SURVEY NOTES N5

BUILDING & STRUCTURAL

SURVEYING

N5

STUDY GUIDE

CONTENTS

1. BASIC PRINCIPLE……………………………………………………………..3

2. LINEAR MEASUREMENT………………………………………………….....6

3. LEVELLING…………………………………………………………………....18

3.1 RISE AND FALL METHOD…………………………………………..19

3.2 COLLIMATION HEIGHT METHOD………………………………....25

3.3 RECIPROCAL LEVELLING…………………………………………..28

4 TACHEOMETRY……………………………………………………………...30

5. TRAVERSING………………………………………………………………....36

5.1 MEAN OBSERVATION ANGLE…………………………………….37

5.2 CO-ORDINATES (JOINING)………………………………………...42

6. SETTING OUT………………………………………………………………..46

6.5 CUT AND FILL……………………………………………………….48

6.6 LENGTH AND SLOPE OF DRAINAGE PIPES……………………..52

6.7 ESTABLISHING HEIGHTS OF SIGHT RAILS……………………..59

7 AREAS AND VOLUMES…………………………………………………….63

8. ANSWERS…………………………………………………………………….

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CHAPTER 1

BASIC PRINCIPLES

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1. BASIC PRINCILPES

1.1 Terminology

1.1.1 SurveyingSurveying is the art of taking measurements upon the surface of the earth either in the horizontal or vertical plane, the results are shown in the formof a map or plan or as calculated figure. The plan is later set out on the ground

1.1.2 Plane surveyingPlane surveying is the surveying whereby the are to be surveyed is small that the curvature (shape) of the earth is not taken into account

1.1.3 Gravity Gravity is the force the keeps the earth in equilibrium, keep us on earth and gives us the sense of balance. It acts towards the centre of the earth

1.1.4 TopographyTopography is a survey done to locate the main natural and artificial features earth surface, e.g hills, rivers, roads, buildings etc

1.1.4.1 PlanimetryIt is the representation, in plan, of the natural and man made features

1.1.4.2 ReliefIt is the indication, in plan, of variations in elevation of the surface of the land. Relief maybe shown in the following methods

a) Colour layeringb) Shading c) Contoursd) Form lines

1.1.5 Contour lineIt is an imaginary line which links up a series of points of the same level on thesurface of the earth

1.1.6 Reduced levelIt is the calculated height of a point above or below a datum as deduced from survey observation

1.1.7 Change pointChange point is the staff position at which a foresight reading was taken and later a backsight was taken

1.1.8 Invert levelIt is the level of the inside bottom surface of the pipe

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1.1.9 Field bookIt is the book in which all the details of the survey are recorded by the surveyor

1.1.10 ChainageIt is the distance to a point along a survey line, even though one talk of taping aline than say chaining a line

1.1.11 BookingIt is the entering of all survey information and measurements in the field book

1.1.12 Setting outIs the transfer of details from a drawing to a piece of ground

1.1.13 Zenith angle (Zenith distance)It is the vertical angle measured from the vertical line (zenith line) downward

1.1.14 Offset It is a distance measured at right angle to a chain line to some feature of the sitesuch as a tree, building etc

1.1.15 HeightIt is the vertical distance of a feature above or below the datum or reference surface

1.1.16 Optical squareIs a hand held instrument used to set out right angles in a building site

1.1.17 DumpyIt is a common site name given to any levelling instrument

1.1.18 Instrument It is a common site name given to a surveying instrument on a tripot

1.1.19 DistometerIt is an used together with a theodolite, it measures distance

1.1.20 Tri-beacon It is the highest point of known height above sea level and of known co-ordinates

1.1.21 GridIt is the representation on a map, of a system of equally spaced straight parallel linesto Y and X of the co-ordinate system

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CHAPTER 2

LINEAR MEASUREMENT

CHAIN SURVEYING

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2. CHAIN SURVEYING

2.1 METHODS OF FIXING A POINT

2.1.1 Offset method (rectangular)

Measure the perpendicular distance DC and the length of AD and BD

C

A B D

2.1.2 Polar co-ordinates (bearing and distance / Length and direction)

Measure angle A and the length of AC

C

A B

2.1.3 Intersecting arcs

Measure the length of AC and BC

C

A B

2.1.4 Triangulation (forward Intersection)

Measure angle A and angle B

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C

A B

2.1.5 Trilateration (Resection)

Measure the angles formed at P by the rays to three known points A , B and C

A

a P b B

c C

2.2 Field problems

2.2.1 Ranging and measuring over a hill(obstacle to line of sight but not to measurement)

C D

A B

Step 1 Rod holders C and D position themselves so that D can see A, and C can see BStep 2 C directs the rod holder of D into line between C and B Step 3 D then directs the rod holder of C into line between A and DStep 4 This will continue until no further movement is possible and they all are in line

2.2.2 Measuring around a pond / Building(obstacles to measurement but not to sight)

C D

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A B E F

a) Bring the survey line to from A to B

b) Set off line BC at right angle to AB

c) Set off line CD at right angle to BC

d) Set off line DE at right angle to CD

e) At E set off line EF at right angle to DE Distance BE = CD

2.2.3 Measuring distance across a river(obstacles to measurement but not to sight)

B

A D C

F

ETo find distance AB

a) Set off line AC at right angle to ABb) Find D midpoint of ACc) Set off line CE at right angle to ACd) Move along line CE until point B and D are in linee) This will be point F Distance AB = CF

2.3 Methods of taping

2.3.1 Surface tapingMeasuring with the tape laid on the ground and fully supported by the ground

2.3.2 Catenary tapingMeasuring with the tape suspended clear off the ground Used when surface is very bad and involves the removal of grass, shrubs etc alongThe line of suvey

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2.4 How step chaining is conducted on site

1. When step chaining is conducted the tape should be stretched out horizontally from point A

2. The distance is limited to the tape being held comfortably at waist height

3. To obtain a true horizontal a plumb bob is held on the tape reading and at the same point a peg is put on the ground.

4. The zero point of the tape is place on the second peg and the horizontal distance measured using the waist height as our level.

5. This is repeated up until point B is reached

2.5 How the 3 – 4 – 5 method of setting out a right angle is applied

1. Three people are needed

2. from the point at which the perpendicular line is required the tape is held at zero and stretch along the existing straight line for 4m and held by the second assistant at that point

3. The third assistant pulls the tape at an acute angle towards the first assistant holding the tape at 7m

4. The remainder of the tape is held by the first assistant at 12m and the whole system is Stretched to form a right angle triangle

2.6 Five requirements to obtain sufficient accuracy when taping

1. Tape must be held horizontal

2. Tape must be held on its correct zero mark

3. The correct tension must be applied to the tape

4. Remove all kinks

5. Tape must be held on correct pegs

6. View tape vertically over the peg

7. Measure the centre of ranging rod

2.7 Care of steel tape

1. Pull tape straight in the direction in which it is curled

2. Ensure that no kinks are present

3. Do not exceed standard pull

4. Wipe tape with oily cloth after use to prevent rust

5. Do not allow vehicle to run over the tape

2.8 CHAINAGE CORRECTIONS

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2.8.1 Constants ( length of tape too short or too long)

CD = correct distanceMD = measured distance (length)ATL = Actual tape lengthGTL = Graduated tape length

2.8.2 Temperature

C = Temperature Correction L = Measured distance (length) e = Coefficient of linear expansion

Tm = Measured temperatureTs = Standard temperature

2.8.3 Sag

for bay For Three bays

C = Sag correction = Mass of the tape in kg/m

L = Measured length (distance) in m T = Measured tension in kgF

2.8.4 Slope

To calculate the horizontal distance To calculate the slope distance

Sc = Slope correction SC = CorrectionL = Measured length of the slope H = Measured horizontal distance

= Slope angle = Slope angle

2.8.5 Height at sea level

CH = Correction L = Measured length (distance) H = Height above or below sea level

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R = Earth radius (if given in km convert to m)

OBSERVATION ANGLE

Z = Zenith distanceθ = Angle of elevation

Vertical Zenith line of sight Zenith distance

Z Angle of elevation

θ Horizontal

Z = 90º – θ

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θ = 90º – Z

Zenith distance

Z θ Angle of depression

line of sight

Z = 90º + θθ = Z – 90º

NB: ALWAYS use angle θ which represents the angle of Elevation or Depression EXAMPLE 1

QUESTION

Calculate the reduced horizontal distance, if the measured distance is 87,281 m at a slope if 86º 22’ 44”the distance is measured is three bays at a temperature of 16 ºC

Standard temperature is 20ºCCo-efficient of expansion is 1,13 x /ºCTension 65NMass of the tape is 0,015 kg/mEarth radius is 6373 kmHeight above sea level 2 280,544 m

SOLUTION

Given : ts = 20ºC e = 1,13 x /ºC T = 65N w = 0,015 kg/m R = 6373 km H = 2280,544 m L = 87,281 m θ = 90 – 86º 22’ 44”= 3º 37’ 16” tm = 16ºC

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( 65 N = = 6,5 kgF )

m

m

CD = 87,281 – 0,0039 – 0,0164 – 0,1743 – 0,0312 = 87,055 m

NB: CD = reduced distance or correct horizontal distance

EXAMPLE 2

QUESTION

Calculate the slope distance if the measured correct horizontal distance to be measured is 98,285 and the slope is 2º 22’ 15”

SOLUTION

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m

Slope distance = 98,285 + 0,0842= 98,369 m

OR

Slope =

= 98,369 m

EXERCISE 1Question 1.1

The standard temperature of a tape is 14ºC and the coefficient of expansion is 0,000012/ºCWhat is the reduced horizontal distance, if the measured distance is 32,40 m on a slope of 5º 40’ and a temperature of 30ºCAnswer 32,248 m

Question 1.2

Given: Length of tape = 100mStandard temperature = 20ºCCoefficient of expansion = 1,13 10–5/ºCTension = 7 kg.fMass of tape = 0,015 kg/mRadius of the earth = 6,373 km

1.2.1 Calculate the reduced horizontal distance, if the measured distance is 94,151 mat a slope of 95º 15’. The distance is measured in three bays at a temperature of 14ºC Answer: 93,732 m

1.2.2 Calculate distance at sea level if the height was 2 017,443 above sea level,use the distance in Question 2.1. Answer: 93,702 m

Question 1.3

A line A-D was measured in three sections

A-B 90,288 @ slope 3º 44’ 20”B-C 72,408 @ slope 4º 32’ 59”C-D 47,652 @ Slope 2º 09’ 07”

Calculate the horizontal distanceA-D Answer: 209,459 m

Question 1.415

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1.4.1 M to R is a distance to be pegged (as viewed on plan)M to R lies on a slope of 0º 34’ 23” Calculate the distance on the slopeM to R to be measured to obtain the correct horizontal distance of 1000 mAnswer: 1000,050 m

1.4.2 Determine the horizontal distance of a measurement of 136,57 m that was done on a Slope of 3,64%Answer: 136,479 m

Question 1.5

A measured distance of 164,81m was measured from T1 to T2 the reduced distance is 160,49 mCalculate the angle of the slope and the zenith distance

Answer: 13º 8′ 50,8″ ; 76º 51′ 7,24″

Question 1.6

A steel tape is used to measure a baseline A-B. Each measurement is done in THREE bays Calculate the correct baseline distance. Give the formula for each of your calculations

A-B 94,01 m /96º 10’ @ 35ºC 83,14 m / 87º 30’ @ 11ºC

Length of tape = 100mStandard temperature = 20ºCCoefficient of expansion = 0,0000112/ºCTension = 70 NMass of tape = 0,015 kg/m

Answer: 176,504 m

Question 1.7

A 100 m steel tape was compared with a 100 m standard base. The length of the base as read on the tape was 99,997 m at a temperature of 22,4ºC. The coefficient of linear expansion is 0,000014/ºC. Calculate the standard temperature of the tape.

Answer: 19,768ºC

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CHAPTER 3

LEVELLING

(Height measurement)

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3. HEIGHT MEASUREMENT ( REDUCED LEVEL)

3.1 RISE AND FALL METHOD

Point BS IS FS Rise Fall Reduced level

A 2,50 50,00B 2,00 0,50 50,50C 2,25 0,25 50,25D 1,75 1,80 0,45 50,70E 1,60 0,15 50,85

4,25 –3,40

0,85

3,40 1,10 –0,25

0,85

0,25 50,85–50,00 0,85

Steps to follow:

a) Enter the backsight, intermediate sight and the foresight in the appropriate columnin different rows. Except that at change point the FS and the BS are entered in the same row ( as in row D)

b) The first reduced level is the height of the O.B.M or other datum which has been used

c) If the IS or FS in smaller than the immediately preceding staff reading, The difference between the two readings is placed in the rise column

Example: immediately preceding reading of 2,00 in row B is 2,50 in row ATherefore 2,50 – 2,00 = 0,50

d) If the IS or FS in larger than the immediately preceding staff reading, The difference between the two readings is placed in the fall column

Example: immediately preceding reading of 2,25 in row C is 2,00 in row BTherefore 2,00 – 2,25 = – 0,25

e) A rise is added to the immediately preceding reduced level entry to obtain the reduced level of a station.

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Example: immediately preceding reduced level of row B is row A = 50,00Reduced level of B= 50,00 + 50 = 50,50

f) A fall is subtracted from the immediately preceding reduced level entry

Example: immediately preceding reduced level of row C is row B = 50,50Reduced level of C = 50,50 – 0,25 = 50,25

Checking on booking

1) Sum up the BS readings and the FS readingsSubtract FS from the BS

Total BS = 2,50 + 1,75 = 4,25Total FS = 1,80 + 1,60 = 3,40Difference = 0,85

2) Sum up Rise and Fall Subtract Fall from the Rise

Total Rise = 0,50 + 0,45 + 0,15 = 1,10 Total Fall = 0,25 Difference = 0,85

3) Subtract the first reduced level from the last reduced level

Last reduced level = 50,85First reduced level = 50,00

0,85

NOTE: If all their differences are the same then the calculation are correct

NOTE: Any entry that is underlined 3,456 or has a line ontop 3,456 means the entry is negative

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Example : Rise and Fall method

NOTE: The numbers in brackets indicate the steps from the calculations below

Point BS IS FS Rise Fall Reduced levelA 2,109 (A1) 130,50B 3,204 (B1) 1,095 (B2) 129,405C 3,119 1,374 (C1) 1,830 (C2) 131,235D 1,281 (D1) 4,400 (D2) 135,635E 2,054 (E1) 0,773 (E2) 136,408F 1,906 2,200 (F1) 4,254 (F2) 132,154G 1,654 1,812 (G1) 3,718 (G2) 135,872H 2,850 (H1) 4,504 (H2) 131,368

5,480–4,6120,868

4,612 10,721–9,8530,868

9,853 131,368 –130,50 0,868

Use the table above with the steps below to understand

A1. The BM is the reduced level in A which is 130,50

B1. To find the Rise or Fall in row B

BS in row A – IS in row B = Rise/Fall 2,109 – 3,204 = – 1,095 ( since its negative it is a Fall)

B2. To find the reduced level at B

Reduced level at A – Fall in B = Reduced level at B 2,109 – 3,204 = 129,405

C1. To find the Rise or Fall in row C

IS in row B – FS in row C = Rise/Fall 3,204 – 1,374 = 1,830 ( since its positive it is a Rise)

C2. To find the reduced level at C

Reduced level at B + Rise in C = Reduced level at C 129,405 + 1,830 = 131,235

D1. To find the Rise or Fall in row D

BS in row C – IS in row D = Rise/Fall 3,119 – ( –1,281) = 4,400 ( since its positive it is a Rise)

D2. To find the reduced level at D

Reduced level at C + Rise in D = Reduced level at D

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131,235 + 4,400 = 135,635

E1. To find the Rise or Fall in row E

IS in row D – IS in row E = Rise/Fall (–1,281) – (–2,054) = 0,773 ( since its positive it is a Rise)

E2. To find the reduced level at E

Reduced level at D + Rise in E = Reduced level at E 135,635 + 0,773 = 136,408

F1. To find the Rise or Fall in row F

IS in row E – FS in row F = Rise/Fall –2,054 – 2,200 = – 4,254 ( since its negative it is a Fall)

F2. To find the reduced level at F

Reduced level at E – Fall in F = Reduced level at F 136,408 – 4,4254 = 132,154

G1. To find the Rise or Fall in row G

BS in row F – FS in row G = Rise/Fall 1,906 – (–1,812) = 3,718 ( since its positive it is a Rise)

G2. To find the reduced level at G

Reduced level at F + Rise in G = Reduced level at C 132,154 + 3,718 = 135,872

H1. To find the Rise or Fall in row H

BS in row G – FS in row H = Rise/Fall –1,654 – 2,850 = –4,504 ( since its negative it is a Fall)

H2. To find the reduced level at H

Reduced level at G – Fall in H = Reduced level at H 135,872 – 4,504 = 131,368

CHECK ON BOOKING

Total BS – Total FS = 5,480 – 4,612 = 0,868

Total Rise – Total Fall = 10,721 – 9,853 = 0,868

Last R/level – First R/Level = 131,368 – 130,50 = 0,868

Everything balances therefore correct

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EXERCISE 2

Question 2.1

Reduce the field notes using the rise and fall method. Do the necessary checks

(If your calculation are correct then the check must give an answer of –1,6)

Point BS IS FS Rise Fall Reduced level RemarksA 2,40 TBM 150,00

B 2,00C 1,90D 2,80 1,40E 2,00F 1,30 2,60G 0,60 3,00H 1,70

7,10–8,70–1,6

8,70

Question 2.2


(If your calculation are correct then the check must give an answer of 4,64)

Point BS IS FS Rise Fall Reduced level RemarksA 0,49

0,27 3,290,39 3,773,72 3,59

1,113,56 0,82

BM 3,89 1,36 BM 1 025,003,72 0,993,69 1,023,86 1,313,90 1,56

B 2,4023,93

–19,294,64

19,29

Question 2.3


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(If your calculation are correct then the check must give an answer of 4,09)

Point BS IS FS Rise Fall Reduced levelBM1 3,141 401,261

A 0,086B 2,111 4,283C 1,406D 2,032E 3,108 3,638F 1,823 2,110G 3,111H 4,123 2,109I 3,281

BM2 3,083

3.2 COLLIMATION HEIGHT METHOD

NOTE:

1. BS, IS and FS reading are entered as in the rise and fall method2. The first reduced level entry is the height of the OBM 3. The first BS reading which is taken with the staff held on the OBM is added to

the first reduced level to give the height of collimation and the entries are place in the same row

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4. The height of collimation is change only when the levelling instrument is moved to a new position. The new height of collimation is obtained by adding the new BS to the reduced level at the change point

EXAMPLE

Point BS IS FS Collimation height Reduced levelA 3,16 (A2) 141,66 (A1) 138,50B 1,95 141,66 (B1) 139,71C 4,16 3,97 (C2) 141,85 (C1) 137,69D 1,62 141,85 (D1) 140,23

NOTE: Collimation height at A is used to find all the reduced level of B and C But since C is a change point, the collimation height change

Step to follow

A1. The reduced level entered is the OBM or TBM etc given

A2 How find the collimation height at A

Reduced level at A + Backsight at A = Collimation height at A 138,50 + 3,16 = 141,66

B1. How to find reduced level at BCollimation height at A – Inter-sight at B = Reduced level at B

141,66 – 1,95 = 139,71

C1 How to find the reduced level at C

Collimation height at A – