MATHS FOR MARINERS

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Ranger Hope © 2014

Introduction

Chapter 1: Logic

1.1 Logic, attributes and sets

1.2 Number, value and counting

1.3 Numbering systems and number base

1.4 Scientific notation - exponents and roots

Chapter 2: Calculations

2.1 Addition, subtraction, multiplication and division

2.2 Fractions and decimals

2.3 Calculating machines

2.4 Elementary algebra

2.5 Trigonometry

Chapter 3: Measuring

3.1 International System of Units

3.2 Derived units and formulas

3.3 Measuring tools

Chapter 4: Shape, size and capacity of tanks

4.1 Definitions

4.2 Areas of common shapes

4.3 Volumes of tanks

4.4 Pumps and flow rate capacity

Chapter 5: Calculating fuel requirements for a voyage

5.1 Endurance

5.2 Specific fuel consumption and fuel coefficients

5.3 Propeller pitch

Chapter 6: Calculating a vessel’s draught and trim

6.1 Displacement and buoyancy

6.2 Moments and longitudinal trim

6.3 Moments and transverse trim

Chapter 7: Lifting equipment

7.1 Definitions

7.2 Strength of rope, wire and chain

7.3 Levers and moments

7.4 Blocks, tackles and purchases

7.5 Friction and efficiency

Chapter 8: Structure and strength

8.1 Stress and strain

8.2 Strength

Chapter 9: Refrigeration

Chapter 10: Electrical calculations

Introduction

This text is intended for those studying to begin a maritime career. It aims to jog the memory of those who may have long since left school and whose arithmetic, algebra, measure and geometry skills could be bettered.

This is a reference source for formulas and methods required to solve mathematical problems common in the mariner’s workplace. Although each section begins with basics, it is not intended to be read from start to finish. However, thumbing back to an earlier chapter may be helpful in refreshing some underpinning concepts.

I learnt arithmetic at school entirely by the rote method of daily chanting the times tables. But as schooling progressed I became entirely confounded by the mystery of slide rules, logarithms and calculus. I never really understood logic or numbers and like many I have survived with imperfect memories and a reliance on calculators. So this text is also a personal journey of re-learning of how to crawl before running. The first chapter that deals with the pre-mathematics skills of logic, sorting, base and numbering systems is provided for those who may also have been dozing at the back of the class.

Chapter 1: Logic

1.1. Logic, attributes and sets

Definitions

Logic refers to consistent rules where actions have predicable results.

Geometry describes how to draw shapes and defines the mathematical rules (relationships) concerning their construction.

Mathematics or maths describes the logical processing information using numbers.

Algebra enables logical solutions by balancing (equating) a set of calculations with a known value against one with a partially known value.

An attribute is that observable feature of a thing that makes it similar to or dissimilar from another thing. It defines what makes an individual.

To sort is to make collections (sets) of individuals determined by their common attributes.

A set is a group (collected) of individuals with one or more common attributes.

A number is a sequenced symbol used to represent an individual unit or the quantity of members of a set. The symbols we commonly use are derived from those of ancient India and Arabia. Those ancients invented the nine basic symbols of 1 representing an individual finger, 2, 3, 4, 5, 6, 7, 8, 9 representing subsequent sets of fingers (digits) and symbol 10 for the full set of fingers.

The value of a number is the size of the collection of individuals that it represents. An individual has a value of one unit.

To count is to use a systematic process to find the value of a collection. Methods have included such as using the fingers, tally marks, counters, number symbols or electronic calculators.

To process is to apply a logical rule to change an individual or to change the members of a set. In logical processes an action (a cause) has a consistent result (an effect). This predictability enables mathematics to be used to provide workplace solutions other than by trial and error.

Basic calculation is to use counters or numbers in the processes of addition (add), subtraction (take away), multiplication (times) and division (share).

Sets and sorting

To understand our world we seek consistent rules where actions have predicable results – we look for a cause leading to a repeatable affect – we apply logic.

Our world is filled with many things - some have similar and some dissimilar features. An attribute is a defining feature of a thing. The picture below shows a collection of things that have the common attribute of being coloured wooden blocks but differing attributes of colour, shape, thickness or size. Attributes define what makes a thing an individual. For instance, there is only one circle that is red, thick and large.

Attributes allow us to compare one thing with another, make rules to classify them into logical collections and sort them into sets (groups of individuals with similar attributes). The chaotic pile of blocks above can be logically sorted and made sense of in a number of ways. All the red blocks have been selected to form a red set, leaving a remainder of those that do not have the red attribute (are yellow or blue).

However we could just as easily have created different sets by using different attributes to separate the wooden blocks into like groups. The logical rule we have used below is to select for each set on the basis of size only – large or small attribute.

Or we could have created other sets by using other attributes of the wooden blocks. The logical rule we have used below is to select on the basis of thickness only – thin or thick attribute.

Other sets are created by using the other attributes of colour or shape.

In summary, attributes define what makes an individual. For instance, there is only one circle that is red, thick and large. Attributes also define the members that can make up a logical group so enabling set size to be compared. For instance, the size of each shape set above compares to a set of 10 of our fingers plus 2 of our toes.

1.2 Number, value, and counting

A number is a concept that represents the value of a set (collection). The number symbols used today are derived from those developed in ancient India and Arabia.

1 represents a finger or individual (a digit), 2, 3, 4, 5, 6, 7, 8, 9 represent the quantity or sequence of digits and 10 represents a full set of digits counted on both hands.

The value of a number describes the quantity of individuals in collection or the sequence of an individual’s position within a set - whether it is first, second, third etc.

Shown above are quantities of six block sets that have been separated from the number lines (each of ten blocks). Note that though both top and bottom sets contain 6 blocks, the blocks themselves are different. Number 6 is not the exclusive name of either set, but a description of a set’s attributes - they each contain six blocks.

Shown above are two number line sequences where the blocks have been rearranged. The name of top row yellow block number 4 is not a constant. If it is re-positioned in the bottom row then it is renamed as block number 3.

Shown below are two number line sequences where the blocks have been rearranged. This time the lower line has been reversed. Unlike the previous number lines, we can give each block a constancy of name if we apply a constancy rule - to reverse the starting point of our sequence from left to right, to right to left.

It is natural to use numbers as names from an object’s place in a group, such as 1^st finger, 2^nd finger and 3^rd finger. However, name and number are different ideas. Constancy of number relates to the logical rule not the name temporarily given to an object. An ability to trust in learnt mathematical rules and before relating back to real objects is needed in order to efficiently count, process and eventually to calculate.

To count is to use a systematic process to find the value of a collection. Methods include using the fingers, measuring, striking tally marks, use of counters, number lines, mental arithmetic, tables and electronic calculators.

Tally - an historical perspective

With the fingers on one hand we can communicate the idea of up to five things. Counting is seen in ancient tally marks scratched onto clay tablets to represent flocks of sheep or chalked as tally marks of present day stevedores. As each collection is gathered (recorded as groups of scratches) the marks are slashed through to reckon a total - below records two sets of 5 sheep and another 4.

Tallying requires only observation skills to accurately strike the marks, unlike processing (to apply a logical rule predicting a result). Processing is explained below.

Processing

Once you have defined the attribute/s of an individual, a rule can be invented to process or logically change that attribute/s. The processing rule used by the colour machine shown below is to paint (change to red) anything passed through the input to the output. All other attributes remain unchanged (the shape, size and thickness).

However we could just as easily have created different output by changing the processing rule. The rule used by the logical thickness machine shown below is to hammer (change to thin) anything passed through the input to the output. All other attributes remain unchanged (the shape, size and colour).

Or we could just as easily have created different output by another processing rule. The rule used by the logical rounding machine shown below is to hammer (change to round) anything passed through the input to the output. All other attributes remain unchanged (the thickness, size and colour).

Similarly the rule used by the logical copy machine shown below is to make two additional copies of anything passed through the input to the output. All attributes remain unchanged as the additional units are exact copies (clones).

We can represent this adding machine as a sum using the processing symbols of + for plus (adding) and = for equals (the result). For instance, for each block inputted:

1 + 2 = 3

1 (red square) + 2 (red squares) = 3 (red squares)

1 (blue circle) + 2 (blue circles) = 3 (blue circles)

1 (yellow triangle) + 2 (yellow triangles) = 3 (yellow triangles)

1 (red circle) + 2 (red circles) = 3 (red circles)

Likewise the removing machine above that takes away two of anything passed through it can be represented by a subtraction using the symbols - for subtract (take away) and = for equals (the result). For instance, for each group of blocks inputted:

3 - 2 = 1

3 (red squares) - 2 (red squares) = 1 (red square)

3 (blue circles) - 2 (blue circles) = 1 (blue circle)

3 (yellow triangles) - 2 (yellow triangles) = 1 (yellow triangle)

3 (red circles) - 2 (red circles) = 1 (red circle)

The logical multiplying machine shown below that trebles anything passed through it can be represented by a multiplication using the symbols x for multiply (times) and = for equals (the answer). For instance, for each block inputted:

1 x 3 = 3

2 (yellow triangles) x 3 = 6 (yellow triangle)

2 (blue circles) x 3 = 6 (blue circle)

2 (red squares) x 3 = 6 (red square)

The logical sharing machine shown below can be represented by a division using the symbols ÷ for divide (share) and = for equals (the answer). For instance, for each group of blocks inputted:

Tables and examples

Add +	Take away -		Times x	Share ÷
1 + 1 = 2	1 - 1 = 0	5 - 5 = 0	1 x 1 = 1	1 ÷ 1 = 1
1 + 2 = 3	2 - 1 = 1	6 - 5 = 1
1 + 3 = 4	3 - 1 = 2	7 - 5 = 2	2 x 2 = 4	2 ÷ 2 = 1
1 + 4 = 5	4 - 1 = 3	8 - 5 = 3	2 x 3 = 6	4 ÷ 2 = 2
1 + 5 = 6	5 - 1 = 4	9 - 5 = 4	2 x 4 = 8	6 ÷ 2 = 3
1 + 6 = 7	6 - 1 = 5		2 x 5 = 10	8 ÷ 2 = 4
1 + 7 = 8	7 - 1 = 6	6 - 6 = 0		10 ÷ 2 = 5
1 + 8 = 9	8 - 1 = 7	7 - 6 = 1	3 x 2 = 6
1 + 9 = 10	9 - 1 = 8	8 - 6 = 2	3 x 3 = 9	3 ÷ 3 = 1
		9 - 6 = 3		6 ÷ 3 = 2
2 + 2 = 4	2 - 2 = 0	7 - 7 = 0		9 ÷ 3 = 3
2 + 3 = 5	3 - 2 = 1	8 - 7 = 1	4 x 2 = 8
2 + 4 = 6	4 - 2 = 2	9 - 7 = 2
2 + 5 = 7	5 - 2 = 3		5 x 2 = 10	4 ÷ 4 = 1
2 + 6 = 8	6 - 2 = 4	8 - 8 = 0		8 ÷ 4 = 2
2 + 7 = 9	7 - 2 = 5	9 - 8 = 1
2 + 8 = 10	8 - 2 = 6	9 - 9 = 0		5 ÷ 5 = 1
	9 - 2 = 7			10 ÷ 5 = 2
3 + 3 = 6	3 - 3 = 0	10 - 1 = 9
3 + 4 = 7	4 - 3 = 1	10 - 2 = 8
3 + 5 = 8	5 - 3 = 2	10 - 3 = 7
3 + 6 = 9	6 - 3 = 3	10 - 4 = 6
3 + 7 = 10	7 - 3 = 4	10 - 5 = 5
	8 - 3 = 5	10 - 6 = 4
	9 - 3 = 6	10 - 7 = 3
		10 - 9 = 2
4 + 4 = 8	4 - 4 = 0	10 - 9 = 1
4 + 5 = 9	5 - 4 = 1
4 + 6 = 10	6 - 4 = 2
	7 - 4 = 3
5 + 5 = 10	8 - 4 = 4
	9 - 4 = 5
Plus	Minus		Multiply	Divide

1.3 Numbering systems and number base

All numbering systems employ a base set of symbols. The commonly used, base ten (1₁₀), imitates a full set of fingers (our digits). Combinations of the nine symbols of 1, 2, 3, 4, 5, 6, 7, 8, 9 (representing each value) and 0 (representing zero or tens of units) can represent any number. Two sets are then 20, three 30, and so on.

Others bases are used when convenient, such as base two or binary system used in computer code. Below the tally marks show eleven marked in sets of two (Base ₂) and twenty three marked off in both sets of five (Base ₅) and in sets of ten (Base ₁₀).

