Download View - Department of Training and Workforce Development

SURVEYING COMPUTATIONS
CPPSIS3010A
CPPSIS4011A
Learner’s Guide
BC1091
CPPSIS3010A
CPPSIS4011A
Surveying Computations
Learner’s Guide
Copyright and Terms of Use
© Department of Training and Workforce Development 2016 (unless indicated
otherwise, for example ‘Excluded Material’).
The copyright material published in this product is subject to the Copyright Act 1968
(Cth), and is owned by the Department of Training and Workforce Development or,
where indicated, by a party other than the Department of Training and Workforce
Development. The Department of Training and Workforce Development supports
and encourages use of its material for all legitimate purposes.
Copyright material available on this website is licensed under a Creative Commons
Attribution 4.0 (CC BY 4.0) license unless indicated otherwise (Excluded Material).
Except in relation to Excluded Material this license allows you to:


Share — copy and redistribute the material in any medium or format
Adapt — remix, transform, and build upon the material for any purpose, even
commercially
provided you attribute the Department of Training and Workforce Development as
the source of the copyright material. The Department of Training and Workforce
Development requests attribution as: © Department of Training and Workforce
Development (year of publication).
Excluded Material not available under a Creative Commons license:
1. The Department of Training and Workforce Development logo, other logos and
trademark protected material; and
2. Material owned by third parties that has been reproduced with permission.
Permission will need to be obtained from third parties to re-use their material.
Excluded Material may not be licensed under a CC BY license and can only be used
in accordance with the specific terms of use attached to that material or where
permitted by the Copyright Act 1968 (Cth). If you want to use such material in a
manner that is not covered by those specific terms of use, you must request
permission from the copyright owner of the material.
If you have any questions regarding use of material available in this product, please
contact the Department of Training and Workforce Development.
Training Sector Services
Telephone: 08 6212 9789
Email: [email protected]
Website: www.dtwd.wa.gov.au
First published 2009
Updated May 2014
ISBN 978-1-74205-268-7
© WestOne Services 2009
All rights reserved. No part of this publication may be reproduced, stored in a retrieval
system or transmitted in any form or by any means, electronic, mechanical, photocopying,
recording or otherwise, without the prior written permission of WestOne Services.
Whilst every effort has been made to ensure the accuracy of the information contained
in this publication, no guarantee can be given that all errors and omissions have been
excluded. No responsibility for loss occasioned to any person acting or refraining from
action as a result of the material in this publication can be accepted by WestOne Services.
Published by and available from
WestOne Services
1 Prospect Place West Perth WA 6005
Tel: (08) 6212 9700 Fax: (08) 9227 8393
Email: [email protected]
Website: www.westone.wa.gov.au
This product contains various images © Jupiter Images 2009, used under licence.
Surveying Computations
Contents
Introduction
7
Chapter 1
Scientific notation; transformation of formulae ..............................9
1.1
Scientific 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
3
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 definition (polar coordinates) ..................................................................... 90
6.3
Point definition ................................................................................................... 92
6.4
The relationship between line definition and rectangular coordinates of
end 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 ..........................................................................143
9.1
Areas of regular rectilinear figures ......................................................................... 143
9.2
Areas of polygons ................................................................................................... 146
9.3
Calculating area using coordinates ......................................................................... 146
Chapter 10 Missing measurements .................................................................151
10.1
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
4
Surveying Computations
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
5
6
Surveying Computations
Introduction
This 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 computations
CPPSIS4011A – 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.
7
Introduction
8
Surveying Computations
Chapter 1 – Scientific notation;
transformation of formulae
1.1 Scientific notation
Some 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 notation
To 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 
2.5
2.5
=
100
102
 2.5  102
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.
9
Chapter 1 – Scientific notation; transformation of formulae
standard position
0.00194  1.94  103

ie
The decimal point is three positions to the left of
the standard position.
0.0194  1.94  102

ie
The decimal point is two positions to the left of the
standard position.
0.194
 1.94  101
1.94
 1.94  100
19.4
 1.94  101
194.0
 1.94 102
1940
(100  1)
ie

The decimal point is two positions to the right of
the standard position.
 1.94  103
To convert a number expressed in scientific notation to
ordinary decimal notation
Example 1
Express 2.76  103 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  103 
2.5
3
10
=
2.76
1000
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  103  0.00276.
Example 2
Express 3.8  104 in ordinary decimal notation.
Solution 3.8  104  3.8  10 000
 38 000
Example 3
Express 4.7  102 in ordinary decimal notation.
Solution 4.7  102  0.047
Example 4
Express 3.85  105 in ordinary decimal notation.
Solution 3.85 105  385 000
10
Surveying Computations
Example 5
Express 3.26  104 in ordinary decimal notation.
Solution 3.26  104  0.000326
Example 6
Express 9.8  103 in ordinary decimal notation.
Solution 9.8  103  9800
1.2 Equation transformation
If 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.
A = r 2
Divide both sides by
A
r 2
=



A
= r2

Find the square root of both sides
A
=

r2
A
= r

Now r can be the subject of this form of the equation r =
11
A
.

Chapter 1 – Scientific notation; transformation of formulae
Example 2
Given Area 
Area =
1
ab sin C, write the equation with b as the subject.
2
1
ab sin C
2
Multiply both sides by 2
2 Area = ab sin C
Divide both sides by a sin C
2 Area
ab sin C
=
a sin C
a sin C
Divide out common terms
2 Area
= b
a sin 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
cos A =
b2 + c2 - a 2
2bc
12
Surveying Computations
Example 4
Change the equation T = 2
1
to make g the subject.
g
Divide each side by 2
T
=
2
1
g
Square each side
æ T ö÷2
çç ÷ = 1
çè 2 ÷ø
g
Multiply each side by g
æ T ö2
g çç ÷÷÷ = 1
çè 2 ø
æ T ö÷2
Divide each side by ççç ÷÷ and divide out
è 2 ø
1
g =
æ T ö÷2
çç ÷
çè 2 ÷ø
Example 5
Given C N =
and
mN ´ L
L
C N = 0.007 m
L = 160.70 m
 L = 2150.53 m
Calculate mN.
13
Chapter 1 – Scientific notation; transformation of formulae
Solution 0.007 =
mN ´ 160.70
2150.53
160.70 mN = 2150.53 ´ 0.007
mN =
2150.53 ´ 0.007
160.70
= 0.094 m
Example 6
Given scale (s) =
and
scale (s) =
f
H - H'
1
13160
H = 2050 m
f = 152 mm
Calculate H '.
Solution
1
0.152
=
13160
2050 - H '
2050 - H ' = 13160(0.152)
2050 - H ' = 2000.3
2050 - 2000.3 = H '
H ' = 49.7 m
Example 7
Given V =
and
D
( A1 + 4 ( Am ) + A2 )
6
V = 1000 m3
D = 20 m
A1 = 39 m 2
Am = 42 m 2
Calculate A2 .
14
Surveying Computations
Solution
1000 =
20
(39 + 4(42) + A2 )
6
= 3.333(39) + 3.333(168) + 3.333( A2 )
= 130 + 560 + 3.333( A2 )
\ 3.333( A2 ) = 1000 - 130 - 560
A2 =
310
3.333
= 93 m 2
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.
15
Chapter 1 – Scientific notation; transformation of formulae
16
Surveying Computations
Chapter 2 – Trigonometric functions
2.1 Angles
An 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.
A
310°
50°
B
C
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:
50°
or
-310°
∠CBA = 310°
or
-50°
∠ABC =
while
17
Chapter 2 – Trigonometric functions
2.2 Coordinate axes
Typically in surveying, the axis coordinates are shown as 0°→360° in a clockwise rotation,
indicating a positive direction.
0°
360°
N
+ve
270° W
E 90°
S
180°
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.
90°
y
+ve
180° , -x
x 0° , 360°
-y
270°
Fig 2.3 Mathematical axis coordinates
18
Surveying Computations
Where directions of lines are considered as northing and easting coordinates, the surveying axes
are shown like this:
N
+ve
-E
E
-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.
N
+N
-E
+N
+E
-E
E
-N
-E
-N
+E
-N
Fig 2.5
19
Chapter 2 – Trigonometric functions
2.3 Sine and cosine
For trigonometric functions we will use a unit circle with a radius length (L) of 1.
N
eastings
L
northings
P
W
E
0
S
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
N
T
P
L
L
Q
L
0
-E
E
L
L
S
R
-N
Fig 2.7
20
Surveying Computations
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,
Example 2
sin θ =
E
L
and
cos θ =
N
L
( where L = unit length )
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).
21
Chapter 2 – Trigonometric functions
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
adjacent
N1
E1
L nuse
te
po
hy
θ
O
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
sin θ = N1E1
=
N1E1
OE1
=
opposite
hypotenuse
similarly,
cos θ =
adjacent
hypotenuse
where OE1 = L (unit circle)
22
Surveying Computations
2.4 Tangents
Using a unit circle with a radius of 1, we can draw in a tangent line through the coordinates (0,1).
N
P
northing
L
-E
E
O
-E
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.
23
Chapter 2 – Trigonometric functions
N
30
31
0°
°
4
1
-E
E
3
2
22
0°
-N
°
160
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.
24
Surveying Computations
Example 2
Given sin θ = 0.6 and cos θ = 0.8, find the tangent value.
0.75
θ
0.8
0.6
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 θ 0.6
=
= 0.75 = tan θ
cos θ 0.8
therefore
Example 3
sin θ
= tan θ.
cos θ
Using the same method from the previous example we can determine the tangent value given sin 90º.
If
sin 90 = 1 and cos90 = 0
then
tan 90 =
sin 90

