) n= number of motor poles T = Performing simple operations Transforming formulas Level 1 Electrician, Maintenance P 9550 speed 83

85 Task 32 categorisation Mathematical subfield Formulae/data provided Change and relationships Types of change Performing simple operations Using formulas Level 1 Mechatronic Task 33 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided v c = π d n Performing simple operations Using formulas Level 1 CNC Operator, Machining Technologist Task 34 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided D d V R = W 2 L Performing simple operations Using formulas Level 1 Mechanic Task 35 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided A = a x f Fc = A x kc h = f x sinx Q = A x vc = a x f x vc volume cutting efficiency P = Fc x vc = Q x kc cutting power table with indicators kc Performing simple operations Using formulas Reading tables Level 1 CNC Operator, Mechanic Task 36 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided ΔL = L x α x Δt ΔL - length increase Performing simple operations Using formulas Level 1 CNC Operator, Designer Task 37 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided Performing simple operations Using formulas Level 1 Mechanic, Operator 84

86 Task 38 categorisation Mathematical subfield Formulae/data provided Change and relationships Types of change Performing simple operations Using formulas Level 1 Mechanic, Locksmith Task 39 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided Performing simple operations Using formulas Level 1 Locksmith, Mechanic Saw Operator Task 40 categorisation Change and relationships Mathematical subfield Types of change Formulae/data provided Level 1 Mechanic Performing simple operations Using formulas 85

87 QUIZ 1. Is it true or false that if two quantities are directly proportional, one quantity is always the product of the other by the same number? a. True b. False 2. Consider the rectangles of an area equal to 24. If the lengths of the rectangles sides are x and y we can say that the lengths of the rectangles sides are inversely proportional. a. Yes b. No c. Not always d. The information given is not enough to give an answer 3. We know that y is inversely proportional to x and y = 8 when x = 3. What is the value of y when x = 4? a. 10 b. 8 c. 6 d. None of the above 4. The stopping distance of a car (in metres) is directly proportional to the square of its speed (in km/h) when the brakes are applied. A car travelling at 50 km/h requires a stopping distance of 20 meters. If the stopping distance is 51.2 meters, what is the speed of the car when the brakes are applied? a km/h b. 80 km/h c km/h d. None of the above 5. Which of the following is the value of log ? a. 2 b. 10 c. 512 d Answers: 1a, 2a, 3c, 4b, 5b 86

88 L8: Different representations of relationships Theory Relationships, and especially functions can be represented in many ways and this depends on the context. Linear functions Linear functions have graphs that are straight lines, their formula can be written in the form y = a + bx, and tables of (x, y) values in which the ratio of change in y to change in x is constant. How to graph a linear function: 1. Find 2 points which satisfy the equation 2. Plot them 3. Connect the points with a straight line Examples: 1. Draw a graph of the linear functiony = x + 2. In this case a = 1 and b = 2. We should find 2 points which satisfy the equation: Let x = 0, then y = 0 +2 = 2. Let x = 1, then y = 1+ 2 = 3. We found 2 points belonging to the line: (0,2) and (1,3). The straight line y = x + 2 is parallel to the straight line y = x since their coefficients a are equal. Coefficient b is responsible for the point of intersection of line y. 2. Draw a graph of the linear function y = 1 x 3. Because b = - 3 it means that one 2 point belonging to the graph is (0, -3). The second will be found by substituting for x e.g. 2: y = = 1 3 = 2. The second point is (2,-2). 2 87

89 Quadratic functions Quadratic functions have graphs that are parabolas, their formula can be written in the form y = ax 2 + bx + c, and tables of (x, y) values in which y values change in a symmetric pattern centred at a maximum or a minimum value. The graph is symmetric about a line called the axis of symmetry. The point where the axis of symmetry intersects the parabola is known as the vertex. How to graph a parabola: 1. Find the vertex. 2. Find the y-intercept (0, f(0)). 3. Solve f(x) = 0 to find the x coordinates of the x-intercepts if they exist. We can have 0, 1, or 2 x-intercepts. 4. Make sure that you have at least one point to either side of the vertex. This is to make sure you get a somewhat accurate sketch. If the parabola has two x-intercepts then you will already have these points. If it has 0 or 1 x-intercept we can either just plug in another x value or use the y-intercept and the axis of symmetry to get the second point. 5. Sketch the graph. Examples: 1. Draw a graph of the quadratic function f(x) = x 2 4x + 3. The coefficients are a = 1, b = -4, c = 3. In order to find a vertex (p,q) we have to calculate p = b = 2. Then q = f(2) = 1. Then we need to find y-intercept: x = 0, y = 3 Now we have to solve f(x) = 0 x 2 4x + 3 = 0 = b 2 4ac = = 4 = 2 x = b 2a = = 1 or x = b 2a = a = 4 2 = 3 88

90 2. Draw a graph of the quadratic function g(x) = 2x 2 2x + 4. The coefficients are a = -2, b = -2, c = 4. In order to find a vertex (p,q) we have to calculate p = b = 2 = 1. Then 2a 4 2 q = f ( 1 ) = Then we need to find y-intercept: x = 0, y = 4 Then we have to solve f(x) = 0 2x 2 2x + 4 = 0 = b 2 4ac = = 36 = 6 x = b 2a = b = 1 or x = = a 4 = 2 Note that the sign of a defines the parabola s orientation: if a>0 then the parabola opens up and if a<0 then the parabola opens down. Exponential functions Exponential functions have curved graphs showing the dependent variable increasing at an increasing rate (for exponential growth) and decreasing at a decreasing rate (for exponential decay) and their formula can be written in the form y = a x, where a is the constant growth or decay factor. In tables of (x, y) values for exponential functions, if successive x values differ by 1, then the ratio of corresponding y values is a. How to graph an exponential function: 89

91 1. Create a table of values e.g. for x = -2, -1, 0, 1, 2 2. The graph passes through the point (0, 1). 3. The graph has a horizontal asymptote: y = 0. Example: Draw a graph of the exponential function f(x) = 2 x. x f(x) 1/4 1/

92 Tasks Task 1 Task 2 a. Investigate how parallel connections of successive resistors of the same resistance (expressed in Ω) affect R T. Consider n = 5, n = 10. Formulate a conclusion. b. Investigate how parallel connections of successive resistors of the same resistance (expressed in Ω) affect I (current flowing). Formulate a conclusion. How does V depend from R, if I is constant? Create a table that demonstrates that dependence. Take, that I = 2.4 A. Task 3 Examine the relationship between I and R when V is constant and equal to 24V. Create a table that demonstrates that dependence. Take, that the smallest resistance R = 5 Ω. Draw a graph of this dependence. Formulate a conclusion. Task 4 You have to make the resistance R T being equal to 20 Ω, by combining parallel connections of other resistors. You can choose from the following types of resistors: R 1 = 100 Ω, R 2 = 25 Ω, R 3 = 50 Ω, R 4 = 35 Ω. Which will you choose? Has the task only one solution? You can freely choose the number of resistors. Task 5 You have to make the resistance R T being equal to 20 Ω, by combining parallel connections of other resistors. You can choose from the following types of resistors: R 1 = 100 Ω, R 2 = 25 Ω, R 3 = 50 Ω, R 4 = 35 Ω. Prices for resistors are the following: R 1 : 5 euros (50 pieces), R 2 : 6 euros (50 pieces), R 3 : 7 euros (50 pieces) and R 4 : 6.5 euros (50 pieces).which combination is the most cost-effective? Task 6 Most materials have an elastic stage where they spring back into shape when the force applied to them is released. A force produces a deformation x. In engineering this force is changed into stress (N/m² or N/mm²) and the deformation into strain. Hooke s Law states that deformation is directly proportional to force below the elastic limit (see graph). 91

93 A test engineer has performed a tensile test on a steel sample using a tensile testing machine and has plotted a graph of the resultant stress and strain. A tensile force produces the maximum stress that a material can withstand while being stretched or pulled before failing or breaking. Using Hooke Law the Modulus of Elasticity can be found. The gradient of the linear part of the graph gives F/x and this was found to be 410x10³ N/mm The test piece had a cross sectional area of 100mm² and a gauge length of 50mm Calculate Young s Modulus of Elasticity for the sample shown. 92

94 SOLUTIONS AND ANSWERS Task 3 We can use the relation: V= IR, Then I = V/R R I /3 1 1/2 24/100 Task 6 It was found from the test that F/x = 410 x 10³ N/mm² If Stress s = F/A then F = sa and Strain e = x/l then x=el Therefore, F/x = sa/el and E = FL/Ax = s/e Putting in the numbers: E = s/e = F/x * L/A = 410 x 10³ x 50/100 = 205,000 N/mm² This can be converted to Pascal Pa = 205,000 MPa or 205 GPa. 93

95 Tasks 1-4 categorisation Change and relationships Mathematical subfields Types of change Different representations of relationships Formulae/data provided Ohm s Law: Vs = I T x R T 1/R T = 1/ R 1 + 1/ R 2 Formula for resistors in parallel Using formulas Identifying regularities Level 3 Electronics Technician PCB Designer Task 5 categorisation Change and relationships Mathematical subfields Types of change Different representations of relationships Formulae/data provided Ohm s Law: Vs = I T x R T 1/R T = 1/ R 1 + 1/ R 2 Formula for resistors in parallel Using formulas Identifying regularities Costing a project Level 3 Electronics Technician PCB Designer Task 6 categorisation Change and relationships Mathematical subfield Different representations of relationships Formulae/data provided Hooke s Law = F/x Stress s=f/a Strain e= x/l Young s Modulus of Elasticity: E = FL/Ax = s/e 1N/mm² = to 1,000,000 N/m², 1N/m² = 1Pascal Pa Performing simple operations Using graphs and diagrams Transforming formulas Level 1 Test Engineer, Metallurgist Designer 94

96 QUIZ 1. The lines which represent the functions f(x) = x+2 and g(x) = 2x+2 are parallel. a. True b. False 2. What is the minimum number of points that you need in order to draw the graph of a linear function? a. 1 b. 2 c. 3 d. More than 3 3. Which of the following is true about the graphs of the functions f(x) = x 2-2x+4 and g(x) = x 2 +4? a. They are both parabolas b. Only the graph of f(x) is a parabola c. Only the graph of g(x) is a parabola d. None of the above is true 4. What kind of function does the graph shown represent? a. Linear b. Quadratic c. Exponential d. Other 5. What is the maximum number of intersection points that the graphs of a linear and a quadratic function can have? a. None b. One c. Two d. Infinite Answers: 1b, 2b, 3a, 4c, 5c 95

97 L9: Approximations and estimations Theory Estimation is the process of finding an estimate, or approximation, which is a value that is usable for some purpose even if the input data may be incomplete, uncertain, or unstable. Examples: The number 43 to the nearest 10 is rounded to 40. The number 173 to the nearest 100 is 200. While approximating a number to the n-th decimal place we have to look at the digit at the n+1-th place. If the digit is a five or greater, you round up the target digit by one. Otherwise, you leave the target as it is. Then you delete the remaining digits. Absolute and relative error of approximation Given some value v and its approximation v approx, the absolute error is ε = v v approx. If v 0 the relative error is η = ε = v v v approx v = 1 v approx. The percent error is δ = 100 η = 100 ε = 100 v v v approx. Example: Give an approximation of the number to the second (to the hundredths place) and then to the third decimal place (to the nearest thousandth). While approximating of the number to the second decimal place we have to look at the digit at the third place. The digit at the second place is 2 and at the third place is 1. The general rule that you round to a given place by looking at the digit one place to the right of the "target" place. If the digit is a five or greater, you round the target digit up by one. Otherwise, you leave the target as it is. Then you delete the rest of digits. In our example in the third decimal number there is 1 which means that the number after rounding to the second decimal place is We can thus write The number after rounding to the third decimal place is (since at the fourth place there is 7). We can write Example: If the exact value is 50 and the approximation is 49.9, then the absolute error is 0.1 and the relative error is 0.1/50 = = 0.2%. v v 96

98 Tasks Task 1 Write a number that is 50, when rounded to the nearest 10. Task 2 The number 490 when rounded to the nearest 1000 is: a. 0 b. 500 c Task 3 Round pi (π) to four places. Task 4 Give an example of the smallest natural number which is greater than 950. Task 5 In the table below the area of the six biggest countries is given (in millions km 2 ) Russia Canada China USA Brasil Australia The area of the land of Earth is equal to mln km 2. Is the area of the six countries bigger than half of the area of the land? Estimate without calculations. Task 6 Which number is greater: 0.13 or 0.1(30)? Task 7 The length of an iron pipe is m. Approximate the length to first decimal place. Calculate approximation error to 0.1%. Task 8 Calculate the absolute value of the approximation:

99 SOLUTIONS AND ANSWERS Task 1 Answer: 45, 46,, 54 Task 2 Answer: a Task 3 π = 3, We count off four places, and look at the number in the fifth place: 9; it is greater than 5, so we are rounding up in the fourth place, truncating the expansion at four decimal places. That is, the 5 becomes a 6, the part disappears, and π, rounded to four decimal places is: Task = = 961 So 30 < 950 < 31. The smallest natural number greater than 950 is 31. Task 5 It is less than half < =65 < /2 Task 6 0.1(30) = > Task The absolute error is = The relative error is = = The percent error is % = % = % 0.2%

100 Task = = = =

101 Tasks 1-3 categorisation Change and relationships Mathematical subfield Approximations and estimations Formulae/data provided - Estimating quantities Level 1 ALL Tasks 4-5 categorisation Change and relationships Mathematical subfield Approximations and estimations Formulae/data provided - Performing simple operations Estimating quantities Level 2 ALL Task 6 categorisation Change and relationships Mathematical subfield Approximations and estimations Formulae/data provided - Understanding basic numerical facts Level 1 ALL Tasks 7-8 categorisation Change and relationships Mathematical subfield Approximations and estimations Formulae/data provided - Performing simple operations Estimating quantities Using formulas Level 2 ALL 100

102 L10: Effective use of calculators Theory The calculator in any way it might be used, does not do the strategic thinking for the user. It is important before any step in a calculation, to decide what operation is appropriate. Equally important is to ask whether the result makes sense and how to interpret the display. Important questions: 1) What numbers do I need to enter? 2) In what order do I key in the numbers? 3) Do I have to add, subtract, multiply or divide these numbers? 4) Does the result make sense? 5) How do I interpret the display? Example: How many 12-seated minibuses are needed for a school consisting of 70 children and 7 teachers? In the sum we have 77 persons for transport. By using a calculator we receive: 77 : 12 = 6.4. We cannot ignore the context of the task. The answer is that 7 minibuses are needed. Calculating percentages On some calculators, the percentage of a value will be displayed after pushing the [%] button while on others you will have to push the [%] button and then the [=] button. Example: Increase the number 120 by 10%. 10% of 120 is 12. The result then is 132. a) Calculator A: Press the following keys: %. The calculator displays the result 132. In some calculators there is a need to press =. b) Calculator B: Press the following keys: %. After pressing = 12 appears. It means that it calculated 10% of 120. In this type of calculators, after pressing % the answer should appear. This means that it calculated the number which decreased by 10% gave us 120. Using calculator s memory Calculators usually have three or four memory keys. M+ - adding a number from the screen to the memory M- - subtracting from a number in the memory the number from the screen. MR - showing at the screen the number saved in the memory MC deleting the memory MRC plays the roles of MR and MC. 101

103 Example: Calculate without paper and pencil: 70/(16-9). Press the following keys: 16-9 = M+ CE 70 / MR The calculator displays the result 10. Calculate: Press the following keys: 4 M- 8 M+ 2 M- 3 M- 5 M+ 10 M- MR. The calculator displays the result -6. Approximations by the calculator Example: Calculate the value of the expression: 10 3 ( ) 2. We can do in two different ways: ( ) ( ) 2 = 1000( ) 2 = 1000(0.2) 2 = = ( ) 2 = 1000 ((2 5) (3 2) 2 ) = 1000( ) 1000( ) = 1000( ) = = 1040 Which result is the correct one? In both cases we made the same approximation of the square roots but only one decimal place and this resulted in a large difference in the final result. The calculator keeps in the memory all digits of the numbers and it operates with the whole number. By using the calculator we receive the final result: no matter in what moment we approximate the square roots. 102

104 Tasks Task 1 How will the result of the multiplication be demonstrated in a calculator? Task 2 Give an example of the smallest natural number which is greater than 950. Task 3 The length of an iron pipe is m. Approximate the length to first decimal place. Calculate approximation error to 0.1%. 103

105 SOLUTIONS AND ANSWERS Task 1 Answer: 5e+32 Task = = 961 So 30 < 950 < 31. The smallest natural number greater than 950 is 31. Task 3 14,764 14,8 The absolute error is = The relative error is = = The percent error is % = % = % 0.2%

106 Task 1 categorisation s Quantity Change and relationships Mathematical subfields Relative size of numbers Effective use of calculators Formulae/data provided - Understanding basic numerical facts Level 1 ALL Task 2 categorisation Change and relationships Mathematical subfield Approximations and estimations Effective use of calculators Formulae/data provided - Performing simple operations Estimating quantities Level 2 ALL Tasks 3 categorisation Change and relationships Mathematical subfield Approximations and estimations Effective use of calculators Formulae/data provided - Performing simple operations Estimating quantities Using formulas Level 2 ALL 105

107 QUIZ 1. When I calculated the square root of 45 the following display was shown. Which of the following sentences is wrong? a. The square root of 45 can be rounded to 7. b. The square root of 45 can be rounded to 6.7 c. The square root of 45 can be rounded to 6.71 d. The square root of 45 can be rounded to 6 2. In order to find the remainder of the division 357: 13 I firstly made the division in the calculator and the following display was shown. Which of the following sentences is wrong? a. The remainder of the particular division has to be a number which is smaller than 13 b. The remainder of the particular division has to be a number which is greater than zero, but smaller than 13 c. The display does not give any information on the remainder of the particular division 3. By the use of a calculator make the operation :21.9. Which of the following is the best estimation of the result in three decimal points? 106

108 a b c d. None of the above 4. If I want to add the results of the multiplications and which of the following is the correct series of buttons to press in a scientific calculator? a = b ( ) = c. ( ) + ( ) = d. ( ) = 5. If I want to find the result of the operation series of buttons to press in a scientific calculator? a = b. ( ) 2 = c. ( ) 2 = d. ( ) = which of the following is the correct Answers: 1d, 2c, 3b, 4c, 5a 107

109 Part 3: Space and shape 108

110 L11: Units of measurement and scales Theory Units of measurement and scale Length is expressed in units of length. Short lengths are measured in centimeters. If one centimeter is split into 10 equal parts, each of these parts represents one millimeter. Therefore, one centimeter = 10 millimeters. Examples: This rectangle is 9 cm long. 1 cm = 10 mm 1 cm 1 cm 1 cm 1 cm 1 cm 1 cm 1 cm 1 cm 1 cm Greater lengths are expressed in other units. The basic unit of length in the SI system is 1 meter. 1 meter = 100 centimeters If we assume that 1 meter is the basic unit of length, we can enter the names of smaller units respectively, using the power series with a negative exponent. 1 = = 100 one m meter 0.1= ,1 = 10 1 one tenth dm decimeter 0.01 = ,01 = 10 2 one hundredth cm centimeter = ,001 = 10 3 one-thousandth mm millimeter =10-6 0, = 10 6 one millionth µm micrometer = , = 10 9 one billionth nf nanofarad = , = one trillionth pf picofard Examples: 1m = 100 cm = 1000 mm = µm = nf = pf 1 mm = 1000 µm 1 cm = 10 mm = µm = µm 1 dm = 10 cm = µm = µm Note: When replacing the unit of lengths pay attention to the number of zeros! In systems other than the SI, there are other units of length. Instead of one cm, a frequently used unit is one inch. 1 in = 2.54 cm 1 cm = in 109

111 A length expressed in inches can be expressed in centimeters by the conversion: x in = x 2.54 cm. See the table below. in cm Conversion to common lengths For converting inches into other units, we use the table below unit of length 1 inch Inches 1 Centimeters 2.54 Meters Kilometers Feet Yards Miles Nautical Miles Scales When you need to enlarge or reduce the actual dimensions of a shape you have to use a scale. The scale tells us the relations between lengths on the drawing and the real distances. It is used on plans or maps. Example: A A - The actual figure Length: 4 cm Width: 2 cm B C C Figure decreased twice Length: 4 cm ½ = 2 cm Width: 2 cm ½ = 1 cm B- Figure increased twice Length: 4 cm 2 = 8 cm Width: 2 cm 2 = 4 cm 110

112 For figure A each 1 cm on the picture represents 1 cm in reality. We will say that figure A is presented in a scale 1 to 1 and mark it as 1 : 1 For figure B each 2 cm on the picture represents 1 cm in reality. We will say that figure B is presented in a scale 2 to 1 and mark it as 2 : 1 For figure C each 1 cm on the picture represents 1 cm in reality. We will say that figure C is presented in a scale 1 to 2, and mark it as 2 : 1 Example: The sketch shows the plan of a flat. 1 cm of the plan represents 2 m of the actual flat. The actual dimensions of the flat are: Length of kitchen: 1.5 cm 200 cm = 3.0 m. Width of kitchen: 1.1 cm 200 cm = 2.2 m Length of bedroom: 1 cm 200 cm = 2.0 m Width of bedroom: 1 cm 200 cm = 2.0 m Length of room 1: 3 cm 200 cm = 8 m Width of room 1: 1.9 cm 200 cm = 3.8 m Length of room 2: 2.9 cm 200 cm = 5.8 m Width of room 2: 1.9 cm 200 cm = 3.8 m The scale is 1cm to 2 m, or 1 cm to 200 cm, so we will write 1 :

113 Tasks Task 1 What is the length of the segment drawn above the ruler? Task 2 Express in millimeters a) 55 µm = b) 9 µm = c) 1036 µm = Hint: 1 µm = mm Task 3 Calculate how much it is: 28 in=. cm 28 cm =. in Task 4 Using the table below, make the conversions: Task 5 a) 1 feet = cm b) 2 yards = m Each centimeter at the picture represents 1 kilometer in reality. The distance between the school and the post office is 4 km. The same distance on the picture is 4 cm. Find other objects whose distance is about 4 km. 112

114 Task 6 Complete the table for the plan of a flat, drawn in the scale 1 : 200 Room Length on plan Actual length Width on plan Actual width Bedroom 3 cm cm = 6 m 2 cm cm = 4 m Kitchen 2.5 cm 1.5 cm cm = 3 m Living room 6 m 4 cm Bathroom 1 cm 2 cm Task 7 A Mechanical/Mechatronics Technician received the dimensions given in microns. On a technical drawing according to the norm (ISO) the dimensions have to be given in millimetres. Thus the units have to be changed. Write in millimetres: a) 12 µm =, b) 3.5 µm =, c) 3455 µm =,, d) 259 µm =,, e) 5.5 µm =,, Hint: 1 µm = mm 113

115 Task 8 A Mechanical/Mechatronics Technician received a technical drawing of an iron cast table top. In order to program a machine he has to change the units into microns. Change millimetres into µm? Task 9 If a machine cuts iron cast table tops from an iron surface of dimensions of µm x µm x µm, then how many table tops (of dimensions 1840 mm x 92 mm x 10 mm) will it cut? Task 10 If a machine cuts iron cast table tops from an iron surface of dimensions of µm x µm x µm, then how many table tops (of dimensions 1840 mm x 92 mm x 10 mm) will it cut? Task 11 a) If a machine cuts iron cast table tops from an iron surface of dimensions of µm x µm x µm, then how many table tops (of dimensions 1840 mm x 92 mm x 10 mm) will it cut? b) Is 96 the maximum number of table tops that can be cut from the iron cast surface? Justify your answer. Task 12 a) If a machine cuts iron cast table tops from an iron surface of dimensions of µm x µm x µm, then how many table tops (of dimensions 1840 mm x 92 mm x 10 mm) will it cut? b) Mr Smith ordered 180 iron cast table tops. How much will he pay if 1m 2 of iron (of the 1cm gauge) costs 4 euros? The cost of production of one table top is 2 euros. The company is selling the table tops at 15% profit. Task 13 Calculate the weight of a cast iron pipe given that its length is 3.5 m, its external diameter is 80 mm and the thickness of the wall is 15 mm (ρ = 7.2 g/cm 3 ). Data: D = 80 mm; 114

116 s = 15 mm; l = 350 cm; ρ = 7.2 g/cm 3 Task 14 What is the maximum length of a cast iron pipe in order to be lifted by manual hoist with a permissible load capacity of 500 kg? The external diameter of the pipe is 80 mm and the thickness of the wall is 15 mm (ρ = 7.2 g/cm 3 ). Task 15 Calculate the weight of a cast iron pipe given that its length is 3.5 m, its external diameter is 80 mm and the thickness of the wall is 20 mm (ρ = 7.2 g/cm 3 ). Task 16 Cast iron pipes have to be transported from one working station to another. The time for transportation of one pipe is 5 min 20 sec. How many pipes can be transported within 2 hours if mounting and unmounting the pipe at the machine takes 15 sec? Task 17 Cast iron pipes have to be transported from one working station to another. The time for transportation of one pipe is 5 min 20 sec. How much time it will take to transport 40 pipes, if mounting and unmounting the pipe at the machine takes 15 sec? Task 18 A company wants to order cast iron pipes of a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 15 mm (ρ = 7.2 g/cm 3 ). The total length of the pipes has to be 9 km. The price of the particular iron pipes is 7.00 euros per kg. How much is the company expected to pay? 115

117 Task 19 A company wants to order cast iron pipes of two kinds: Type 1, with a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 15 mm and Type 2, with a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 20 mm (ρ = 7.2 g/cm 3 ). The company knows that the price of a Type 1 iron pipe is euros per piece. Is euros a fair price for a Type 2 pipe? Task 20 A company wants to order cast iron pipes of a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 15 mm (ρ = 7.2 g/cm 3 ). The total length of the pipes has to be 9 km. They received two offers: Offer 1: 7.00 euros per kg of the pipe Offer 2: euros per pipe. Which is the best offer? Justify your opinion. Task 21 In a designed element of a machine with the shape of a right triangle with a hypotenuse of c = 45 mm and a vertical side of a = 27 mm, a Mechanical Technician has to calculate the second vertical side and the angles of designed part. Calculate them. Task 22 In a designed element of a machine with the shape of a right triangle with a hypotenuse of 45 mm and a vertical side of 27 mm, a Mechanical Technician has to drill 3 holes in all three vertexes of the triangle at the distance of 3mm from both edges. What are the distances between the vertexes and the holes? Task 23 In a designed element of a machine with the shape of a right triangle with a hypotenuse of 45 mm and a vertical side of 27 mm, a Mechanical Technician has to drill a hole in its centroid. Calculate the distances of the hole from all three sides of the triangle knowing that the hole has a diameter of 2 mm. 116

118 SOLUTIONS AND ANSWERS Task 1 Answer: 7 cm Task 2 Answer: Task 3 a) 55 µm = mm b) 9 µm = mm c) 1036 µm = mm 28 in = cm = cm 28 cm = in = cm Task 4 a) 1 in = feet 1 foot = 1/ in = in = cm = cm b) 1 in = yards 1 yard = 1/ in = in = m = m 2 yards = m 2 = m Task 5 Answer: Cinema Kindergarten, Cinema Art Gallery, Patisserie Toyshop Task 6 Answer: room Length on plan Actual length Width on plan Actual width Bedroom 3 cm cm = 6 m 2 cm cm = 4 m Kitchen 2.5 cm cm = 5 m 1.5 cm cm = 3 m Living room 600 cm : 200 = 3 cm 6 m 4 cm cm = 8 m Bathroom 1 cm cm = 2 m 2 cm cm = 4 m Task 7 Answer: a) 12 µm = mm b) 3.5 µm = mm 117

119 c) 3455 µm = mm d) 259 µm = mm e) 5.5 µm = mm Task mm = µm 92 mm = µm 10 mm = µm Task µm = 9200 mm = 9.20 m µm = 1250 mm = 1.25 m µm = 10 mm 1840 mm =1.84 m 92 mm = m 9.20 : 1.84 = : x 13 = 65 table tops The surface left is too small to cut more table tops. Task 10 Surface dimensions: µm = 8850 mm = 8.85 m µm = 1930 mm = 1.93 m µm = 10 mm Table top: 1840 mm =1.84 m 92 mm = m 8.85 : :

120 4 x 20 = 80 table tops The surface area left which can be used for cutting table tops: 1.49 m x 1.93 m 1.49 : extra table tops = 96 in total Task 11 Answer: a) see task 10. Task 12 a) 65 table tops (see task 9). b) 180/65 = Mr Smith has to pay for 3 iron cast surfaces Total area of the surfaces: 11.5 m 2. Task x 3 = 34.5 m 2 area of 3 surfaces 34.5 x 4 = 138 euros cost of material 2 x 180 = 360 euros cost of production = 498 euros x 498 = euros V D - Volume of cylinder of diameter D V d Volume of cylinder of diameter d d = D 2 s = = 50 mm V D = π = 5600π cm 3 V d = π = π cm 3 V p Volume of pipe V p = 5600π π = π cm 3 w weight of pipe w = ρ V 119

121 w = = g = kg Task kg = g : 7.2 = (4) maximum volume V p = π 4 2 l π l = 9.75 π l 9.75 π l = π l l max cm = m Task 15 V D - Volume of cylinder of diameter D V d Volume of cylinder of diameter d d = D 2 s = = 40 mm V D = π = 5600π cm 3 V d = π = 1400 π cm 3 V p Volume of pipe V p = 5600π 1400π = 4200π cm 3 w weight of pipe w = ρ V w = = g = kg Task 16 Total time needed for one pipe is 15 sec + 5 min 20 sec + 15 sec = 5 min 50 sec = 350 sec. 2 hours = 120 min = 7200 sec : Within 2 hours it is possible to transport 20 pipes. Task 17 Total time needed for one pipe is 15 sec + 5 min 20 sec + 15 sec = 5 min 50 sec = 350 sec. 120

122 = sec = 233.(3) min = 3 h 53 min 20 sec. Task 18 The company has to order 2572 pipes (9000:3.5 = ). The weight of one pipe is kg (task 13). The total price of the order is euros. Task 21 a = 27 mm c = 45 mm b =?, α =?, β =? 27 mm = 2.7 cm 45 mm = 4.5 cm a 2 + b 2 = c 2 (2.7) 2 + b 2 = (4.5) 2 \ (2.7) 2 b 2 = b = = 3.6 sin α = a c sin α = 2.7 = 0.6 then α = Then β= 180 ( ) = 54 Task 22 a = 27 mm c = 45 mm 27 mm = 2.7 cm 45 mm = 4.5 cm sin α = a c sin α = 2.7 = 0.6 then α = Then β= 180 ( ) =

123 A a C β b c α B A β 2 3 β 2 3 x the distance from the vertex A sin β 2 = 3 x x = 3 3 = 6.61 mm sin y the distance from the vertex B sin α 2 = 3 y y = 3 3 = 9.71 mm sin z the distance from the vertex C z = mm 122

124 Task 1 categorisation Space and shape Mathematical subfield Units of measurement and scales Formulae/data provided Performing simple measurements Level 1 ALL Task 2 categorisation Space and shape Mathematical subfield Units of measurement and scales Formulae/data provided Performing simple operations Level 1 ALL Task 3 categorisation Space and shape Mathematical subfield Units of measurement and scales Formulae/data provided 1 inch = 2.54 cm Performing simple operations Level 1 ALL Tasks 4 categorisation Space and shape Mathematical subfield Units of measurement and scales Formulae/data provided Performing simple operations Level 1 ALL Task 5 categorisation Space and shape Mathematical subfield Units of measurement and scales Formulae/data provided Performing simple measurements Estimating distances Level 2 ALL Tasks 6 categorisation Space and shape Mathematical subfield Units of measurement and scales Formulae/data provided Performing simple operations Level 1 ALL Tasks 7-8 categorisation Space and shape Mathematical subfield Units of measurement and scales Formulae/data provided 1 µm = mm Performing simple operations Level 1 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Task 9 categorisation Space and shape Mathematical subfield Units of measurement and scales 123

125 Spatial visualisation Formulae/data provided 1 µm = mm Performing simple operations Estimating areas Level 1 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Task 10 categorisation Space and shape Mathematical subfield Units of measurement and scales Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided 1 µm = mm Performing simple operations Estimating areas Level 2 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Task 11 categorisation Space and shape Mathematical subfield Units of measurement and scales Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided 1 µm = mm Performing simple operations Estimating areas Level 3 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Task 12 categorisation Space and shape Mathematical subfield Units of measurement and scales Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided 1 µm = mm Performing simple operations Estimating areas Level 2 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Tasks categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Spatial visualisation Formulae/data provided V d Volume of cylinder of diameter d: V d = π ( d 2 )2 l Weight in relation to density ρ and volume V: w = ρ V Performing simple operations Calculating measures of weight and volume Level 1 CNC Operator Tasks categorisation Space and shape Mathematical subfield Units of measurement and scales 124

126 Formulae/data provided 2D-3D shapes and properties Application of numerical skills to solve geometrical problems V d Volume of cylinder of diameter d: V d = π ( d 2 )2 l Weight in relation to density ρ and volume V: w = ρ V Performing simple operations Calculating measures of weight, volume and time Comparing quantities Costing a project Level 2 Sales manager Task 21 categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Trigonometry and geometry Formulae/data provided Pythagorean Theorem Trigonometric functions (definitions and table) Performing simple operations Use of formulas Calculating lengths and angles Level 1 Mechanical Technician/Engineer, CNC Operator Task 22 categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Trigonometry and geometry Formulae/data provided Pythagorean Theorem, Trigonometric functions (definitions and table) Performing simple operations Use of formulas Calculating lengths and angles Level 2 Mechanical Technician/Engineer, CNC Operator Task 23 categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Trigonometry and geometry Formulae/data provided Pythagorean Theorem, Trigonometric functions (definitions and table), Law of cosines, The centroid divides each median into parts in the ratio 2:1. Performing simple operations Use of formulas Calculating lengths and angles Level 2 Mechanical Technician/Engineer, CNC Operator 125

127 L12: 2D-3D shapes and properties Theory Types of geometric figures We can classify geometric figures according to their different properties. One of the properties is their dimensionality. We have figures: Zero - dimensional: points: A B C One-dimensional: straight lines and segments. Straight lines are limitless; segments are limited parts of straight lines (you can measure their length). Two-dimensional shapes: polygons, circles, curves, Three-dimensional shapes: solids Two-dimensional shapes Polygons have straight sides. They can have three, four, five,... sides. Triangles Quadrangles Pentagons Hexagons Heptagons 126

128 Triangles Triangles have 3 sides and 3 angles. a b c If all sides are of equal length, the triangle is called the equilateral triangle - figure a If the two sides are of equal length, the triangle is called the isosceles triangle - figure b If each side has a different length the triangle is called the scalene triangle - figure c Triangles may have different angles. If 0 0 < α < 90 0, 0 0 < β < 90 0, 0 0 < γ < 90 0, the triangle is called acute-angled triangle (fig.a) If γ = 90 0 the triangle is called a rectangular (fig.b) If 90 0 < γ < 180 0, the triangle is called obtuse triangle (fig.c) The sum of the measures of the interior angles of any triangle is Each triangle has three heights. These are segments perpendicular to the base, wherein one end of it is the vertex of the triangle opposite to the base. Heights (or their extensions) intersect at one common point. Quadrangles Among polygons having 4 sides there are: Square: - 4 equal sides - 4 right angles - 2 pairs of parallel sides Rhombus: - 4 equal sides - 2 pairs of equal angles - 2 pairs of parallel sides Rectangle: -2 equal sides -2 other equal sides - 4 right angles 2 pairs of parallel sides Trapezium: -4 sides -only 1 pair of parallel sides Parallelogram: -2 equal sides -2 other equal sides -2 pair of equal angles 2 pairs of parallel sides Deltoid: - 2 equal sides - 2 other equal sides - 2 equal angles 127

129 Example: Fill in the table with the properties of typical polygons Shape Equal sides One pair of parallel sides Right angle Square Rectangle An angle larger than a right angle An angle smaller than a right angle Parallelogram? Rhombus? Trapezium?? Regular hexagon We can build a trapezium with two equal sides: We can also build a trapezium with a right angle: A parallelogram can have a right angle; in that case all its angles are right and it is a rectangle. A rhombus can have a right a right angle; in that case all its angles are right and it is a square. Line symmetry Some of polygons are symmetrical, since they possess a line of symmetry. As if the figure was cut with a sheet folded in half. Folding line Relations of shape to shape Line of symmetry of the figure Shapes can be arranged in various ways. The following are some of the geometrical transformations: The parallel shift (translation): all elements of a figure are moved in the same way as the vector 128

130 Axial symmetry: both figures are on the opposite sides of the axis of symmetry, and the segments linking the corresponding points are perpendicular to the axis. The axis bisects these segments into half. Rotation on Point O is stationary. The points of the figure move on a semicircle. Such arrangements can be found in a number of mosaics, such as follows. Perimeter The perimeter of a shape is the distance all the way around it. Example: In the triangle the lengths of the sides are 4.5 cm, 4.5cm and 4cm, so its perimeter is equal to 4.5 cm cm + 4 cm = 13 cm. The perimeter of parallelogram is equal to 5 cm cm + 5 cm+ 5.5 cm = 21 cm. Example: The length of rectangle is 20 cm, the width is 6.2 cm. How much is the perimeter? Answer: The perimeter is 20 cm cm + 20 cm cm = 52.4 cm. The same result can be reached by the following calculations: 2 20 cm cm = 40 cm cm = 52.4 cm. 129

131 Or in the following manner: (20 cm cm) 2 = 26.2 cm 2 = 52.4 cm. For specific shapes we can use formulas: O = a + b + c O = a + 2b O = 3a O = 2a + 2b O = 4a O = 2a + 2b O = 4a O = a + 2b + c O = 2a + 2b Perimeter of circles The perimeter of a circle is called its circumference. It is a bit more than 3 times its diameter (see the green line at the sketch). By making the exact calculations we can have that circumference: diameter = = π. Therefore, circumference = diameter π = π 2r = 2 πr Area The area is the amount of surface covered by a flat shape. Typically we use square units. When the side of square is 1 cm, the unit is square centimeters (cm 2 ). This is 1 cm 2 Example: The area of each one of these shapes is 4 cm 2 Sometimes it is necessary to cut the unit, to be able to calculate the area of more complicated shapes. 130

132 Area of typical shapes Area of each square is depended on the length of its sides. Here the area is 4 4 cm 2 = 16 cm 2. When the length is a, the area of square is a a = a 2 Area of rectangle is a b Area of triangle. Look at the sketches: h h h a a a P = ½ a h P = ½ a h P = ½ a h Area of triangle is ½ a h Area of parallelogram. Look at the sketches: Area of parallelogram is: S = a h 1 or S = b h 2 Area of trapezium. Look at the sketches: b a S + S = 2S = (a + b) h S = ½ (a + b) h = Area of trapezium is: S = ½ (a + b) h 131

133 Three- dimensional shapes (3-D) We describe 3D shapes by saying how many faces, edges and vertices they have. For the cube: face edge vertex 6 faces (squares) 12 edges (segments the same lengths) 8 vertices (points) Other 3-D shapes are: Square-based pyramid Triangular prism Cylinder Cuboid Volume Volume is the space taken up by a cubic unit, having 1 cm edges. This is one cubic unit: = 1 cm 3 Each of these 3-D shapes has a 4 cm 3 volume Volume of cuboid is = length width height Nets For making a solid (3-D shape) from paper, we draw its net on the plane. The net is the system of its faces, which can be folded into the solid. Here are some solids and their nets: 132

134 Tasks Task 1 Find the angle γ, if the triangle is isosceles and one base angle is Task 2 Is a square a rhombus? Task 3 How many lines of symmetry does a square have? Task 4 Draw the missing elements of the floor shown. What types of transformations can you identify? Task 5 How to work out the total perimeter for the complex shape shown in the image? Task 6 What are the areas of shapes 1,2,3 and 4? 133

135 Task 7 Describe a triangular prism. Task 8 Find the volume of each cuboid. Complete the table. length width height volume 6m 3m 5m 15 cm 4 cm 46 mm 2.4 cm 58 mm 6 cm Task 9 Is this a net of cube? Task 10 Calculate the weight of a cast iron pipe given that its length is 3.5 m, its external diameter is 80 mm and the thickness of the wall is 15 mm (ρ = 7.2 g/cm 3 ). Data: D = 80 mm; s = 15 mm; l = 350 cm; ρ = 7.2 g/cm 3 Task 11 What is the maximum length of a cast iron pipe in order to be lifted by manual hoist with a permissible load capacity of 500 kg? The external diameter of the pipe is 80 mm and the thickness of the wall is 15 mm (ρ = 7.2 g/cm 3 ). 134

136 Task 12 Calculate the weight of a cast iron pipe given that its length is 3.5 m, its external diameter is 80 mm and the thickness of the wall is 20 mm (ρ = 7.2 g/cm 3 ). Task 13 Cast iron pipes have to be transported from one working station to another. The time for transportation of one pipe is 5 min 20 sec. How many pipes can be transported within 2 hours if mounting and unmounting the pipe at the machine takes 15 sec? Task 14 Cast iron pipes have to be transported from one working station to another. The time for transportation of one pipe is 5 min 20 sec. How much time it will take to transport 40 pipes, if mounting and unmounting the pipe at the machine takes 15 sec? Task 15 A company wants to order cast iron pipes of a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 15 mm (ρ = 7.2 g/cm 3 ). The total length of the pipes has to be 9 km. The price of the particular iron pipes is 7.00 euros per kg. How much is the company expected to pay? Task 16 A company wants to order cast iron pipes of two kinds: Type 1, with a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 15 mm and Type 2, with a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 20 mm (ρ = 7.2 g/cm 3 ). The company knows that the price of a Type 1 iron pipe is euros per piece. Is euros a fair price for a Type 2 pipe? Task 17 A company wants to order cast iron pipes of a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 15 mm (ρ = 7.2 g/cm 3 ). The total length of the pipes has to be 9 km. They received two offers: Offer 1: 7.00 euros per kg of the pipe Offer 2: euros per pipe. Which is the best offer? Justify your opinion. Task 18 In a designed element of a machine with the shape of a right triangle with a hypotenuse of c = 45 mm and a vertical side of a = 27 mm, a Mechanical Technician has to calculate the second vertical side and the angles of designed part. Calculate them. 135

137 Task 19 In a designed element of a machine with the shape of a right triangle with a hypotenuse of 45 mm and a vertical side of 27 mm, a Mechanical Technician has to drill 3 holes in all three vertexes of the triangle at the distance of 3mm from both edges. What are the distances between the vertexes and the holes? Task 20 In a designed element of a machine with the shape of a right triangle with a hypotenuse of 45 mm and a vertical side of 27 mm, a Mechanical Technician has to drill a hole in its centroid. Calculate the distances of the hole from all three sides of the triangle knowing that the hole has a diameter of 2 mm. Task 21 The image shows a block of material 60mm wide x 80mm long x 20mm high. A pillar drill is used to drill a 10mm diameter hole through the block. Calculate the volume of material remaining. Task 22 The image shows a block of material 60mm wide x 80mm long x 20mm high. A pillar drill is used to drill a hole through the block. If the volume of the remaining material is 94,429mm³, what was the diameter of the drilled hole? Task 23 The image shows a block of material 60mm wide x 80mm long x 20mm high. A pillar drill is used to drill a 10mm diameter hole through the block. The process is repeated for a number of material blocks, so that the total volume of the remaining material is approximately 25505mm³. How many blocks were drilled? 136

138 Task 24 The image shows a block of material 60mm wide x 80mm long x 20mm high. A pillar drill is used to drill a hole through the block. What is the maximum diameter of the drilled hole so that the volume of the remaining material will be around 94.5mm³? Task 25 The image shows a block of material 60mm wide x 80mm long x 20mm high. A pillar drill is used to drill a 10mm diameter hole through the block. The process is repeated for a number of material blocks and the material taken from the holes is placed in a container, which has a volume of 10m 3. The drilling process is stopped automatically when the container reaches 90% of its capacity. What is the maximum number of blocks that can be drilled before the process is stopped? Task 26 In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 30 o. The drill should reach hole 2 at the latest after 0.8 s. What is the minimum average movement speed in m/min? Task 27 In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 45 o. The drill should reach hole 2 at the latest after 0.8 s. What is the minimum average movement speed in m/min? 137

139 Task 28 In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 30 o. The drill reaches hole 2 in 0.8 s and the drilling of a hole lasts 0.5 s. The process is repeated for a number of holes. How many holes can be bored within 2 min? What is the distance between first and the last hole? Task 29 In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 30 o. The drill reaches hole 2 in 0.8 s. and the drilling of a hole lasts 0.5 s. The process is repeated for a number of holes for 2 min. Given that the distance of the centre of hole 1 is 2 cm from both edges of the block, calculate the minimum dimensions of the block in order for the drilling process to be completed. Task 30 A numerically controlled machine tool moves with a constant speed of 6.5 m/min. The drilling of a hole lasts 0.5 s. Find the shortest routes that the drill should follow in order to complete the patterns shown in the images below. Task 31 A machine produces steel cylinders. The length of the metal plate, from which the cylinders are cut off, is 2000 mm, and its thickness is 100 mm. What is the weight of one cylinder if the length of cylinder is 240 mm and its width is equal to the thickness of the metal plate (ρ = 7.85 kg/dm 3 )? 138

140 Task 32 A client ordered steel cylinders from a company. The length of the metal plate, from which the cylinders are cut off, is 2000 mm, and its thickness is 100 mm. The length of cylinder is 240 mm and its width is equal to the thickness of the metal plate (ρ = 7.85 kg/dm 3 ). How many track containers are needed in order to transport the ordered materials to the client? The total mass of transport vehicle cannot be more than 40 t, the weight of an empty track container is around 20 t. 10 cylinders are packed in a box which weights 20 kg. Task 33 A company which produces steel cylinders transports the materials at first cost. How much will a client pay for cylinders and how much will the company earn from the client? The firm of the client is located in a distance of 200 km from the company. The cost of transport of 1 track container is 80 euros per hour. The cost of production of 1 cylinder is 53 euros. The profit margin comes to 10% for one cylinder. Task 34 Calculate the area of the following polygons circumscribed on a circle of a given diameter d: a) an equilateral triangle b) a square c) hexagon Hint: In the drawings, consider what is the relationship between the diameter d and the length of the respective sides of the polygons. a r Task 35 Balls with a diameter of 20 cm must be individually packed in the appropriate box. Design a box for that purpose. Calculate how much material is needed for its production. Consider the following shapes for boxes: 1. A box in the shape of a hexagonal prism. 2. The box in the shape of a cube. 3. A box in the shape of a cylinder. 4. Propose your own box. Task 36 Balls with a diameter of 20 cm must be individually packed in the appropriate box. Design a box for that purpose. Estimate for which box the less possible material is need (which surface is the smallest). 139

141 Task 37 Balls with a diameter of 20 cm will be individually packed in cylinder boxes. The components of the bases of the boxes are cut from a sheet measuring 130 cm x 75 cm. Examine the following arrangements for the cut-out bases. A. How many bases can be cut from a single sheet for each of these arrangements? B. Which of these arrangements is the optimal, if we take into account the amount of material used? Take into account the fact that in the arrangement of the elements it is necessary to provide 0.5 mm for a cutting line. Task 38 Balls with a diameter of 20 cm will be individually packed in prismatic boxes. The components of the bases of the boxes are cut from a sheet measuring 130 cm x 75 cm. Examine the following arrangements for the cut-out bases: with a square or with a hexagonal base. A. How many bases can be cut from a single sheet for each of these arrangements? B. Which of these arrangements is the optimal, if we take into account the amount of material used? Take into account the fact that in the arrangement of the elements it is necessary to provide 0.5 mm for the cutting line. Task 39 Balls with a diameter of 20 cm will be individually packed in cylinder boxes. The components of the bases of the boxes are cut from a sheet. You have to cut out 24 circles. In the Image you can see two possible arrangements. 140

142 The distance from the edge of the sheet must be at least 2 mm and it is necessary to provide 0.5 mm for the cutting line. A. How to design the most advantageous arrangement? B. What should be the dimensions of the sheet to get the least waste? C. What percentage of the surface of a rectangular sheet will be the waste? Task 40 A CNC Operator is supposed to cut the rod so that its diameter mildly decreases from the value of the bigger diameter D = 400 mm at the distance L = 200 mm to the value of the smaller diameter d = 300 mm. An operator is then expected to calculate the taper of the cone C and the angle of a tool slide setup. What is the taper of the cone C, if D is 400 mm, d is 300 mm, and the length L is 200 mm? Task 41 A CNC Operator is meant to make an element of a conical shape and dimensions L = 120 mm and d= 20 mm on a rod of a diameter D = 24 mm and a total length L w = 300 mm. An operator has to reset the axle of a tailstock by a certain value, which he/she has to calculate. Thus, the problem is the following: a cone of dimensions L = 120 mm and d = 20 mm is about to be machined on a rod of dimensions D =24 mm and L w = 300 mm by means of a tailstock shift. What is the cone taper? 2. What is the tailstock retraction? 3. What is the maximum tailstock shift? Równolegle do osi 141

143 Task 42 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. This can be done in two ways, shown in Diagrams A and B below. A client came with a metal sheet of the dimensions 950x1200 mm. He wants to receive 27 pairs of elements. Is it possible? Diagram A above - see enlarged parts below in working out Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm 142

144 Task 1 SOLUTIONS AND ANSWERS In the isosceles triangle the two base angles have the same size. Therefore, we calculate the angle γ: = This angle is larger than a right angle, so the triangle is obtuse. Task 2 Yes, because it has 4 equal sides, 2 pair of equal angles (even more, but it doesn t matter), 2 pair of equal angles (even more, but it doesn t matter), 2 pairs of parallel sides (even more, but it doesn t matter). Task 3 Answer: Four. Task 4 Answer: Translations. Task 5 Starting from left top corner, going down: 2 cm + 3 cm + 3 cm + 6 cm + 2 cm + 5 cm + (2 cm + 3 cm + 2 cm) + x cm = 21 cm + 7 cm + x cm = 28 cm + x cm. x - horizontal segments + vertical segments. Horizontal segments are: 3 cm + 6 cm + 5 cm = 14 cm. Vertical segments are: 2 cm + 2 cm = 4 cm. So, x = 14 cm + 4 cm = 18 cm. The total perimeter is: 28 cm + 18 cm = 46 cm. Task 6 Answer: For shape 1 the area is 6 ¾ cm 2. For shape 2 the area is 6 cm 2. For shape 3 the area is 6 ¼ cm 2. For shape 4 the area is 12 ½ cm 2. Task 7 Answer: Triangular prism has 5 faces 2 faces are triangles; 3 faces are rectangles. It has 6 vertices and 9 edges. 143

145 Task 8 Answer: We have to use the same units for all sides. length width height volume 6m 3m 5m 90 m 3 15 cm 4 cm 4,6 cm 276 cm cm 5.8 cm 6 cm 69.6 cm 3 Task 9 Answer: Yes. Task 10 V D - Volume of cylinder of diameter D V d Volume of cylinder of diameter d d = D 2 s = = 50 mm V D = π = 5600π cm 3 V d = π = π cm 3 V p Volume of pipe V p = 5600π π = π cm 3 w weight of pipe w = ρ V w = = g = kg Task kg = g : 7.2 = (4) maximum volume V p = π 4 2 l π l = 9.75 π l 9.75 π l = π l l max cm = m Task 12 V D - Volume of cylinder of diameter D V d Volume of cylinder of diameter d 144

146 d = D 2 s = = 40 mm V D = π = 5600π cm 3 V d = π = 1400 π cm 3 V p Volume of pipe V p = 5600π 1400π = 4200π cm 3 w weight of pipe w = ρ V w = = g = kg Task 13 Total time needed for one pipe is 15 sec + 5 min 20 sec + 15 sec = 5 min 50 sec = 350 sec. 2 hours = 120 min = 7200 sec : Within 2 hours it is possible to transport 20 pipes. Task 14 Total time needed for one pipe is 15 sec + 5 min 20 sec + 15 sec = 5 min 50 sec. = 350 sec = sec = 233.(3) min = 3 h 53 min 20 sec. Task 15 The company has to order 2572 pipes (9000:3.5 = ). The weight of one pipe is kg (task 10). The total price of the order is: euros. Task 18 a = 27 mm c = 45 mm b =?, α =?, β =? 27 mm = 2.7 cm 45 mm = 4.5 cm a 2 + b 2 = c 2 (2.7) 2 + b 2 = (4.5) 2 \ (2.7) 2 b 2 = b = =

147 sin α = a c sin α = 2.7 = 0.6 then α = Then β= 180 ( ) = 54 Task 19 a = 27 mm c = 45 mm 27 mm = 2.7 cm 45 mm = 4.5 cm sin α = a c sin α = 2.7 = 0.6 then α = Then β= 180 ( ) = 54 A a β c A β 2 3 β 2 3 C b α B x the distance from the vertex A sin β 2 = 3 x x = 3 3 = 6.61 mm sin y the distance from the vertex B sin α 2 = 3 y y = 3 3 = 9.71 mm sin z the distance from the vertex C z = mm 146

148 Task 21 The volume of the block V1: V1 = l x w x h V1 = 80 x 60 x 20 = 96,000 mm³ The volume of the hole V2: V2 = πr 2 h π = h=20mm r= 5mm V2 = x 5²x20 V2 = 1571mm³ The volume of material remaining: V1 V2 = 96,000 1,571 = 94,429 mm³ Task 22 Answer: 10 mm. Task 23 Answer: 270. Task 26 Answer: v = 6.3 m/min Task 27 Answer: m/min Task 28 Answer: 92 holes Distance: 7,644 m Task 29 Answer: Min. dimensions of the block are: a = 3.85 m, b = 6.67 m. Task 31 h = 240 mm = 0.24 m r = 50 mm = 0.05 m V volume of one cylinder V = πr 2 h = π(0.05) [m 3 ] // π

149 If 7.85 kg 1 dm 3 Then x kg 1000 dm 3 (1 m 3 ) x = 7850 kg The average density of steel on 1 m 3 is 7850 kg. y average weight of one cylinder 7850 kg 1 m 3 y kg m 3 y = kg. Task 32 The weight of loading is = 20 t. 1 cylinder weights kg (why?). We will need 1000 boxes ( : 10). The total weight of 1 box with 10 cylinders is kg +20 kg = kg : One track can be loaded with 118 boxes : We need 9 tracks. Task 33 The client will pay = 58.3 euros per 1 cylinder = euros per cylinders. The client will pay euros The average time for the distance of 200 km (if the average speed is 60 km/h) comes to 3h 20 min. The cost of the transport for 1 track is = euros for 8 tracks. 80 = = euros The company earns 5.3 euros on every cylinder, euros for cylinders = euros The company will earn euros. Task 35 a r 148

150 1. In order to calculate the area of the base, you have to calculate the area of 6 triangles of the height r =10cm (picture). Because r = (a 3)/2, a = 2/3 r 3, then 6P = 6 a 2 3/4 = 2 r the area of the base. Then two bases have the area: 4 r 2 3, and the lateral area is: 6a2r = 8r 2 3. The total area: 12 r The box in the shape of a cube: the area equals: 6 4 r 2 = 24 r 2 3. A box in the shape of a cylinder: the area equals: 2 Π r Πr 2r = 6 Π r 2 Task 40 C = D d L C = Task 41 Answer: 400 mm 300 mm 200 mm 1. C = 1:30 2. V R = 5 mm 3. V R MAX = 6 mm Task 42 = 100 mm 200 mm = 1 2 = 1 2. We try to put the elements like in a diagram B. The width: = elements in row The height: = rows 8 3 = 24 elements There is a piece of sheet in the shape of rectangle left. Its dimensions are 205x1200 mm (205 = ). We check if it is possible to put there the figure of shape L. Note that the triangles ADE and GFC are isosceles. 149

151 Thus AD = DE = FG = CF = 20 mm. AB = BC = =255 mm. Then AC = mm So, h = = mm = mm < 205 mm So, if we turn the figure in such a way, the figure will fit to the remaining part of the sheet. We need 3 more figures = = 1123 mm < 1200 mm It is possible to put 3 figures in the part which was left. In the area inside figures L, it is possible to put small elements. So, it is possible to cut 27 pairs of elements. 150

152 Task 1 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Formulae/data provided Performing simple operations Level 1 ALL Tasks 2-3 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Formulae/data provided Understanding basic properties of shapes Level 1 ALL Task 4 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Formulae/data provided Understanding basic properties of shapes Level 1 ALL Tasks 5-6 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Formulae/data provided Performing simple operations Working with complex shapes Level 2 ALL Task 7 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Formulae/data provided Understanding basic properties of shapes Level 1 ALL Task 8 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Formulae/data provided Volume of cuboid is = length width height Performing simple operations Level 1 ALL Task 9 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Formulae/data provided Understanding basic properties of shapes Level 2 ALL Tasks categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties 151