Base Five (1₅)

Base five numbering system imitates a set of fingers on one hand only and (example below) uses combinations of four symbols and a zero to express all numbers.

In base₅, we have an individual digit, or we can build a square or a cube of with groups of individuals. The total value of each pile is logically numbered as shown below using the invented symbols of slashes on sticks.

In base₅ using our invented symbols we can express sets, values and a number to represent each value as:
Base Sets	Digits Value	Number Symbol
Two sets are written (digits of both hands)
Three sets are written
Five sets are written

However, we could have used the Indus-Arabic symbols of 1, 2, 3, 4, and 0 to express our numbers. To represent the thumb a zero is now partnered with symbol 1 so the number symbol of 10 now represents a full set of five digits.

Take care that you only use one base system at a time, as each base attributes a different value to its number symbols. Base₅ number 1000₅ does not have the same value as Base₁₀ number 1000₁₀. We use base₁₀ so universaly that we rarely annotate with the base system symbol or think about the bases of our numbering systems.

In base five using conventional Arabic symbols we can express sets, values and a number to represent each value as:
Base Sets	Digits Value	Base 5	Base 10
Two sets are written (digits of both hands)		20	10
Three sets are written		30	15
Five sets are written		100	25

Base Two (1₂)

The Base-two number system (binary numeral system) is used for computer code because electronic switches can easily read and store two states, on or off (twin voltage may be used by some systems). To these states are given the symbols 1 and 0.

A conversion from a binary number’s value to its equivalent base₁₀ value can be made with the table shown above. Binary values are derived from sets of two and progressive multiples. These values of 64, 512, 1028, and 2056 megabytes are used to specify the Random Access Memory power of a computer’s micro-processor.

1.4 Scientific notation - integral exponents and roots

Understand Chapter Two before reading this section. To reduce the zeros required to write the very large and very small numbers that engineers use, a short hand termed of numbers to a power (integral exponents) is used. These symbols are written by the right hand top of a number. Shown in the examples below, ten is the number and the power is to two and to three, commonly called squared or cubed.

Described as ten squared and written as 10² is calculated 10 x 10 = 100

Described as ten cubed and written as 10³ is calculated 10 x 10 x 10 = 1000

Similarly,

Ten to the fourth power is 10⁴ is calculated as 10 x 10 x 10 x 10 = 10,000

Ten to the fifth power is 10⁵ is calculated as 10 x 10 x 10 x 10 x 10 = 100,000

Any number can be written to a power as in the examples below:

Three squared is written as 3² and calculated as 3 x 3 = 9

Four cubed is written as 4³ and calculated as 4 x 4 x 4 = 32

Five to the fourth power is 5⁴ and calculated as 5 x 5 x 5 x 5 = 625

Six to the fifth power is 6⁵ and calculated as 6 x 6 x 6 x 6 x 6 =1296

When calculating with multiples of 10, the power defines the number of zeros (0):

Ten to the power of two has two zeros (100 = a hundred),

Ten to the power of three has three zeros (1000 = a thousand)

Ten to the power of six has six zeros (1,000,000 = a million)

Ten to the power of twelve has twelve zeros (1,000,000,000,000 = a billion)

* Note France & US use one thousand times one million to define a US billion)

For instance, below both numbers can be written to a power.

257 = 2.57 x 100 = 2.57 x 10²

2,570,000 = 2.57 x 1,000,000 = 2.57 x 10⁶

The convenience of exponents is further seen in expressing the concept of infinity (going on forever without end). The simple expression of ten to the power of infinity as below would otherwise require an unending task of writing 0 after 0 without end.

Exponents also simplify otherwise tedious calculations and lessen the chance of error in the positioning of a decimal point. As below, exponents can be written as powers of ten or as decimals by use of the minus sign.

10⁵ = 100,000 10ˉ⁵ = 0.00001

10⁴ = 10,000 10ˉ⁴ = 0.0001

10³ = 1000 10º = 1 10ˉ³ = 0.001

10² = 100 10ˉ² = 0.01

10¹ = 10 10ˉ¹ = 0.1

For consistency using decimal place, the following rule is used:

Place the decimal point immediately to the right of the first non zero digit.

Compare the number of digits between the new decimal point and the original.

This number is the value of the exponent.

0.00257 = 2.57 x 0.001 = 2.57 x 10ˉ³

0.257 = 2.57 x 0.1 = 2.57 x 10ˉ¹

Roots (inverse powers)

The root of a number is that value that when multiplied by itself for specified repeat instances will be equal to the number. It is the inverse (reversed process) of a power, for instance:

All numbers have a root as in the examples below:

Some numbers roots and number are the same, for example numbers 0 and 1 whose roots are 0 and 1:

The roots of some numbers are called irrational numbers because they never resolve as a whole number, for example the roots of 2, 5 and 7:

Fractional exponents

Some engineering problems require solving problems that involve the use of fractional powers. While these are easily solved following the stepped instructions with a scientific calculator, the following examples are shown in explanation.

Exponents can be written as fractions. In these cases exponent ½ indicates the power of 1 and square root of the number. Similarly exponent ⅓ indicates the cube root, ¼ the fourth root, ⅛ the eighth root, and so on.

In mathematical terms, the fractional exponents indicate that the numerator will raise a number to that power and the denominator takes it to that root.

Chapter 2: Calculations

Using the fingers to count or share more than ten units is difficult. This section describes other ways to solve arithmetic problems when using base ten, the commonly used numbering system. (See more about other bases in Section 1.3).

The four principle operations are denoted by symbols (signs) and are named:

+ Plus to add or to sum by addition

- Minus to subtract or to take away by a subtraction

x Times to multiply or times by a multiplication

÷ Divide by to divide or to share by a division

A calculation is a arithmetical problem with its solution (its answer). Calculations can be expressed (written down in maths shorthand) using an equation. The problem is written on the left, the solution on the right with an equals symbol (=) between them. In a balanced equation, each side of the equals sign must have the same total. (See more on equations and algebra in Section 2.4). Simple equations are shown below:

1 + 1 = 2 or 3 - 1 = 2 or 2 x 2 = 4 or 4 ÷ 2 = 2

1 plus 1 equals 2 3 minus 1 equals 2 2 times 2 equals 4 4 divide by 2 equals 2

2.1 Addition, subtraction, multiplication and division

Various methods used to find solutions are descibed below.

Using the Number Line method and mental arithmetic

A scale or ruler is a number line when used as a visual aid in addition (counting out), subtraction (finding a remainder), division (sharing) and multiplication (times).

In your mind cut the stick below into 12 slices and then count the slices out to share with ten persons. Ten can be evenly shared but two slices are left over. We have used the number line below to record our sharing and the remainder that was left.

What we have done can be expressed by the equations:

Addition Subtraction

1+1+1+1+1+1+1+1+1+1=10 shared orange 12 -10 = 2 remainder brown

Multiplication and division could also be visualised using the number line. The 12 slices could be counted out in sets (See 1.1 Sets and sorting):

2 sets of six slices can be written as 2 x 6 = 12 or 12 ÷ 6 = 2

3 sets of four slices can be written as 3 x 4 = 12 or 12 ÷ 3 = 4

4 sets of three slices can be written as 4 x 3 = 12 or 12 ÷ 4 = 3

6 sets of two slices can be written as 6 x 2 = 12 or 12 ÷ 2 = 6

While this scale is marked in twelve units, each unit could be further divided into 8 sub units to solve more complex problems (by counting up the units), as below:

1 units x 8 sub-units = 8 sub-units

4 units x 8 sub-units = 32 sub-units

12 units x 8 sub-units = 96 sub-units

The number line works well for small numbers but as it relies on counting out every unit it is slow for complex problems.

Using the Times Tables method

Fast solution to maths problems can be looked up or learnt as the times tables below. (also see addition tables Section1.2). They can also be used to find division solutions. It can be seen that one side of each equation is equal to the other. For example, in the times x 4 table equation of below 8 x 4 = 32, it is also true that 32 shared among 4 will have a solution of 8.

8 x 4 = 32 or 32 ÷ 4 = 8

Times x 1	Times x 2	Times x 3	Times x 4	Times x 5	Times x 6
1 x 1 = 1	1 x 2 = 2	1 x 3 = 3	1 x 4 = 4	1 x 5 = 5	1 x 6 = 6
2 x 1 = 2	2 x 2 = 4	2 x 3 = 6	2 x 4 = 8	2 x 5 = 10	2 x 6 = 12
3 x 1 = 3	3 x 2 = 6	3 x 3 = 9	3 x 4 = 12	3 x 5 = 15	3 x 6 = 18
4 x 1 = 4	4 x 2 = 8	4 x 3 = 12	4 x 4 = 16	4 x 5 = 20	4 x 6 = 24
5 x 1 = 5	5 x 2 = 10	5 x 3 = 15	5 x 4 = 20	5 x 5 = 25	5 x 6 = 30
6 x 1 = 6	6 x 2 = 12	6 x 3 = 18	6 x 4 = 24	6 x 5 = 30	6 x 6 = 36
7 x 1 = 7	7 x 2 = 14	7 x 3 = 21	7 x 4 = 28	7 x 5 = 35	7 x 6 = 42
8 x 1 = 8	8 x 2 = 16	8 x 3 = 24	8 x 4 = 32	8 x 5 = 40	8 x 6 = 48
9 x 1 = 9	9 x 2 = 18	9 x 3 = 27	9 x 4 = 36	9 x 5 = 45	9 x 6 = 54
10 x 1 =10	10 x 2 = 20	10 x 3 = 30	10 x 4 = 40	10 x 5 = 50	10 x 6 = 60
Times x 7	Times x 8	Times x 9	Times x 10	Times x 11	Times x 12
1 x 7 = 7	1 x 8 = 8	1 x 9 = 9	1 x 10 = 10	1 x 11 = 11	1 x 12 = 12
2 x 7 = 14	2 x 8 = 16	2 x 9 = 18	2 x 10 = 20	2 x 11 = 22	2 x 12 = 24
3 x 7 = 21	3 x 8 = 24	3 x 9 = 27	3 x 10 = 30	3 x 11 = 33	3 x 12 = 36
4 x 7 = 28	4 x 8 = 32	4 x 9 = 36	4 x 10 = 40	4 x 11 = 44	4 x 12 = 48
5 x 7 = 35	5 x 8 = 40	5 x 9 = 45	5 x 10 = 50	5 x 11 = 55	5 x 12 = 60
6 x 7 = 42	6 x 8 = 48	6 x 9 = 54	6 x 10 = 60	6 x 11 = 66	6 x 12 = 72
7 x 7 = 49	7 x 8 = 56	7 x 9 = 63	7 x 10 = 70	7 x 11 = 77	7 x 12 = 84
8 x 7 = 56	8 x 8 = 64	8 x 9 = 72	8 x 10 = 80	8 x 11 = 88	8 x 12 = 96
9 x 7 = 63	9 x 8 = 72	9 x 9 = 81	9 x 10 = 90	9 x 11 = 99	9 x 12 =108
10 x 7 = 70	10 x 8 = 80	10 x 9 = 90	10 x 10 =100	10 x 11 =110	10 x 12 =120

Using Short calculations - simple addition, subtraction, multiplication, division

Short addition calculations are written as columns to sum in an answer box below:

UNIT	UNIT	UNIT	UNIT
5	4	7	6
+3	+5	+2	+3
8	9	9	9

Lists can also be added. Note - for numbers of 10 or more the first step is to add the Units column (right hand in blue below), and the second step the left hand Tens:

		TENS UNITS	TENS UNITS
3	4	4	1 2
2	1	1 2	1 3
+3	+2	+1 2	+1 0
8	7	2 8	3 5

Check your progress addition -try these sums yourself:

2	4	5	2
1	1	1	3
+2	+2	+3	+5
?	?	?	??
5	6	12	21
5	10	14	13
+5	+12	+10	+15
??	??	??	??
Answers addition - as below:
2	4	5	2
1	1	1	3
+2	+2	+3	+5
5	7	9	10
5	6	12	21
5	10	14	13
+5	+12	+10	+15
15	28	36	49

Short subtraction calculations can also be written in the same format as below. Note - for numbers of 10 or more the first step is to add the Units column (U), and the second step adds the Tens column (T):

		T U	T U
5	6	17	32
-3	-5	-12	-11
2	1	5	21

Check your progress subtraction -try these calculations:

		T U	T U
6	5	15	25
-2	-3	- 3	-12
?	?	??	??
Answers subtraction - as below:
6	5	15	25
-2	-3	- 3	-12
4	2	12	13

Short multiplication calculations can be written as below. If we have 10 or more on the top line we multiply the Units first (U) and the Tens last (T):

		T U	T U
3	4	14	12
x 2	x 3	x 2	x 4
6	12	28	48

Check your progress multiplication -try these calculations:

		T U	T U
5	12	24	33
x 2	x 3	x 2	x 3
??	??	??	??
Answers multiplication - as below:
5	12	24	33
x 2	x 3	x 2	x 3
10	36	48	99

Short division calculations are written differently. The divisor (how many to share with) is written on the left while he quantity to share is boxed off to the right. The answer is written above the division box (in blue above).

If the quantity does not share equally we write down a remainder (in red above)

Check your progress division -try these calculations:

Answers division - as below:

Using Long calculations - addition, subtraction, multiplication and division

Long addition calculations - With larger numbers the Unit columns can add up to ten or more. To solve this kind of problem we first add the Unit column (blue) but in the answer box only write the unit part of the answer. We carry over the ten value to the Ten’s column, then sum that column with the carried over as the full answer.

In example one, the step one adds the units column 5 + 7 = 12. We write the 2 from the 12 in our answer box and then pencil in 1 (shown red) next to column of tens. We then add the new tens column 1 ten + 2 tens + the carried over ten and find the result is 4 lots of ten. The solution is 42. For larger sums we may need a step three with a Hundreds column (H).

Example :		Example :
Step 1	Step 2	Step 1	Step 2
T U	T U	T U	T U
1 5	¹1 5	3 5	¹3 5
+2 7	+2 7	+4 9	+4 9
? 2	4 2	? 4	8 4
Example :		Example :
Step 1	Step 2 & 3	Step 1	Step 2 & 3
H T U	H T U	H T U	H T U
2 1 5	2¹1 5	3 6 5	¹4¹6 5
+4 2 7	+4 2 7	+5 8 9	+5 8 9
? 3 2	6 4 2	? ? ? 4	1 0 5 4

Check your progress with long addition sums - try these calculations:
Question 1:		Question 2:
Step 1		Step 1
T U		T U
5 6		7 8
+3 9		+5 9
? ?		? ?
Question 3:		Question 4:
Step 1		Step 1
H T U		H T U
6 3 5		4 6 5
+2 6 7		+5 8 9
? ? ?		? ? ? ?
Answers with long addition sums - as below: - try these calculations:
Answer 1:		Answer 2:
Step 1	Step 2	Step 1	Step 2
T U	T U	T U	T U
5 6	¹5 6	7 8	¹7 8
+3 9	+3 9	+5 9	+5 9
? 5	9 5	? 7	1 3 7
Answer 3:		Answer 4:
Step 1	Step 2 & 3	Step 1	Step 2 & 3
H T U	H T U	H T U	H T U
6 3 5	¹6¹3 5	4 6 5	¹4¹6 5
+2 6 7	+2 6 7	+5 8 9	+5 8 9
? ? 2	9 0 2	? ? ? 4	1 0 5 4

Long subtraction calculations – either of two methods described below are used:

First method - When the top line units are greater than the bottom line, we have to borrow from the next greater column (tens or hundreds). The first Step borrows from the ten column by pencilling 1 (ten) next in the unit’s column enabling completion of the units part of the answer. To pay back the borrowed ten we cross out the bottom ten’s numeral and add to it a ten payback correction.

In example nine, subtracting the units 5 - 6 not possible, so the first Step is to borrow as 15 – 6 = 9. But the ten has to be paid back so in Step two the 2 (tens) bottom numeral is crossed and replaced with 3, so enabling the solution of 19 to be written in the answer box. In example eleven we have to borrow from both the ten’s and from the hundred’s column to get our answer.

Example :		Example :
Step 1	Step 2	Step 1	Step 2
Example :
Step 1	Step 2	Step 3
Example :
Step 1	Step 2	Step 3
Note in example twelve the hundreds column of 6 hundreds minus 6 hundreds gives an answer of zero hundreds (0). If the taking away number is larger than the number to be taken away from then the arrangement above will not work. In such cases we reverse position of the lines so we can continue taking away a smaller number from a larger, but call the answer we call a minus number. See example thirteen where we have 637- 834 = -197. For our calculation to work we must take away 637 (the smaller) from 834 (the larger) and call the answer a minus number (-197)
Example :
Step 1	Step 2	Step 3

Long subtraction calculations

Second method - When the top line units are greater than the bottom line, we have to borrow from a higher column. The first Step borrows from the ten’s column by pencilling 1 (ten) next in the top line unit so enabling completion of the units part of the answer. To pay back the borrowed ten we cross out top line ten’s numeral and subtract a ten from it for a payback correction.

In example fourteen, Step One finds subtracting the units 5 - 6 not possible, so ten is borrowed to calculate 15 – 6 = 9. But the ten has to be paid back so in Step Two the 4 (tens) top numeral is crossed out and replaced with ten less (3), so enabling the answer box to be completed with the same solution as the first method of 19. In example sixteen we have to borrow from the ten’s and from the hundred’s column to get our answer.

Example :			Example :
Step 1	Step 2		Step 1	Step 2
Example :
Step 1	Step 2		Step 3
Example :
Step 1	Step 2		Step 3
Check your progress with long subtraction - try these calculations:
Question 1:		Question 2:
T U		T U
7 7		4 2
- 5 8		- 3 9
? ?		? ?
Question 3:		Question 4:
H T U		H T U
6 2 3		8 5 3
- 4 8 9		- 5 6 5
? ? ?		? ? ?

Answers long subtraction:
Answer 1:		Answer 2:
Step 1	Step 2	Step 1	Step 2
T U	T U	T U	T U
Answer 3:
Step 1	Step 2	Step 3
H T U	H T U	H T U
Answer 4:
Step 1	Step 2	Step 3
H T U	H T U	H T U

Long multiplication calculations – Times tables are not readily available for multipliers of over twelve, so greater calculations are done in steps. The first Step is to multiply the units, then the tens and then hundreds. Finally all the answer lines are added. Remember to add a zero (0) as will be required for the ten line to be the correct value, as in Example 18 below:

Example :
Step 1	Step 2	Step 3	Step 4
H T U	H T U	H T U	H T U

Example :
Step 1		Step 2		Step 3		Step 4
H T U		H T U		H T U		T H T U
Check your progress multiplication -try these calculations:
Question 1:	Question 2:		Question 3:		Question 4:
4 2 3	2 3 7		5 6 8		7 7 7
x 2 5	x 5 2		x 3 1 2		x 4 3 6
? ? ?	? ? ?		? ? ?		? ? ?
Answers multiplication - as below:
Answer 1:	Answer 2:		Answer 3:		Answer 4:
4 2 3	2 3 7		5 6 8		7 7 7
x 2 5	x 5 2		x 3 1 2		x 4 3 6
2 1 1 5	4 7 4		1 1 3 6		4 6 6 2
8 4 6 0	1 1 8 5 0		5 6 8 0		2 3 3 1 0
1 0,5 7 5	1 2,3 2 4		1 7 0 4 0 0		3 1 0 8 0 0
			1 7 7,2 1 6		3 3 8 7 7 2

Long division calculations – In example 20 below, Step 1 estimates how many divisors (5) fit in the left hand boxed numeral (1). If it will not fit, Step 2 estimates how many fit into the first + the second numerals (1 & 3), in the example giving 2. As 2 x 5 =10 we subtract 10 from 1 & 3 to leave 3. Step 3 drops the third numeral (5) to give 35. Step 4 finds that the divisor fits 7 times into 35 giving a final answer of 27.

Example :
Step 1	Step 2	Step 3	Step 4

Example :
Step 1	Step 2	Step 3	Step 4
Example :
Step 1	Step 2	Step 3	Step 4
Example :
Step 1	Step 2	Step 3	Step 4
Example :
If the quantity does not share equally we write down a remainder (in red below), or write it as a fraction (see next Section 2.3)
Step 1	Step 2	Step 3	Step 4

Check your progress long division -try these calculations:
Question 1:	Question 2:	Question 3:	Question 4:
Answers long division - as below:
Answer 1:	Answer 2:	Answer 3:	Answer 4:

2.3 Fractions and decimals

Fractions

To share a cake we slice it into parts. Each part can be called a fraction of the whole cake, for example, slice number 1 below is from a cake with 8 slices. It is one eighth of the cake. All fractions are parts of a whole.

Fraction’s are expressed (written) separated by a line – as above	1	of a cake.
	8

A numerator - a number on top showing us the parts of the whole we have.

A denominator - a bottom number showing how many parts make up the whole.

The cake could have been cut into any other fractions, such as:

1	=	one slice of a whole cake cut into two	=	half a cake
2
1	=	one slice of a whole cake cut into four	=	quarter of a cake
4
1	=	one slice of a whole cake cut into ten	=	tenth of a cake
10

Calculating with fractions

Addition – with a common denominator

To add fractions the denominators (the number below the line) must be the same (called common). The answer is found by totalling the numerators (number above the line) of each fraction and expressing with the common denominator, as below:

1		+		2	+	3	+	4		=		10	half a
20				20		20		20				20	cake
Expression and Substitution However, we could have received several cake slices expressed as fractions below:
4	=		four slices of a whole cake cut into eight						=		1	half of cake
8											2
4	=		four slices of a whole cake cut into twelve						=		1	third of cake
12											3

You will notice that both 4 slices of an 8 slice cake and10 slices of a 20 slice cake both are half of each cake. So fractions can be expressed with differing numerator and denominator and still be the same value as a part of the whole, for instance:

10	=	1	=	5	=	2	=	10
20		2		10		4		20

The rule is that we must change both the top and the bottom line of a fraction with the same factor (divisor or multiplier) if the fraction is to retain its value. Applying this rule is called substitution.

10 x 4	=	40 ÷ 10	=	4 x 3	=	12 ÷ 12	=	1	ü
20 x 4		80 ÷ 10		8 x 3		24 ÷ 12		2
Note below that substitution does not work by adding or subtracting to numerator and denominator. Only by multiplying and dividing equally gives a correct answer.
10 + 4	=	14		X Wrong!!		10 - 4	=	6	X Wrong !!
20 + 4		24				20 - 4		16

Addition – fractions without a common denominator

We need a common denominator to add fractions. But in the example below different cakes have been cut into different numbers of slices. Each cake’s denominator is different. We have to change each fraction to the same denominator but still express its same value (part of a whole cake). All fractions must be of same sized slices.

To find our common denominator we could multiply all denominators, as below:

10	+	3	+	6	+	2	=	?
20		5		10		4		?
20	x	5	x	10	x	4	=	4000

The common denominator in this case is 4000, a very large number of tiny slices of a cake and a complex number for calculations. It will be easier for us to find the lowest common denominator, the smallest number that all denominators will divide into without leaving any remainder. In the example below it is 20.

Step 1 - draw a long line above our chosen lowest common denominator 20.

Step 2 - divide the lowest common denominator with each fraction’s denominator.

Step 3 - multiply the answer with each old numerator. Mathematical balance is regained with the value being equal to each old fraction’s value. Finally total the numerators and express the answer with the lowest common denominator.

10	+	4	+	6	+		2		=	?
20		5		10			4			?
Step 3:
1 x 10 = 20		4 x 4 = 12		2 x 6 = 12			5 x 2 = 10
Step 2:
20 ÷ 20 = 1		20 ÷ 5 = 4		20 ÷ 10 = 2			20 ÷ 4 = 5
Step 1:
10 + 16 + 12 + 10						=		48	=		2	8	=	2	2
20								20				20			5

Unlike earlier examples our answer is an improper fraction. The numerator 48 is greater than the denominator 20. Remember, the slices are from more than one cake. To express properly we divide the numerator with the denominator. The 48 pieces from 20 sliced cakes could be reassembled as 2 wholes with 8 slices left over. Eight twentieths can be substituted as the same as two fifths of a cake.

Sometimes we have to add whole numbers with fractions as below. The method is to convert each whole number with fraction into an improper fraction, find the common denominator and then continue as normal addition. The answer, an improper fraction, is converted back to whole number with any remainder fraction. The remainder can be substituted with lowest denominator for the simplest expression.

1	1		+	2	2		+	1	1		=	?
	2				3				6
2	+	1	+	6	+	2	+	6	+	1	=	?
2		2		3		3		6		6
	3		+		8		+		7		=	?
	2				3				6
	9 + 16 + 7										=	32	=	5	2	=	5	1
				6								6			6			3

Subtraction

As with adding, subtracting fractions requires a common denominator. The answer is found by subtracting the numerators and expressing with the common denominator:

19	-	10	-	5	-	3	=	1
20		20		20		20		20

If the denominator is not common, we have to substitute to express all with the same denominator.

Step 1 - draw a long line above our chosen lowest common denominator 20.

Step 2 - divide the lowest common denominator with each fraction’s denominator.

Step 3 - multiply the answer with each old numerator. Mathematical balance is regained with the value being equal to each old fraction’s value. Finally subtract the numerators and express the answer with the lowest common denominator.

19	-	1	-	2	-	1	=	?
20		2		10		5		?
Step 3:
1 x 19 = 19		10 x 1 = 10		2 x 2 = 4		4 x 1 = 4
Step 2:
20 ÷ 20 = 1		20 ÷ 2 = 10		20 ÷ 10 = 2		20 ÷ 4 = 5
Step 1:
19 - 10 - 4 - 4							=	1
20								20

Sometimes we have to subtract whole numbers with fractions. The method is to convert each number with fraction into an improper fraction, find the common denominator and then continue as normal subtraction. The answer, an improper fraction, is converted back to whole number with any remainder. The remainder can be substituted with lowest denominator for the simplest expression.

5	1		-	1	2		-	2	3		=	?
	2				3				6
10	+	1	-	3	+	2	-	12	+	3	=	?
2		2		3		3		6		6
	11		-		5		-		15		=	?
	2				3				6
	33 - 10 - 15										=	8	=	1	2	=	1	1
					6							6			6			3

Check your progress adding and subtracting fractions -try these calculations:

Question 1:																			Question 2:
1		+		2			+		1			=		?					2			+		3		+	4	=			?
6				6					6					?					10					10			10				?
Question 3:
1		+		2			+		5			=		?
2				3					6					?
Question 4:																			Question 5:
9		-		2			-		5			=		?					19			-		4		-	4	=			?
10				10					10					?					22					22			22				?
Question 6:
9		-		1			-		2			=		?
12				3					6					?
Question 7:
1	1		+		3	3		+		2	3	=				?
	2					4					8					?
Question 8:
8	2		-		2	1		-		3	4		=		?
	3					2					6				?
Answers adding and subtracting fractions - as below:
Answer 1:																			Answer 2:
1		+		2			+		1			=		4					2			+		3		+	4	=			9
6				6					6					6					10					10			10				10
Answer 3:
1		+		2			+		5			=		3 + 4 + 5										=		12	=	2
2				3					6					6												6
Answer 4:																			Answer 5:
9		-		2			-		5			=		2					19			-		4		-	4	=			11		=		1
10				10					10					10					22					22			22				22				2
Answer 6:
9		-		1			-		2			=		9 - 4 - 4										=		1
12				3					6					12												12
Answer 7:
1	1		+		3	3		+		2	3	=				3	+	15		+	19		=		12 + 30 + 19				=	61		=		7		5
	2					4					8					2		4			8				8					8						8
Answer 8:
8	2		-		2	1		-		3	4		=		26		-	5		-	22		=		52 - 15 - 22				=	15		=		2		3
	3					2					6				3			2			6				6					6						6

Multiplication with fractions

Common denominators are not required to multiply fractions, as shown below:

Example :												Example :
1		x			1		=		1			1		x		2		x		4		=			8	=	2
2					2				4			2				3				6					36		9
With whole number and fraction combinations, each can be multiplied separately. Note that the answer is usually expressed as the simplest fraction (substituted with a lowest common denominator). In 8 over 36 above, both are divided by 4.
Example :												Example :
2	1		x	3		3	=		6		3	3	2		x	4	4		x	6	1			=	72	8	=	72	1
	2					4					8		3				6				4					72			9
Check your progress multiplying fractions -try these calculations:
Question 1:												Question 2:
3		x		2			=	?				1		x		5		x		2			=		?
5				3								2				6				3
Question 3:												Question 4:
2	5		x	2				?				5	2		x	4	5		x	3	1			=	?
	6												3				6				5
Answers multiplying fractions - as below:
Answer 1:												Answer 2:
3		x		2			=	6				1		x		5		x		2			=		10	=	5
5				3				15				2				6				3					36		18
Answer 3:												Answer 4:
2	5		x	2		3	=	4		15		5	2		x	4	5		x	3	1			=	60	10	=	60	1
	6					4				24			3				6				5					90			9

Division with fractions

Dividing fractions with fractions is easiest achieved by inverting the second fraction’s numerator and denominator and then multiplying. Note that the answer is usually expressed by substitution as the simplest fraction.

NUM	÷	num	=	NUM	x	den	=	Nd
DEN		den		DEN		num		Dn
Example :
3	÷	1	=	3	x	2	=	6	=	1	2	=	1	1
4		2		4		1		4			4			2
Example :
6	÷	1	=	6	x	3	=	18	=	2	2	=	2	1
8		3		8		1		8			8			4

With whole number and fraction combinations it is easiest to covert each to an improper fraction. Then invert the dividing fraction and multiplying with it. Note that the answer is usually expressed as the simplest.

Example :
1	1	÷		1	=	2 + 1		÷		1		=	3	÷	1	=	3		x		4		=	12			=	6
	2			4		2				4			2		4		2				1			2
Example :
2	1	÷	1	2	=	6 + 1		÷		6 + 2		=	7	÷	8	=	7		x		6		=	42			=	1	18
	3			6		3				6			3		6		3				8			24					24
Check your progress dividing fractions -try these calculations:
Question 1:
1		÷		1			=		?
2				4
Question 2:
5		÷		2			=		?
6				3
Question 3:
2	1	÷	1	1			=			?
	2			4
Question 4:
2	5	÷	1	6			=			?
	6			9
Answers dividing fractions - as below:
Answer 1:
1		÷		1			=		1		x		4		=	4		=		2
2				4					2				1			2
Answer 2:
5		÷		2			=		5		x		3		=	15		=		1		3			=	1		1
6				3					6				2			12						12						4
Answer 3:
2	1	÷	1	1	=	4 + 1		÷		4 + 1		=	5	÷	5	=	5		x		4		=	20			=	2
	2			4		2				4			2		4		2				5			10
Answer 4:
2	5	÷	1	6	=	12 + 5		÷		9 + 6		=	17	÷	15	=	17		x		9		=	153			=	1	21
	6			9		6				9			6		9		6				15			90					30

Decimals

A dot or point (·) following a numeral is another way to express a fraction of its whole. A decimal point expresses a fraction that is a division of ten or its multiples when using the base ten numbering system (see more about base in Section 1.3), for instance:

1· 0 One whole only.

1· 1 One whole and one tenth of a whole.

1· 25 One whole and twenty five hundredths whole (¼).

1· 5 One whole and five tenths whole (½).

1· 75 One whole and seventy five hundredths whole (¾).

1· 9 One whole and nine tenths of a whole.

The decimal place determines the fractions multiple of ten. If it is placed immediately to the right of a whole number it expresses a tenth (1 x 10), if it is placed two spaces to the right it expresses one hundredth (10 x 10), etc, as below:

1· 1 One whole and one tenth of a whole.

1· 01 One whole and one hundredth of a whole.

1· 001 One whole and one thousandths of a whole.

1· 0001 One whole and the thousandths of a whole.

1· 00001 One whole and one hundred thousandths of a whole.

1· 000001 One whole and one millionths of a whole.

Adding and subtracting

Writing sums in a table format (or columns) ensures that the decimal place of the answer is correctly positioned above the decimal place of the question.

Example:	Example:	Example:
1· 1	1· 52	106· 325
+ 3· 6	+ 33· 34	+ 521· 520
4· 7	34· 86	627· 845
Example:	Example:	Example:
3· 6	33· 34	521· 520
- 1· 1	- 1· 52	- 106· 325
2· 5	31· 82	415· 195

Multiplying

It is easiest to multiply each whole numeral and decimal numeral as in long multiplication. After completing the calculation, the answer’s decimal place is determined from counting the places found in the question, as below:

Example:	Example:	Example:
3·6	3 3·3 4	2 1·5 1 1
x 1·1	x 2·1 2	x 6·3 2 5
3 6	6 6 6 8	1 0 7 5 5 5
3 6 0	3 3 3 4 0	4 3 0 2 2 0
3·9 6	6 6 6 8 0 0	6 4 5 3 3 0 0
	7 0·6 8 0 8	1 2 9 0 6 6 0 0 0
		1 3 6·0 5 7 0 7 5

As a double check, ensure that the answer’s decimal place roughly meets the expected units. Above 3 x 1 should roughly be 3 units and 33 x 2 should be 66.

Dividing

It is easiest to multiply each numeral and decimal by a multiple of ten until they become whole numbers, before calculating as in long division. If both the divisor and the divider are multiplied by the same amount then the proportions and answer will be correct.

Example:	Example:
(2 4 ÷ 0·2) x 1 0 = (2 4 0 ÷ 2) =	(3 6 ÷ 0·0 6) x 1 0 0 = (3 6 0 0 ÷ 6) =
1 2 0	6 0 0
2 )2 4 0	6 )3 6 0 0
Example:	Example:
4·8 ÷ 2 x 1 0 = (4 8 ÷ 2 0) =	(1 5·7 5 ÷ 0·75) x 1 0 0 = (1 5 7 5 ÷ 75) =
2·4	2 1
2 0 )4 8·0	75 )1 5 7 5
4 0	1 5 0
8 0	7 5

In the examples above the numerals and decimals were multiplied to become whole numbers, but we could just have easily moved the decimal places to the right to achieve the same result.

Converting fractions to decimals

To convert fractions to decimals the numerator is divided by the denominator. Don’t

always expect a simple decimal answer unless the fraction is a division of ten.

Example:			Example:			Example:
1	=	0·5	3	=	0·7 5	5	=	0· 6 2 5
2		2)1·0	4		4)3·0 0	8		8)5·0 0 0
					2 8			4 8
					2 0			2 0
								1 6
								4 0

Check your progress decimals -try these calculations:
Question 1:	Question 2:	Question 3:
5· 2	11· 23	345· 345
+ 3· 6	+ 24· 62	+ 123· 123
?	?	?
Question 4:	Question 5:	Question 6:
5· 2	24· 62	573· 267
- 3· 6	- 1· 52	- 106· 325
?	?	?
Question 7:	Question 8:	Question 9:
5·7	2 7·3 2	4 1·2 2
x 2·1	x 1·4 2	x 2·3 3
?	?	?
Question 10:	Question 11:
7·2 ÷ 4 = ?	3 4·2 5 ÷ 0·25 = ?

Answers decimals - as below:
Answer 1:	Answer 2:	Answer 3:
5· 2	11· 23	345· 345
+ 3· 6	+ 24· 62	+ 123· 123
8· 8	35· 85	468· 468
Answer 4:	Answer 5:	Answer 6:
5· 2	24· 62	573· 267
- 3· 6	- 1· 52	- 106· 325
1· 6	23· 10	466· 942

Answer 7:	Answer 8:		Answer 9:
5·7	2 7·3 2		4 1·2 2
x 2·1	x 1·4 2		x 2·3 3
5 7	5 4 6 4		1 2 3 6 6
1 1 4 0	1 0 9 2 8 0		1 2 3 6 6 0
1 1·9 7	2 7 3 2 0 0		8 2 4 4 0 0
	3 8·7 9 4 4		9 6·0 4 2 6
Answer 10:		Answer 11:
(7·2 ÷ 4) x 1 0 = 72 ÷ 4 0 =		(3 4·2 5 ÷ 0·25) x 1 0 0 = 3 4 2 5 ÷ 25 =
1·8		1 3 7
4 0 )72·0		25 )3 4 2 5
4 0		2 5
3 2 0		9 2 5

2.3 Calculating machines

The abacus – a counting machine

The abacus is an ancient counting machine that is constructed in a variety of ways across the world. It consists of a frame supporting a number of rails upon which beads can slide, so recording counting operations (addition or subtraction). Each rail supports a group of beads that are attributed with levels of value.

In the decimal based Abacus shown below, the beads on the bottom rail each have a value of one unit, the rail above contains beads of value ten, the rail above contains beads of value one hundred and the rail above contains beads of value one thousand.

When the maximum number of beads for each rail is slid to the left, their combined value will equal one of that above. The operation is to slide one of the above to the left and slide the ten below back to the right. The status shown on the abacus is the value of one thousand, four hundreds, two tens and three units (1423 units in base ten system). This Abacus is limited to recording a maximum value of 10,999 units.

The calculator

Addition, subtraction, multiplication and division of fractions and decimals, as well as conversions from fraction form to decimal form, can all be undertaken on a standard calculator. A scientific calculator has additional functions including but not limited to exponents, roots, constants and trigonometry.

The keypads and operation can differ from one calculator to another, as described in the user’s manuals, but the keying below is common. If in doubt, as with algebra, use the processing order of bemdas.

Addition and subtraction

Press all addition values, equal, press all negative values, equal. Example:

2

+

3

+

4

+

5

+

6

=

20

-

6

-

5

-

4

-

3

-

2

=

0

Multiplication and division

Press all multiplication values, equal, press all divisions (÷ key is / ), equal. Example:

2

x

3

x

4

x

5

x

6

=

720

/

6

/

5

/

4

/

3

/

2

=

0

Mixed operators

The problem below is set out in the bemdas sequence. Example,

2 x 3 x 4 = 24 ÷ 2 ÷ 3 = 4 + 2 + 3 = 9 - 4 - 3 = 2

The problem is set out using brackets to clarify the bemdas sequence. Example:

(

2

x

3

x

4

)

=

24

(

/

2

/

3

)

=

4

(

+

2

+

3

)

=

9

(

-

4

-

3

)

=

2

Fractions

Standard calculators may not have a dedicated fractions keypad. Each fraction must be resolved (as if a division in brackets) into a decimal fraction. Only then can the decimal fractions be re-entered for the division operation. Example:

1			÷		1		=
2					4
(	1	/		2		)		=	·5	(	1	/	4	)	=	·25	·5	/	·25	=	2

If a MS (memory store) and a MR (memory recall) are available in your calculator, then rearranging the processing order can simplify keying entries, as below.

1

/

4

=

·25

MS

1

/

2

=

·5

/

MR

=

2

Some scientific calculators may have a fraction keypad to press between numerator and denominator entries to enable fractional to be calculated directly. Example:

2	-	(	4	x	1	)	=
3			6		2

2

a

b

3

-

4

a

b

6

x

1

a

b

2

=

0·33

c

c

c

Check your progress - Remember bemdas with these calculations:

Question 1. 12·50 + 7·25 + 2·75 = ?

Question 2. 23·12 - 5·70 - 8·22 = ?

Question 3. 15·23 x 3·50 x 4·23 = ?

Question 4. 54·16 ÷ 0·54 ÷ 2.34 = ?

Question 5. 12·50 x 7·25 + 2·75 = ?

Question 6. 26·12 ÷ 6·53 - 3·22 = ?

Question 7. 15·23 + 3·50 x 4·23 = ?

Question 8. 54·16 - 4·68 ÷ 2.34 = ?

Question 9.	1	-	(	3	x	1	)	= ?
	8			4		2

Question 10.	7	÷	(	1	x	9	)	= ?
	8			4		8

Answers

Answer 1. 12·5 + 7·25 + 2·75 = 22·50

Answer 2. 23·12 - 5·70 - 8·22 = 9·20

Answer 3. 15·23 x 3·50 x 4·23 = 225·48

Answer 4. 54·16 ÷ 0·54 ÷ 2.34 = 42·86

Answer 5. 12·5 x 7·25 + 2·75 = 93·375

Answer 6. 26·12 ÷ 6·53 - 3·22 = 0·78

Answer 7. 15·23 + 3·50 x 4·23 = 30·035

Answer 8. 54·16 - 4·68 ÷ 2.34 = 52·16

Answer 9.	1	-	(	3	x	1	)	= - 0.25
	8			4		2

Answer 10.	7	÷	(	1	x	9	)	= 3·111
	8			4		8

2.4 Elementary algebra

Balanced equations

A number has a value that can also be expressed as calculations equal to that value. These alternative balanced equations behave like weighing scales where the same known weight (value) is in balance each side of the pivot point (the equals sign).

Balanced equations
10	=	5 + 5	=	15 - 5	=	2 x 5	=	50 ÷ 5	=	10
	▲		▲		▲		▲

If weights are changed equally on each side the scales will still balance. Below, each side of the equations has undergone the operation of multiplication by 2, the total value increases to 20 but the equations are still balanced. Note that the brackets (parenthesis) recommend completing that part first before multiplying by 2. Also, multiplication can be indicated by placing a multiplying number to the left of the multiplied, for instance 2(5 + 5) means 2 x (5 + 5), and 2a means 2 x a.

Operations with equations
2 x 10	=	2(5 + 5)	=	2(15 - 5)	=	2 (2 x 5)	=	2(50 ÷ 5)	=	20
	▲		▲		▲		▲		▲

The equations above have known values as they were written with numbers. But algebra can express a value that is unknown (called variable) by substituting letters for values. Re -assembling balanced equations using algebra can make calculations easier to understand and simpler to solve, as below:

Balance (equality) is shown if adding c to each side of this equation.

If a = b then a + c = b + c

It is also shown if subtracting c from each side of this equation.

If a + c = b + c then a + c - c = b + c – c then a = b

It is also shown if multiplying both sides of this equation by c.

If a = b then ac = bc

It is also shown if dividing both sides of this equation by c.

If ac = bc then ac = bc then a = b

c c

A basic rule or truth is that when a value is multiplied by 1 it is not changed and when a value is multiplied by its opposite (its reciprocal) the result is 1.

x 1						x reciprocal
b	x	1	=	b		b	x	1	=	1
								b
					or
y	x	1	=	y		y	x	1	=	1
								y

Note above that some operations are paired, in that they can reverse the operation of the other. Above, the multiplication by a reciprocal has the same result as division. This property was used earlier when simplify fraction operations by the process of substitution. This pairing is seen with the operations:

Paired operations
+	←	add and subtract	→	-
x	←	multiply and divide	→	÷
a²	←	power or exponents and roots	→	√
		(see section 1.4)

Algebraic notation

Names are given to the parts within an algebraic expression. This assists in when describing algebra and when deciding the order (steps) when calculating.

Algebraic notation
1	2	3	4	1	2	2	4		5
4	z	²	-	6	z	y	+	=	c
{				{
6				6

1 – a coefficient - known number or value.

2 – a variable - unknown number or value.

3 – an exponent - power or inverse power (root).

4 – an operator - arithmetical process.
5 – a constant - standard value that can be found..
6 – a term - grouped entity of 1, 2 and 3.

Constants are often used in a formula. Formulae are specific calculations used to solve common arithmetical problems, and they can be found in reference material.

Rules of Algebra

Re-assembling equations as simpler arrangements relies on some basic rules that must be followed for consistent answers. The order in which operations are processed could alter the answer, as shown in the example below:

3 + 4 x 5 could be 3 + 4 = 7 x 5 = 35 but at the same time,

3 + (4 x 5) will be 4 x 5 = 20 + 3 = 23

Note that adding brackets clears up any doubt as to which part of the calculation is meant as the first step to solve. Order rules of elementary algebra include:

Order of operations:

The general rule for order in calculating is by:

Step 1- calculations within brackets (parenthesis).

Step 2- calculations of exponents.

Step 3- calculations of multiplication.

Step 4- calculations of division.

Step 5- calculations of addition.

Step 6- calculations of subtraction.

A simple aid to memory for order of calculations is bemdas or pemdas. Other rules apply with re-assembling equations including:

Commutative rule:

The commutative rule allows swapping around the notation positions on both sides of an equation to provide an easier calculation alternative, for instance with addition:

z + y = y + z or

z + 3 = 3 + z

It also allows swapping the notation positions on each side of an equation to provide an easier calculation alternative, for instance with multiplication:

ab = ba or

y3 = 3y

Associative rule:

The associative rule provides an alternative bracketed group to operate addition within an equation, for instance:

(a + b) + c = a + (b + c) or

(a + 4) - 2c = a + (4 - 2c)

It also provides an alternative bracketed group to operate multiplication within an equation, for instance:

c(ba) = c(ba) or

(a x 2)3b = a(2 x 3b)

Distributive rule:

The distributive rule allows multiplying to the sum of addition within a bracket (parenthesis) or to each quantity followed by adding, for instance:

a(b + c) = ab + ac or

z(3 + 2y) = 3z + 2zy

Operations with zero:

When zero is added to any number it is not changed.

b + 0 = b

y + 0 = y

When zero is subtracted from any number it is not changed.

b - 0 = b

y - 0 = y

When any number is multiplied by zero the result is zero.

b x 0 = 0

y x 0 = 0

If multiplying values equals zero then at least one of the values must be zero.

If ac = 0 then a = 0 or c = 0

When zero is divided by any number it is the result is zero.

0 = 0

y

A number cannot be divided by zero.

y = the denominator cannot be 0.

0

When a number is added to its opposite pair the result is zero.

b - b = 0

y - y = 0

Operations with negative quantities:

Care must be taken when calculating with negative values and subtraction. Subtraction can be considered as addition with a negative number, just as division can be thought of as inverse multiplication.

(-1) a	=	- a			(- 1) 3	=	-3
- (- a)	=	a			- (- 3)	=	3
(- a) b	=	- (ab)	=	a(- b)	(- 3) y	=	- (3y)	=	3(- y)
(- a) (- b)	=	ab			(- 3) (- y)	=	3y
- (a + b)	=	(- a)	+	(- b)	- (y + 3)	=	(- y)	+	(- 3)	=	- y- 3

Fractions

The same rules as in fractions applies to algebraic calculations with fractions.(see section 2.3). These include:

When adding or subtracting with the same denominator, add or subtract the numerator variables.

When	a	±	c	=	a ± c
	b		b		b

When adding or subtracting with the different denominators, find a common denominator and add or subtract the numerator variables.

When	a	±	c	=	ad ± bc
	b		d		bd

Balance (equivalence) is shown by cross multiplying unless b and d equal zero.

a

=

c

Multiply all by

bd

=

abd

=

cbd

=

ad

=

cb

b

d

1

b

d

Negative signs can go anywhere in the fraction and two negatives equal a positive.

-	a	=	a	=	a	and	- a	=	a
	b		b		- b		- b		b

If multiplying both top and bottom of a fraction by the same quantity maintains the fraction’s same value, then the quantity must have been zero.

If	a	=	ac	then	c	=	0
	b		bc

When multiplying fractions multiply the numerators and multiply the denominators.

When	a	x	c	=	ac
	b		d		bd

When dividing fractions, reverse the dividing fraction (the reciprocal) and multiply.

When	a	÷	c	=	a	x	d	=	ad
	b		d		b		c		bc

Solving problems using algebra

The formula for the energy potential of mass is:
Example:
e = mc²	where e is energy in joules
	m is mass in kilograms
	c constant, the speed of light (300,000,000 mtrs per sec)
e = 1 kilo x 300,000,000 x 300,000,000 = 90,000 million, million joules.
Or 1 gram = 90,000 million, million ÷ 1000 = 90 million, million joules

This is the incredible amount of energy available in 1 gram of matter available to unleash by nuclear weapons, equivalent to the energy of a 20 kiloton TNT bomb.

The formula for the volume of a cylindrical tank is given by v = pr^2h.

Example:

v = p r²h

where v is volume of a tank

p the constant pi, 22 ÷ 7 = approximately 3·124.

r is the radius of a circle

h is height of a tank

If we knew the radius of the tank was 0·6 mtrs and its height was 1·5 mtrs, then

v = p r²h

=

3·142 x 0·6 x 0·6 x 1·5

=

3·142 x 0·36 x 1·5

=

1·696 cubic mtrs

For practical purposes the tank’s volume is rounded to approx.

1.7

If the tank was only a third full (a sounding of 0.5 mtrs) then

v = p r²h

=

3·142 x 0·6 x 0·6 x 1·5

=

3·142 x 0·36 x 0·5

=

0·5 6 mtrs³

If we knew the volume of the tank (2 mtrs³) and its radius (0.4 mtrs) then its height is:

v = p r²h

=

v x 1

=

p r² h

=

v

=

h

=

2

=

3·9783577 mtrs

p r²

p r²

p r²

3·142 x 0·4²

The tank’s height is rounded to approx. 3·98 mtrs.

Note the process of rounding to 3 mtrs and 98 cms is 1·64 millimetres greater that our calculated answer, and for the purposes of refuelling will be within the required accuracy level. Note also that pi is one of a few irrational numbers. It cannot be determined as a final number, for example:

p = 22 ÷ 7 = 3·1428571428571428571428571???????? and so on.

2.5 Trigonometry

Trigonometry is a mathematical means of calculating the relationships between the length of the sides of a triangle and the angles between those sides.

Properties of triangles

A circle can be drawn around any triangle to touch each of its three points. The sum of the internal angles of the points always adds up to 180º.

If we know any two angles then we can calculate the other. For example:

A 63º + B 75º = 138º therefore C = 180º - 138º = 42º

Triangles come in different shapes. They can have a 90º angle (right angle triangle), a 90º angle and two equal length sides (equilateral triangle), equal length sides and angles (isosceles triangle) or different length sides and angles (irregular triangle). An angle is designated by the symbol of θ.

Opposite, adjacent and hypotenuse

In this right angle triangle using θ ABC as a reference point:

The opposite side is opposite θ.

The adjacent side is adjacent (next to) to θ.

The hypotenuse is the longest of the sides.

Trigonometry calculates an unknown side or angle of right angle triangles using the functions of Tangent, Sine and Cosine and formulas. In the past, the answers to all the sides’ lengths and angles were measured and tabulated so solutions could be looked up in the books of tables. More often nowadays we use the function keys on an electronic calculator to do the same thing.

Sine (sin) of θ	=	Opposite	=	b
		Hypotenuse		a
Cosine (cos) of θ	=	Adjacent	=	c
		Hypotenuse		a
Tangent (tan) of θ	=	Opposite	=	b
		Adjacent		c

These formulas can be remembered by the mnemonic of:

SOH-CAH-TOA

Alternately the reciprocal functions of SOH-CAH-TOA above can be used:

1	θ	=	Secant (sec) θ	=	Hypotenuse	=	a
Sin					Adjacent		c
1	θ	=	Cosecant (cosec) θ	=	Hypotenuse	=	a
Cos					Opposite		b
1	θ	=	Cotangent (cot) θ	=	Adjacent	=	c
Tan					Opposite		b

Finding a length of a side in the triangle ABC below:

If we know the length of a side and two angles of a right angle triangle we can calculate the other sides by the rules of trigonometry.

Example 1: (Finding the hypotenuse with the sine function of a scientific calculator)

Sin

31

=

0.515

Therefore, 0.6 = Opposite ÷ Hypotenuse or Hyp = 3cms ÷ 0.515 = 5.8cms

Example 2: (Finding the hypotenuse with the cosine function of a scientific calculator)

Cos

31

=

0.857

Therefore, 0.857 = Adjacent ÷ Hypotenuse or Hyp = 5cms ÷ 0.867 = 5.8cms

Example 3: (Finding the adjacent with the tangent function of a scientific calculator)

Tan

31

=

0.6008

Therefore, 0.6 = Opposite ÷ Adjacent or Adjacent = 3cms ÷ 0.6 = 5cms

Finding an angle in the triangle ACB below:

If we know the length of two sides and two angles of a triangle we can calculate the other angle by the rules of triangles.

Example: We can find the last of the triangles angles (ACB) by the rules of triangles:

A 90º + B 31º = 121 therefore C = 180º - 121 = 59º

We could also find the last angle by the rules of trigonometry if we referenced our unknown angle θ, (angle ACB) with the Opposite, Adjacent and Hypotenuse lengths.

Example 1: (Finding the θ with the inverse tangent function of a scientific calculator)

Opposite ÷ Adjacent = Tangent θ

5

÷

3

=

1.666

Shift

Tan

=

59º

Example 2: (Finding the θ with the inverse sine function of a scientific calculator)

Opposite ÷ Hypotenuse = Sine θ

5

÷

5.8

=

0.862

shift

Sin

=

59º

Example 3: (Finding the θ with the inverse cosine function of a scientific calculator)

Adjacent ÷ Hypotenuse = Cos θ

3

÷

5.8

=

0.862

shift

Cos

=

59º

Radians and properties of circles

Up until now we have described angular measures by using degrees, there being 360º in a full circle. The dimensions in circles are found with the constant p (pi) that has an approximate value of 22 ÷ 7 = 3·124. These dimensions include:

The area of a circle = p x radius².

The circumference of a circle = 2p x radius (2pr).

For solving rotational problems the radian as a measure is often preferred by mathematicians over the degree. A radian (rad) is the angle formed at the centre of any circle and an arc on its circumference that is the same length as the circle’s radius. A radian is equal to 57.295º.

If the circle above is rolled through half a revolution its circumference will move through 180º. As the length of the circumference of a circle is 2p x radius, then it will also have moved through have this distance, or p radians = 3.142 radians.

If the circle above is rolled in one complete revolution its circumference will move through 360º. As the circumference is 2p r, then it will also have moved through 2p radians = 2 x 3.142 = 6.286 radians.

*Note - 360º ÷ 6.286 = 57.3º

Likewise,

360º = 2p radians, 180º = p radians, 90 º = p/2 radians

Summary:

1 radian per second = approximately 57.295 degrees per second

1 radian per second = approximately 9.5493 revolutions per minute (rpm)

0.1047 radian per second = approximately 1 rpm

One use of the unit radian per second is to calculate the power transmitted by a shaft (P) being the product of w (in radians per sec.) times the torque (T) in newton-meters applied to the shaft, or:

P = w x T, in watts.

As the radian is a dimensionless unit, the radian per sec. is dimensionally equivalent to the hertz, both being defined as one s⁻¹. Consequently care must be taken to avoid confusing angular frequency (w) and frequency (v).

Chapter 3: Measuring

3.1 International System of Units

Measuring developed from comparison of commonly found weights and dimensions of local materials. During history the greatest empires spread their measurement systems furthest in support of their navigation, trade and colony building. Two main systems survive in the Imperial System and the Metric System; the qualities being Imperial with human scale (feet) and the Metric for easier calculations (decimal). Consequently, system conversions may be required in international trading and machine parts are not always interchangeable. To solve this, the metric based International System of Units (SI Units) has been adopted as a worldwide standard.

The SI base measurements and symbols
Length	m	= metre
Mass	Kg	= kilogram
Time	s	= second
Electric current	A	= Ampere
Luminous intensity	cd	= candela
Temperature	K	= Kelvin	= The thermodynamic Kelvin scale uses the degree Celsius for its unit increments absolute zero or - 273·15º Celsius. Conversion K = °C + 273·15 °C = K−273·15
Amount of substance	mol	= mole	In chemistry it expresses the amount of substance cwith as many atoms, molecules, ions, electrons as there are atoms in 12 grams of carbon-12 (relative atomic mass 12).

Measures are also found in multiples/derivations of SI measures, for example:

Density is measured as an object’s Mass per unit of Volume

Electrical Power is measured in Volts/Amps = Watts

Energy is measured in Joules

Force is measured in Newtons

Pressure is measured in Pascals

Stress is measured in Newtons/ metre²

Torque is measured in Newton/metres

Speed is measured in Metres/second

Velocity is measured in Metres/second in a specified direction.

Multiples of Units
Length	millimetre	mm	x 1000 =	metre	m	x 1000 =	kilometre	k
Mass	gram	gm	x 1000 =	kilogram	kg	x 1000 =	tonne	T
Force	Newton	N
Pressure	Pascal	Pa	x 1000 =	hecta pascal	hPa	x 1000 =	mega pascal	mPa
Current	Ampere	A	x 1000 =	kilo amp	kA	x 1000 =	mega amp	mA
Power	Watts	W	x 1000 =	kilo watt	kW	x 1000 =	mega watt	mW
Energy	joule	j	x 1000 =	kilo joule	kj	x 1000 =	mega joule	mj
Luminous	candela	cd	x 1000 =	kilo candela	kcd	x 1000 =	megacandela	cd
Time	second	s	x 60 =	minute	m	x 60 =	hour	h
Nautical distance	metres	m	x 1852 =	mile	nm	nautical miles per hour are termed knots		kt

3.2 Derived units and formulas

Standard formulas simplify working with derived units and measures. However, some non SI unit measures commonly persist, for instance, nautical miles. When calculating, pay attention to whether the units are SI units and whether they are consistent for all variables or constants used within a formula.

Density is measured as an object’s Mass per unit of Volume.

The SI units for density are kg/m³. The imperial units are lb/ft³ (slugs/ft³). Though pounds per cubic foot are used as an equivalent measure of density, pounds are actually a measure of force, not mass. Slugs are the correct measure of mass. Multiply slugs x 32·2 for an approximate value in imperial pounds.

*Note - although equivalent for sea level calculations, technically weight unlike mass is an object’s measured downward pressure at a specified gravitational locality.

The formulas of mass, density and volume are expressed within the triangle as:

The formulas of mass, density and volume are expressed within the triangle as:
d	=	m	m	=	d x V	V	=	m
		V						d
Where: d = density (meters³/kilogram, feet³/slug) m = mass (kilograms, slugs) V = volume (metres³, feet³)

Specific Gravity or Relative Density of a substance is a comparison of the mass of a volume of one substance to the mass of an equal volume of pure water. Specific Gravity is the mass compared to pure water valued as 1 (it is a ratio so it has no units). Relative Density is expressed as a percentage of pure water’s mass.

S.G.	=	mass of substance	=	R.D.%
		mass of fresh water

Example 1:

What is the mass of a cubic meter of fresh water? What is the mass of a cubic meter of salt water? How much salt is in a cubic meter of salt water?

Fresh water SG 1·000- 1m³ = 1000 litres x SG 1·000 = 1000 kilograms =1 tonne

Salt water SG 1·025- 1m³ = 1000 litres x SG 1·025 = 1025 kilograms =1·025 tonne

Salt water 1025 kilograms - Fresh water 1000 kilograms = 25 kilograms

Example 2:

What is the mass of fuel in a tank that measures 4 metres in length, 2 metres wide and 1·8 metres deep if the fuel has a specific gravity 80% that of fresh water?

Volume = Length x Width x Depth

= 4 metres x 2 metres x 1·8 metres

= 6·4m³ (cubic metres)

Mass (tonnes) = Volume (cubic metres) x Specific Gravity

= 6·4m³ x 0·8

= 5·12 tonnes

Example 3:

If oil weighs 860 kgs/m³, what is its specific gravity?

* Though the terms are used

S.G. (R.D)	=	mass of substance
		mass of fresh water
	=	860
		1000
	=	0·86

Electrical Power- Electricity is measured in units described by Ohm’s law:

the current through a conductor between two points is directly proportional to the potential difference across the points, and inversely proportional to the resistance between them”.

I	=	V	V	=	I x R	R	=	V
		R						I
Where:  I = current (ampere, A) V = potential difference (volt, V) R = resistance (ohm, Ω)

Example 1:

If a 6 Ω resistor is placed in a 12V circuit, what will the current be?

I	=	V	=	12	=	2 Amps
		R		6

Example 2:

If a 3 amp motor has 4 Ω resistance, what will the voltage be?

V

=

I x R

=

3 x 4

=

12 Volts

Example 3:

In a 12V circuit what resistance will a 1 Amp radio develop?

R	=	V	=	12	=	12 Ohms
		I		1

Electrical power is termed watts and is calculated as P = V x I

Where:

P = power (watts)

I = current (ampere, A)

V = potential difference (volt, V)

Example 4:

In a 12V circuit, what power will a 2 amp globe use?

P

=

V x I

=

12 x 2

=

24 Watts

Energy is measured in Joules

The joule (j) is defined as the energy of work or heat equal expended in applying a force of one newton through a distance of one metre (Nm), or in passing an electric current of one amp through a resistance of one ohm for one second.

j	=	Kg x m²	=	N x m	=	Pa x m³	=	W x s	=	C x V
		s²

Where:

J is a joule,

kg is kilogram,

m is metre,

s is second,

N is newton,

Pa is pascal,

W is watt,

C is coulomb,

V is volt.

In electrical terms, the joule can also be defined as the work required to move one coulomb of electric charge through a potential difference of one volt (a coulomb volt). Similarly a volt is defined as the work required to produce one watt of power for one second (a watt second).

A microjoule equals one millionth (10^-6) of a joule.

A kilojoule equals one thousand (10³) joules.

A megajoule equals one million (10⁶) joules.

Converting joules into watts:

You can calculate watts from joules and seconds, but you can't convert joules to watts, since joule and watt units represent different quantities.

Power (W) = Energy (J) ÷ time (s) or Watt = joule ÷ second

Where:

P is power in watts (W), E energy in joules (J), time is seconds (s)

Example:

What is the power consumption of an electrical circuit that has energy consumption of 120 joules for time duration of 10 seconds?

Watt = joule ÷ second

Watts = 120 ÷ 10 = 12 W

Converting watts into joules:

Energy (j) = Power (W) x time (s) or joules = Watts x seconds

Example:

What is the energy consumption of an electrical circuit that has power consumption of 12 watts for time duration of 10 seconds?

joules = Watts x seconds

joules = 12 x 10 = 120 j

Force is measured in Newtons and describes motive energy.

One Newton of force is required to accelerate one kilogram by one metre per second per second. A force over a given area is described as stress.

Pressure is measured in Pascals

Stress is measured in Newtons/ metre²

Torque is measured in Newton/Metres and describes angular/rotating motion.

The power of a rotating body is expressed as:

P = T w

where

P = power

T = torque or moment (N/m)

w = angular velocity (rad/s)

1 rad = 360^o/ 2p = 57.295^o

Torque of a body in angular motion is expressed as:

T = I a

where

I = moment of inertia (kg m²)

a = angular acceleration (rad/s²)

The Newton metre as a unit of torque (a moment) is equal to one Newton applied perpendicular to a moment arm of one metre long. But it is also a unit of energy/work equivalent to the joule. For energy/work usage, the metre is the distance moved in the direction of the force rather than the perpendicular distance from a fulcrum of torque. Since torque is energy expended per angle of revolution, a Newton metre of torque is equivalent to a joule per radian. Seen below, Newton metres and joules are dimensionally equivalent as can be calculated by the same expression:

1 Nm	=	Kg x m²	1 j	=	Kg x m²
		s²			s²

However, Nm and j must be distinguished in order to avoid misunderstanding torque and energy the same. Similar dimensionally equivalent units are Pa versus j/m³.

Conversion factors

1 newton metre = 0.7375621 pound-feet (often "foot-pounds")

1 kilogram-force metre = 9.80665 Nm

1 pound-foot (foot-pound) ≡ 1 pound-force-foot = 1.3558 N·m

1 inch-ounce-force = 7.0615518 mNm

1 dyne-centimetre = 10⁻⁷ Nm

Speed is the rate at which distance is travelled.

The formulas of speed, distance and time are expressed within the triangle as:

s	=	d	d	=	s x t	t	=	d
		t						s

Where:

s = speed (metres/second, kilometres/hour, knots)

d = distance (metres, kilometres, nautical miles)

t = time (seconds, hours)

When calculating use compatible units, for example:

m/s = metres ÷ seconds

k/hr = kilometres ÷ hours

knots = nautical miles ÷ hours

Example 1:

A vessel travels 15 nautical miles in three hours. What is its speed?

s	=	d	=	15	=	5 Knots
		t		3

Example 2:

How far does a vessel travel if she steams at 10 knots for 4 hours?

d

=

s x t

=

10 x 4

=

40 nautical miles

Example 3:

How long will a vessel take to travel 60 nm at 5 knots?

t	=	d	=	60	=	12 hours
		s		5

Velocity is speed in a specific direction. Acceleration is a measure of the change in speed of a body expressed as:

acceleration m/s² = dv ÷ dt    = (v₂ - v₁) ÷ (t₂ - t₁)    

Where:

dv = change in velocity

v₂ = final speed (m/s, ft/s)

v₁ = initial speed (m/s, ft/s)

dt = time taken (s)

t₂ = final time (s)

t₁ = initial time (s)

Temperature is measured in scales Kelvin, Celsius (centigrade) or Fahrenheit.

The thermodynamic Kelvin scale uses the degree Celsius for its unit increments absolute zero or - 273·15º Celsius. Conversion:

K = °C + 273·15 °C = K−273·15

3.3 Measuring Tools

Some illustrations in this section contain enhanced drawings courtesy of ANTA Publications

Measuring accurately

Measuring tools will be damaged if dropped. Tools that give readings of less than 0.1 mm can be put out of adjustment by poor handling. They must be calibrated regularly by checks against a standard of measurement (a precisely made gauge) to ensure accuracy. Steel expands with temperature rise so gauge accuracy is also affected by

such change. If using screw pitch, radius or form gauges you should sight against a light background to see differences between the work piece and the gauge clearly.

Care and use

When using all measuring instruments the following precautions should be observed:

• Never drop the instrument
• When not in use replace in its case or clean rag, never on a hard steel bench
• Never allow dirt, filings, cutting oils or any other foreign substances to come in contact with the instrument
• Do not put the instrument on top of or under other tools
• Never measure moving objects
• Zero the instrument before use

Storage Procedures

To ensure long and reliable service:

• Clean the instrument thoroughly during and after use
• Lightly oil or wrap the instrument in oiled paper
• Store the instrument in its own case protected from outside damage
• Store the instrument in a dry place away from corrosive chemicals or solvents

Standard Gauges

Screw Pitch Gauge

A screw pitch gauge is a collection of blades of differing teeth sizes enabling both the pitch and form (shape) of threads to be checked. Each blade has a thread form stamped on it being in Metric, Whitworth, BSF, UNF or UNC.

Before measuring, assess the approximate pitch with a ruler. For metric threads:

• Line up a major division on the ruler with a crest of the thread.
• Count the number of crests to another major division, (distance 20 - 30 mm).
• Divide the chosen length by the number of crests counted.

That result is roughly the thread pitch. Next choose the gauge blade closest to this for a fine check.

Use the same method over a distance of one inch for imperial threads as they are measured in threads per inch (TPI).

Radius Gauge

This gauge is a set of blades shaped with matching convex and concave radii. They are used to check internal and external radii. Radius gauges of less than 90º may be called fillet gauges.

Feeler Gauge

This gauge is a set of differing thickness blades (typically 0.05 mm-1mm) used to measure the width of small grooves or set clearances between mating parts. The blades can be doubled up to measure a greater thickness if required. Care should be taken with thinner blades to pull through any gap rather than pushing, in order to avoid blade the kinking with consequent damage.

Thickness Gauge

This gauge is used to measure the thickness of sheet materials such as paper, plastics, cardboard or leather and sheet metals. Exposure to dirt and moisture should be avoided by returning to their storage box immediately after use.

Form or Profile Gauge

These can be fixed or adjustable types, as below, used to check work pieces shape. The adjustable type is set to a master shape to compare with the work piece.

Callipers

Various models all transfer a measure from a standard piece to a work piece. Two arms are hinged enabling a chosen gap to be set. Some are spring‑loaded joint with an adjusting screw. Accurate measurement when using callipers depends upon the tightness of the calliper’s arms against the work piece. Skill is needed to obtain the correct light pressured feel on the callipers as they slip over the work.

Using outside callipers

Outside callipers are used for outside diameters/dimensions or to check whether external surfaces are parallel. To use:

• Open the arms until they pass around the diameter to be measured. The work piece must be stationary.
• Gently tap the back of one arm against a solid part of the work piece to slightly close up the arms.
• Try the new setting with the callipers at right angles to the work. Continue to check and adjust the callipers until you feel the arms just slip over the work.

· Lastly compare the arms’ gap with a steel rule on a flat smooth surface

Using inside callipers

Inside callipers are used for inside diameters/dimensions or to check whether internal surfaces are parallel. To use:

· Place an arm of the callipers just inside the bottom of the hole to be measured.

· With the adjusting screw, open the other arm so it touches the top of the hole.

· Rock the callipers slightly on the lower arm and adjust the screw until you feel the callipers tight within the hole.

· Wriggle the callipers within the hole to ensure that the arms are a tight fit all around.

Graduated Measuring Tools

Steel Rules

Steel rules can measure accurately to ± 0.5 mm if maintained and read properly.

The ruler must be maintained with square and sharp edges. A common error called parallax is caused by sighting at an angle to the scale, rather than at right angles.

Measuring Tapes

The tape is subject to damage so it must be cleaned as it self-retracts into the housing. Salt water or corrosive/sticky contaminants can ruin a tape and obliterated the scale markings rendering reading difficult.

Depth Gauge

This gauge is used to measure the depth of holes or the distance between surfaces.

Vernier Callipers & Micrometers

Verniers

Used for internal, external or depth measurement, they have scales of up to 250 mm. Standard callipers measure to within 0.02 mm (0.001 in). Digital callipers are have accuracy to 0.01 mm (0.0005 in). They must be stored in a clean, dry place (preferably in their protective case). The jaws must be protected against damage.

Reading a Vernier – They have a main scale providing a course measure and a vernier scale for the remainder fine measure. An experienced user can roughly estimate the fractional remainder, and consequently where to expect the vernier scale markings to align. Sighting across rather than over the marks will facilitate taking a reading. Move to a position where the light comes from behind the vernier scale at the same angle as your line of sight.

Read the main scale to the left from the metric vernier zero to find the whole number (as below 7 mm). To read the remaining fraction, look at the vernier scale. Note which of the vernier marks align with a main scale mark. Each of the vernier scale marks represents 0.02 mm. Finally multiply the number of marks on the vernier scale by 0.02 and add the result to the reading of the main scale, in this case 7.50 mm.

The drawing above shows a metric vernier reading just over 37 marks on the main scale to the left of the vernier scale’s zero, or 37+ millimetres. The 33rd mark on the vernier scale aligns with a main scale mark giving 33 x 0.02 = 0.66mm

Now add 0.66 mm to the main scale reading of 37 mm to give a total of 37.66 mm.

Some metric verniers with a 49 mm long scale have each fifth mark of the vernier scale numbered from 1 to 10. As each mark on the vernier scale represents 0.02 mm, then the fifth mark of (5 x 0.02 = 0.1 mm) is numbered 1. The tenth mark is marked 2, the fifteenth mark marked 3, and so on.

With this type of scale read the main scale as before, read the numbered divisions of the vernier scale as tenths of a millimetre then complete the reading by adding the extra 0.02 lines. In the main scale above reads 60 millimetres. The vernier shows the fifth mark which represents 0.5 mm, plus 3 extra divisions which represent:

3 x 0.02 = 0.06 mm Total reading is 60 + 0.5 + 0.06 = 60.56 mm

Some metric verniers have their main scale divided into millimetres and half millimetres. The vernier scale is divided into 25 equal marks (24.5 mm long). The length of each vernier division is therefore one twenty‑fifth of 24.5 mm (0.98 mm), as shown below.

Example -The tool below has a vernier scale 24.5 mm long with 0.02 mm vernier scale readings (the 25 marks represent from 0mm - 0.5 mm).

There are 37 major divisions on the main scale to the left of the zero, which equals 37 mm. There is also one half‑millimetre division which equals 0.5 mm.

37 + 0.5 = 37.5 mm

The eighth mark on the vernier’s scale aligns with a mark on the main scale. Multiply 8 by 0.02 which represents 0.16 and add this to the main scale reading.

Total reading 37.50 + 0.16 = 37.66 mm

Check your progress reading a vernier - try these:

Drawing courtesy of Anta publications

Answers reading a vernier - as below:

1. 15mm 2. 66mm + (42 x 0.02=0.84) = 66.84mm

3. 29.5mm + (8 x 0.02=0.16) = 29.66mm 4. 35mm + (20 x 0.02=0.4) = 35.4mm

5. 9mm + (16 x 0.02=0.32) = 9.32mm 6. 21.5mm + (16 x 0.02=0.32) = 21.82mm

7. 42mm + (22 x 0.02=0.44) = 42.44mm 8. 13.5mm + (2 x 0.02=0.04) = 13.54mm

9. 36mm + (28 x 0.02=0.56) = 36.56mm 10. 2 + (13 x 0.02=0.26) = 2.26mm

Micrometers

Outside micrometers enable very accurate measurement of diameters, thickness and length. All having a measuring range limited by the size of frame and length of the spindle’s thread, typically 0 to 25 millimetres. Larger opening frames (50 mm) can use a removable 25 mm spacer insert to double the range while still using a standard 25 mm thread length.

In the drawing below, the main parts of a micrometer are:

Frame – Anvil - Spindle & Thread - Sleeve or Barrel - Thimble.

In use, the anvil is held against the work piece while the spindle is wound inwards to make a tight fit. A knurled collar or lever locks the spindle in the barrel. The dimension is read from the scale.

Skill is needed to get accurate measures with a micrometer. Excessive pressure taking readings will give lower readings, strain the thread and distort the frame. The anvil is set against a work piece and the finger grip turned to tighten the spindle. A fine adjuster may be fitted, to limit over tightening. Turn it until the spring‑loaded torque ratchet audibly clicks. Finally, tighten the spindle lock to stop movement when reading the scale. Remember to loosen it before using again.

