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BSQ3113 Land Surveying Chapter 1 : Introduction to Surveying

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Page 1: Land Surveying Chapter 1 Introduction to Surveying

BSQ3113 Land Surveying

Chapter 1 : Introduction to Surveying

Page 2: Land Surveying Chapter 1 Introduction to Surveying

Introduction

Surveying

Science and techniques used to determine the relative and absolute spatial location of points on the earth’s surface.

Involves staking out the lines and grades needed for construction of buildings, roads, dams, & other engineering structures.

Includes the computation of areas, volumes, and other quantities (field measurements), as well as the preparation of necessary maps and diagrams.

Page 3: Land Surveying Chapter 1 Introduction to Surveying

Contents

1.1 Units of Measurement

1.2 Measurement Conversions

1.3 Trigonometric Ratios

1.4 Rules of Sine and Cosine

1.5 Area of Triangles

1.6 Scale

1.7 Drawing to Scale

Page 4: Land Surveying Chapter 1 Introduction to Surveying

Système Internationale (SI)

Most common system used in the measurement of distance and angle.

Basic and derived units agreed internationally.

1.1 Units of Measurement

Page 5: Land Surveying Chapter 1 Introduction to Surveying

Table below shows the basic unit of prime interest :

1.1 Units of Measurement

Quantity Unit SI Symbol

Length metre m

Area square metre m2

Volume cubic metre m3

Mass kilogram kg

Capacity litre l

Page 6: Land Surveying Chapter 1 Introduction to Surveying

1.1 Units of Measurement

Prefix Multiplication factor Derived unit SI recommended unit

kilo 1000 kilometre kilometre

hecto 100 hectometre

deca 10 decametre

1 metre metre

deci 0.1 decimetre

centi 0.01 centimetre

mili 0.001 milimetre milimetre

Page 7: Land Surveying Chapter 1 Introduction to Surveying

Angle

Measured in degree (sexagesimal unit – numeral system with sixty as its base).

Degree is subdivided into minutes and seconds. (similar to time).

• 1° (degree) = 60’ (minutes)• 1’ (minute) = 60” (seconds)

1.1 Units of Measurement

Page 8: Land Surveying Chapter 1 Introduction to Surveying

Linear Volume

1 km = 1,000 m1 m = 100 cm1 cm = 10 mm1 foot = 12 inches1 inch = 2.54 cm1 km = 0.631371 mile

1 m3 = 1,000,000 cm3

1 litre = 1,000 cm3

1 litre = 0.264 U.S. gal

Weight

1 metric ton = 1,000 kg1 kg = 1,000 g1 kg = 2.2 Ib1 Ib = 16 ounces

Area

1 km2 = 1,000,000 m2

1 km2 = 247.1 acres1 hectare = 10,000 m2

Angle

1 degree = 60 minutes1 minutes = 60 secondsπ radian = 180 degrees

1.2 Measurement Conversions

Page 9: Land Surveying Chapter 1 Introduction to Surveying

Challenge your mind…

Convert the following measurements to the required unit in brackets :

(a) 3.3 feet [m] Solution 1.2a

(b) 7200’’ [° ’ ’’] Solution 1.2b

1.2 Measurement Conversions

Page 10: Land Surveying Chapter 1 Introduction to Surveying

Let R (x, y) be a point in XY coordination system.

θ is an angle formed by lineOR and the x-axis (O is an origin).

If a line is drawn perpendicularto the x-axis at S, a right angletriangle ORS is formed.

OR is called hypotenuse (r),RS is opposite (y), andOS is adjacent (x).

1.3 Trigonometric Ratios

0 x S

r

R

y

(x, y)

y

x

Page 11: Land Surveying Chapter 1 Introduction to Surveying

1.3 Trigonometric Ratios

Note:

Page 12: Land Surveying Chapter 1 Introduction to Surveying

Challenge your mind…

If coordinate R is (4, 5),determine :

(a) length r(b) sin θ(c) cos θ(d) tan θ(e) angle θ Solution 1.3a, 1.3b-d,

1.3e

1.3 Trigonometric Ratios

y

x0

R (4, 5)

r

Page 13: Land Surveying Chapter 1 Introduction to Surveying

1.4 Rules of Sine and Cosine

a

b

c-x

x

C

B

A

Page 14: Land Surveying Chapter 1 Introduction to Surveying

Challenge your mind…

If a = 5 cm, b = 6 cm,

c = 7 cm, determine :

(a) angle A

(b) angle B

(c) angle C Solution 1.4a-c

1.4 Rules of Sine and Cosine

a

b

c

C

B

A

Page 15: Land Surveying Chapter 1 Introduction to Surveying

1.5 Area of Triangles

b

h

b

h

For a right angle triangle or an arbitrary triangle shown above, if the base (b) and the altitude or height (h) of the triangle are given, then

Page 16: Land Surveying Chapter 1 Introduction to Surveying

For an arbitrary triangle shown on the right,

1.5 Area of Triangles

a

b

c-x

x

C

B

A

Page 17: Land Surveying Chapter 1 Introduction to Surveying

Challenge your mind…

Determine area of triangle :

(a) If a = 4 m, b = 6 m,and C = 65° 35’ 56”.

Solution 1.5a

(b) If a = 45 cm, b = 50 cm,and c =35 cm

Solution 1.5b

1.5 Area of Triangles

a

b

c

C

B

A

Page 18: Land Surveying Chapter 1 Introduction to Surveying

1.6 Scale

Scale Ratio between drawing of an object and actual object itself.

Scale of a map or plan can be shown in 3 ways:• Expressed in words.

e.g. 1 centimetre represents 1 metre.By definition of scale, this simply means that one centimetre on the plan represents 1 metre on the ground.

Page 19: Land Surveying Chapter 1 Introduction to Surveying

1.6 Scale

• By a drawn scale.A line is drawn on the plan and is divided into convenient intervals that distances on the map can be easily obtained from it. Scale in the diagram below : Scale of 1 cm represents 1 m.

Note : Diagram is not to scale.

1 0 1 2

Page 20: Land Surveying Chapter 1 Introduction to Surveying

1.6 Scale

• By a representative fraction (RF).

A fraction is used, in which the numerator represents the number of units on the map (always unit 1) and the denominator represents the number of the same units on the ground.

For a scale of 1 cm represents 1 m, its RF is 1/100, shown as 1:100, since there are 100 cm in 1 m.

Page 21: Land Surveying Chapter 1 Introduction to Surveying

1.6 Scale

• By a representative fraction (RF).

RF-related simple formula for calculating plan area is as follows :

Plan area = Ground area x (RF)2

Page 22: Land Surveying Chapter 1 Introduction to Surveying

1.6 Scale

Challenge your mind…

An area of 250 cm2 was measured on a plan,

using a planimeter. Given that the plan scale

is 1:500, calculate the ground area in m2.

Solution

1.6

Page 23: Land Surveying Chapter 1 Introduction to Surveying

1.7 Drawing to Scale

The whole idea behind creating scale drawings is to allow the drafter to create a drawing which is proportionately the same as the artifact it represents.

Using a measuring device called a scale, accurate drawings of both very large objects or very small objects can be created and fitted on a standard size piece of paper.

Page 24: Land Surveying Chapter 1 Introduction to Surveying

1.7 Drawing to Scale

A surveyor’s main objective is to achieve accuracy in field operations.

Unless results can be depicted accurately, legibly and pleasingly on paper, proficiency in the field is robbed of much of its value.

Some of equipment required for plotting include paper (A4, A3, A1, …), scale rule (usually manufactured with eight scales – 1:1, 1:5, 1:50, 1:100, 1:200, 1:250, 1:1250, 1:2500), two set squares (45° & 60°), protractor, varying grade of pencils, etc.

