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CHAIN SURVEYING

OBJECTIVES

Explain the methods of measurement of distance

Describe the construction of standard chain and tape

Explain the method of ranging and measuring the length of a survey line

List the different errors that can occur during the measurement of distance using chain or tape

Apply the necessary corrections to measured distances

Describe the equipment and methods for chain triangulation or traversing

Explain the methods of laying out chain angles

Describe the method of booking field notes

Explain obstacles to chaining and methods to overcome the same

Explain the methods of plotting chain survey data

Explain the limits of precision in chain surveying

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THE METRIC CHAIN AND TAPE

ACCESSORIES FOR CHAIN SURVEY

RANGING A LINE

MEASURING ALONG SLOPE

CORRECTIONS TO MEASUREMENTS

CHAIN TRIANGULATION

PROBLEMS IN CHAINING

OBSTACLES TO CHAINING

CHAIN SURVEY

CROSS STAFF SURVEY

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METRIC CHAIN AND TAPE

Metric chains come in lengths of 5 m, 10 m,

20 m and 30m

Older chains were in 100

feet(engineers),66 feet (Gunter’s) and 33

feet (revenue)

Chains have tallies and rings to identify

intermediate values

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METRIC CHAINS

BIS standard for chains – is 1492- 1964

Made of 4mm galvanized iron wire

Made of links 200 mm long and connected

by circular or oval rings

End links shorter for providing handles

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5M AND 10M CHAINS

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20M AND 30 M CHAINS

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HANDLES,RINGS AND TALLIES

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METRIC TAPE

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TAPES

Cloth tape

Metallic tape

Steel tape

Invar tape

Cloth or linen tapes are not good for field

work as they shrink, tear easily and not

used for survey work.

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TAPES

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METALLIC TAPES

Lengths of 2m, 5m, 10 m, 20m, 30m, 50m etc

Yarn interwoven with metal fibres

Metal ring to hold at the outer end

16 mm wide, marked in cm and m

Rolled out by pulling and rolled back using rotating handle

Commonly used for ordinary survey work

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TAPES

Tapes come with a handle to pull out and in

a case with a handle to pull in.

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STEEL TAPE

Steel tapes are made of galvanized steel or stainless steel

Lengths from 1 m to 50 m Marked in meters, decimeters and

centimeters with end section in millimeters

Costly but very accurate Can be pulled out with the handle and

rolled back automatically Used for accurate survey work

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INVAR TAPE

Made of an alloy of steel and nickel

About 6 mm wide and in lengths of 30m,

50 m and 100m

Very low thermal coefficient

Used for very precise work as in base line

measurement

Should be handled very carefully

16 BACK

ACCESSORIES

1. Ranging rods

2. Ranging poles

3. Arrows

4. Offset rod

5. Wooden pegs

6. Laths and whites

7. Other equipment for clearing bushes, cleaning

ground

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ACCESSORIES

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RANGING

Ranging required when line is longer than a chain/tape length

Placing a line along the shortest distance between points

When end stations are inter-visible, direct ranging can be done

When end stations not inter-visible, indirect ranging is done

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DIRECT RANGING

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INDIRECT RANGING

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LINE RANGER

Line ranger is an instrument for ranging

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BACK

MEASURING ALONG SLOPE

For plotting, horizontal distances are required

For a measured distance along slope, horizontal

distance can be calculated.

Horizontal length is less than length along slope

For a given horizontal distance, slope distance can

be calculated

The increase in length along slope is called

hypotenusal allowance

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METHOD OF STEPPING

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MEASURING ALONG SLOPE

Slope angle measured using clinometer

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MEASURING ALONG SLOPE

When slope angle is known

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MEASURING ALONG SLOPE

When slope is in gradients

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HORIZONTAL DISTANCE

Horizontal distance = L COS θ, Where L is

the slope distance and θ is the slope

angle.

If slope is in gradient, 1:n, then

Horizontal distance = L n/[√(1+n²)]

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HYPOTENUSAL

ALLOWANCE Is the additional distance measured along

the slope to give a chain length

horizontally.

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HYPOTENUSAL

ALLOWANCE Hypotenusal allowance is given by

L [sec θ – 1], exactly and

L θ²/2, where θ is in radians.

