surveying computations cppsis3010a cppsis4011a

SURVEYING COMPUTATIONSCPPSIS3010ACPPSIS4011A

Learner’s Guide

BC1091

CPPSIS3010ACPPSIS4011A

Surveying Computations

Learner’s Guide

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3


Contents

Introduction 7

Chapter 1 Scientific notation; transformation of formulae ..............................9 1.1 Scientifi c notation ................................................................................................. 9

1.2 Equation transformation ..................................................................................... 11

Chapter 2 Trigonometric functions ..................................................................17 2.1 Angles ................................................................................................................ 17

2.2 Coordinate axes .................................................................................................. 18

2.3 Sine and cosine ................................................................................................... 20

2.4 Tangents ............................................................................................................. 23

2.5 Inverse trigonometric functions ......................................................................... 26

2.6 Domain of angles ............................................................................................... 27

2.7 The reciprocal trigonometric functions (cosec, sec, cot) ................................... 34

Chapter 3 Right-angled triangles .....................................................................35 3.1 Introduction ........................................................................................................ 35

3.2 Sin and cos functions ......................................................................................... 36

3.3 Tan functions ...................................................................................................... 37

3.4 Pythagoras’ theorem ........................................................................................... 38

3.5 Areas .................................................................................................................. 38

Chapter 4 Scalene triangles ..............................................................................43 4.1 Elements ............................................................................................................. 43

4.2 Sine rule, cosine rule and the area formula ........................................................ 45

4.3 Loss of precision when computing angles from lengths .................................... 50

4.4 Using the formulae ............................................................................................. 53

4.5 Special situations ................................................................................................ 60

4.6 Applications ....................................................................................................... 63

4.7 Solving triangles when area is given .................................................................. 69

4

Chapter 5 Angles and bearings ........................................................................75 5.1 Angles ................................................................................................................ 75

5.2 Bearings .............................................................................................................. 77

Chapter 6 Distance, line definition (polar coordinates), point definition,(rectangular coordinates) ...............................................................89

6.1 Distance .............................................................................................................. 89

6.2 Line defi nition (polar coordinates) ..................................................................... 90

6.3 Point defi nition ................................................................................................... 92

6.4 The relationship between line defi nition and rectangular coordinates ofend points of a line ............................................................................................. 94

6.5 Calculators: use of rectangular→polar (R→P) function key ............................... 106

Chapter 7 Vectors ............................................................................................107 7.1 N and E as the components of a vector ....................................................... 108

7.2 Vectors in surveying – summing vectors to prove a survey ................................... 110

Chapter 8 Adjustment of misclose in closed traverses ...............................115 8.1 Traverse surveying .................................................................................................. 115

8.2 Traverse closure ...................................................................................................... 116

8.3 Misclose .................................................................................................................. 117

8.4 Angular misclose .................................................................................................... 119

8.5 Linear misclose and adjustment .............................................................................. 127

Chapter 9 Areas of polygons ..........................................................................1439.1 Areas of regular rectilinear figures ......................................................................... 143

9.2 Areas of polygons ................................................................................................... 146

9.3 Calculating area using coordinates ......................................................................... 146

Chapter 10 Missing measurements .................................................................15110.1 Summing vectors to prove a survey ........................................................................ 151

10.2 The sum, difference and translation of vectors ....................................................... 152

10.3 Closing vectors ....................................................................................................... 153

10.4 Partially defined vectors ......................................................................................... 153

10.5 Missing vectors ....................................................................................................... 154

10.6 Calculating missing measurements ......................................................................... 158

5


Chapter 11 Simple circular curves ...................................................................175 11.1 Introduction ............................................................................................................. 175

11.2 The geometry of curves .......................................................................................... 175

11.3 Elements of simple circular curves ......................................................................... 180

11.4 Derived formulae .................................................................................................... 181

11.5 Areas ....................................................................................................................... 184

Chapter 12 Roads ..............................................................................................189 12.1 Introduction ............................................................................................................. 189

12.2 Single-width roads .................................................................................................. 190

12.3 Computations relating to roads ............................................................................... 194

Appendix A Calculator requirements ................................................................199

7


7


IntroductionThis resource is designed to help you develop an understanding of the mathematical principles and methods commonly used in surveying, in particular the knowledge and skills related to the following units of competency.

CPPSIS3010A – Perform basic spatial computationsCPPSIS4011A – Perform surveying computations

The resource is divided into chapters. At the end of each chapter, you will be required to complete set activities in your tutorial book to practice knowledge and methods discussed in each chapter.

You will require a scientific calculator. For more information, please see the Appendix A – Calculator requirements at the back of this Learner’s Guide.

88

Introduction

9


9


Chapter 1 – Scientifi c notation; transformation of formulae

1.1 Scientific notationSome calculators may give an answer of:

7.29000 12

This means 7.29 1012, which is a number in scientific notion.

In ordinary decimal notation 7.29 1012 is written as 7 290 000 000 000. Calculator displays are often in scientific notation because very large or very small numbers can be expressed compactly when this notation is used.

To express a rational number in scientific notationTo express a number in scientific notation is to express it as the product of a decimal number between 1 and 10 and an integral power of 10.

Example 1 2 500 000 2.5 1 000 000

2.5 106

Example 2 0.025 22.5 2.5100 10

=

2.5 102

The effect of multiplying or dividing a number by 10 in the decimal system is the shift in the position of the decimal point. Therefore, changing from one form to the other becomes a matter of counting the number of places you must shift the decimal point. The following have all been expressed in scientific notation.

1010

Chapter 1 – Scientific notation; transformation of formulae

standard position

0.00194 1.94 103 ie The decimal point is three positions to the left of the standard position.

0.0194 1.94 102 ie The decimal point is two positions to the left of the standard position.

0.194 1.94 101

1.94 1.94 100 (100 1)

19.4 1.94 101

194.0 1.94 102 ie The decimal point is two positions to the right of the standard position.

1940 1.94 103

To convert a number expressed in scientific notation to ordinary decimal notationExample 1 Express 2.76 103 in ordinary decimal notation.

Solution The negative index in 10−3 indicates that the decimal point must be shifted three times to the left and zeroes added if necessary. This is because:

2.76 103 32.5 2.76

100010=

A negative index in scientific notation indicates that we have divided by a power of ten, whereas a positive index indicates that we have multiplied by a power of ten, and so the decimal point is shifted to the right.

Hence, 2.76 103 0.00276.

Example 2 Express 3.8 104 in ordinary decimal notation.

Solution 3.8 104 3.8 10 000

38 000


Solution 4.7 102 0.047


Solution 3.85 105 385 000

11


11



Solution 3.26 104 0.000326


Solution 9.8 103 9800

1.2 Equation transformationIf we want to transform an equation (change the subject of an equation), the following laws tell us what we may do to the equation without ‘upsetting the balance’.

1. Add the same number to both sides.

2. Subtract the same number from both sides.

3. Multiply both sides by the same number/variable.

4. Divide both sides by the same number/variable.

5. Square both sides or, in general, raise both sides to the nth power.

6. Take the square root of both sides or, in general, take the nth root of both sides.

All the above rules may be applied to any given example. In some cases, several of them have to be applied to the same equation in succession to achieve the desired form.

Although there is no set procedure or rigid method of solution, the following examples will give a sufficient basis of knowledge to enable any equation to be transformed.

Example 1 The area of a circle is given by A r 2. In this form the formula is useful for calculating the area. If, however, we require the radius, then we need to manipulate the formula so that we can make r the subject.

2A = r

Divide both sides by

2A r=

2A = r

Find the square root of both sides

2A r

=

A r

=

Now r can be the subject of this form of the equation Ar =

.

1212


Example 2 Given Area 12

ab sin C, write the equation with b as the subject.

1Area sin2

ab C=

Multiply both sides by 2

2Area sinab C=

Divide both sides by a sin C

2Area sin

sin sinab C

a C a C=

Divide out common terms

2Areasin

ba C

=

Example 3 Given a2 b2 c2 2bc cos A, rewrite with cos A as the subject.

a2 b2 c2 2bc cos A

Add 2bc cos A to both sides

2bc cos A a2 b2 c2

Subtract a2 from both sides

2bc cos A a2 a2 b2 c2 a2

Or 2bc cos A b2 c2 a2

Divide both sides by 2bc

2 2 2cos

2b c aA =

bc+ -

13


13


Example 4 Change the equation 12Tg

= to make g the subject.

Divide each side by 2

12T

g=

Square each side

2 12T

gæ ö÷ç =÷ç ÷çè ø

Multiply each side by g

21

2Tg

æ ö÷ç =÷ç ÷çè ø

Divide each side by 2

2Tæ ö÷ç ÷ç ÷çè ø

and divide out

21

T2

g

=æ ö÷ç ÷ç ÷çè ø

Example 5 Given Nm LC N =L´

and 0.007mC N =

160.70 mL =

2150.53 mL =

Calculate mN.

1414


Solution 160.700.007

2150.53Ńm=

160.70 2150.53 0.007Ńm =

2150.53 0.007160.70

Ńm =

0.094 m=

Example 6 Given scale (s) 'f

=-H H

and scale (s) 1

13160=

2050 m=H

f 152 mm=

Calculate 'H .

Solution '1 0.152

13160 2050 H=

-

'2050 13160(0.152)H- =

'2050 2000.3H- =

'2050 2000.3 H- =

' 49.7 mH =

Example 7 Given 1 m 24( )( )6

+ += A ADV A

and 31000 m=V

20 m=D

21 39 m=A

2m 42 m=A

2Calculate A .

15


15


Solution 2201000 (39 4(42) )6

+ += A

23.333(39) 3.333(168) 3.333( )+ += A

2130 560 3.333( )+ += A

23.333( ) 1000 130 560\ = - -A

2310

3.333=A

293 m=

After having studied this chapter and all the given examples, you should go on to complete the relevant tutorial exercises in your Tutorial book. If you encounter difficulties in the tutorial exercises you can refer back to the examples in this chapter, in the first instance, or seek help from your lecturer/tutor.

Note: The layout of the solutions to problems is important. You may need to refer back to the given examples in this chapter for the correct layout.

1616


17


Chapter 2 – Trigonometric functions

2.1 AnglesAn angle is defined as a measure of the amount of rotation one line makes about a point to coincide with another line that also passes through that point.

C

B

A

50°310°

Fig 2.1

In the above, the angle ABC = ∠ABC = 50°.

In surveying we read an angle as a rotation:

∠ABC or ∠CBA

In addition the notation used can be positive (+ve) or negative (-ve):

• A positive direction or rotation is one which is clockwise.

• A negative direction or rotation is one which is anticlockwise.

Unless otherwise specified, angles will be assumed positive, ie clockwise rotation.

Thus, in the above example:

∠ABC = 50° or -310°

while ∠CBA = 310° or -50°

18


2.2 Coordinate axesTypically in surveying, the axis coordinates are shown as 0°→360° in a clockwise rotation,indicating a positive direction.

270° W

180° S

E 90°

0°360°

N

+ve

Fig 2.2 Surveying axis coordinates

This is different from the mathematical axis coordinates, where the x and y axes are equated with 0°–180° and 90°–270° axes respectively, and where the positive rotation is in the opposite direction.

+ve

x 0° , 360°180° , -x

-y270°

90°y

Fig 2.3 Mathematical axis coordinates

19


Where directions of lines are considered as northing and easting coordinates, the surveying axes are shown like this:

+ve

E-E

-N

N

Fig 2.4

In the four quadrants we have the following rotations of N, E, -N, -E. These will be referred to throughout the text.

E

+E

-E

-N

N

+N-E

+N

+E-N

-E-N

Fig 2.5

20


2.3 Sine and cosineFor trigonometric functions we will use a unit circle with a radius length (L) of 1.

eastings

EW

S

N

0

north

ings

L

P

Fig 2.6

Using a unit circle it is now possible to locate any point on the circle in terms of its distance from the coordinate axes; in other words as an ordered pair of coordinates.

Looking at the above circle (with a radius unit of 1), we see the coordinates of north (0,1), east (1,0), south (0,-1) and west (-1,0). The easting coordinates are always stated first.

This is the notation used for mapping coordinates.

Let us now select a number of points on the unit circle and view their coordinates.

Example 1

E-E

-N

N

0

L

L

P

L

Q

R

L

L

T

S

Fig 2.7

21


Point Coordinates (E,N)

P (0.6,0.8)

Q (0.8,0.6)

R (0.4,-0.9)

S (-0.8,-0.6)

T (-0.6,+0.85)

If we were to draw in rays through these points from the origin (0), we would have rotational angles for these coordinate values.

At point P we have (E, N) coordinates (0.6,0.8).

• The easting of P(0.6) is called the sine of ∠NOP and is opposite the angle.

• The northing of P(0.8) is called the cosine of ∠NOP and is adjacent to the angle.

Thus, the E coordinate of a point is always the sine of the angle and the N coordinate of a point is always the cosine of the angle.

Or, ( )E Nsin and cos where unit lengthLL L

= = =θ θ

Example 2 Using the above information we can obtain the sine and cosine for each point on the unit circle.

Point Sine (sin) Cosine (cos)

P ∠ NOP = 0.6 ∠ NOP = 0.8

Q ∠ NOQ = 0.8 ∠ NOQ = 0.6

R ∠ NOR = 0.4 ∠ NOR = -0.9

S ∠ NOS = -0.8 ∠ NOS = -0.6

T ∠ NOT = -0.6 ∠ NOT = -0.85

Note: 1. Examples given are positive rotation angles.

2. We can determine in which quadrant each ray lies by the coordinate signs (positive and negative).

22


From the coordinate values, the angle rotation from North can now be derived using a calculator.

∠NOP = approximately 37°

∠NOQ = approximately 53°

∠NOR = approximately 156°

∠NOS = approximately 233°

∠NOT = approximately 323°

We can also consider the trigonometric functions (sin, cos) in terms of the sides of a right-angled triangle.

opposite

adja

cent

hypotenuse

N1 E1

L

Oθ

Fig 2.8

We can describe the rotation of sides as opposite, adjacent and hypotenuse to the angle θ (as shown in Figure 2.8).

Thus, in the triangle ON1 E1

we have 1 1 1sin where (unit circle)N E OE L= =θ

1 1

1

N EOE

=

oppositehypotenuse

=

similarly, adjacentcoshypotenuse

=θ

23


2.4 TangentsUsing a unit circle with a radius of 1, we can draw in a tangent line through the coordinates (0,1).

north

ing

N P

-E

-E E

L

O

Fig 2.9

We can now make an angle that cuts the tangent at point P. The coordinates of P are (E0.6,N1).

The tangent of ∠NOP = 0.6 and the easting is equal to the tangent value where the northing is equal to 1.

Note: All northing values will be equal to 1 when using a unit circle. However, at this stage, we are more interested in the size of the angle than in the coordinate value.

Looking at the four quadrants of the unit circle we can now determine the values of tangents subtended by the angle of rotation.

24


N

E

-N

2

14

3

-E

220°

310°

160°

30°

Fig 2.10

Example 1 Using the above unit circle, find the coordinate or tan values for the following.

(a) θ = 30º (b) θ = 160º

(c) θ = 220º (d) θ = 310º

Note that this diagram is not to scale so you have to use your calculator.

(a) When θ = 30º, tan θ = 0.56 and it is an easting value.

(b) When θ = 160º we can see that the ray does not meet the tangent. This will happen in the 2nd and 3rd quadrants. To overcome this, we simply draw in the opposite ray until it cuts the tangent.

Thus, when θ = 160º, tan θ = -0.36 and it is a negative (-ve) easting value.

(c) When θ = 220º, tan θ = 0.84 and it is a positive (+ve) easting value.

(d) When θ = 310º, tan θ = -1.20 and it is a negative (-ve) easting value.

Let us now look at the reverse procedure where, given the coordinate values, we have to find the angle of rotation.

25


Example 2 Given sin θ = 0.6 and cos θ = 0.8, find the tangent value.

0.75

0.6

0.8

θ

Fig 2.11

By using the unit circle and two coordinate values; sin θ = 0.6 cos θ = 0.8 we can plot these coordinates on the circle and produce the ray which is rotating the angle through the tangent.

Thus we find that the tangent value for θ = 0.75.

Using ratios from simple mathematics, we see that:

sin 0.6 0.75 tancos 0.8

= = =θ

θθ

therefore sin tancos

=θ

θ.θ

Example 3 Using the same method from the previous example we can determine the tangent value given sin 90º.

If sin90 1 and cos90 0

sin90 1tan90 undefined or0cos90

∞

= =

= = =then

And if E(1) sin

N(2) cos

sin E(3)cos N

θ

θ

θθ

L

L

=

=

=

and

then

It is important to memorise these three equations.

26


2.5 Inverse trigonometric functionsConsider the trigonometric function sin X.

If we are have a relationship such as:

sin X = 0.5 (this is actually 0.5000000)

we interpret it to mean ‘X is the angle whose sine value is equal to 0.5’. … (1)

Knowing the value of the sine of angle X, we may deduce the value of X, in degrees, by construction and measurement using tables or a calculator.

An alternative way of stating the relationship in (1) is to use the inverse trigonometric function notation:

X = sin–1 (0.5)

which also reads, ‘X is the angle whose sine value is equal to 0.5’. … (2)

Thus, for the sake of simplicity, the two expressions

sin X = 0.5 and X = sin−1 (0.5)

can be considered to be giving us the same information.

Using one of the methods mentioned above, we deduce that the value of angle X is 30 degrees.

In general terms:

if y = sin X, then X = sin–1(y).

Inverse trigonometric functionsAll calculators have an inverse function (or second function) button to change sin, cos, tan to sin−1, cos−1, tan−1.

Example 1 If X = sin−1(y)

then the value of ‘y’ is keyed in, followed by the inverse function (or second function) button and sin button, to create sin−1.

key in 0.5

press inverse (second function) key

press sin−1

displayed is 30 degrees

The logic of the operation may be considered in the following way.

27


If a value for an angle X (say, 30 degrees) is keyed into a calculator and the ‘sin’ function is pressed, the calculator responds with a sine value of 0.5.

On the other hand, if a sine value is keyed in to the calculator and the sin−1 is pressed, the calculator responds with the value of the angle (say, 30 degrees).

In other words, the second operation is the inverse of the first.

2.6 Domain of anglesIt is important to note here that calculators only show sine and tangent inverse trigonometric functions between -90º and +90º, and cosine between 0º and 180º. We need to consider therefore, angular values greater than 90º, ie between 0º and 360º. This is best illustrated by way of graphs.

Sine graphLet a line be rotated clockwise to form a circle of radius = 1 unit. Then the height on each side of the vertical axis represents the sine of the angle of rotation.

At 90º it reaches a maximum = 1

At 180º it returns to the axis

At 270º it reaches a minimum = -190°

360°

180°

90°

-1 +10

360°330°

300°

270

240°

210°180°

150°

120°

90°

60°

270°-ve

-ve

+ve

+ve

30°

Fig 2.12 The sine graph

28


From the graph we can see the following:

sin 30º = sin (180º − 30º) = sin 150º

= -sin (180º + 30º) = -sin 210º

= -sin (360º − 30º) = -sin 330º

Thus, the sine of all angles 0º to 180º is +ve (positive) and the sine of all angles 180º to 360º is -ve (negative).

Cosine graphThis is the same as the sine graph but displaced by 90º.

90°

360°

270°

90°

-90

+ve

-ve

-ve

-ve -1 +1 +ve+ve0°

-180°

-180°

Fig 2.13 The cosine graph

From the graph we can see the following:

cos 30º = -cos (180º − 30º) = -cos 150º

= -cos (180º + 30º) = -cos 210º

= cos (360º − 30º) = sin 330º

Thus, the cosine of all angles 0º to 90º and 270º to 360º is +ve (positive) and the cosine of all angles 90º to 270º is -ve (negative).

29


Tangent graphThis is dissimilar to the other two graphs as it is discontinuous.