153 Formulae/data provided Spatial visualisation V d Volume of cylinder of diameter d: V d = π ( d 2 )2 l Weight in relation to density ρ and volume V: w = ρ V Performing simple operations Calculating measures of weight and volume Level 1 CNC Operator Tasks categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Application of numerical skills to solve geometrical problems Formulae/data provided V d Volume of cylinder of diameter d: V d = π ( d 2 )2 l Weight in relation to density ρ and volume V: w = ρ V Performing simple operations Calculating measures of weight, volume and time Comparing quantities Costing a project Level 2 Sales manager Task 18 categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Trigonometry and geometry Formulae/data provided Pythagorean Theorem, Trigonometric functions (definitions and table) Performing simple operations Use of formulas Calculating lengths and angles Level 1 Mechanical Technician/Engineer, CNC Operator Task 19 categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Trigonometry and geometry Formulae/data provided Pythagorean Theorem, Trigonometric functions (definitions and table) Performing simple operations Use of formulas Calculating lengths and angles Level 2 Mechanical Technician/Engineer, CNC Operator Task 20 categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Trigonometry and geometry Formulae/data provided Pythagorean Theorem, Trigonometric functions (definitions and table), Law of cosines, The centroid divides each median into parts in the ratio 2:1. Performing simple operations 152

154 Use of formulas Calculating lengths and angles Level 2 Mechanical Technician/Engineer, CNC Operator Tasks categorisation Space and shape, including measurements Mathematical subfield 2D-3D shapes and properties Formulae/data provided Volume of rectangular cuboid = l x w x h or V = lwh Volume of a cylinder = πr2h Calculating volumes Using formulas Level 1 Machinist, Buyer, Designer Tasks categorisation Space and shape, including measurements Mathematical subfield 2D-3D shapes and properties Formulae/data provided Volume of rectangular cuboid = l x w x h or V = lwh Volume of a cylinder = πr2h Calculating volumes Using formulas Level 2 Machinist, Buyer, Designer Tasks categorisation s Space and shape, including measurements Change and relationships Mathematical subfields 2D-3D shapes and properties Trigonometry Formulae/data provided Trigonometric table, v=s/t Transformation of formula Switching between units Calculating distance by using trigonometric functions Level 1 Machinist, CNC machinist Task 29 categorisation s Space and shape, including measurements Change and relationships Mathematical subfields 2D-3D shapes and properties trigonometry Formulae/data provided Trigonometric table, v=s/t Transformation of formula Switching between units Calculating distance by using trigonometric functions Level 2 Machinist, CNC machinist Task 30 categorisation s Space and shape, including measurements Change and relationships Mathematical subfields 2D-3D shapes and properties Trigonometry Formulae/data provided Trigonometric table, v=s/t Transformation of formula Switching between units Trial and error Level 3 Machinist, CNC machinist 153

155 Task 31 categorisation s Space and shape, including measurements Mathematical subfields 2D-3D shapes and properties Formulae/data provided Density of steel ρ = 7,85 kg/dm 3 volume of cylinder: V = πr 2 h Transformation of formula Switching between units Calculating distance by using trigonometric functions Level 2 Machinist Task 32 categorisation Space and shape, including measurements s Quantity Mathematical subfields 2D-3D shapes and properties Basic operations Formulae/data provided Density of steel ρ = 7,85 kg/dm 3 volume of cylinder: V = πr 2 h Transformation of formula Switching between units Level 2 Machinist, buyer Task 33 categorisation s Space and shape, including measurements Quantity Mathematical subfields Formulae/data provided 2D-3D shapes and properties Basic operations Performing simple operations Costing a project Level 2 Machinist, Buyer Tasks categorisation Space and shape Mathematical subfields 2D-3D shapes and properties Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided - Estimating areas Level 2 ALL Tasks categorisation Space and shape Mathematical subfields 2D-3D shapes and properties Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided - Estimating areas Level 3 ALL Task 40 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Formulae/data provided Formulae for calculating tapers and angles of setup C = D d L 154

156 Performing simple operations Using formulas Level 1 Mechanic, CNC Operator Task 41 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Formulae/data provided Cone taper formulae C = D d Tailstock shift Maximum tailstock shift Performing simple operations Using formulas Level 1 Mechanic, CNC Operator Task 42 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided Performing simple operations Estimating areas Costing a project Level 3 Buyer, Estimator, Sales Engineer, Technical Support, Designer, CAD Operator, Laser Operator, Welder L 155

157 QUIZ 1. Is it true or false that the sum of the interior angles of any 4-sided polygon is 360 degrees? a. True b. False 2. The area of a rectangle is cm 2 and its length is 8.6 cm. What is the perimeter of the rectangle in centimetres? a. 5.2 b c How many lines of symmetry does the circle have? a. None b. One c. Two d. Infinite 4. Is it true or false that a trapezium can have two right angles? a. True b. False 5. Is it true or false that the volume of a cylinder is directly proportional to its height? a. True b. False Answers: 1a, 2b, 3d, 4a, 5a 156

158 L13: Trigonometry and geometry Theory When we draw triangles in any scale, they all have the same shape the are only larger or smaller. These figures, which differ only in size are called similar figures. Consider some right triangles similar to each other. When we superimpose them each other so that their vertex is touching, you can see that the corresponding angles are equal. We say that in the similar triangles the size of the angle does not depend on the length of the sides. But can the angle be determined by the length of the sides of a triangle? It turns out that yes. That is what trigonometry does. B A= A C C A B AB = = 1 2, A C AC = = 1 2, B C BC = = 1 2 B These are the two similar right-angled triangles in the scale of 1: 2 The side AC is 5 cm, the side BC is 2.4 cm, the side AB 5.54 cm. Then, the side A C is 2.5 cm, the side B C is 1.2 cm, the side A B is 2.77 cm. These sizes are due to the scale of 1: 2. What happens when we calculate the other ratios of measures designed by sides of these triangles? B C = 1,2 = 0.48 BC = 2,4 = 0.48 these ratios are the same A C 2,5 AC 5 157

159 B C = 1.2 = BC = 2.4 = these ratios are the same A B 2.77 AB 5.54 A C = 2.5 = AC = 5 = these ratios are the same A B 2.77 AB 5.54 Regardless of whether we will create a ratio in a large or a small triangle, the value of the ratio is the same. Such ratios in some way describe the size of the acute angle in a rightangled triangle. Example: This sketch can be interpreted as set of similar right triangles: ABC, AB 1 C 1, AB 2 C 2, AB 3 C 3 having a common angle α. Complete the table below and compare ratios. AB = AB 1 = AB 2 = AB 3 = AC = AC 1 = AC 2, = AC 3 = BC = B 1 C 1 = B 2 C 2 = B 3 C 3 = AB/AC = AB 1 / AC 1 = AB 2 / AC 2 = AB 3 / AC 3 = BC/ AB = B 1 C 1 / AB 1 = B 2 C 2 / AB 2 = B 3 C 3/ AB 3 = Answer: AB = 3,4 cm AB 1 = 4,2 cm AB 2 = 4,8 cm AB 3 = 5,4 AC = 4 cm AC 1 = 5 cm AC 2, = 5,7 cm AC 3 = 6, 4 BC = 2,2 cm B 1 C 1 = 2,7 cm B 2 C 2 = 3 cm B 3 C 3 = 3,4 AB/AC =0.85 AB 1 / AC 1 = 0,84 AB 2 / AC 2 = 0, AB 3 / AC 3 =0, BC/ AB = 0, B 1 C 1 / AB 1 = B 2 C 2 / AB 2 = B 3 C 3/ AB 3 = Ratios in the rows are almost the same the difference are caused by the measurement uncertainty. Every change in the angle causes a change in the ratio. Here, the nominator is still the same, but denominators vary 158

160 Therefore we can assume that the ratio of the length of all possible sides of a triangle depends on the size of angle. The ratio is a function of the angle. The sides of a right triangle have their names associated with an acute angle. hypotenuse Opposite vertical side We define : The sine of an angle α in a right triangle is the ratio of length of the side of the triangle opposite the angle and the hypotenuse sin α = opposite/ hypotenuse The cosine of an angle α in a right triangle is the ratio of length of the side of the triangle adjacent to the angle and the hypotenuse. cos α = adjacent / hypotenuse The tangent of an angle α in a right triangle is the ratio of length of the side of the triangle opposite the angle and the adjacent to the angle. tg α = opposite/ adjacent The cotangent of an angle α in a right triangle is the ratio of length of the side of the triangle adjacent the angle and the opposite to the angle. ctg α = opposite/ adjacent adjacent vertical side All values of trigonometric functions are calculated and collected in the tables. Such arrays can be used in two ways: knowing the angle you can find the value of the corresponding trigonometric function, or by knowing the appropriate function one can find the size of the angle. The illustration shows how to read that tan 15 0 is α α α α 159

161 The values of trigonometric functions of certain angles can be known by heart. They are summarized in the following table: α sinα cosα tanα cotα /2 3/2 3/ /2 2/ /2 1/2 3 3/3 Knowledge of trigonometric functions is helpful in solving many practical problems. Example: Construct angles α, β, knowing that tan α = 2, cot β = 0.6. We can assume that 2 = 2b/b, for any b. Then t is enough to construct the right triangle, where one vertical side is b, and the second vertical side is 2b. Cot β = 0.6 = 3/5 = 3b/5b. Both constructions are displayed at the sketch. Example: By using some measuring instruments an observer stated that from the window situated at a height of 20 m above the river he can see the edges of the river under angles 180 o and 420 o. What is the width of the river? 20m/b = tan 18 0 b = 20/tan 18 0 b = 20/ m Let us consider two right angled triangles. The acute angle in the first triangle is 420, in the second one 180. We focus on the length of the sides, so we use the tangent function. In the first foundation triangle is denoted by a. In the second by b. The width of the river is b a. 20m/a = tan 42 0 a = 20/tan 42 0 a = 20/ m The width of the river is: m m = m 160

162 Tasks Task 1 How to describe sin γ, cos γ, tan γ, cot γ in the triangle on the sketch? Task 2 sin α = How much is α? How is tan 13 0? Use the table below. Task 3 The drawing shows a hammer handle that is to be machined on a lathe. The machine operator needs to know the angle between the two diameters to be able to machine it accurately. Use trigonometry to find out the angle. 161

163 Task 4 In a designed element of a machine with the shape of a right triangle with a hypotenuse of c = 45 mm and a vertical side of a = 27 mm, a Mechanical Technician has to calculate the second vertical side and the angles of designed part. Calculate them. Task 5 In a designed element of a machine with the shape of a right triangle with a hypotenuse of 45 mm and a vertical side of 27 mm, a Mechanical Technician has to drill 3 holes in all three vertexes of the triangle at the distance of 3mm from both edges. What are the distances between the vertexes and the holes? Task 6 In a designed element of a machine with the shape of a right triangle with a hypotenuse of 45 mm and a vertical side of 27 mm, a Mechanical Technician has to drill a hole in its centroid. Calculate the distances of the hole from all three sides of the triangle knowing that the hole has a diameter of 2 mm. Task 7 In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 30 o. The drill should reach hole 2 at the latest after 0.8 s. What is the minimum average movement speed in m/min? Task 8 In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 45 o. 162

164 The drill should reach hole 2 at the latest after 0.8 s. What is the minimum average movement speed in m/min? Task 9 In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 30 o. The drill reaches hole 2 in 0.8 s and the drilling of a hole lasts 0.5 s. The process is repeated for a number of holes. How many holes can be bored within 2 min? What is the distance between first and the last hole? Task 10 In order to bore hole 2 shown in the image below, the drill of a numerically controlled machine tool should start moving from hole 1 in the direction of an angle of 30 o. The drill reaches hole 2 in 0.8 s. and the drilling of a hole lasts 0.5 s. The process is repeated for a number of holes for 2 min. Given that the distance of the centre of hole 1 is 2 cm from both edges of the block, calculate the minimum dimensions of the block in order for the drilling process to be completed. Task 11 A numerically controlled machine tool moves with a constant speed of 6.5 m/min. The drilling of a hole lasts 0.5 s. Find the shortest routes that the drill should follow in order to complete the patterns shown in the images below. 163

165 SOLUTIONS AND ANSWERS Task 1 Answer: sin γ = t/s cos γ = r/s tan γ = t/r cot γ = r/t Task 2 Answer: α = 9 0. tan 13 0 = Task 3 Firstly, find H = (Dia. 22/2) (Dia.15.8 / 2) [to give the difference between radii] = 3.1mm or written as H = Dia.22mm Dia.15.8mm = 3.1 mm 2 2 Tangent Angle X = opposite/adjacent = H/X Tangent Angle X = 3.1 / 7 Tangent Angle X = Therefore, Angle X = Since Angle X is not a critical dimension and the tolerance is +/- 1, the nearest whole degree = 21 Task 4 a = 27 mm c = 45 mm b =?, α =?, β =? 27 mm = 2.7 cm 45 mm = 4.5 cm a 2 + b 2 = c 2 (2.7) 2 + b 2 = (4.5) 2 \ (2.7) 2 b 2 =

166 b = = 3.6 sin α = a c sin α = 2.7 = 0.6 then α = Then β= 180 ( ) = 54 Task 5 a = 27 mm c = 45 mm 27 mm = 2.7 cm 45 mm = 4.5 cm sin α = a c sin α = 2.7 = 0.6 then α = Then β= 180 ( ) = 54 A β c a α C b x the distance from the vertex A B A β 2 3 β 2 3 sin β 2 = 3 x x = 3 3 = 6.61 mm sin y the distance from the vertex B sin α 2 = 3 y y = 3 3 = 9.71 mm sin z the distance from the vertex C z = mm 165

167 Task 7 Answer: v = 6.3 m/min Task 8 Answer: m/min Task 9 Answer: 92 hols Distance: m Task 10 Answer: Min. dimensions of the block are: a = 3.85 m, b = 6.67 m. 166

168 Task 1 categorisation Space and shape Mathematical subfield Trigonometry and geometry Formulae/data provided Understanding basic properties of shapes Level 1 ALL Task 2 categorisation Space and shape Mathematical subfield Trigonometry and geometry Formulae/data provided Reading tables Level 1 ALL Task 3 categorisation Space and shape Mathematical subfield Trigonometry and geometry Formulae/data provided Tangent θ = opposite side/adjacent side Performing simple operations Using formulas Level 1 Machinist, Designer Task 4 categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Trigonometry and geometry Formulae/data provided Pythagorean Theorem, Trigonometric functions (definitions and table) Performing simple operations Use of formulas Calculating lengths and angles Level 1 Mechanical Technician/Engineer, CNC Operator Task 5 categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Trigonometry and geometry Formulae/data provided Pythagorean Theorem, Trigonometric functions (definitions and table) Performing simple operations Use of formulas Calculating lengths and angles Level 2 Mechanical Technician/Engineer, CNC Operator Task 6 categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Trigonometry and geometry Formulae/data provided Pythagorean Theorem, Trigonometric functions (definitions and table), Law of cosines, The centroid divides each median into parts in the ratio 2:1. 167

169 Performing simple operations Use of formulas Calculating lengths and angles Level 2 Mechanical Technician/Engineer, CNC Operator Tasks 7-9 categorisation Space and shape, including measurements s Change and relationships 2D-3D shapes and properties Mathematical subfields Trigonometry Types of change Formulae/data provided Trigonometric table, v=s/t Transformation of formula Switching between units Calculating distance by using trigonometric functions Level 1 Machinist, CNC machinist Task 10 categorisation Space and shape, including measurements s Change and relationships 2D-3D shapes and properties Mathematical subfields Trigonometry Types of change Formulae/data provided Trigonometric table, v=s/t Transformation of formula Switching between units Calculating distance by using trigonometric functions Level 2 Machinist, CNC machinist Task 11 categorisation Space and shape, including measurements s Change and relationships 2D-3D shapes and properties Mathematical subfields Trigonometry Types of change Formulae/data provided Trigonometric table, v=s/t Transformation of formula Switching between units Trial and error Level 3 Machinist, CNC machinist 168

170 QUIZ 1. Two triangles ABC and A B C are similar in the scale of 1: 3. The side AC is 3, the side BC is 6 and the side CA is 4. What is the perimeter of the triangle A B C? a. 38 b. 41 c. 39 d Calculate sinα if the lengths of the sides of a right triangle ABC are: AB = 5, BC = 10 and AC = 5 5. The angle α lies at the vertex A. a. 1 2 b c. 5 d Calculate tanα if the lengths of the sides of a right triangle ABC are: AB = 5, BC = 10 and AC = 5 5. The angle α lies at the vertex A. a. 1 2 b c. 5 d Is it true or false that if sinα = 3 2 a. True b. False then α= 45? 5. Is it true or false that sinα + cosα+ tanα + cotβ = a. True b. False 6 if α = 30 and β = 45? Answers: 1c, 2b, 3d, 4b, 5a 169

171 L14: Spatial visualisation Theory Often a worker must be able to imagine how an object will look like if she puts it in another way, or if she will look at it from another side. This ability is associated with making mental manipulations - you need to be able to move or rotate the solid mentally, or its element thereof. Many times this ability involves such manipulations, which in mathematics are described by transformations: axial symmetry, rotation, parallel shift. Intuitively, the effects of such transformations describe how the figure was lying before and after the manipulation. In parallel transformation: In mirror symmetry (line symmetry) In rotation In rotation on on any angle Exercises in acquiring such skills can start from manipulating in 2D. The introduction to 3D can be done by experimenting with a cube. Manipulation in 2-D Example: This is a template of a figure (in black). Cross out all that figures, which are its symmetrical reflection (and after reflection their location on the plane might also change). 170

172 Answer: Example: Find the squares which are coloured in the same way: Answer: 2 and 4 Example: Adjust the item to the card, from which it has been cut. Fill in the pairs in a table a b c d e 5 Answer: 1-b, 2-d, 3-e, 4-a, 5-c Manipulation in 3-D Example: Which net was used for making this cube? a) b) c) 171

173 Answer: c) Only this net has three walls coloured in such a way to be visible at a cube which, when assembled, will form a common vertex. Example: From the displayed net a cube has been sized and placed on the wall with a dot. Draw how it looks like from all four sides. Answer: Example: How will the cube look like when placed on each of its six walls? Mark its base, then paint the view of the cube lying on the ground. Is there only one solution to this task? Answer: By putting the cube on the one side you can get four different views. A sample of the solution is shown below. 172

174 Example: The structure shown consists of 4 cubes. Draw how it will look like, if a red cube will be placed in 4 different positions. Answer: Sample of the solution: Example: Which arrangement shows the model which is presented? 1) 2) 3) Answer: 2). 173

175 Tasks Task 1 If a machine cuts iron cast table tops from an iron surface of dimensions of µm x µm x µm, then how many table tops (of dimensions 1840 mm x 92 mm x 10 mm) will it cut? Task 2 If a machine cuts iron cast table tops from an iron surface of dimensions of µm x µm x µm, then how many table tops (of dimensions 1840 mm x 92 mm x 10 mm) will it cut? Task 3 Task 4 Task 5 a) If a machine cuts iron cast table tops from an iron surface of dimensions of µm x µm x µm, then how many table tops (of dimensions 1840 mm x 92 mm x 10 mm) will it cut? b) Is 96 the maximum number of table tops that can be cut from the iron cast surface? Justify your answer. a) If a machine cuts iron cast table tops from an iron surface of dimensions of µm x µm x µm, then how many table tops (of dimensions 1840 mm x 92 mm x 10 mm) will it cut? b) Mr Smith ordered 180 iron cast table tops. How much will he pay if 1m 2 of iron (of the 1cm gauge) costs 4 euros? The cost of production of one table top is 2 euros. The company is selling the table tops at 15% profit. Calculate the weight of a cast iron pipe given that its length is 3.5 m, its external diameter is 80 mm and the thickness of the wall is 15 mm (ρ = 7.2 g/cm 3 ). Task 6 What is the maximum length of a cast iron pipe in order to be lifted by manual hoist with a permissible load capacity of 500 kg? The external diameter of the pipe is 80 mm and the thickness of the wall is 15 mm (ρ = 7.2 g/cm 3 ). 174