cos90
=
1
= undefined or ∞
0
E
L
N
(2) cos θ =
L
sin θ
E
(3)
=
cos θ N
And if (1) sin θ =
and
then
It is important to memorise these three equations.
25
Chapter 2 – Trigonometric functions
2.5 Inverse trigonometric functions
Consider 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
… (1)
0.5’.
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 functions
All 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.
26
Surveying Computations
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 angles
It 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 graph
Let 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 = -1
90°
360°
-ve 270°
180°
90°
+ve
-ve -1
+1 +ve
330°
0
360°
300°
30°
60°
270
90°
240°
120°
210°
180°
150°
Fig 2.12 The sine graph
27
Chapter 2 – Trigonometric functions
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 graph
This is the same as the sine graph but displaced by 90º.
90°
360°
+ve
270°
-ve -180°
90°
-ve -1
0°
+ve
+ve +1
-90
-ve -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).
28
Surveying Computations
Tangent graph
This is dissimilar to the other two graphs as it is discontinuous.
+00
+ve
360°
-ve
-00
270°
+ve
+00
+ve
+00
180°
-ve
-00
90°
0°
-ve
-00
-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).
29
Chapter 2 – Trigonometric functions
Sign of the function
The sign of the function can be seen from Figure 2.15. Let the rotating area be positive θº.
1st Quadrant (θ1 = θ)
+
1st Quadrant 0º–90º
θ1
+
4
+
sin θ1 =
+
+
=+
cos θ1 =
+
+
=+
tan θ1 =
+
+
=+
1
90°
+
−
3
2
−
2nd Quadrant θ2 = (180º − θ)
+
θ
2nd Quadrant 90º–180º
−
−
+
−
4th Quadrant 270º–360º
+
= -
tan θ2 =
+
= -
= -
cos θ3 =
+
= -
−
tan θ3 =
-
=+
+
4th Quadrant θ4 = (360º − θ)
θ3
θ4
cos θ2 =
+
+
−
sin θ3 =
−
+
=+
θ
−
+
+
3rd Quadrant θ3 = (θ − 180º)
+
+
sin θ2 =
+
θ2
+
3rd Quadrant 180º–270º
−
+
θ
sin θ4 =
+
= -
cos θ4 =
+
+
=+
tan θ4 =
+
= -
+
−
−
Fig 2.15
30
Surveying Computations
The previous table is summarised in the figure shown below.
sin θ cos θ +
tan θ -
sin θ +
cos θ +
tan θ +
360°
0°
N
C
A
E 90°
270° -E
T
sin θ cos θ tan θ +
S
-N
180°
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.
31
Chapter 2 – Trigonometric functions
Examples showing the use of sin, cos, tan with correct
signs
Example 1
Evaluate the value(s) of A for which cos A = 0.553452, for the domain
0 £ A £ 360.
This means that the domain of the variable A (the range of A) is to be from
0º to 360º.
360°
0°
5"
3'4
°2
56
°
56
A
C
5"
'4
3
2
270°
90°
S
T
180°
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"
32
Surveying Computations
Example 2
Evaluate angle A (0 £ A £ 180 ) 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°
'
45
7°
"
48
2
A 1
C
90°
270°
T
S
180°
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
= +0.8660
with a +ve sign for the 2nd Quadrant
(b) sin 270º = sin 180º + 90º
= sin 90º
= -1
with a -ve sign for the 3rd Quadrant
(c) tan 315º = tan 360º − 45º
= tan 45º
= -1
with a -ve sign for the 4th Quadrant
33
Chapter 2 – Trigonometric functions
2.7 The reciprocal trigonometric functions
(cosec, sec, cot)
The inverse trigonometric function notation should not be confused with the reciprocal
trigonometric function.
æ 1 ö÷
That is, sin−1 is in no way related to cosec X ççç
÷ .
è sin X ø÷
1
.
In other words, sin−1 X does not mean
sin X
1
Calculators only have sin, cos and tan function keys. Therefore the key must be used to obtain
x
cosec, sec and cot values.
æ1ö
Conversely, when cosec, sec or cot values are given, then the inverse çç ÷÷÷ must be obtained in
çè x ø
order to calculate the angle using a sin, cos or tan function key.
Trigonometric function
Reciprocal
sin
æ 1 ö÷
cosec çç=
çè sin ø÷÷
cos
æ 1 ö÷
s ec çç=
çè cos ÷÷ø
tan
æ 1 ö÷
cot çç=
çè tan ø÷÷
Note: cosec, sec and cot are rarely used now as calculators can easily convert to sin, cos, tan.
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.
34
Surveying Computations
Chapter 3 – Right-angled triangles
3.1 Introduction
We 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 sin
=
adjacent cos
nu
hy
po
te
1
se
northing
easting
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).
35
Chapter 3 – Right-angled triangles
3.2 Sin and cos functions
The 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.
B
c
a
90°
θ
A
C
b
Fig 3.2
In the diagram above:
sin θ =
a
c
cos θ =
b
c
tan θ =
a
b
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 θ =
thus, the ratio of
as would
and
a
c
a
would equal the value of sin θ
c
b
= cos θ
c
a
= tan θ
b
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 θ =
a
c
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
b
similarly, cos θ = , and b = cos θ.
c
36
Surveying Computations
Also, if we knew one side, but didn’t know the hypotenuse, then we would have
sin θ =
a
where a is unknown
c
so by transposition we would get:
c=
a
.
sin θ
In fact, all that is needed to solve a right-angled triangle is one acute angle and one side.
3.3 Tan functions
Ratios of the sides of a triangle are also applied when using the tan function.
Example 1
A
c
B
b
θ
a
C
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,
tan θ =
b opposite
=
a adjacent
If we have values for sides a and b we can apply these to the tan ratio.
Thus if
then
a = 100 cm and b = 256 cm
tan θ =
256
= 2.56 cm .
100
37
Chapter 3 – Right-angled triangles
3.4 Pythagoras’ theorem
Pythagoras’ 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 Areas
The basic formula for area calculation is
1
base × perpendicular height
2
Using the triangle in Fig 3.3 this is:
Area =
a
´b
2
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.
38
Surveying Computations
For the triangle ABC, calculate (a) side AB
(b) ∠A (c) ∠B
(d) the area.
B
253.75 cm
Example 1
A
325.56 cm
C
Fig 3.4
(a)
Using Pythagoras’ theorem we get:
AB 2 = 253.752 + 325.562
AB = 253.752 + 325.562
= 412.77 cm
(b)
Using the tan function we get:
tan A =
253.75
325.56
æ 253.75 ö÷
or ÐA = tan-1 çç
çè 325.56 ø÷÷
= tan−1 (0.779426)
= 37° 56' 02" (c)
(using a calculator)
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"
39
Chapter 3 – Right-angled triangles
(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.
Y
25
.5
12
m
44° 27'
X
Z
Fig 3.5
(a)
Using the sin function we get:
sin 44° 27 ' =
YZ
12.525
∴ 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'
40
Surveying Computations
Example 3
The area of triangle ABC equals 6.0 m2 and the sides AC and CX are 3.0 m and
2.0 m respectively. The line ACX is straight and BX is perpendicular to ACX.
Calculate the side BX.
B
Area ABC = 6.0 m2
90
°
A
C
3.0 m
X
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
1
base ´ perpendicular height
2
3.0
6.0 =
´ BX
2
Area ∆ ABC =
BX =
6.0 ´ 2
3.0
=4m
Example 4
From a point on a horizontal plane, the angle of elevation of the top of the
vertical 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
22.00 m
90
'
22° 41
B
15.00 m
°
C
X
Fig 3.7
41
Chapter 3 – Right-angled triangles
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 =
BC =
=
AC
BC
AC
tan ÐB
22
tan 22° 41'
= 52.64 m
Now
BC = XC + 15
∴
XC = BC − 15
(using a calculator)
C = 52.64 − 15.00
= 37.64 m
In ∆ AXC
æ AC ö÷
tan ∠X = ççç
÷
è XC ø÷
æ AC ö÷
tan ∠X = tan-1 ççç
÷
è XC ø÷
æ 22 ö÷
= tan-1 ççç
÷
è 37.64 ø÷
= 30° 18'
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.
42
Surveying Computations
Chapter 4 – Scalene triangles
4.1 Elements
A triangle has seven elements:
•
three sides
•
three angles
•
area
A
b
c
B
a
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.
43
Chapter 4 – Scalene triangles
A
A
c
b
B
B
a
C
a
C
(i)
(ii)
A
A
b
b
B
a
B
C
(iii)
a
(iv)
Fig 4.2
In this chapter we will look at solving triangles given these combinations.
44
C
Surveying Computations
4.2 Sine rule, cosine rule and the area formula
Sine rule
In any triangle, the sides are proportional to the sines of the opposite angles.
In any triangle ABC
a
b
c
=
=
sin A
sin B
sin C
Proof
a)
For an acute-angled triangle
A
c
B
b
a
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:
c
b
=
sin C
sin B
45
C
Chapter 4 – Scalene triangles
b)
For an obtuse-angled triangle
BCA and ACX are supplementary
 sin BCA  sin ACX
A
c
b
a
B
C
X
Fig 4.4
AX  c sin B and AX  b sin ACX
or
equating
AX  b sin C
c sin B  b sin C
Dividing both sides by sin B, sin C we get:
c
b
=
sin C
sin B
Similarly, by drawing a different altitude, it may be proved that each of
these is equal to:
a
sin A
46
Surveying Computations
Cosine rule
In any triangle ABC
a 2 = b 2 + c 2 - 2bc cos A
or
b2 + c2 - a 2
2bc
cos A =
Proof
a)
For Fig 4.5
A
c
b
B
a
X
C
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
47
Chapter 4 – Scalene triangles
b)
For Fig 4.6
A
c
b
B
a
C
X
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 formula
In any triangle ABC
Area ABC 
1
base  perpendicular height
2
or
Area ABC =
1
perpendicular height  base
2
or
Area ABC 
1
ab sin C
2
48
Surveying Computations
Proof
a)
A
P
A
Q
b
b
B
X
B
C
C
Fig 4.7
X
Fig 4.8
Area rectangle  length (BC)  breadth (AX)

Area triangle 
1
length (BC)  breadth (AX)
2
Area ABC  Area AXB  Area AXC
Area ABC  Area AXB  Area AXC

1
1
AX  BX  AX  XC
2
2

1
1
AX  BX  AX  CX
2
2

1
AX (BX  XC )
2

1
AX (BX  XC )
2

1
AX (BC )
2

1
AX (BC )
2

1
perpendicular height  base
2
1
Therefore, for both triangles Area ABC  base  perpendicular height.
2
Proof
b)
Now AX  b sin C
Substituting for AX we get
Area ABC 
Therefore Area ABC 
1
(b sin C)  BC
2
1
ab sin C
2
49
Chapter 4 – Scalene triangles
4.3 Loss of precision when computing angles
from lengths
Look 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.
1.5
B
A
100
C
Fig 4.9
We can calculate the value of angle A using the tan function:
tan A =
A tan1
opposite
adjacent
1.5
100
= 0' 34"
Similarly
B tan1
 A = tan-1
or
opposite
adjacent
(both are exact values)
to the nearest second
100
1.5
= 89' 26"
to the nearest second
Using Pythagoras’ theorem we can calculate the length of side AB.
c 2 a2b2

1.521002

2.25 10 000
\ c =
10 002.25
\ side AB = 100.0112494
50
Surveying Computations
We can see from this calculation that:
side AB  100.011
to 3 decimal places
 100.0112
to 4 decimal places
 100.01125
to 5 decimal places
 100.011249
to 6 decimal places
 100.0112494 to 7 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:
cos A =
adjacent
hypotenuse
 A = cos-1
or
A cos1
100.000
100.011
= 0' 59"
(to 3dp)
A cos1
100.0000
100.0112
= 0' 57"
(to 4 dp)
A cos1
100.00000
= 0' 34"
100.01125
(to 5dp)
adjacent
hypotenuse
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.
sin A =
opposite
hypotenuse
or  A = sin-1
opposite
hypotenuse
 A sin1
1.50000
= 0' 34"
100.011
(to 3dp)
 A sin1
1.5000
100.01
= 0' 34"
(to 2dp)
 A sin1
1.500
100.0
= 0' 34"
(to 1dp)
51
Chapter 4 – Scalene triangles
Note how accuracy is maintained using three (3), two (2) and one (1) decimal places in the
calculation of angleA.
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 sin1
100.000
100.011
= 89' 01"
B sin1
100.0000
100.0112
= 89' 33"
B sin1
100.00000
= 89' 26"
100.01125
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 cos1
1.50000
= 89' 26"
100.011
B cos1
1.5000
100.01
= 89' 26"
B cos1
1.500
100.0
= 89' 26"
Looking at these calculations we can see that a minimum of one (1) decimal place is required
for accuracy.
Summary
From 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 90using 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 0and 90
As a result, we can make a set of general guidelines for solving triangles which contain very
small angles, say less than 5or angles close to 90
•
When determining small angles, it is best to use the sin or tan functions rather than the
cos function.
•
When determining angles close to 90it is best to use the cos function.
52
Surveying Computations
4.4 Using the formulae
The 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
B
a
c
C
b
A
Fig 4.10
(a) Firstly we must select the best formula to use from those we have looked
at to solve for the missing elements: ie A, B, C.
The formulae are:
(i)
a
b
c
=
=
sin A
sin B
sin C
(sine rule)
(ii)
a2  b2  c2  2bc cos A
(cosine rule)
(iii)
æ b2 + c2 - a 2
ç
A = cos-1 çç
çè
2bc
(cosine rule)
(iii) is the best formula to use, and by using it we can find the value for A.
æ b 2 + c 2 - a 2 ÷ö
ç
÷÷
 A = cos-1 çç
÷÷
2bc
çè
ø
æ 250.002 + 175.002 - 200.002 ö÷
ç
÷÷
 A = cos-1 çç
÷÷
2 ´ 250.00 ´ 175.000
çè
ø
Therefore  A = 52 37' 00"
53
Chapter 4 – Scalene triangles
Using the same formula we can also find the value of B.
æ a 2 + c 2 - b 2 ö÷
ç
÷÷
 B = cos-1 çç
÷
2ac
çè
ø÷
æ 200.002 + 175.002 - 250.002 ö÷
ç
 B = cos-1 çç
÷÷÷
2 ´ 200.00 ´ 175.00
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
2ö
æ
ç 200.00 + 250.00 - 175.00 ÷÷
C = cos-1 çç
÷÷
2 ´ 200.00 ´ 250.00
çè
ø÷
Therefore C = 44' 55"
(b) Check
The three computed angles are now summed as a double-check of our
calculations.