Or

Hypotenusal allowance = √(L² + h²) – L (exact

value) or h²/2L approximately.

30 BACK

CORRECTIONS TO

MEASUREMENT 1. Wrong length

2. Slope

3. Alignment

4. Temperature

5. Pull

6. Sag

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INCORRECT LENGTH

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INCORRECT LENGTH

1. Chain or Tape long or short

Correction = (L’/L) x measured length

Where L is designated length

L’ is actual length of chain/tape

Correction to area = (L’/L)²x measured area

Correction to volume = (L’L)³x measured volume

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SLOPE MEASUREMENT

Correction = h²/2L or = Lθ/2

Where h is the height for length L and

Θ is the slope angle in radians.

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INCORRECT ALIGNMENT

Correction = d²/2L

Where d is the deviation over length L

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TEMPERATURE

Correction = ± Lα t

Where L = Length

α = Coefficient of thermal expansion

(12 x 10^(-6) for steel tape)

t = difference in temperature.

t = (T – T’), T is the ambient temperature and T’ is the standardising temperature.

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PULL

Correction for pull = ± (P – P’)L/AE

Where P is the pull applied during measurement

P’ is the pull while standardising the tape

L is the length

A is the area of cross section of tape

E is the Young’s modulus of elasticity (200 GN/m²)

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SAG

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SAG

Correction for sag = Lw²/24n²P²

L is the length

w is the weight per meter

N is the number of spans

P is the pull applied.

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NORMAL TENSION

Tape sags due to self weight; tension

straightens the sag; the two effects are

opposite.

When tension applied is such that it

neutralizes the effect of sag, it is called

normal tension.

Normal tension = 0.204(wl)√(AE)/√(P- Po)

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SUMMARY OF

CORRECTIONS 1. Incorrect length = (L’/L) x length (±)

2. Slope = h²/2L or Lθ/2 (negative)

3. Incorrect alignment = d²/2L (negative)

4. Temperature = Lα t (±)

5. Pull = (P – P’)L/AE (±)

6. Sag = Lw²/24n²P² (negative)

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LIMITS OF PRECISION

Limits of precision in chain/tape surveys depend upon many factors like

Nature of terrain, equipment and accessories, time and resources, weather conditions and importance of the job.

The limit of precision expressed as 1:n. Higher value of n shows more precision

Chaining on rough terrain – 1:250

Under good conditions – 1:500

Tested chain under excellent conditions = 1:1000

Steel tape under good conditions – 1:2000

Invar tape with accessories – 1:10000

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Problem 1

P and Q are the two points 517m apart on

the same bank of a river. The bearings of a

tree on the other bank observed from P

and Q are N 33 ̊ 40’ E and N 43 ̊ 20’ W. Find the width of the river if the bearings of PQ are N 78 ̊ 0’ E.

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Problem 2

A survey line BAC crosses a river. A and C

being on the near and distant banks

respectively. Standing at D, a point 50

metres measured perpendicular to AB

from A, the bearings of C and B are 320 ̊ and 230 ̊ respectively. AB being 25 m, find the width of the river.

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Errors and mistakes.

• Compensating errors

• Cumulative errors

• Mistakes

Compensating errors:

1. Incorrect holding the chain

2. Incorrect measurement of right angles

3. Horizontally and vertically – sloping

done not properly

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Contd..

Cumulative errors:

Positive errors

Negative errors

Mistakes:

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CHAIN TRIANGULATION

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CHAIN TRIANGULATION

Main stations – vertices of triangles

Base Line – long line and accurately

measured

Tie line – run between lines to locate

details

Check lines – to check accuracy of

measurement

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OFFSETS

Offsets are run to locate details

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OFFSETS

1. Offsets generally at right angles to chain

line

2. Oblique offsets are also taken

3. Number of offsets depend upon the

detail.

4. Very long offsets are avoided.

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NUMBER OF OFFSETS

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NUMBER OF OFFSETS

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INSTRUMENTS FOR

SETTING RIGHT ANGLES For taking right-angled offsets, instruments

used are cross staff, optical square, prism

square.