176 Task 7 Calculate the weight of a cast iron pipe given that its length is 3.5 m, its external diameter is 80 mm and the thickness of the wall is 20 mm (ρ = 7.2 g/cm 3 ). Task 8 Calculate the area of the following polygons circumscribed on a circle of a given diameter d: a) an equilateral triangle b) a square c) hexagon Hint: In the drawings, consider what is the relationship between the diameter d and the length of the respective sides of the polygons a r Task 9 Balls with a diameter of 20 cm must be individually packed in the appropriate box. Design a box for that purpose. Calculate how much material is needed for its production. Consider the following shapes for boxes: 1. A box in the shape of a hexagonal prism. 2. The box in the shape of a cube. 3. A box in the shape of a cylinder. 4. Propose your own box. Task 10 Balls with a diameter of 20 cm must be individually packed in the appropriate box. Design a box for that purpose. Estimate for which box the less possible material is need (which surface is the smallest). 175

177 Task 11 Balls with a diameter of 20 cm will be individually packed in cylinder boxes. The components of the bases of the boxes are cut from a sheet measuring 130 cm x 75 cm. Examine the following arrangements for the cut-out bases. A. How many bases can be cut from a single sheet for each of these arrangements? B. Which of these arrangements is the optimal, if we take into account the amount of material used? Take into account the fact that in the arrangement of the elements it is necessary to provide 0.5 mm for a cutting line. Task 12 Balls with a diameter of 20 cm will be individually packed in prismatic boxes. The components of the bases of the boxes are cut from a sheet measuring 130 cm x 75 cm. Examine the following arrangements for the cut-out bases: with a square or with a hexagonal base. A. How many bases can be cut from a single sheet for each of these arrangements? B. Which of these arrangements is the optimal, if we take into account the amount of material used? Take into account the fact that in the arrangement of the elements it is necessary to provide 0.5 mm for the cutting line. Task 13 Balls with a diameter of 20 cm will be individually packed in cylinder boxes. The components of the bases of the boxes are cut from a sheet. You have to cut out 24 circles. In the Image you can see two possible arrangements. The distance from the edge of the sheet must be at least 2 mm and it is necessary to provide 0.5 mm for the cutting line. 176

178 A. How to design the most advantageous arrangement? B. What should be the dimensions of the sheet to get the least waste? C. What percentage of the surface of a rectangular sheet will be the waste? Task 14 A laser cutter operator has to cut a number of two different shaped parts out of a piece of sheet metal ready for a welder to join together. Material is often costly and the operator has to make optimal use of the stock material to minimise the amount of scrap. The two shaped parts to be cut are Parts 1 and 2. The CAD designer inputs the shapes onto a 500x855mm sheet (see diagram B) and uses nesting to maximise the space, taking into account the spacing required for the laser to cut. Diagram A above - see enlarged parts below in working out Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm a) Using this method instead of individual parts placed on the sheet (Diagram A), how much material in mm² is saved? b) If the cost of the material is 0.03 per 100mm² how much has been saved from minimising the scrap? Task 15 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. Estimate what part of the whole sheet are the elements from the sheet shown in Diagrams A and B. Give your answers in percentage. 177

179 50mm. Diagram A above - see enlarged parts below in working out Task 16 Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together (see the images below). By the end of a day in the machine a metal sheet of 500 x 580 mm was left. What is the maximum number of pairs of elements that can be cut? 235 mm Part 1 Part 2 20 mm 20mm 235mm All gaps between shapes and edges of plate are 10mm. Task 17 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. Material is often costly and the operator has to make optimal use of the stock material to minimise the amount of scrap. At the sheet of 178

180 dimensions 500 x 835 mm we put 12 elements of a shape L, in the way shown on diagram B. What is the maximum number of figures of the second kind we could cut from this sheet? Diagram A above - see enlarged parts below in working out Task 18 Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. Material is often costly and the operator has to make optimal use of the stock material to minimise the amount of scrap. Mr Smith wants to order 12 big and 12 small elements. The company X offers the price of 0.08 for 1 mm 2 and uses the method of cutting shown in diagram B and the company Y offers the price of 0.04 for 1 mm 2 and uses the method of cutting shown in diagram A. Which offer is better for Mr Smith? Both companies are paid for the whole material. Diagram A above - see enlarged parts below in working out 179

181 Task 19 Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. Material is often costly and the operator has to make optimal use of the stock material to minimise the amount of scrap. What will be the dimensions of the sheets in case of A and B shown in the diagrams if we prolong by 1 mm every dimension of both elements? Diagram A above - see enlarged parts below in working out Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm 180

182 Task 20 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. This can be done in two ways, shown in Diagrams A and B below. A client came with a metal sheet of the dimensions 950x1200 mm. He wants to receive 27 pairs of elements. Is it possible? Diagram A above - see enlarged parts below in working out Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm 181

183 SOLUTIONS AND ANSWERS Task µm = 9200 mm = 9.20 m µm = 1250 mm = 1.25 m µm = 10 mm 1840 mm =1.84 m 92 mm = m 9.20 : 1.84 = : x 13 = 65 table tops The surface left is too small to cut more table tops. Task 2 Surface dimensions: µm = 8850 mm = 8.85 m µm = 1930 mm = 1.93 m µm = 10 mm Table top: 1840 mm =1.84 m 92 mm = m 8.85 : : x 20 = 80 table tops The surface area left which can be used for cutting table tops: 1.49 m x 1.93 m 1.49 : extra table tops = 96 in total Task 3 Answer: a) see task

184 Task 4 Task 5 a) 65 table tops (see task 1). b) 180/65 = Mr Smith has to pay for 3 iron cast surfaces Total area of the surfaces: 11.5 m x 3 = 34.5 m 2 area of 3 surfaces 34.5 x 4 = 138 euros cost of material 2 x 180 = 360 euros cost of production = 498 euros x 498 = euros V D - Volume of cylinder of diameter D V d Volume of cylinder of diameter d d = D 2 s = = 50 mm V D = π = 5600π cm 3 V d = π = π cm 3 V p Volume of pipe V p = 5600π π = π cm 3 w weight of pipe w = ρ V w = = g = kg Task kg = g : 7.2 = (4) maximum volume V p = π 4 2 l π l = 9.75 π l 9.75 π l = π l l max cm = m Task 7 Answer: V D - Volume of cylinder of diameter D V d Volume of cylinder of diameter d 183

185 50mm d = D 2 s = = 40 mm V D = π = 5600π cm 3 V d = π = 1400 π cm 3 V p Volume of pipe V p = 5600π 1400π = 4200π cm 3 w weight of pipe w = ρ V w = = g = kg Task 9 a r 1. In order to calculate the area of the base, you have to calculate the area of 6 triangles of the height r =10cm (picture). Because r = (a 3)/2, a = 2/3 r 3, then 6P = 6 a 2 3/4 = 2 r the area of the base. Then two bases have the area: 4 r 2 3, and the lateral area is: 6a2r = 8r 2 3. The total area: 12 r The box in the shape of a cube: the area equals: 6 4 r 2 = 24 r 2 3. A box in the shape of a cylinder: the area equals: 2 Π r Πr 2r = 6 Π r 2 Task 14 a) Firstly, calculate the area required to place the parts individually, as in diagram A W = (235 x 6) + (7 x 10gaps) = 1,480mm wide H = (235 x 2) + (1x50) + (4x10 gaps) = 560mm high Therefore, Area A = W x H = 1,480 x 560 = 828,800mm² The nested Area B = 500 x 855 = 427,500mm² The difference in area = A-B = 828, ,500 = 401,300mm² b) The cost saving is 401, 300mm²/100mm² x 0.03 = mm Part 1 Part 2 20m m 20mm 235mm All gaps between shapes and edges of plate are 10mm. 184

186 Task 15 The area of one figure of a shape L: (235 20) 20 = = 9000 mm 2 Approximation of area of the smaller figure: = mm 2 The area of all elements: ( ) 12 = mm 2 Area of the sheet A is mm 2 Area of the sheet B is mm 2 Thus, / % % / % % Task 16 Firstly, we check how many big elements we can put vertically (how many rows of elements): = 500 We can put 2 elements. Then, we check how many big elements we can put horizontally (how many columns of elements) = 560 In one row we can put 4 elements 2 4 = 8 We can cut 8 pairs of elements. Task 17 We calculate the dimensions of area W: The height of area W: 235 (2 20) = = 195 mm The width of area W: ( ) 2 20 = = 225 mm We subtract the spacing required for the laser to cut: 185

187 The height: = 175 mm The width: = 205 mm The dimensions of area W is 175 x 205. Option 1 (Elements put horizontally) The height: = 170 < el. The width: = 200 < elements. 3 7 = 21 elements 21 6 = 126 Option 2 (Elements put vertically) The height: = = 170 < el. The width: = 170 < el. 6 3 = 18 elements = 35 > 30 = we can put one more row horizontally = 21 elements 21 6 = 126 The maximum number of elements we can put is 126. Task 18 The area of the sheet A is mm 2 The area of the sheet B is mm 2 The cost of the order in company X: = The cost of the order in company Y: = It is better to order in company Y. Task 19 Case A The width: 10 x x (235+10) = 1540 mm The height: 10 x 4 + ( ) + 2 x ( ) = 590 mm Case B The width: 10 x x ( ) + 3 x ( ) = 895 mm The height: 10 x x ( ) = 520 mm In case A it is 590x1540mm and in case B 520x895mm. 186

188 Task 20 Answer: We try to put the elements like in a diagram B. The width: = elements in row The height: = rows 8 3 = 24 elements There is a piece of sheet in the shape of rectangle left. Its dimensions are 205x1200 mm (205 = ). We check if it is possible to put there the figure of shape L. Note that the triangles ADE and GFC are isosceles. Thus AD = DE = FG = CF = 20 mm. AB = BC = =255 mm. Then AC = mm So, h = = mm = mm < 205 mm So, if we turn the figure in such a way, the figure will fit to the remaining part of the sheet. We need 3 more figures = = 1123 mm < 1200 mm It is possible to put 3 figures in the part which was left. In the area inside figures L, it is possible to put small elements. So, it is possible to cut 27 pairs of elements. 187

189 Task 1 categorisation Space and shape Mathematical subfield Units of measurement and scales Spatial visualisation Formulae/data provided 1 µm = mm Performing simple operations Estimating areas Level 1 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Task 2 categorisation Space and shape Mathematical subfield Units of measurement and scales Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided 1 µm = mm Performing simple operations Estimating areas Level 2 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Task 3 categorisation Space and shape Mathematical subfield Units of measurement and scales Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided 1 µm = mm Performing simple operations Estimating areas Level 3 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Task 4 categorisation Space and shape Mathematical subfield Units of measurement and scales Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided 1 µm = mm Performing simple operations Estimating areas Level 2 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Tasks 5-7 categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Spatial visualisation Formulae/data provided V d Volume of cylinder of diameter d: V d = π ( d 2 )2 l Weight in relation to density ρ and volume V: w = ρ V Performing simple operations Calculating measures of weight and volume Level 1 CNC Operator 188

190 Tasks 8-9 categorisation Space and shape Mathematical subfields 2D-3D shapes and properties Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided - Estimating areas Level 2 ALL Tasks categorisation Space and shape Mathematical subfields 2D-3D shapes and properties Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided - Estimating areas Level 3 ALL Tasks categorisation Space and shape Mathematical subfield Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided Performing simple operations Estimating areas Costing a project Level 1 Buyer, Estimator, Sales Engineer, Technical Support, Designer, CAD Operator, Laser Operator, Welder Tasks categorisation Space and shape Mathematical subfield Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided Performing simple operations Estimating areas Costing a project Level 2 Buyer, Estimator, Sales Engineer, Technical Support, Designer, CAD Operator, Laser Operator, Welder Task 20 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided Performing simple operations Estimating areas Costing a project Level 3 Buyer, Estimator, Sales Engineer, Technical Support, Designer, CAD Operator, Laser Operator, Welder 189

191 QUIZ 1. Which one of the following pairs of figures are symmetrically reflected? a. b b. e c. c d. a 2. Is it true or false that a square has 6 symmetrical axes? a. True b. False 3. Which shape represents a rotation of 180 of the shape shown below? a. 1 b. 2 c. 3 d

192 4. Which image is not a net of a cube? a. A b. B c. C d. D 5. If you cut off all the corners of a cube, the new solid is called cube octahedron. How many faces does it have? a. 8 b. 12 c. 14 d. None of the above Answers: 1c, 2b, 3a, 4a, 5c 191

193 L15: Application of numerical skills to solve geometrical problems Theory Scale The scale determines how the actual dimensions were changed. The form 1: 1000 is interpreted: 1cm on the sketch is 1,000 cm (10 m) in reality. Scale 2: 5000 indicates that 2 cm on the picture is 5,000 cm (50 m) in reality. Sometimes the scale should be turned to a simpler record. We do this by reducing to lower terms: 1/2500 or 2/5000 = 2: 5000 is the same as 1: 2500 Example: Write each scale as the simplest ratio. 4 : 400, 28 : 7000, 500 : Answer: 4 : 400 = 4/400 = (reducing by 4) 1/100. The ratio is 1: : 700 = 28/700 = (reducing by 7) = 4/100 (reducing by 4) = 1/25. The ratio is 1 : : = 500/20000 = (reducing by 100) = 5/200 = (reducing by 5) = 1/400. The ratio is 1: 400 The concept of the scale can be used in different ways. You can calculate the scale while having the real lengths and the lengths on the drawing, but usually you calculate the actual dimensions, having given the scale and dimensions of the figure. Example: In a catalogue a picture of drawer and its dimensions are given. In what scale is the drawer presented? Example: In a catalogue a picture of drawer and its dimensions are given. In what scale is the drawer presented? 9 cm 5,3 cm Depth: 43.5 cm Width: 135 cm Height: 79.5 cm 192

194 We have to calculate the scale. The image provides information only on the width and height. Actual width: 135 cm The width on the picture: 9 cm We calculate the scale: 9/135 = 1/15, thus 1:15 scale Checking the other dimensions: Actual height: 79.5 cm The height on the photo: 5.3 cm Calculating the scale: 5.3 / 79.5 = 1/15 Example: These screws and nails are drawn in scale 2:1. What is their real length? Answer: The scale informs us that the real length is enlarged two times. In order to calculate the real lengths, we have to divide the length from the sketch by 2. The nail from sketch c) is 2 cm, this means that its real length is 1 cm. The nail from sketch h) is 2.8 cm, thus its real length is 1.4 cm. Perimeters, areas, volume of solids Many technical drawings need to be further interpreted to get the eligible information. There are similar situations in other practical circumstances. Example: Write down how to calculate the field of painted figures, using the symbols from the image a) b) c) d) Answer: a) This is a parallelogram. The base is a 2x, the height is h 2y. The area is: (a 2x)( h 2y). 193

195 b) From the area of the rectangle subtract 4 identical areas of triangles. It is: 4a 3a 4 ½ 2a 1 ½ a = 12 a 2 6a 2 = 6a 2 c) From the area of the big triangle subtract the area of the trapezium. It is: ½ a h [ ½ (a + ½ a) ½ h] = ½ah ¼ 3/2 ah = ½ah 3/8 ah = 1/8 ah d) The base of the triangle is r + s + t. Its high is k- p. The area: ½ (r + s + t) (k p) Use of the Pythagorean Theorem The Pythagorean theorem allows you to calculate the length of the third side in a right triangle if you know the length of two other sides. This is expressed in the following form: If the triangle is rectangular, then the sum of the squares of the length of the vertical sides is equal to the square of the length of the hypotenuse. a b According to the markings as shown, you can write: c a 2 + b 2 = c 2 Example: What is the third side of a right triangle, knowing that the vertical sides are equal to 5 cm and 12 cm? If we denote x - the length of the hypotenuse, then Example: x 2 = x 2 = x 2 = 169 x = 13 How to indicate the length of the diagonal of a square? DAB is an isosceles rectangular triangle. The sides of the triangle have a length a. The diagonal d of the square is the hypotenuse of the triangle. a Example: d 2 = a 2 + a 2 d 2 = 2a 2 d = a 2 How to calculate the length of the height of an equilateral triangle? ADC is a rectangular triangle. One vertical side has a length of ½ a, and the hypotenuse has a length a. The altitude h of the triangle is a vertical side of the triangle. a 2 = h 2 + (½ a) 2 h 2 = a 2 ¼ a 2 h = a 3/2 194

196 Trigonometry In a right-angled triangle, where a, b are the vertical sides, and c the hypotenuse (see figure), the following trigonometric relationships are established: sin α = a/c cos α = b/c tg α = a/b ctg α = b/a a b c α By using simple calculations, we can find interesting relationships between trigonometric functions. Example: sin α : cos α = a/c : b/c = a/c c/b = a/b = tan α cos α : sin α = b/c : a/c = b/c c / a = b/ a = cot α By using the Pythagorean formula: a 2 + b 2 = c 2 and dividing by c 2 a 2 / c 2 + b 2 / c 2 = 1 (a/c) 2 +(b/c) 2 = 1 (sin α) 2 +(cos α) 2 = 1 Trigonometric compounds are useful for the calculation of lumps. Example: Calculate the volume of a cuboid knowing that the diagonal of the cuboid is 10 cm and the angle between the base and its diagonal is 60 o, and one edge of the base is 4 cm in length. The above relationships are presented in the image below: Triangle AC C is a right triangle. Let CC = h. h/10 = sin 60 0 = 3/2, thus h = 5 3. AC/10 = cos 60 0 = ½, thus AC = 5. Triangle ABC is a right triangle. By using the Pythagorean formula AB = 5 2. AB 2 = = 9 AB = 3. Tasks V = =

197 TASKS Task 1 From a rectangular we cut out an area, as shown. a) What is the length of the fence surrounding the area now? b) How long are the shorter sections of the fence, formed after cutting out of the area? It is known that they are of equal length. 4.7 m 1.9 m 0.8 m 2.9 m Task 2 A cuboid with dimensions of 20cm, 12cm, 5cm must be tied with string. Which should be the minimum length of the string, if for tying a knot 20 cm are needed? Task 3 If a machine cuts iron cast table tops from an iron surface of dimensions of µm x µm x µm, then how many table tops (of dimensions 1840 mm x 92 mm x 10 mm) will it cut? Task 4 a) If a machine cuts iron cast table tops from an iron surface of dimensions of µm x µm x µm, then how many table tops (of dimensions 1840 mm x 92 mm x 10 mm) will it cut? b) Is 96 the maximum number of table tops that can be cut from the iron cast surface? Justify your answer. Task 5 a) If a machine cuts iron cast table tops from an iron surface of dimensions of µm x µm x µm, then how many table tops (of dimensions 1840 mm x 92 mm x 10 mm) will it cut? b) Mr Smith ordered 180 iron cast table tops. How much will he pay if 1m 2 of iron (of the 1cm gauge) costs 4 euros? The cost of production of one table top is 2 euros. The company is selling the table tops at 15% profit. 196

