survey questions and problems

8/10/2019 Survey Questions and Problems

1/676

Problems

Solutions

Shepherd

52-

X-

ARNOLD

Surveying

Problems

and

Solutions

F.

A. Shepherd

ii^ ^iiA

*


2/676

\

Thfs

new

book gives

a

presentation

concentrating

on

mathematical

problems,

an

aspect

of

the

subject

which

usually

causes

most

difficulty.

Summaries

of

basic

theory

are

followed

by

worked

examples

and

selected

exer-

cises.

The book

covers

three

main

branches

of

surveying:

measurement,

surveying

techniques

and

industrial

appli-

cations.

It

is

a

book

concerned

mainly

with

engineering

surveying

as

applied,

for

example,

in the

construction

and

mining

industries.

Contents

Linear

Measurement

Surveying

Trigonometry

Co-ordinates

Instrumental

Optics

Levelling

Traverse

Surveys

Tacheometry

Dip

and Fault

Problems

Areas

Volumes

Circular

Curves

Vertical

and

Transition

Curves

Values

in

both

imperial

and

metric

(S.

units

are

given

in

the

problems


3/676

SURVEYING

PROBLEMS

&

SOLUTIONS

Shop

l>ord


4/676

Surveying

Problems

and

Solutions

F. A.

Shepherd

c.En

g

.,

A.R.i.c.s.,M.i.Min.E.

Senior

Lecturer

in

Surveying

Nottingham

Regional

College

of

Technology


5/676

HARRIS

Co


6/676

PREFACE

This

book

is

an

attempt

to

deal

with

the

basic

mathematical

aspects

of

'Engineering

Surveying',

i.e.

surveying

applied

to

construction

and

mining

engineering

projects,

and

to

give

guidance

on

practical

methods

of

solving

the

typical

problems

posed

in

practice

and,

in theory,

by

the

various

examining

bodies.

The

general

approach

adopted

is

to

give

a

theoretical

analysis

of

each

topic,

followed

by

worked

examples

and, finally,

selected

exer-

cises

for

private

study.

Little

claim

is

made

to

new

ideas,

as

the

ground

covered

is

elementary

and

generally

well

accepted.

It is

hoped

that

the

mathematics

of

surveying,

which

so often

causes

trouble

to

beginners,

is

presented

in

as

clear

and

readily

understood

a

manner

as

possible.

The

main

part

of

the

work

of the

engineering

surveyor,

civil

and

mining

engineer,

and

all

workers

in

the

construction

industry

is

confined

to

plane

surveying,

and

this

book

is

similarly

restricted.

It

is

hoped

that

the

order

of the

chapters

provides

a

natural

sequence,

viz.:

(a)

Fundamental

measurement

(i)

Linear

measurement

in

the

horizontal

plane,

(ii)

Angular

measurement

and

its

relationship

to

linear

values,

i.e.

trigonometry,

(iii)

Co-ordinates

as

a

graphical

and

mathematical

tool.

(b)

Fundamental

surveying

techniques

(i)

Instrumentation.

(ii)

Linear

measurement

in

the

vertical

plane,

i.e.

levelling,

(iii)

Traversing

as a

control

system,

(iv)

Tacheometry

as

a

detail

and

control

system.

(c)

Industrial

applications

(i)

Three-dimensional

aspects

involving

inclined

planes,

(ii)

Mensuration,


7/676

physical properties affords a

more

complete

understanding

of

the

con-

struction

and

use

of

instruments.

To

facilitate

a

real

grasp

of the

sub-

ject,

the

effects of

errors

are

analysed

in

all

sections. This may

appear

too advanced for students who

are

not familiar

with

the

element-

ary

calculus,

but

it

is

hoped that

the

conclusions

derived

will

be

beneficial

to

all.

With the introduction

of

the

Metric System in

the

British

Isles

and

elsewhere,

its

effect

on

all

aspects of

surveying

is

pin-pointed

and

conversion

factors

are

given.

Some examples

are duplicated

in

the

proposed

units

based

on

the

International

System

(S.I.)

and

in order

to

give

a

'feel'

for the

new

system, during

the

difficult

transition period,

equivalent

S.I.

values are

given in

brackets

for

a

few

selected examples.

The

book is

suitable

for

all students

in

Universities

and

Technical

Colleges,

as

well

as

for

supplementary

postal

tuition,

in

such

courses

as

Higher

National

Certificates, Diplomas

and Degrees

in

Surveying,

Construction,

Architecture,

Planning, Estate

Management,

Civil and

Mining

Engineering,

as

well

as

for

professional

qualification

for the

Royal

Institution of

Chartered Surveyors,

the

Institution

of

Civil

Engineers,

the

Incorporated

Association

of

Architects

and

Surveyors,

the

Institute

of

Quantity

Surveyors,

and

the Institute

of

Building.

ACKNOWLEDGMENTS

I

am

greatly

indebted

to

the

Mining

Qualifications

Board

(Ministry

of Power) and

the

Controller

of

H.M.

Stationery

Office,

who have

given

permission

for the

reproduction

of

examination

questions.

My

thanks

are

also

due

to

the Royal

Institution

of

Chartered

Surveyors, the

Institution

of

Civil

Engineers,

to

the

Senates

of the Universities

of

London

and

Nottingham, to

the East

Midlands

Educational

Union and

the

Nottingham

Regional College

of

Technology,

all

of

whom

have

allowed

their

examination

questions

to

be used.

My

special

thanks are due

to many

of

my colleagues at

Nottingham,


8/676

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10/676

Vll

Mass

per

unit

length

1

lb/ft

=

1-488

16

kg/m

Mass

per

unit

area

lib/ft

2

=

4-88243

kg/m

2

Density

1

ton/yd

3

=

1328-94

kg/m

3

1

lb/ft

3

=

16-018

5

kg/m

1 kg/m

3

=

0-062428

lb/ft

3

1

lb/gal

=

99-776

3

kg/m

3

0-09978

kg/1

Force

Hbf

=

4-448

22

N

IN

=

0-224

809

lbf

Ikgf

=

9-80665

N

1

kgf

=

2-20462

lbf

Force

(weight)

/unit

length

1

lbf/ft

=

14-593

9

N'm

Pressure

1

lbf/ft

2

=

47-880

3

N/m

2

1

N/m

2

=

0-000

145

038

lbf/in

2

1

lbf

/in

2

=

6894-76

N/m

2

1

kgf

/cm

2

=

98-066

5

kN/m

2

lkgf/m

2

=

9-80665

N/m

2

Standard

gravity

32-1740

ft/s

2

=

9-80665

m/s

2

N.B.

lib

=

0-453

592

kg

1

lbf

=

0-453

592

x

9-80665

=

4-448

22

N

1

newton

(N)

unit

of

force

=

that

force

which

applied

to

a mass


11/676

Vlll

CONTENTS

Chapter

Page

1 LINEAR

MEASUREMENT

1

1.1

The

basic

principles

of

surveying

1

1.2

General

theory

of

measurement

2

1.3

Significant

figures in

measurement

and

computation

3

1.4

Chain surveying

6

1.41

Corrections to

ground

measurements

6

1.42

The maximum length of

offsets from

chain

lines

13

1.43

Setting

out

a right

angle

by

chain

15

1.44

To

find

the point

on

the

chain

line

which

produces

a

perpendicular from

a point

outside

the

line

16

1.45

Obstacles

in

chain

surveying 17

Exercises

1(a)

22

1.5

Corrections

to

be

applied

to

measured

lengths

23

1.51

Standardisation

23

1.52

Correction

for slope

23

1.53

Correction for

temperature

26

1.54

Correction

for tension

27

1.55

Correction

for

sag

32

1.56

Reduction

to

mean

sea

level

38

1.57

Reduction of

ground

length

to

grid

length

39

1.6

The effect

of

errors

in

linear measurement

45

1.61

Standardisation

45

1.62

Malalignment

and

deformation of the

tape

45

1.63

Reading

or

marking

the tape

46

1.64

Errors

due

to

wrongly

recorded

temperature

46

1.65

Errors

due

to

variation

from

the recorded

value

of

tension

47


12/676

2.12

Trigonometrical

ratios

58

2.13

Complementary

angles

60

2.14 Supplementary

angles

60

2.15

Basis

of

tables of

trigonometrical

functions

63

2.16

Trigonometric

ratios

of

common

angles

64

2.17

Points

of

the

compass

65

2.

18

Easy

problems

based

on

the

solution

of

the

right-

angled

triangle

67

Exercises

2(a)

71

2.2

Circular

measure

72

2.21 The radian

72

2.22 Small

angles and

approximations

73

2.3

Trigonometrical

ratios

of

the

sums and

differences

of

two

angles

77

2.4

Transformation

of

products

and

sums

79

2.5

The

solution

of

triangles

80

2.51

Sine

rule

80

2.52

Cosine

rule

81

2.53

Area

of

a

triangle

82

2.54

Half-angle

formulae

82

2.55

Napier's

tangent

rule

83

2.56

Problems

involving

the

solution of triangles

83

2.6

Heights

and

distances

91

2.61

To

find

the

height

of an

object

having

a

vertical face

91

2.62

To find

the

height

of

an

object

when

its

base

is

inaccessible

92

2.63

To find

the height

of

an

object

above

the

ground

when

its base

and

top

are

visible

but not

accessible

95

2.64

To find the

length

of

an

inclined

object

on

the

top

of

a

building

98


13/676

3.14

Plotting

accuracy

114

3.15

Incorrect

scale

problems

114

3.2

Bearings

115

3.21

True

north

115

3.22

Magnetic

north

115

3.23

Grid north

116

3.24

Arbitrary

north

116

3.25

Types

of

bearing

117

3.26

Conversion

of

horizontal

angles into

bearings

121

3.27

Deflection

angles

124

Exercises

3(a)

126

3.3 Rectangular

co-ordinates

127

3.31

Partial

co-ordinates,

AE, AN

128

3.32

Total

co-ordinates

128

Exercises

3(b)

(Plotting)

131

3.4 Computation processes

133

3.41

Computation

by

logarithms

134

3.42

Computation by

machine

134

3.43

Tabulation process

135

3.44

To

obtain the bearing and distance

between two

points

given their

co-ordinates

136

3.5

To

find

the co-ordinates

of

the

intersection

of two lines

146

3.51

Given

their bearings from two known

co-ordinate

stations

146

3.52

Given

the

length and

bearing

of

a line

AB

and

all

the angles

A,

B

and

C

149

Exercises

3(c) (Boundaries)

157

3.6

Transposition

of

grid

158

3.7

The National

Grid

Reference

system

160

Exercises 3(d)

(Co-ordinates)

163

Appendix (Comparison

of

Scales)

169

4

INSTRUMENTAL

OPTICS

170


14/676

XI

4.2 Refraction

at

plane

surfaces

177

4.21

Laws

of

refraction

177

4.22 Total

internal

reflection

177

4.23

Relationships

between

refractive

indices

178

4.24 Refraction

through

triangular

prisms

179

4.25 Instruments

using

refraction

through

prisms

180

Exercises

4(a)

184

4.3

Spherical

mirrors

184

4.31 Concave

or

converging

mirrors

184

4.32

Convex

or

diverging

mirrors

186

4.33

The

relationship

between

object

and

image

in

curved

mirrors

186

4.34

Sign

convention

lg7

4.35

Derivation

of

formulae

Igg

4.36 Magnification

in

spherical

mirrors

190

4.4

Refraction

through

thin

lenses

191

4.41

Definitions

191

4.42

Formation

of

images

192

4.43

The

relationship

between

object and

image

in

a

thin

lens

193

4.44 Derivation

of

formulae

193

4.45

Magnification

in

thin

lenses

195

4.5

Telescopes

196

4.51

Kepler's

astronomical

telescope

196

4.52

Galileo's

telescope

196

4.53

Eyepieces

I97

4.54

The

internal

focussing

telescope

198

4.55

The

tacheometric

telescope

(external

focussing)

201

4.56

The

anallatic

lens

203

4.57

The

tacheometric

telescope

(internal

focussing)

207


15/676

Xll

4.8

Angular

error

due

to

defective centring

of

the

theodolite

234

4.9

The

vernier

237

4.91

Direct reading vernier

237

4.92

Retrograde

vernier 238

4.93

Special

forms used

in

vernier

theodolites

238

4-94

Geometrical

construction

of

the

vernier

scale

238

Exercises

4(b)

240

LEVELLING

244

5.

1

Definitions

244

5.2

Principles

245

5.3

Booking,

of

readings

246

5.31 Method

1,

rise

and

fall

246

5.32

Method

2,

height

of

collimation

247

Exercises

5

(a)

(Booking)

254

5.4

Field

testing

of the

level

257

5.41

Reciprocal

levelling

method

257

5.42

Two-peg

method

259

Exercises

5 (b)

(Adjustment)

264

5.5

Sensitivity

of

the

bubble

tube

267

5.51

Field

test

267

5.52

O-E

correction

268

5.53

Bubble

scale

correction

268

Exercises

5(c)

(Sensitivity)

270

5.54 Gradient

screws (tilting

mechanism)

271

5.6

The

effect

of

the

earth's

curvature

and

atmospheric

refraction

272

5.61

The

earth's

curvature

272


16/676

Xlll

5.82

The use

of sight

rails

and

boning

(or

travelling)

rods

284

5.83 The

setting

of

slope

stakes

286

Exercises

5(g)

(Construction

levelling)

288

Exercises

5

(h)

(General)

289

TRAVERSE

SURVEYS

298

6.

1

Types

of

traverse

298

6.11 Open

298

6.12 Closed

298

6.2

Methods

of

traversing

299

6.21 Compass

traversing

300

6.22 Continuous

azimuth

method

301

6.23 Direction

method

302

6.

24

Separate

angular

measurement

304

Exercises

6(a)

304

6.3

Office

tests

for

locating

mistakes

in

traversing

306

6.31

A

mistake

in

the

linear

value

of

one

line

306

6.32

A

mistake

in the

angular

value

at

one

station

307

6.33 When the

traverse

is

closed

on to fixed

points

and

a

mistake

in the

bearing

is

known

to exist

307

6.4

Omitted

measurements

in

closed

traverses

308

6.41

Where

the

bearing

of

one

line

is missing

308

6.42

Where

the

length

of

one

line is

missing

309

6-43

Where

the

length

and

bearing

of

a

line

are

missing

309

6.44

Where

the

bearings

of

two

lines

are

missing

309

6.45

Where two

lengths

are

missing 314

6.46

Where

the

length

of

one

line and

the

bearing

of

another

line

are

missing

315


17/676

XIV

7

TACHEOMETRY

359

7.1

Stadia

systems

fixed

stadia

359

7.2

Determination

of

the

tacheometric

constants

m and K

360

7.21

By physical

measurement of

the

instrument

360

7.22

By

field

measurement

361

7.3

Inclined

sights

362

7.31

Staff

normal

to the

line of

sight

362

7.32

Staff

vertical

363

7-4

The

effect

of

errors

in

stadia

tacheometry

367

7.41

Staff

tilted

from

the normal

367

7.42

Error

in the

angle

of

elevation

with the

staff

normal

367

7.43

Staff

tilted

from

the vertical 368

7.44

Accuracy

of

the vertical

angle

to

conform

to

the

overall

accuracy 371

7.45

The

effect

of the

stadia

intercept

assumption

372

Exercises

7(a)

380

7.5

Subtense

systems

383

7.51

Tangential

method

383

7.52

Horizontal

subtense bar

system

388

7.6

Methods used

in

the field

392

7.61

Serial

measurement

392

7.62

Auxiliary

base

measurement

393

7.63

Central

auxiliary

base

395

7.64

Auxiliary

base perpendicularly

bisected

by

the

traverse

line

397

7.65

Two

auxiliary

bases

398

7-66

The auxiliary

base

used

in

between

two


18/676

XV

8.24 Given

two

apparent

dips,

to find

the

rate

and

direction

of

full

dip

416

8.25 Given

the rate

of

full

dip

and

the rate

and

direction

of an

apparent

dip,

to

find

the

direction

of

full

dip

421

8.26

Given

the

levels

and

relative

positions

of three

points

in

a

plane

(bed

or

seam),

to find

the

direction

and

rate

of

full dip

422

8.3

Problems

in

which

the

inclinations

are

expressed

as

angles

and a

graphical

solution

is required

427

8.31 Given

the

inclination

and

direction

of

full

dip,

to

find

the

rate

of

apparent

dip in

a

given

direction

427

8.32

Given

the

inclination

and

direction

of

full dip,

to

find

the

direction

of a

given

apparent

dip

428

8.33

Given

the

inclination

and

direction

of

two

apparent

dips,

to

find

the

inclination

and

direction

of

full

dip

429

Exercises

8(a)

429

8.4

The

rate

of

approach

method

for

convergent

lines

432

8.5

Fault

problems

437

8.51

Definitions

437

8.52

To

find

the

relationship

between the

true

and

apparent

bearings

of

a

fault

443

8.53

To

find

the

true

bearing

of

a

fault

when

the

throw

of

the

fault

opposes

the

dip of the

seam

444

8.54

Given

the

angle

8

between

the

full dip

of

the

seam

and

the

true bearing

of

the

fault,

to find

the

bearing

of

the

line

of

contact

446

8.55

To find

the

true

bearing

of

a

fault

when

the

downthrow

of

the

fault

is

in

the

same

general

direction

as

the dip

of

the

seam

449

8.56

Given

the

angle

8

between

the

full

dip

of

the


19/676

XVI

9.14

Surface areas

461

9.2

Areas

of

irregular

figures

471

9.21

Equalisation

of the boundary

to

give

straight

lines

471

9.22

The

mean

ordinate

rule

472

9.23

The

mid-ordinate

rule

473

9.24

The

trapezoidal

rule

473

9.25

Simpson's

rule

474

9.26

The

planimeter

477

9.3

Plan areas

481

9.31

Units

of

area

481

9.32

Conversion

of

planimetric

area

in

square inches

into

acres

482

9.33 Calculation

of

area

from

co-ordinates

482

9.34

Machine

calculations

with

checks

488

9.4

Subdivisions

of

areas

490

9.41

The

subdivision

of

an

area

into

specified

parts

from a

point

on

the

boundary

490

9.42

The

subdivision

of

an

area

by a

line

of

known

bearing

491

9.43

The

sub-division

of an

area

by

a

line

through

a

known

point

inside

the figure

492

Exercises

9

497

10 VOLUMES

501

10.

1

Volumes

of

regular

solids

501

10.2

Mineral

quantities

509

Exercises

10

(a)

(Regular

solids)

511

10.3

Earthwork

calculations

513

10.31

Calculation

of volumes from

cross-sectional

areas


20/676

XV11

10.64

Free-haul

and

overhaul

546

Exercises

10

(c)

(Earthwork

volumes)

552

11

CIRCULAR

CURVES

559

11.1

Definition

559

11.2 Through

chainage

559

11.3

Length of curve

L

560

11.4

Geometry

of

the

curve

560

11.5

Special

problems

561

11.51

To

pass

a

curve tangential

to

three

given

straights

561

11.52 To

pass

a

curve

through

three

points

563

Exercises

11(a)

566

11.53

To

pass

a

curve

through

a given

point

P

567

Exercises

11(b)

(Curves

passing

through

a

given

point) 571

11.54

Given

a curve

joining

two tangents, to

find

the

change

required in the

radius

for

an

assumed

change

in

the

tangent

length

572

11.6

Location

of

tangents

and

curve

575

11.7

Setting

out of

curves

576

11.71 By

linear

equipment

only 576

11.72

By

linear and

angular equipment

580

11.73

By

angular

equipment

only

580

Exercises

11(c)

588

11.8 Compound curves

591

Exercises

11(d)

(Compound curves;

599

11.9

Reverse

curves

600

Exercises

11(e)

(Reverse

curves)

605

12

VERTICAL

AND

TRANSITION CURVES

607


21/676

XV111

12.6 Transition

curves

627

12.61

Superelevation

627

12.62

Cant

628

12.63

Minimum

curvature

for standard

velocity

628

12.64

Length

of

transition

629

12.65

Radial

acceleration

629

12.7

The ideal

transition

curve

630

12.8

The

clothoid

632

12.81

To find

Cartesian

co-ordinates

632

12.82

The

tangential

angle

633

12.83 Amount

of

shift

633

12.9

The

Bernouilli

lemniscate

634

12.91

Setting

out

using

the

lemniscate

635

12.

10

The

cubic

parabola

636

12.11

The

insertion

of

transition

curves

637

12.12

Setting-out

processes

640

12.

13

Transition

curves

applied

to

compound

curves

644

Exercises

12(b)

649

Abbreviations

used

for

Examination

Papers

E.M.E.U. East

Midlands

Educational

Union

I.C.E.

Institution

Of

Civil

Engineers

L.U.

London

University

B.Sc. (Civil

Engineering)

L.U./E

London

University

B.Sc.

(Estate

Management)

M.Q.B./S

Mining

Qualifications

Board

(Mining

Surveyors)

M.Q.B./M

Mining

Qualifications

Board

(Colliery

Managers)

M.Q.B./UM

Mining

Qualifications

Board

(Colliery

Undermanagers)

N.R.C.T.

Nottingham

Regional

College

of

Technology


22/676

LINEAR

MEASUREMENT

1.1

The

Basic

Principles

of

Surveying

Fundamental

rule 'Always

work

from

the

whole to

the

part*.

This

implies

'precise

control

surveying'

as the first

consideration,

followed

by

'subsidiary

detail

surveying'.

A

point

C

in

a

plane may be fixed

relative

to

a

given line

AB in

one

of

the

following

ways:

1. Triangulation

Angular

measurement

from

a fixed

base

line. The

length

AB

is

known.

The

angles a

and

/3

are

measured.

B

a.

Xe

li

.V

Fig.

1.1(a)

2.

Trilateration

Linear

measurement

only. The

lengths

AC

and

BC

are

measured

or

plotted. The

position of

C

is always

fixed

provid-

ed

AC

+

BC

>

AB.

Uses:

(a)

Replacing

triangulation with

the use

of

microwave

mea-

suring

equipment.

(b)

Chain

surveying.

A


23/676

SURVEYING

PROBLEMS

AND

SOLUTIONS

3.

Polar

co-ordinates

Linear

and

angular

measurement.

Uses:

(a)

Traversing.

(b)

Setting

out.

(c)

Plotting

by

protractor.

,-

c

(s,6)

BhT

Fig. 1.1(c)

4. Rectangular

co-ordinates

Linear

measurement

only

at

right-angles.

Uses:

(a)

Offsets.

(b)

Setting

out.

(c) Plotting.

A

A

Bit

90

OC


24/676

LINEAR

MEASUREMENT

3

mined.

(3)

The

degree

of

accuracy,

or

its

precision,

can

only

be quoted

as

a relative

accuracy,

i.e.

the estimated

error

is

quoted as

a fraction

of

the

measured

quantity.

Thus

100

ft measured

with

an

estimated

error

of

1

inch

represents

a

relative

accuracy of

1/1200.

An

error of

lcm

in 100 m

=

1/10000.

(4)

Where

readings

are

taken

on

a graduated

scale

to the

nearest

subdivision,

the

maximum error

in

estimation will

be

l

/

2

division.

(5)

Repeated

measurement

increases

the

accuracy

by

y/n, where

n

is the

number

of

repetitions.

N.B.

This cannot

be

applied

indefinite-

ly-

(6)

Agreement

between repeated

measurements

does

not

imply

accuracy

but

only

consistency.

1.3

Significant

Figures in

Measurement

and

Computation

If

a

measurement

is recorded

as

205

ft to

the

nearest

foot,

its

most

probable

value

is

205

0*5

ft,

whilst

if measured

to the

nearest

0*1

ft

its

most

probable

value

is

205-0

0-05

ft.

Thus

the

smallest recorded

digit

is subject

to

a

maximum error

of

half

its

value.

In

computation,

figures are

rounded

off

to the

required

degree

of

precision,

generally

by

increasing

the

last

significant

figure

by

1

if the

following

figure

is

5

or

more.

(An alternative

is

the

rounding

off

with

5

to

the

nearest

even

number.)

Thus

205-613 becomes

205-61 to

2

places,

whilst

205-615 becomes

205-62

to

2

places,

or

205-625

may

also

be

205*62,

giving

a

less

biased

value.

It

is

generally

better to

work to

1

place

of decimals

more

than

is

required

in the

final

answer,

and

to

carry

out the rounding-off

process

at the end.

In

multiplication

the

number of

significant figures

depends

on

the


25/676

4

SURVEYING

PROBLEMS

AND

SOLUTIONS

Thus

the relative

accuracy

of the

product

is the

sum of

all

the

relative

accuracies

involved

in

the

product.

Example

1.1

A rectangle

measures

3-82 in. and

7-64

in.

with errors

of

0*005

in.

Express

the

area

to the

correct

number

of

significant

figures.

P

=

3-82

x

7-64

=

29*184

8

in

2

relative

accuracies

~

_i_

3-82

~

750

0-005

..

1

7-64

1500

500

SP =

290-

+

-L-)

=

\750 1500/

=

0-06

.-.

the

area

should

be

given as

29-2in

2

.

As

a

general

rule

the

number

of

significant figures

in

the

product

should

be

at

least

the same

as,

or

preferably

have

one

more

significant

figure

than,

the

least

significant

factor.

The

area

would

thus

be

quoted as

29-18 in

2

In

division

the

same

rule

applies.

Q

=

-

y

Q

+

8Q

=

x

+

8x

=

*

+

f

-

rf^

+

...

y

+

8y

y

y

y

2

Subtracting

Q

from both

sides

and

dividing

by

Q

gives

SQ

=

Q

(?I

-

*)

(1.2)


26/676

LINEAR

MEASUREMENT

5

nSR

=

Sx

8R

_

8x_

R

n

~

nx

8R

=

-8x

(1.4)

Example

1.2

If

R

=

(5-01

0-005)

2

5-01

2

=

25-1001

8R =

2

x 0-005 =

0-01

.'.

R

should

be

given

as 25*10

Example

1.3

If

R

=

V

25

*

10

*

01

v'25-10

=

5-009

9

8R =

^

=

0-005

.*.

R

should

be given

as

5-01

Example

1.4

A

rectangular

building

has

sides

approximately

480

metres

and

300

metres.

If

the

area

is

to

be

determined

to

the

nearest

10

m

2

what

will

be

the

maximum

error

permitted

in each

line,

assuming

equal

precision

ratios

for

each

length?

To

what

degree

of

accuracy

should the

lines

be

measured?

A =

480

x

300

-

144

000 m

2

8A

=

10

m

2

8A

=

_1

=

x

Sy

A

14400

x

+

y

but

8x

=

8y

.

8x

8y

_

28x

x

y

x

y

~

x

8x_

1


27/676

SURVEYING

PROBLEMS

AND

SOLUTIONS

1.4

Chain

Surveying

The

chain

There

are

two

types

(a)

Gunter's

chain

1

chain*

=

100

links

=

66

ft

1

link

=

0-66

ft

=

7-92

in.

Its

advantage

lies

in

its

relationship

to

the

acre

10

sq

chains

=

100

000

sq

links

=

1 acre.

(b)

Engineer's

chain

100

links

=

100

ft

(Metric

chain

100

links

=

20

m

1

link

=

0-2

m)

Basic

figures

There are

many

combinations

of chain

lines

all

dependent

on

the

linear

dimensions

forming

trilateration,

Fig.

1

.2.

Tie line

C

A


28/676


29/676

SURVEYING PROBLEMS

AND

SOLUTIONS

or

A

T

=

A

M

(l

y)

(1.8)

Effect

of

standardisation

on

volumes

Based

on

the

principle

of

similar

volumes,

,

/

true

length

of

tape

V

true

volume

V

T

=

apparent volume

x

(

apparent

length

of

tapJ

ue.

V

r

=

V(l

^)

(110)

N.B.

Where

the

error

in

standardisation is

small

compared

to

the

size

of

the

area,

the

%

error

in

area

is

approximately

2

x

%

error

in

length.

Example

1.6

A

chain is found

to be

0*8

link

too long and

on using

it

an area of

100

acres

is

computed.

. .

inn A00-8\

2

The true

area

= 1UU

I

-

TqTT)

=

100

x

1-008

2

= 101-61

acres

alternatively,

linear

error

=

0*8%

area

error

=

2

x

0*8

=

1*6%

acreage

=

100 +

1*6 acres

= 101*6 acres

This

is

derived

from

the

binomial

expansion

of

(1

+ x)

z

=

1

+

2x

+

x

2

i.e .if

x

is small

x

z

may

be

neglected

/.

(1

+

x)

2

a

1

+

2x

Correction

for

slope

(Fig.

1.3)


30/676

LINEAR

MEASUREMENT

h

Fig.

1.3

(1)

Given

the

angle

of

inclination

a

AB

=

AC

cos

a

i.e.

h

=

/

cos

a

(1.11)

c

=

I

-

h

=

I

-

I

cos a

=

/(1-cosa)

=

/

versine

a

(1-12)

N.B.

The

latter

equation

is

a

better

computation

process.

Example

1.7

If

AC

=

126-3

m,

a

=

234\

byEq.(l.ll) AB

=

126-3

cos

234'

=

126-3

x

0-999

=

126-174

m

or by

Eq.

(1.12)

c

=

126-3

(1

-

0-999)

=

126-3

x

0-001

=

0-126

m


31/676

10

SURVEYING

PROBLEMS

AND SOLUTIONS

cos

a

=

1

-

0-001

=

0-999

a

=

234'

(i.e.

1

in

22)

Also, if

-j

=

=

=

1

-

cos a

3000

cos

a

=

1

-

0-00033

=

0-99967

a

=

129'

(i.e.

1

in

39)

If

the

difference

in level

,

d,

is known

h

=

(I

2

-

d

2

y

=

j(/-d)x

(/+

d)}*

or

I

2

=

h

2

+ d

2

=

(/

-

cf

+

d

2

=

I

2

-

2lc

+ c

2

+ d

2

.-.

c

2

-

2lc

=

-d

2

c(c-2l)

=

-d

2

c

=

-d

2

c-2l

c

~

d

z

as

c

is

small

compared

(1.13)

Rigorously,

using

the binomial expansion,

c

-

I

-

(I

2

-

d

2

y

-'-


32/676

MEASUREMENT

11

For

setting

out

purposes

Here

the

horizontal

length

(h)

is

given

and

the

slope

length

(/)

is

required.

/

=

h

sec

a

c

=

h

sec

a

-

h

=

h(sec

a

-

1)

(1.16)

Writing

sec

a

as

a

series

1

+

^-

+

^+

,

where

a

is

in

radians,

i

x

see

p.

72.

ho.

2

-s-

(

in

radians)

(1.17)

~

^(0-017

45a)

2

2i

1-53 ft

x

10~

4

x

a

2

(a

in

degrees)

(1.18)

-

1*53 x

10

2

x a

2

per

100

ft

(or

m)

(1.19)

Example

1.9

If

h

=

100

ft

(orm),

a

=

5,

by

Eq.

(1.16)

c

=

100(1-003

820-1)

=

0-382

Oft

(orm)

per 100

ft

(orm)

or

by

Eq.

(1.18)

c

= 1-53

x

100

x

10

4

x

5

2

=

1-53 x

25

x

10~

2

=

0-382

5

ft

(or

m)

per

100

ft

(or

m)

Correction

per

100ft

(orm)

1

0-015

ft (orm)

6 0-551

ft

(orm)

2

0-061

ft

(orm)

7

0-751 ft

(orm)

3

0-137

ft

(orm)

8

0-983 ft

(orm)

4


33/676


34/676

14

SURVEYING

PROBLEMS

AND SOLUTIONS


35/676

If

the

maximum length

PP

%

represents

the minimum

plotable

point,

12

i.e.

0*01

in

which

represents

^

xft,

where x

is

the representative

fraction

1/x, then

0-000

83

x

=

la

I

206 265

171-82

x

a

Assuming

the

maximum

error

a

=

4,

i.e.

14400

,

=

171-82*

o-012

x

(1.25)

14400

If

the

scale

is

1/2500,

then

x

=

2500,

and

/

=

2500

x

0-012 =

30ft

(^10

m)

If the

point

P

lies

on

a

fence

approximately

parallel to

ABC,

Fig.

1.6,

then the

plotted

point

will

be

in

error

by

an

amount

P^P

2

=

1(1- cos

a). (Fig.

1.5).

Boundary

line

12

(1

- cos

a)

=

4,

Eq.

fl

C

Fig.

1

.6

,

=

0-01

(1>26)

LINEAR

MEASUREMENT

15


36/676

the

direction

of

the

measurement

being

ignored,

Fig.

1.7.

Fig.

1.7

1.43

Setting

out

a

right

angle

by

chain

From

a

point

on

the

chain line (Fig.

1.8)

(a)

(i)

Measure

off

BA

=

BC

(ii)

From

A

and

C

measure

off AD

=

CD

(Proof: triangles

ADB

and

DCB

are

congruent, thus

ABD

=

DBC

=

90

c

as ABC

is

a

straight

line)

Fig.

1.8

Z

/

/

/

7k

\

\

\

8

B

Fig.

1.9

(b)

Using

the

principle

of

Pythagoras,

z

z

=

x

z

+

y

2

(Fig.

1.9)


37/676

LINEAR

MEASUREMENT

17


38/676

(2)

When

the

point

is

not

accessible

(Fig.

1.12).

From

D

set

out

lines

Da

and

Db

and,

from

these

lines,

perpendicular

ad

and

be.

The

inter-

section

of

these

lines

at X

gives

the

the

line

DX

which

when

produced

gives

B,

the

required

point.

To

set out

a

line

through

a

given

point

parallel

to

the

given

chain

line

(Fig.

1.13).

Given

the

chain

line

AB

and

the

given

point

C.

From

the

given

point

C

bisect

the

line

CB

at

X.

Measure

AX

and

produce

the

line

to D

such

that AX

=

XD.

CD will

then

be

parallel

to

AB.

1.45

Obstacles

in

chain

surveying

18

SURVEYING

PROBLEMS

AND SOLUTIONS


39/676

(1)

Obstacles

to

ranging

(a) Visibility

from

intermediates

(Fig.

1.14). Required to

line

C and

D on

the line

AB.

Place ranging pole

at d,

and

line

in

c,

on

line

Ad

}

.

From

B

ob-

serve

c,

and

move

d

2

on

to

line Be,

.

Repetition

will produce

c

2

,

c

3

and

d

2

,

d

3

etc

until

C and

D

lie on

the

line AB.

(b) Non-visibility

from

intermediates

(Fig.

1.15).

Required

to

measure

a

long

line

AB

in

which

A and

B

are not

inter-

visible and intermediates

on

these

lines

are not

possible.

Set out

a 'random

line'

AC

approximately on

the

line

AB.

From

B find the

perpendicular

BC

to line

AC

as

above.

Measure

AC

and

BC.

Calculate

AB.

(2)

Obstacles

to

chaining

(a)

No

obstacle to

ranging

(i)

Obstacle can

be

chained

around.

There

are

many

possible

variations

depending on

whether

a

right angle

is set out

or

not.

o*=

A

B

Fig.

1.15

LINEAR

MEASUREMENT

19


40/676

Fig.

1.17

VI

Set

out

line

be so

that

bB

=

Be.

Compute

BC

thus,

BC*

=

CO

2

Be

+

jCcf

bB

_

be

but

bB

=

Be,

.

RC

z

BbjbC

2

-

Cc

2

)

_.

DK

-

^

=

2fib

-BbxBb

=

UbC

z

+ Cc

2

)

-

Bb

2

(1.30)

(1.31)

20

SURVEYING PROBLEMS

AND

SOLUTIONS


41/676

d

2

=

q

z

x +

p

2

y

x

+

y

-

xy

--/

Q

x +

p

y

x

+

y

-

xy

(1.32)

If

x

=

y,

(ii) Obstacle cannot be

chained

around.

A

river or

stream

represents

. this

type of obstacle. Again

there

are many variations

depending on

whether a

right

angle is set

out or

not.

By

setting

out

right

angles

(Fig.

1.19).

A random line

DA^

is

set out and

from perpendiculars

at

C and

B

points

C

and

B

are

obtained.

By similar

triangles DC,C and

C

1

B

1

B

2

,

DC

CB

DC

CC

X

Bfi,

-

CC

y

CB

x

CC,

BB.

-

CC,

LINEAR

MEASUREMENT

21

.


42/676

Without

setting

out

a

right

angle

(Fig.

1.20).

A

point

F is

chosen.

From

points

B

and

C

on

line

AE,

BF

and

CF

are

measured

and

produced

to

G

and

H.

BF

=

FG

and

CF

=

FH.

The

intersection

of

DF

and

GH

produce

to

intersect

at

J.

Then

HJ

=

CD.

(iii)

Obstacles

which

obstruct

ranging

and

chaining.

The

obstruction,

e.g.

a

building,

prevents

the

line

from

being

ranged

and

thus

produced

beyond

the

obstacle.

By

setting

out

right

angles

(Fig.

1

.21)

On

line

ABC

right

angles

are

set

out

at B

and

C

to

produce

B

and

C,

,

where

SB,

=

CC,

'

B,

C,

is

now

produced

to

give

D,

and

E

t

where

right

angles

are

set out

to

give

D

and

E,

where

D^D=E^E=BB

y

=

CC,

.

D

and

E

are

thus

on

the

line

ABC

produced

and

0,0,

=

DC.

22

SURVEYING

PROBLEMS

AND

SOLUTIONS


43/676

Exercises

1

(a)

1.

The

following

measurements

were

made

on

inclined ground.

Reduce the slope

distances

to

the

horizontal

giving

the answer in

feet.

(a)

200-1

yd

at

1

in

2^

(b)

485*5 links

at

1

in

5-75

(c)

1/24

th

of

a

mile at

1

in 10-25

(Ans.

(a)

557-4 ft

(b) 315-7

ft (c)

218-9

ft)

2.

Calculate

the

acreage

of an area of 4

in

2

on

each

of the plans

drawn

to

scale, 2

chains

to

1

in.,

1/63

360, 1/2500

and

6

in.

to

1

mile

respectively.

(Ans.

1-6,

2560,

3-986,

71-1 acres)

3.

A

field

was measured

with a

chain

0-3

of

a link too

long. The

area

thus

found

was

30

acres.

What is

the

true

area?

(I.C.E.

Ans.

30-18

acres)

4.

State

in acres and

decimals

thereof

the

area

of an

enclosure

mea-

suring

4

in.

square

on

each

of

three

plans

drawn

to

scale

of

1/1584,

1/2500,

1/10560

respectively.

(Ans.

6-4,

15-9,

284-4

acres)

5.

A

survey

line was

measured

on

sloping

ground and

recorded as

386-6

ft

(117-84

m).

The

difference

of

elevation between

the ends

was

19-3 ft (5-88 m).

The tape

used was

later found

to

be

100-6

ft

(30-66

m)

when

com-

pared with

a

standard

of 100

ft (30-48

m).

Calculate

the

corrected

horizontal

length

of the

line.

(Ans.

388-4

ft

(118-38

m))

6.

A plot

of

land in

the

form of a

rectangle

in

which

the

length is

twice the

width

has

an

area

of

180000

ft

2

.

Calculate

the

length

of

the

sides

as

drawn

on

plans

of

the

follow-

ing

scales.

(a)

2

chains to 1

inch,

(b)

1/25000.

(c)

6

inch

to

1

mile.

LINEAR

MEASUREMENT

23


44/676

8.

Find,

without

using

tables,

the

horizontal

length

in

feet

of

a

line

recorded

as

247*4

links

when

measured

(a) On

ground

sloping

1

in

4,

(b) on

ground

sloping

at

1826'

(tanl826'

=

0-333).

(Ans.

(a)

158-40

(b)

154-89

ft)

9.

Show

that

for

small

angles

of

slope

the

difference

between

the

horizontal

and

sloping

lengths

is

h

z

/2l

(where

h

is

the

difference

of

vertical

height

of

the

two

ends

of

a

line

of sloping

length

/).

If errors

in

chaining

are

not

to

exceed

1

part

in

1000,

what

is

the

greatest

slope

that

can

be

ignored?

(L.U./E

Ans.

1 in

22*4)

1.5

Corrections

to

be

Applied

to

Measured

Lengths

For

every

linear

measurement

the

following

corrections

must

be

considered,

the

need

for

their

application

depending

on

the

accuracy

required.

1.

In

all

cases

(a)

Standardisation.

(b)

Slope.

2.

For

relative

accuracies

of

1/5000 plus

(a)

Temperature.

(b)

Tension,

(c)

Sag.

(where

applicable)

3.

For

special

cases,

1/50000

plus

(a)

Reduction

to

mean

sea

level.

(b)

Reduction

to

grid.

Consideration

has

already

been

given,

p.

6/9,

to

both

standardisa-

tion

and

reduction

to the

horizontal

as

they

apply

to

chain

surveying

but

more

care

must

be

exercised

in

precise

measurement

reduction.

1.51

Standardisation

The

measuring

band

in

the

form

of

a

tape

or

wire

must

be

compared

with

a

standard

24

SURVEYING PROBLEMS

AND

SOLUTIONS


45/676

the

measured vertical angle

6

the

slope

of

the measured line

a

the

length of the

measured

line

/

Fig.

1.23

In

Fig.

1.23,

a

=

d

+

8$.

In

triangle

A

A

B

Z

B^

by

the sine rule

(see

p.

80),

(h,

-h

z

)

sin

(90 +

0)

sin

86

86

I

(fc,

-

h

2

)

cos

6

/

206 265

(h

}

-h

z

)

cos

6

_

(1.34)

(1.35)

N.B.

The

sign

of

the correction

conforms

precisely

to

the

equation.

(1)

If

/i

i

=

h

z

,

86

=

a

=

6

(2)

If h\

h

z

and

6

is

-ve,

86

is

-ve

(Fig.

1.24d)

if

a is +ve,

86

is +ve

(Fig.

1.24b)

(4)

If

/i,

4

-/>,

(a)

(b)

(c)

(d)

Fig.

1.24

Correction

to

measured

length

(by

Eq.

1.12),

26

SURVEYING PROBLEMS

AND

SOLUTIONS


47/676

1.53

Correction

for temperature

The

measuring

band

is

standardised

at

a

given

temperature

(t

8

).

If

in the field

the temperature of the

band

is recorded as

(*

m

)

then the

band

will

expand

or

contract

and

a

correction to

the measured

length

is given as

c

=

la(t

m

-

t

a )

(1.36)

where

/

=

the

measured

length

a

=

the

coefficient

of linear

expansion

of

the

band

metal.

The coefficient of

linear

expansion

(a)

of

a solid is

defined

as

'the

increase

in

length

per

unit

length

of

the

solid

when

its

temper-

ature

changes

by

one degree'.

For

steel the

average

value

of

a

is

given as

6-2

x

10

6

per

F

Since

a change

of

1

F

=

a change

of

5/9

C,

using

the value

above

gives

a

=

6-2

x

10

6

per

5/9

C

=

11-2

x

10

6

per

C

The range of linear coefficients

a

is

thus

given as

Steel

Invar

per 1C

-6

10-6

to

12-2

(x

10

)

5-4

to

7-2

(xl0~

7

)

per 1F

5-9

to

6-8

3

to

4

To

find

the

new

standard

temperature t'

8

which

will

produce the

nominal

length

of

the band.

Standard length

at t

a

=

I

81

To

reduce

the

length by

5/:

81

=

(/

5/).a.r

where t

=

number

of

degrees

of

temperature

LINEAR

MEASUREMENT

27


48/676

Example

1.14

A

traverse

line

is

500

ft

(152*4

m)

long. If

the

tape

used

in

the

field

is

100

ft

(30-48

ra)

when

standardised

at 63

F

(17-2

C),

what

correction

must

be

applied if

the

temperature

at

the

time

of mea-

surement

is

73

F

(22-8

C)?

(Assume a

=

6-2

x

10~

6

per

deg

F

=

11-2

x

10~

6

per

deg

C)

From

Eq.

(1.36)

c

m

=

500

x

6-2

x

10~

6

x

(73

-

63)

=

+0-031

Oft

or

c

(m)

=

152-4

x

11-2

x

10

6

x

(22-8

-

17-2)

=

+0-009

6

m

Example

1.15

If

a

field

tape

when

standardised

at

63

F

measures

100-005

2

ft,

at

what

temperature

will

it

be

exactly

the

nominal

value?

(Assume

a

=

6-5

x

10

6

per

deg

F)

SI

=

+0-0052

ft

'.

from

Eq.

(1.37)

t'

=63

0-0052

100 x

6-5

x

10

6

=

63F-8F

=

55

F

In

its

metric form the

above

problem

becomes:

If

a

field

tape

when

standardised

at 17-2 C

measures

100-005

2 m,

at

what

temperature

will

it

be

exactly

the

nominal

value?

(Assume

a

=

11-2

x

10

6

per

deg C)

81

=

+0-005

2

m

.'.

from

Eq.(1.37)

t'

s

=

17#2

_

0-0052

100 x

11-2

x

10

6

28

SURVEYING

PROBLEMS

AND

SOLUTIONS


49/676

where

L

=

the

measured

length

(the

value

of

c

is

in

the

same

unit

as

L),

A

=

cross-sectional

area

of

the

tape,

E

= Young's

modulus of elasticity

i.e.

stress/strain.

The

units

used

for

T,

A

and E must

be

compatible,

e.g.

T(lbf)

A

(in

2

)

E

(lbf/in

2

)

or

T,(kgf)

4,

(cm

2

)

E,

(kgf/cm

2

)

(metric)

or

T

2

(N)

A,(m

2

)

E

2

(N/m

2

)

(new

S.I.

units)

Conversion

factors

lib

=

0-453592

kg

1

in

2

=

6-451

6

x

10~

4

m

2

.'.

lib/in

2

=

703-070

kg/m

2

Based

on

the

proposed

use

of

the

International

System

of

Units

(S.I.

units)

the

unit

of

force

is

the

Newton (N), i.e.

the

force

required

to

accelerate

a

mass of

1 kg

1

metre

per second

per

second

The

force

1

lbf

=

mass

x

gravitational

acceleration

=

0-453592 x

9-80665

m/s

2

(assuming

standard

value)

=

4-448 22N

lkgf

=

9-806

65

N

(1

kg

=

2-204

62

lb)

whilst for

stress

1

lbf/in

2

=

6894-76

N/m

2

For steel,

E

~

28

to

30

x

10

6

lbf/in

2

(British

units)

~

20

to

22

x

10

5

kgf/cm

2

(Metric

units)

ot

19-3

to

20-7

x

10

10

N/m

2

(S.I.

units)

For

invar,

E

~

20 to

22

x

10

6

lbf/in

2

^

14

to 15-5

x

10

5

kgf/cm

2

~

13-8

to

15*2

x 10

,0

N/m

2

LINEAR

MEASUREMENT

29

Example

1.16

A

tape

is

100

ft

at

a

standard


50/676

tension

of

251bf

and

mea-

sures

in

cross-section

0-125

in.

x

0-05

in. If

the

applied

tension

is

20

lbf

and

E

=

30 x

10

5

lbf/in

2

,

calculate

the

correction

to

be

applied.

Rv

F1C.Q

.

10

X

(20

-

25)

By

t-q.

l.iy

c

=

.

i

=

-0*009

7 ft

(0-125

x

0-05)

x

(30

x

10

6

)

27

Converting

the

above

units

to

the

metric

equivalents

gives

c

=

30-48

m

x

(9-072

-

11-340)

kgf

(40-32

x

10~

7

)m

2

x

(21-09

x

10

9

)

kgf/m

2

=

-0-008

13

m

(i.e.

-0-002

7 ft)

Based

on

the

International

System

of

Units,

2-268

kgf

=

2-268

x

9-806

65

N

=

22-241

or

5

lbf

=

5

x

4-448

22

N

=

22-241 N.

For

stress,

(21-09

x

10

10

)

kgf/m

2

=

21-09

x

10

9

x

9-80665

=

20-684

x

10

,o

N/m

2

or

(30

x

10

6

)

lbf/in

2

=

30

x

10

6

x

6894-76

=

20-684

x

10'

N/m

2

Thus,

in

S.I.

units,

30-48

x

22-241

c

=

(40-32

x

10

-7

)

x

(20-684

x

10

,o

)N/m

2

=

-

0-008

13

m

Measurement

in

the

vertical

plane

Where

a

metal

tape

is

freely

suspended

it

will

elongate

due

to

the

applied

tension

produced

by

its

own

weight.

The

tension

is

not

uniform

and

the

stress

varies

along

its

length.

Given

an

unstretched

tape

AB and

a

stretched

tape

AB,

,

Fig.

1.25,

let

P

and

Q

be

two

close

points

on the

tape

which

become

^Q,

under

'

tension.

30

SURVEYING

PROBLEMS AND SOLUTIONS

if

the tension

at

is

T

+


51/676

Q

dT

T-(T

+ dT)

=

wdx

i.e.

dT

=

-wdx

(2)

x

+

s

dx+ds

Fig.

1.2

5

Elongation

in

a

suspended tape

In

practice

the

value

w is a function

of

x

and by

integrating the

two equations the tension

and

extension are

derived.

Assuming

the

weight

per

unit

length

of the

tape is

w with

a

sus-

pended

weight

W,

then

from

(2)

dT

=

-wdx

LINEAR

MEASUREMENT

31

Therefore

putting

constants

into

equations

(3)

and


52/676

(4)

gives

T

=

-wx

+ W

+

wl

T

=

W

+

w(l-x).

(i.4i)

and

EAs

=

-

1

wx

2

+

Wx

+

wlx

If

x

=

/,

then

and

if

W

=

0,

=

Wx

+

w(2lx-

x

2

)

=

ml

Wx

+

i

w

&

lx

-*

2

)]

d-42)

S=

A[^

+

I

w/2

]

(1-45)

S

~

2E4

(1.44)

Example

1.17 Calculate

the

elongation

at

(1)

1000

ft

and

(2)

3000

ft

of

a

3000

ft

mine-shaft

measuring

tape

hanging

vertically

due

to

its

own

weight.

The

modulus

of

elasticity

is

30

x 10

6

lbf/in

2

;

the

weight

of

the

tape

is 0*05

lbf/ft

and

the

cross-sectional

area

of

the

tape

is

0*015

in

2

.

From

Eq.(1.42)

s

=

ljz[

w

*+

i (2lx-x

z

)]

As

W

=

0,

s=

m

[2lx

-

x

*

]

when

x

=

1000

ft

/ =

3000

ft

32

SURVEYING

PROBLEMS

AND

SOLUTIONS

Example

1.18

If

the

same tape is

standardised

as

3000

ft

at

451bf


53/676

tension

what

is the

true

length of

the

shaft recorded

at

2998*632

ft?

IEq.(1.44)

s-

-S?

-i:

2EA

EA

i.e. T

=

\W

where

W

=

total

weight

of tape

=

3000 x

0-05

=

150

Ibf

Applying

the tension

correction,

Eq.(1.39),

L(J

m

-

T

s

)

C

=

EA

3000(75-45)

=

30

x

10

2

x

30

=

.

200ft

30

x

10

6

x

0-015

30

x

10

6

x 0-015

.'.

true

length

=

2998-632 +

0-2

=

2998-832

ft

1.55

Correction

for

sag

The

measuring band may

be

standardised in

two

ways,

(a) on

the

flat or

(b)

in catenary.

If the

band is

used

in

a

manner

contrary to the

standard

conditions

some correction

is

necessary.

(1)

//

standardised

on the

flat

and used

in

catenary

the general

equation

for

correction

is

applied,

viz.

w

2

/

3

c

=

-

r-^

d-45)

(1.46)

24

T

2

W

2

l

24T

where

w

=

weight

of

tape

or wire per unit

length

W

=

wl

=

total weight of tape in

use,

LINEAR

MEASUREMENT

33

of

the

tape

or


54/676

(b)

the

length

of

the

tape

in

catenary

may

be

given.

(i)

If

the

tape

is

used

on the

flat

a

positive

sag

correction

must

be

applied

(ii)

If

the

tape

is

used

in

catenary

at

a

tension

T

m

which

is

different

from

the

standard

tension

T

s

,

the

correction

will

be

the

difference

between

the

two

relative

corrections,

i.e.

W

z

l

r

1

IT

C

--24[n;-Tl\

(1-47)

If

T

m

> Ts

the

correction

will

be

positive.

(iii)

If

standardised

in

catenary

using

a

length

l

a

and

then

applied

in

the

field

at

a

different

length

l

m

,

the correction

to

be

applied

is

given

as

c

=

?i

.

//,

3

w

2

HW'

h

\24I

2

24T

Alternatively,

the

equivalent

tape

length

on

the

flat

may

be

com-

puted

for

each

length

and

the

subsequent

catenary

correction

applied

for

the

new

supported

condition,

i.e.

if

/,

is

the

standard

length

in

catenary,

the

equivalent

length

on

the

ground

=

/.

+

c

where

c

=

the

catenary

correction.

s

If

l

m

is

the

applied

field

length,

then

its

equivalent

length

on

the

flat

=

W

+ c

s

)

Applying

the

catenary

correction

to

this

length

gives

lm

+ C =

(l

s

+

C

j

_

Cm

34

SURVEYING

PROBLEMS

AND

SOLUTIONS

The

sag

correction is

an

acceptable

approximation

based

on

the


55/676

assumption

that

the

measuring

heads

are at

the

same

level.

If

the

heads

are at

considerably

different

levels, Fig. 1.27,

the

correction

should

be

c

=

c,

cos

2

(l^sin*)

(1.49)

the sign

depending on

whether

the

tension

is

applied

at the

upper or

lower end of

the

tape.

Measuring

head

Fig. 1.27

For

general

purposes

c

=

c,

cos

2

w

2

Pcos

2

g

24T

2

(1.50)

The

weight

of

the

tape

determined

in

the

field

The

catenary

sag

of

the

tape

can be

used

to

determine

the

weight

of

of the tape,

Fig.

1.28

LINEAR

MEASUREMENT

35

Example

1.19

Calculate

the

horizontal

length

between

two


56/676

supports,

approximately

level,

if the

recorded

length

is

100.237

ft,

the

tape

weighs

15

ozf

and

the

applied

tension

is 20

lbf

From

Eq.(1.46)

c

=

-

W

*

1

24T

2

The

value

of

/

is

assumed

to

be

100

for

ease

of

computation.

Then

5\

2

jjx

100

24

x

20

2

= -0-0092

ft

True

length

=

100-2370

-

0-0092

=

100-227

8

ft

Example

1.20

A

100

ft

tape

standardised

in

catenary

at

25

lbf

is

used

in

the

field

with

a

tension

of

20

lbf.

Calculate

the

sag

correction

if

w

= 0-021

lbf/ft.

From

Eq.

(1.47)

c

=

-[1^1

T

2

T

2

)

-

100

3

x

0-021

2

/

1

_1_

24

(20

5

25

1

=

-0-01656

i.e.

-0-016

6

ft.

Example

1.21 A

tape

100

ft

long

is

suspended

in

catenary

with

a

ten-

sion

of

30 lbf.

At

the

mid-point

the

sag

is

measured

as

0-55

ft.

Cal-

culate

the

weight

per

ft

of

the

tape.

From

Eq.

(1.51),

36

SURVEYING

PROBLEMS

AND

SOLUTIONS

Thus

there is no significance in changing

the

weight

W

and

tension T


57/676

into units

of

force,

though

the

unit of

tension

must be

the

newton.

30-552

2 /

0-425

\

2

24

\

9-072/

=

-0-002

8 m

(-0-009

2

ft).

1.20(a) A

30*48

m

tape

standardised

in

catenary

at

111-21

N

is

used

in

the field with a tension

of

88-96

N.

Calculate

the

sag

correction

if

w=

0-0312kgf/m.

Conversion

of

the mass/

unit

length

w

into

a

total

force

gives

30-48

x

0-0312

x

9-80665

=

9-326

N.

Eq.

(1 .47)

becomes

c

=

24

'--

-)

T

2

T

z

\88-96

2

111-21

2

/

=

-30-48

x

9-326

24

=

-0-00504

m

(-0-016

6

ft).

1.21(a)

A tape

30-48

m

long is

suspended

in catenary

with

a tension

of

133-446

N.

At the mid-point

the

sag is

measured

as

0-168

m.

Cal-

culate

the weight

per

metre

of the

tape.

Eq.

(1.51)

becomes

n (/

,

8 x

T

x

y

0-816

Ty

w (kgf/m)

=

-

=

9-80665 /

2

I

2

0-816

x

133-446

x

0-168

30-48

2

=

0-019

6

kgf/m

(0-013

2

lbf/ft)

LINEAR

MEASUREMENT

37

True

length

of

sub-length

on

flat


58/676

49-964

=

1Q0

x 99-9747

=

49-952

Sag

correction

for

50

ft

(c oc

/

3

)

=

l/8c,

= -0-005

True

length

between

supports

=

49-

947

ft

Alternatively,

by

Eq.(1.48)

50

x

0-01

2

=

100

x

24 x

10

(5

10

')

= -0-018

ft

.

true

length

between

supports

=

49-964

-

0-018

=

49-946

ft.

Example

1.23

A

survey questions and problems

Documents