+00

+00

+00

-00

-00

-00

+ve

+ve

+ve

-ve

-ve

-ve

360°

270°

180°

90°

0°

-90°

Fig 2.14 The tangent graph

We can see from this graph that:

tan 30º = tan (180º + 30º) = tan 210º

= -tan (180º − 30º) = -tan 150º

= -tan (360º − 30º) = -tan 330º

Thus, the tangents of all angles 0º to 90º and 180º to 270º are +ve (positive) and the tangents of all angles 90º to 180º and 270º to 360º are -ve (negative).

30


Sign of the functionThe sign of the function can be seen from Figure 2.15. Let the rotating area be positive θº.

1st Quadrant 0º–90º

1st Quadrant (θ1 = θ)

sin θ1 = ++

= +

cos θ1 = ++

= +

tan θ1 = ++

= +

2nd Quadrant 90º–180º

2nd Quadrant θ2 = (180º − θ)

sin θ2 = ++

= +

cos θ2 = -+

= -

tan θ2 = -+

= -

3rd Quadrant 180º–270º

3rd Quadrant θ3 = (θ − 180º)

sin θ3 = -+

= -

cos θ3 = -+

= -

tan θ3 = --

= +

4th Quadrant 270º–360º

4th Quadrant θ4 = (360º − θ)

sin θ4 = -+

= -

cos θ4 = ++

= +

tan θ4 = -+

= -

Fig 2.15

θ1

θ2

θ

θ3

θ

θ4

θ

+

+

+

+

+

+

+

+

++

+

+

+

++

−

−

−

−

−−

−

−

−

−

−

−

90°

1

23

4

31


The previous table is summarised in the figure shown below.

sin θ -cos θ +tan θ -

sin θ -cos θ -tan θ +

C A

T S

E-E

-N

N0°

360°

180°

270° 90°

sin θ +cos θ +tan θ +

sin θ +cos θ -tan θ -

Fig 2.16

It must be remembered that these trigonometric ratios or functions will always remain +ve or -ve within the quadrants, ie regardless of whether angle rotations are +ve (clockwise) or -ve (anticlockwise).

A useful mnemonic is CAST, as shown in Figure 2.16.

CAST C = Cos is positive here.

A = All are positive here.

S = Sine is positive here.

T = Tan is positive here.

32


Examples showing the use of sin, cos, tan with correct signsExample 1 Evaluate the value(s) of A for which cos A = 0.553452, for the domain

0 360A£ £ .

This means that the domain of the variable A (the range of A) is to be from 0º to 360º.

360°

180°

270° 90°

0°

ST

C A56°2

3'"

45

56°23'45"

Fig 2.17

A = cos−1 (0.553452)

= 56º 23' 45" (using a calculator)

As cos was given as a positive value, then possible angles lie in 1st or 4th quadrants.

1st 0º + 56º 23' 45" = 56º 23' 45"

4th 360º − 56º 23' 45" = 303º 36' 15"

Therefore A = 56º 23' 45" or 303º 36'15"

33


Example 2 Evaluate angle A (0 180A £ £ ) given cos A = -0.612401.

A=cos−1(-0.612401)

= 127.763314º (using a calculator)

As cos is negative, then it only lies in the 2nd Quadrant.

Therefore A = 127º 45' 48"

0°

90°

180°

270°

C A

T S

127°4

5 48'"

Fig 2.18

Example 3 Evaluate to four decimal digits.

(a) sin 120º (b) sin 270º (c) tan 315º

(a) sin 120º = sin 180º − 60º

= sin 60º with a +ve sign for the 2nd Quadrant

= +0.8660

(b) sin 270º = sin 180º + 90º

= sin 90º with a -ve sign for the 3rd Quadrant

= -1

(c) tan 315º = tan 360º − 45º

= tan 45º with a -ve sign for the 4th Quadrant

= -1

34


2.7 The reciprocal trigonometric functions (cosec, sec, cot)The inverse trigonometric function notation should not be confused with the reciprocal trigonometric function.

That is, sin−1 is in no way related to cosec X 1

sin Xæ ö÷ç ÷ç ÷çè ø

.

In other words, sin−1 X does not mean 1sin X

.

Calculators only have sin, cos and tan function keys. Therefore the 1x

key must be used to obtain cosec, sec and cot values.

Conversely, when cosec, sec or cot values are given, then the inverse 1x

æ ö÷ç ÷ç ÷çè ø must be obtained in

order to calculate the angle using a sin, cos or tan function key.

Trigonometric function Reciprocal

sin1cosec

sinæ ö÷ç= ÷ç ÷çè ø

cos1sec

cosæ ö÷ç= ÷ç ÷çè ø

tan1cot

tanæ ö÷ç= ÷ç ÷çè ø

Note: cosec, sec and cot are rarely used now as calculators can easily convert to sin, cos, tan.



35


Chapter 3 – Right-angled triangles

3.1 IntroductionWe have already seen from the previous trigonometric functions that certain relationships exist within the right-angled triangle. You will need to remember these relationships.

sin = easting coordinate = opposite

hypotenuse

cos = northing coordinate = adjacent

hypotenuse

tan = easting coordinate (where northing = 1) = opposite sinadjacent cos

=

easting

north

ing

hypo

tenuse1

θ

Fig 3.1

Often we are asked to ‘solve’ a triangle. This means that when we are given some elements (or parts) of a triangle we can use these to find the missing elements.

It is essential to know at least three elements.

When solving a triangle, all the elements of a triangle must be determined: three sides, three angles, the area.

In the following examples (and for surveying in general), all distances are in metres and area calculations are in square metres. If hectares are given these must be converted to square metres (1 ha = 10 000 m2).

36


3.2 Sin and cos functionsThe right-angled triangle is the easiest to solve of all triangles since it has been established that the ratios of the sides are dependent upon the trigonometric function of the angle opposite or adjacent to the side.

A

B

C

ac

bθ 90°

Fig 3.2

In the diagram above:

sin ac

=θ

cos bc

=θ

tan ab

=θ

Thus, if we knew the three sides of the triangle, it would be a simple matter to solve for the missing angles.

For example, sin ac

=θ

thus, the ratio of ac

would equal the value of sinθ

as would cosbc

= θ

and tanab

θ=

Similarly, if we knew only one side of the triangle and the three angles, we could simply solve for the missing sides.

For example, knowing θ and side c then sin ac

θ =

and by cross-multiplying we get c sin θ = a

therefore, side a can be found by multiplying sin θ by c, ie a = c sin θ

similarly, cos , and cosb bc

= =θ θ.

37


Also, if we knew one side, but didn’t know the hypotenuse, then we would have

sin ac

θ = where a is unknown

so by transposition we would get:

.sin

acθ

=

In fact, all that is needed to solve a right-angled triangle is one acute angle and one side.

3.3 Tan functionsRatios of the sides of a triangle are also applied when using the tan function.

Example 1

A

B Ca

bc

θ

Fig 3.3

Suppose sides a and b were known and we wanted to solve angles A and B and side c. We would use the tan function to find angle B.

Putting B = θ we can solve for θ using the ratios shown on the previous page.

Thus, oppositetanadjacent

ba

= =θ

If we have values for sides a and b we can apply these to the tan ratio.

Thus if a = 100 cm and b = 256 cm

then 256tan 2.56 cm100

θ = = .

38


3.4 Pythagoras’ theoremPythagoras’ theorem may be used to solve the third side, given any two sides of a right-angled triangle. We can apply this theorem to the right-angled triangle in Fig 3.3.

Example 1 c2 = a2 + b2

Thus if a = 100.000 cm, b = 256.000 cm

then c is calculated from

c2 = a2 + b2

= 1002 + 2562

= 75 536

therefore c = 75 536

= 274.838 cm

3.5 AreasThe basic formula for area calculation is

12

base × perpendicular height

Using the triangle in Fig 3.3 this is:

Area2a b= ´

Note: 10 000 m2 = 1 hectare = 1 ha

For areas greater than 10 000 m2 show the final result in ha to four decimal places.

39


Example 1 For the triangle ABC, calculate (a) side AB (b) ∠A (c) ∠B (d) the area.

A C

B

325.56 cm

253.

75 c

m

Fig 3.4

(a) Using Pythagoras’ theorem we get:

2 2 2253.75 325.56AB = +

2 2253.75 325.56AB = +

= 412.77 cm

(b) Using the tan function we get:

253.75tan325.56

A =

1 253.75or tan325.56

A = - æ ö÷çÐ ÷ç ÷çè ø

= tan−1 (0.779426)

= 37° 56' 02" (using a calculator)

(c) We know that the total of angles in a triangle is 180°.

We also know that ∠C is 90°, and we have just determined ∠A so:

∠B = 90° 00' − ∠A

∠B = 90° 00' − 37° 56' 02"

= 52° 03' 58"

40


(d) Using the basic formula for area calculation we get:

Area = 0.5 (325.56) (253.75)

= 41305.4 m2

= 4.1305 ha

Example 2 For the triangle XYZ calculate (a) ∠Y (b) side XZ (c) side YZ.

X Z

Y

12.525 m

44°27 '

Fig 3.5

(a) Using the sin function we get:

sin 44 27 '12.525

YZ=°

∴ YZ = 12.525 sin 44° 27'

= 12.525 × 0.700287

= 8.771 m2

(b) Using Pythagoras’ theorem (transformed) we get:

XZ 2 = (12.525)2 − (8.771)2

XZ 2 = 79.945

XZ = 8.941 m

(c) Total of angles in ∆ XYZ is 180°.

∠Y = 90° 00' − 44° 27'

= 45° 33'

41


Example 3 The area of triangle ABC equals 6.0 m2 and the sides AC and CX are 3.0 m and2.0 m respectively. The line ACX is straight and BX is perpendicular to ACX. Calculate the side BX.

Area ABC = 6.0 m2

90°3.0 mA X

B

C 2.0 m

Fig 3.6

In the figure above, there are three triangles, namely ABC, CBX and ABX. The side BX is the perpendicular height for all three.

Therefore Area ∆ ABC = 1 base perpendicular height2

´

6.0 = 3.02

BX´

BX = 6.0 23.0´

= 4 m

Example 4 From a point on a horizontal plane, the angle of elevation of the top of thevertical tower 22.00 m high standing on the plane is 22º 41'. Calculate the angle of elevation of a point 15.00 m nearer to the tower.

A

B CX

22.00 m

15.00 m

90°22°41'

Fig 3.7

42


Let AC be the height of the vertical tower. B is the point subtending an elevation of 22º 41'. X is the point 15.00 m nearer.

In ∆ ABC

tan ∠B = ACBC

BC = tan

ACBÐ

= 22

tan 22 41'°

= 52.64 m (using a calculator)

Now BC = XC + 15

∴ XC = BC − 15

C = 52.64 − 15.00

= 37.64 m

In ∆ AXC

tan ∠X = ACXC

æ ö÷ç ÷ç ÷çè ø

tan ∠X = 1tan ACXC

- æ ö÷ç ÷ç ÷çè ø

= 1 22tan37.64


= 30° 18'



43


Chapter 4 – Scalene triangles

4.1 ElementsA triangle has seven elements:

• three sides

• three angles

• area

A

B Ca

b c

Fig 4.1

If three independent elements are known, then all the other elements can be computed.

Note: The three angles are not independent elements as the third angle is dependent on the other two.

Without considering area, there are four possible combinations of elements:

(i) three sides

(ii) two angles and one side

(iii) two sides and the angle included between them

(iv) two sides and an angle opposite one of them.

44


AA

AA

BB

B BC C

CCa

a a

a

b

bb

c

(i) (ii)

(iii) (iv)

Fig 4.2

In this chapter we will look at solving triangles given these combinations.

45


4.2 Sine rule, cosine rule and the area formula

Sine ruleIn any triangle, the sides are proportional to the sines of the opposite angles.

In any triangle ABC

sin sin sina b c

A B C= =

Proof

a) For an acute-angled triangle

A

B Ca

bc

X

Fig 4.3

AX c sin B and AX b sin C

equating c sin B b sin C

Dividing both sides by sin B, sin C we get:

sin sinc b=

C B

46


b) For an obtuse-angled triangle

BCA and ACX are supplementary

sin BCA sin ACX

A

B C Xa

bc

Fig 4.4

AX c sin B and AX b sin ACX

or AX b sin C

equating c sin B b sin C

Dividing both sides by sin B, sin C we get:

sin sinc b=

C B

Similarly, by drawing a different altitude, it may be proved that each of these is equal to:

sina

A

47


Cosine ruleIn any triangle ABC

2 2 2 2 cosa b c bc A= + -

or

2 2 2cos

2b c aA =

bc+ -

Proof

a) For Fig 4.5

a

bc

A

B CX

Fig 4.5

AX b sin C

BX a XC

a b cos C

In ABX

c 2 BX 2 AX 2

(a b cos C )2 (b sin C )2

a2 2ab cos C b2 cos2 C b2 sin2 C

c2 a2 b2 (sin2 C cos2 C ) 2ab cos C

Now sin2X cos2X = 1.

Therefore c2 = a2 b2 2ab cos C

48


b) For Fig 4.6A

B C Xa

b

c

Fig 4.6

CX b cos ACX

CX -b cos C

AX b sin ACX

AX b sin C

In ABX

c2 BX 2 AX 2

(a + CX )2 AX 2

(a b cos C )2 (b sin C )2

c2 a2 – 2ab cos C b2 cos2C b2 sin2C

c2 a2 b2 (sin2C cos2C ) 2ab cos C

Therefore c 2 a2 b2 2ab cos C

Area formulaIn any triangle ABC

Area ABC 12

base perpendicular height

or

Area ABC = 12

perpendicular height base

or

Area ABC 12

ab sin C

49


Proof

a)

A

B

P Q

CX

b

A

B C X

b

Fig 4.7 Fig 4.8

Area rectangle length (BC) breadth (AX)

Area triangle 12

length (BC) breadth (AX)

Area ABC Area AXB Area AXC Area ABC Area AXB Area AXC

12

AX BX 12

AX XC 12

AX BX 12

AX CX

12

AX (BX XC ) 12

AX (BX XC )

12

AX (BC ) 12

AX (BC )

12

perpendicular height base

Therefore, for both triangles Area ABC 12

base perpendicular height.

Proof

b)

Now AX b sin C

Substituting for AX we get

Area ABC 12

(b sin C) BC

Therefore Area ABC 12

ab sin C

50


4.3 Loss of precision when computing angles from lengthsLook at the right-angled triangle ABC, with sides AC and BC having lengths of 100 and 1.5 exactly. Each side has an infinite number of decimal places, eg 1.500000.

A

B

C

1.5

100

Fig 4.9

We can calculate the value of angle A using the tan function:

1tan or tanopposite oppositeA = A = adjacent adjacent

-

A tan1 1.5100

(both are exact values)

= 0' 34" to the nearest second

Similarly B tan1 1001.5

= 89' 26" to the nearest second

Using Pythagoras’ theorem we can calculate the length of side AB.

c 2 a2b2

1.521002

2.25 10 000

10 002.25

side 100.0112494

c =

AB =

\

\

51


We can see from this calculation that:

side AB 100.011 to 3 decimal places

100.0112 to 4 decimal places




To how many decimal places should the value of line AB be if we had to work out the angle of A using the cos function and achieve the same level of accuracy?

Let’s investigate this question by working through the solution.

Using the cos function:

1cos or cosadjacent adjacentA = A = hypotenuse hypotenuse

-

A cos1 100.000100.011

= 0' 59" (to 3dp)

A cos1 100.0000100.0112

= 0' 57" (to 4 dp)

A cos1 100.00000100.01125

= 0' 34" (to 5dp)

From these calculations it can be seen that the length of AB needs to be known to five (5) decimal places in order to achieve the required accuracy. That is, the value obtained here is exactly the same as the value we obtained using the tan function at the beginning – both are 0°51' 34".

We can also use the sin function instead of the cos function to derive the value of angle A.

1sin or sinopposite oppositeA = A = hypotenuse hypotenuse

-

A sin1 1.50000100.011

= 0' 34" (to 3dp)

A sin1 1.5000100.01

= 0' 34" (to 2dp)

A sin1 1.500100.0

= 0' 34" (to 1dp)

52


Note how accuracy is maintained using three (3), two (2) and one (1) decimal places in the calculation of angleA.

Now we shall see how many decimal places are needed to solve for B using the sin function (sin–1), to achieve exact values (to the nearest second).

B sin1 100.000100.011

= 89' 01"

B sin1 100.0000100.0112

= 89' 33"

B sin1 100.00000100.01125

= 89' 26"

By comparing these calculations to the values of B we obtained using the tan function, we can see that the value of side AB needs to be known to five (5) decimal places.

We can also use the cos function instead of the sin function to solve for B.

B cos1 1.50000100.011

= 89' 26"

B cos1 1.5000100.01

= 89' 26"

B cos1 1.500100.0

= 89' 26"

Looking at these calculations we can see that a minimum of one (1) decimal place is required for accuracy.

SummaryFrom this investigation we can see that when solving triangles involving small angles using the cos function (cos–1) the answers we obtain lack accuracy, unless quite a number of decimal places are given in the original data. The same is also true when solving triangles with angles close to 90using the sin function (sin–1). A detailed explanation will not be given here but the reason lies in the rate of change of sine and cosine in the region of 0and 90

As a result, we can make a set of general guidelines for solving triangles which contain very small angles, say less than 5or angles close to 90

• When determining small angles, it is best to use the sin or tan functions rather than thecos function.

• When determining angles close to 90it is best to use the cos function.

53


4.4 Using the formulaeThe following examples make use of the sine and cosine formulae in the solution of triangles.

Example 1 Solve a triangle for the missing elements given the lengths of the three sides.

Given a 200.00 cm

b 250.00 cm

c 175.00 cm

A

B

C

a

b

c

Fig 4.10

(a) Firstly we must select the best formula to use from those we have lookedat to solve for the missing elements: ie A, B, C.

The formulae are:

(i) sin sin sin

a b cA B C

= = (sine rule)

(ii) a2 b2 c2 2bc cos A (cosine rule)

(iii) 2 2 2

1cos2

b c aA =bc

+- -æçççççè

(cosine rule)

(iii) is the best formula to use, and by using it we can find the value for A.

2 2 21cos

2b c aA

bc+-

æ ö- ÷ç ÷ç = ÷ç ÷ç ÷çè ø

2 2 21 250.00 175.00 200.00cos

2 250.00 175.000A +-

´ ´

æ ö- ÷ç ÷ç = ÷ç ÷ç ÷çè ø

Therefore A = 52 37' 00"

54


Using the same formula we can also find the value of B.

2 2 21cos

2a c bB

ac+-

æ ö- ÷ç ÷ç = ÷ç ÷ç ÷çè ø

2 2 21 200.00 175.00 250.00cos

2 200.00 175.00B +-

´ ´

æ ö- ÷ç ÷ç = ÷ç ÷ç ÷çè ø

Therefore B = 83' 04"

To find C we will again use the cosine formula, and not sum A and B, and subtract the total from 180º.

2 2 21 200.00 250.00 175.00cos

2 200.00 250.00C +-

´ ´

æ ö- ÷ç ÷ç = ÷ç ÷ç ÷çè ø

Therefore C = 44' 55"

(b) Check

The three computed angles are now summed as a double-check of ourcalculations.

A = 52° 37' 00"

B = 83° 20' 04"

C = 44° 02' 55"

å = 179° 59' 59"

You will notice that the total of the three angles does not add up to 180°. The difference between the totalled angles and 180° is called the misclose. The misclose for our added angles (above) is -1".

Any angle may now be adjusted by 1" to make the three angles sum to 180°. If the misclose was 3" then each angle would be adjusted by 1".

Usually, the misclose is adjusted equally (if possible) between the three angles.

Therefore our angles will now become:

A = 52' 00"

B = 83' 05"

C = 44' 55"

å =180' 00"

55


(c) The final element we have to find for this triangle is the area.

1Area sin2

ABC ab CD =

0.5 200 250 sin´ ´= 44' 00"

217 382m=

= 1.7382 ha

Example 2 Solve a triangle for the missing elements given two angles and the length of one side.

Given A 33 30' 30"

B 70 21' 30"

c 550.00 m

A

B

C

550.0

0 m

33° 30

' 33"

70°21' 30"

Fig 4.11

(a) In this example two angles are given so we can easily calculate the thirdone. Whenever two angles of a triangle are given, the third may becomputed by summing the two known angles and subtracting them from180. However, you should always double-check this calculation.