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Chain line

offset

CROSS STAFF

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OPTICAL SQUARE

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OPTICAL SQUARE

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OPTICAL SQUARE

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INDIAN OPTICAL

SQUARE

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PRISM SQUARE

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OBSTACLES TO CHAINING

Chaining not possible, No intervisibility

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OBSTACLES TO CHAINING

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OBSTACLES TO CHAINING

Chaining not possible, stations intervisible

i) Erect perpendiculars at A and B of equal length using

optical square. CD = AB

ii) Using optical square, set a right angle of sufficient

length to get C. Measure CB. AB = √(BC² - AC²).

Iii) Similar method with right angle at C. AB = √(AC²+

BC²)

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OBSTACLES TO CHAINING

iv) Select point C such that A and B are visible. Range AC

and make CD = AC. Range BC and make CE = BC.

Measured ED which is equal to AB.

v) When it is not possible to set a right angle, select

point C. Range CA and make AD = CA.

Cosα = (BC² +CD² -BD²)/(2BCxCD)

AB² = BC² + CA² + 2 CA x BC x cos α.

vi) Select a point C approximately midway of AB and

such that A and B are visible. Measure CA and CB.

CD and CE are made proportional to CA and CB.

Measure DE and calculate AB from similar triangles.

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OBSTACLES TO CHAINING

Chaining not possible, intervisible (River)

1. Erect AC perpendicular to AB. Obtain point D on

chain line by making CD perpendicular to CB.

Calculate AB = AC x AC/BD.

2. Erect perpendicular to AB at A to get C at convenient

distance. Bisect AC at D. Range EDB to get E. AB is

then equal to CE.

3. Lay out line CD at right angle to CB, making CA =

AD. Set a right angle at D to get point E on the chain

line. Measure AE which is equal to AB.

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OBSTACLES TO CHAINING

4. Select a convenient distance CA. Erect perpendiculars at

A and C and make AC = DF. Extend the line CE to

obtain E in line with DB. AB = AC x DF/ FE.

5. When the chain line crosses the river obliquely, make a

line DAC such that CB and DE are perpendicular to this

line. Point E is so chosen that EAB is in line. AE = AB

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OBSTACLES TO CHAINING

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OBSTACLES TO CHAINING

Chaining and ranging obstructed

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OBSTACLES TO CHAINING

Chaining and ranging obstructed [Ex. Building]

i) Set right angles at A and C and make distances CD and AE

equal and to clear the building. Range DE and extend it

beyond building. Select points B and F and set right angles

to get G and H such that BG and FH equal to CD. GH is

the extension of the chain line. Confirm the field work by

measuring the diagonals of both the squares made on

either side.

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OBSTACLES TO CHAINING

II) Select DA along the chain line. Erect perpendicular at A

making AC = AD. Extend the line DC to E. From E,

range a line EF making EF = DE. Make FG = DC. From F

and G, swing arcs equal in length to DC and get point H.

HF is the continuation of the chain line and CG = AB.

iii) Set equilateral triangle CAD. Range CD and extend it

to E. Make EFG an equilateral triangle equal in size to

CDA. Range EG and extend it to H such that CE = EH.

Set an equilateral triangle HBJ equal in size to the

previous triangles. ECH is also an equilateral triangle. AB

= CE – AC – BH.

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OBSTACLES TO CHAINING

iv) Line DAE is set approximately at right angles such that

the lines can be ranged beyond the building. Select a

point C on the chain line. CF and CG are set

proportional to CD and CE. Divide FG in the same

ratio as AD to AE. Extend the lines CF and CG in

proportion to CD and CE. Divide the line HI in the

same proportion to get J. BJ is in continuation of the

chain line. Distance AB can be calculated from the

triangles.

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OBSTACLES TO CHAINING

Distance between points past an obstacle

To find AB, Take a convenient point C and measure CA and CB.

Select point D on CA and find E as a proportional distance of CD x CB/CA.

Measure DE. AB can be found as DE x CD/CA.

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CHAIN SURVEY

PROCEDURE

1. Reconnaissance

2. Equipment – chain/tape, arrows, ranging rods, ranging poles, offset rod, pegs etc

3. Marking stations

4. Triangulation

5. Locating details

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FIELD BOOK

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CROSS STAFF SURVEY

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CROSS STAFF SURVEY

BOOKING READINGS The reading of cross staff survey are booked by chainages along

the central line and marking the offsets to points on the left and right sides.

The distances to the points are noted along with the point name.

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END