EASA Part 66 CAT B1 Module 1 Mathematics PDF

Revision: 3 MTO Maintenance Training 1. Ausgabedatum: Organisation 15.05.2010 Part – 147 Revisionsdatum: 12....

Revision: 3 MTO Maintenance Training 1. Ausgabedatum: Organisation 15.05.2010 Part – 147 Revisionsdatum: 12.06.2024 Coverpage Module 1 EASA Part 66 CAT B1 MODULE 1 Mathematics AIR-TECH BILDUNG GmbH Flughafen Kiel Boelckestraße 100 24159 Kiel Deutschland Mobile +49(0)174 / 312 235 68 MailTo [email protected] Internet www.airtech-bildung.de  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES Index 1 ARITHMETIC................................................................................. 1-3 1.1 INTRODUCTION...................................................................... 1-3 1.2 ARITHMETIC TERMS............................................................... 1-3 1.3 DIRECTED NUMBERS............................................................... 1-5 1.4 FACTORS............................................................................... 1-7 1.4.1 Prime Numbers...................................................... 1-7 1.4.2 Highest Common Factor (HCF).............................. 1-8 1.4.3 Lowest Common Multiple (LCM)............................ 1-8 1.5 ARITHMETICAL PRECEDENCE.................................................. 1-9 1.5.1 Bodmas Example................................................... 1-9 1.6 FRACTIONS........................................................................... 1-10 1.6.1 Addition of Fractions............................................... 1-10 1.6.2 Subtraction of Fractions......................................... 1-10 1.6.3 Multiplication of Fractions....................................... 1-11 1.6.4 Division of Fractions............................................... 1-11 1.7 DECIMAL FRACTIONS.............................................................. 1-12 1.7.1 Addition & Subtraction............................................ 1-12 1.7.2 Multiplication & Division.......................................... 1-13 1.8 WEIGHTS AND MEASURES...................................................... 1-14 1.9 RATIO AND PROPORTION........................................................ 1-15 1.10 AVERAGES AND PERCENTAGES............................................... 1-16 1.10.1 Averages................................................................ 1-16 1.10.2 Percentage............................................................. 1-17 1.11 POWERS AND ROOTS............................................................. 1-18 1.11.1 Powers................................................................... 1-18 1.11.2 Roots...................................................................... 1-19 2 ALGEBRA..................................................................................... 2-19 2.1 INTRODUCTION...................................................................... 2-19 2.1.1 Operation................................................................ 2-19 2.1.2 Basic Laws............................................................. 2-21 2.2 EQUATIONS........................................................................... 2-22 2.2.1 Solving Linear Equations........................................ 2-22 2.3 TRANSPOSITION IN EQUATIONS............................................... 2-26 2.3.1 Construction of Equations...................................... 2-28 2.4 SIMULTANEOUS EQUATIONS................................................... 2-29 2.5 QUADRATIC EQUATIONS.......................................................... 2-31 Page 1 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 3 NUMBERS..................................................................................... 3-33 3.1 INDICES AND POWERS............................................................ 3-33 3.1.1 Standard Form........................................................ 3-35 3.2 NUMBERING SYSTEMS............................................................ 3-35 3.2.1 Decimal System of Numeration.............................. 3-35 3.2.2 Binary System of Numeration................................. 3-37 3.2.3 Octal System of Numeration................................... 3-38 3.2.4 Conversion to other bases...................................... 3-39 3.3 LOGARITHMS.......................................................................... 3-42 4 GEOMETRY.................................................................................. 4-43 4.1 ANGULAR MEASUREMENT........................................................ 4-43 4.1.1 Angles associated with parallel lines...................... 4-44 4.2 GEOMETRIC CONSTRUCTIONS................................................ 4-45 4.2.1 Triangle................................................................... 4-45 4.2.2 Similar & Congruent Triangles................................ 4-46 4.2.3 Polygon................................................................... 4-46 4.2.4 Quadrilaterals......................................................... 4-47 4.2.5 Parallelogram......................................................... 4-47 4.2.6 Rectangle............................................................... 4-48 4.2.7 Rhombus................................................................ 4-48 4.2.8 Square.................................................................... 4-48 4.2.9 Trapezium............................................................... 4-48 4.2.10 Circles..................................................................... 4-49 4.3 AREA AND VOLUME................................................................ 4-53 4.3.1 Area........................................................................ 4-53 4.3.2 Volumes.................................................................. 4-57 5 GRAPHS........................................................................................ 5-58 5.1 CONSTRUCTION..................................................................... 5-58 5.1.1 Graphs and Mathematical Formulae...................... 5-61 5.1.2 Function and Shape............................................... 5-62 5.2 NOMOGRAPHS....................................................................... 5-65 6 TRIGONOMETRY.......................................................................... 6-67 6.1.1 Trigonometrical Calculations & Formula................. 6-68 6.1.2 Construction of Trigonometrical Curves................. 6-70 6.2 VALUES IN 4 QUADRANTS........................................................ 6-72 7 CO-ORDINATE GEOMETRY........................................................ 7-73 Polar / Rectangular Co-Ordinates....................................... 7-75 Page 2 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 1 ARITHMETIC 1.1 INTRODUCTION Mathematics is the basic language of science and technology. It is an exact language that has a vocabulary and meaning for every term. Since mathematics follows definite rules and behaves in the same way every time, scientists and engineers use it as their basic tool. Long before any metal is cut for a new aircraft design, there are literally millions of mathematical computations made. Aviation maintenance technicians perform their duties with the aid of many different tools. Like the wrench or screwdriver, mathematics is an essential tool in the maintenance, repair and fabrication of replacement parts. With this in mind, you can see why you must be competent in mathematics to an acceptable level. These notes cover the complete mathematics syllabus required to comply with the EASA Part-66 B1 and B2 licence level. Arithmetic is the basic language of all mathematics and uses real, non-negative numbers. These are sometimes known as counting numbers. Only four operations are used, addition, subtraction, multiplication and division. Whilst these operations are well known to you, a review of the terms and operations used will make learning the more difficult mathematical concepts easier. 1.2 ARITHMETIC TERMS The most common system of numbers in use is the decimal system, which uses the ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. These ten whole numbers from zero to 9 are called integers. Above the number nine, the digits are re-used in various combinations to represent larger numbers. This is accomplished by arranging the numbers in columns based on a multiple of ten. With the addition of a minus (-) sign, numbers smaller than zero are indicated. To describe quantities that fall between whole numbers, fractions are used. Common fractions are used when the space between two integers is divided into equal segments, such as quarters. When the space between integers is divided into ten segments, decimal fractions are typically used. Page 1-3 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES Students will be familiar with this system and the basic operations, which may involve Addition, Subtraction, Multiplication and Division. When numbers are added, they form a sum. When numbers are subtracted, they create a difference. When numbers are multiplied, they form a product. When one number (the dividend) is divided by another (the divisor), the result is a quotient. It is useful if a student is proficient at simple mental arithmetic, and this is only possible if one has a “feel” for numbers, and the size of numbers. A knowledge of simple “times tables” is also useful. TIMES TABLE The following simple tests for divisibility may be useful. A number is divisible by: 2 if it is an even number. 3 if the sum of the digits that form the number is divisible by 3. 4 if the last two digits are divisible by 4. 5 If the last digit is 0 or 5. 10 if the last digit is 0 Page 1-4 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 1.3 DIRECTED NUMBERS Directed numbers are numbers which have a + or – sign attached to them. Directed numbers can be added, subtracted, etc. etc, but care should be taken to ensure a correct solution. The following rules should assist. To add several numbers of the same sign, add them together and ensure sign of the sum is the same as the sign of the numbers. To add 2 numbers with different signs, subtract the smaller from the larger. The sign of the resultant (the difference) is the same as the sign of the large number. eg. -12 + 6 = (12 - 6) = 6 ® -6 If there are more than 2 numbers, carry out the operation 2 numbers at a time, or produce two numbers by adding up all the numbers with like signs. And then apply the rules above. eg. -15 - 8 + 13 - 19 + 6 = (-15 - 8) = -23 + 13 = -10 - 19 = -29 + 6 = - 23 or -15 + (-8) + (-19) = -42 and +13 +6 = +19 -42 + 19 = - 23 To subtract directed numbers, change the sign of the number to be subtracted and add the resulting numbers. eg. -10 - (-6) = - 10 + 6 = -4 7 - (+18) = 7 - 8 = -11 A minus in front of brackets should be taken to mean –1. Using the above example –(-6) should be read as –1(-6) i.e. minus 1 times minus six. Similarly, a positive sign in front of brackets should be read as +1, so +(-6) should be read as +1(-6) i.e. plus 1 times minus 6. The product of two numbers with like signs is positive (+ve), the product of numbers with unlike signs is negative (-ve). When dividing numbers with like signs, the quotient of the result is +ve. When dividing numbers with unlike signs, the quotient is –ve. This can be summarised as follows: (+) x (+) = (+) (-) x (+) = (-) (+) x (-) = (-) (-) x (-) = (+) Page 1-5 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES Page 1-6 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 1.4 FACTORS We know that 2 x 6 = 12. 2 and 6 are factors of 12. We could also state that, as 3 x 4 = 12, 3 and 4 are also factors of 12. Similarly 12 and 1. This may seem obvious, but it is sometimes useful to "factorise" a number, i.e. determine the factors that make up the number. More commonly it is necessary to find the factors of an algebraic expression. Example Find the possible factors of 60. (in other words, find the integers that divide into 60). The factors will be: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60 Check them yourself. 1.4.1 PRIME NUMBERS A prime number is a number whose only factors are 1 and itself. The prime numbers between 1 and 30 are: 1, 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29. Check them yourself. It is sometimes useful to express the factors of a given number in terms of prime numbers. For example, let us look at the factors of 60 again, taking 4 and 15 as 2 factors. (4 x 15 = 60), but 4 has factors of 2 and 2, and 15 has factors of 5 and 3. Hence the number 60 can be expressed as 2 x 2 x 3 x 5, which are all factors of 60. Note: we have now factorised the number 60 in terms of prime numbers. Page 1-7 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 1.4.2 HIGHEST COMMON FACTOR (HCF) The highest common factor is the biggest factor (number) that will divide into the numbers being examined. Suppose that we take 3 numbers, 1764, 2100 and 2940. The highest common factor of these numbers is 84. In some instances you will be able to identify this value simply by looking at the numbers, in others you will need to calculate it. To calculate the HCF, we must identify the factors of each number in terms of prime numbers: 2 ´ 2 ´ 3 ´ 3 ´ 7 ´ 7 = 1764 2 ´ 2 ´ 3 ´ 5 ´ 5 ´ 7 = 2100 2 ´ 2 ´ 3 ´ 5 ´ 7 ´ 7 = 2940 We then select the common prime factors and multiply them together to produce the High Common Factor, in this case: 2 ´ 2 ´ 3 ´ 7 = 84 1.4.3 LOWEST COMMON MULTIPLE (LCM) The lowest common multiple of a set of numbers is the smallest number into which each of the given numbers will divide exactly. The LCM can be found by multiplying together all of the factors common to each of the individual numbers. Consider the previous three numbers, 1764, 2100 and 2940 and their factors. 2 ´ 2 ´ 3 ´ 3 ´ 7 ´ 7 = 1764 2 ´ 2 ´ 3 ´ 5 ´ 5 ´ 7 = 2100 2 ´ 2 ´ 3 ´ 5 ´ 7 ´ 7 = 2940 The Lowest Common Multiple of these three numbers will be: 2 x 2 (in all) x 3 x 3 (in 1764) x 5 x 5 (in 2100) x 7 x 7 (in 1764 and 2940) 2 ´ 2 ´ 3 ´ 3 ´ 5 ´ 5 ´ 7 ´ 7 = 44,100 So: 2 x 2 x 3 x 3 x 5 x 5 x 7 x 7 = 44,100 is the L.C.M 1764 x 25 = 44,100 2100 x 21 = 44,100 2940 x 15 = 44,100. Page 1-8 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 1.5 ARITHMETICAL PRECEDENCE The term Arithmetic Precedence means the order in which we carry out arithmetic functions. Sometimes it doesn’t matter what order we carry them out. Consider the expression 2 + 3 = 5. It makes no difference if we write 3 + 2 = 5. Again, consider 3 x 4 = 12, there is no difference if we write 4 x 3 = 12. However, if I write 2 + 3 x 4, what is the answer? If we first add 2 + 3, we will get 5 and then 5 x 4 = 20. Alternatively, multiplying 3 x 4 = 12, adding 2 we get 14. If we are going to agree on the answer we must first agree on the rules we use. This introduces the topic known as arithmetical precedence, and is most easily remember by the term BODMAS. BODMAS indicates the precedence, or the order in which we perform our calculations: B stands for Brackets O stands for "Of" D stands for Division M stands for Multiplication A stands for Addition S stands for Subtraction 1.5.1 BODMAS EXAMPLE Find the value of: 64 ÷ (-16) + (-7 -12) - (-29 +36)(-2 +9) This expression becomes: 64 ÷ (-16) + (-19) - (7)(7) B = (-4) + (-19) - (7)(7) D = (-4) + (-19) - 49 M = - 23 - 49 A = - 72 S Page 1-9 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 1.6 FRACTIONS 0,5 is an example of a Proper Fraction, generally abbreviated to fraction. It has the same meaning as 1 ÷ 2, that is, 1 divided by 2. The number above the line is the Numerator; the number below the line is the Denominator. 5,75 is also a fraction, but because 23 is greater than 4, it is called an Improper fraction. It will normally be written as 5 ¾ = 23/4, 1.6.1 ADDITION OF FRACTIONS The important thing to remember here is that only fractions with the same (a common) denominator can be added or subtracted. Example 1 5/1 + 3/4 = (5*4 + 3*1) / 1*4 = (20 + 3) / 4 = 23 / 4 If the denominators are not the same, then it is necessary to find the lowest Common Denominator (LCD) and to put each fraction in terms of this value. Finding the Lowest Common Denominator is essentially the same as finding the Lowest Common Multiple, which was covered in a previous topic. Having found the LCD, each fraction now needs to be expressed in terms of the LCD. This is achieved by dividing the LCD by the denominator and multiplying the result by the numerator. 1.6.2 SUBTRACTION OF FRACTIONS The basic procedure is very similar to that used for addition; find the LCM, convert the individual fractions, but subtract the numerators instead of adding. There may be one difference which is important. Here are some Taks: 7 27 2× a 5 a ×b 49 , 54 3× a 6 ×b 3×a 5 ×a × b a+b 3 ×a 2 9 ; 7 ; 6 ×a ; c ; 13 x- y - x² - x²- y ² x+ y x+ y ; x²- y ² ; x²- y ² ; x- y Page 1-10 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 3 4 + 56 + 157 = a 3 + 2×b 5 = 7a 3b - 45a = 3a 2b - ba = 1.6.3 MULTIPLICATION OF FRACTIONS These calculations are generally easier to perform than addition and subtraction. Example 1 ¾ * ½ = (3 * 1) / (4 * 2) = 3/8 Simply multiply the numerators together and multiply the denominators together. So convert them to a mixed number or simplify as necessary. 1.6.4 DIVISION OF FRACTIONS To divide two fractions we invert the divisor (the number we are dividing by) and multiply. Example 5¾ ÷ 5½ Firstly, convert into improper fractions. Then invert the second fraction and multiply. so 23 / 4 ÷ 11 / 2 = 23 / 4 x 2 / 11 = 26 / 44 = 9 / 11. Note. Every opportunity should be taken to simplify by "cancelling" numbers above and below the line wherever possible. For example 28 / 4 ÷ 2 /4 = 7 / 1 ÷ 1 / 2 = 7 x 2 = 14 Here are some Tasks: 3×a 5×b ×b = 7×u 5×v × 4v = 2x 3y × 15 y = x+ y 3 ÷ x +4 y = x -3 10 a ÷ x5-a3 = Page 1-11 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 1.7 Decimal fractionsDecimal fractions are fractions where the Denominator is equal to some power of 10, i.e. 100, 1000, 10000 etc. For example, 3 / 100 is a decimal fraction. Decimal fractions are usually re-written as decimals. This is very easily done by using a Decimal Point. Take the example 0.03. Place a decimal point to the right of the numerator (top number). Then move the decimal point to the left, by a number of places equal to the number of "noughts" in the denominator (bottom number). Remove one nought from the denominator for each move. Any fraction can be formed into a decimal, by dividing the numerator by the denominator. For example 857 / 1000 becomes 0.875. Found by a process of long division. 1.7.1 ADDITION & SUBTRACTION The main thing to remember when adding or subtracting decimal numbers is to ensure they are correctly lined up using the decimal point as a reference. Example 1 2.683 + 34.41 2 · 6 8 3 3 4 · 4 1 0 3 7 · 0 9 3 the answer is 37×093 Page 1-12 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 1.7.2 MULTIPLICATION & DIVISION Multiplication of Decimals is the same as ordinary "long" multiplication, but the number of decimal places in the answer must equal the sum of decimal places in the numbers being multiplied. Example 1 6.24 x 3.121 6 ·2 4 3 1· 2 1 6 2 4 1 2 4 8 0 6 2 4 0 0 1 8 7 2 0 0 0 1 9 4 7 5 0 4 There are two digits after the decimal place in the first number and 3 in the second. Therefore, there must be 5 digits after the decimal place in the answer, so the answer becomes 19·47504. (Common sense helps here. A number slightly greater then 6 is multiplied by another number slightly greater then 3. Logically the answer should be approximately 18). Division is also the same as ordinary long division, but again a simple rule helps to simplify the process ‘Do not try to divide by a fraction’. Multiply both the divisor and dividend by a power of ten (move the decimal place to the right) so that the divisor becomes a whole number. Examples 3650 ÷ 45.56 - Multiply both numbers by 100 (102) to give 365000 ÷ 4556 769 ÷ 0×364 - Multiply both numbers by 1000 (103) to give 796000 ÷ 364 Page 1-13 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 1.8 WEIGHTS AND MEASURES A wide number of different weights and measures are used during the maintenance of aircraft. The ones that come to mind first are probably fuel capacities, tyre pressures, temperatures and speeds. There are however very many others, which you will meet as you progress through your course. Firstly, the most commonly used system in aviation today is the Systeme Internationale (SI). This system is based on multiples of 10 and has been accepted widely, with one or two exceptions. It consists of a standard set of units for length (metre), mass (kilogram), time (second), temperature (Kelvin), current (ampere) and light (candela). There are several other units which, whilst not being part of the basic S.I. ones above, are in common use and still use the metric system for calculations. An older system that is still used in some countries today, is the Imperial System, which uses a mixture of old units such as feet and inches for length, pounds for weight, gallons for capacity and Fahrenheit for temperature. You will occasionally meet a mixture of systems, which will require conversion from one to another. A good example is the amount of fuel put into an aircraft's tanks. You will find this being measured in imperial gallons, American gallons, imperial pounds, SI kilograms or metric litres. Changing a quantity in one unit to a quantity in another unit requires a conversion factor. When the quantity in the first unit is multiplied by the conversion factor, the result is the quantity in the second units. For example, to convert imperial gallons to litres, they must be multiplied by 4.546 Example 1 Convert 25 gallons into litres. 25 x 4.546 = 113.65 Litres. Example 2 Convert 1500 miles into kilometres using the conversion factor 1×6094 1500 x 1×6094 = 2413×9 Kilometres. Note: You will normally be given the conversion factor, however, you may have to transpose a formula in order to use it. Page 1-14 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 1.9 RATIO AND PROPORTION This topic is an extension of several previous topics. Ratio and proportion are essentially statements that link two or more "quantities" together. For example, a ‘3 to1 mix of sand and cement’, which may be written as a 3:1 mix of sand and cement, means ‘mix 3 parts of sand to 1 part of cement". This is a commonly used statement which you will notice has no formal units, although volume is inferred. Parts could be represented by shovels full, buckets full, wheelbarrows full etc. The mixture simply has a total of 4 parts, of which 3 parts, ¾ , is sand, and 1 part ¼ is cement. A ratio therefore simply provides a means of comparing one value with another. For example, if an engine turns at 4000rpm and the propeller turns at 2400rpm, the ratio of the two speeds is 4000 to 2400, or ‘5 to 3’ when reduced to its lowest terms. This relationship can also be expressed as 5/3 or 5:3. The use of ratios is common in aviation, such as when considering the compression ratio in an engine. This is the ratio of cylinder displacement, when the piston is at the bottom of its stroke compared with the displacement when it is at the top. For example, if the volume of the cylinder at the bottom of its stroke is 240 cm2 and at the top becomes 30 cm2 the ratio is 240:30 or, reduced to its lowest terms, 8:1. Another typical ratio is that of different gear sizes. For example, the ratio of a drive gear with 15 teeth to a driven gear with 45 teeth is 15:45 or 1:3 when reduced. This means that for every one tooth of the drive gear there are three teeth on the driven gear. However, when working with gears, the ratio of teeth is opposite the ratio of revolutions. In other words, since the drive gear has one third as many teeth as the driven gear, the drive gear must complete three revolutions to turn the driven gear once. This results in a revolution ratio of 3:1, which is the opposite of the ratio of teeth. A proportion is a statement of equality between two or more ratios and represents a convenient way to solve problems involving ratios. For example, if an engine has a reduction gear ratio between the crankshaft and the propeller of 3:2, and the engine is turning at 2700rpm, what is the rotational speed of the propeller? In this problem let Vp represent the unknown value, which in this case is the speed of the propeller. Next, set up a proportional statement using the fractional form, 3/2 = 2700/Vp. To solve this equation, cross multiply to arrive at the equation 3Vp = 2 x 2700, or 5400rpm. To solve for Vp divide 5400 by 3. Thus, the propeller speed is 1800rpm. Page 1-15 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES Example Divide £240 between 4 men in the ratio of 9:11:13:15. The normal procedure for this type of problem is to: A. Add all of the individual proportions to find the total number of parts. B. Divide the total amount by the number of parts to find the value of each part. C. Multiply each ratio by the value of each part. So. 9 + 11 + 13 + 15 = 48 £240 divided by 48 = £5. Therefore each part is worth £5. 9 x 5 = 45 11 x 5 = 55 13 x 5 = 65 15 x 5 = 75 The proportions are therefore £45, £55, £65 and £75 A useful check is to add the individual parts together, to ensure the total is the amount you started with. 1.10 AVERAGES AND PERCENTAGES 1.10.1 AVERAGES When working with numerical information, it is sometimes useful to find the average value. When estimating the time a particular journey would be no point in basing the time on the slowest speed or the highest speed, always use an average speed. We would also use average fuel consumption to estimate how much fuel an aircraft would use for a particular flight. In both of these types of calculation, we can only work out the average by dividing the total distance or fuel used by the time. Example 1 An aircraft travels a total distance of 750 km in a time of 3 hours 45 minutes. What is the average speed in km/hr? 750 Average speed = Total Distance/Time = = 200km / hr 3.75 Page 1-16 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES Example 2 An aircraft uses 300 gallons of fuel for a flight of duration 4 hours. What is the average fuel consumption? 300 Average Fuel Consumption = = 75 gallons / hour 4 We often need to calculate averages based on many data items. Example 3 The weight of six items are as follows: 9.5, 10.3, 8.9, 9.4, 11.2, 10.1 What is the average weight? To calculate this we simply add the total weights and divide by the number of items. 59.4 The total weight is 59.4 kg The average is = 9.9 kg 6 1.10.2 PERCENTAGE Percentages are special fractions whose denominator is 100. The decimal fraction 0.33 is the same as 33/100 and is equivalent to 33 percent or 33%. You can convert common fractions to percentages by first converting them to decimal fractions and then multiplying by 100. For example, 5/8 expressed as a decimal is 0.625, and is converted to a percentage by moving the decimal point two places to the right, the same as multiplying by 100. This becomes 62.5%. To find the percentage of a number, multiply the number by the decimal equivalent of the percentage. For example, to find 10% of 200, begin by converting 10% to its decimal equivalent, which is 0.1. This is achieved by dividing the percentage figure by 100. Now multiply 200 by 0.1 to arrive at the value of 20. If you want to find the percentage one number is of another, you must divide the first number by the second and multiply the quotient by 100. For instance, an engine produces 85hp from a possible 125hp. What percentage of the total horsepower available is being developed? To solve this, divide the 85 by 125 and multiply the quotient by 100. Example: 85 ÷ 125 = 0.68 x 100 = 68% power. Page 1-17 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES Another way that percentages are used, is to determine a number when only a portion of the number is known. For example, if 4180rpm is 38% of the maximum speed, what is the maximum speed? To determine this, you must divide the known quantity, 4180rpm, by the decimal equivalent of the percentage. Example: 4180 ÷ 0.38 = 11,000rpm maximum A common mistake made on this type of problem is multiplying by the percentage instead of dividing. One way of avoiding making this error is to look at the problem and determine what exactly is being asked. In the problem above, if 4180rpm is 38% of the maximum then the maximum must be greater than 4180. The only way to get an answer that meets this criterion is to divide by 0.38. 1.11 POWERS AND ROOTS 1.11.1 POWERS When a number is multiplied by itself, it is said to be raised to a given power. For example, 6 x 6 = 36; therefore 62 = 36. The number of times the base number is multiplied by itself is expressed as an exponent and is written to the right and slightly above the base number. A positive exponent indicates how many times a number is multiplied by itself. Example: 32 is read "3 squared" or "3 to the power of 2". Its value is found by multiplying 3 by itself. 3x3=9 23 is read "2 cubed" or "2 to the power of 3". Its value is found by multiplying 2 by itself 3 times. 2x2x2=8 If the exponent is a negative integer, the minus sign indicates the inverse or reciprocal of the number with its exponent made positive. Example: 2-3 is the same as the reciprocal of 23 which is 1 / 23 so 1 1 2-3 = = 2´ 2´ 2 8 Page 1-18 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES Any number, except zero, that is raised to the zero power equals 1. When a number is written without an exponent, the exponent value is assumed to be 1. Furthermore, if the exponent does not have a sign, (+ or -) preceding it, the exponent is assumed to be positive. 1.11.2 ROOTS The root of a number is that value which, when multiplied by itself a certain number of times, produces that number. For example, 4 is a root of 16 because when multiplied by itself, the product is 16. However, 4 is also a root of 64 because 4 x 4 x 4 = 64. The symbol used to indicate a root is the radical sign ( x ) placed over the number. If only the radical sign appears over a number, it indicates you are to extract the square root of the number under the sign. The square root of a number is the root of that number, when multiplied by itself, equals that number. When asked to extract a root other than a square root, an index number is placed outside the radical sign. For example, the cube root of 64 is expressed as 3 64 = 4 Another way of indicating roots is by showing the root of a number is by showing an exponent as in powers. In the case of roots, however, the exponent is shown as a fraction. 1 The cube root of 64 can also be expressed as 64 3 1 The square root of 16 would be expressed as 16 2 Page 1-19 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 2 ALGEBRA 2.1 INTRODUCTION Very often students will claim that they never have and never will understand Algebra. They say they can understand and work with numbers, but not with letters, and yet Algebra is designed to make matters simple and clear. For example, suppose a room is 5 metres long by 3 metres wide and we need to know how much carpet is needed to cover the floor. No one would have any hesitation in calculating the answer, 15 square metres (m2). But that answer only applies to that room. The general answer is that the area is found by multiplying length by width (or breadth). i.e. Area = length x breadth. But it is easier to write A = L x b, where the letters A, L, b represent in this case Area, Length and breadth, and that is what Algebra is all about; letters represent some variable and only when particular values. i.e. numbers are known, do we resort to them instead. So when using Algebra, it is important to state what the letters represent. Some letters are often used, particularly x and y, but g often represents acceleration due to gravity, r represents density, and so on. This is what Algebraic notation is about. 2.1.1 OPERATION Algebraic operations are in essence the same as when using numbers. So Adding a and b is written a+b Subtracting a and b is written a - b Multiplying a and b is written ab Dividing a by b is written a/b Squaring a a2 We are not restricted to 2 letters only. a multiplied by b and divided by c becomes, logically, a*b / c Page 2-19 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES Note also that the order in which letters appear is basically unimportant. a x b x c x d = abcd = bdac = cadb etc. etc. (3 x 4 is obviously the same as 4 x 3 etc.) When symbols such as x and y are multiplied together we do not need to include the multiplication sign. This is the same if a number and a symbol are multiplied together. 3 x y, 4 x z, s x p, a x b, y x z x m can all be written without the multiplication sign as 3y, 4z, sp, ab and yzm The same is not true of numbers on their own: 7 x 8, 4 x 5 and 6 x 7 cannot be written as 78, 45 and 67. Like Terms are terms comprised of the same algebraic quantity - this is important. 7x, 5x and -3x are all terms containing x 7a, 4b, 3a and -6b can be split into two groups of like terms, 7a and 3a, and 4b and -6b. If like terms contain numerical coefficients, they can be simplified. 7x + 5x - 3x = (7 + 5 - 3)x = 9x 7a + 3a + 4b - 6b = 10a - 2b. Terms like ab + cb - db may be simplified as (a + c - d) b. (b is a common factor of the 3 terms) Page 2-20 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES When dealing with algebraic terms and expressions the ability to factorise is a great asset. Similarly, the ability to divide numerator and denominator by the same terms (i.e. cancelling top and bottom) allows simplification. 3a 2b 3 a/ a b/ a = = 6ab 2 6 a/ b/ b 2b 2.1.2 BASIC LAWS Algebra obeys the same laws of procedure as Arithmetic, i.e. BODMAS. Note that Brackets appear rather more often in Algebra, and are only removed when there is a good reason to do so, for example, when further operations ultimately lead to greater simplification. (3x + 7y) - (4x + 3y) = 3x + 7y - 4x - 3y = -x + 4y Note especially that when removing brackets, all the terms inside the brackets are multiplied by what is immediately outside the brackets. The basic procedure is as follows. a (x + y) = ax + ay a + b (x + y) = a + bx + by (both x and y are multiplied by b) (a + b) (x + y) = ax + ay + bx + by (x and y are multiplied by (a+b) (a + b)2 = (a + b) (a + b) = (a x a) + (a x b) + (b x a) + (b x b) = a2 + ab + ab + b2 = a2 + 2ab + b2 When factorising, examine each term is order to look for common factors. the common factors of a2b and -2ab2 are a and b (they appear in both), hence a2b - 2ab2 can be written (ab)(a - 2b). (ab) and (a - 2b) are both factors of the complete expression a2b and -2ab2. Page 2-21 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES As already stated, the ability to "see" factors is an asset. ax + bx + ay + by = x (a + b) + y (a + b) = (x + y) (a + b) or = a (x + y) + b (x + y) = (x + y) (a + b) Algebra can be extended to include fractions. e.g. a / b + c / d = (ad + cb)/bd (bd is the LCD, ad + cb is the Numerator) 2.2 EQUATIONS The statement a – 4 = 5 is an equation. What we are saying is that an unknown quantity minus 4 equals 5. It does not take a genius to work out that the unknown quantity in this case is 9, there is only one value that will be correct. The value of a can be calculated using guesswork or elimination. The process of establishing that a = 9 is called solving the equation. 2.2.1 SOLVING LINEAR EQUATIONS A linear equation is one containing only the first power of the unknown quantity. 5y – 5 = 3y + 9 or 5(m – 2) = 15 These are both linear equations. When we solve linear equations, the appearance of the equation may change. For example, the first equation could be re-written as 5y – 3y = 9 + 5 and the second as 5m – 10 = 15. Both of these look different from the original form, but equality has been maintained and they are therefore the same. The general rule for all equations is: Whatever you do to one side of the equation, you must do the same to the other side. By convention we name each side of the equation Left Hand Side (LHS) or Right Hand Side (RHS) Page 2-22 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 2.2.1.1 Equations Requiring Multiplication or Division x Solve the equation =4 5 x If we multiply both sides by 5 we get ´5 = 4´5 5 So the solution is x = 20 Solve the equation 4b = 20 Dividing both sides by 4 we get 4b/4 = 20/4 So the solution is b = 5 2.2.1.2 Equations Requiring Addition or Subtraction The simplest type of linear equation is of this type: x - 6 =9 To solve all equations we must manipulate the equation to get the unknown on one side and the known values on the other side. In this case we must eliminate the value of –6 from the LHS. This can be done by adding 6 to the LHS, but we must also add 6 to the RHS. So the equation becomes x - 6 + 6 = 9 + 6 We then Simplify the equation to obtain x = 9 + 6 = 15 So the solution is x = 15 A simpler way of solving this type of equation is to switch values from one side to another. When we do this, we must, however change the sign. Example: Solve y + 4 = 14 If we switch the + 4 to the RHS and change the sign it becomes Y = 14 – 4 So the solution is y = 10 Page 2-23 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES If the equation has multiples of the unknown quantity, such as: Solve 5x – 12 = 3 the first stage is the same, i.e. 5x = 3 + 12 So 5x = 15 It seems obvious that x = 3, but how mathematically is this achieved? If we divide both sides by 5 we will get the solution x = 3. 2.2.1.3 Equations Containing Unknowns on both Sides In equations of this type we should group the unknown quantities on one side and the other terms on the other side. For example, solve 8y + 4 = 5y + 22 If we subtract 4 from both sides, and also subtract 5y from both sides we will get: 3y = 18 The solution can then be obtained by dividing each side by 3. 3y/3 = 18/3 = y=6 Note: As in all cases of solving equations, we can and should check our solution is correct by substituting the solution in the original equation. i.e. LHS (8 x 6) + 4 = 48 + 4 = 52 RHS (5 x 6) + 22 = 30 + 22 = 52 2.2.1.4 Equations Containing Brackets The first step is to remove the brackets and then solve as normal 3(2y + 3) = 21 first expand the brackets to obtain 6y + 9 = 21 then subtract 9 from both sides 6y = 12 then divide both sides by 6 The solution is y=2 To check the solution is correct, we substitute y = 2 in the original equation. LHS 3(2 x 2 + 3) = 3(4 + 3) = 3 x 7 = 21 RHS = 21 Page 2-24 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 2.2.1.5 Equations Containing Fractions In this case we must multiply each term by the LCM of the denominators. y 3 3y Example 1. + = -2 4 5 2 The LCM of the denominators 4, 5 and 2 is 20, so we must multiply each term in the equation by 20 y 3 3y ´ 20 + ´ 20 = ´ 20 - 2 ´ 20 4 5 2 5y + 12 = 30 y - 40 5y - 30 y = -40 - 12 so - 25 y = -52 Note in this case we have negative values on both sides. If we swap them around and change the signs i.e. swap the LHS for the RHS We get 52 = 25 y Note this is exactly the same as 25 y = 52. This can be proved by taking the equation - 25 y = -52 and adding 25y to both sides, and then adding 52 to both sides. x-4 2x - 1 Example 2. Solve the equation - = 4 3 2 The LCM of 3 and 2 is 6, so we multiply all of the terms by 6 x-4 2x - 1 ´6 - ´6 = 4´6 3 2 2(x - 4) - 3(2x - 1) = 24 2x - 8 - 6 x + 3 = 24 29 - 4 x - 5 = 24 so - 4 x = 29 and the solution is x = - = - 7.25 4 Page 2-25 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 2.3 TRANSPOSITION IN EQUATIONS Consider a formula (equation) given in a certain form. 6a + 11 = 25 - a This contains one algebraic quantity, "a", within an equation. Think of an equation as a statement of ‘balance’. In this one, 6a + 11 on the LHS equals, or balances, 25 - a on the RHS. As we have one equation and one unknown ‘a’, there is only one numerical value which can produce a balance. What is it? By manipulating (transposing is the word) the equation, it is possible to isolate the ‘a’ on the LHS and balance it with an actual number on the RHS. This will then be the unique value of ‘a’. Look again at the equation. 6a + 11 = 25 - a To remove the ‘a’ on the RHS, we must add ‘a’ to both sides. 6a + 11 + a = 25 - a + a therefore 7a + 11 = 25 To remove + 11, we must subtract 11 from both sides 7a + 11 – 11 = 25 - 11 so 7a = 14 and if 7a = 14 then a=2 We have found that a = 2. This is the unique value which satisfies 6a + 11 = 25 - a. Study it again to see how we worked to isolate the required term ‘a’ on one side, and remember, what you do to one side of an equation, you must do to the other side if the balance is to be maintained. Here is another a formula involving several algebraic symbols. Find N, if C = N/2p – n/2p Remember, we want N on one side by itself. It is important to get a 'feel' for the form of the equation. To help, we will put brackets around (N - n). So C = (N - n) / 2p To remove the 2p we must multiply both sides by 2p Page 2-26 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES C x 2p = (N - n) / 2p x 2p which gives 2Cp = (N – n) To remove the -n, we must add n to both sides 2Cp + n = (N – n) + n = N That's it, N = 2Cp + n Here's another example. V = 1/3 x r2ph (the volume of a cone). Find r (the radius), step by step. Vx 3 = 1/3 x r2ph x3 (multiply both sides by 3) V / ph = r2ph / ph = r2 (divide both sides by ph) Remember, to find r, take the square root of r2 and do the same to both sides. 3V \ = r2 = r. ph 3V r = ph. This is what transposition is all about. We are re-arranging formulas expressed as equations, which then allows us to find a particular numerical value for one (unknown) quantity if the other numerical values are given. One important point, it is only possible to find an unknown quantity if all the other values are known. This is known as 'solving an equation'. The rule is, One unknown quantity can be deduced from one equation, Two unknowns require two different equations, Three unknowns required three different equations, and so on. Page 2-27 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 2.3.1 CONSTRUCTION OF EQUATIONS As already stated, Maths serves as a "tool" for Engineers at the design stage. Design is the creation of a component or mechanism on paper, i.e. before it take shape in metal or plastic. The design engineer hopefully makes it strong enough - his knowledge of materials and their strengths allow him to do this by calculation. He uses formulas and equations. To do this, he must allocate letters to represent some variable or known quantity. He can then construct a formula or equation by using the letters within some ‘reasonable’ statement about the situation. He studies the situation and then makes the statement. How do we construct equations from the facts contained within a scenario? Example 1 Think of a number, double it, add 6 and divide the result by 3. What is the answer? Let the number you think of be A. Doubling this number gives 2A. If 6 is then added, we have 2A + 6, which must then be divided by 3, making the answer = (2A + 6) / 3 This formula can be used to calculate the answer no matter what number you think of. Example 2 If one side of a rectangular field is twice as long as the other, and the short side is 100m. Calculate the area of the field. Let the short side of the field be L. The long side is therefore 2 x L or 2L. To calculate the area we multiply one side by the other, so: Area = 2L x L = 2L2 where L equals 100m Area = 2(100 [m])2 = 20,000 [m2] Example 3 A certain type of motor car cost seven times as much as a certain make of motor cycle. If two cars and three motor cycles cost £8500, find the cost of each vehicle. Let the cost of a car be C (at present C is an unknown). Let the cost of a motor cycle be M (another unknown). We know that 2C + 3M = £8500 (this has two unknowns within one equation). Page 2-28 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES But we also know that C = 7 x M, therefore, we can substitute for C in the first equation. 2 (7M) + 3M = £8500 14M + 3M = 17M = £8500 M = £8500 / 17 = £500 The cost of a motor cycle is therefore £500, and the cost of a car must be 7 X £500 = £3500. Here 2 equations were constructed from the facts, and then combined to allow a solution to be found. In the next example, we form equations from the facts, and then transpose to produce a solution. Example 4 Three electric radiators and five convector heaters together cost £740. A convector cost £20 more than a radiator. Find the cost of each." Let R represent the cost of a radiator, and C represent the cost of a convector. Then 3R + 5C = £740 And C = R + 20 \ 3R + 5 (R + 20) = 3R + 5R + 100 = 740 \ 8R = 740 - 100 = 640 R = 640 / 8 = £80 (the cost of a radiator) and C = 80 + 20 = £100 (the cost of a convector) 2.4 SIMULTANEOUS EQUATIONS Consider the equation 4x - 3y = 1. There are 2 unknowns (x and y) in one equation, and so the equation cannot be solved to give a single value for x and y. There are an infinite number of values of x for which there are corresponding values of y. For example: if x = 4, then y = 5 if x = 7, then y = 9 if x = 1, then y = 1 However, if a second equation exists, for example x + 3y = 19, then these two equations can be evaluated simultaneously to give single values for x and y. Page 2-29 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES The process is simple and involves modifying the equations, whilst still preserving the equalities. 4x – 3y = 1 (1) x + 3y = 19 (2) The method of solution of all simultaneous equations is to: first manipulate one or both of the equations so that the coefficient of one of the unknowns is the same in both equations. then add or subtract one of the equations from the other to produce a third equation with only one unknown. The other having become zero. solve the new equation to find the unknown. put the solution into one of the original equations to find the other unknown. put both solutions into the equation not used in the stage above to check your answers. Using the two equations above as an example: We do not need to manipulate either of the equations because the co-efficient of y is the same in both equations. Therefore, we can eliminate the “y” value simply by adding the two equations. The result is: 5x = 20 So x = 4 If we then substitute x = 4 in the second equation we get: 4 + 3y = 19 So 3y = 19 - 4 = 15 So y = 5 Our solutions are x = 4 and y = 5 Example 1 2x + 3y = 8 (1) 3x + 5y = 11 (2) Multiply equation (1) by the coefficient of x in equation (2). (2x + 3y = 8) x 3 = 6x + 9y = 24 Multiply equation (2) by the coefficient of x in equation (1). (2x + 5y = 8) x 2 = 6x + 10y = 22 So 6x + 9y = 24 (3) Page 2-30 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 6x + 10y = 22 (4) Subtract equation (4) from (3) 0x - 1y = 2. so -y = 2 and y = -2 substitute y = - 2. in either equation (1) or (2) to solve for x. I have selected (1). 2x + 3(-2) = 8 therefore 2x = 14 and x = 7 Check your answer by substituting both values in equation (2). Do not use equation (1) because it will not highlight an error. If you had used equation (2) to find x, then the check should be carried using equation (1). 3x + 5y = 11 3(7) + 5(-2) = 11 therefore 21 +(-10) = 11 - correct The same result would be found if y was eliminated as shown below. (2x + 3y = 8) x 5 10x + 15y = 40 (3) (3x + 5y = 11) x 3 9x + 15y = 33 (4) x = 7 etc. 2.5 QUADRATIC EQUATIONS Any equation of the form y = ax2 + bx + c, where a, b and c are numbers, is known as a quadratic equation. An equation of this type will produce a curve called a parabola. The actual value for coefficients a, b and c will determine the exact shape and position of the curve. Page 2-31 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES It will be noted that one of the curves cuts the x-axis at points P and S. P and S are known as the roots of the equation. Alternatively, P and S are the values of x which satisfy the condition y = ax2 + bx + c = o. It can be shown that the Roots are found to be equal to: -b ± b 2 - 4ac 2a This equation gives two values, one for P the other for S. Example Find the roots of y = 6x2 - 5x - 6 (a = 6, b = -5, c = -6) - (- 5 ) ± (- 5)2 - 4 (6 )(- 6 ) x = 2 (6 ) 5 ± 25 + 144 5 ± 13 x = = 12 12 18 -8 1 -2 æ -2 1ö x = or = 1 or ç in this case, points P & S are and 1 ÷ 12 12 2 3 è 3 2ø Note - depending on a, b and c, it is possible that b2 - 4ac results in a negative value. It has been considered impossible to find the square root of a negative value. The equation concerned is then said to have no real roots. When b2 - 4ac is negative, the equation is said to have complex roots, where the roots comprise both a real and imaginary component. This concept is not considered in these notes. Page 2-32 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 3 NUMBERS 3.1 INDICES AND POWERS It is often to necessary to multiply a number by itself once, twice or several times. To indicate this, a method of notation has evolved, which is both convenient and capable of being extended to introduce other concepts. 3 x 3 is written as 32 2 x 2 x 2 x 2 x 2 is written as 25 4 x 4 x 4 is written as 43 etc, etc. In the above examples, the number being multiplied by itself is known as the base and the number of times it is multiplied by itself is known as the power or index. Alternatively, the number 2 has been raised to power 5. Power 2 and power 3 are generally referred to as the square and the cube. 3 x 3 = 32 = 9 9 is the square of 3 or 3 squared equals 9 4 x 4 x 4 = 43 = 64 64 is the "cube" of 4. or 4 cubed equals 64 But put another way, 3 is said to be the square root of 9, 4 is the cube root of 64 and 2 is the fifth root of 32. The method of notation used is that: 1 3 = 9 2 or 9 1 5 2 = 32 5 or 32 1 3 4 = 64 3 or 64 It is possible to re-write the above, so that 3 = 90.5, 2 = 320.2 and 4 = 640.333. Where the power is expressed as a decimal, instead of a fraction. To allow the use of numbers involving powers and indices, some rules have evolved, which are reproduced, using the symbol N to represent any base number. Rule 1. N2 x N3 = N5 \ Na x Nb = N(a + b) Page 3-33 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES Rule 2. N5 ÷ N2 = N3 \ Na ÷ Nb = N(a - b) Rule 3 (N2)3 = N2 x N2 x N2 using rule 1 this equals N6 \ (Na)b = N(a x b) or Nab Rule 4 N2 ÷ N2 = N(2 – 2) = N0 Any number divided by itself equals 1 so N0 = 1 Therefore 1/N2 = N0 ÷ N2 using rule 2 this equals N-2 \ 1/Na = N-a also 1/N-a = Na because 1 ÷ N2 is the same as N0 - N2 = N(0 – 2) = N-2 Rule 5 If N1/3 x N1/3 x N1/3 = N1 = N then N1/3 must be the third root of N, because the only number that can be multiplied by itself 3 times to make N is the third root of N. therefore N1/3 = 3,N similarly if N2/3 x N2/3 x N2/3 = N2 then N2/3 must be the third root of N2 therefore N2/3 = 3,N 2 so Na/b = b,N a Page 3-34 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 3.1.1 STANDARD FORM If the number 8.347 is multiplied by 10,000 then the product is 83470. This calculation can be written as 8.347 x 104 = 83470. When 83470 is written as 8.347 x 104, it is known as Standard Form. A number in standard form has two parts. The first part is a number between 1 and 10 (but does not equal 10), and the second part is 10 raised to some whole number power. The first part is called the Mantissa, the second part the Exponent. To express a number in standard form, move the decimal point left or right to create a number between 1 and 10 (the mantissa), and then create the exponent. The value of which equals the number of places by which the decimal point has been moved. If the point was moved Left, the power is positive, if the point was moved Right, it is negative. Examples 526 = 5.26 x 102 0.3716 = 3.716 x 10-1 0.002 = 2.0 x 10-3 3.2 NUMBERING SYSTEMS The most widely used system of numbers is the decimal system, based on the hindu-arabic symbols 0, 1, 2, 3 etc but roman symbols such as V, X, L and C are also well known and understood. To-day, the practice of engineering requires a measure of competence in handling several different systems of numerals. In general a system of numeration consists of a set of symbols together with a rule by which the symbols can be combined together. Number is the property associated with a set or collection of things. It is independent of the nature of the individual items in the set. The number fourteen may be written as 14 or XIV. In this case the number is the same but the system or numeration is different. 3.2.1 DECIMAL SYSTEM OF NUMERATION In the decimal system, the symbols are combined by arranging them in a horizontal line, the contribution that each digit makes being governed by its position. A decimal point enables numbers less than one to be represented. Page 3-35 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES Example 1 Decimal 368 is really: (3 ´ 102) + (6 ´ 101) + (8 ´ 100) or in column form: 102 101 100 (hundreds) (tens) (units) 3 6 8 Example 2 Decimal 452.64 is really: (4 ´ 102) + (5 ´ 101) + (2 ´ 100) + (6 ´ 10-1) + (4 ´ 10-2) or in column form: 102 101 100 10-1 10-2 4 5 2 6 4 Ten is known as the base or radix of the decimal system. The index indicates the power to which the base is raised. The base, and the particular index to which it is raised is called the weight. e.g. least significant weight = 100 = 1 next most significant weight = 101 = 10 The numbers by which weight is multiplied are called digits. In practice only the digits of the system are written, the weight being implied e.g. 368, 53.24. Note: 0 is counted as a digit, so that there are ten digits in the decimal system, 0 to 9 inclusive. Page 3-36 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 3.2.2 BINARY SYSTEM OF NUMERATION Only the symbols 0 and 1 are used and the base is two, otherwise the system of numeration is the same as before. The two digits 0 and 1 are referred to as bits, an abbreviation of binary digits. Example 1 101101 is really: (1 ´ 25) + (0 ´ 24) + (1 ´ 23) + (1 ´ 22) + (0 ´ 21) + (1 ´ 20) or in column form: 25 24 23 22 21 20 1 0 1 1 0 1 (= 45 in decimal) Example 2 110.11 is really: (1 ´ 22) + (1 ´ 21) + (0 ´ 20) + (1 ´ 2-1) + (1 ´ 2-2) or in column form: 22 21 20 2-1 2-2 1 1 0 1 1 (= 6.75 in decimal) Note: All digits to the right of the binary point refer to negative powers. Page 3-37 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES The binary system is very suitable for use with electrical switching circuits. A switch is either off or on corresponding, for example, to 0 and 1 respectively. There is no ambiguity. 3.2.3 OCTAL SYSTEM OF NUMERATION In the octal system of numeration the symbols 0 to 7 are used and the base is 8. Again the system of numeration is the same as that used for decimal and binary, with each column increasing by a power of one as you move from right to left. Example 1 3768 is really: (3 ´ 82) + (7 ´ 81) + (6 ´ 80) or in column form: 83 82 81 80 0 3 7 6 (= 254 in decimal) Example 2 37·13 is really: (3 ´ 81) + (7 ´ 80) + (1 ´ 8-1) + (3 ´ 8-2) or in column form: 82 81 80 8-1 8-2 0 3 7 1 3 in decimal = (3 x 8) + (7 x 1) + (1 x 0·125) + (3 x 0·015625) Page 3-38 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES = 31·140625 Note: All digits to the right of the octal point refer to negative powers. 3.2.4 CONVERSION TO OTHER BASES Conversion from decimal to any other base can be achieved by dividing the decimal number repeatedly by the new base and recording the remainder. The remainder gives the number in the new base and should be read from bottom to top. Example – convert 2910 to binary. 2 29 2 14 Rem 1 2 7 Rem 0 Result 1 1 1 0 12 2 3 Rem 1 2 1 Rem 1 0 Rem 1 Example 2 – convert 5710 to octal 8 57 8 7 Rem 1 Result 7 18 7 Rem 7 Example 3 – convert 6310 to hexadecimal 16 63 16 3 Rem 15(F) Result 3 F16 0 Rem 3 Page 3-39 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES To convert binary numbers to decimal. The easiest way to convert from binary to decimal is to remember the weightings, or if necessary write the weightings above each binary digit, and add them up. Example 1 – convert 1 0 1 1 0 1 to decimal. 25 24 23 22 21 20 (32) (16) (8) (4) (2) (1) 1 0 1 1 0 1 (1 x 25) + (0 x 24) + (1 x 23) + (1 x 22) + (0 x 21) + (1 x 20) = 4510 An alternative method for long binary numbers is to take the left-hand digit, double it and add the result to the next digit to the right as shown below (double and add to next digit to the right). 1 0 1 1 1 0 1 2 5 11 23 46 To convert binary to octal or vice versa. Each octal digit can be represented by 3 binary digits. Therefore, to convert from binary to octal: i. split the binary number into groups of 3 digits starting from the right. ii. weight the numbers in each group 4 – 2 – 1 iii. find the total of each group of 3 digits, the result is the octal value. Example 1 – convert 1 0 1 1 1 0 0 1 to octal Binary No 1 0 1 1 1 0 0 1 Weighting 4 2 1 4 2 1 4 2 1 Octal No (sum) 2 7 1 Answer 1 0 1 1 1 0 0 12 is equal to 2718 The reverse process should be used to convert octal to binary. Convert each digit into a 3 digit binary number keeping the order of digits the same. Work from the bottom to the top of the table shown above to convert 2718 to binary. Page 3-40 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES To convert binary to hexadecimal or vice versa. The process for converting a binary number to a hexadecimal one, is the same as that used to convert binary numbers to octal. Each hexadecimal digit can be represented by 4 binary digits, therefore the binary number is split into groups of 4 digits starting from the right. The weightings this time are 8 – 4 – 2 – 1. Again, the reverse process is used to convert from hexadecimal to binary. Convert each hexadecimal digit into its binary equivalent keeping the order the same. Example 1 – convert A716 to binary. Hexadecimal No A 7 Weightings 8 4 2 1 8 4 2 1 Binary No 1 0 1 0 0 1 1 1 Answer A716 is equal to 1 0 1 0 0 1 1 12 Page 3-41 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 3.3 LOGARITHMS Logarithms are a mathematical concept that was developed to simplify multiplication and division of large numbers. Logarithms enable multiplication and division to be performed using addition and subtraction. The use of logarithms is no longer so widespread as the electronic calculator has become so readily available. Remembering that when, for example, 25 is written as 52, 5 is known as the base and 2 as the power, then the logarithm of 25 can be expressed as 2, to the base 5. The general definition is, that if y = ax then x = loga y So logarithms can be calculated for any base a, but generally only logarithms to the base of 10 or e (2.71) are used, and are commonly available in tabular form. However, logarithms are more easily obtained from the calculator. An example of the function of logarithms is shown below. Example Calculate 6.412 x 23.162 From the calculator the log10 of 6.412 is 0.80699 and the log10 of 23.162 is 1.36478. So 6.412 x 23.162 = 100.80699 x 101.36478 and using the laws of indices 6.412 x 23.162 = 10(0.80669 + 1.36478) = 10(2.17177) It is now necessary to find the base 10 number whose logarithm is 2.17177. The calculator shows this to be 148.51474 (this is the anti-log of 2.17177). If the calculator is used to solve 6.412 x 23.162, the product is 148.51474. It is important to realise that this example shows how logarithms can be used, in practice, the calculator is used as normal. If a division is to be performed, the powers of logs are subtracted. It is the concept of a logarithm that is important at this stage, because it re- appears later. Page 3-42 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4 GEOMETRY 4.1 ANGULAR MEASUREMENT If two straight lines are drawn, we can see that they make an "angle". But how are 'angles' expressed or measured. Consider a single line, and rotate it through a complete revolution. Then the angle that this line has turned through is 360º. A degree is 1/360° of a revolution. Note that half a revolution is therefore 180º and a right angle (¼ of a revolution) is 90º. Note that 1 degree can be sub-divided into 60 minutes and 1 minute can be sub- divided into 60 seconds (very small). A few definitions are included here: An Acute angle - less than 90º An Obtuse angle - between 90º and 180º A Reflex angle - greater than 180º Complementary angles - their sum is 90º Supplementary angles - their sum is 180º Page 4-43 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4.1.1 ANGLES ASSOCIATED WITH PARALLEL LINES Now consider 2 parallel lines, cut by a transversal. A = C, B = D (they are opposite and equal), similarly L = P, and M = Q. Also A = L, D = Q, etc. etc. (they are corresponding angles) D = M, C = L (they are alternate angles) D + L = 180 (= C + M) (these are interior angles, and are supplementary) Page 4-44 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4.2 GEOMETRIC CONSTRUCTIONS There are many different shapes associated with geometry. The more common ones are described in the following text. 4.2.1 TRIANGLE A triangle obviously has 3 sides and 3 (internal) angles. The sides are often represented by the 3 (small) letters a, b and c; the angles by the (large) letters A, B and C. The 3 angles add up to 180º. The construction of a dotted line parallel to AB and an extension of BC proves this. The area of a triangle = ½ base x vertical height 4.2.1.1 Triangle Types There are many different types of triangle. The main types and features are summarised as follows: Acute-angled triangle has all of it’s angles less than 90º. Obtuce-angled triangle has one angle greater than 90º. Scalene triangle has three sides of different lengths. Right-angled triangle has one of it’s angles equal to 90º. The longest side is opposite the 90º angle (right-angle) and is called the hypotenuse. Isosceles triangle has two sides and two angles equal. The equal angles lie opposite to the equal sides. Equilateral triangle has all it’s sides and angles equal. Page 4-45 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4.2.2 SIMILAR & CONGRUENT TRIANGLES You may study two triangular shapes and estimate whether they are the same or not. We need to be more precise. If they have the same shape, we are really saying that their angles are the same, they are then described as similar triangles. Similar triangles do not have to be the same size. One triangle may have sides twice or ten times as large as another triangle and still be classified as similar. If they are exactly the same shape and size, their sides are the same length, then they are described as Congruent triangles. It is sometimes necessary to determine whether triangles are Congruent. A simple criteria exists to assist us. Two triangles are congruent if: Their corresponding sides are of equal length. (side, side, side) They have two angles and the common side equal. (angle, side, angle) They have two sides and the included angle is equal. (side, angle, side) The hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and the corresponding side of another right-angled triangle. 4.2.3 POLYGON A polygon is a geometric closed figure bounded by straight lines. The term poly means multi. A triangle has the least number of sides. Other multi-sided figures have names indicating the number of sides. Hence: Pentagon – 5 sided, Hexagon – 6 sided, Octagon – 8 sided Page 4-46 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4.2.4 QUADRILATERALS A quadrilateral is any four-sided shape. There are various types, some are common and you are probably familiar with their names. Some are not so common. Since a quadrilateral has four sides, it can be divided into two triangles. The sum of it’s angles must therefore be 360º. 4.2.5 PARALLELOGRAM A parallelogram has both pairs of opposite sides parallel. The following properties apply to parallelograms: Each pair of opposite sides is equal in length. Each pair of opposite angles are equal The diagonals bisect each other The diagonals bisect the parallelogram and form two congruent triangles Page 4-47 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4.2.6 RECTANGLE A rectangle is a parallelogram with it’s angle equal to 90º. It has the same properties as a parallelogram with the addition that the diagonals are equal in length. 4.2.7 RHOMBUS A rhombus is a parallelogram with all of it’s sides equal in length. It also has all of the properties of a parallelogram and the following additional properties: The diagonals bisect at right angles 4.2.8 SQUARE A square is a rectangle with all the sides equal in length. It has all the properties of a parallelogram, rectangle and rhombus. 4.2.9 TRAPEZIUM A trapezium is a quadrilateral with one pair of sides parallel. Page 4-48 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4.2.10 CIRCLES Circles are not just particular mathematical shapes but are involved in our everyday life, for example, wheels are circles, gears are basically circular and shafts revolve in a circular fashion. Hence, we must be aware of some important definitions and properties. If the line OP is fixed at O and rotated around O, the point P traces a path which is circular - it forms a circle. Page 4-49 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES The length OP is the Radius of the circle. Note that OP = OA = OB and that the length of the line AB is clearly equal to twice the radius. AB = 2OP. AB is the Diameter of the circle (D = 2R). We already know that if OP is rotated through 1 complete revolution, it will have rotated through 360 degrees, but what is the distance travelled by P in tracing this circular path? Put another way, how far will a wheel whose radius is R, roll along a surface, during one revolution? The distance, known as the Circumference is obviously dependent on the length of the length of the diameter, but can be calculated precisely from the equation C = pD (= 2pR). The value p is actually the ratio between the circumference of a circle and it’s diameter. p (Greek letter, pronounced "pi") can be approximated to 3.142. It will certainly be found on a scientific calculator, but the fraction Fehler! is a very good approximation. Page 4-50 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES The line AP drawn so that it touches the circle at point P is known as the Tangent to the circle. It should be noted that AP is always at right-angles to the radius OP. Example: A wheel, diameter 715 mm, makes 30 revolutions. How far does it move from its start point? The distance moved in 1 rev. = the length of the circumference. \ distance in 1 rev. = p x diameter = (p) (715) mm \ distance in 30 revs. = (30) (p) (715) = 67410 mm = 67.4 metres 4.2.10.1 Radian Measure We already know that an angle of 360º represents 1 complete revolution. But there is another important unit of angular measurement, known as the Radian. Consider a circle of radius R and consider an arc AB, where length is also equal to R. The angle at the centre of the circle, AOB is then equal to I Radian. Page 4-51 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES It can be deduced that I revolution is equivalent to 2p Radians, i.e. I rev = 6.2832 rads. Therefore 360º = 2p rads, and we can derive conversion factors, as that; 1º = 2p/360 radians, or 360/(2p) = 1 radian (approx. 57.3º) One final and useful point concerning radian measure. If an arc of a circle, radius r, subtends an angle, equal to q Radians, the length of the arc is r.q. Note also that if a point P is moving with speed N, then the rotational speed w is equal to (r x N)/r2 (N = r.w). w is expressed in Radians per second. Page 4-52 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4.3 AREA AND VOLUME 4.3.1 AREA We are already familiar with the concept of length, e.g. the distance between 2 points, we express length in some chosen unit, e.g. in meters. If we want to fit a picture-rail along a wall, all we need to known is the length of the wall, so that we can order sufficient rail. But if we wish to fit a carpet to the room floor, the length of the room is insufficient. Obviously we also need to know the width. This two- dimensional concept of size is termed Area. 4.3.1.1 Rectangular Area Consider a room 4m by 3m as shown above. Clearly it can be divided up into 12 equal squares, each measuring 1m by 1m. Each square has an area of 1 square meter. Hence, the total area is 12 square meters (usually written as 12m2 for convenience). So, to calculate the area of a rectangle, multiply length of one side by the length of the other side. 4m x 3m = 12m2 (Don't forget the m2). Page 4-53 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4.3.1.2 Area of Triangles This concept can be extended to include non-rectangular shapes. Consider the triangles ABC and ADC which together form a rectangle ABCD. Inspection reveals the 2 triangles are congruent. Hence their areas are equal and the area of ABC = ½ area of ABCD. If we consider this diagram, the area of the triangle can be seen to equal 1/2 x base x perpendicular height. This is true for any triangle, but remember its the perpendicular height. Note again that base (in meters) x height (in meters) gives m2. A theorem exists stating that triangles with the same base and drawn between the same parallels will have the same area. Page 4-54 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4.3.1.3 Area of Circular Shapes The area of a circle is given by the formula: A = pr2 (where r = radius) or 2 æ dö pd2 A = pç ÷ = (if the diameter is given r = d/2) è 2ø 4 Remember that any area is so many square units. So the area of a circle must include a 'squared' term; Example: What is the area of a semi-circle where the diameter is 30cm? 2 æ 30 cm ö Area of circle = pç ÷ è 2 ø 2 æ1ö æ 30cm ö semi - circle = ç ÷ (p ) ç ÷ è2ø è 2 ø æ1ö ç ÷ (p )(15 cm ) = 353.43 cm 2 = 2 2 è ø 4.3.1.4 Area of Other Shapes The table below indicates the areas of many common shapes. SHAPE AREA Circle pd2 pr 2 or 4 Triangle ½ base x height Rectangle Base x height Square Side2 Parallelogram Base x vertical height Trapezium ½(sum of length of parallel sides) x vertical height Page 4-55 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4.3.1.5 Calculation of Areas of Shapes Sometimes an area calculation must be made where the object or shape is not one of the common shapes listed. Sometimes it is made up from a combination of shapes. Example: An office 8.5m by 6.3m is to be fitted with a carpet, so as to leave a surround 600mm wide around the carpet. What is the area of the surround? With a problem like this, it is often helpful to sketch a diagram. The area of the surround = office area - carpet area. = (8.5 m x 6.3 m) - (8.5 m - 2 x 0.6 m) (6.3 m - 2 x 0.6 m) = 53.55 m2 - (7.3 m) (5.1 m) = 53.55 m2 - 37.23 m2 = 16.32 m2 Note that 600mm had to be converted to 0.6m. Don't forget to include units in the answer e.g. m2. We may need to find the area of an object that is a combination of shapes: In this case the shape comprises a rectangle and a semi-circle. The rectangle has dimensions 150mm x 100mm The semi-circle has a diameter of 100mm Total area is the sum of the two individual areas. Area = (100 x 150) + pr 2 = 15000 + p ´ 2500 = 15000 + 7854 = 22854mm 2 Page 4-56 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 4.3.2 VOLUMES Solids are objects that have three dimensions: length, width and height. Having the ability to calculate volume enables you to determine the capacity of a fuel tank or reservoir, calculate the capacity of a cargo area or work out the volume of a cylinder. Volumes are calculated in cubic units such as cubic centimetres, cubic metres, cubic inches etc. However, volumes are easily converted to other terms, such as litres. For example, a cubic metre contains 1000 litres of liquid. Instead of squares, we now consider cubes. This is a 3-dimensional concept and the typical units of volume are cubic metres (m3). If we have a box, length 4m, width 3m and height 2m, we see that the total volume = 24 cubic metres (24m3). Each layer contains 4 x 3 = 12 cubes. There are 2 layers. Hence the volume is 12 x 2 = 24m3. Basically, therefore, when calculating volume, it is necessary to look for three dimensions, at 90º to each other, and then multiply them together. For a box - type shape, multiplying length x width x height = volume. For irregular or particular shapes, different techniques or approximations can be used, or sometimes a specific formula may exist. For example: Volume of cylinder = pR2h (R = radius, L = height) Volume of cone = 1/3 x pR2h Volume of sphere = 4/3 x pR3 Page 4-57 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES Note that all these formulae contain 3 dimensions so that when multiplied, a volume will result. e.g. R2h = R x R x h or R3 = R x R x R If you have not got 3 dimensions, you have not got a volume! Example: What is the cubic capacity of a 2 cylinder engine, with a bore of 77mm and a stroke of 89mm? bore = diameter = 77mm stroke = height = 89mm Volume of cylinder = area of circle x height. 2 æ 77 mm ö \ Volume of 1 cylinder = pç ÷ x (89 mm ) è 2 ø Volume of 1 cylinder = 414440 mm3 Volume of 2 cylinders = 828880 mm3 Note that in this example, the dimensions have been given in mm. The volume would normally be given in cm3. Note, to convert mm3 to cm3, divide by (10)3. \ 828880 mm3 becomes 828.88 cm3. When calculating areas or volumes, remember the basic formulas, but be ready to spot when an area or solid body is a combination of basic shapes that can be added or subtracted. Page 4-58 - 80 Pages  EASA 66 CAT B1 Revision 3 Issue Date: MODULE 01 15.05.2010 MATHEMATICS Revision Date: 12.06.2024 ENGINES 5 GRAPHS Graphs are a pictorial method of displaying numerical data that enables you to quickly visualise certain relationships, complete complex calculations and predict trends. The data can be presented in many different ways as shown below, and most data can be presented in any format. However, care should be taken when selecting a format to use, some formats are better suited to particular types of data or data sets. For example, if have a whole amount divided into known proportions, then this is better presented as a pie chart; if we have a list of scores in a test, then a bar graph is better. If we are plotting temperature with respect to time then a continuous line graph is better, 5.1 CONSTRUCTION In order to construct graphs effectively, some simple rules should be followed. First of all, present the data in a clear, tabular form. The data will data will generally comprise 2 variables, one that is being varied, the independent variable, and the one that changes as a result of the variation, the dependent variable (its value depends on the value of the other). Page 5-58 - 80 Pages  EASA 66 CAT B1 Rev