C 180 A B)

180 52' 00")

08' 00"

56


(b) We now need to select the best formula to use to solve for the missingelements: sides a and b.

The formulae are:

(i) sin sin sin

a b cA B C

= = (sine rule)

(ii) a2 b2 c2 2bc cos A (cosine rule)

(iii) 2 2 2

1cos2

b c aA =bc

+- -æ ö÷ç ÷ç ÷ç ÷ç ÷çè ø

(cosine rule)

For this example, formula (i) is the best to use to find the length of sidesa and b.

sin sina c

A C=

sinsinc Aa

C=

550.00 sin33 30' 30"sin 76 08' 00"

=

Therefore 312.75 ma =

(For our checks, store the complete calculated value in your calculator NOT the rounded off value of 312.75.)

sin sinb c

B C=

sinsinc Bb

C=

550.00 sin 70 21' 30"sin 76 08' 00"

=

Therefore 533.55 mb =

(For our checks, store the complete calculated value in your calculator NOT the rounded off value of 533.55.)

57


(c) Check

Once the values of the sides have been computed, we then have to double- check our calculations.

sin sin sina b c

A B C= =

312.75 566.526 msin sin33 30'30"

aA

= =

533.55 566.514 msin sin 70 21'30"

bB

= =

550.00 566.510 msin sin 76 08'00"

cC

= =

Note: These values are all different because rounded off values for a and b have been used for the first two calculations (values for c and sin C are given or calculated from given values, so are exact). If you recalculate

sina

Aand

sinb

B using your stored values for a and b, all three answers

will be exactly the same. Try this now to prove it to yourself!

This shows the importance of using stored values for further calculations, rather than using rounded off values.

(d) Area is the final element to be calculated.

1Area sin2

ABC bc AD =

0.5 533.55 550.00 sin´ ´ ´= 33' 30"

281001 m=

= 8.1001 ha

58


Example 3 Solve a triangle for the missing elements given the lengths of two sides and theangle included between them.

Given a 252.640 m

c 138.760 m

B 54 24' 12"

B

A C

252.640 m

138.

760

m

54°24'12"Fig 4.12

(a) Firstly, we must select the best formula to use to solve for the missingelements: side b, A, C.

The formulae are:

(i) sin sin sin

a b cA B C

= =

(ii) a2 b2 c2 2bc cos A

(iii) 2 2 2

1cos2

b c aA =bc

+- -æ ö÷ç ÷ç ÷ç ÷ç ÷çè ø

In this case, we would select formula (ii) for the initial calculation.

Side b can be computed by rearranging this formula to get:

2 2 2 cosb a c ac B+= -

2 2252.640 138.760 2(252.640)(138.760) cos54 24'12"b += -

205.598 mb =

59


(b) Since the lengths of all three sides are now known, angles A and C can becomputed using the cosine rule.

2 2 21cos

2b c aA =

bc+-

æ ö- ÷ç ÷ç ÷ç ÷ç ÷çè ø

2 2 21 205.598 138.760 252.640cos

2(205.598)(138.760)= +-


Therefore 92 18'45"A =

If we used the sine rule to calculate A we would have found two values.

1sin sinaA = Bb


Therefore 87 41' 09" or 92 18' 51"A =

We will look at this ambiguity in the following section, but in general it is always safer to use the cosine rule as this will never give ambiguous results.

(c) Finally, we calculate C using, again, the cosine rule.2 2 2

1cos2

a b cC =ab

+-æ ö- ÷ç ÷ç ÷ç ÷ç ÷çè ø

2 2 21 252.640 205.598 138.760cos

2(252.640)(205.598)= +-


Therefore 37 17' 03"C =

(d) Check

To double-check our calculations for the angles we can simply sum all ofthem.

A = 9218' 45"

B = 5424' 12"

C = 3317' 03"

å =18000' 00"

(e) Now we calculate the area of the triangle.1Area sin2

ABC ac BD =

0.5 252.640 138.760 sin54 24'12"´ ´ ´=

214 252.7 m=

1.4253 ha=

60


4.5 Special situations

AmbiguousA condition of any triangle is that the three angles total 180. That is, if is any angle in a triangle then 0

This results in ambiguous values of when sin–1x (where 0 x 1). That is, there are two possible values for since sin is positive in the first and second quadrants. The two values are and (180 ).

Ambiguity does not exist for cosine and tangent because they are both negative in the second quadrant.

When the known elements of a triangle are the lengths of two sides and an angle opposite one of them, an ambiguous situation may arise because there are a couple of possible line lengths and positions.

Let’s investigate this further by looking at ABC where a, b and A are the given elements, and A is acute.

Let CD be perpendicular to AX so that CD b sin A a.

A X

C

D

a

(CD = b sin A)

Fig 4.13

If a b sin A, then as a general rule, an ambiguous situation may occur.

A 90

a b

A B

C

XB'

a a

b

Fig 4.14

In Fig 4.14 a > b sin A. The circle cuts AX at two points: B and B' and two triangles are determined:

• AB ' C in which B ' is obtuse

• ABC in which B (180° B ' ) is acute.

61


A special ambiguous case occurs when the circle cuts AX at B and B ' (where B ' falls on A). One triangle (isosceles) is determined.

A 90

a b A B

A B X

C

D

ab

(B')

Fig 4.15

Non-ambiguousAmbiguous situations cannot occur in the following circumstances.

a) If a b sin A, then no ambiguous situation will occur as no triangle is formed and there isno solution. CD is the shortest distance possible from C to AX so CD is at 90 to AX.

A 90

a b

A

C

D X

ab

Fig 4.16

b) If a b sin A, then the subscribed circle will touch at D, thus forming a right-angledtriangle with only one solution.

A 90

a b

A X

C

D

ab

Fig 4.17

62


c) If the circle cuts AX at B and B ' (where B ' falls on X A produced), then only one triangle isdetermined. The triangle AB 'C does not contain the given angle A, only triangle ABC does.

A 90

a b

C

B' B XD

A

ab

a

Fig 4.18

d) If angle A is obtuse, ie A 90, and a b or a b, then no triangle can be formed.

A 90

a b

a

b

A

C

X

Fig 4.19

e) If angle A is obtuse and a b, then only one triangle is formed since AB 'C does notcontain A.

A 90

a b

A XB(B')

C

ab

Fig 4.20

63


4.6 ApplicationsExample 1 Solve a triangle for the missing elements given the lengths of the two sides and

an angle opposite one of them.

Given b 327.00

c 1217.00

C 80 04' 00"

(a) As no diagram is given, construct the triangle to determine if it is likely tobe ambiguous.

The process is:

1. Draw in a baseline CX.

2. At C, mark out angle C, length CA = b = 327.00 cm.

3. From A, scribe an arc of length

AB = c = 1217.00 cm.

Looking at our drawing we can conclude that only one solution exists.

A

C X

B80°04' 00"

b =

327.

00 c

m 1217.00 cm

Fig 4.21

(b) To compute B we will use the sine rule.

sin sinb c=

B C

sin sinbB = Cc

1 327.00sin sin80 04' 00"1217.00

B = - æ ö÷ç ÷ç ÷çè ø

Therefore B can either be:

(i) 15 20'49.5" or (ii) 164 39'10.5" .

(To maintain accuracy in later parts of this calculation, store these values in your calculator and record each to 0.1".)

From the diagram, it must be 15 20'49.5" .

64


This can also be proven by calculating A .

Using value (i)

A180C + B)

18080 04' 00" + 15 20' 49.5")

18095 24' 49.5")

84 35' 10.5"

Using value (ii)

A 18080 04' 00" 164 39' 10.5")

180244 43' 11") which is impossible

Therefore, only one solution exists, A 84 35' 10.5".

(c) Next we will calculate side a by either

sin sinorsin sin

b A c Aa = a =B C

327.00 sin84 35'10.5" 1217.00 sin84 35'10.5"orsin15 20'49.5" sin80 04'00"

a = a =

a 1230.010 cm or a 1230.010 cm

Thereforea 1230.010 cm

(d) Area is calculated.

1Area sin2

= ab C

0.5 1230.010 327.00 sin 80 04'00"= ´ ´ ´

2198 091.9 m=

Therefore Area 19.8092 ha=

Example 2 Solve a triangle for the missing elements given the lengths of two sides and an angle.

Given b 859.00 m

c 31.80 m

C 19 07'

65


(a) Draw the triangle as in the previous example.

C

A

B

B X

b = 859.00 m

19° 07 '

31.80 m

1

Fig 4.22

Calculate AB'

from AB' = AC sin 197'

= 859.00 sin 1907'

= 281.316 m

As this is larger than the given side AB (= 31.80 m) no solution is possible.

(b) As another check, calculate B

sin sinbB = Cc

1 859.00sin sin19 07 '31.80

B = - æ ö÷ç ÷ç ÷çè ø

( )1sin 8.85B = -

No solution is possible since B 1 and for a sine value this is impossible.

Example 3 Solve a triangle for the missing elements given the lengths of two sides and an angle.

Given b 859.000 cm

c 318.000 cm

C 19 07' 00"

(a) Draw the triangle and test for the number of possible solutions.

859.000 cm

318.000 cm 318.000 cm

19° 07' 00"

C

A

BXXB'

Fig 4.23

66


From the previous example

AX = 281.316 cm

As this is less than

AC (= 859.000 cm) two solutions are possible.

Therefore it is an ambiguous case.

(b) Compute B.

sin sinbB Cc

=

1 859.000sin sin19 07 '00"318.000

B - æ ö÷ç= ÷ç ÷çè ø

Therefore (i) 62 12'27"B = or (ii) 117 47 '33"B = (at B').

(c) Compute A.

180 ( )A B C+ = -

180 81 19'27" or 180 136 54'33"= - - (for ACB')

Therefore (i) A 98 40'33"= or (ii) A 43 05'27"= (for ACB').

(d) Compute side a.

sin sinorsin sin

b A c Aa aB C

= =

Using value (i)

959.904 cma = or 959.903 cma =

Using value (ii)

663.355 cma = or 663.355 cma =

Therefore (i) a = 959.904 cm or (ii) a = 663.355 cm

= 959.903 cm or = 663.355 cm

67


(e) Listed below is a summary of the calculations for the two solutions.

Value (i) Value (ii)

62 12'27B =

98 40'33"A =

959.904 cma =

117 47 '33"B =

43 05'27A =

663.355 cma =

(f) Compute area.

A

BC a

b c

a

b c

A

BC

Fig 4.24 Fig 4.25

Using value (i) Using value (ii)

1Area sin2

= ab C 1Area sin2

= ab C

0.5 959.904859.00 0.5 663.355 859.00

sin 1907' 00" sin 1907' 00"

135 018.31m2 93 306.3m2

= 13.5018 ha = 9.3306 ha

68


Task

The two large belts of trees shown on the diagram below are known to be on, or near, the boundary line (BC), between Fred’s block and Ron’s block. These trees mean it is not possible to see between corners B and C.

Between these two belts of trees is a small spring, but Fred and Ron do not know in whose property the spring occurs.

From old survey information (shown on the diagram) and distance AX measured to the centre of the spring, you have been asked to determine in whose property the spring occurs, and by how much.

Fred’s block

Ron’s block

Large belt of treeson or near boundary line BC

Estimated position of the spring

To spring620.00 m

1262.30 m

1381

.70

m

68°17'00"

57°4

3'00

"

A

X

B

C

Fig 4.26

Method

1. Assume ‘X’ is on the boundary line BC.

2. Solve triangle ABC for the length CB, B and C.

3. In triangle ACX, calculate X and length AX.Check by calculating X and length AX from triangle ABX.

4. If AX is greater than 620 meters then the spring is in Fred’s property and vice versa.

69


4.7 Solving triangles when area is givenIn this section we will look at solving the missing elements of a triangle when the area of the triangle is one of the given elements. If a diagram is not given, always consider ambiguous situations.

When considering area, there are three possible combinations of elements. We will examine:

• two sides and the area

• one angle, one adjacent side and the area

• two angles (three angles) and the area.

Formulae usedThe formula for area commonly used when given two sides and the included angle is

Area 12

ab sin C

The formula used for solving a side, given three angles and the area is

a sin A 2Areasin sin sinA B C

Example 1 Solve a triangle for the missing elements given the lengths of two sides and the area.

Given a 313.90 m

b 475.80 m

Area 2.3875 ha

A

BB

C

'

2.3875 ha313.90 m

313.90 m

475.80 m

Fig 4.27

Draw the triangle from the information given. Unless the shape of the triangle is known, the solution is ambiguous so there are two possible positions for B (B and B').

70


(a) Select the best formula to solve for the missing elements.

To compute angle C it will be Area 12

ab sin C.

Rearranging the formula we get:

1 2AreasinC

ab- æ ö÷ç = ÷ç ÷çè ø

(The area must be converted to square metres, ie 23 875 m2.)

1 2 23 875sin313.90 475.80

´-

´

æ ö÷ç ÷= ç ÷ç ÷çè ø

Therefore (i) 18 38'44"C = or (ii) 161 21'16"C = :

(b) The triangle now has the lengths of two sides and the angle includedbetween them. The length of the third side (c) can now be computedusing the cosine rule.

Using (i)

2 2 2 2 cosc a b ab C+= -

2 2(313.90) (475.80) 2(313.90)(475.80) cos18 38'44"c = + -

Therefore 204.669 mc =

or

Using (ii)

2 2(313.90) (475.80) 2(313.90)(475.80) cos161 21'16"c = + -

Therefore 779.710 mc = .

(c) The triangle now has the lengths of the three sides defined. We will usethe cosine rule to compute angles A and B.

Compute A using (i).2 2 2

1cos2

b c aA =bc

+-æ ö- ÷ç ÷ç ÷ç ÷ç ÷çè ø

2 2 21 475.80 204.67 313.90cos

2 475.80 204.67= +-

´ ´



or

Compute A using (ii).2 2 2

1 475.80 779.71 313.90cos2 475.80 779.71

= +-

´ ´


Therefore A 7 23'42"=

71


Compute B using (i).2 2 2

1cos2

a c bB =ac

+-æ ö- ÷ç ÷ç ÷ç ÷ç ÷çè ø

2 2 21 313.90 204.67 475.80cos

2 313.90 204.67= +-

´ ´


Therefore 131 59'30"B =

or

Compute B using (ii).

2 2 21 313.90 779.71 475.80cos

2 313.90 779.71B = +-

´ ´


Therefore B 11 15'01"=

(d) Check

Perform a check of the calculations for each solution.

(i) (ii)

29 21'47" or 7 23'42"

131 59'30" or 11 15'01"

18 38'44" or 161 21'16"

180 00'01" or 179 59'59"

A

B

C

=

=

=

å =

Example 2 Solve a triangle for the missing elements given one angle, length of one of the adjacent sides and the area.

Given C 65 25' 20"

a 152.76 m

Area 1.5655 haA

B C

1.5655 ha

152.76 m

65°25'"

20

Fig 4.28

72


(a) Select the best formula to use to solve for the missing elements.

To compute side b it will be Area 12

ab sin C.

2Areasin

ba C

= (Area must be converted to m2.)

2 15655152.76 sin 65 25'20"

´=

Therefore 225.382 mb = .

(b) We now have the lengths of two sides and the angle included between them. Thecosine rule is now used to compute the length of the third side (c).

2 2 2 2 coc a b ab sC+= -

2 2(152.76) (225.38) 2 152.76 225.38 cos 65 25'20"c + ´ ´ ´= -

Therefore 213.290 mc = .

(c) The triangle now has the lengths of its three sides defined. We will use thecosine rule to compute the two remaining angles.

2 2 21cos

2b c aA =

bc+-


2 2 21 225.38 213.29 152.76cos

2 225.38 213.29= +-

´ ´



2 2 21cos

2b c bB =

ac+-


2 2 21 152.76 213.29 225.38cos

2 152.76 213.29= +-

´ ´


Therefore 73 56'12"B = .

73


(d) Check

Perform a check of the calculations.

40 38'28"

73 56'12"

65 25'20"

180 00'00"

A

B

C

=

=

=

å =

Example 3 Solve a triangle for the missing elements given two angles and the area.

Given B 57 30' 00"

C 65 45' 00"

Area 1.5865 haA

B C

1.5865 ha

57°30'00" 65°45'00"

Fig 4.29

(a) Select the best formula to use to solve for the missing elements.

To compute the third angle (A) sum the other angles, then subtract them from180Using the three angles and the area, the sides can then be computed usingthe following formula.

2areasinsin sin sin

a = AA B C

Compute A.

180 ( )+A = B C -

180 (123 15'00")= -


74


(b) Compute the three sides.

2areasinsin sin sin

a = AA B C

(Area must be converted to m2.)

The part of the equation contained by the square root sign is common to the next three calculations. It can therefore be calculated (= 222.128 m) and multiplied by sin A, sin B, sin C to obtain the following results for a, b, c.

sin56 45'00" 222.128 185.762 m

sin57 30'00" 222.128 187.340 m

sin 65 45'00" 222.128 202.527 m

a =

b =

c =

´

´

´

=

=

=

Therefore 185.762 m187.340 m202.527 m

a =b =c =

(c) Check

sin sin sina b c

A B C= =

185.76 187.34 202.53sin56 45'00" sin57 30'00" sin 65 45'00"

= = =

222.127 222.127 222.127= = =



Answer

Task (page 68)

AX 602.79 metres; therefore, the spring is in Ron’s property.

BC 2356.44

75


Chapter 5 – Angles and bearings

5.1 Angles

1. Angle definitionAn angle is defined as a measure of the amount of rotation one line makes about a point tocoincide with another line which also passes through that point.

Consider any two lines AB and BC meeting at point B, as shown in Fig 5.1.

A

B

C

50°

Fig 5.1

The angle ABC ABC 50

is the symbol for angle, and the sequence of letters indicates the sense in which the angle is given.

ABC is read as ‘the angle at B from the line BA to the line BC’. Thus, the line BA is defined as the initial line (ray) of this angle and BC the final line of the angle, ie the amount of rotation the line BA makes to coincide with line BC.

Similarly, the angle CBA, denoted by CBA, is read as ‘the angle at B from the line BC to the line BA’. BC is the initial line, BA is the final line and CBA gives the amount of rotation that BC makes to coincide with the line BA.

If ABC 50 then CBA 310.

76


2. Angle conventionThe previous definition of an angle is expanded to include a convention relating to direction of rotation.

In surveying, angles are considered as clockwise (positive) rotations. However, some calculations require anticlockwise (negative) rotations. This may be required for traverse calculations. See Fig 5.2.

An angle is a measure of the amount of rotation one line makes about a point to coincide with another line passing through that point. The direction of rotation is given by a sign – a positive value indicating clockwise rotation and a negative value indicating anticlockwise.

-310° 50°B

A

C

Fig 5.2

From the above figure we can see:

• ABC 50

• ABC -310 (if expressed as a negative rotation).

Note: ABC is read as ‘the angle at B from the initial line BA to the final line BC’.

77


5.2 Bearings0°

50°

50°

135°

90°

135°

315°

290°

270°

290°

180°

reverse

bearing

W

S

E

N

Fig 5.3

A bearing is the ‘direction’ of a line in a survey, with respect to a common orientation for the survey. The 0 value is a northerly direction. It can be true or magnetic north at a point, grid north, and in the absence of these, or if true orientation is unimportant, assumed north may be used.

A datum direction of 0 (orientation) must be established at some point in the survey. In Fig 5.4, the ‘datum direction’ has been established at ‘A’ and the orientation at any other point in the survey is defined as being parallel to this.

N N N

A

B

D

C θCD

θ AB

θ BC

' "

Fig 5.4

78


AN, BN ' and CN " are all parallel.

The Greek symbol theta (is used for bearing notation. For the above figure, ABis read as ‘the bearing at A from A to B’.

The subscript AB after the symbol refers to line AB. The order of the letter is critical. AB is NOT the same as BA.

If a number of lines meet at a point and the bearings of these lines are known, then any angle may be computed as a difference between two bearings.

Computing angles from bearingsIf a number of lines of known bearings meet at a point, the angles between these lines may be computed.