Checking a micrometer with scale to 0.01 of a millimetre

Keeping the marks on the sleeve towards you, hold the micrometer’s frame with one hand. Use the other hand to screw the knurled part of the thimble anti‑clockwise. This moves the spindle to uncover the marks on the sleeve.

The gap between the anvils should equal to the uncovered length of the datum line. The datum line on the sleeve is marked in millimetres and half millimetres, from zero

to 25 mm, and usually each fifth millimetre is numbered. Turn the thimble until zero is level with the datum line. Note the position of the mark on the sleeve.

Turn the thimble one complete turn. The thimble will move along one graduation of the sleeve scale. This is because the pitch of the thread on the spindle is half a millimetre. There are 50 marks and each fifth mark is numbered.Two turns of the thimble move the spindle one millimetre.

Now wipe the face of the anvils with a piece of clean cloth. Screw the thimble inwards towards the frame until the anvils are touching. Both scales should both read zero. If they are not then you need to get the tool repaired. If the error is slight and consistent then you can mathematically correct all measures from that tool.

Reading a metric micrometer

Example 1 - Read the number of fully visible millimetre and half millimetre marks on the barrel. Read the thimble scale number that aligns with the datum line. Total the readings, as below:

4.0 mm + 0.5 mm + 0.05 mm = 4.55 mm

Example 2 - Read the number of fully visible millimetre and half millimetre marks on the barrel. Read the thimble scale number that aligns with the datum line. Total the readings, as below:

5.0 mm + 0.5 mm + 0.12 mm = 5.62 mm

Check your progress reading a micrometer - try these:

Drawing courtesy of Anta publications

Answers reading a micrometer - as below:

1. 12.5mm + 0.19mm = 12.69mm 2. 23.5mm + 0.49 = 23.99mm

3. 1.5mm + 0.39mm = 1.89mm 4. 5.5mm + 0mm = 5.5mm

5. 0.5mm + 0mm = 0.5mm 6. 7.0mm + 0.22mm = 7.22mm

7. 19.5mm + 0.05mm = 19.55mm 8. 2.5mm + 0.25mm = 2.75mm

9. 21mm + 0.14mm = 21.14mm 10. 9mm + 0.10mm = 9.10mm

Chapter 4: Shape, size and capacity of tanks

Tanks store fuel or water and can provide a second skin to increase watertight spaces. Below are day tanks are fitted from which fuel is gravity fed to the motors. Port and starboard tanks 1 are for the fuel oil needs of the passage, and are regularly pumped to press up (fill) the day tanks. Tanks 2-3 are used for ballast or fresh water cargo, and tanks 4 & 5 are for oil cargo. The latter tanks are separated by a void that can be filled with water to limit the spread of fire (a coffer dam).

Sounding rods or sight gauges enable tank capacity to be measured. However, measurements at sea are inaccurate due to the rolling and pitching of the vessel. Using the vessel’s plans of tanks and spaces and standard formulas enables the condition of loading of each tank can be calculated.

4.1 Definitions

Length is the longer horizontal measure.

Width is the shorter horizontal measure.

Height is the vertical measure.

Area is the space occupied by a shape, calculated by multiplying length and width.

Perimeter is the distance around the outside (boundary) of a shape. It is calculated by adding all lengths and all widths.

Volume is the holding capacity, calculated by multiplying length, width and height.

4.2 Areas of Common Shapes:

Area is the measurement of the footprint for a two dimensional object.

Rectangles - The area is measured by multiplying the Length by the Width.

Example:

Find the area of a rectangle 10·2 metres long and 6·0 metres wide.

Area = L x W

= 10·2 x 6·0

= 61·2 m² (square metres)

Triangles -The area of a triangle is calculated by multiplying half of the base of the triangle by the height of the triangle. Or equivalently, the base can be multiplied by the height and the result then divided by two.

Example: What is the area of a triangle with a base of 3.8 m and 1.1 m high?

Area = ½ x B x H

= ½ x 3·8 x 1·1

= 2·09 m² (square metres)

Trapeziums - A trapezium is a four sided figure that has only two parallel sides. Its area is calculated by multiplying half its height by the sum of both parallel sides. Where A & B are the parallel sides and H is the perpendicular (shortest) distance between them, the height. Note: Do not measure up one of the sides.

Example:

What is the area of a trapezium having parallel sides of 2·12 m and 3·1 m which are 1·2 m apart.

Area = ½ x (A + B) x H

= ½ x (2·12 + 3·1) x 1·2

= ½ x (5·22) x 1·2

= 3·132 m²

Circles - The area of a circle is given by using the formula:

Where p or pi = approximately 3·14.

r or radius = half of the diameter of a circle.

Example:

Find the area of a 2·6 cm diameter circle. Give your answer to 2 decimal places.

Area = p x (½ x 2·6)²

= p x 1·3²

= 5·309291585²

= 5·31 cms²

An alternative formula of Area = p x diameter² can be also used.

4

4.3 Volumes of tanks

Volume is the capacity measurement for three dimensional objects. Tanks can be considered to be “regular” or “irregular” in shape:

Regular shaped tanks:

Irregular shaped tanks:

Tanks that taper also fit into this category.

In practical situations you may need to make calculations based on an approximate shape. For example, this curved tank can be approximated as a triangular tank or a quarter of a cylinder depending on the lengths of A and B and the curvature.

Alternatively, tanks may be considered as composite shapes and the capacity of section each calculated separately. For instance, the tank below is calculated as the composite of a rectangular top section added to the triangular bottom section to give the overall tank volume.

Rectangular Tanks:

To calculate the volume (and capacity) of rectangular tanks the formula is Length multiplied by the Width multiplied by the Height of the tank.

Example:

A tank is 3·1 m long, 2·24 m wide and 1·1 m deep, what is the volume.

Volume = L x W x H

= 3·1 x 2·24 x 1·1

= 7·6384 m³

= 7·64 m³ (in cubic metres to 2 decimal places)

Cylindrical Tanks:

The volume of a cylindrical tank is measured by multiplying the area of the circle by the height or length of the tank.

Example:

A cylinder has a circular base of 1.8m in diameter and stands 2.2m high. What is the capacity of the cylinder?

Volume = p r² x h

= p x 0·9² x 2·2

= 5·595 m³

Trapezoidal tanks:

Given the shape of some vessels and the limited space available below decks, it is often necessary to make fuel tanks in an irregular shape.

The area of a trapezium is calculated by multiplying half its height H by the sum of the two parallel sides A and B.

Area = ½(A+B) x H x L

Once you have calculated the area of the side ends, you can calculate the volume of the tank by multiplying it by the length L.

Example:

Referring to the above shape, calculate the volume if the dimensions of the tank are:

A = 1·5 B = 3 H = 2 L = 4

Area = ½ (A+B) x H x L

= ½ x (1·5 + 3) x 2 x 4

= 4·5 x 4

=18 m³ (the tank has a volume of 18 cubic metres)

Measuring tank content

Four methods are available to measure the contents of a tank.

Gauges

Sight glass

Sounding

Ullage

Sounding Depth of fuel	Ullage Depth to fuel	Volume of fuel	SG 0.8	Weight of fuel
0·0 m	1·4 m	0·0 m³ = 0 ltr	x 0.8
0·2 m	1·2 m	0·2 m³ = 200 ltr
0·4 m	1·0 m	0·4 m³ = 400 ltr
0·6 m	0·8 m	0·6 m³ = 600 ltr
0·8 m	0·6 m	0·8 m³ = 800 ltr
1·0 m	0·4 m	1·0 m³ = 1000 ltr
1·2 m	0·2 m	1·2 m³ = 1200 ltr
1·4 m	0·0 m	1·4 m³ = 1400 ltr

Check your progress

a. What is the volume of a tank measuring 7·1m long, 3m wide and 4500mm deep? Give you answer in cubic metres.

b. What is the volume of a ballast tank measuring 3080mm long, 1420mm wide and 64cm deep? Give your answer in m³. What is this tank’s capacity in litres, to 2 decimal places. Remember that 1 Litre = 0·001 m³ or 1000 Litres = 1 m³

c. What is the capacity of a cylindrical fresh water tank if it has a diameter of 6·2m and a length of 424·2cm? Give your answer to 2 decimal places.

d. What volume of fuel is in a cylindrical tank of 1·2 metres diameter and a height of 3 metres when it is half full?

e. What weight of fuel is in a cylindrical tank of 3 metres diameter and a height of 2 metres when there is an ullage taken of 400mm? Use a specific gravity for fuel of 0·8.

f. What weight of fuel is in a trapezoid tank of 2 metres bottom width and 2·4 metres top width, a height of 1·6 metre and a length of 4 metres? Use a specific gravity for fuel of 0·8.

g. What weight of fuel is used by a vessel travelling 300nm that maintains 10 knots with her single 200 kW engine consuming 125cc per kW per hour? Use SG 0·8.

h. Allowing in the above case for a 25% reserve, what weight of additional fuel would be prudent to load?

Answers to check your progress:

a. 95·85 m³

b. 2·799 m³ 2799 litres

c. 12·80 m³ 12800 litres

d. 1·69 m³ 1690 litres

e. 11·3 m³ 11304 litres x 0·8 = 9043·2 kilos = 9·043 tonnes

f. 14·8 m³ 14800 litres x 0·8 = 11264 kilos = 11·264 tonnes

g. 750 litres x 0·8 = 600 kilos = 0·6 tonnes

h. An additional 150 kilos = 0·15 tonnes

4.4 Pumps and flow rate capacity

Pumping calculations are used when transferring or bunkering fuel. The time needed, the volume to move and the flow rate of the pump must be known. Flow rate is how quickly a pump transfers a volume of liquid. For calculating, the flow rate is found with the pump’s manufacturer specifications. If not available you can calibrate the pump by transferring a sample volume while recording the time taken.

Flow rate	=	Volume transferred Time taken
Where: Fr = flow rate (litres/minute) V = volume of liquid (litres) T = time (minutes)

Example 1. Flow Rate:

What is the flow rate of a ballast pump is rated to pump out a 1000 litre tank in 20 minutes?

V ÷ T = Fr

1000 ÷ 20 = 50 litres per minute

Example 2. Time:

How long will it take a 60 litres/minute pump to empty a 6000 litre tank?

V ÷ Fr = T

6000 ÷ 60 = 100 minutes ÷ 60 = 1 hours 40 minutes

Example 3. Volume:

What volume of liquid can be transferred by a 15 litres/minute pump in 20 minutes?

Fr x T = V

15 x 20 = 300 litres

Example 4. Transfer time:

Fuel is transferred from a main tank to a daily service tank at a rate of 7.5 litres/minute. If the pump operates for 12 minutes, how many litres will be pumped?

7.5 x 12 = 90 litres

If the daily service tank holds 210 litres, how much longer will be required to complete the transfer?

210 - 90 = 120 litres ÷ 7.5 = 16 minutes

Transfers with multiple pumps

The principles and formula from above apply when calculating for pumps operating in parallel with different flow rates.

Example 1. Paired pumps:

How long will it take to empty a 2 cubic metres tank using a pair of pumps. Pump A has a flow rate of 6 litres/minute and pump B has a flow rate of 4 litres/minute.

Pump A flow rate + pump B flow rate = 6 + 4 = 10 litres/minute

V ÷ Fr = T

2000 ÷ 10 = 200 minutes ÷ 60 = 3.333 hours = 3 hours 20 minutes

Example 2. Paired pumps:

An older pump (A) has the capacity of 4 litres/minute.

A new pump (B) has a capacity of 8 litres/minute. What are the transfer capacities when operating in parallel to fill a 960 litre tank?

Pump A capacity = 960 ÷ 4 = 240 minutes = 4 hours

Pump B capacity = 960 ÷ 8 = 120 minutes = 2 hours

A + B capacities/hour = ¼ + ½ = ¾ of 960 litres = 75% of tank

How long will it take to fill the 960 litre tank?

75% tank transfer took 60 minutes, therefore combined transfer rate is:

(960 litres x 75%) ÷ 60 = 12 litres/minute

Pumping time = 960 ÷ 12 = 80 minutes = 1 hours 20 minutes

Example 3. Paired pumps:

A ballast pump A can empty out a 20000 litre tank in 2 hours. A feed pump B can empty out the same tank in 4 hours. If a crew member used both pumps at the same time, how long would it take to empty the 20000 litre tank.

Capacities per hour = ½ (A) + ¼ (B) = ¾ (Both) of tank per hour

Total time required = 1 hour 60 ÷ ¾ = 1⅓ = 1 hour 20 minutes

60

Example 4. Paired pumps:

Calculate the time to empty a 12,000 litre tank