Page 25: Land Surveying Chapter 1 Introduction to Surveying

1.7 Drawing to Scale

Triangular scale :

Page 26: Land Surveying Chapter 1 Introduction to Surveying

1.7 Drawing to Scale

Set square :

Page 27: Land Surveying Chapter 1 Introduction to Surveying

1.7 Drawing to Scale

Pencil’s grade:

Page 28: Land Surveying Chapter 1 Introduction to Surveying

End of Chapter 1

Revise…1.1 Units of Measurement1.2 Measurement Conversions1.3 Trigonometric Ratios1.4 Rules of Sine and Cosine1.5 Area of Triangles1.6 Scale1.7 Drawing to Scale

Page 29: Land Surveying Chapter 1 Introduction to Surveying

Solution(s)

Solution 1.2(a)

3.3 feet = 3.3 ft × (12 in. ÷ 1 ft)= 39.6 in.= 39.6 in. × (2.54 cm ÷ 1 in.)= 100.584 cm= 100.584 cm × (1 m ÷ 100 cm)= 1.00584 m

Page 30: Land Surveying Chapter 1 Introduction to Surveying

Solution(s)

Solution 1.2(b)

7200’’ = 7200’’ × (1’ ÷ 60’’)= 120’ 0’’= 120’ × (1° ÷ 60’)= 2° 0’ 0’’

Page 31: Land Surveying Chapter 1 Introduction to Surveying

Solution(s)

Solution 1.3(a)

r² = x² + y² r² = 4² + 5² r = (4² + 5²)½

= 6.403 unit0 4

6.403

R

5

(4, 5)

y

x

Page 32: Land Surveying Chapter 1 Introduction to Surveying

Solution(s)

Solution 1.3(b),(c),(d)

sinθ = 5 ÷ 6.403 = 0.781cosθ= 4 ÷ 6.403 = 0.625tanθ= 5 ÷ 4 = 1.250

0 4

6.403

R

5

(4, 5)

y

x

Page 33: Land Surveying Chapter 1 Introduction to Surveying

Solution(s)

Solution 1.3(e)

θ = sin-1 0.781 = 51° 21’ 08”

θ = cos-1 0.625 = 51° 19’ 04”

θ = tan-1 1.250 = 51° 20’ 25”

0 4

6.403

R

5

(4, 5)

y

x

Page 34: Land Surveying Chapter 1 Introduction to Surveying

Solution(s)

Solution 1.4(a),(b),(c)

c² = a² + b² - 2abcosC7² = 5² + 6² - 2(5)(6)cosC2(5)(6)cosC = 5² + 6² - 7²

cosC = (5² + 6² - 7²) ÷ ( 2 × 5 × 6) C = cos-1 0.2

= 78° 27’ 47”

a

b

c

C

B

A

Page 35: Land Surveying Chapter 1 Introduction to Surveying

Solution(s)

Solution 1.4(a),(b),(c)

A = sin-1 [5 sin(78° 27’ 47”) ÷ 7]= 44° 24’ 55”

B = sin-1 [6 sin(78° 27’ 47”) ÷ 7]= 57° 07’ 18”

a

b

c

C

B

A

Page 36: Land Surveying Chapter 1 Introduction to Surveying

Solution(s)

Solution 1.5(a)

Area= ½ ab sin C

= ½ (4)(6) sin 65° 35’ 56”= 10.928 m2

Page 37: Land Surveying Chapter 1 Introduction to Surveying

Solution(s)

Solution 1.5(b)

s = (45 + 50 + 35) ÷ 2= 65 cm

= [65(65 – 45)(65 – 50)(65 – 35)]½

= 1099.545 cm2

Page 38: Land Surveying Chapter 1 Introduction to Surveying

Solution(s)

Solution 1.6

Plan area = 250 cm2

RF (scale) = 1:500Plan area = Ground area x (RF)2

Ground area = Plan area ÷ (RF)2

= 250 cm2 ÷ (1/500) 2

= 62,500,000 cm2

= 6250 m2

Page 39: Land Surveying Chapter 1 Introduction to Surveying

Extra(s)

Planimeter

Precision tool for measuring the areas of irregular shaped objects.

Simply trace the outline of the object and the planimeter will display the area.

This video features the Planix 7 digital Planimeter but the general operation is the same for all digital roller planimeters.