Consider the diagram below.

50°

88°

310°

260°

190°

DX

E

C

B

A

Fig 5.5

Lines XA, XB, XC, XD and XE all radiate out from point X, with bearings as shown.

The angles between the lines can be computed from the differences in the bearings.

79


Example 1 AXB is the difference in bearings between lines XA and XB.

AXB = XB – XA (final bearing – initial bearing)

= 88 – 50

= 38

BXC = XC – XB

= 190 – 88

= 102

In all these calculations, you can see the rule for calculating the angles between the lines is:

‘final bearing – initial bearing’.

The way the angle is written indicates the final bearing and the initial bearing.

For example, for AXB:

• the final bearing is from X to B

• the initial bearing is from X to A

• ‘X’ is the pivot point of the two lines XA, XB.

Note: That in all cases the bearings must start from ‘X’, the common origin for both lines.

We can apply this rule to calculate other angles.

AXC = XC – XA

= 190 – 50

= 140

AXE = XE – XA

= 310 – 50

= 260

The next example is a bit different because the final bearing appears to be smaller than the initial bearing.

80


Example 2 CXB = XB – XC

= 88 – 190

= (88 + 360 – 190 (88is the same as 88 + 360 = 448, one extra rotation of 360. See Fig 5.6.)

= 448 – 190

= 258

0°

88° + 360° = 448° = 88°

88°

90° 270°

180°

N

Fig 5.6

Example 3 EXA = XA – XE

= 50 – 310

= (50 + 360 – 310

= 410 – 310

= 100

81


Forward and reverse bearingsThe bearing of a line from the initial point to the end point is the forward bearing of the line. The reverse bearing of a line is the bearing from the end point to the initial point.

Example 1 Consider the line AB – from the initial point A to the end point B – shown in Fig 5.7.

N

N

B'

B

A

50°

50°

230°180°θ AB

= 50°

θ BA= 230°

00°

'

Fig 5.7

We can see:

• the forward bearing AB 50

• the reverse bearing BA 230.

The computation for the reverse bearing is shown below.

AN || BN ', thus NAB AB N 'BB ' (BB ' is the continuation of the line segment AB.)

and B 'BA

then N 'BA N 'BB '

or BA AB

50

Therefore BA

It can be shown that this relationship exists for any line; that isthe reverse bearing is equal to the forward bearing plus 180.

82


Using Fig 5.7 again we see that for line AB:

AB

BA

However, for line BA:

BA this is the forward bearing)

The reverse bearing is:

AB (

or

AB –

In general, we can add 180 to the forward bearing to get the reverse bearing. But, we can also subtract 180 from the forward bearing if this makes the computation easier, (as shown in the example above).

Computing bearings from a starting bearing and an observed angle Given an initial bearing at a point, and an angle at that point, the bearing of the second line can be calculated.

Example 4 In Fig 5.8 the bearing of line AB (AB) is known and angle BAC is known.

A

B

C

Fig 5.8

Note: This is a positive (clockwise) rotation of line AB to meet line AC.

Then the bearing of AC (AC) is calculated from

AC AB + BAC

If AB = 30and BAC = 60, then AC = 30 + 60

= 90

83


In these types of calculations, the initial bearing (AB) is better known as the ‘back’ bearing and the computed bearing (AC) is the ‘forward’ bearing.

Example 5 Similarly, in Fig 5.9 we can calculate the bearing for line YX.

Given YZ = 210 (= back bearing)

Angle ZYX = 110

Z

X

Y

Fig 5.9

YX = 210110

= 320 (= forward bearing)

Note: This is a positive (clockwise) rotation of line YZ to meet line YX.

It is important to note that in this type of calculation, the back bearing must always start at the common point, that is where the angle is measured between the two adjoining lines.

In Fig 5.8 the back bearing must start at A.

In Fig 5.9 the back bearing must start at Y.

If the back bearing is IN to the common point, a reverse bearing will have to be computed first.

84


Example 6 Compute the forward bearing GH.

Given FG = 150 (back bearing)

Angle FGH = 120

G

F

H

Fig 5.10

GH = GF + FGH

= (FG + 180°) + FGH

= (150° + 180°) + 120°

= 330° + 120°

= 450°

= 90°

Note: The reverse bearing of FG is computed from (FG + 180°) and then the angle to H is added.

In traverse calculations, you may find the given angle is negative, as in Example 7.

Example 7 Compute the bearing LM.

Given NL = 30 (back bearing)

Angle MLN = 120

N

L M

Fig 5.11

85


In this case, this is a negative rotation of line LN to meet line LM.

LM = LN – MLN

= (NL + 180°) – MLN

= (30° + 180°) – 120°

= 210° – 120°

= 90°

The principle of reversing bearings is used to ‘carry bearings forward’ (to establish the datum direction from point to point) over lines measured in a survey. Consider the lines measured between, and the angles observed at the points A, B, C and D as shown in Fig 5.12.

Y

D

C

BX

A

58° 1

0 '"

15

125° 40' "00

156° 50 ' "30

50° 0

0 '"

45

212°

30'

"00

Fig 5.12

The bearing from A to X (AX) has been established as 21230'00" (known or assumed). The bearings of the other lines in the survey have been calculated as shown in the following table.

86


These calculations are extensions of the calculations in the previous section.

Bearing Comment

AX 21230'00" back bearing away from A (given)

- BAX 58 10'15"

154 19'45"-

AB=

negative rotation forward bearing (AB)

180 00' 00"334 19'45"

+

BA=

reverse bearing (BA)

ABC 50 00'45"

384 20' 30"+

positive rotation

360 00' 00"24 20' 30"

-

BC=

to normalise, ie 0 < < 360

180 00' 00"

204 20' 30"+

CB= reverse bearing (CB)

BCD 125 40' 00"

330 00' 30"+

CD=

positive rotation

180 00' 00"

510 00' 30"+

reverse bearing (DC)

360 00' 00"150 00' 30"-

DC=

to normalise reverse bearing (DC)

-YDC 156 50' 30"6 50' 00"

--

negative rotation

360 00' 00"353 10' 00"

+

DY=

to normalise

87


With practice, this process may be shortened by mental application of the 180 and 360 values. For example:

( )

( )

( )

212 30' 00"

58 10'15"154 19'45"

50 00'45"180 360 24 20' 30"

125 40' 00"180 330 00' 00"

156 50' 30"180 353 10' 00"

AX

AB

BC

CD

DY

-- BAX =

+ ABC +=+

+ BCD +=+

- YDC -=+

=

- =

=

=



88


89


Chapter 6 – Distance, line defi nition (polar coordinates), point defi nition, (rectangular coordinates)

6.1 Distance

Definition of distanceDistance, in surveying, can be defined as a measure of length of a line segment joining two points.

In Fig 6.1, let A and B be two points. Then the distance AB, denoted in mathematics as AB, is the number of unit lengths, whole and fractional, that will fit into the line segment between A and B. (AB is a segment of the infinite line containing A and B). In Fig 6.1, below, let the small subdivisions be unit lengths.

A

B

Fig 6.1

If AB is any line, then 5AB BA .

A is the initial point and B is the end point of the line segment AB, and B is the initial point and A is the end point of the line segment of BA.

The sequence of letters indicates the sense in which the distance is given; that is, AB is read as ‘the distance from A to B’ and BA as ‘the distance from B to A’.

The convention used throughout this text will be simply AB BA 5.

90

Chapter 6 – Distance, line definition (polar coordinates), point definition, (rectangular coordinates)

6.2 Line definition (polar coordinates)

Mathematics convention for surveyingIn mathematics, any line may be considered as a radius vector length ‘r’ forming a rotation of degrees with the positive x-axis. (r, ) are the polar coordinates of a line. See Fig 6.2.

2nd 1st

3rd 4th

−x

−y

+x 0°

+y

P1 (x1 , y1)

P2 (x2 , y2)

(0, 0)

r 1

r 2

θ 1θ 2

Fig 6.2

In surveying, any line may be considered as a radius vector of length ‘L’ forming a rotation of degrees with the positive N-axis. (, L) are the polar coordinates of a line and is the bearing of the line (Fig 6.3). In surveying calculations distances are always in metres.

1st

2nd

4th

3rd

+N

+E

−N

−E

P1(N, E)

P2(−N, E)

L 2

L 1

θ1

θ2

0°

Fig 6.3

91


Line definition in surveyingWe are now able to adequately define any line in a survey.

A line denoted ‘AB’ is read as ‘the line from A to B’. We have seen in the previous chapter that the direction of the line is given by the bearing as stated in the line definition – it may be either a forward or reverse bearing.

The horizontal relationship of the initial and end points of a line is given by a bearing and the distance, which is complete and unambiguous.

Line AB is defined as:

and the bearing the distance

from toAB

AB

A BL AB

The line BA is defined as:

and the bearing the distance

from toBA

BA

B AL BA

For example, look at Fig 6.4.

N

L = 426.53 m

A

B

θ 121°

(0, 0)

Fig 6.4

121426.53 m

AB

ABL

This is the conventional form for line definition.

Using the same figure we can also see:

301426.53 m

BA

BAL

92


6.3 Point definition

Mathematics and surveying conventions comparedIn mathematics, a point is related to the origin by distances along the x-axis and the y-axis. The distances, (x, y) are the Cartesian coordinates of that point.

y+

x+

y

x

P1(x, y)

(0, 0) 0

Fig 6.5

In surveying, a point is related to the origin by distances along the N-axis and the E-axis.

N

EE

A(NA, EA)

NAEA

0(0, 0)

Fig 6.6

93


A line does not have coordinates. It is the end points of a line which have coordinates. A line is defined by the differences between the coordinates of its end points. The figures below graphically illustrate these statements and are explained in detail in the next section.

N

N

EE

A (NA , EA)

B (NB , EB)

C ( NC , EC)

( ND , ED)

ND ECED

NC

NA

EA

EBNB

D

Fig 6.7

N

EE

N

D

(3, 2)

( 3, 2)

3

3

22

A

Fig 6.8

94


6.4 The relationship between line definition and rectangular coordinates of end points of a line

Rectangular coordinatesAs already discussed, in surveying, a line is defined by its bearing and distance (polar coordinates). These two quantities give the relationship of the two points A and B, as shown in Fig 6.9, where point B is distance ‘L’ units from A, and at a bearing of ‘’ degrees.

NB

EA

Lθ

Fig 6.9

However, this system has limitations when the desired points are separated by more than one line, so several bearings and distances are required.

A

B

C

D

Fig 6.10

To define the relationship of control points A and E, four bearings and four distances are necessary: AB, BC, CD and DE.

An alternative definition of relativity is in terms of rectangular coordinates – simply referred to as coordinates.

Coordinates of a point are defined as distances, north and east of a given point called the origin, the coordinate values of the origin being zero.

By definition, coordinates can only be given for a point. A line cannot have coordinates.

Note: N and E coordinates are vectors; that is they have positive (ve) and negative (-ve) values.

95


Example 1

A

B

E

N

1

2

2

1

(0, 0) Origin

Fig 6.11

Looking at the above figure we can see that point A has coordinates of:

2 1A AN E- -

and point B has coordinates of:

2 1B BN E

Note: The subscript letter denotes the point, and the ‘’ signs are optional.

Differences in rectangular coordinatesAs stated previously, a line cannot have coordinates. The initial and end points of a line have coordinates and the relativity of these is given by differences in coordinates.

‘’, the Greek letter delta, is a prefix commonly used to denote difference.

Thus N is read as the difference in the ‘northing’.

E is read as the difference in the ‘easting’.

N and E may also be referred to as partial coordinates (based on the line definition).

96


In Fig 6.12, let AB be any line, and 0 the origin of the coordinating system.

NB

EA

EB

(NA, EA)

(NB, EB)

NA

ΔEAB

ΔNAB

A

B

0

Fig 6.12

The relationship between the two points A and B is such that:

AB B AN N N

AB B AE E E

Difference can be determined using the following formula.

or

Difference end value initial value

AB B AX X X

where X is any quantity, A is the initial value and B is the end value

From this formula, given the initial value and a difference, we can also find the end value by rearranging this formula.

B A ABX X X where A indicates an initial value, B the end or fi nal value, and X represents any quantity.

Therefore, using the two points from Fig 6.12 we can see:

B A AB

B A AB

A B BA

A B BA

N N NE E EN N NE E N

97


Example 1 Given the coordinates of K and L, compute NKL and EKL.

L

K NK 4321.94EK 292.43

NL 5397.21EL 510.19

Fig 6.13

(end value minus initial value)

5397.21 4321.94

1075.27

510.19 292.43

217.76

KL L K

KL L K

N N N

E E E

Example 2 Given 596.215107.005

-AB

AB

NE

and the coordinates for B, as shown in Fig 6.14, compute the coordinates of A.

A

B NB 100.000EB 100.000

Fig 6.14

98


Because we don’t know NA we have to rearrange the formula for difference so that:

A B BAN N N

We are given the value for NAB , but the formula requires NBA , because we are working from the given data at B to the required data at A.

The direction of the line is very important.

If -596.215 then 596.215

Similarly, if +107.005 then -107.005

AB BA

AB BA

N N

E E

These coordinates are illustrated below.

−107.005

+107.005

+E

+N

−E

−NB

BA

A

+596.215

−596.215

Fig 6.15

We can now return to the formula and continue with the solution.

100.000 ( 596.215)

696.215

100.000 (-107.005)

- 7.005

A B BA

A B BA

N N N

E E E

Therefore, the coordinates of A are: 696.215-7.005

A

A

NE

99


Differences in coordinates as an alternate form of line definitionWe have now seen that two points can be defined relative to each other in two ways:

1. Given a bearing and a distance (or length).

A

L

B

LL

X

C

θ

θ

θ

Fig 6.16

Note: An extended survey traverse between A and X will result in several bearings and distances defining one point relative to another.

2. Given a series of end point line coordinates, with the differences in the coordinatesshowing their relationship with each other.

Points A and X, with coordinates given, are related by way of the series of N and Evalues, as shown below.

ΔE

ΔE

ΔN ΔNΔN

ΔEB

A

X

CNAEA

NXEX

Fig 6.17

100


Differences in coordinates from line definition (bearing and distance)As already discussed in Chapter 2, the definitions of principle trigonometric functions using the surveying convention are as follows:

sin cos tan E N EL L N

N E1

E

P1 (N1, E1)

P2 (N2, E2)

N2

E2

N1 1

2

L 1

L 2

0

180

90 27090 270

Fig 6.18

Or in the general case, where we consider any coordinate system, they can be:

sin cos tan

E N E

L L N

0 E1

E2

N2

N1

P1(N1, E1)

P2(N2, E2)

L 2

L 11

2

Fig 6.19

101


Thus, having defined a line by and L, we can calculate the differences in N and E (N, E) using a simple relationship. The relationship between these quantities is such that, for all cases:

cossin

AB AB AB

AB AB AB

N LE L

where A is the initial point and B is the end point of any line

By knowing, or assuming, coordinates for any point in the survey, we can then obtain thecoordinates for all points by computing the differences for all defined lines by using the equations:

and B A AB B A ABE E E N N N

Example 1 Compute the coordinates of point Y given the information below.

Y

X

490.321260°19' "28

NX = 790.428EX = 200.000

Fig 6.20

We will use the formula and Y X XY Y X XYN N N E E E

( cos )

790.428 (490.321 cos260 19' 28")

790.428 (-82.408)

708.020

( sin )

200.005 (490.321 sin 260 19' 28")

200.005 (-483.346)

-283.341

.

.

Y X XY

X XY XY

X XYY

X XY XY

N N N

N L

E E E

E L

the coordinates are: NY EY -

Note: The bearing must be shown in the correct direction (from X to Y) as given in the formula and the correct signs applied (adding negative values when appropriate).

102


Example 2 An example of a traverse is shown in Table 6.1, below. This form of calculation is in a special table called a close table.

Assume the starting coordinate for A is N = 0, E = 0.

Table 6.1

Line Bearing()

Distance(or length)

(L)

N E N E

A 0 0

AB 30 1000 866 500

B 866 500

BC 120 1000 -500 866

C 366 1366

CD 210 1000 -866 -500

D -500 866

DE 300 1000 500 -866

E 0 0

EF 210 1000 -866 -500

F -866 -500

FG 120 1000 -500 866

G -1336 -366

GH 210 1000 -866 -500

H -2232 -134

Note: Points A and E, having the same coordinates, must occupy the same position, ie bothpoints are coincidental.

Close tables like the one above will be used again in the next chapter.

103


Line definition from differences in coordinatesThe definition of a line between any two points in a survey defined by coordinates is an equally simple computation. Again, let A be the initial point and B the end point of any line to be defined, then:

AB B A

AB B A

N N N

E E E

By using the relationships shown in Fig 6.12, we find that LAB is the hypotenuse of a right-angled triangle formed from ABN and ABE .

2 2 AB AB ABL N E

and AB (the bearing of the line AB) is given by the tan function between ABE and ABN .

1tan

AB

ABAB

EN

There is ambiguity in this latter formula. AB can lie in one of two quadrants: in the first or third if the tangent is positive, and in the second or fourth if it is negative. This ambiguity is resolved by examining whether the differences in coordinates are positive or negative.

Tangent is ve If N or E is ve, then must lie in the first quadrant and not the third.

Tangent is -ve If N -ve or E is ve, then must lie in the second quadrant and not the fourth.

As a check on your calculated bearing you should always recalculate the tan value from your

bearing. You should get the same result as your value from EN

.

Further manipulation of the formulae yields other solutions.

To summarise:

From cos

sin

AB AB AB

AB AB AB

N L

E L

and

then 2 2

1

1 1

sin cos

tan

sin cos

AB ABAB AB AB

AB AB

ABAB

AB

AB AB

AB AB

E NL N E

EN

E NL L

104


Example 1 Find the (a) length and (b) bearing of line CF. Use the values for points C and F found in Table 6.1.

(a) Difference of line CF end point F – initial point C.

-866 366

-1232

CF F CN N N

-500 1366

-1866

CF F CE E E

2 2

2 2

+

-1232 -1866

2236

CF CF CFL N E

(b) 1

1

tan

-1866tan 1.514610-1232

56 33'57"

CFCF

CF

EN

+

+

−

−

56°3

4'

Fig 6.21

(Because both E and N are negative means that this bearing falls in the third quadrant.)

56 33'57" 180

236 33'57"

CF

Check this answer by finding the tan value of 23633' 57". You should get 1.514614.

105


Example 2 Given 106.132-110.296

A

B

NN

-11.063103.000

A

B

EE

Calculate, (a) ABL and (b) AB

Remember that AB signifies that B is the final point and A the initial point.

(a)

-110.296 106.132

-216.428

103.000 -11.063

+114.063

AB B A

AB B A

N N N

E E E

2 2

2 2216.428 114.063

244.646

AB AB ABL N E

-

(b) 1

1

tan

114.063tan 0.527025216.428

27 47 '25"

ABAB

AB

EN

--

-

The required bearing must fall in the second quadrant as E is ve and N is -ve.

AB = 180 27 47 '25"

= 152 12'35"

Check this answer by finding the tan value of 15212' 35". You should get -0.527023.

+

+

−

−

' "27°47 25

Fig 6.22

106


6.5 Calculators: use of rectangular → polar (R → P) function keyEfficiencies may be achieved in the conversion between rectangular and polar values through the use of the R → P function key on the calculator. Most calculators should have this function key.

Example 1 Rectangular → polar conversions

The bearing () and the length (L) are entered and, with the press of a key, the partial coordinates (N and E) are displayed in turn. Most calculators work on the mathematical notation for polar coordinates where r L and the bearing is measured anticlockwise from the x-axis, see the figure below.

x

y

X

Y

r

(r, θ)

θ

Fig 6.23

It is, therefore, vital to know your calculator and to check:

• which of the polar variables is entered first, ie r (L) or

• which of the partial coordinates is displayed first, ie x or y (N or E).

Note: 1. The alternative coordinate may be obtained by pressing the (x↔y) key. You should read your calculator manual to check these operations.

2. Calculators with multi-line displays will display both N and Evalues. These may also be shown in matrix form. You will needto know your calculator and understand whether N or E isdisplayed first.