 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"
54
Surveying Computations
(c) The final element we have to find for this triangle is the area.
Area D ABC =
1
ab sin C
2
= 0.5 ´ 200 ´ 250 sin 44' 00"
= 17 382 m 2
= 1.7382 ha
Solve a triangle for the missing elements given two angles and the length of one
side.
A  33 30' 30"
Given
B  70 21' 30"
c  550.00 m
B
.00
m
1' 30"
70° 2
550
Example 2
°
33
'
30
"
33
C
A
Fig 4.11
(a) In this example two angles are given so we can easily calculate the third
one. Whenever two angles of a triangle are given, the third may be
computed by summing the two known angles and subtracting them from
180. However, you should always double-check this calculation.
C  180 A  B)
 180 52' 00")
  08' 00"
55
Chapter 4 – Scalene triangles
(b) We now need to select the best formula to use to solve for the missing
elements: sides a and b.
The formulae are:
(i)
a
b
c
=
=
sin A
sin B
sin C
(sine rule)
(ii)
a2  b2  c2  2bc cos A
(cosine rule)
(iii)
æ b 2 + c 2 - a 2 ÷ö
ç
÷÷
A = cos-1 çç
2bc
÷÷ø
çè
(cosine rule)
For this example, formula (i) is the best to use to find the length of sides
a and b.
a
c
=
sin A
sin C

a =
=
Therefore
c sin A
sin C
550.00 sin 33 30' 30"
sin 76 08' 00"
a = 312.75 m
(For our checks, store the complete calculated value in your calculator NOT
the rounded off value of 312.75.)
b
c
=
sin B
sin C
b =
=
Therefore
c sin B
sin C
550.00 sin 70 21' 30"
sin 76 08' 00"
b = 533.55 m
(For our checks, store the complete calculated value in your calculator NOT
the rounded off value of 533.55.)
56
Surveying Computations
(c) Check
Once the values of the sides have been computed, we then have to doublecheck our calculations.
a
b
c
=
=
sin A
sin B
sin C
a
312.75
=
= 566.526 m
sin A
sin 33 30'30"
b
533.55
=
= 566.514 m
sin B
sin 70 21'30"
c
550.00
=
= 566.510 m
sin C
sin 76 08'00"
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
a
b
and
using your stored values for a and b, all three answers
sin A
sin B
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.
Area D ABC =
1
bc sin A
2
= 0.5 ´ 533.55 ´ 550.00 ´ sin 33' 30"
= 81 001 m 2
= 8.1001 ha
57
Chapter 4 – Scalene triangles
Solve a triangle for the missing elements given the lengths of two sides and the
angle included between them.
Given
a  252.640 m
c  138.760 m
B  54 24' 12"
B
252
.64
4'
°2
54
0m
"
12
138.760 m
Example 3
C
A
Fig 4.12
(a) Firstly, we must select the best formula to use to solve for the missing
elements: side b, A, C.
The formulae are:
(i)
a
b
c
=
=
sin A
sin B
sin C
(ii)
a2  b2  c2  2bc cos A
(iii)
æ b 2 + c 2 - a 2 ÷ö
ç
÷÷
A = cos-1 çç
÷÷
çè
2bc
ø
In this case, we would select formula (ii) for the initial calculation.
Side b can be computed by rearranging this formula to get:
b =
a 2 + c 2 - 2ac cos B
b =
252.6402 + 138.7602 - 2(252.640)(138.760) cos54 24'12"
b = 205.598 m
58
Surveying Computations
(b) Since the lengths of all three sides are now known, angles A and C can be
computed using the cosine rule.
æ b 2 + c 2 - a 2 ö÷
ç
 A = cos-1 çç
÷÷÷
2bc
÷ø
çè
æ 205.5982 + 138.7602 - 252.6402 ÷ö
ç
÷÷
= cos-1 çç
÷÷
2(205.598)(138.760)
çè
ø
Therefore  A = 9218'45"
If we used the sine rule to calculate A we would have found two values.
æa
ö
A = sin-1çç sin  B÷÷÷
çè b
ø
Therefore  A = 87 41' 09" or
9218' 51"
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.
æ a 2 + b 2 - c 2 ö÷
ç
÷÷
C = cos-1 çç
çè
2ab
÷÷ø
æ 252.6402 + 205.5982 - 138.7602 ÷ö
ç
÷÷
= cos-1 çç
çè
2(252.640)(205.598)
÷÷ø
Therefore C = 3717' 03"
(d) Check
To double-check our calculations for the angles we can simply sum all of
them.

 A = 9218' 45"
 B = 5424' 12"
C = 3317' 03"

å =18000' 00"
(e) Now we calculate the area of the triangle.
1
Area D ABC =
ac sin B
2
= 0.5 ´ 252.640 ´ 138.760 ´ sin 54 24'12"
= 14 252.7 m 2
= 1.4253 ha
59
Chapter 4 – Scalene triangles
4.5 Special situations
Ambiguous
A 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.
C
a
(CD = b sin A)
A
X
D
Fig 4.13
If a b sin A, then as a general rule, an ambiguous situation may occur.
C
A  90
a  b
b
a
B'
A
a
B
X
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.
60
Surveying Computations
A special ambiguous case occurs when the circle cuts AX at B and B ' (where B ' falls on A). One
triangle (isosceles) is determined.
C

A  90
a  b A  B
b
a
D
A
(B')
B
X
Fig 4.15
Non-ambiguous
Ambiguous 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 is
no solution. CD is the shortest distance possible from C to AX so CD is at 90 to AX.
C
A  90
a
a  b
b
A
D
X
Fig 4.16
b)
If a  b sin A, then the subscribed circle will touch at D, thus forming a right-angled
triangle with only one solution.
C
A  90
a  b
b
a
A
D
Fig 4.17
61
X
Chapter 4 – Scalene triangles
c)
If the circle cuts AX at B and B ' (where B ' falls on X A produced), then only one triangle is
determined. The triangle AB 'C does not contain the given angle A, only triangle ABC does.
C
A  90
a
a  b
a
b
D
A
B'
B
X
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
C
a
a  b
b
A
X
Fig 4.19
e)
If angle A is obtuse and a  b, then only one triangle is formed since  AB 'C does not
contain A.
C
A  90
a  b
a
b
(B')
A
Fig 4.20
62
B
X
Surveying Computations
4.6 Applications
Solve a triangle for the missing elements given the lengths of the two sides and
an angle opposite one of them.
b  327.00
Given
c  1217.00
C  80 04' 00"
(a) As no diagram is given, construct the triangle to determine if it is likely to
be 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.
.00 cm
Looking at our drawing we can conclude that only one solution exists.
b = 327
Example 1
A
°
80
'
04
"
00
1217
.00 c
m
B
C
X
Fig 4.21
(b) To compute B we will use the sine rule.
b
c
=
sin B
sin C


sin B =
b
sin C
c
æ 327.00
ö
B = sin-1 çç
sin 80 04' 00"÷÷÷
çè1217.00
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" .
63
Chapter 4 – Scalene triangles
This can also be proven by calculating A .
Using value (i)
A180C + B)

18080 04' 00" + 15 20' 49.5")

18095 24' 49.5")

84 35' 10.5"
Using value (ii)
A 18080 04' 00" 164 39' 10.5")
180244 43' 11") which is impossible
Therefore, only one solution exists, A 84 35' 10.5".
(c) Next we will calculate side a by either
a =
b sin A
sin B
a =
327.00 sin 84 35'10.5"
sin15 20'49.5"
or
a =
a  1230.010 cm
c sin A
sin C
1217.00 sin 84 35'10.5"
sin 80 04'00"
or
a =
or
a 1230.010 cm
Thereforea 1230.010 cm
(d) Area is calculated.
Area =
1
ab sin C
2
= 0.5 ´ 1230.010 ´ 327.00 ´ sin 80 04'00"
= 198 091.9 m 2
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'
64
Surveying Computations
(a) Draw the triangle as in the previous example.
A
31.80 m
b=
.00
859
m
B
19° 07 '
C
X
B1
Fig 4.22
Calculate AB'
from AB' = AC sin 197'
= 859.00 sin 1907'
= 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
b
sin C
c
æ 859.00
ö
sin19 07 '÷÷÷
B = sin-1 çç
çè 31.80
ø
sin B =
B = sin-1 (8.85)
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.
A
19° 07
C
m
0c
.00
859
318.000 cm
318.000 cm
' 00"
X
B'
Fig 4.23
65
B
X
Chapter 4 – Scalene triangles
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 B =
b
sin C
c
æ 859.000
ö
B = sin-1 çç
sin19 07 '00"÷÷÷
çè 318.000
ø
Therefore (i)  B = 6212'27"
or
(ii) B = 117 47 '33" (at B').
(c) Compute A.
 A = 180 - ( 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.
a =
b sin A
sin B
c sin A
sin C
or
a =
or
a = 959.903 cm
or
a = 663.355 cm
Using value (i)
a = 959.904 cm
Using value (ii)
a = 663.355 cm
Therefore (i) a = 959.904 cm
or
(ii) a = 663.355 cm
= 959.903 cm
or
= 663.355 cm
66
Surveying Computations
(e) Listed below is a summary of the calculations for the two solutions.
Value (i)
Value (ii)
 B = 6212'27
 B = 117 47 '33"
 A = 98 40'33"
 A = 43 05'27
a = 959.904 cm
(f)
a = 663.355 cm
Compute area.
A
A
c
b
C
B
a
a
C
Fig 4.24


B
Fig 4.25
Using value (i)
Area =
c
b
Using value (ii)
1
ab sin C
2
Area =
0.5 959.904859.00
 sin 1907' 00"
1
ab sin C
2
0.5 663.355 859.00
 sin 1907' 00"
135 018.31m2
93 306.3m2
= 13.5018 ha
= 9.3306 ha
67
Chapter 4 – Scalene triangles
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.
A
1262.30
1.70
138
°1
7'
00
X
m
C
"
m
.00
620
ing
spr
To
m
Fred’s block
57°43' 00 "
68
Large belt of trees
on or near boundary
line BC
Estimated
position of
the spring
Ron’s block
B
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.
68
Surveying Computations
4.7 Solving triangles when area is given
In 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 used
The formula for area commonly used when given two sides and the included angle is
Area 
1
ab sin C
2
The formula used for solving a side, given three angles and the area is
a  sin A
Example 1
2 Area
sin A sin B sin C
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
B
B'
313.
90 m
90 m
2.3875 ha
A
313.
C
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').
69
Chapter 4 – Scalene triangles
(a) Select the best formula to solve for the missing elements.
1
To compute angle C it will be Area  ab sin C.
2
Rearranging the formula we get:
æ 2 Area ö÷ (The area must be converted to square
C = sin-1 çç
çè ab ø÷÷  metres, ie 23 875 m2.)