198 Task 6 Cast iron pipes have to be transported from one working station to another. The time for transportation of one pipe is 5 min 20 sec. How many pipes can be transported within 2 hours if mounting and unmounting the pipe at the machine takes 15 sec? Task 7 Cast iron pipes have to be transported from one working station to another. The time for transportation of one pipe is 5 min 20 sec. How much time it will take to transport 40 pipes, if mounting and unmounting the pipe at the machine takes 15 sec? Task 8 A company wants to order cast iron pipes of a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 15 mm (ρ = 7.2 g/cm 3 ). The total length of the pipes has to be 9 km. The price of the particular iron pipes is 7.00 euros per kg. How much is the company expected to pay? Task 9 A company wants to order cast iron pipes of two kinds: Type 1, with a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 15 mm and Type 2, with a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 20 mm (ρ = 7.2 g/cm 3 ). The company knows that the price of a Type 1 iron pipe is euros per piece. Is euros a fair price for a Type 2 pipe? Task 10 A company wants to order cast iron pipes of a length of 3.5 m, external diameter of 80 mm and thickness of the wall of 15 mm (ρ = 7.2 g/cm 3 ). The total length of the pipes has to be 9 km. They received two offers: Offer 1: 7.00 euros per kg of the pipe Offer 2: euros per pipe. Which is the best offer? Justify your opinion. Task 11 Calculate the area of the following polygons circumscribed on a circle of a given diameter d: a) an equilateral triangle b) a square c) hexagon a r Hint: In the drawings, consider what is the relationship between the diameter d and the length of the respective sides of the polygons. 197

199 Task 12 Balls with a diameter of 20 cm must be individually packed in the appropriate box. Design a box for that purpose. Calculate how much material is needed for its production. Consider the following shapes for boxes: 1. A box in the shape of a hexagonal prism. 2. The box in the shape of a cube. 3. A box in the shape of a cylinder. 4. Propose your own box. Task 13 Balls with a diameter of 20 cm must be individually packed in the appropriate box. Design a box for that purpose. Estimate for which box the less possible material is need (which surface is the smallest). Task 14 Balls with a diameter of 20 cm will be individually packed in cylinder boxes. The components of the bases of the boxes are cut from a sheet measuring 130 cm x 75 cm. Examine the following arrangements for the cut-out bases. A. How many bases can be cut from a single sheet for each of these arrangements? B. Which of these arrangements is the optimal, if we take into account the amount of material used? Take into account the fact that in the arrangement of the elements it is necessary to provide 0.5 mm for a cutting line. Task 15 Balls with a diameter of 20 cm will be individually packed in prismatic boxes. The components of the bases of the boxes are cut from a sheet measuring 130 cm x 75 cm. Examine the following arrangements for the cut-out bases: with a square or with a hexagonal base A. How many bases can be cut from a single sheet for each of these arrangements? B. Which of these arrangements is the optimal, if we take into account the amount of material used? Take into account the fact that in the arrangement of the elements it is necessary to provide 0.5 mm for the cutting line. 198

200 Task 16 Balls with a diameter of 20 cm will be individually packed in cylinder boxes. The components of the bases of the boxes are cut from a sheet. You have to cut out 24 circles. In the Image you can see two possible arrangements: The distance from the edge of the sheet must be at least 2 mm and it is necessary to provide 0.5 mm for the cutting line. A. How to design the most advantageous arrangement? B. What should be the dimensions of the sheet to get the least waste? C. What percentage of the surface of a rectangular sheet will be the waste? Task 17 A laser cutter operator has to cut a number of two different shaped parts out of a piece of sheet metal ready for a welder to join together. Material is often costly and the operator has to make optimal use of the stock material to minimise the amount of scrap. The two shaped parts to be cut are Parts 1 and 2. The CAD designer inputs the shapes onto a 500x855mm sheet (see diagram B) and uses nesting to maximise the space, taking into account the spacing required for the laser to cut. Diagram A above - see enlarged parts below in working out 199

201 Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm a) Using this method instead of individual parts placed on the sheet (Diagram A), how much material in mm² is saved? b) If the cost of the material is 0.03 per 100mm² how much has been saved from minimising the scrap? Task 18 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. Estimate what part of the whole sheet are the elements from the sheet shown in Diagrams A and B. Give your answers in percentage. Diagram A above - see enlarged parts below in working out Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm 200

202 50mm Task 19 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together (see the images below). By the end of a day in the machine a metal sheet of 500 x 580 mm was left. What is the maximum number of pairs of elements that can be cut? 235 mm Part 1 20 mm 20mm Part 2 235mm All gaps between shapes and edges of plate are 10mm. Task 20 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. Material is often costly and the operator has to make optimal use of the stock material to minimise the amount of scrap. At the sheet of dimensions 500 x 835 mm we put 12 elements of a shape L, in the way shown on diagram B. What is the maximum number of figures of the second kind we could cut from this sheet? Diagram A above - see enlarged parts below in working out Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm 201

203 Task 21 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. Material is often costly and the operator has to make optimal use of the stock material to minimise the amount of scrap. Mr Smith wants to order 12 big and 12 small elements. The company X offers the price of 0.08 for 1 mm 2 and uses the method of cutting shown in diagram B and the company Y offers the price of 0.04 for 1 mm 2 and uses the method of cutting shown in diagram A. Which offer is better for Mr Smith? Both companies are paid for the whole material. Diagram A above - see enlarged parts below in working out Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm Task 22 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. Material is often costly and the operator has to make optimal use of the stock material to minimise the amount of scrap. What will be the dimensions of the sheets in case of A and B shown in the diagrams if we prolong by 1 mm every dimension of both elements? 202

204 Diagram A above - see enlarged parts below in working out Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm Task 23 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. This can be done in two ways, shown in Diagrams A and B below. A client came with a metal sheet of the dimensions 950x1200 mm. He wants to receive 27 pairs of elements. Is it possible? Diagram A above - see enlarged parts below in working out 203

205 Diagram B -Parts shown nested to optimise material Note: Not to Scale All measurements in mm 204

206 SOLUTIONS AND ANSWERS Task 1 a. The square had a fence length of 2 4,7 m m = 9.4 m m = 15.2 m. Now the fence has been extended by m = 3.8 m, that makes up 15.2 m m = 19 m. b. (2.9 m 0.8 m) : 2 = 1.05 m Task 2 The length of the cord must be equal to the circumference of a rectangle of 20 cm x 5 cm + circumference of a rectangle of 12 cm x 5 cm + 20 cm per knot. ( ) + ( ) + 20 = = 104 You need at least 104 cm cord. Task 3 Surface dimensions: µm = 8850 mm = 8.85 m µm = 1930 mm = 1.93 m µm = 10 mm Table top: 1840 mm =1.84 m 92 mm = m 8.85 : : x 20 = 80 table tops The surface area left which can be used for cutting table tops: 1.49 m x 1.93 m 1.49 : extra table tops = 96 in total Task 4 Answer: a) see task

207 Task 5 a) 65 table tops. b) 180/65 = Mr Smith has to pay for 3 iron cast surfaces Task 6 Total area of the surfaces: 11.5 m x 3 = 34.5 m 2 area of 3 surfaces 34.5 x 4 = 138 euros cost of material 2 x 180 = 360 euros cost of production = 498 euros x 498 = euros Total time needed for one pipe is 15 sec + 5 min 20 sec + 15 sec = 5 min 50 sec = 350 sec. 2 hours = 120 min = 7200 sec : Within 2 hours it is possible to transport 20 pipes. Task 7 Total time needed for one pipe is 15 sec + 5 min 20 sec + 15 sec = 5 min 50 sec. = 350 sec = sec = 233.(3) min = 3 h 53 min 20 sec. Task 8 The company has to order 2572 pipes (9000:3.5 = ). The weight of one pipe is kg (task 13). The total price of the order is: euros. Task 12 a r 1. In order to calculate the area of the base, you have to calculate the area of 6 triangles of the height r =10cm (picture). Because r = (a 3)/2, a = 2/3 r 3, then 6P = 6 a 2 3/4 = 2 r

208 50mm the area of the base. Then two bases have the area: 4 r 2 3, and the lateral area is: 6a2r = 8r 2 3. The total area: 12 r The box in the shape of a cube: the area equals: 6 4 r 2 = 24 r 2 3. A box in the shape of a cylinder: the area equals: 2 Π r Πr 2r = 6 Π r 2 Task 17 a) Firstly, calculate the area required to place the parts individually, as in diagram A W = (235 x 6) + (7 x 10gaps) = 1,480mm wide H = (235 x 2) + (1x50) + (4x10 gaps) = 560mm high Therefore, Area A = W x H = 1,480 x 560 = 828,800mm² The nested Area B = 500 x 855 = 427,500mm² The difference in area = A-B = 828, ,500 = 401,300mm² b) The cost saving is 401, 300mm²/100mm² x 0.03 = mm Part 1 Part 2 20 mm 20mm 235mm All gaps between shapes and edges of plate are 10mm. Task 18 The area of one figure of a shape L: (235 20) 20 = = 9000 mm 2 Approximation of area of the smaller figure: = mm 2 The area of all elements: ( ) 12 = mm 2 Area of the sheet A is mm 2 Area of the sheet B is mm 2 Thus, / % % / % % 207

209 Task 19 Firstly, we check how many big elements we can put vertically (how many rows of elements): = 500 We can put 2 elements. Then, we check how many big elements we can put horizontally (how many columns of elements) = 560 In one row we can put 4 elements 2 4 = 8 We can cut 8 pairs of elements. Task 20 We calculate the dimensions of area W: The height of area W: 235 (2 20) = = 195 mm The width of area W: ( ) 2 20 = = 225 mm We subtract the spacing required for the laser to cut: The height: = 175 mm The width: = 205 mm The dimensions of area W is 175 x 205. Option 1 (Elements put horizontally) The height: = 170 < el. The width: = 200 < elements. 3 7 = 21 elements 21 6 = 126 Option 2 (Elements put vertically) The height: = = 170 < el. The width: = 170 < el. 208

210 6 3 = 18 elements = 35 > 30 = we can put one more row horizontally = 21 elements 21 6 = 126 The maximum number of elements we can put is 126. Task 21 The area of the sheet A is mm 2 The area of the sheet B is mm 2 The cost of the order in company X: = The cost of the order in company Y: = It is better to order in company Y. Task 22 Case A The width: 10 x x (235+10) = 1540 mm The height: 10 x 4 + ( ) + 2 x ( ) = 590 mm Case B The width: 10 x x ( ) + 3 x ( ) = 895 mm The height: 10 x x ( ) = 520 mm In case A it is 590x1540mm and in case B 520x895mm. Task 23 We try to put the elements like in a diagram B. The width: = elements in row The height: = rows 8 3 = 24 elements There is a piece of sheet in the shape of rectangle left. Its dimensions are 205x1200 mm 209

211 (205 = ). We check if it is possible to put there the figure of shape L. Note that the triangles ADE and GFC are isosceles. Thus AD = DE = FG = CF = 20 mm. AB = BC = =255 mm. Then AC = mm So, h = = mm = mm < 205 mm So, if we turn the figure in such a way, the figure will fit to the remaining part of the sheet. We need 3 more figures = = 1123 mm < 1200 mm It is possible to put 3 figures in the part which was left. In the area inside figures L, it is possible to put small elements. So, it is possible to cut 27 pairs of elements. 210

212 Tasks 1-2 categorisation Mathematical subfield Formulae/data provided Space and shape Trigonometry and geometry Performing simple operations Understanding basic properties of shapes Level 2 ALL Task 3 categorisation Space and shape Mathematical subfield Units of measurement and scales Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided 1 µm = mm Performing simple operations Estimating areas Level 2 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Task 4 categorisation Space and shape Mathematical subfield Units of measurement and scales Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided 1 µm = mm Performing simple operations Estimating areas Level 3 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Task 5 categorisation Space and shape Mathematical subfield Units of measurement and scales Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided 1 µm = mm Performing simple operations Estimating areas Level 2 Mechanical Technician, Mechanical Engineer, Mechatronics Technician, Mechatronics Engineer Tasks 6-10 categorisation Space and shape Mathematical subfield Units of measurement and scales 2D-3D shapes and properties Application of numerical skills to solve geometrical problems Formulae/data provided V d Volume of cylinder of diameter d: V d = π ( d 2 )2 l Weight in relation to density ρ and volume V: w = ρ V Performing simple operations Calculating measures of weight, volume and time Comparing quantities Costing a project Level 2 Sales manager 211

213 Tasks categorisation Space and shape Mathematical subfields 2D-3D shapes and properties Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided - Estimating areas Level 2 ALL Tasks categorisation Space and shape Mathematical subfields 2D-3D shapes and properties Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided - Estimating areas Level 3 ALL Tasks categorisation Space and shape Mathematical subfield Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided Performing simple operations Estimating areas Costing a project Level 1 Buyer, Estimator, Sales Engineer, Technical Support, Designer, CAD Operator, Laser Operator, Welder Tasks categorisation Space and shape Mathematical subfield Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided Performing simple operations Estimating areas Costing a project Level 2 Buyer, Estimator, Sales Engineer, Technical Support, Designer, CAD Operator, Laser Operator, Welder Task 23 categorisation Space and shape Mathematical subfield 2D-3D shapes and properties Spatial visualisation Application of numerical skills to solve geometrical problems Formulae/data provided Performing simple operations Estimating areas Costing a project Level 3 Buyer, Estimator, Sales Engineer, Technical Support, Designer, CAD Operator, Laser Operator, Welder 212

214 Part 4: Uncertainty and data 213

215 L16: Probability Theory In the mathematical field of probability the chance of an event happening is assigned a numerical value that predicts how likely that event is to occur. Example When we toss a coin, we don t know whether the coin will land with the head side or the tail side. However, heads and tails have equal chances of happening whenever we toss a fair coin. We can describe this situation by saying that the probability of heads is P(heads) = P(H) = 1 2 and the probability of tails is P(tails) = P(T) = 1 2. Empirical probability may be defined as the most accurate scientific estimate, based on a large number of trials, of the cumulative relative frequency of an event happening. The cumulative relative frequency of an event can be found by dividing the total number of occurrences of the event by the total number of trials. Note: The probability of an event is usually written as a fraction. A standard calculator display, however, is in decimal form. Therefore, if you use a calculator when working with probability, it is helpful to know fraction-decimal equivalents. Example: The probability of getting a 4 in rolling a fair die is: P(4) = 1 6. Theoretical probability An outcome is a result of some activity or experiment. A sample space is a set of all possible outcomes for the activity. An event is a subset of the sample space. The theoretical probability of an event is the number of ways that the event can occur, divided by the total number of possibilities in the sample space. This is represented as P(E) = n(e) n(s) where: P(E) represents the probability of event E; n(e) represents the number of outcomes in event E; n(s) represents the number of outcomes in sample space S. Example: The probability of getting a number greater than 4 while rolling a die is P(x > 4) = 2 6 = 1 3. Random selection We are making a random selection when we select an object from a collection of objects without knowing any of the special characteristics of the object. Impossible events, certain events and probability of any event The probability of an impossible event is

216 The probability of an event that is certain to occur is 1. The probability of any event E must be equal to or greater than 0, and less than or equal to 1: 0 P(E) 1. Probability of A and B Event (A and B) consists of the outcomes that are in event A and in event B. Event (A and B) may be regarded as the intersection of sets, namely A B. Example: The probability of obtaining a number that is greater than 3 and less than 6 when rolling a fair die once is P(3 < x < 6) = P(A B) = n(a B) = 2 = 1. n(s) 6 3 Probability of A or B Event (A or B) consists of the outcomes that are in event A or in event B. Event (A or B) may be regarded as the union of sets, namely A B and is equal to: P(A B) = P(A) + P(B) P(A B) Example: A standard deck of 52 cards is shuffled, and one card is drawn at random. The probability that the card is a queen or an ace can be found by the following procedure: There are four queens in the deck, so P(queen) = There are four aces in the deck, so P(ace) = These are mutually exclusive events. The set of queens and the set of aces are disjoint sets, having no elements in common. Thus P(queen or ace) = P(queen) + P(ace) = = 8 52 =

217 Tasks Task 1 The English alphabet contains 26 letters. There are 5 vowels (A, E, I, O, U). The other 21 letters are consonants. If a person turns 2 6 tiles from a word game face down and each tile represents a different letter of the alphabet, what is the probability of turning over: a. the D? b. a vowel? c. a consonant? Task 2 A standard deck of cards contains 52 cards. There are four suits: hearts, diamonds, spades, and clubs. Each suit contains 13 cards: 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, and ace. The diamonds and hearts are red; the spades and clubs are black. In selecting a card from the deck without looking, find the probability of drawing: a. the 9 of hearts b. a 9 c. a heart Task 3 What is the probability of getting a number less than 4 while rolling a die? Task 4 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. In a company which produces series of elements 3 welders were working. It was calculated that John did 34% of the elements, Peter did 40% of the elements and Danny did 26% (he was a beginner). From the elements produced by these workers, 5%, 2% and 10% did not pass the quality control. One defective element was chosen. What is the probability that the element was cut by Peter and what is the probability that it was cut by Danny? 216

218 SOLUTIONS AND ANSWERS Task 1 Answer: a. P(D) = 1 26 Task 2 b. P(vowel) = 5 26 c. P(consonant) = Answer: a. P(9 of hearts) = 1 52 Task 3 b. P(9) = 4 52 c. P(heart) = = 1 4 Answer: P(x < 4) = 3 6 = 1 2. Task 4 B J the event that the element cut by John was chosen B P the event that the element cut by Peter was chosen B D the event that the element cut by Danny was chosen 1) B J B P B D = Ω 2) B J, B P, B D are pairwise disjoint 3) P(B J ), P(B P ), P(B D ) > 0 The conditions of total probability are fulfilled. A the event that a defective element was chosen. P(B J ) = 0.34 P(B P ) = 0.4 P(B D ) = 0.26 P(A B J ) = 0.05 P(A B P ) = 0.02 P(A B D ) = 0.1 P(A) = = = P(B P A) = ( ) /0.051 = 0.008/0.051 = 8/51 P(B D A) = ( )/0.051 = 0.026/0.051 = 26/

219 Tasks 1-3 categorisation Mathematical subfield Formulae/data provided Uncertainty and data Probability Performing simple operations Using formulas Level 1 ALL Task 4 categorisation Uncertainty and data Mathematical subfield Probability Comparison of different datasets Formulae/data provided Performing simple operations Estimating areas Costing a project Level 3 Buyer, Estimator, Sales Engineer, Technical Support 218

220 QUIZ 1. The probability of getting a number larger than 4 while rolling a die is: a. 1/2 b. 1/3 c. 1/4 d. None of the above 2. A standard deck of cards contains 52 cards. In selecting a card from the deck without looking, what is the probability of drawing a 10? a. 10/52 b. 1/52 c. 1/13 d. None of the above 3. Which one of the following statements is false? a. There are some events whose probability is 100%. b. There are some events whose probability is zero. c. There are some events whose probability is over 100%. 4. A standard deck of cards contains 52 cards. In selecting a card from the deck without looking, what is the probability of drawing a 10? a. 10/52 b. 1/52 c. 1/13 d. None of the above 5. The probability of getting a number larger than 2 and less than 6 while rolling a die is: a. 5/6 b. 2/3 c. 1/2 d. None of the above Answers: 1d, 2c, 3c, 4c, 5c 219

221 L17: Introduction to types of data Theory Statistics is the study of numerical data. There are three main steps in a statistical study: collection of data; organization of data into tables, charts, and graphs; drawing conclusions from an analysis of these data. Data can be either qualitative or quantitative. Quantitative data can be either discrete or continuous. Discrete Data can only take certain values, while continuous data can take any value, within a particular range. Examples A hotel owner may ask the customers on their opinion on the accommodation; the opinion can vary from very poor to excellent. This type of data is qualitative. A shop owner may measure the number of customers that entered her shop; the data collected in this case is quantitative and discrete. The police may measure the speed of cars in a particular point in a motorway; the data collected in this case is quantitative and continuous. 220

222 Tasks Task 1 For each of the following data, write if they are qualitative or quantitative and, in case they are quantitative, if they are discrete or continuous. a. A student is making a research on her fellow-students favourite music bands. b. The weights of the new-born babies in the last year in a particular hospital. c. The number of voters who voted for a particular party. 221

223 SOLUTIONS AND ANSWERS Task 1 Answer: a. Qualitative b. Quantitative and continuous c. Quantitative and discrete 222

224 Task 1 categorisation Uncertainty and data Mathematical subfield Introduction to types of data Formulae/data provided Categorising data Level 1 ALL 223

225 QUIZ 1. From the following data, which one is not quantitative? a. Number of people who voted for a political party b. Time needed to reach a destination by different transport means c. Height of the students in a training course d. Colour of eyes of students in a training course 2. Which of the following quantitative data is continuous? a. Number of people who voted for a political party b. Weight of new born babies during a year in a clinic c. Number of courses offered in a training course d. The population of a city during the last 10 years 3. One thousand students were asked what their favourite course is. Is it true or false that the different courses offered constitute qualitative data? a. True b. False 4. Is it true or false that continuous quantitative data can take any value within a specified range? a. True b. False 5. Is it true or false that from a particular population you can acquire only one type of data either qualitative or quantitative? a. True b. False Answers: 1d, 2b, 3a, 4a, 5b 224

226 L18: Lists and tables Theory When we have a large number of data, it is usually convenient to organize these into groups or intervals. In order to do this, you have to follow some rules: 1. The intervals must cover the complete range of values. 2. The intervals must be equal in size. 3. The number of intervals should generally be between 5 and 15. We usually use a large number of intervals only when we have a very large set of data. 4. Every data value must fall into one and only one interval. When an interval ends with a counting number, the following interval begins with the next counting number. 5. The intervals must be listed in order, either highest to lowest or lowest to highest. Example: An adult educator had marked a set of 32 test papers. The scores achieved by the trainees were the following: 90, 85, 74, 86, 65, 62, 100, 95, 77, 82, 50, 83, 77, 93, 73, 72, 98, 66, 45, 100, 50, 89, 78, 70, 75, 95, 80, 78, 83, 81, 72, 75 The above data can be grouped in intervals as shown in the following table: Interval Frequency The data included in the table above can be also visually represented in a histogram. 225

227 Tasks Task 1 Connect the given Brinell hardness numbers to the material. Material Softwood (e.g. pine) Copper pure Lead Mild steel Hardwood Pure Aluminium Glass Rhenium diboride Hardness 4600 HB 5.0 HB 15 HB HB 1550 HB 35 HB 1.6 HB 120 HB Check your solution in any sources (internet, handbooks) Task 2 Ten measurements with the Brinell hardness test were done. The achieved results are collected in the table below. Calculate the average and the median. No result Task 3 added to steel to make it more corrosion resistant. Calculate the percentage of Ferrous in grade 316 steel, if the composition of the materials is given in the table below. Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron??? 226

228 Task 4 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. Draw a pie chart of the grade 316 steel, if the composition of the materials is given in the table below. Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% Task 5 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. In the pie chart of the grade 316 steel shown, fill in the missing percentage. Composition of Grade 316 Stainless Steel 0,08% 60,87% 2,00% 1,00% 0,05% 3,00%??? 13,00% 3,00% C - Carbon Mn - Manganese Si - Silicon P - Phosphorus S - Sulphur Cr - Chromiun Ni - Nickel Mo - Molybdenum Fe - Ferrous - Iron Task 6 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. Which of the following statements, which are based on the pie chart shown are correct? 227

229 Composition of Grade 316 Stainless Steel 0,08% 60,87% 2,00% 1,00% 0,05% 3,00% 17,00% 13,00% 3,00% C - Carbon Mn - Manganese Si - Silicon P - Phosphorus S - Sulphur Cr - Chromiun Ni - Nickel Mo - Molybdenum Fe - Ferrous - Iron Task 7 A. The weight of Sulphur is the same with the weight of Molybdenum. B. The weight of Molybdenum is half the weight of Manganese. C. Chromium is the second element concerning its weight in Grade 316 stainless steel. D. If I double the weight of Molybdenum, I will have to triple the weight of Silicon in order to preserve the properties of Grade 316 stainless steel. Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. Calculate the amount of Ferrous, Carbon and Phosphorus in grade 316 steel of the weight of 100 kg, if the composition of the materials is given in the table below. Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% Task 8 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. A piece of grade 316 steel contains 4.5 g of Sulphur. How many g of Chromium does it contain? The composition of the materials is given in the table below. 228

230 Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% Task 9 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. To 100 kg of grade 316 steel we add 20 kg of Ferrous. What are the weights of the other ingredients that have to be added to in order for the material to not lose its original properties? The composition of the materials is given in the table below. Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% Task 10 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. The following table shows the weights of the chemical composition of a piece of metal weighing 145 kg: Chemical composition Weight (kg) Carbon 0.12 Manganese 2.90 Silicon 1.45 Phosphorus 0.07 Sulphur 4.35 Chromiun Nickel Molybdenum 4.35 Ferrous - Iron

231 Does the above composition refer to that of the grade 316 steel, which is given in the table below? Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% Task 11 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. The table below shows the weights of the chemical compositions of five pieces of metal with different weights. Which of them have the same composition with the grade 316 stainless steel? Chemical Weight (kg) composition Metal 1 Metal 2 Metal 3 Metal 4 Carbon Manganese Silicon Phosphorus Sulphur Chromiun Nickel Molybdenum Ferrous - Iron Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% 230

232 Task 12 Gears/bearings on shafts can sometimes wear or get damaged. This can cause vibration in motors, gearboxes etc. which can subsequently lower the efficiency and ultimately cause costly damage to the assembly. To avoid this, we need to check the run-out of the gear/shaft is within the permitted limits as stated by the manufacturer. The photograph shows a DTI test to check the run-out or concentricity of the mounted bearing/gear assembly. DTI Test Typical shaft, bearing, gear assembly What you have to do is measure 12 marked points on the gear using the Dial Test Indicator (DTI). This will measure the concentricity of the shaft rotation. Then fill in the table of results shown and plot a graph. Are the results within the manufacturer s permitted upper and lower tolerance of +/-0.2mm (indicate the BS notation for the tolerance)? Position Reading DTI 231

233 SOLUTIONS AND ANSWERS Task 2 Answer: Average: Median: Task 3 Answer: To calculate the % of Ferrous we add all the other values together = 39.13% then 100% 39.13% = 60.87% Task 4 To draw a pie chart manually calculate the angle by using 360 x the figure/100 i.e. 360 x 60.87/ 100 = 219, 360 x 17/ 100 = 61 and so forth Note: 0.08 x 360/100 = 0.29 this can be shown by a line due to its very small proportion By using Microsoft Excel you can get the following pie chart: Composition of Grade 316 Stainless Steel 0,08% 2,00% 1,00% 0,05% 3,00% 17,00% 13,00% 60,87% 3,00% C - Carbon Mn - Manganese Si - Silicon P - Phosphorus S - Sulphur Cr - Chromiun Ni - Nickel Mo - Molybdenum Fe - Ferrous - Iron Task 5 Answer: 17% Task 6 Answer: A and C are correct. 232

234 Task 7 Answer: Ferrous: kg Carbon: 8 kg Phosphorus: 5 kg Task 8 Answer: g of Chromium Task 9 Answer: Chemical composition Weight Added (kg) Carbon 0.03 Manganese 0.66 Silicon 0.33 Phosphorus 0.02 Sulphur 0.99 Chromiun 5.59 Nickel 4.27 Molybdenum 0.99 Ferrous - Iron Task 10 Answer: Yes. Task 11 Metals 1 and 3 have the same composition with the grade 316 stainless steel. In Metal 2 the weight of Sulphur is 6 kg instead of 9 kg and in Metal 4 the weight of Manganese is 6 kg instead of 16 kg, and the weight of Ferrous is kg instead of kg. Task 12 Below you see some possible results and the corresponding graph. They both show that the results are within the manufacturer s permitted upper and lower tolerance of +/-0.2mm. Position Reading DTI

235 DTI measurement 0,18 0,16 0,14 0,12 0,1 0,08 0,06 0,04 0,

236 QUIZ 1. If you have a big amount of data, it is better to organise them in a table than in a list. a. True b. False 2. The following numbers refer to the lifespan, in hours, of 50 flashlight batteries: 73, 81, 92, 80, 108, 76, 84, 102, 58, 72, 82, 100, 70, 72, 95, 105, 75, 84, 101, 62, 63, 104, 97, 85, 106, 72, 57, 85, 82, 90, 54, 75, 80, 52, 87, 91, 85, 103, 78, 79, 91, 70, 88, 73, 67, 101, 96, 84, 53, 86. In order to group them, you need to create a table with intervals. If the first interval is 50-59, how many equal-length intervals will be needed in total? a. 5 b. 6 c. 7 d The following data consist of the weights, in kilograms, of a group of 30 students: 70, 43, 48, 72, 53, 81, 76, 54, 58, 64, 51, 53, 75, 62, 84, 67, 72, 80, 88, 65, 60, 43, 53, 42, 57, 61, 55, 75, 82, 71. In order to group them, you need to create a table with intervals. If the first interval is 40-49, how many equal-length intervals will be needed in total? a. 5 b. 6 c. 7 d Based on the table you have created in the previous question, what is the highest frequency that an interval has? a. 6 b. 7 c. 8 d. None of the above 5. Based on the table you have created in the question 3, how many students weigh less than 70 kilograms? a. Less than 16 b. 16 c. 17 d. 18 Answers: 1a, 2b, 3a, 4c, 5d 235

237 L19: Graphs and charts Theory Histograms A histogram is a vertical bar graph in which each interval is represented by the width of the bar and the frequency of the interval is represented by the height of the bar. Example: The following table shows the ages of the trainees in an adult training centre. Age Number of trainees The following histogram represents the above data: 236

238 Example: An adult educator had marked a set of 32 test papers. The scores achieved by the trainees are presented in the following table: Interval Frequency The following histogram represents the above data: 237

239 Tasks Task 1 The following graph shows the ages of the trainees in an adult training centre. Which one of the following statements is true? Task 2 A. Most trainees are younger than 27. B. Most trainees are younger than 25. C. Most trainees are 26 or older. D. None of the above is true. Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. Calculate the percentage of Ferrous in grade 316 steel, if the composition of the materials is given in the table below. Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron??? Task 3 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. Draw a pie chart of the grade 316 steel, if the composition of the materials is given in the table below. 238

240 Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% Task 4 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. In the pie chart of the grade 316 steel shown, fill in the missing percentage. Composition of Grade 316 Stainless Steel 0,08% 2,00% 1,00% 0,05% 3,00%??? 13,00% 60,87% 3,00% C - Carbon Mn - Manganese Si - Silicon P - Phosphorus S - Sulphur Cr - Chromiun Ni - Nickel Mo - Molybdenum Fe - Ferrous - Iron Task 5 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. Which of the following statements, which are based on the pie chart shown are correct? 239

241 Composition of Grade 316 Stainless Steel 0,08% 2,00% 1,00% 0,05% 3,00% 17,00% 13,00% 60,87% 3,00% C - Carbon Mn - Manganese Si - Silicon P - Phosphorus S - Sulphur Cr - Chromiun Ni - Nickel Mo - Molybdenum Fe - Ferrous - Iron Task 6 A. The weight of Sulphur is the same with the weight of Molybdenum. B. The weight of Molybdenum is half the weight of Manganese. C. Chromium is the second element concerning its weight in Grade 316 stainless steel. D. If I double the weight of Molybdenum, I will have to triple the weight of Silicon in order to preserve the properties of Grade 316 stainless steel. Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. Calculate the amount of Ferrous, Carbon and Phosphorus in grade 316 steel of the weight of 100 kg, if the composition of the materials is given in the table below. Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% Task 7 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. A piece of grade 316 steel contains 4.5 g of Sulphur. How many g of Chromium does it contain? The composition of the materials is given in the table below. Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% 240

242 Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% Task 8 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. To 100 kg of grade 316 steel we add 20 kg of Ferrous. What are the weights of the other ingredients that have to be added to in order for the material to not lose its original properties? The composition of the materials is given in the table below. Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% Task 9 An engineer performs an inspection and uses a Vernier calliper to measure a batch of manufactured rods that have been cut to length from a bar on a band saw. The nominal length required is 100mm. The results are varied and s/he needs to check that the permitted tolerance of +/- 2mm is not exceeded. The engineer inspects and measures ten rods as they are produced on the band saw every 1min, and finds that the lengths are:

243 Draw a simple SPC (Statistical Process Control) graph to show the variation in lengths. Then answer the following questions: Task 10 What is the median length? What is the mode? What is the range? What is the mean length? An engineer performs an inspection and uses a Vernier calliper to measure a batch of manufactured rods that have been cut to length from a bar on a band saw. The nominal length required is 100mm. The results are varied and s/he needs to check that the permitted tolerance of +/- 2mm is not exceeded. The engineer inspects and measures ten rods as they are produced on the band saw every 1min, and finds that the lengths are: Answer the following questions: Task 11 Is the batch of products within the accepted tolerance of +/-2mm? What is the trend? Refer to the standard deviation. What could this mean to the manufacture? An engineer performs an inspection and uses a Vernier calliper to measure a batch of manufactured rods that have been cut to length from a bar on a band saw. 242

244 DIMENSIONS (MM) Statistical Process Control Graph 102,00 101,50 101,00 100,50 100,00 99,50 99,00 98,50 98, ROD The nominal length required is 100mm. The results are varied and s/he needs to check that the permitted tolerance of +/- 2mm is not exceeded. The results of measuring 30 rods are shown in the diagram above: A) Estimate the following: the median length the mode the range the mean length. B) Answer the following questions: Task 12 Is the batch of products within the accepted tolerance of +/-2mm? Has the accuracy of production improved compared to the previous results? What is the trend? Prepare a short report for the engineer to give to his supervisor, including any relevant numerical data and measures. An engineer performs an inspection and uses a Vernier calliper to measure a batch of manufactured rods that have been cut to length from a bar on a band saw. The nominal length required is 100mm. The results are varied and s/he needs to check that the permitted tolerance of +/- 2mm is not exceeded. After measuring 30 rods the engineer proposed some improvements in the production line, in order to increase the accuracy of 243

245 the length of the rods. The table below includes the measurements taken before the implementation of the improvements (cells 1 to 30) and after (cells 31-60). Do you see any improvement in the accuracy? Justify your answer by using mathematical concepts Hint: You may draw a diagram and also calculate the standard deviation for the two groups of data. You may also group the data. Task 13 Gears/bearings on shafts can sometimes wear or get damaged. This can cause vibration in motors, gearboxes etc. which can subsequently lower the efficiency and ultimately cause costly damage to the assembly. To avoid this, we need to check the run-out of the gear/shaft is within the permitted limits as stated by the manufacturer. The photograph shows a DTI test to check the run-out or concentricity of the mounted bearing/gear assembly. DTI Test 244

246 Typical shaft, bearing, gear assembly What you have to do is measure 12 marked points on the gear using the Dial Test Indicator (DTI). This will measure the concentricity of the shaft rotation. Then fill in the table of results shown and plot a graph. Are the results within the manufacturer s permitted upper and lower tolerance of +/-0.2mm (indicate the BS notation for the tolerance)? Position Reading DTI 245

247 SOLUTIONS AND ANSWERS Task 1 Answer: A is true. Task 2 Answer: To calculate the % of Ferrous we add all the other values together = 39.13% then 100% 39.13% = 60.87% Task 3 To draw a pie chart manually calculate the angle by using 360 x the figure/100 i.e. 360 x 60.87/ 100 = 219, 360 x 17/ 100 = 61 and so forth Note: 0.08 x 360/100 = 0.29 this can be shown by a line due to its very small proportion By using Microsoft Excel you can get the following pie chart: Composition of Grade 316 Stainless Steel 0,08% 2,00% 1,00% 0,05% 3,00% 17,00% 13,00% 60,87% 3,00% C - Carbon Mn - Manganese Si - Silicon P - Phosphorus S - Sulphur Cr - Chromiun Ni - Nickel Mo - Molybdenum Fe - Ferrous - Iron Task 4 Answer: 17% Task 5 Answer: A and C are correct. 246

248 DIMENSIONS (MM) Task 6 Answer: Ferrous: kg Carbon: 8 kg Phosphorus: 5 kg Task 7 Answer: g of Chromium Task 8 Answer: Chemical composition Weight Added (kg) Carbon 0.03 Manganese 0.66 Silicon 0.33 Phosphorus 0.02 Sulphur 0.99 Chromiun 5.59 Nickel 4.27 Molybdenum 0.99 Ferrous - Iron Task 9 102,00 101,50 101,00 100,50 Statistical Process Control Graph 100,00 99,50 99, ROD 98,50 98,00 247

249 Median (middle number): 99.5, , , , , , , , , = /2 =100.55mm Mode = mm Range = = 2mm Mean = /10 = mm Task 10 The batch of products is within the accepted tolerance. The trend is increasing. Standard deviation: σ = Task 11 A) Median: mm Mode = mm Range = = 2mm Mean = B) Note: If needed the following table of data can be provided: 99, , ,4 100, , , ,76 99, ,5 101,5 100,2 100,3 100, ,2 100,1 100,4 100,2 100,7 100,2 Task 13 Below you see some possible results. They show that they are within the manufacturer s permitted upper and lower tolerance of +/-0.2mm. 248

250 Position Reading DTI DTI wyniki 0,18 0,16 0,14 0,12 0,1 0,08 0,06 0,04 0,

251 Task 1 categorisation Mathematical subfield Formulae/data provided Level 1 ALL Uncertainty and data Graphs and charts Classifying data Creating and interpreting graphs Task 2 categorisation Uncertainty and data Mathematical subfield Lists and tables Formulae/data provided Performing simple operations Level 1 Metallurgist, Designer, Test Engineer, Quality Assessor Tasks 3-4 categorisation Mathematical subfield Formulae/data provided Uncertainty and data Lists and tables Graphs and charts Performing simple operations Reading tables Creating and interpreting graphs Level 1 Metallurgist, Designer, Test Engineer, Quality Assessor Tasks 5 and 7-8 categorisation Mathematical subfield Formulae/data provided Uncertainty and data Lists and tables Graphs and charts Performing simple operations Reading tables Creating and interpreting graphs Level 2 Metallurgist, Designer, Test Engineer, Quality Assessor Task 6 categorisation Mathematical subfield Formulae/data provided Uncertainty and data Lists and tables Graphs and charts Performing simple operations Reading tables Creating and interpreting graphs Level 1 Metallurgist, Designer, Test Engineer, Quality Assessor Task 9 categorisation Mathematical subfield Formulae/data provided Uncertainty and data Graphs and charts Measures of average Median: put the numbers in order to find the middle number, Mode = Most often number, Range = Largest Smallest, Mean = the sum of numbers added together/number of numbers, Standard deviation: 250

252 N σ = 1 N (x i μ) 2 Reading tables Grouping data Calculating statistical measures Using formulas Level 1 Quality Control, Inspection, Machine Operator Task 10 categorisation Uncertainty and data Mathematical subfield Graphs and charts, measures of average Formulae/data provided Median: put the numbers in order to find the middle number, Mode = Most often number, Range = Largest Smallest, Mean = the sum of numbers added together/number of numbers, Standard deviation: i=1 N σ = 1 N (x i μ) 2 Reading tables Grouping data Calculating statistical measures Using formulas Level 2 Quality Control, Inspection, Machine Operator Task 11 categorisation Uncertainty and data Mathematical subfield Graphs and charts Measures of average Formulae/data provided Median: put the numbers in order to find the middle number, Mode = Most often number, Range = Largest Smallest, Mean = the sum of numbers added together/number of numbers, Standard deviation: i=1 N σ = 1 N (x i μ) 2 Reading and interpreting diagrams Grouping data Calculating statistical measures Using formulas Level 2 Quality Control, Inspection, Machine Operator Task 12 categorisation Uncertainty and data Mathematical subfield Graphs and charts Measures of average Formulae/data provided Median: put the numbers in order to find the middle number, Mode = Most often number, Range = Largest Smallest, Mean = the sum of numbers added together/number of numbers, Standard deviation: i=1 251

253 N σ = 1 N (x i μ) 2 Reading and interpreting diagrams Grouping data Calculating statistical measures Using formulas Level 3 Quality Control, Inspection, Machine Operator Task 13 categorisation Uncertainty and data Mathematical subfield Lists and tables Graphs and charts Formulae/data provided - Using graphs and diagrams Level 1 Maintenance Fitter, Test Engineer, Quality Manager i=1 252

254 QUIZ 1. The histogram below represents the of the test scores of some students: 12 Test scores How many students had an excellent mark (from 91 to 100)? a. 8 b. 10 c. Over 10 d. Cannot tell by just looking at the histogram 2. Based on the histogram of question 1, how many students had scores less than 61? a. Less than 5 b. 5 c. 6 d. None of the above 3. Based on the histogram of question 1, how many students in total took the test? a. Less than 30 b. 30 c. 36 d Based on the histogram of question 1, is it correct to say that the majority of students did it quite well in the test, since their scores were over 70? a. Correct b. Incorrect 5. Based on the histogram of question 1, is it correct to say that the number of students who had a score over 80 is double than those who had 80 or less? a. Correct b. Incorrect Answers: 1a, 2b, 3d, 4a, 5b 253

255 L20: Measures of average Theory Average (mean) The arithmetic average is a measure of central tendency and is also called the arithmetic mean or the numerical average. It is calculated by adding the numbers and dividing the sum by n. The symbol used for the mean is x. Example: The results of a test taken by five students are the following: 93, 80, 86, 72, and 94. In order to find the mean we firstly add the five data values: = 425. Then we divide this sum by five: 425 : 5 = 85. Median The median is another measure of central tendency and is the middle value for a set of data arranged in numerical order. In order to find the median of a set of n numbers we need to do the following: 1. Arrange the numbers in numerical order. 2. If n is odd, find the middle number. This number is the median. 3. If n is even, find the mean (arithmetic average) of the two middle numbers. This average is the median. Example: The ages of seven employees of a company are the following: 17, 19, 20, 17, 46, 17, and 18. In order to find the median we firstly have to arrange the ages in numerical order: 17, 17, 17, 18, 19, 20, 46. Then we need to find the middle number: 17, 17, 17, 18, 19, 20, 46, which is 18. Mode The mode is another measure of central tendency and is the data value that appears most often in a given set of data. In order to find the mode for a set of data we need to do the following: 1. If one number appears most often in the data, that number is the mode. 2. If two or more numbers appear more often than all other data values, and these numbers appear with the same frequency, then each of these numbers is a mode. 3. If each number in a set of data occurs with the same frequency, there is no mode. Example: The mode for the following data: 2, 9, 3, 7, 3 is the most frequent value, which is

256 Standard deviation The standard deviation is a measure of dispersion that tells you how tightly all data are clustered around the mean. It is the square root of the variance, which is the average of the squared differences from the mean. It is calculated by the following formulas: 1. For a population of size N, which has the average μ: N σ = 1 N (x i μ) 2 i=1 2. For a sample of size n, which has the average x Example: s = 1 n n 1 (x i x ) 2 The number of job applications received weekly by a company for a period of six weeks were the following: 17, 15, 23, 7, 9, 13. In order to calculate the standard deviation we firstly have to calculate the mean: x = 6 i=1 x i 6 = 84 6 standard deviation. i=1 = 14. Then s = (x i 14) 2 6 i=1 = , which is a high 255

257 Tasks Task 1 A student has the marks of 75, 82, and 90 on three tests. What mark must he obtain on the next test to have an average of exactly 85 for the four tests? Task 2 A student has had the following marks of 75, 74, and 73 on three tests. He has one more test to take, and the best mark he can get in it is 100. Is it possible to get such a mark in the fourth test, so that his average in the four tests will be 85? Task 3 Find the mode of the following data: 3, 4, 5, 4, 3, 7, 2. Task 4 Calculate the standard deviation of the following set of data, which are a sample of some trainees results in a test: 76, 68, 89, 94, 90, 77. Task 5 Ten measurements with the Brinell hardness test were done. The achieved results are collected in the table below. Calculate the average and the median. No Result Task 6 An engineer performs an inspection and uses a Vernier calliper to measure a batch of manufactured rods that have been cut to length from a bar on a band saw. The nominal length required is 100mm. The results are varied and s/he needs to check that the permitted tolerance of +/- 2mm is not exceeded. The engineer inspects and measures ten rods as they are produced on the band saw every 1min, and finds that the lengths are:

258 Draw a simple SPC (Statistical Process Control) graph to show the variation in lengths. Then answer the following questions: Task 7 What is the median length? What is the mode? What is the range? What is the mean length? An engineer performs an inspection and uses a Vernier calliper to measure a batch of manufactured rods that have been cut to length from a bar on a band saw. The nominal length required is 100mm. The results are varied and s/he needs to check that the permitted tolerance of +/- 2mm is not exceeded. The engineer inspects and measures ten rods as they are produced on the band saw every 1min, and finds that the lengths are: Answer the following questions: Is the batch of products within the accepted tolerance of +/-2mm? What is the trend? Refer to the standard deviation. What could this mean to the manufacture? Task 8 An engineer performs an inspection and uses a Vernier calliper to measure a batch of manufactured rods that have been cut to length from a bar on a band saw. 257

259 DŁUGOŚĆ (MM) Statistical Process Control Graph 102,00 101,50 101,00 100,50 100,00 99,50 99,00 98,50 98, PRĘT The nominal length required is 100mm. The results are varied and s/he needs to check that the permitted tolerance of +/- 2mm is not exceeded. The results of measuring 30 rods are shown in the diagram above: A) Estimate the following: the median length the mode the range the mean length. B) Answer the following questions: Is the batch of products within the accepted tolerance of +/-2mm? Has the accuracy of production improved compared to the previous results? What is the trend? Prepare a short report for the engineer to give to his supervisor, including any relevant numerical data and measures. Task 9 An engineer performs an inspection and uses a Vernier calliper to measure a batch of manufactured rods that have been cut to length from a bar on a band saw. The nominal length required is 100mm. The results are varied and s/he needs to check that the permitted tolerance of +/- 2mm is not exceeded. After measuring 30 rods the engineer proposed some improvements in the production line, in order to increase the accuracy of 258

260 the length of the rods. The table below includes the measurements taken before the implementation of the improvements (cells 1 to 30) and after (cells 31-60). Do you see any improvement in the accuracy? Justify your answer by using mathematical concepts Hint: You may draw a diagram and also calculate the standard deviation for the two groups of data. You may also group the data. 259

261 SOLUTIONS AND ANSWERS Task 1 Let x be the student s mark on the fourth test. The sum of the four test marks divided by 4 should be 85: x 4 Solving for x gives us x=93. Task 2 Let x be the student s mark on the fourth test. The sum of the four test marks divided by 4 should be 85: x 4 = 85. = 85. Solving for x gives us x=118, thus it is impossible to him to get a mark in the fourth test that will make his average 85 (since the maximum he can get is 100). Task 3 We may arrange the data in numerical order: 2, 3, 3, 4, 4, 5, 7. Then we observe that both 3 and 4 appear twice. Thus, there are two modes, 3 and 4. Task 4 The standard deviation of the sample is equal to: s = 1 6 (x 6 1 i=1 i x ) 2 = 10.13, which is a high standard deviation. Task 5 Answer: Average: Median:

262 DŁUGOŚĆ (MM) Task 6 Statistical Process Control Graph 102,00 101,50 101,00 100,50 100,00 99,50 99, PRĘT 98,50 98,00 Task 7 Median (middle number): 99.5, , , , , , , , , = /2 =100.55mm Mode = mm Range = = 2mm Mean = /10 = mm The batch of products is within the accepted tolerance. The trend is increasing. Standard deviation: σ = Task 8 A) Median: mm Mode = mm Range = = 2mm Mean = B) Note: If needed the following table of data can be provided:

263

264 Tasks 1-2 categorisation Mathematical subfield Formulae/data provided Level 2 ALL Tasks 3-4 categorisation Mathematical subfield Formulae/data provided Level 1 ALL Uncertainty and data Measures of average Performing simple operations Using formulas Uncertainty and data Measures of average Performing simple operations Using formulas Task 5 categorisation Uncertainty and data Mathematical subfield Lists and tables Measures of average Formulae/data provided 2F HB = πd(d (D 2 d 2 ) where D = Ball Diameter in mm d = impression diameter in mm F = Load in kgf HB = Brinell Hardness No. Performing simple operations Using formulas Level 1 Metallurgist, Designer, Test Engineer, Quality Assessor Task 6 categorisation Mathematical subfield Formulae/data provided Uncertainty and data Graphs and charts Measures of average Median: put the numbers in order to find the middle number, Mode = Most often number, Range = Largest Smallest, Mean = the sum of numbers added together/number of numbers Standard deviation: N σ = 1 N (x i μ) 2 Reading tables Grouping data Calculating statistical measures Using formulas Level 1 Quality Control, Inspection, Machine Operator Task 7 categorisation Uncertainty and data Mathematical subfield Graphs and charts, measures of average Formulae/data provided Median: put the numbers in order to find the middle number, Mode = Most often number, Range = Largest Smallest, Mean = the sum of numbers added together/number of i=1 263

265 numbers Standard deviation: N σ = 1 N (x i μ) 2 Reading tables Grouping data Calculating statistical measures Using formulas Level 2 Quality Control, Inspection, Machine Operator Task 8 categorisation Uncertainty and data Mathematical subfield Graphs and charts Measures of average Formulae/data provided Median: put the numbers in order to find the middle number, Mode = Most often number, Range = Largest Smallest, Mean = the sum of numbers added together/number of numbers, Standard deviation: i=1 N σ = 1 N (x i μ) 2 Reading and interpreting diagrams Grouping data Calculating statistical measures Using formulas Level 2 Quality Control, Inspection, Machine Operator Task 9 categorisation Uncertainty and data Mathematical subfield Graphs and charts Measures of average Formulae/data provided Median: put the numbers in order to find the middle number, Mode = Most often number, Range = Largest Smallest, Mean = the sum of numbers added together/number of numbers, Standard deviation: i=1 N σ = 1 N (x i μ) 2 Reading and interpreting diagrams Grouping data Calculating statistical measures Using formulas Level 3 Quality Control, Inspection, Machine Operator i=1 264

266 QUIZ 1. The ages of seven students participating in a special training course are 17, 19, 20, 17, 46, 17, and 18 years old. Which of the following measures provides the best representation of this data? a. Mean b. Median c. Mode d. All the previous are good enough 2. Which of the following measures is mostly affected by extreme cases included in a dataset? a. Mean b. Median c. Mode d. All the previous are equally affected 3. Is it true or false that a set of numerical data may have two or more numbers as modes? a. True b. False 4. Is it true or false that a set of numerical data may have the same mean and median? a. True b. False 5. The following are the scores of a student in seven different course exams 13, 14, 14, 15, 16, 16, 17. Which of the following is the closest estimation to the standard deviation of these scores? a. 1 b. 2 c. 3 d. 4 Answers: 1b, 2a, 3a, 4a, 5a 265

267 L21: Comparison of different datasets Theory In order to compare different datasets, we can use the measures of central tendency (average, median, mode) and the measures of dispersion (standard deviation). Correlation In statistics we often want to study the relationship between two different sets of values. Although cases involving two-valued or bivariate statistics require complex statistical methods, we can investigate some of the properties of simple cases by looking at graphs. A graph that shows the pairs of values in the data as points in the plane is called a scatter plot. When a given relationship involves two sets of data: Example: In some cases a straight line, a line of best fit, can be drawn to approximate the relationship between the data sets. If a line of best fit has a positive slope, the data has positive linear correlation. If the line of best fit has a negative slope, the data has negative linear correlation. A line of best fit can be drawn through(x, y ), the point whose coordinates are the means of the given data. Any data point that appears to lie on or near the line of best fit can be used as a second point to write the equation. When the graphed data points are so scattered that it is not possible to draw a straight line that approximates the given relationship, the data has no correlation. A driver recorded the number of litres of gasoline used and the number of km driven each time he filled the tank. The results are shown in the table: Gasoline (litres) Distance (km)

268 This data has positive linear correlation. The points in the scatter plot below approximate a straight line that has a positive slope In order to draw the line of best fit, we need to firstly calculate the means of the two datasets; let x be the gasoline in litres and y the distance in kilometers. These are equal to: x = 22.1 and y = The line of best fit is drawn in the diagram in red. Example: A long-distance truck driver travels 800 km each day. As he passes through different areas on his trip, he records his average speed and the length of time he drives each day. The table below shows a record of average speed and time for a 10-day period. Speed (km/h) Time (h) This data has negative linear correlation. The points in the scatter plot approximate a straight line that has a negative slope. 267

269 10,5 10 9,5 9 8,5 8 7,5 7 6, In order to draw the line of best fit, we need to firstly calculate the means of the two datasets; let x be the speed in km/h and y the time in hours. These are equal to: x = 99 and y = 8.22 The line of best fit is drawn in the diagram in red. 268

270 Tasks Task 1 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. The following table shows the weights of the chemical composition of a piece of metal weighing 145 kg: Chemical composition Weight (kg) Carbon 0.12 Manganese 2.90 Silicon 1.45 Phosphorus 0.07 Sulphur 4.35 Chromiun Nickel Molybdenum 4.35 Ferrous - Iron Does the above composition refer to that of the grade 316 steel, which is given in the table below? Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% Task 2 Different compositions are added to materials to change their properties. For instance, Chromium is added to steel to make it more corrosion resistant. The table below shows the weights of the chemical compositions of five pieces of metal with different weights. Which of them have the same composition with the grade 316 stainless steel? Chemical Weight (kg) composition Metal 1 Metal 2 Metal 3 Metal 4 Carbon Manganese Silicon Phosphorus

271 Sulphur Chromiun Nickel Molybdenum Ferrous - Iron Chemical composition Weight Max % Carbon 0.08% Manganese 2.00% Silicon 1.00% Phosphorus 0.05% Sulphur 3.00% Chromiun 17.00% Nickel 13.00% Molybdenum 3.00% Ferrous - Iron 60.87% Task 3 A laser cutter operator has to cut a number of two differently shaped parts out of a piece of sheet metal ready for a welder to join together. In a company which produces series of elements 3 welders were working. It was calculated that John did 34% of the elements, Peter did 40% of the elements and Danny did 26% (he was a beginner). From the elements produced by these workers, 5%, 2% and 10% did not pass the quality control. One defective element was chosen. What is the probability that the element was cut by Peter and what is the probability that it was cut by Danny? 270

272 SOLUTIONS AND ANSWERS Task 1 Answer: Yes. Task 2 Metals 1 and 3 have the same composition with the grade 316 stainless steel. In Metal 2 the weight of Sulphur is 6 kg instead of 9 kg and in Metal 4 the weight of Manganese is 6 kg instead of 16 kg, and the weight of Ferrous is kg instead of kg. Task 3 B J the event that the element cut by John was chosen B P the event that the element cut by Peter was chosen B D the event that the element cut by Danny was chosen 1) B J B P B D = Ω 2) B J, B P, B D are pairwise disjoint 3) P(B J ), P(B P ), P(B D ) > 0 The conditions of total probability are fulfilled. A the event that a defective element was chosen. P(B J ) = 0.34 P(B P ) = 0.4 P(B D ) = 0.26 P(A B J ) = 0.05 P(A B P ) = 0.02 P(A B D ) = 0.1 P(A) = = = P(B P A) = ( ) /0.051 = 0.008/0.051 = 8/51 P(B D A) = ( )/0.051 = 0.026/0.051 = 26/

273 Tasks 1-2 categorisation Mathematical subfield Formulae/data provided Uncertainty and data Lists and tables Comparison of different datasets Performing simple operations Reading tables Creating and interpreting graphs Level 2 Metallurgist, Designer, Test Engineer, Quality Assessor Task 3 categorisation Mathematical subfield Formulae/data provided Uncertainty and data Probability Comparison of different datasets Performing simple operations Estimating areas Costing a project Level 3 Buyer, Estimator, Sales Engineer, Technical Support 272

274 QUIZ 1. The following are the scores of two students in the same seven course exams: student A: 10, 11, 12, 15, 18, 19, 20, and student B: 13, 14, 14, 15, 16, 16, 17. Which of the following measures should be used to better compare these two datasets? a. Mean b. Median c. Standard deviation d. All of the above can be equally used 2. The table shows two sets of data: A B Which of the following best represents the correlation between these sets of data? a. Data A and data B are positive correlated b. Data A and data B are negatively correlated c. There is no correlation between data A and data B d. We cannot say anything about the correlation between data A and data B 3. What does the following scatter plot show about the correlation of the data represented? a. Positive correlation 273

275 b. Negative correlation c. Positive and negative correlation d. No correlation 4. Is it true or false that if the slope of the line of best fit is positive, the data have positive correlation? a. True b. False 5. The table shows two sets of data: A B Which of the following best represents the correlation between these sets of data? a. Data A and data B are positive correlated b. Data A and data B are negatively correlated c. There is no correlation between data A and data B d. We cannot say anything about the correlation between data A and data B Answers: 1c, 2a, 3b, 4a, 5c 274

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