107


Chapter 7 – Vectors

In surveying a line is defined by a length and a direction.

A

B

Fig 7.1

For example, the above line (AB) is defined by its length LAB and bearing AB .

This line can be considered as a ‘vector’, that is the locus of a point moving in rectilinearmotion (in a straight line) between A and B. The notation for the vector AB is AB.

The vector AB is said to be the ‘position vector’ of B with respect to A and is the direct path (shortest distance) between A and B.

There can, however, be an infinite number of paths between A and B.

In the three examples in Fig 7.2, the total motion is from A to B, even though different paths have been taken.

A A

A

B B

B

X

Z

PQ

(i) (ii)

(iii)

Fig 7.2

→

108


For each example, the sum of the component vectors is equal to the total vector between A and B.

That is:

(i) AX XB AB

(ii) AP PQ QB AB

(iii) AZ ZB AB

In each case, the total vector AB of a series of component vectors is indicated by the first letterof the first component vector (A) and the second letter of the final component vector (B).

For example, the total vector of the components

AB BC CD is AD

and XY YZ ZA XA.

7.1 N and E as the components of a vectorAs discussed in earlier notes, length and direction (vector) line definition does not lend itself to mathematical computation.

The vectors, as defined by L and , are broken down into component vectors as horizontal and vertical movements parallel to the E and N axes of the coordinate system.

For example, in the survey traverse ABCDX (Fig 7.3), the lengths and bearings of the traverse lines are measured quantities or derived from measured quantities.

+N

+E

A

B

CD

X

Fig 7.3

→ → →

→ → → →

→ → →

→

→ → → →

→ → → →

109


The total ‘resultant’ movement from A through B, C and D to X is the straight line (‘vector’) from A to X.

That is, AB BC CD DX AX ↑ ↑ first last component component

By computing the horizontal and vertical components of the vectors AB, BC, CD and DX, we can sum the horizontal and vertical movements (algebraically) to obtain the total horizontal and total vertical movements from A to X.

B

CD

X

A

Fig 7.4

Using Fig 7.4 as an example we can show the total horizontal and vertical movements through the following calculations:

DXAB BC CD AX

DXAB BC CD AX

N N N N N

E E E E E

The sum of the component vectors (NAX and EAX) can then be used to resolve the total vector, A to X.

The component vectors (Ns and Es) are computed by converting polar coordinates (L and ) to rectangular coordinates. The total vector AX is computed by converting the rectangular coordinates (NAX and EAX) to polar coordinates (L and .

These computations are carried out in a close tabulation, similar to the close tables used in the previous chapter.

→ → → → →

110


7.2 Vectors in surveying – summing vectors to prove a surveyThe principle of summing vectors is used to ‘prove’ a survey; that is, in a closed (loop) traverse, shown below, the sum of the vectors should total zero because the total motion is, in theory, zero. In this way any misclose can be computed.

B

A

C

D

Fig 7.5

Using this example we can see that

AB BC CD DA 0 (as the initial and end point are the same)

If you were to replace each vector by its bearing and distance and then compute the N and E components in a close table, the sum of the N and E values should (theoretically) be zero.

In a closed traverse between two points (say, X and Y, as shown in Fig 7.6), the positions of the initial and end points must be known relative to each other. That is, the vector XY must be known.

A

B

C

X

Y

Fig 7.6

XA AB BC CY XY

In this example, if you replace each vector by its bearing and distance and then compute the N and E components in a close table, the sum of the N and E values will be the N and E components of the vector XY.

→ → → →

→ → → → →

111


Example 1 Look at the closed (loop) traverse below and the tabled data. Note that the N and E values for each line are computed in the close table below.

A

B

C

D

E

Values are not to scale and are onlyintended to illustrate concepts.

3.5

3.04.0 m

2.7 m 4.0 m

3.9 m

3.2 m

263

60

310

162

112

3.0

2.0

-2.5

2.5

1.0

1.0

4.00.5

Fig 7.7

Line Bearing (θ) Distance (L) N E

AB 60 4.0 2.0 3.5

BC 112 2.7 - 1.0 2.5

CD 162 3.2 - 3.0 1.0

DE 263 4.0 - 0.5 - 4.0

EA 310 3.9 2.5 - 3.0

0 0

This example shows clearly that for a closed (loop) traverse:

AB BC CD DE EA 0

or

0

0

DEAB BC CD EA

DEAB BC CD EA

N N N N N

E E E E E

→ → → → →

112


Example 2 This is also a closed traverse but the start and end points are not the same.

A

B

C

D

E


3.5

3.04.0 m

2.7 m 4.0 m

3.2 m

263

60

162

112

3.0

2.0

-2.5

2.5

1.0

1.0

4.00.5

Fig 7.8

Line Bearing (θ) Distance (L) N E

AB 60 4.0 2.0 3.5

BC 112 2.7 - 1.0 2.5

CD 162 3.2 - 3.0 1.0

DE 263 4.0 - 0.5 - 4.0

-2.5 + 3.0

For this example

AB BC CD DA AE

or

-2.5

3.0

DEAB BC CD AE

DEAB BC CD AE

N N N N N

E E E E E

That is, the sum of the N and E values (-2.5, +3.0) form the N and E components of the line E. These values can be converted back to polar form for the bearing and distance of line AB.

The use of vectors and vector equations will be used again in another chapter.

You should therefore make sure you understand this chapter thoroughly before moving on.

→ → → → →

113




114


115


Chapter 8 – Adjustment of misclose in closed traverses

8.1 Traverse surveyingA traverse is a series of related points or stations connected together by angles and distances.

Types of traversesTraverses are classified as either closed or open. Look at the examples below.

A

D

B

C

E

A

B

C

D

a) b)

T1T1

T2T2

T3

T4

A

A

B

B

C

CD

Fig 8.1

a) A traverse closing back on to its starting point with the internal angles measured.

b) A traverse closing back on to its starting point with the external angles measured.

c) A closed traverse between two fixed coordinated points T2 and T3.

d) An open traverse from the coordinated point T2 (not closing onto another coordinated point).

Note: a) and b) are loop traverses.

c) is a point-to-point traverse.

c) d)

116


The closed traverseIf a traverse proceeds from one coordinated (fixed) point to another, it is known as a closed traverse. A closed traverse may either close back to its starting point or to any other coordinated point. It is therefore able to be checked and adjusted to fit accurately between these known points. Look at examples a) to c) in Fig 8.1.

The open traverseAn open traverse does not close on to a known point. Looking at example d) in Fig 8.1 we can see that the end of the traverse, point D, is left ‘swinging’ with no accurate means of checking angular or linear errors that may have occurred between T2 and D. The only check would be to repeat the whole traverse or resurvey in the opposite direction.

8.2 Traverse closureIt has previously been shown that in a closed traverse the sum of the vectors should equal zero. As each vector consists of a bearing () and a distance, the following examples outline the equations and conditions of closure for both angular and linear values.

Example 1 The closed traverse

θ

θANAEA

BNBEB

Fig 8.2

For this type of traverse the following is true:

angular misclose 0

0

0

A B

A B

N N N

E E E

Where: A and B are the initial and finalstations of the traverse.

∑∆ N and ∑∆ E are the sums of the differences in coordinate values for the distances between A and B.

117


Example 2 The loop traverse

The general equations for these traverses are somewhat simplified and the sum of the angles should total to a predetermined mathematical sum.

A

Fig 8.3

0

0

N

E

Where: A is both the initial and final station of the traverse.

∑ ∆N and ∑ ∆E are the sums of the differences in coordinate values for the distances measured from A in a clockwise direction.

8.3 MiscloseIf all the component angles and distances of a closed traverse are measured, the physical limitations of the observations lead to the situation where they do not satisfy the conditions of closure for the equations.

The reason the observed values do not usually satisfy these equations is because of the small random ‘accidental errors’ to which any physical observation is subject.

It is usual practice to ‘close’ a survey, that is to measure all angles and distances. The purpose of this is to ensure that no ‘gross’ errors have occurred. As random errors are small, the errors (or miscloses) in the equations will also be small. If these miscloses are greater than those expected from random errors, it means that a gross error has occurred in the measurements.

Before further computation is done a recheck of the calculations must be carried out to find the gross error. If it is not found a resurvey needs to be done.

Assuming that the observations are within expected limits, then an adjustment or correction to the results has to be made to satisfy the equations for closure. Again, this needs to be done before further computation.

118


There are many ways of adjusting the results. One way would simply be to discard the unwanted observations; however this often leads to complications.

By looking at the figures below and assuming all angles and distances have been measured with equal care, we can see that it is difficult to decide which values should be discarded.

Fig 8.4 Fig 8.5

A

B

Fig 8.6

It is normal practice to use all the observations as they are equally reliable, and adjust the values so that small corrections are applied to each observation. Even though several methods are available, the one used for the majority of surveying work firstly adjusts the angular (bearing) misclose, then the differences in the northing and easting values are computed and finally, the linear misclose is adjusted.

The equations used to describe the conditions of closure for angular and linear values are useful in demonstrating the linkage of vector principles to traverse closure. More practical methods, however, will now be used to calculate the actual miscloses.

119


8.4 Angular misclose

For a closed polygon

A

B

C

D

EF

Fig 8.7

In the above traverse the lines close to form a polygon. The traverse is virtually self-checking as the sum of the interior angles of any polygon conforms to the following formula.

int. angles (2n 90 (where n is the number of stations) or

int. angles (n 180

We can use Fig 8.7 as an example to show the computation, using the above formula.

int. angles (2n 490

26490

12490

720

If exterior angles are used the formula changes to become:

ext. angles (2n 90 (where n is the number of stations)

Due to small random errors inherent in all angular measurements, it is unlikely that the angles will sum exactly to this amount. They should be close to it, however, and certainly within the following limits of closure:

• for urban surveys, 20" the square root of the number of angles, or 1' 30", whichever is the lesser.

• for rural surveys, 30" the square root of the number of angles or 2', whichever is the lesser.

120


Between coordinated points

θA θBA1

A2

A3

A4 A5A B

Fig 8.8

In the closed traverse above, the survey begins from A , a known bearing and ends at B , also a known bearing. The survey is self-checking in that B , computed from the traverse angles, should agree with the known B . The difference between the two values is the traverse angular (bearing) misclose.

In other words, to compute the misclose in angles (mA), we use the initial bearing A and the observed angles to compute the final bearing B .

A (calculated) (known) B Bm

From the initial bearing, we apply observed angles to compute the bearings of the lines in the traverse through to the final (known) line.

The angular (bearing) misclose can then be computed using the above formula.

The limits of closure formulae or other regulated limits for a closed polygon are used to assess the acceptability of the misclose.

121


Angular adjustmentIf all the angles have been observed with equal care (they are equally reliable), each is equally likely to contain random errors of the same magnitude. Thus, all angles should receive an equal correction or adjustment. Under these circumstances, the correction (cA) to an angle may be found from the following formula:

A -

Amck

Where:

mA is the angular misclose

k is the number of angles

The correction is applied algebraically to the observed angles.

Example 1 Calculate the angular misclose of ABCDE and the adjusted bearings of the lines, assuming that the bearing from C to D is 18000' 00".

A

B

C

DE

91 08 14' "

9210

32'"

131 26 52' "

88 0419'"137 10 56'

"

180

000'

"

Fig 8.9

(a) We firstly calculate the angular misclose.

The internal angles of a closed loop traverse should equal to (2 4) 90º n ,where n is the number of stations.

In Figure 8.9 the sum of the five angles in the traverse should equal:

int. angles (2n 90

(10 90

540

We can check this calculation by summing all the internal angles in thistraverse.

122


A 91 08'14"

B 92 10'32"

C 131 26'52"

D 88 04'19"

E 137 10'56"

540 00'53"

There is, therefore, an excess of 53" in the final amount; that is, the angular misclose is 53".

If the misclose is distributed evenly between the five angles then the correction per angle is 10.6 " (53 5 = 10.6" ).

As the given angles are to the nearest second, this converts to:

-10" for two angles

-11" for three angles.

We assume that all the angles are measured with the same accuracy, underthe same conditions and are all equally liable to error.

Corrections of -10" and -11" are therefore made to alternate angles.

(b) Now we have calculated the angular misclose we can apply the corrections to each angle.

A 91 08'14" 11" 91 08' 03"

B 92 10'32" 10" 92 10' 22"

C 131 26'52" 11" 131 26' 41"

D 88 04'19" 10" 88 04' 09"

E 137 10'56" 11" 137 10' 45"

-53" 540 00' 00"

check

One second of arc is 0.0005 m in 100 m so it does not matter if the adjustment is

-11, -10, -11, -10, -11

or -11, -11, -11, -10, -10

or another variation.

However, it is generally better to even out the distribution of the adjustment, in this case, -11, -10, -11, -10, -11.

123


(c) The adjusted bearings can now be computed using these adjusted angles. You may need to refer back to Chapter 5 to revise this type of calculation. Start from the known bearing of CD.

Work clockwise around the traverse from the starting bearing ofCD 180' 00" and add the angles to calculate the bearings of the lines.

180 00'00"

360 00'00"

88 04'09"

271 55'51"

91 55'51" 451 55'51"

137 10'45"

314 45'06"

134 45'06"

91 08'03"

43 37 '03"

223 37 '03"

92 10'22"

131 26'41"

CD

DC

DE

ED

EA

AE

AB

BA

BC

C

EDC

or

ED

BAE

CBA

311 26'41"

131 26'41"180 00'00"

B

CD

DCB

As your calculated bearing of CD is the same as your starting bearing of CD your calculations must be correct. If they are not the same you have obviously made an error in your calculations, so you will need to redo the work to correct the error.

It would also have been possible to go anticlockwise around the traverse, in which case, the angles would all be positive rather than negative. Details of this calculation are shown below.

check

124


180 00' 00"

131 26' 41"

311 26' 41"

131 26' 41"

92 10' 22"

223 37 ' 03"

43 37 ' 03"

91 08' 03"

134 45' 06"

314 45' 06"

137 10' 45"

451 55' 51"

91 55' 51"271 55' 51"

CD

CB

BC

BA

AB

AE

EA

ED

DE

DCB

CBN

BAE

AED

ED 88 04' 09" C360 00' 00" DC

The previous methods calculate the angular misclose and adjust this misclose before computing the bearings. However, it is also possible to use the original (unadjusted) angles to compute the bearings and obtain a bearing misclose. This misclose is then adjusted proportionately directly to the bearings.

If corrected angles are required, they are computed as the angles between the adjusted bearings.

check

125


Alternative method by bearings

Unadjusted bearings Correction Adjusted bearings

The difference between the known bearing of CD and the calculated bearing of CD is -53", so the traverse error is -53".

Note: Using unadjusted bearings (based on algebraically negative angles) results in a bearing misclose of -53". This should not be seen to conflict in any way with the previously calculated angular misclose of 53". Had the bearings been applied in the opposite direction (anticlockwise) then the error (misclose) would have been 53".

The full 53" correction is applied to the final line CD, a correction of 11" less to the next line, 10" to the next, and so on.

Adjusted angles can be computed (if required) from these adjusted bearings. These will be exactly the same as computed in the previous method.

180 00'00" 0 180 00'00"

360 00'00"

88 04'19"

271 55'41" 11" 271 55'51"

451 55'41"

137 10'56"

314 44'45" 21" 314 45'06"

134 44'45"

91 08'14"

43 36'31" 32" 430 37 '03"

223 36'31"

CD

DC

DE

ED

EA

AE

AB

BA

EDC

ED

BAE

92 10'32"

131 25'59" 42" 131 26'41"

311 25'59"

131 26'52"

calculated 179 59'07" 53" 180 00'00"

known 180 00'00"

BC

CB

CD

CD

CBA

DCB

126


Example 2 The traverse ABCD has been run between the known points A and D. DYθ and AXθ are known. Compute the misclose in angles between X and Y and the adjusted bearings between A and D.

A

B

C

D

Y

X

227°50' 11

"

115°30'00"130°

20'10

"114°5

0'25"

64°0

0'40

"147°10'30"

Fig 8.10

The angular (bearing) misclose is obtained by comparing the given bearings usingunadjustedangles.Eachbearingisthenadjustedbyanamountwhichreduces the traverse error between successive lines.

Unadjusted bearings Correction Adjusted bearings

115 30'00" 0 115 30'00"130 20'10"

245 50'10" 4 245 50'14"425 50'10"

114 50'25"

310 59'45" 8 310 59'53"130 59'45"64 00'40"

195 00'25" 12 195 00'37"375 00'25"

147 10'30"

AX

AB

BA

BC

CB

CD

DC

DY

XAB

BCD

CBA

YDC

θ ° °

∠ + °

θ ° + °

θ °∠ − °

θ ° + °

θ °

∠ °

θ ° + °

θ °

∠ − °

θ 227 46'55" calculated 16 227 50'11"227 50'11" given

DY

° + °

θ °

127


The calculated bearing is less than the given bearing by 16", therefore, the traverse (bearing) error is -16".

The error is then adjusted to each bearing from DY to AB . There are four bearings so each has to be adjusted equally: 16" 4 4". Therefore, each bearing is adjusted by 4" so the adjustments are 16", 12", 8" and 4" respectively.

If adjusted angles are required, then these are obtained by subtracting the adjoining adjusted bearings of the traverse. Generally, only the seconds column needs to be considered in this mental calculation. The adjusted angles are:

13020'14" 11450'21" 6400'44" 14710'26"

8.5 Linear misclose and adjustmentThe total linear misclose is derived from the misclose in the northing and easting values. Let these be represented by Lm , Nm and Em respectively. Then:

2 2 (from Pythagoras' theorem)

N A B

E A B

L N E

m N N N

m E E E

m m m

The limits of acceptable misclose are specified for various standards of work. It is inadequate to state the magnitude of the misclose because the longer the traverse, the greater the accumulation of random error and the acceptable misclose. The usual method of stating misclose is as a proportional misclose, or rate of error.

• The linear misclose in an urban survey should not be less than 1:12 000.

• The linear misclose for a rural survey should not be less than 1:6 000.

The proportional misclose (or rate of error) can be computed using the following formula.

1: RL

Lmm

Where:

the rate of error

the sum of the distances in the traverseRm

L

128


Example 1 The traverse ABCD has been measured between the known points A and D (Fig 8.11).

A

B

C

D

Y

X

22750' 11

"

115 30'00"130

20'10

"114 50

'25"

6400

' 40"

147 10'30"

345.

76 m

183.85 m

422.60 m

NAEA

NDED

Fig 8.11

From a close table computation, the misclose at D is 0.06 in the northing and -0.09 in the easting. Refer back to Chapter 6 for a review of close tables. We canshow the misclose for this below.

C

Y

D

D'

mE -0.09

mN 0.06mL 0.11 m

Fig 8.12

129


Example 2 Angles and distances have been measured and are shown in Fig 8.13.

A B

C

D

366.230 m

569.820 m

287.270 m

512.0

60 m36 20' "00

7049'

"55

4100 '

"30

14810'

"50

Fig 8.13

From a close table calculation, the misclose at D is -0.50 in the northing and 0.20 in the easting, then we can show the misclose for this below.

A C

D

D'

mL

m

mE

mN -

Fig 8.14

In each case, differences in coordinates have been computed using adjusted bearings, but due to the small random errors in measurement, the computed position at D(D') does not coincide with the known position D. DD' represents the total linear misclose with its components mN and mE.

The linear misclose is adjusted by correcting the differences in coordinates by the Bowditch adjustment method. The Bowditch adjustment is generally used because it has a sound theoretical basis.

Note: 1. You may see the Bowditch adjustment method called the ‘Compass’ method. This is the name used in the USA and in some software packages.

2. Some countries prefer to use the ‘Transit’ adjustment method but this is not used in Australia.

130


Bowditch adjustment methodThere are two ways in which the Bowditch adjustment method can be applied, either:

• the accumulated length method

• the partial coordinates method.

Both of these methods are based on the theory that any linear misclose is adjusted proportionally to the length of the traverse.

The accumulated length method could be considered more accurate for manual calculations as rounding errors are not carried forward through the calculation. However, it does not have a built-in closing check so you must always double check your work.