æ 2 ´ 23 875 ÷ö
÷
= sin-1 çç
çè 313.90 ´ 475.80 ÷÷ø
Therefore (i) C = 18 38'44"
or
(ii) C = 161 21'16":
(b) The triangle now has the lengths of two sides and the angle included
between them. The length of the third side (c) can now be computed
using the cosine rule.
Using (i)
c 2 = a 2 + b 2 - 2ab cos C
c =
(313.90)2 + (475.80)2 - 2(313.90)(475.80) cos18 38'44"
Therefore c = 204.669 m
or
Using (ii)
c =
(313.90)2 + (475.80)2 - 2(313.90)(475.80) cos161 21'16"
Therefore c = 779.710 m .
(c) The triangle now has the lengths of the three sides defined. We will use
the cosine rule to compute angles A and B.
Compute A using (i).
æ b 2 + c 2 - a 2 ö÷
ç
÷÷
 A = cos-1 çç
÷÷
2bc
çè
ø
æ 475.802 + 204.67 2 - 313.902 ÷ö
ç
÷÷
= cos-1 çç
÷÷
2 ´ 475.80 ´ 204.67
çè
ø
Therefore  A = 29 21'47"
or
Compute A using (ii).
æ 475.802 + 779.712 - 313.902 ÷ö
ç
÷÷
= cos-1 çç
÷÷
çè
2 ´ 475.80 ´ 779.71
ø
Therefore A = 7 23'42"
70
Surveying Computations
Compute B using (i).
æ a 2 + c 2 - b 2 ö÷
ç
÷÷
 B = cos-1 çç
çè
2ac
÷÷ø
æ 313.902 + 204.67 2 - 475.802 ÷ö
ç
÷÷
= cos-1 çç
çè
2 ´ 313.90 ´ 204.67
÷÷ø
Therefore  B = 131 59'30"
or
Compute B using (ii).
æ 313.902 + 779.712 - 475.802 ÷ö
ç
÷÷
 B = cos-1 çç
çè
2 ´ 313.90 ´ 779.71
÷÷ø
Therefore B = 1115'01"
(d) Check
Perform a check of the calculations for each solution.
(i)
Example 2
(ii)
A =
29 21'47"
or
7 23'42"
B =
131 59'30"
or
1115'01"
C =
18 38'44"
or
161 21'16"
å =
180 00'01"
or
179 59'59"
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 ha
A
1.5655 ha
Fig 4.28
71
0"
152.76 m
2
5'
°2
65
B
C
Chapter 4 – Scalene triangles
(a) Select the best formula to use to solve for the missing elements.
To compute side b it will be Area 
b=
=
2 Area
a sin C
1
ab sin C.
2
(Area must be converted to m2.)
2 ´ 15655
152.76 sin 65 25'20"
Therefore b = 225.382 m .
(b) We now have the lengths of two sides and the angle included between them. The
cosine rule is now used to compute the length of the third side (c).
c 2 = a 2 + b 2 - 2ab cosC
c =
(152.76) 2 + (225.38) 2 - 2 ´ 152.76 ´ 225.38 ´ cos 65 25'20"
Therefore c = 213.290 m .
(c) The triangle now has the lengths of its three sides defined. We will use the
cosine rule to compute the two remaining angles.
æ b 2 + c 2 - a 2 ö÷
ç
÷÷
 A = cos-1 çç
÷÷
2bc
çè
ø
æ 225.382 + 213.292 - 152.762 ÷ö
ç
÷÷
= cos-1 çç
÷÷
2 ´ 225.38 ´ 213.29
çè
ø
Therefore  A = 40 38'28"
æ b 2 + c 2 - b 2 ö÷
ç
÷÷
 B = cos-1 çç
÷÷
2ac
çè
ø
æ152.762 + 213.292 - 225.382 ÷ö
ç
÷÷
= cos-1 çç
÷÷
2 ´ 152.76 ´ 213.29
çè
ø
Therefore  B = 73 56'12" .
72
Surveying Computations
(d) Check
Perform a check of the calculations.
 A = 40 38'28"
B =
73 56'12"
C = 65 25'20"
å = 180 00'00"
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 ha
A
1.5865 ha
B
30
57°
"
'00
65°
45 '
00 "
C
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 from
180Using the three angles and the area, the sides can then be computed using
the following formula.
a = sin A
2area
sin A sin B sin C
Compute A.
 A = 180 - ( B + C )
= 180 - (12315'00")
Therefore  A = 5645'00"
73
Chapter 4 – Scalene triangles
(b) Compute the three sides.
a = sin A
2area
sin A sin B sin 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.
a = sin 56 45'00" ´ 222.128 = 185.762 m
b = sin 57 30'00" ´ 222.128 = 187.340 m
c = sin 65 45'00" ´ 222.128 = 202.527 m
Therefore a = 185.762 m
b = 187.340 m
c = 202.527 m
(c)
Check
a
sin A
=
=
=
c
sin C
185.76
187.34
=
sin 56 45'00"
sin 57 30'00"
=
202.53
sin 65 45'00"
=
=
222.127
222.127
b
sin B
=
222.127
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.
Answer
Task (page 68)
AX  602.79 metres; therefore, the spring is in Ron’s property.
BC  2356.44
74
Surveying Computations
Chapter 5 – Angles and bearings
5.1 Angles
1. Angle definition
An angle is defined as a measure of the amount of rotation one line makes about a point to
coincide 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
50°
C
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.
75
Chapter 5 – Angles and bearings
2. Angle convention
The 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.
A
-310°
B
50°
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’.
76
Surveying Computations
5.2 Bearings
0°
N
31
5°
°
e
rs
ve ng
re ari
be
290°
50
50°
135°
E 90°
270° W
°
290
13
5°
S 180°
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"
θ BC
θ AB
B
C
A
D
Fig 5.4
77
θCD
Chapter 5 – Angles and bearings
AN, BN ' and CN " are all parallel.
The Greek symbol theta (is used for bearing notation. For the above figure, ABis 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 bearings
If a number of lines of known bearings meet at a point, the angles between these lines may be
computed.
Consider the diagram below.
A
E
310°
50°
88°
260°
B
X
D
190°
C
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.
78
Surveying Computations
Example 1
AXB is the difference in bearings between lines XA and XB.
AXB
BXC
=
XB – XA
=
88 – 50
=
38
=
XC – XB
=
190 – 88
=
102
(final bearing – initial bearing)
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
AXE
=
XC – XA
=
190 – 50
=
140
=
XE – XA
=
310 – 50
=
260
The next example is a bit different because the final bearing appears to be smaller than the
initial bearing.
79
Chapter 5 – Angles and bearings
Example 2
CXB
=
XB – XC
=
88 – 190
=
(88 + 360 – 190 (88is the same as 88 + 360 = 448, one
extra rotation of 360. See Fig 5.6.)
=
448 – 190
=
258
N
0°
88° + 360° = 448°
= 88°
88°
90°
270°
180°
Fig 5.6
Example 3
EXA
=
XA – XE
=
50 – 310
=
(50 + 360 – 310
=
410 – 310
=
100
80
Surveying Computations
Forward and reverse bearings
The 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.
Consider the line AB – from the initial point A to the end point B – shown in Fig 5.7.
N'
N
B'
50
°
0°
= 23
B
θ BA
50
°
00°
Example 1
18
0°
°
= 50
θ AB
23
0°
A
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 '
and
B 'BA 
(BB ' is the continuation
of the line segment AB.)
then N 'BA N 'BB '  
BA  AB  
or


  50  
Therefore BA  
It can be shown that this relationship exists for any line; that is
the reverse bearing is equal to the forward bearing plus 180.
81
Chapter 5 – Angles and bearings
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.
B
A
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 = 30and BAC = 60, then AC = 30 + 60
= 90
82
Surveying Computations
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
X
Y
Z
Fig 5.9
YX = 210110
= 320
Note:
(= forward bearing)
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.
83
Chapter 5 – Angles and bearings
Example 6
Compute the forward bearing GH.
Given FG
=
Angle FGH =
150 (back bearing)
120
F
G
H
Fig 5.10
GH =
Note:
GF + FGH
=
(FG + 180°) + FGH
=
(150° + 180°) + 120°
=
330° + 120°
=
450°
=
90°
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
=
Angle MLN =
30
(back bearing)
120
L
M
N
Fig 5.11
84
Surveying Computations
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
125° 40' 00 "
156° 50 ' 30 "
C
58° 10 '15 "
21
2°
30
'0
0"
50° 00 ' 45 "
A
B
X
Fig 5.12
The bearing from A to X (AX) has been established as 21230'00" (known or assumed). The
bearings of the other lines in the survey have been calculated as shown in the following table.
85
Chapter 5 – Angles and bearings
These calculations are extensions of the calculations in the previous section.
 Bearing
 AX
- BAX


Comment

21230'00"

back bearing away from A (given)

-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)
-156 50' 30"

-6 50' 00"

negative rotation

to normalise
 -YDC 


+360 00' 00"

35310' 00" =  DY
86
Surveying Computations
With practice, this process may be shortened by mental application of the 180 and 360 values.
For example:
 AX

212 30' 00"
- BAX
=
-5810'15"
15419'45" =  AB
+ ABC
=
(+180 - 360)
+50 00'45"
24 20' 30" =  BC
+ BCD
(+180)
=
+125 40' 00"
330 00' 00" = CD
-YDC
(+180)
=
-15650' 30"
35310' 00" =  DY
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.
87
Chapter 5 – Angles and bearings
88
Surveying Computations
Chapter 6 – Distance, line definition
(polar coordinates), point definition,
(rectangular coordinates)
6.1 Distance
Definition of distance
Distance, 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.
B
A
Fig 6.1
If AB is any line, then AB  BA  5.
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.
89
Chapter 6 – Distance, line definition (polar coordinates), point definition, (rectangular coordinates)
6.2 Line definition (polar coordinates)
Mathematics convention for surveying
In 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.
+y
2nd
1st
P1 (x1 , y1)
r1
θ2
θ1
−x
+x 0°
r
2
(0, 0)
3rd
4th
P2 (x2 , y2)
−y
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.
+N
0°
4th
1st
P1(N, E)
L1
θ1
−E
+E
L
3rd
2
θ2
P2(−N, E)
2nd
−N
Fig 6.3
90
Surveying Computations
Line definition in surveying
We 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:
 AB  the bearing from A to B
and
LAB  the distance AB
The line BA is defined as:
 BA  the bearing from B to A
and
LBA  the distance BA
For example, look at Fig 6.4.
N
A
θ 121°
L=
426
.53
m
B
(0, 0)
Fig 6.4
 AB  121
LAB  426.53 m
This is the conventional form for line definition.
Using the same figure we can also see:
 BA  301
LBA  426.53 m
91
Chapter 6 – Distance, line definition (polar coordinates), point definition, (rectangular coordinates)
6.3 Point definition
Mathematics and surveying conventions compared
In 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+
P1(x, y)
y
(0, 0)
x+
x
0
Fig 6.5
In surveying, a point is related to the origin by distances along the N-axis and the E-axis.
N
A(NA, EA)
NA
EA
E
0(0, 0)
Fig 6.6
92
E
Surveying Computations
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
A (NA , EA)
EA
B (NB , EB)
NA
EB
NB
E
N
D
E
ED
EC
D
NC
(ND , ED)
C (NC , EC)
N
Fig 6.7
N
A
(3, 2)
3
2
2
E
E
3
(3, 2)
D
N
Fig 6.8
93
Chapter 6 – Distance, line definition (polar coordinates), point definition, (rectangular coordinates)
6.4 The relationship between line definition and
rectangular coordinates of end points of a line
Rectangular coordinates
As 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.
N
B
L
θ
E
A
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.
B
D
A
C
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.
94
Surveying Computations
Example 1
N
1
B
2
(
(0, 0)Origin
E
2
A
1
1
Fig 6.11
Looking at the above figure we can see that point A has coordinates of:
N A  -2
E A  -1
and point B has coordinates of:
NB   2
Note:
EB   1
The subscript letter denotes the point, and the ‘’ signs are optional.
Differences in rectangular coordinates
As 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).
95
Chapter 6 – Distance, line definition (polar coordinates), point definition, (rectangular coordinates)
In Fig 6.12, let AB be any line, and 0 the origin of the coordinating system.
ΔEAB
B
(NB, EB)
ΔNAB
NB
A (NA, EA)
NA
0
EA
EB
Fig 6.12
The relationship between the two points A and B is such that:
 N AB  N B  N A
 E AB  EB  E A
Difference can be determined using the following formula.
Difference  end value  initial value
or
 X AB  X B  X A
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.
X B  X A   X AB
where A indicates an initial
value, B the end or final value,
and X represents any quantity.
Therefore, using the two points from Fig 6.12 we can see:
N B  N A   N AB
EB  E A   E AB
N A  N B   N BA
E A  EB   N BA
96
Surveying Computations
Example 1
Given the coordinates of K and L, compute NKL and EKL.
L
NL 5397.21
EL 510.19
K
NK 4321.94
EK 292.43
Fig 6.13
 N KL  N L  N K
(end value minus initial value)
 5397.21  4321.94
 1075.27
 EKL  EL  EK
 510.19  292.43
 217.76
Example 2
Given  N AB  -596.215
 E AB  107.005
and the coordinates for B, as shown in Fig 6.14, compute the coordinates of A.
A
B
Fig 6.14
97
NB 100.000
EB 100.000
Chapter 6 – Distance, line definition (polar coordinates), point definition, (rectangular coordinates)
Because we don’t know NA we have to rearrange the formula for difference so
that:
N A  N B   N BA
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
 N AB  -596.215
then
 N BA  596.215
Similarly, if
 E AB  +107.005
then
 EBA  -107.005
These coordinates are illustrated below.
−107.005
+N
A
+596.215
−E
B
+E
A
−596.215
B
−N
+107.005
Fig 6.15
We can now return to the formula and continue with the solution.
N A  N B   N BA
 100.000  (596.215)
 696.215
E A  EB   EBA
 100.000  (-107.005)
 - 7.005
Therefore, the coordinates of A are: N A  696.215
E A  -7.005
98
Surveying Computations
Differences in coordinates as an alternate form of line
definition
We 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).
B
θ
θ
L
L
θ
L
X
C
A
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 coordinates
showing their relationship with each other.
Points A and X, with coordinates given, are related by way of the series of N and E
values, as shown below.
ΔE
B
ΔE
NA
EA
X
ΔN
ΔN
NX
EX
ΔN
A
C
Fig 6.17
99
ΔE
Chapter 6 – Distance, line definition (polar coordinates), point definition, (rectangular coordinates)
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  
E
L
N
L
cos  
0q
N
tan  
E1
P1 (N1, E1)
L
1
N1 T1
E2
90q 270q
E 90q 270q
T2
N 2
E
N
L2
P2 (N2, E2)
r180q
Fig 6.18
Or in the general case, where we consider any coordinate system, they can be:
N
L
cos  
T
'N1
'E1
'E2
'N2
T2
L2
P2(N2, E2)
Fig 6.19
100
E
N
P1(N1, E1)
1
0q
tan  
L
E
L
1
sin  
Surveying Computations
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:
 N AB  LAB cos  AB
 E AB  LAB sin  AB
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 the
coordinates for all points by computing the differences for all defined lines by using the equations:
EB  E A   E AB
Example 1
N B  N A   N AB
and
Compute the coordinates of point Y given the information below.
28"
19'
°
0
26
.321
490
X NX = 790.428
EX = 200.000
Y
Fig 6.20
We will use the formula NY  N X   N XY
and
EY  E X   E XY
NY  N X   N XY
 N X  ( LXY cos  XY )
 790.428  (490.321. cos 260 19' 28")
 790.428  (-82.408)
 708.020
EY  E X   E XY
 E X  ( LXY sin  XY )
 200.005  (490.321. sin 260 19' 28")
 200.005  (-483.346)
 -283.341
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).
101
Chapter 6 – Distance, line definition (polar coordinates), point definition, (rectangular coordinates)
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
A
AB
30
1000
866
120
1000
-500
210
1000
-866
300
1000
500
210
1000
-866
120
1000
-500
210
1000
-866
1366
-500
866
0
0
-866
-500
-1336
-366
-2232
-134
-500
H
Note:
366
866
G
GH
500
-500
F
FG
866
-866
E
EF
0
-500
D
DE
0
866
C
CD
E
500
B
BC
N
Points A and E, having the same coordinates, must occupy the same position, ie both
points are coincidental.
Close tables like the one above will be used again in the next chapter.
102
Surveying Computations
Line definition from differences in coordinates
The 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:
 N AB  N B  N A
 E AB  EB  E A
By using the relationships shown in Fig 6.12, we find that LAB is the hypotenuse of a rightangled triangle formed from  N AB and  E AB .
 LAB 
 N 2 AB   E 2 AB
and  AB (the bearing of the line AB) is given by the tan function between  E AB and  N AB .
1   E AB 

  AB  tan 
  N AB 
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
E
bearing. You should get the same result as your value from
.
N
Further manipulation of the formulae yields other solutions.
To summarise:
From
 N AB  LAB cos  AB
and
 E AB  LAB sin  AB
then
LAB 
 N 2 AB   E 2 AB 
 E AB
 N AB

sin  AB
cos  AB
  E AB 
 AB  tan 1 

  N AB 
  E AB 
1   N AB 
 sin 1 
  cos 

 LAB 
 LAB 
103
Chapter 6 – Distance, line definition (polar coordinates), point definition, (rectangular coordinates)
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.
 NCF  N F  NC
  -866    366 
 -1232
 ECF  EF  EC
  -500    1366 
 -1866

 LCF 

 N 2CF +  E 2CF
 -1232 2   -1866 2
 2236
+
  ECF 
CF  tan 1 

  NCF 
+
 56 33'57"
34
 -1866 
 tan 1 
  1.514610
 -1232 
'
−
56
°
(b)
−
Fig 6.21
(Because both E and N are negative means that this bearing falls in the
third quadrant.)

CF  5633'57"  180
 23633'57"


Check this answer by finding the tan value of 23633' 57". You should get
1.514614.
104
Surveying Computations
Example 2
Given
N A  106.132
E A  -11.063
N B  -110.296
EB 
Calculate,
(a) LAB
and
103.000
(b)  AB
Remember that AB signifies that B is the final point and A the initial point.
(a)
 N AB  N B  N A
  -110.296   106.132
 -216.428
 E AB  EB  E A
 103.000   -11.063
 +114.063

LAB 

 N 2 AB   E 2 AB
 -216.4282
 114.0632
 244.646
(b)
 AB  tan 1
 E AB
 N AB
 114.063 
 tan 1 
  -0.527025
 -216.428 
 -2747 '25"
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 15212' 35". You
should get -0.527023.
105
−
27°47 ' 25 "
Fig 6.22
Chapter 6 – Distance, line definition (polar coordinates), point definition, (rectangular coordinates)
6.5 Calculators: use of rectangular → polar
(R → P) function key
Efficiencies 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.
Y
x
r
(r, θ)
y
θ
X
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 E
values. These may also be shown in matrix form. You will need
to know your calculator and understand whether N or E is
displayed first.
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.
106
Surveying Computations
Chapter 7 – Vectors
In surveying a line is defined by a length and a direction.
B
A
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 rectilinear
→
motion (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.
B
B
A
A
X
P
(i)
Z
B
(iii)
A
Fig 7.2
107
Q
(ii)
Chapter 7 – Vectors
For each example, the sum of the component vectors is equal to the total vector between A and B.
That is:
→
→
→
AX  XB  AB
→
→
→
→
(ii) AP  PQ  QB  AB
→
→
→
(iii) AZ  ZB  AB
(i)
→
In each case, the total vector AB of a series of component vectors is indicated by the first letter
of 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 vector
As 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
B
A
C
D
X
+E
Fig 7.3
108
Surveying Computations
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
C
D
A
X
Fig 7.4
Using Fig 7.4 as an example we can show the total horizontal and vertical movements through
the following calculations:
 N AB   N BC   NCD   N DX   N AX
 E AB   EBC   ECD   EDX
  E AX
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.
109
Chapter 7 – Vectors
7.2 Vectors in surveying – summing vectors to
prove a survey
The 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
C
A
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
X
C
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.
110
Surveying Computations
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.
3.5
2.5
B
112
q
2.0
60q
1.0
2.7
m
.0 m
C
4
3.2 m
31
0q
3.
9
-2.5
3.0
162q
Values are not to scale and are only
intended to illustrate concepts.
A
1.0
m
D
263q
E
3.0
4.0 m
4.0
0.5
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:
→
→  CD
→  DE
→  EA
→  0
AB  BC
or
 N AB   N BC   N CD   N DE   N EA  0
 E AB   EBC   ECD   EDE   EEA  0
111
Chapter 7 – Vectors
Example 2
This is also a closed traverse but the start and end points are not the same.
3.5
2.5
B
112
q
2.0
60q
1.0
2.7
m
.0 m
C
4
3.0
162q
Values are not to scale and are only
intended to illustrate concepts.
A
1.0
3.2 m
-2.5
D
263q
E
3.0
4.0 m
4.0
0.5
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
 N AB   N BC   N CD   N DE   N AE  -2.5
 E AB   EBC   ECD   EDE   E AE  3.0
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.
112
Surveying Computations
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.
113
Chapter 7 – Vectors
114
Surveying Computations
Chapter 8 – Adjustment of misclose in
closed traverses
8.1 Traverse surveying
A traverse is a series of related points or stations connected together by angles and distances.
Types of traverses
Traverses are classified as either closed or open. Look at the examples below.
E
B
A
A
D
a)
C
C
T1
b)
B
T1
D
C
T2
B
T3
D
C
T2
A
c)
d)
T4
B
A
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.
115
Chapter 8 – Adjustment of misclose in closed traverses
The closed traverse
If 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 traverse
An 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 closure
It 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
θ
A
NA
EA
B
NB
EB
Fig 8.2
For this type of traverse the following is true:
angular misclose  0
N A    N  NB  0
E A    E  EB  0
Where:
A and B are the initial and final
stations of the traverse.
∑∆ N and ∑∆ E are the sums of
the differences in coordinate values
for the distances between A and B.
116
Surveying Computations
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
N  0
 E  0
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 Misclose
If 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.
117
Chapter 8 – Adjustment of misclose in closed traverses
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
B
A
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.
118
Surveying Computations
8.4 Angular misclose
For a closed polygon
B
A
C
D
F
E
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
number of stations)
or
(where n is the
int. angles  (n 180
We can use Fig 8.7 as an example to show the computation, using the above formula.
int. angles  (2n 490

26490

12490

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.
119
Chapter 8 – Adjustment of misclose in closed traverses
Between coordinated points
θA
A1
A4
A
A2
A5
θB
B
A3
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 .
mA   B (calculated)   B (known)
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.
120
Surveying Computations
Angular adjustment
If 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:
m 
cA  -  A 
 k 
Where:
mA is the angular misclose
k is the number of angles
The correction is applied algebraically to the observed angles.
Calculate the angular misclose of ABCDE and the adjusted bearings of the lines,
assuming that the bearing from C to D is 18000' 00".
0'32
"
B
92q1
A
"
6'52
1q2
91q 08'14"
0
7q1
C
13
'56
"
88
q0
13
E
4'1
9"
180q0' 00 "
Example 1
D
Fig 8.9
(a) We firstly calculate the angular misclose.
The internal angles of a closed loop traverse should equal to (2n  4)  90º ,
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 this
traverse.
121
Chapter 8 – Adjustment of misclose in closed traverses
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, under
the 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
D
E
13126'52"  11"  131 26' 41"
8804'19"  10" 
88 04' 09"
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.
122
Surveying Computations
(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 of
CD  180' 00" and add the angles to calculate the bearings of the
lines.
 CD

18000'00"
  DC

36000'00"
EDC
  8804'09"
 DE

27155'51"
 ED

9155'51"
 ED
  13710'45"
 EA

31445'06"
 AE

13445'06"
BAE
  9108'03"
 AB

4337 '03"
 BA

22337 '03"
CBA
  9210'22"
 BC

13126'41"
 CB

31126'41"
DCB
 CD
or 45155'51"
  13126'41"
 18000'00"
check
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.
123
Chapter 8 – Adjustment of misclose in closed traverses
 CD
DCB

180 00' 00"
 131 26' 41"
 CB

311 26' 41"
 BC

131 26' 41"
CBN

 92 10' 22"
 BA

223 37 ' 03"
 AB

43 37 ' 03"
BAE

 91 08' 03"
 AE

134 45' 06"
 EA

314 45' 06"
AED
 137 10' 45"
 ED

451 55' 51"
 DE


91 55' 51"
271 55' 51"
EDC

 88 04' 09"
 DC
 360 00' 00"
check
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.
124
Surveying Computations
Alternative method by bearings
Unadjusted bearings
 CD
18000'00"
 DC
36000'00"
EDC
 8804'19"
 DE
27155'41"
 ED
45155'41"
 ED
Correction
0
Adjusted
bearings
18000'00"
11"
27155'51"
21"
31445'06"
32"
43037 '03"
42"
13126'41"
53"
18000'00"
 13710'56"
 EA
31444'45"
 AE
13444'45"
BAE
 9108'14"
 AB
4336'31"
 BA
22336'31"
CBA
 9210'32"
 BC
13125'59"
 CB
31125'59"
DCB
 13126'52"
 CD
calculated
17959'07"
 CD
known
18000'00"
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.
125
Chapter 8 – Adjustment of misclose in closed traverses
Example 2
The traverse ABCD has been run between the known points A and D. θ DY
and θ AX are known. Compute the misclose in angles between X and Y and the
adjusted bearings between A and D.
7°
10
'30
"
0
°5
5"
°5
0
'2
A
13
0°
20
'10
"
14
11
4
64°00'40"
C
B
115
°
D
30 '
00"
X
1"
'1
7
22
Y
Fig 8.10
The angular (bearing) misclose is obtained by comparing the given bearings
using unadjusted angles. Each bearing is then adjusted by an amount which
reduces the traverse error between successive lines.
Unadjusted bearings
θ AX
115°30'00"
∠XAB
+130°20'10"
θ AB
245°50'10"
θ BA
∠CBA
425°50'10"
−114°50'25"
θ BC
310°59'45"
θ CB
130°59'45"
∠BCD
64°00'40"
θ CD
195°00'25"
θ DC
375°00'25"
∠YDC
−147°10'30"
θ DY
227°46'55" calculated
θ DY
227°50'11" given
126
Correction
Adjusted
bearings
0
115°30'00"
+4
245°50'14"
+8
310°59'53"
+12
195°00'37"
+16
227°50'11"
Surveying Computations
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:
13020'14"
11450'21"
6400'44"
14710'26"
8.5 Linear misclose and adjustment
The total linear misclose is derived from the misclose in the northing and easting values. Let
these be represented by mL , mN and mE respectively. Then:
mN  N A    N  N B
mE  E A    E  E B
2
 mE2 (from Pythagoras' theorem)
mN
mL 
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.
mR  1:
L
mL
Where:
mR
 the rate of error
L
 the sum of the distances in the traverse
127
Chapter 8 – Adjustment of misclose in closed traverses
The traverse ABCD has been measured between the known points A and D
(Fig 8.11).
7q
10
'30
"
A
NA
EA
422.60 m
'10
"
m
.76
m
64q00'40"
.85
183
14
11
4q
50
'2
5"
C
345
0q
20
B
13
Example 1
115
q
30 '
00"
X
D
27
0
q5
2
1"
'1
ND
ED
Y
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 can
show the misclose for this below.
C
-0.09
mE
D
m
L
0.
11
m
Y
Fig 8.12
128
mN
D'
0.06
Surveying Computations
Angles and distances have been measured and are shown in Fig 8.13.
C
148q
.270
10 ' 5
287
366.230 m
36q 2
0 ' 00 "
9.8
20
30
"
.06
B
70q 49' 5
56
'
00
0m
0"
A
q
41
m
512
Example 2
m
5"
D
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
mN
mE
m
L
m
-
D'
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.
129
Chapter 8 – Adjustment of misclose in closed traverses
Bowditch adjustment method
There 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 method
With this method the adjustment of the misclose is performed on the calculated coordinates of
the points in the traverse.
The formulae are:
cN  
mN  L accum
L
cE  
mE  L accum
L
Where:
cE = correction to the eastings
Where:
cN = correction to the northings
mN = misclose in the northings
mE = misclose in the eastings
L accum = accumulated length
L accum = accumulated length
 L = total length of traverse
 L = total length of traverse
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.
130
Surveying Computations
Look at the traverse below and the following data. Compute adjusted coordinates
for A, B and C.
Given: N D = 1000.000
C
ED = 1000.000
B
"
' 54 m
39 .230
q
53 66
3
A
512.060 m
160q 50' 02 "
21
q5
28 0' 3
7.2 8 "
70
m
Example 1
D
569.820 m
270q 00' 00"
Fig 8.15
Use a close table to calculate unadjusted coordinates.
Line
Bearing
()
Distance
(L)
Accum.
distance
N
E
D
DA
27000'00"
569.820
A
AB
5339'54"
CD
1223.320
16050'02"
1000.000
1000.000
430.180
1216.993
725.203
1483.637
832.090
999.960
1000.203
-483.677 168.113
512.060
D
1000.000
266.644 106.887
287.270
C
E
216.993 295.023
366.230
936.050
2150'38"
N
-569.820
569.820
B
BC
0.00
Unadjusted
coordinates
1735.380
  1735.380
131
Chapter 8 – Adjustment of misclose in closed traverses
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 


Note:
 - 0.203
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
cN A =
=
cN  LDA
L
0.040  569.820
1735.380
= 0.013
Point B
cN B =
cN   L (D  B)
L
=
cN  (LDA  LAB )
L
=
0.040  (569.820  366.230)
1735.380
=
0.040  (936.050)
1735.380
= 0.022
132
Surveying Computations
Point C
cNC =
cN   L (D  C)
L
=
cN  (LDA  LAB  LBC )
L
=
0.040  (569.820  366.230  287.270)
1735.380
=
0.040  (1223.320)
1735.380
= 0.028
Point D
cN D =
=
cN   L
L
0.040  (1735.380)
1735.380
= 0.040
(c) The eastings are calculated in the same way.
Point A
cE A =
=
cE  LDA
L
-0.203  569.82
1735.380
= -0.067
Point B
cEB =
cE   L (D  B)
L
=
-0.203  (569.820  366.23)
1735.380
=
-0.203  (936.05)
1735.380
= -0.109
133
Chapter 8 – Adjustment of misclose in closed traverses
Point C
cEC =
cE   L (D  C)
L
=
-0.203  (569.820  366.230  287.270
1735.380
=
-0.203  1223.320
1735.380
= -0.143
Point D
cED =
=
cE   L
L
-0.203(1735.380)
1735.380
= -0.203
A summary of the calculations we have done can be added to the table to
show final adjusted coordinates.
Line
Bearing
Accum.
Distance (L)
()
distance
N
E
N
D
569.820
A
0.00
53º39'54"
366.230
216.993
287.270
C
CD 160º50'02"
266.644
D
-483.677
0
cE
0
N
E
1000.000 1000.000
1000.000 430.180 +0.013 -0.067
1000.013 430.113
1216.993 725.203 +0.022 -0.109
1217.015 725.094
1483.637 832.090 +0.028 -0.143
1483.665 831.947
999.960 1000.203 +0.040 -0.203
1000.000 1000.000
106.887
1223.320
512.060
cN
Adjusted
coordinates
295.023
936.050
21º50'38"
Corrections
-569.820
569.820
B
BC
E
1000.000 1000.000
DA 270º00'00"
AB
Unadjusted
coordinates
168.113
1735.380
=

= 1735.380
134
Surveying Computations
Partial coordinates method
With this method the corrections are proportional to the length of the line.
The formulae are:
c N  
mN  L
L
c E  
mE  L
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.
C
B
"
54 m
9' 30
3
q 2
53 66.
3
A
512.060 m
160q 50' 02 "
21
q5
28 0' 3
7.2 8 "
70
m

Given: ND  
ED  
D
569.820 m
270q 00' 00"
Fig 8.16
Line
Bearing (θ)
Distance (L)
DA
27000'00"
569.820
0.0000
AB
5339'54"
366.230
 216.9932
 295.0226
BC
2150'38"
287.270
 266.6443
 106.8871
CD
16050'02"
512.060
- 483.6769
 168.1134
 1735.380
 - 0.0394
 + 0.2031
Note:
∆N
∆E
- 569.820
N, E are written to four decimal places (4 dp) to minimise rounding
off errors.
135
Chapter 8 – Adjustment of misclose in closed traverses
(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.)
m2 N + m2E
Linear misclose (mL ) =
(-0.0394) 2 + (0.2031) 2
=
= 0.2069
Rate of error (mR ) =
1: ∑ L
mL
= 1:
1735.38
0.2069
= 1: 8387
Note: For this calculation it is best to use the computed (stored) value
of mL in the calculator.
(b) Now we can calculate the adjustments for the northings.
m × L
c∆ N =N
∑L
 −0.039(569) 
c∆ N DA =
−
0.0129
 =
 1735.38 
 −0.039(366) 
c∆ N AB =
−
0.0083
 =
 1735.38 
 −0.039(287) 
c∆ N BC =
−
0.0065
 =
 1735.38 
 −0.039(512) 
c∆ NCD =
−
0.0116
 =
 1735.38 
∑ =0.0393 Check
136
Surveying Computations
(c) The eastings adjustments are also calculated.
c∆ E
c∆ E DA
mE × L
=
∑L
 +0.203(569) 
=
−
− 0.0666
 =
 1735.38 
 +0.203(366) 
c∆ E AB =
−
− 0.0429
 =
 1735.38 
 +0.203(287) 
c∆ E BC =
−
− 0.0336
 =
 1735.38 
 +0.203(512) 
c∆ E CD =
−
− 0.0600
 =
 1735.38 
∑ = − 0.2031 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
D
DA
A
AB
B
BC
C
CD
D
Bearing
(θ)
Distance
(L)
∆ N
∆ E
Corrections
cN
cE
270º00'00"
569.820
0.00
-569.820
0.0129
-0.0666
53º39'54"
366.230
216.9932
+295.0226
0.0083
-0.0429
21º50'38"
287.270
266.6443
+106.8871
0.0065
-0.0336
160º50'02"
512.060
∑ = 1735.38
-483.6769
∑ = -0.0394
+168.1134
0.0116
Adjusted
coordinates
N
E
1000.000
1000.000
1000.013
430.113
1217.014
725.093
1483.665
831.947
1000.000
1000.000
-0.0600
∑ = 0.2031 ∑ = 0.0393 ∑ = -0.2031
Note: 1. The adjusted coordinates are calculated from
coordinate + DN + cN , eg NA=ND + DNDA + cDA
= 1000.000 + 0 + 0.0129
=1000.013
coordinate + DE + cE ,
eg EA=ED + DNDA + cDA
= 1000.000 – 569.820 – 0.0665
=430.113
2. DN , DE , cN , cE values are all written to 4 dp to minimise rounding off errors. The
final adjusted coordinates are rounded back to 3 dp.
137
Chapter 8 – Adjustment of misclose in closed traverses
Example 2
Linear adjustment between coordinated points
Look at the following open traverse and the accompanying data.
C
7q
22
'
50
25
"
50
'
'1
0"
11
4q
' 30
"
20
B
115
q
0q
10
A
m
422.600
30 '
13
7q
345
.760
m
64q 00' 40"
0m
.85
183
14
871.650
NA =
EA = 10 060.080
D
"
11
00"
X
ND = 485.260
ED = 9 446.290
Y
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.
N
E
Line
Bearing ()
AB
24550'14"
422.600
- 172.983
- 385.574
BC
31059'53"
183.850
+ 120.612
- 138.757
CD
19500'37"
345.760
- 333.962
- 89.549
 = 950.000
 = - 386.333
 = - 613.880
Distance (L)
(a) The difference in the coordinates from A to D are computed.
ND  N A 

ED  E A
485.260  871.650
-386.390
 9 446.290  10 060.080

-613.790
138
Surveying Computations
(b) We now use the formula looked at on page 116 in N and E:
N A    N  ND  0
E A    E  ED  0
to calculate the misclose in N and E.
mN  N A    N  N D
 871.650  (-386.333)  485.260
  0.057
mE  E A    E  ED
 10 060.008  (-613.880)  9446.290
 -0.090
(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.
c N

- mN  L
L
c N AB 
-(0.057)(422)
 -0.025
950
c N BC 
-(0.057)(183)
 -0.011
950
c NCD 
-(0.057)(345)
 -0.021
950
 
139
0.057
check
Chapter 8 – Adjustment of misclose in closed traverses
c E

-mE  L
L
 -0.090 
c E AB  - 
 (422)  0.040
 950 
 -0.090 
c EBC  - 
 (183)  0.017
 950 
 -0.090 
c ECD  - 
 (345)  0.033
 950 
  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)
Adjusted
N
Adjusted
E
A
AB
E
871.650 10 006.080
245 50' 14" 422.600
-173.008
-385.534
B
BC
N
698.642 9 674.546
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
140
Surveying Computations
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.
141
Chapter 8 – Adjustment of misclose in closed traverses
142
Surveying Computations
Chapter 9 – Areas of polygons
9.1 Areas of regular rectilinear figures
Rectilinear figures are bounded by straight lines and conform to specific geometrical shapes, eg
triangle, square.
The square and the rectangle
For these figures the formula is:
Area = L  B
B
C
B
C
A
D
A
D
Area = AB  BC
Area = AB  BC
Fig 9.1
The triangle
For these figures the formula is:
Area 
1
base  perpendicular height
2
B
A
Area 
B
C
D
A
1
AC BA
2
Area 
Fig 9.2
143
1
AC  BD
2
C
Chapter 9 – Areas of polygons
For a scalene triangle the formula is:
1
ab sin C
2
Area 
B
c
a
A
b
C
Fig 9.3
Area =
1
a b sin C
2
Area =
1 a 2 sin B ´ sin C
2
sin A
Area =
where s =
s ( s - a )( s - b)( s - c )
1
(a + b + c)
2
The parallelogram
The formula is:
Area  a b  sin A
B
o
a
A
Area =
C
b
D
1
AC ´ BD ´ sin BOC
2
Fig 9.4
144
Surveying Computations
The trapezium
A trapezium has two parallel sides and the angles and the sides may all be different.
The formula is:
Area 
1
∑ parallel sides  perpendicular height
2
B
C
h
A
D
AD + BC
´ h
2
Area =
Fig 9.5
For the figure above, A + B = 180
C + D = 180
and
However, if EF is parallel to AD and BC and midway between them (as shown in the figure
below)
C
B
h
2
F
E
h
2
D
A
Area = EF ´ h
Fig 9.6
then
EF =
AD + BC

2
145
Chapter 9 – Areas of polygons
9.2 Areas of polygons
These 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 coordinates
Once 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 )
EC
C(NCEC )
A (NAEA )
EA
NB
NC
NA
ED
D(NDED )
ND
Fig 9.7
146
Surveying Computations
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.
B
B
C
A
C
A
2
1
3
D
a
b
d
c
a
b
D
4
d
c
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 =
1
parallel sides perpendicular height
2
(N A + NB )
´ ( EB - E A )
or Area (aABb) =
2
or 2Area (aABb)  (NA  NB) (EB  EA)  (NA  NB)(EB  EA)
This rearrangement of the formula can then be applied to the remaining trapeziums
so 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
147
Chapter 9 – Areas of polygons
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.
–
NA
EA
NB
EB
NC
–
–
+
EC
ND
ED
NA
EA
–
+
+
+ (repeated)
Fig 9.9
Then 2(Area) is the algebraic sum of the products NAEB , NBEC , etc, and the
products -NBEA , -NCEB , etc.
Task
Find the area of the polygon ABCD (Fig 9.10), given the coordinates of A and the line information
shown.
N
B
21
q
28 ' 3
8.2 9"
30
m
127
q
278 ' 49
.16 "
0m
C
NA = 450.650
32
3q
5"
m
'4
30
0
2.
35
EA = 249.630
m
196q
' 3
4"
A
D
0
(0,0)
Fig 9.10
148
E
Surveying Computations
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.
149
Chapter 9 – Areas of polygons
Answers
Task (page 148)
Find the area of the polygon ABCD (Fig 9.10), given the coordinates and the tabled data.
N
NB = 719.626
B
127
EB = 353.208
21
q
28 ' 3
8.2 9"
30
q
278 ' 49
.16 "
0
C
A
NC = 548.385
NA = 450.650
3q
'4
0
03
2.
32
35
EA = 249.630
196q
' 3
4"
EC = 572.410
5"
ND = 168.415
D
0
ED = 460.030
E
(0,0)
Line
Bearing
()
N
Distance
(L)
E
A
N
E
450.650 249.630
AB
2103' 39"
288.230
268.976
-193 693.97 m2
103.578
B
719.626 353.208
BC
12759' 49"
278.160 -171.241
-96 402.43 m2
-207 312.06 m2
219.202
C
-179 640.24 m2
548.385 572.410
CD
19628' 34"
159 173.18 m2
396.240 -379.970 -112.380
D
168.415 460.030
DA
32317' 45"
352.030
252 273.55 m2
283.234 -210.403
A
450.649 249.627
= 0.00 

= 0.00
411 921.12 m2
42 040.93 m2
 = 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.
150
Surveying Computations
Chapter 10 – Missing measurements
10.1 Summing vectors to prove a survey
In 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.
3.5
2.5
B
112
q
2.0
60q
1.0
2.7
m
m
4.0
C
3.2 m
31
0q
3.
9
-2.5
3.0
162q
Values are not to scale and are only
intended to illustrate concepts.
A
1.0
m
D
3.0
E
263q
4.0 m
4.0
0.5
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
151
Chapter 10 – Missing measurements
10.2 The sum, difference and translation of vectors
A 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).
B
B
X
AX
XB
A
X
A
+
Fig 10.2
X
Fig 10.3
The line joining A and B represents the vector AX + XB. That is:
→
→
→
AX  XB  AB thesum or resultant vector)
+X
B
The difference of the vectors AX and XB is obtained by adding (- XB) to AX.
A
X
−
-XB
AX
XB
B
Fig 10.4
152
Surveying Computations
10.3 Closing vectors
A 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.
→
BA closing vector
Therefore,
B
A
X
Fig 10.5
For example, in Fig 10.6:
C
→ → → → →
AB  BC  CD  DE  EA 0.
B
The vector EA is undefined so,
→
→
→
→
→
AB  BC  CD  DE - EA
→
AE = ‘sum vector’
D
A
→
EA ‘closing vector’
and
Therefore, LEA  LAE
E
and EA AE 80
Fig 10.6
10.4 Partially defined vectors
Partially 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
00'
70°
m
.00
120
B
ng
issi
m
n
m
ctio
dire 175.00
1
leng 25° 00
th m '
issin
g
C
Fig 10.7
153
D
Chapter 10 – Missing measurements
10.5 Missing vectors
Taking 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.
N

E

Line
Bearing (θ
Distance (L)
AB
60
4.0
2.0
3.5
BC
112
2.7
- 1.0
2.5
DE
263
4.0
- 0.5
- 4.0
EA
310
3.9
2.5
- 3.0
 = 3.0
 = - 1.0
CD
(a) To calculate the bearing of the missing side.
We must remember that the sum of all the vectors equals zero.
→
→
→
→
→
→
→
→
→ -CD
→
AB  BC  CD  DE  EA

0
AB  BC  DE  EA
N  -NCD
That is
E  -ECD
-N  -NCD
or
-E  -ECD
-3.0 = NCD
That is
1.0 = ECD
æ DE ÷ö
These values are then used in the formula q = tan-1 çç
to calculate
çè DN ÷÷ø
the bearing.
CD
æ +1.0 ÷ö
÷÷
çè -3.0 ÷÷ø
= tan-1 ççç
= -18
= 180 - 18 (as the bearing must lie in the second quadrant)
= 162
154
Surveying Computations
(b) To calculate the length of the missing side.
LCD =
2
2
DN CD + DE CD
LCD =
(-3.0)2
+
(1.0)2 = 3.2
therefore, the solution is CD = 162
LCD = 3.2
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)
AB
60º
4.0
BC
∆N
2.0
3.5
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.
→
→
→
→
→ 0
→  DE
→  EA
→ - BC + CD)
AB
→ →
or - N  BC  CD
AB  BC  CD  DE  EA

→
→
and - E  BC  CD
→
→
→
→
- 4.0  NBC NCD
That is, - 4.0  BC  CD
and
∆E
3.5  BC  CD
or
155
3.5 EBC ECD
Chapter 10 – Missing measurements
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
m
5.3 9q
13
-4
æ +3.5 ö÷
÷
 = tan 1 çç
çè - 4.0 ø÷÷
= 139
L
D
3.5
=
(- 4.0)2
+
(3.5)2
= 5.3 m
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
2.7 m
m
162q
5.3
B
q D
ĺ
C
ĺC -4
C
q
Fig 10.9
156
D
Surveying Computations
This triangle is now solved for the LCD ( 3.2 m) and BC ( 112).
To calculate angle C, use the sine rule.
BC
BD
=
sin BDC
sin DCB
2.7
5.3
=
sin 23
sin C
æ 5.3sin 23 ÷ö
C = sin-1 çç
çè 2.7 ø÷÷
= 50 or 130
(obtuse angle adopted as per Fig 10.1)
B = 180 - (130 + 23)
= 27
 BC = 139 - 27 = 112
Use the sine rule again to calculate CD.
CD
BD
=
sin CBD
sin DCB
CD
5.3
=
sin 27
sin130
CD =
5.3sin 27
sin130
= 3.2 m
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.
157
Chapter 10 – Missing measurements
10.6 Calculating missing measurements
A 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 measurements
The 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.
158
Surveying Computations
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.)
C
674
.
405
B
116q 45'05"
m
0.
126
46
9
m
77q52'57" D
348.625 m
"
A
' 43
q 41
63
Example 1
E
Fig 10.10
To start this calculation you will have to know the bearings of the lines
BC, 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.
Line
Bearing (θ)
Distance (L)
∆N
∆E
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
 = - 512.319
 = + 84.277
AB
159
Chapter 10 – Missing measurements
Therefore,
 N AB = +512.319
It is important to note
the sign change.
 EAB = -84.277
The computed values for N and E- form the rectangular
components for missing line AB.
æ -84.277 ÷ö
÷
çè +512.319 ÷÷ø
 AB = tan-1 ççç
2
(84.277)2 + (512.319) = 519.205 m
LAB =
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.)
674
B
.
405
11
3
°2
9' 4
5
m
116°45' 05"
C
59
8.8
02
'3
10
5°
69
4
0.
63
D
m
43"
A
m
77° 52' 57"
126° 41'
0"
"
10
Example 2
= 35039'30"
E
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.
160
Surveying Computations
Line
Bearing (θ)
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, ∆ N
Distance (L)
∆ N EA = +547.752
∆ EEA = - 431.097
=
-1  ∆E 
=
tan


 ∆N 
321°47 '46"
LEA=
∆E2 + ∆N 2
θ EA
=
 - 431.097 
tan -1 

 547.752 
431.097 2 + 547.7522
=
= 697.049 m
The triangle to solve now is shown below.
B
1°
32
"
350° 39' 30
49
6"
7.0
'4
47
69
A
275°50' 0
0"
E
Fig 10.12
Solving this triangle using the sine rule gives the missing lengths:
AB = 519.205 m
EA = 348.625 m
161
∆ E
Chapter 10 – Missing measurements
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
B
B
519.205
A
519.205
348.625
A
348.625
E
Fig 10.13
E
Fig 10.14
As you do not know which solution is correct, you have to solve for both.
162
Surveying Computations
28q' 44"
B
105q' 30"
28q' 44"
A
6"
'4
m
1q
49
32
7.0
69
105q' 30"
45q' 46"
A
E
45q' 46"
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.
Missing non-adjacent measurements
For this example we have to find the bearing of one line, and the distance of
another line.
"
' 45
B
C
57q 74 m
.6
405
12
A
0q
59
8.8
02
' 40
"
m
D
350q ' 30"
Example 4
m
275q ' 00
"
Fig 10.16
163
E
m
Chapter 10 – Missing measurements
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
35.433
- 346.820
DE
EA
630.469
348.625
275º 50' 00"
 = - 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
NAB + NDE – (-47.701) = 0
therefore
EAB + EDE – 510.448 = 0
and
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.
(AB + DE)
Fig 10.17
164
A
47.701
-510.448
E
Surveying Computations
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.
(B, D)
m
I
D
69
' 30"
63
A
m
350q 0"
E
0.4
0.
63
350q ' 3
46
9
B
I (A, E)
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

æ D E ö÷
÷
 AE = tan-1 çç
çè D N ø÷÷
æ -510.448 ÷ö
 AE = tan-1 çç
çè 47.701 ÷÷ø
= 275 20'19"
and
LAE =
D E2 + D N 2

LAE =
510.4482 + 47.7012
=
262 833
= 512.672 m
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"
165
Chapter 10 – Missing measurements
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:
sin A
sin B
=
A
B
æ sin 7519'11" . 512.672 ö÷
I = sin-1 çç
÷
çè
ø÷
630.469
Therefore
= sin-1 (0.786614)
= 51 52'13"
E = 180 00'00" - 75190'11" - 51 52'13"
and
= 52 48'36"
Again from sine rule
AI
æ 630.469 . sin 52 48'36"÷ö
= AB = çç
÷÷÷
sin 7519'11"
èç
ø
= 519.203 m
and
 DE =  BA + I
= 170 39'30" + 51 52'13"
= 222 31'43"
Missing non-adjacent distances
For this example we have to find the missing distances (lengths) AB and CD.
B
45"
09'
º
m
7
5
674
.
5
40
C
º
'
"
º'"
Example 5
A
D
m
º'
"
Fig 10.19
166
m
"
'
º
E
Surveying Computations
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"
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:
→
→
→
→
→
→
→
→
AB  BC  CD  DE  EA 0
→
→
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
example,
æ D E ö÷
÷ , as from the previous
 AD = tan-1 çç
çè D N ø÷÷
then
 AD = 6410'03"
and
LAD = 480.115
167
Chapter 10 – Missing measurements
The vector can be shown graphically:
I (B, C)
12
0q
24
' 40
"
350q 39
209.206 m
'30"
432.138 m
A
D
m
15
1
.
"
0
48
' 03
0
1
64q
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
AB =
480.115
x sin 5614'37" = 519.206 m
sin 5014'50"
CD =
480.115
x sin 73 30'33" = 598.802 m
sin 5014'50"
Note: With two missing distances, the solution is always unique.
That is, it is never ambiguous.
168
Surveying Computations
Missing non-adjacent bearings
For this example we have to find the missing bearings BC and DE.
50"
'
50q
C
m
º
'
"
m
40
5.
67
4
B
D
m
m
"
º'
10
5q
'3
0"
Example 6
m
A
º'"
E
Fig 10.21
Line
Bearing(θ)
Distance (L)
∆N
∆E
AB
350º 39' 30"
519.206
+ 512.320
- 84.278
- 303.114
+ 516.416
+ 35.433
- 346.820
 = + 244.639
 = + 85.318
BC
CD
405.674
120º 24' 40"
DE
EA
598.802
630.469
275º 50' 00"
348.625
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
æ - 85.318 ÷ö
÷
 BE = tan-1 çç
çè - 244.639 ÷÷ø
and
= 19913'34"
and
LBE =
85.3182 + 244.6392
= 259.090 m
169
Chapter 10 – Missing measurements
The sum of these two vectors can be shown graphically.
I'
I
4
q 259 ' 34"
.09
0 m
-244.639
B
9
6
0.4
m
63
199
E
(C, D)
m
74
05.6
-85.318
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"
170
= 42° 31' 43"
Surveying Computations
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.
"
' 15
B
C
66q 74 m
.6
405
12
9q
59
8.8
' 10
02
"
m
D
0q' 00"
Example 7
m
A
m
285q ' 30
"
E
Fig 10.23
Line
Bearing(θ)
AB
0º 0' 0"
BC
66º 30' 15"
405.674
161.735
372.039
CD
129º 45' 10"
598.802
- 382.920
460.365
DE
Distance (L)
∆N
0
630.469
EA
285º 10' 30"
∆E
- 495.935
348.625
91.259
 = - 129.926
4 + 519.207
deduced
3 231º 52' 12"
calculated
- 336.469
 = 495.935
2 - 389.281
calculated
As
θAB
then
Δ EAB = 0
and
Δ EDE must therefore equal - 495.935, if ΣΔ E is equal to 0.
= 0° 00' 00"
Now, knowing Δ EDE = - 495.935
and
LDE
= 630.469
Δ NDE can be calculated using Pythagoras’ theorem
171
…(1)
1
Chapter 10 – Missing measurements
thus
LDE 2 =
or
DN
2
DN
2
+ DE
2
=
LDE 2 - D E 2
=
630.4692 - 495.9352
=
389.291
…(2)
Knowing both Δ NDE and Δ EDE we can calculate θDE from
æ D E ö÷
÷
 DE = tan-1 çç
çè D N ø÷÷
=
- 495.935
- 389.281
= 231 52'12"
…(3)
Finally, knowing that ΣΔ N = 0, ΔNAB must be 519.207 to satisfy
this 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 lines
On 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.
Line
AB
Bearing(θ)
270º 00' 00"
BC
CD
Distance (L)
1000.000
∆N
∆E
0.000
- 1000.000
3464.100
- 2000.000
3000.000
330º 00' 00"
DE
4000.000
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
172
Surveying Computations
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.
Line
Bearing(θ)
Distance (L)
∆N
∆E
AB
61º 13' 30"
728.968
350.904
638.953
BC
90º 00' 00"
780.290
0.000
780.290
- 69.908
409.901
CD
578.874
DE
99º 40' 43"
EF
415.820
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.
ΔNCD + EF = - 687.429
and ΔECD + EF = 527.403
æ D E ö÷
÷
CF = tan -1 çç
çè D N ø÷÷
æ 527.403 ÷ö
÷
= tan -1 çç
çè - 687.429 ÷÷ø
= - 37 29' 45" (+180)
= 142 30' 15"
LCF =
687.4292 + (- 527.4032 )
= 866.437 m
173
Chapter 10 – Missing measurements
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"
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.
174
Surveying Computations
Chapter 11 – Simple circular curves
11.1 Introduction
Horizontal, 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.
I (= IP)
A (= TP)
B (= 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 curves
In 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.
175
Chapter 11 – Simple circular curves
Theorem 1
If 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 
C
B
A
O
Fig 11.2
Theorem 2
Chords which are equidistant from the centre are equal.
Chords AC and DF are equal because BO = OE.
C
B
A
O
D
E
F
Fig 11.3
176
Surveying Computations
Theorem 3
The angle at the centre of the circle is double the angle at the circumference, standing on the
same arc.
AOC  2ADC because both stand on arc AC.
B
C
A
O
D
Fig 11.4
Theorem 4
Angles 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.
C
D
B
A
P
E
Fig 11.5
177
Chapter 11 – Simple circular curves
Theorem 5
The 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 6
The angle in a semicircle is a right angle.
The angles AEB, BDA and BCA standing on the diameter AB, are 90º.
B
E
D
C
A
Fig 11.6
Theorem 7
The 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.
P
T
Q
O
Fig 11.7
178
Surveying Computations
Theorem 8
Two 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
P
B
Fig 11.8
Theorem 9
If 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.
A
01
03
B
Fig 11.9
179
02
Chapter 11 – Simple circular curves
11.3 Elements of simple circular curves
I
'
E
TD
TD
C
Arc
M
A
TP
R
B
D
90º
Long
TP
chord
'
2
'
2
R
O
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)
180
Surveying Computations
11.4 Derived formulae
Using Fig 11.10 we can show the following formula derivations.
Tangent distance