The partial coordinates method does have a built-in closing check but you need to take care that rounding errors are not carried forward.

Accumulated length methodWith this method the adjustment of the misclose is performed on the calculated coordinates of the points in the traverse.

The formulae are:

accum

N

Nm Lc

Laccum

EE

m LcL

Where:

accum

= correction to the northings= misclose in the northings= accumulated length

= total length of traverse

N

N

cm

LL

Where:

accum

= correction to the eastings= misclose in the eastings= accumulated length

= total length of traverse

E

E

cm

L

L

That is, the corrections applied to the coordinates of a point are proportional to the accumulated length of the traverse to that point.

The adjusted coordinates are found by adding the corrections to the computed coordinates. It is important to strictly apply the signs of the corrections algebraically.

131


Example 1 Look at the traverse below and the following data. Compute adjusted coordinates for A, B and C.

Given: = 1000.000= 1000.000

D

D

NE

366.230 m

569.820 m28

7.270

m

512.

060

m

53 39'"

54

270 00' 0"0

160

50'

"02

2150

'"

38

B

C

A D

Fig 8.15

Use a close table to calculate unadjusted coordinates.

Line Bearing()

Distance(L)

Accum. distance N E

Unadjusted coordinates

N E

D 1000.000 1000.000

DA 27000'00" 569.820 0.00 -569.820

A 569.820 1000.000 430.180

AB 5339'54" 366.230 216.993 295.023

B 936.050 1216.993 725.203

BC 2150'38" 287.270 266.644 106.887

C 1223.320 1483.637 832.090

CD 16050'02" 512.060 -483.677 168.113

D 1735.380 999.960 1000.203

1735.380

132


Looking at the table we can see that the misclose for ∆ N is - 0.040 and for ∆ E is 0.203.

(a) Using the data we can calculate the total corrections for both the northings and eastings.

the correction for ∆ N is known (given) N calculated N

1000.00

0.040

and the correction for ∆ E is known (given) E calculated E

1000.00

- 0.203

Note: The correction will always be the same magnitude but of opposite sign to the misclose.

(b) Now we can calculate the corrections in the northings for each point of the traverse.

Point A =

0.040 569.820=1735.380

= 0.013

N DAA

c LcNL

Point B (D B)

=

( )=

0.040 (569.820 366.230)=1735.380

0.040 (936.050)=1735.380

= 0.022

NB

N DA AB

c LcNL

c L LL

133


Point C (D C)

=

( )=

0.040 (569.820 366.230 287.270)=1735.380

0.040 (1223.320)=1735.380

= 0.028

NC

N DA AB BC

c LcNL

c L L LL

Point D =

0.040 (1735.380)=1735.380

= 0.040

ND

c LcNL

(c) The eastings are calculated in the same way.

Point A =

-0.203 569.82=1735.380

= -0.067

E DAA

c LcEL

Point B (D B)=

-0.203 (569.820 366.23)=1735.380

-0.203 (936.05)=1735.380

= -0.109

EB

c LcEL

134


Point C (D C)=

-0.203 (569.820 366.230 287.270=1735.380

-0.203 1223.320=1735.380

= -0.143

EC

c LcEL

Point D =

-0.203(1735.380)=1735.380

= -0.203

ED

c LcEL

A summary of the calculations we have done can be added to the table to show final adjusted coordinates.

Line Bearing () Distance (L) Accum.

distance N EUnadjusted coordinates Corrections Adjusted

coordinates

N E cN cE N E

D 1000.000 1000.000 0 0 1000.000 1000.000

DA 270º00'00" 569.820 0.00 -569.820

A 569.820 1000.000 430.180 +0.013 -0.067 1000.013 430.113

AB 53º39'54" 366.230 216.993 295.023

B 936.050 1216.993 725.203 +0.022 -0.109 1217.015 725.094

BC 21º50'38" 287.270 266.644 106.887

C 1223.320 1483.637 832.090 +0.028 -0.143 1483.665 831.947

CD 160º50'02" 512.060 -483.677 168.113

D 1735.380 999.960 1000.203 +0.040 -0.203 1000.000 1000.000

== 1735.380

135


Partial coordinates methodWith this method the corrections are proportional to the length of the line.

The formulae are:

Nm Lc N

L

Em Lc E

L

The adjusted differences in coordinates are found by adding the corrections to the computed values. It is important to strictly apply the signs when doing the computations. The adjusted differences are then used to compute coordinates or areas. Care must be taken with manual calculations so that rounding off errors are not carried forward.

Example 1 A closed polygon

Look at the traverse below and the following data.

Given: ND ED

366.230 m

569.820 m

287.2

70 m

512.

060

m

53 39'"

54

270 00' 0"0

160

50'

"02

2150

'"

38

B

C

A D

Fig 8.16

Line Bearing (θ) Distance (L) ∆ N ∆ E

DA 27000'00" 569.820 0.0000 - 569.820

AB 5339'54" 366.230 216.9932 295.0226

BC 2150'38" 287.270 266.6443 106.8871

CD 16050'02" 512.060 - 483.6769 168.1134

1735.380 - 0.0394 + 0.2031

Note: N, E are written to four decimal places (4 dp) to minimise rounding off errors.

136


(a) Using the data we can calculate the linear misclose and rate of error. (For the purposes of the Bowditch adjustment calculation, the nearest whole number for ∑L and L will suffice.)

2 2N E

2 2

Linear misclose ( ) =

= (-0.0394) (0.2031)

= 0.2069

1:Rate of error ( ) =

1735.38= 1:0.2069

= 1:8387

L

RL

m m m

Lmm

+

+

∑

Note: For this calculation it is best to use the computed (stored) value of mL in the calculator.

(b) Nowwecancalculatetheadjustmentsforthenorthings.

0.039(569) 0.01291735.38

0.039(366) 0.00831735.38

0.039(287) 0.00651735.38

0.039(512) 0.01161735.38

0.0393

N

DA

AB

BC

CD

m Lc NL

c N

c N

c N

c N

×∆ =

∑− ∆ = − =

− ∆ = − =

− ∆ = − =

− ∆ = − =

∑ = Check

137


(c) The eastings adjustments are also calculated.

0.203(569) 0.06661735.38

0.203(366) 0.04291735.38

0.203(287) 0.03361735.38

0.203(512) 0.06001735.38

0.2031

E

E DA

E AB

E BC

E CD

mE LcL

c

c

c

c

×∆ =

∑+ ∆ = − = −

+ ∆ = − = −

+ ∆ = − = −

+ ∆ = − = −

∑ = − Check

(d) The above corrections are then added to the ∆N and ∆E values, and the adjusted coordinates can now be calculated, as in the following table.

Line Bearing (θ)

Distance (L)

∆ N ∆ E Corrections Adjusted coordinates

cN cE N ED 1000.000 1000.000DA 270º00'00" 569.820 0.00 -569.820 0.0129 -0.0666A 1000.013 430.113AB 53º39'54" 366.230 216.9932 +295.0226 0.0083 -0.0429B 1217.014 725.093BC 21º50'38" 287.270 266.6443 +106.8871 0.0065 -0.0336C 1483.665 831.947CD 160º50'02" 512.060 -483.6769 +168.1134 0.0116 -0.0600D 1000.000 1000.000

∑ = 1735.38 ∑ = -0.0394 ∑ = 0.2031 ∑ = 0.0393 ∑ = -0.2031

Note: 1. The adjusted coordinates are calculated from coordinate + ∆N + cN , eg NA = ND + ∆NDA + cDA = 1000.000 + 0 + 0.0129 = 1000.013 coordinate + ∆E + cE , eg EA = ED + ∆NDA + cDA = 1000.000 – 569.820 – 0.0665 = 430.113

2. ∆N , ∆E , cN , cE values are all written to 4 dp to minimise rounding off errors. The final adjusted coordinates are rounded back to 3 dp.

138


Example 2 Linear adjustment between coordinated points

Look at the following open traverse and the accompanying data.

A

B

C

D

Y

X

22750' 11

"

115 30'00"

130

20' 10

"114

50' 2

5"

6400

' 40"

147 10 '30"

345.

760

m422.600 m

183.850 m

ND = 485.260ED = 9 446.290

NA = 871.650EA = 10 060.080

Fig 8.17

The angular misclose has already been computed (see page 126) and adjusted bearings found for the traverse lines between A and D. These are tabulated below and the N and E for each line computed.

Line Bearing () Distance (L) N E

AB 24550'14" 422.600 - 172.983 - 385.574

BC 31059'53" 183.850 + 120.612 - 138.757

CD 19500'37" 345.760 - 333.962 - 89.549

= 950.000 = - 386.333 = - 613.880

(a) The difference in the coordinates from A to D are computed.

485.260 871.650

-386.390

9 446.290 10 060.080

-613.790

D A

D A

N N

E E

139


(b) We now use the formula looked at on page 116 in N and E:

00

A D

A D

N N NE E E

to calculate the misclose in N and E.

871.650 (-386.333) 485.260

0.057

N A Dm N N N

10 060.008 (-613.880) 9446.290

-0.090

E A Dm E E E

(c) We can use the misclose values for N and E to calculate the adjusted coordinates for B and C by following two steps.

Step 1. Adjust the N and E values according to the Bowditch adjustment method. For this example we will use the method of partial coordinates.

Step 2. Apply the adjusted N and E values to the given coordinates at A to obtain adjusted values at B and C.

Step 1. -

-(0.057)(422) -0.025950

-(0.057)(183) -0.011950

-(0.057)(345) -0.021950

N

AB

BC

CD

m Lc NL

c N

c N

c N

0.057 check

140


-

-0.090- (422) 0.040950

-0.090- (183) 0.017950

-0.090- (345) 0.033950

E

AB

BC

CD

m Lc EL

c E

c E

c E

0.090 check

Step 2.

The above corrections are then applied to the N and E values to produce the adjusted values, shown in the table below. Coordinates for B and C are calculated with a check at point D.

Point line

Bearing()

Distance(L)

AdjustedN

AdjustedE N E

A 871.650 10 006.080

AB 245 50' 14" 422.600 -173.008 -385.534

B 698.642 9 674.546

BC 310 59' 53" 183.850 120.601 -138.740

C 819.243 9 535.806

CD 195 00' 37" 345.760 -333.983 -89.516

D 485.260 9 446.290

-386.390 -613.790

The sum of the ∆ N and ∆ E values are now correct, as are the coordinates for D.

Therefore, the adjusted coordinates for B and C are:

NB = 698.642 EB = 9674.546

NC = 819.243 EC = 9535.806

141




142


143


Chapter 9 – Areas of polygons

9.1 Areas of regular rectilinear figuresRectilinear figures are bounded by straight lines and conform to specific geometrical shapes, eg triangle, square.

The square and the rectangleFor these figures the formula is:

Area = L B

A

BB C

DD

C

A

Area = AB BC Area = AB BC

Fig 9.1

The triangleFor these figures the formula is:

Area 12

base perpendicular height

A

B B

CDC A

Area 12

AC BA Area 12

AC BD

Fig 9.2

144


For a scalene triangle the formula is:

Area 12

ab sin C

A

B

C

a

b

c

Fig 9.3

1Area sin2

ab C=

21 sin sinArea2 sin

a B CA´=

( )( )( )Area s s a s b s c= - - -

( )1where2

s a b c+ +=

The parallelogramThe formula is:

Area a b sin A

A

B C

D

a

b

o

1Area sin2

AC BD BOC´ ´=

Fig 9.4

145


The trapeziumA trapezium has two parallel sides and the angles and the sides may all be different.

The formula is:

Area 12

∑ parallel sides perpendicular height

A

B C

D

h

Area2

AD BC h+´=

Fig 9.5

For the figure above, 180

180

A B

C D

+ =

+ =

and

However, if EF is parallel to AD and BC and midway between them (as shown in the figure below)

A

B C

D

E F

h2

h2

Area EF h´=

Fig 9.6

then 2

AD BCEF +=

146


9.2 Areas of polygonsThese are figures that have straight sides but do not belong to any of the previous categories. To compute the areas of these figures it may be necessary to:

a) divide the figure into a number of triangles

b) divide the figure up into a number of trapeziums.

The first method is cumbersome to use when calculating area. However, by means of constructions and scaled-off distances, an approximation of area can be found.

The second method is the best to use when computing area for the most complicated rectilinear figures. The lengths and bearings of the sides must be known or be calculated as area is computed using these coordinates.

9.3 Calculating area using coordinatesOnce the coordinates of the points in a closed figure have been computed, they may be used to calculate the area.

In Fig 9.7 the coordinates of the four corners of the polygon ABCD have been determined. We will use this information to compute the area.

EB B(NBEB )

(NAEA )

D(NDED )

C(NCEC )

ND

ED

EA

EC

NA

A

NC

NB

Fig 9.7

147


Example 1 The area of the polygon ABCD is found by dividing the figure into four trapeziums (shown in Fig 9.8), calculating each of these areas and then summing two and subtracting two of the areas.

A

B

C

D3

4

a b cd

A

B

C

D

1

2

a b cd

Fig 9.8

Area (ABCD) Area (1) Area (2) Area (3) Area (4)

Area (aABb) Area (bBCc) Area (aADd) Area (dDCc)

To compute the trapeziums we use the formula:

Area = 12parallel sides perpendicular height

or Area (aABb) = ( ) ( )

2A B

B AN N E E+

´ -

or 2Area (aABb) (NA NB) (EB EA) (NA NB)(EB EA)

This rearrangement of the formula can then be applied to the remaining trapeziumsso that:

2Area (bBCc) (NB NC) (EC EB) (NB NC)(EC EB)

2Area (aADd) (NA ND) (ED EA) (NA ND)(ED EA)

2Area (dDCc) (ND NC) (EC ED) (ND NC)(EC ED)

Therefore,

2Area (ABCD) (NA NB)(EB EA) + (NB NC)(EC EB)

– (NA ND)(ED EA) – (ND NC)(EC ED)

Expanding this by multiplying and summing we get:

2Area (ABCD) NAEB NBEA NBEC NCEB – NAED NDEA NDEC NCED

148


Regrouping we get:

2Area (ABCD) (NAEB NBEC NCED NDEA NBEA NCEB NDEC NAED)

When this form is employed, computations can be made conveniently from the coordinates appearing in the close tabulation that follows.

NAEA

EB

EC

ED

EA

NB

NC

ND

NA

–

–

–

–

+

+

+

+ (repeated)

Fig 9.9

Then 2(Area) is the algebraic sum of the products NAEB , NBEC , etc, and the products -NBEA , -NCEB , etc.

TaskFind the area of the polygon ABCD (Fig 9.10), given the coordinates of A and the line information shown.

N

0 E(0,0)

21' 3

9"

127

m

49"'

323 45"'

196

34"

'

288.2

30 m 278.160 m

352.030 m

NA = 450.650

EA = 249.630

A

B

C

D

Fig 9.10

149




150


AnswersTask (page 148)Find the area of the polygon ABCD (Fig 9.10), given the coordinates and the tabled data.

N

0 E(0,0)

21' 3

9"

127

49"'

323 45"'

196

34"

'

288.2

30

278.160

352.030

NA = 450.650

EA = 249.630

NB = 719.626

EB = 353.208

NC = 548.385

EC = 572.410

ND = 168.415

ED = 460.030

A

B

C

D

Line Bearing()

Distance(L)

N E N E

A 450.650 249.630 -179 640.24 m2

AB 2103' 39" 288.230 268.976 103.578 -193 693.97 m2

B 719.626 353.208 -96 402.43 m2

BC 12759' 49" 278.160 -171.241 219.202 -207 312.06 m2

C 548.385 572.410

CD 19628' 34" 396.240 -379.970 -112.380 159 173.18 m2

D 168.415 460.030 411 921.12 m2

DA 32317' 45" 352.030 283.234 -210.403 252 273.55 m2

A 450.649 249.627 42 040.93 m2

= 0.00 = 0.00 = 188 360.08 m2

2Area 188 360.08 m2

Area m2

ha

If the coordinates had not been given for A, B, C and D, then they would have needed to be calculated from the N and E values using assumed coordinates for A; say NA 1000 and EA 1000.

151


Chapter 10 – Missing measurements

10.1 Summing vectors to prove a surveyIn Chapter 7 we looked at closed polygon traverses and saw that the sum of the vectors should equal zero because the total motion was, in theory, zero. The simple example we examined is repeated here.

The sum of the component vectors (∆N and ∆E values) given for each line adds up to zero; ie the addition of the ∆N and ∆E columns give a result of zero.

A

B

C

D

E


3.5

3.04.0 m

2.7 m 4.0 m

3.9 m

3.2 m

263

60

310

162

112

3.0

2.0

-2.5

2.5

1.0

1.0

4.00.5

Fig 10.1

Line Bearing ( Distance (L) N E

AB 60 4.0 2.0 3.5

BC 112 2.7 - 1.0 2.5

CD 162 3.2 - 3.0 1.0

DE 263 4.0 - 0.5 - 4.0

EA 310 3.9 + 2.5 - 3.0

= = 0 = 0

152


10.2 The sum, difference and translation of vectorsA vector is a line segment which has a definite direction sense and which can be arbitrarily displaced, parallel to itself, thus achieving a translation of vectors from one position to another that may form a triangle built on the closing vector.

Look at the two vectors AX and XB, below. We can construct the sum of two vectors by moving each vector parallel to itself until an intersection is made (at X).

A

B

X

XAX + XB

A

B

X

Fig 10.2 Fig 10.3

The line joining A and B represents the vector AX + XB. That is:

→AX →XB →AB thesum or resultant vector)

The difference of the vectors AX and XB is obtained by adding (- XB) to AX.

A

B

AX − XB -XB

+XB

X

Fig 10.4

153


10.3 Closing vectorsA closing vector in a closed loop traverse is one which will cause the sum of all vectors in the traverse to total zero.

For example, in Fig 10.5:

→AX →XB →BA 0.

Therefore, →BA closing vector

A

B

X

Fig 10.5

For example, in Fig 10.6:

→AB →BC →CD →DE →EA 0.

The vector EA is undefined so,

→AB →BC →CD →DE -→ EA

→AE = ‘sum vector’

and →EA ‘closing vector’

Therefore, LEA LAE and EA AE 80

A

B

C

D

E

Fig 10.6

10.4 Partially defined vectorsPartially defined vectors are those which have a definite length or direction but not both. In the figure below, AB is a fully defined vector. BC and CD are partially defined.

A

B

C

D

length missingdirection missin

g

70°00' 125°00' 175.00 m

120.00 m

Fig 10.7

154


10.5 Missing vectorsTaking the example in Fig 10.1 one step further, it will now be shown that when certain elements, ie a vector or partially defined vectors, are missing in the traverse, then it is possible to solve for these elements.

Example 1 Look at the table data below and compute the length and bearing of the missing side, CD.

Line Bearing (θ Distance (L) N E

AB 60 4.0 2.0 3.5

BC 112 2.7 - 1.0 2.5

CD

DE 263 4.0 - 0.5 - 4.0

EA 310 3.9 2.5 - 3.0

= = 3.0 = = - 1.0

(a) To calculate the bearing of the missing side.

We must remember that the sum of all the vectors equals zero.

→AB →BC →CD →DE →EA 0

→AB →BC →DE →EA -→CD

That is N -NCD

E -ECD

or -N -NCD

-E -ECD

That is -3.0 = NCD 1.0 = ECD

These values are then used in the formula 1= tan D-

D

æ ö÷çq ÷ç ÷çè øEN

to calculatethe bearing.

CD

+1 --

1.0tan 183.0

æ ö÷- ç ÷ç ÷ç ÷ç ÷è ø= =

180 18 (as the bearing must lie in the second quadrant)

162

= -

=

155


(b) To calculate the length of the missing side.

LCD = 2 2CD CDN ED + D

LCD = ( ) ( )2 23.0 1.0- + = 3.2

therefore, the solution is 162

L 3.2

=

=

CD

CD

We can verify these results by comparing them with the values in Fig 10.1.

To summarise, in the case of a traverse where the length and bearing of one line is missing, the sum of the N and E values gives the rectangular definition of the closing vector (line) from which the bearing and length for that line may be computed.

In the case of a traverse with two lines partially defined, by summing all the known N and E values, the result is the sum of the N and E values of the partially defined lines. Once again, these sums give the rectangular definition of a closing vector. The closing vector is the sum of the two partially defined vectors and these three vectors define a triangle which may be solved for the missing measurements.

Example 2 Consider the same traverse ABCDE with the bearing of BC missing and the length of CD missing.

Line Bearing () Distance (L) ∆N ∆E

AB 60º 4.0 2.0 3.5

BC 2.7

CD 162º

DE 263º 4.0 - 0.5 - 4.0

EA 310º 3.9 + 2.5 - 3.0

= = 4.0 = - 3.5

The sum of the vectors must total zero.


→AB →DE →EA - BC + CD)

or - N →BC →CD

and - E →BC →CD

That is, - 4.0 →BC →CD

or

- 4.0 NBC NCD

and 3.5 →BC →CD 3.5 EBC ECD

156


These values (- 4.0 and 3.5) are termed the closing vector, and they may be graphically shown in either rectangular or polar form as below.

B

D

-4

3.5

5.3 m139

( ) ( )

1

2 2

3.5tan4.0

139

4.0 3.5

5.3 m

L

-

-

+

+


=

=

=

Fig 10.8

The vector BD is the sum of the vectors BC + CD and these are brought together in one triangle BCD (as shown in Fig 10.9).

B

D

C

-4

5.3 m

2.7 m

162

BC CD

Fig 10.9

157


This triangle is now solved for the LCD ( 3.2 m) and BC ( 112).

To calculate angle C, use the sine rule.

sin sin=

BC BDBDC DCB

( )

1

2.7 5.3sin 23 sin

5.3sin 23sin2.7

50 or 130 (obtuse angle adopted as per Fig 10.1)

180 130 23

27

139 27 112

-

=

æ ö÷ç = ÷ç ÷çè ø

=

= - +

=

= - = BC

C

C

B

Use the sine rule again to calculate CD.

sin sin=

CD BDCBD DCB

5.3sin 27 sin130

5.3sin 27sin130

3.2 m

=

=

=

CD

CD

The results in this example have been rounded off for the purposes of the computation. We can verify the results by comparing them with the values in Fig 10.1.

In the last example, the two partially defined vectors were adjoining lines, and the resultant triangle between these lines and the closing vector was easily brought together and graphically illustrated. The same principles, however, apply when the partially defined vectors are not adjoining. The separated lines will still form a resultant triangle from which the missing elements are determined. We will only look at two partially defined vectors as adjoining lines.

From this simple approach to solving missing measurements using rounded-off data we will now approach the solution of these problems in a more formal manner using realistic measurements.

158


10.6 Calculating missing measurementsA closed traverse gives rise to three equations of closure:

angular misclose = 0 … (1)

NA N– NB … (2)

EA E– EB … (3)

Each of the equations allows us to solve for a misclose or a missing measurement; thus, in any closed rectilineal figure or traverse, we can solve for as many as three missing values. These may be either angles or distances, provided that, if there are three missing values, at least one is an angle. This is because equation 1 is a function of angles only and can therefore be used to solve for an angle only. Equations 2 and 3, on the other hand, are functions of angles and distances and may be used to solve for either.

Remember that in a closed figure, the initial values (A, NA, EA) are equal to the final values (B, NB, EB).

Guidelines for calculating missing measurementsThe following points are a guide to help you when you are calculating adjacent missing measurements of a figure.

1. Methodically and systematically tabulate the measured angles, distances and known bearings. If no bearings are known, an arbitrary datum bearing may be assumed.

2. Calculate all possible bearings.

3. Sum all known differences in the coordinates and, from these, deduce the sum of the values for the partially defined line(s). A partially defined line is one for which only a bearing or a distance is known. Consequently, the differences in coordinates for these lines cannot be computed.

4. These last sums will give the rectangular definition of a closing vector (line). The polar definition (bearing and distance) may then be computed.

5. The closing vector must be the sum of the partially defined vectors, and the known parameters for these three vectors must define a triangle which may be solved for the missing polar definitions. The missing measurements may then be deduced.

159


Example 1 Missing bearing and distance for one line

We will firstly look at the simplest combination of missing measurements; that is, for one line, a missing bearing and a missing distance (or length).

(Refer back to ‘Missing vectors’ Example 1, for details of this method.)

A

B

C

D

E

77 52'57"

348.625 m

12641'43"

116

45'0

5"

405.674 m

630.4

69 m

Fig 10.10

To start this calculation you will have to know the bearings of the linesBC, CD, DE and EA. If these are not given (as in this case) assume a bearing for one line and use given angles to calculate remaining bearings.

Assume BC = 57º 09' 45"

then calculate remaining bearings from angles at C, D, E.

CD = 120º 24' 40"

DE = 222º 31 ' 43"

EA = 275º 50' 00"

Use these bearings with given distances to solve for the line AB. Put your data into the close table shown below.


AB

BC 57º 09' 45" 405.674 + 219.980 + 340.852

CD 120º 24' 40" 598.802 - 303.114 + 516.416

DE 222º 31' 43" 630.469 - 464.618 - 426.171

EA 275º 50' 00" 348.625 + 35.433 - 346.820

= = - 512.319512.319 = + 84.277= + 84.277

160


Therefore, AB

AB

+512.319

84.277

-

N

E

=

=

The computed values for N and E- form the rectangularcomponents for missing line AB.

( ) ( )

1

22

84.277tan 350 39'30"512.319

84.277 512.319 519.205 m

-AB

ABL

æ ö- ÷ç ÷ç ÷ç ÷÷ç+è ø

+

= =

= =

Example 2 Two missing adjacent distances

In Fig 10.11 the distances for AB and EA are missing.

(Refer back to ‘Missing vectors’ Example 2, for details of this method.)

A

B

C

D

E

405.674 m 598.802 m

630.4

69 m

77°52' 57"

113°29' 45"

105°

10'30

"

116°

45'0

5"

126°41' 43"

Fig 10.11

When any two adjacent elements are missing, then they may be isolated from the original figure by means of a closing line calculation (described earlier). We can close BCDEB for the missing bearing and distance EB. Then we can solve the triangle ABE for the missing distances AB and EA, using the sine rule.

It is important to note the sign change.

161



AB 350° 39' 30"

BC 57° 09' 45" 405.674 219.980 340.852

CD 120° 24' 40" 598.802 - 303.114 516.416

DE 222° 31' 43" 630.469 - 464.618 - 426.171

EA 275° 50' 00"

∑ = - 547.752 ∑ = 431.097

Therefore, +547.752

431.097EA

EA -

N

E

=

=

∆

∆

1 1

2 2

2 2

431.097tan tan547.752

321 47 '46"

431.097 547.752697.049 m

EA

EA

EN

L E N

θ - -∆ - = = ∆ = °

= ∆ + ∆

= +

=

The triangle to solve now is shown below.

A

B

E

697.049

350°39' 30"

275°50' 00"

321°47' 46"

Fig 10.12

Solving this triangle using the sine rule gives the missing lengths:

AB = 519.205 m

EA = 348.625 m

162


Example 3 Two missing adjacent bearings

Using the same diagram as in Example 1 and Example 2, assume bearings AB and EA are not given.

As before, close BCDEB for EB , EB and then construct the triangle to solve for AB , EA .

Use the cos rule to solve for angles at A, B, E then check they sum to 180

But here is where you must be careful. Usually, if you are given the diagram to work from, you can calculate bearings AE and EA correctly. But if you do not have a diagram, it is possible to have two solutions. That is, it may be an ambiguous solution.

In our case, if we did not have a diagram to work from, the triangle to solve could be either:

Solution 1 or Solution 2

A

B

E348.625

519.205

A

B

E

348.625

519.205

Fig 10.13 Fig 10.14

As you do not know which solution is correct, you have to solve for both.

163


A

A

B

E697.049 m

28 '44"

105 '30"

105 '30"

321' 46"

45 '46"

45 '46"

28 '44"

Fig 10.15

Solution 1 has AB = 350º 39' 30"

EA = 275º 50' 00"

Solution 2 has AB = 292º 56' 02"

EA = 7º 45' 32"

Note that even with a diagram, if there is any doubt at all about the calculated bearings, calculate the second solution. A typical example of when this is necessary is where a calculated bearing is 88º. You will not be able to see from the diagram that the bearing is 88º and not 92º (the second solution) so solve for both.

Example 4 Missing non-adjacent measurements

For this example we have to find the bearing of one line, and the distance of another line.

A

B

C

D

E

405.674 m 598.802 m

m

m

57'45"

350

'30"

120' 40"

275 '00"

Fig 10.16

164


Line Bearing(θ) Distance (L) ∆ N ∆ E

AB 350º 39' 30"

BC 57º 09' 45" 405.674 + 219.980 + 340.852

CD 120º 24' 40" 598.802 - 303.114 + 516.416

DE 630.469

EA 275º 50' 00" 348.625 35.433 - 346.820

= = - 47.701 = = 510.448

In this example we are given a diagram so we know the shape of the figure. If a diagram is not given, you can sketch an approximation of the shape from the information given.

As this is a closed figure then:

NAB + NBC + NCD + NDE + NEA = 0

and EAB + EBC + ECD + EDE + EEA = 0

ie N = 0

and E = 0

but NBC + NCD + NEA = - 47.701

and EBC + ECD + EEA = 510.448

therefore NAB + NDE – (-47.701) = 0

and EAB + EDE – 510.448 = 0

therefore NAB + NDE = 47.701

and EAB + EDE = 510.448

The sum of the two vectors AB and DE can be shown graphically.

Starting from A.

A

E

(AB + DE)

-510.448 47.701

Fig 10.17

165


Then knowing that LDE = 630.469, point D must lie on the circumference of a circle of radius 630.469, centre E.

Also, θAB = 350° 39' 30" so B must lie in that direction from A.

I (B, D)

A

E

630.4

69 m 350

'30"

I (A, E)

DB

630.4

69 m

350'30"

Fig 10.18

Note: ‘I’ is the intersection point for B and D.

In the graphic the point of intersection (I) of the line with a bearing of 350° 39' 30" from A with the arc of radius 630.469, centre E, represents coincident points B and D.

From Fig 10.17 1

1

2 2

2 2

tan

510.448tan47.701

275 20'19"

510.448 47.701

262 833

512.672 m

AE

AE

AE

AE

EN

-

L E N

L

-

-

+

+

D

D

D D


æ ö÷ç= ÷ç ÷çè ø

=

=

=

=

=

and

Triangle EAI must now be solved for the bearing of ED and the length of AB.

A = θAI – θAE

= 350º 39' 30" – 275º 20' 19"

= 75º 19' 11"

166


Triangle EAI is now defined by the two sides EI and EA and the angle A, so the remaining sides and angles can be computed using the sine rule:

Therefore 1

1

sin A sin BA B

sin 75 19'11" 512.672I sin630.469

sin (0.786614)

51 52'13"

E 180 00'00" 75 190'11" 51 52'13"

52 48'36"

-

-

=

æ ö ÷ç = ÷ç ÷çè ø

=

=

= - -

=

.

and

Again from sine rule

and

sin630.469 52 48'36"sin 75 19'11"

519.203 m

I

170 39'30" 51 52'13"

222 31'43"

AI AB

DE BA +

æ ö ÷ç ÷= = ç ÷ç ÷ç è ø

=

=

= +

=

.

Example 5 Missing non-adjacent distances

For this example we have to find the missing distances (lengths) AB and CD.

A

B

C

D

E

405.674 m 57º 09' 45"

m

º'

"

º'

"

m

º'

"

º ' "

Fig 10.19

167



AB 350º 39' 30"

BC 57º 09' 45" 405.674 + 219.980 + 340.852

CD 120º 24' 40"

DE 222º 31’ 43” 630.469 - 464.619 - 426.170

EA 275º 50' 00" 348.625 + 35.433 - 346.820

= = - 209.206 = - 432.138

In this example we are given a diagram so we know the shape of the figure. Again, if a diagram is not given, you can sketch an approximation of the shape from the information given.

ABCDE is a closed figure so that:


→BC + →DE + →EA = - (→AB + →CD )

= - (→AD)

or - (→BC + →DE + →EA ) = →AD

This is broken up into N and E components so that:

-N = →AD

and -E = →AD

NAB + NCD = 209.206

similarly EAB + ECD = 432.138

Using Pythagoras’ theorem and 1tan

64 10'03"

480.115

AD

AD

AD

EN

L

- D

D


=

=

, as from the previous example,

then

and

168


The vector can be shown graphically:

A

D

I (B, C)

480.115 m

432.138 m

209.

206

m

120 24'40"

35039'30"

64 10' 03"

Fig 10.20

In the graphic the vector AD represents the components of the missing vectors AB + CD. The bearing of AB is known to be 350° 39' 30", and of CD 120° 24' 40". These vectors can be added to A and D in the sketch, to form the triangle shown. Again, I is the intersection point where AB and CD intersect.

In the triangle AID, the bearings of the three sides and the distance AD are known. Thus it is defined and may be solved for the missing sides AI and DI (= AB and CD).

A = θAD – θAB = (360° + 64° 10' 03") – 350° 39' 30" = 73° 30' 33"

I = θBA – θCD = 170° 39' 30" – 120° 24' 40" = 50° 14' 50"

D = θDC – θDA = 300° 24' 40" – 244° 10' 03" = 56° 14' 37"

= 180° 00' 00" check

Now by the sine rule

480.115 x sin56 14'37" 519.206 msin 50 14'50"

480.115 x sin 73 30'33" 598.802 msin 50 14'50"

AB

CD

= =

= =

Note: With two missing distances, the solution is always unique.That is, it is never ambiguous.

169


Example 6 Missing non-adjacent bearings

For this example we have to find the missing bearings BC and DE.

A

B

C

D

E

m

m

º'

"

m

º'

"

m

º ' "

405.6

74 m

50'50"

105

' 30"

Fig 10.21


AB 350º 39' 30" 519.206 + 512.320 - 84.278

BC 405.674

CD 120º 24' 40" 598.802 - 303.114 + 516.416

DE 630.469

EA 275º 50' 00" 348.625 + 35.433 - 346.820

= = + 244.639 = = + 85.318

In this example we are given a diagram so we know the shape of the figure. Again, if a diagram is not given, you can sketch an approximation of the shape from the information given.

As with the previous examples, ABCDE is a closed figure so that:

ΣΔ N = 0, and Δ NBC + Δ NDE = - 244.639

similarly Δ EBC + Δ EDE = - 85.318

and 1

2 2

85.318tan244.639

199 13'34"

85.318 244.639

259.090 m

BE

BE

--

L

-

+


=

=

=

and

170


The sum of these two vectors can be shown graphically.

B

I '

I(C, D)

E

405.674 m

630.469 m

259.

090

m

-85.318

-244

.63919

9'3

4"

Fig 10.22

In the graphic BE represents the closing vector defined by the sum of the components of the missing lines BC, DE. The known values for these lines are their lengths. Thus, if arcs of radii 405.674 and 630.469 are swung about B and E respectively, the intersection of these arcs must define points C, D.

The bearing of BI appears to be approximately 50°, which appears correct from the given diagram. The alternative bearing of BI appears to be approximately 350°. From the given diagram, this is obviously wrong. Therefore, I’ can be ignored.

Note: If a diagram is NOT given, you will not know if I or I’ is correct. You will, therefore, have to compute two solutions: one for triangle BEI, the other for triangle BEI’. This solution is therefore ambiguous.

The triangle BIE is now defined by the three sides, and the angles may be computed using the cosine rule.

B = cos–1 (-0.788690) = 142° 03' 48"

I = cos–1 (0.967558) = 14° 38' 03"

E = cos–1 (0.918429) = 23° 18' 19"

= 180° 00' 00" check

From which

θBC = θBE – B = 199° 13' 34" – 142° 03' 48" = 57° 09' 46"

θED = θEB + E = 19° 13' 24" + 23° 18' 19" = 42° 31' 43"

171


Example 7 Missing lines on a cardinal bearing

The solutions where a distance is missing and a bearing is known; either two missing distances or a missing distance with a missing bearing, are simplified when the line with the distance missing falls on a cardinal direction; that is it has a bearing of 0°, 90°, 180° or 270°.

As an example consider the problem below where the line AB has a known bearing of 0° 00' 00" with the length missing, and the bearing DE missing.

A

B

C

D

E

405.674 m 598.802 m

m

m

66'15"

0'0

0"

129' 10"

285 '30"

Fig 10.23


AB 0º 0' 0" 0

BC 66º 30' 15" 405.674 161.735 372.039

CD 129º 45' 10" 598.802 - 382.920 460.365

DE 630.469 - 495.935

EA 285º 10' 30" 348.625 91.259 - 336.469

= - 129.926 = = 495.935

As θAB = 0° 00' 00"

then Δ EAB = 0

and Δ EDE must therefore equal - 495.935, if ΣΔ E is equal to 0. …(1)

Now, knowing Δ EDE = - 495.935

and LDE = 630.469

Δ NDE can be calculated using Pythagoras’ theorem

3 231º 52' 12" calculated

2 - 389.281 calculated

1

4 + 519.207 deduced

172


thus 2 2 2DE

2 2 2DE

2 2630.469 495.935

389.291

L N E

N L E

= D + D

D = - D

= -

=

or

…(2)

Knowing both Δ NDE and Δ EDE we can calculate θDE from

1tan

495.935389.281

231 52'12"

DEEN

--

- D

D


=

= …(3)

Finally, knowing that ΣΔ N = 0, ΔNAB must be 519.207 to satisfythis condition. …(4)

The graphic indicates that there is to be a negative ΔN, that is, - 389.281. (In the absence of a diagram both +ve and –ve values would hold, resulting in two possible solutions).

Indeterminate solutions – parallel linesOn some occasions in a missing measurement close, the presence of parallel lines means that a finite solution is not always possible, and other factors have to be considered. The following examples illustrate this point. They have been partially worked. A solution for the missing measurements is required, or where this is not possible, the extent to which this is so must be determined.

Example 7

For this example there are two missing bearings.


AB 270º 00' 00" 1000.000 0.000 - 1000.000

BC 3000.000

CD 330º 00' 00" 4000.000 3464.100 - 2000.000

DE 3000.000

EF 60º 00' 00" 866.000 433.000 750.000

FA 150º 00' 00" 4500.000 - 3897.100 2250.000

= = 0.000 = 0.000

173


There are two important points to note in this problem:

(a) the sum of both the Δ N and Δ E columns add to zero

(b) the two lines of missing bearing have the same length.

Deduction

As ΣΔ N and ΣΔ E must both remain as zero after the inclusion of the missing Δ N and Δ E values for BC and D, then it follows that Δ NBC and Δ EBC must be equal to Δ NDE and Δ EDE, but of opposite sign.

An infinite number of solutions is therefore possible, the only necessary condition being that θDE = θBC ± 180°. That is, the missing measurements are parallel, but with reverse bearings.

Example 8 Special case

For this example there are two missing bearings.


AB 61º 13' 30" 728.968 350.904 638.953

BC 90º 00' 00" 780.290 0.000 780.290

CD 578.874

DE 99º 40' 43" 415.820 - 69.908 409.901

EF 287.563

FG 260º 16' 40" 1391.275 - 234.947 - 1371.294

GA 303º 03' 48" 1175.624 641.380 - 985.253

= 687.429 = 527.403

From the close table calculations, the closing vector is CF.

ΔNCD + EF = - 687.429 and ΔECD + EF = 527.403

1

1

2 2

tan

527.403tan687.429

37 29' 45" ( 180 )

142 30' 15"

687.429 ( 527.403 )

866.437 m

-CF

-

CF

EN

-

-

L -

+

+

D

D



=

=

=

=

174


Observation

The two distances with missing bearings sum exactly to the length of the closing vector, ie 578.874 + 287.563 = 866.437 m.

Deduction

No triangle of solution can be formed. Therefore, the three lines of the ‘triangle’ are parallel and will have the same bearing of 142° 30' 15".