IAO + OBI

AOB + BIA = 180º
But ∆ + BIA
= 180º
= 180º
Therefore AOB = ∆
Triangles IAO and IOB are congruent

AOI and IOB =

and AI = IB
2

2
TD = R tan
R =
tan
TD

tan
2

TD
=
2
R
Long chord
AB = 2AD
AD = DB

AD = R sin

2
AB = 2R sin

2
Crown secant
In right triangle IAO
IO =
E =
R
cos
 ,
2
R

cos
2
OC = R, Crown secant (E) = IC = IO – OC
- R
or
181
æ
ö÷
çç
÷÷
1
ç
E = R çç
- 1÷÷÷

çç
÷÷
çè cos 2
ø÷
Chapter 11 – Simple circular curves
Substituting
R =
TD
for R, it can be shown that:

tan
2
E = TD . tan

4
Mid-ordinate
Mid-ordinate (M)  DC  OC – OD
 R - R cos
or

2
æ
ö
M = R çç1 - cos ÷÷÷
çè
2ø
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
O
R
1 radian
0.1 radian
0.1 R
Fig 11.11
Angle at centre (in radians) =
182
arc length
radius
Surveying Computations
1
When the arc length is equal to the radius, the ratio = = 1, and the angle at the centre contains one
1
radian.



R
1 radian
=
(circumference of circle)
360
2R
=
or, 1 radian
=
360
2
180

= 57.2957795
= 5717 '44.806"
or, in minutes, 1 radian
= 3437.74677 '
or
= 206 264.806"
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) 
a)
arc length  radius  angle at centre (in radians), and
b)
radius

arc length
we can get:
radians
arc length
angle 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:


180
a)
degrees to radians by multiplying x by
b)
radians to degrees by multiplying x radians by
180
.

Note: Most calculators have a button to convert between degrees and radians.
183
Chapter 11 – Simple circular curves
11.5 Areas
Sector
I
Δ
A
B
R
R
Δ
O
Fig 11.13
Sector area =

(in radians) ´ R 2
2
Since ∆ R equals the arc AB of the sector, the formula can also be written as:
Sector area =
184
1
arc ´ R
2
Surveying Computations
Segment
I
¨
A
B
R
R
¨
O
Fig 11.14
Segment area  area sector AOB – area triangle AOB

1
1
 (radians) R 2 - R 2 sin 
2
2

1 2
R [  (radians) - sin  ]
2
Segment area  1 R 2 [  (radians) - sin  ]
2
185
Chapter 11 – Simple circular curves
Truncation
I ¨
A
B
R
¨ ¨
2 2
R
O
Fig 11.15
Truncation area = 2 ´ area triangle IAO - sector AOB
æ1
ö
= 2 ´ çç AI ´ AO÷÷÷ - sector AOB
çè 2
ø
= AI ´ AO - sector AOB

1
= R tan
´ R - R 2  (in radians)
2
2
æ 
ö

= R 2 çç tan - (in radians)÷÷÷
çè
ø
2
2
æ
1 ö
or = R ççTD - arc÷÷÷
çè
2 ø
æ 
ö

Truncation area = R 2 çç tan - (in radians)÷÷÷
çè
ø
2
2
æ
1 ö
or = R ççTD - arc÷÷÷
çè
2 ø
186
Surveying Computations
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
x
TPB
arc
140°00'000"
Fig 11.16
TPA
(a) Firstly, write down the formulae for calculating the truncation area.
æ
1 ö
Truncation area = R ççTD - arc÷÷÷
çè
2 ø
æ 
ö

or
R 2 çç tan - (in radians)÷÷÷
çè
ø
2
2
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.
R
Crown secant (E) 
or

cos

-R
2
æ
ö÷
çç
÷÷
1
ç
- 1÷÷÷
E  R çç

çç
÷÷
çè cos 2
÷ø
æ 1
ö
- 1÷÷÷ the angle between the
10  R çç
÷ø
çè cos 20
straights is 140, so the
deflection angle is 40
R  155.817 m
187
Chapter 11 – Simple circular curves
(c) The truncation area may now be calculated.
æ 
ö

in radians)
Truncation area = R2 çç(tan
tan – (in radians)÷÷÷
çè
ø
2
2
= (155.817)2 (tan 20 – 20 (in radians))
= (155.817)2 (0.363970 – 0.349066)
= 361.853 m2
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.
188
Surveying Computations
Chapter 12 – Roads
12.1 Introduction
Boundaries 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.
ted
d
d
Peg
Calcula
ted
Peg
nd
Calc. a
ed
u
Meas r
Calcula
Measure
Measure
ed
Measur
Calc. and
Peg
Peg
Fig 12.1
Note: Sometimes the centre line of the road boundary is surveyed and the boundaries laid out
either side (via calculations).
189
Chapter 12 – Roads
12.2 Single-width roads
Traditionally, 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 rightangled triangle; the opposite peg lies on the bisection of the angle as measured.
‘cotangent’
distance
A
B
b'
G
‘secant’
distance
‘secant’
distance
D
E
b"
e'
roa
a'
H
‘cotangent’
distances
dw
C
idth
f'
‘secant’
distance
‘secant’
distance
f"
F
‘cotangent’
distances
Fig 12.3
190
‘cotangent’
distance
Surveying Computations
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 
1
).
tan
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 
1
.
sin
‘cotangent’ distances
T T
2 2
idth
road
road w
width
‘secant’
distance
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.
191
Chapter 12 – Roads
Calculating boundaries
Given 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
7
m
th
7º3
B
D
º12
157
20
wid
13
98º1
3'
1269
.85 m
5'
5º2
d
roa
2'
E
'
F
C
Fig 12.5
(a) Firstly calculate the secant distances, AD and BC.
secant distance 
AD 
road width

sin

20
15712'
sin
2
 20.403 m
BC 
20
13732'
sin
2
 21.457 m
192
Surveying Computations
(b) Then calculate the cotangent distances, AE and FC.
cotangent distance 
AE 

road width

tan

20
15712'
tan
2
20
tan 786'
 4.033 m
FC =
=
20
137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
193
Chapter 12 – Roads
12.3 Computations relating to roads
Single-width roads
As 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.
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
b'
A
0m
70.00 "
0
75º0'0
a
"
0'00
330º
idth
oad w
20 m
r
b"
50.
0
120 00 m
º0'0
0"
E
D
C
180º0'00"
Example 1
f
F
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
At B 
 7500'00"
255 -120
 6730'00"

At F  360 – 300  6000'00"
Then compute the bearings and lengths of secants AD, BE, CF using the half
angle, already calculated at A, B, F.
194
Surveying Computations
The formula is:
road width

sin

secant distance 

LAD 
20
sin 75
LBE 
20
 21.648 m,
sin 67 '
BE  120’ 1870’
LCF 
20
sin 60
CF  180(given)
 20.705 m,
 23.094 m,
AD  150(given)
(b) To calculate the length of boundary DEF compute the cotangent distances.
The formula is:
cotangent distance 
Aa 
road width

sin

20
tan 75
 5.359
Bb 
20
tan 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
195
Chapter 12 – Roads
(c) Finally compute the area of the portion of road ABCFEDA.
æ sum of boundary lengths ö÷
Area  road width çç
÷ formula for a trapezium
çè
ø÷
2
æ120.000 + 109.620 ö÷
= 20çç
÷÷
çè
ø
2
= 2296 m2
Alternative calculation methods
The 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’.
•
•
 AB +  BC
+ 90 .
2
With this ‘formula’ you do not need to calculate the half angles.
Firstly, the bearing for BE can be calculated directly from
•
By using ‘missing measurement’ software – this is available for some calculators and also
from surveying programs on computers, you can solve triangles AaD and CFf for distances
DA and Aa, and CF and Ff.
•
If you then ‘close’ DABED with calculated LDA, calculated θBE and known bearing and
distance 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 the
bearing 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. This
is an easy task and missing information can then be easily extracted from the diagram
constructed.
196
Surveying Computations
Two-width roads
Sometimes 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
ș2
A
W2 = 15 m
ș1
0m
W1
B
=2
Fig 12.7
The formulae below are used for the calculations. Use one for a check.
æ W sin  - W sin  ö

2
 ÷÷
AB = tan–1 ççç 1
÷
è W1 cos  - W2 cos  ÷ø
LAB =
W1
sin ( AB - 1)
=
W2
sin ( AB -  2)
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 the
diagram.
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.
197
Chapter 12 – Roads
Example 2
In the situation shown below, compute the bearing and length of the join AB.
19′20″
θ 2 = 75°
A
W 2 = 15
0″
′3
51
1°
=4
θ1
0m
W
=1
m
B
1
Fig 12.8
Note that both bearings ‘flow’ from left to right.
(a) To calculate the bearing for AB.


æ 10sin 7519'20" - 15sin 4151'30" ö÷
÷÷
AB  tan–1 ççç
è10cos 7519'20" - 15cos 4151'30"ø÷
 213'32''
However, from the diagram, θAB must be 18213'32''.
(b) To calculate the length.
LAB 
10
15
or
sin (18213'32" - 4151'30")
sin (18213'32" - 7519'20")
 15.677 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.
198
Surveying Computations
Appendix A
Calculator requirements
Because 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.
199
Appendix A – Calculator requirements
Examples of use
Below 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 
b 2 + c 2 - 2bc cos A
2
2
  13.3 + 20.1 - 2(13.3)(20.1) (cos17 54'27")
Example A
Example B
Calculator with the basic functions previously
listed (including parentheses).
Calculator with operation ‘stacking’ register and
reverse notation.
Key
Display
Key
Display
13.3
x2
13.3
13.3
13.3
176.8
Enter
13.3

176.89

176.89
20.1
x2
20.1
20.1
20.1
401.01
Enter
20.1

580.9

401.01

580.9
(
2
2
2
2

2
Enter
2
13.3
13.3
13.3
13.3

26.6

26.6
20.1
20.1
20.1
20.1


534.66
17.5427
17.5427
DMS  Dec. Deg.
17.9075
Cos
17.5427
0.951554
)
508.7579

72.1420
√¯¯
8.4936
534.66
Shift key
H  HMS
17.9075
 H Menu key
Cos

508.7579

72.1420
√
a = 8.494
0.951554
¯¯
8.4936
a = 8.494
200
Surveying Computations
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 calculators
work 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 regarded
as standards for all calculators. Each calculator model has its own handbook which must be
studied carefully to discover its particular advantages and mode of operation.
The calculators used in examples A and B were:
Example A
Example B
Sharp scientific calculator.
Hewlett Packard RPN scientific calculator.
201
SURVEYING COMPUTATIONS
CPPSIS3010A
CPPSIS4011A
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 PACKAGE
CPP07 – 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
BC1091
SURVEY COMPUTATIONS
ISBN 978-1-74205-268-7
ORDERING INFORMATION:
Contact WestOne Services on Tel: (08) 6212 9700 Fax: (08) 9227 8393 Email: [email protected]
Orders can also be placed through the website: www.westone.wa.gov.au
9 781742 052687