θCD = 142° 30' 15"

θEF = 142° 30' 15"



175


Chapter 11 – Simple circular curves

11.1 IntroductionHorizontal, circular or simple curves are curves of constant radii connecting two ‘straights’ set out on the ground.

Curves, in surveying, are used for a variety of purposes. They are found in surveys of road and boundary lines, including road truncations, railways, kerb lines and pipelines.

A B

I (= IP)

(= TP) (= TP)

Fig 11.1

AB is a simple curve. The curve is tangent to the straights at A and B, referred to as tangent points (TP). The straights intersect at I, known as the intersection point (IP).

11.2 The geometry of curvesIn the design, setting out and computational stages of surveying work involving curves, there are a number of geometrical theorems. Knowledge of these is desirable. These are stated on the following pages with a short explanation after each.

176


Theorem 1If a straight line drawn from the centre of a circle bisects any chord which does not pass through the centre, it will cut that chord at right angles.

Chord AC is bisected at B. B

A

BC

O

Fig 11.2

Theorem 2Chords which are equidistant from the centre are equal.

Chords AC and DF are equal because BO = OE.

A

B

C

O

EF

D

Fig 11.3

177


Theorem 3The angle at the centre of the circle is double the angle at the circumference, standing on the same arc.

AOC 2ADC because both stand on arc AC.

A

B

C

D

O

Fig 11.4

Theorem 4Angles in the same segment of a circle are equal.

Angles at C and D are equal because they are formed on the same side of the chord AB.

D

B

EP

A

C

Fig 11.5

178


Theorem 5The opposite angles of a quadrilateral inscribed in a circle are together equal to two right angles ( a cyclic quadrilateral).

BCA + AEB 180 in quadrilateral ACBE.

Theorem 6The angle in a semicircle is a right angle.

The angles AEB, BDA and BCA standing on the diameter AB, are 90º.

EB

D

C

A

Fig 11.6

Theorem 7The tangent at any point of a circle is perpendicular to the radius drawn to the point of contact.

OTQ = 90º because QTP is a tangent and OT a radius.

T

P

Q

O

Fig 11.7

179


Theorem 8Two tangents can be drawn to a circle from an external point.

Corollary – the two tangents are equal, and subtend equal angles at the centre.

PA and PB are tangents and are equal. BOP = POA.

A

O

B

P

Fig 11.8

Theorem 9If two circles touch each other, the centres and point of contact are in one straight line.

Centres O1 and O2 and point of contact A are in one straight line.

Centres O1 and O3 and point of contact B are in one straight line.

01

02

03

B

A

Fig 11.9

180


11.3 Elements of simple circular curves

E

I

O

R

TP

TD

Arc

Long90º

chord

TD

TP

R

M

2 2

A

C

DB

Fig 11.10

I intersection point of tangents

O centre of circular arc

TP tangent points

∆ deflection angle of the tangents

R radius of circular arc

TD tangent distance

Arc distance from beginning to end of curve along circular arc

E Crown secant (ex-secant in USA) – distance from I to midpoint of curve

M Mid-ordinate – distance from midpoint of curve to midpoint of LC

Long chord

straight line distance from beginning to end of curve ( between tangent points)

181


11.4 Derived formulaeUsing Fig 11.10 we can show the following formula derivations.

Tangent distance

IAO + OBI = 180º

AOB + BIA = 180º

But ∆ + BIA = 180º

Therefore AOB = ∆

Triangles IAO and IOB are congruent

AOI and IOB = 2

and AI = IB

tan2

tan2

tan2

TD R

TDR

TDR

=

=

=

Long chord

AD = DB AB = 2AD

AD = R sin 2

AB = 2R sin 2

Crown secant

In right triangle IAO

IO = cos

2

R , OC = R, Crown secant (E) = IC = IO – OC

cos

2

RE R= - or 1 1

cos2

E R

æ ö÷ç ÷ç ÷ç ÷ç= - ÷ç ÷ç ÷÷ç ÷ç ÷çè ø

182


Substituting tan

2

TDR =

for R, it can be shown that:

tan4

E TD=.

Mid-ordinate

Mid-ordinate (M) DC OC – OD

cos2

R R-

or 1 cos2

M Ræ ö÷ç= - ÷ç ÷çè ø

Note: The crown secant and mid-ordinate distances are rarely required now in curve calculations.

Arc lengths and circular measure of angles

In circular measure of angles, the magnitude of an angle is expressed in radians. An angle expressed in radians is defined as the ratio of the arc subtended by that angle at the centre of a circle to the radius of that circle.

R

R

0.1 R0.1 radian

1 radianO

Fig 11.11

arc lengthAngle at centre (in radians)radius

=

183


When the arc length is equal to the radius, the ratio = 11 = 1, and the angle at the centre contains one

radian.

1 radian (circumferenceof circle)

360 2R

R =

360or, 1 radian2

180

57.2957795

57 17 '44.806"

or, in minutes, 1 radian 3437.74677 '

or 206264.806"

=

=

=

=

=

=

When calculating lengths and measures of arcs it is useful to remember the following points.

1. By transposing the basic formula angle at centre (in radians) arc length

radians we can get:

a) arc length radius angle at centre (in radians), and

b) radius arc lengthangle at centre (in radians)

2. Referring back to Fig 11.11 you can see that the radian measure of an angle is directly proportional to the number of degrees in the angle; that is, 150º in radians 90º in radians + 60º in radians.

3. As 1 radian 180

you can convert:

a) degrees to radians by multiplying by180

x

b) radians to degrees by multiplying x radians by 180

.

Note: Most calculators have a button to convert between degrees and radians.

184


11.5 Areas

SectorΔ

Δ

A B

RR

I

O

Fig 11.13

2Sector area (in radians)2

R= ´

Since ∆ R equals the arc AB of the sector, the formula can also be written as:

1Sector area arc2

R= ´

185


Segment

A B

I

O

RR

Fig 11.14

Segment area area sector AOB – area triangle AOB

2 21 1(radians) sin2 2

R R-

[ ]21 (radians) sin2

R -

Segment area [ ]21 (radians) sin2

R -

186


Truncation

A B

I

O

RR22

Fig 11.15

2

2

Truncation area 2 area triangle sector 12 sector 2

sector 1tan (in radians)

2 2

tan (in radians)2 2

1or TD arc2

IAO AOB

AI AO AOB

AI AO AOB

R R R

R

R

= ´ -æ ö÷ç= ´ ´ -÷ç ÷çè ø

= ´ -

= ´ -

æ ö÷ç= - ÷ç ÷çè ø


2Truncation area tan (in radians)2 2

1or TD arc2



R

R

187


Example 1 Calculate the shaded area, given the crown secant = 10 m and the angle between the straights is equal to 140º00'00".

10 m

140°00'000"

x

arc

TPA

TPB

Fig 11.16

(a) Firstly, write down the formulae for calculating the truncation area.

2

1Truncation area TD arc2

or tan (in radians)2 2


æ ö÷ç - ÷ç ÷çè ø

R

R

Both of these formulae require information which we do not have at this stage. However, the second of the two requires us to find only one piece of missing information, ie the radius, and is the better one to use.

(b) To calculate the radius, we need to use the information available, ie crown secant and deflection angle. The radius may be found from transposing the formula for the crown secant.

Crown secant (E) 2

cos

R R-

or E

2

1 1cos

R

æ ö÷ç ÷ç ÷ç ÷ç - ÷ç ÷ç ÷÷ç ÷ç ÷çè ø

10 1 1cos20

R

æ ö÷ç ÷-ç ÷ç ÷çè ø the angle between the

straights is 140, so the deflection angle is 40 R 155.817 m

188


(c) The truncation area may now be calculated.

Truncation area = R2 (tan – in radians)

= (155.817)2 (tan 20 – 20 (in radians))

= (155.817)2 (0.363970 – 0.349066)

= 361.853 m2



2 tan (in radians)2 2 æ ö÷ç - ÷ç ÷çè ø

R

189


Chapter 12 – Roads

12.1 IntroductionBoundaries of reserves for roads, railways, pipelines and easements are usually formed by a series of pairs of parallel straight lines. For the purpose of this chapter, all reserves will be termed roads.

Because the boundaries of these reserves are usually parallel, it is economical to survey only one side of the boundary and to calculate the other sides (for plan and diagram purposes).

Thus, we will see that the pegs positioned on the calculated side can be calculated from the survey information gathered from the surveyed side.

Fig 12.1

Note: Sometimes the centre line of the road boundary is surveyed and the boundaries laid out either side (via calculations).

Measured Measured

Peg

Peg

Peg

Calc

. and

Mea

sure

dCal

c. a

ndM

easu

red

PegCalculatedCalculated

190


12.2 Single-width roadsTraditionally, road widths have been 20 m, although this changes according to circumstances. In modern subdivisions, the width of roads may be less than 20 m, with 16 m and 15 m common in some town planning schemes. On the other hand, in rural areas, a width considerably greater than 20 m may be necessary in order to allow a nature area to be retained for protection against erosion, as well as for beautification.

Before metrication, most roads were surveyed at one hundred links (or multiples thereof). Consequently, a large number of these roads will always exist. The conversion calculation for these roads is shown below.

1 foot 0.3048 m exactly

100 links 1 chain = 66 feet x 0.3048 m 20.1168 m 20.117 m

Fig 12.2

For single-width roads the calculation of the opposite corner peg is a simple solution to a right-angled triangle; the opposite peg lies on the bisection of the angle as measured.

‘cotangent’distance

road widthA

D

E

a'

b'e'

b"

f "f 'H

G

B

C

F

‘cotangent’distance

‘cotangent’distances

‘cotangent’distances

‘secant’distance




Fig 12.3

191


In Fig 12.3, boundary ABCD of the road is to be surveyed and boundary EFGH is to be calculated. It can be seen that the distances EF, FG and GH can be found by the addition or subtraction of the cotangent distances, and the distances of the pegs E, F, G and H from their respective opposites can be found by calculating the secant distances.

Aa', Bb' (Bb"), Ff ' (Ff ") and Ee' are the cotangent distances from right-angled triangles AHa',

BGb' etc. (Cotangent 1tan

).

Similarly, AH, BG, CF and DE are the secant distances from the same right-angled triangles.

These secant distances actually use the cosecant trigonometric function where the cosecant 1sin

.

‘cotangent’ distances

road width road width‘secant’distance

2 2

Fig 12.4

Given the width of the road and the angle as measured, the secant cotangent distances can be easily calculated.

With calculators, it is easier to use sin and tan rather than cosecant and cotan. We will therefore rewrite the formulae as:

secant distance road width

sin

and

cotangent distance road width

tan

Note: The terms ‘secant’ distance and ‘cotangent’ distance are traditional.

192


Calculating boundariesGiven the measurements for one side of a road and the road width, we can calculate the measurement for the other side.

AB is one section of a single-width road. Calculate the opposite side of the road, ie section CD.

Example 1 A

E

B

CF

D20 m road width

1269.85 m

98º13'75º25' 13

7º32'

157º12'

Fig 12.5

(a) Firstly calculate the secant distances, AD and BC.


sin

AD 20157 12'sin

2

20.403 m

BC 20137 32'sin

2

21.457 m

193


(b) Then calculate the cotangent distances, AE and FC.


tan

AE 20157 12'tan

2

20tan 78 6'

4.033 m

FC = 20137 32'tan

2

= 20

tan 68 46'

= 7.771 m

(c) Therefore CD = AB – AE + FC

= 1269.85 – 4.033 + 7.771

= 1273.588 m

194


12.3 Computations relating to roads

Single-width roadsAs mentioned in the previous section, one boundary is normally surveyed and the opposite boundary calculated. It is therefore required to compute:

• the information necessary to set out the boundary that has not been directly surveyed in the field

• the lengths of the lines making up this boundary

• the area of the portion of the road reserve concerned.

These computations are best illustrated by the following example.

Example 1 Given the following single-width road compute:

(a) LAD, LBE, LCF necessary to set out points D, E, F

(b) the length of boundary DEF

(c) the area of the portion of road ABCFEDA.

B

C

F

A75º0'00"

180º

0'00

"

70.000 m120º0'00"

330º0'00"50.000 m

D

E

b"b'

f

a

road width 20 m

Fig 12.6

(a) Firstly, calculate the half angles (those angles adjacent to right angles and marked with a cross in the diagram). Calculate these angles from the given bearings.

At A 150 – 75 7500'00"

At B 255 120-

6730'00"

At F 360 – 300 6000'00"

Then compute the bearings and lengths of secants AD, BE, CF using the half angle, already calculated at A, B, F.

195


The formula is:


sin

LAD 20

sin 75 20.705 m, AD 150(given)

LBE 20

sin 67 ' 21.648 m, BE 120’ 1870’

LCF 20

sin 60 23.094 m, CF 180(given)

(b) To calculate the length of boundary DEF compute the cotangent distances.

The formula is:


sin

Aa 20tan 75

5.359

Bb 20tan 67 '

8.284

Ff 20

tan 60

11.547

then LDE LAB – Aa + Bb'

LDE 70.000 – 5.359 – 8.284

56.357 m

and LEF LBC – Bb'' + Ff

LEF 50.000 – 8.284 + 11.547

53.263 m

Boundary DEF Length (DEF) 109.620 m

196


(c) Finally compute the area of the portion of road ABCFEDA.

Area sum of boundary lengthsroad width2

æ ö÷ç ÷ç ÷çè øformula for a trapezium

= 120.000 109.620202+æ ö÷ç ÷ç ÷çè ø

= 2296 m2

Alternative calculation methodsThe calculations in Example 1 for the single-width road have been done in the traditional way. However, they can also be done using a calculator and software to calculate ‘missing information’.

• Firstly, the bearing for BE can be calculated directly from 902

AB BC ++

.

• With this ‘formula’ you do not need to calculate the half angles.

• By using ‘missing measurement’ software – this is available for some calculators and alsofrom surveying programs on computers, you can solve triangles AaD and CFf for distancesDA and Aa, and CF and Ff.

• If you then ‘close’ DABED with calculated LDA, calculated θBE and known bearing anddistance for AB, you can calculate LBE and LED.

• If you use this calculated LBE with known bearing and distance for BC and calculated LCF,you can calculate θFE and LFE. You have a check on your work as you know already thebearing of FE.

• You can use this method for any number of sections of road.

Yet another modern method is to simply draw the road using surveying software. Thisis an easy task and missing information can then be easily extracted from the diagramconstructed.

197


Two-width roadsSometimes different sections of road have differing widths. When this occurs traditional cosecant and cotangent formulae cannot be used. Special formulae are available to calculate the connections between the surveyed boundary and computed boundary, eg AB in the diagram below.

Example 1

W 1 = 20 m

W2 = 15 m

2

1

B

A

Fig 12.7

The formulae below are used for the calculations. Use one for a check.

AB = tan–1 1 2

1 2

W sin W sinW cos W cos

æ ö- ÷ç ÷ç ÷ç ÷ç -è ø

LAB = ( ) ( )

21

1 2sin sinAB AB

W W=- -

Where:

AB is the connection between surveyed and calculated points (boundaries).

W1 and W2 are the road widths.

θ1 and θ2 are the corresponding bearings.

Note: 1. In certain instances, the formulae give the reverse bearing for AB. Always check thediagram.

2. LAB may sometimes be calculated to be a negative quantity. Ignore this as LAB is always positive.

3. Bearings 1 and 2 must both be in the same direction – this can be left to right or right to left.

198


Example 2 In the situation shown below, compute the bearing and length of the join AB.

W 1 = 10 m

W2 = 15 m

θ2 = 75°19′20″

θ 1 = 41°51′3

0″

B

A

Fig 12.8

Note that both bearings ‘flow’ from left to right.

(a) To calculate the bearing for AB.

AB tan–1 10sin 75 19'20" 15sin 41 51'30"10cos75 19'20" 15cos41 51'30"æ ö- ÷ç ÷ç ÷ç ÷ç -è ø

213'32''

However, from the diagram, θAB must be 18213'32''.

(b) To calculate the length.

LAB ( ) ( )

10 15orsin 182 13'32" 41 51'30" sin 182 13'32" 75 19'20"- -

15.677 m



199


199


Appendix A

Calculator requirementsBecause the majority of calculations involve the use of trigonometric functions, you should not purchase a calculator until you ensure it provides the following basic functions:

• a display of at least 8 digits

• addition, subtraction, multiplication and division

• square, square root and nth root

• Pi ( )

• reciprocal

• trig. and inverse trig. functions

• summation function and registers (desirable)

• parenthesis (bracketing): unless you have a Hewlett Packard calculator

• polar to rectangular and the converse (highly desirable, but if your calculator does not have this as a keyboard or menu function then you may be able to write it as a small program)

• easy conversion between degrees, minutes and seconds and decimal degrees

• ability to handle values in the range 1099 to 10−99

• ability to set different numbers of decimal places

• data storage register

There are suitable low-cost models, with all these features.

Hewlett Packard calculators use a reverse Polish system of notation. It will be noted that there is no ‘equal’ sign on these calculators. This system, it is claimed, is the most efficient for evaluating mathematical expressions, and is such that the operands are entered into the calculator (one to the Y register, one to the X register) and the arithmetic performed by using the appropriate keys ( , , , ). The result of the calculation is returned to the X register (displayed).

This system of operation eliminates the need for a separate parenthesis function.

The range of suitable calculators with all or most of the features previously listed is large, and you should weigh up the advantages of the various models carefully before making a purchase.

200200

Appendix A – Calculator requirements

Examples of useBelow are examples of computations using the cosine rule, a basic formula which will be used many times during this course. The calculator you purchase should enable you to solve such an exercise without the need to write down or store intermediate results in a separate recall memory.

Given data: b 13.300, c 20.100, A 17 54' 27"

a2 b2 c2 2bc cos A

a 2 2 2 cos+ -b c bc A

2 213.3 20.1 2(13.3)(20.1)(cos17 54'27")+ -

Example A Example BCalculator with the basic functions previously listed (including parentheses).

Calculator with operation ‘stacking’ register and reverse notation.

Key Display Key Display

13.3 13.3 13.3 13.3x2 176.8 Enter 13.3

176.89 176.89

20.1 20.1 20.1 20.1x2 401.01 Enter 20.1 580.9 401.01( 580.92 2 2 2 2 Enter 213.3 13.3 13.3 13.3 26.6 26.620.1 20.1 20.1 20.1 534.66 534.6617.5427 17.5427 17.5427

DMS Dec. Deg. 17.9075Cos 0.951554 Shift key) 508.7579 H HMS

72.1420 H Menu key 17.9075

√¯¯ 8.4936 Cos 0.951554 508.7579 72.1420√¯¯ 8.4936

a = 8.494 a = 8.494

201


201


Note the following with respect to examples A and B.

1. In both examples the angles must be converted to decimals of a degree (most calculatorswork this way).

2. The positioning of the parentheses in the formula is critical to the solution.

3. The two examples are only intended as a general rule to usage and should not be regardedas standards for all calculators. Each calculator model has its own handbook which must bestudied carefully to discover its particular advantages and mode of operation.

The calculators used in examples A and B were:

Example A Sharp scientifi c calculator.Example B Hewlett Packard RPN scientifi c calculator.

SURVEYING COMPUTATIONSCPPSIS3010ACPPSIS4011A

Learner’s Guide

This Learner’s Guide includes content relevant to Units of Competency from the Property Services Training Package (CPP07) – CPPSIS3010A and CPPSIS4011A. Topics covered include; trigonometric functions, angles, bearings, missing measurements, simple circular curves and simple road calculations.

TRAINING PACKAGECPP07 – Property Services Training Package

COURSE/QUALIFICATION• CPP30107 Certificate III in Spatial Information Services• CPP40107 Certificate IV in Surveying

UNITS OF COMPETENCY• CPPSIS3010A – Perform basic spatial computations• CPPSIS4011A – Perform surveying computations

9 7 8 1 7 4 2 0 5 2 6 8 7

BC1091 SURVEY COMPUTATIONSISBN 978-1-74205-268-7

ORDERING INFORMATION:Contact WestOne Services on Tel: (08) 6212 9700 Fax: (08) 9227 8393 Email: [email protected] can also be placed through the website: www.westone.wa.gov.au

surveying computations cppsis3010a cppsis4011a

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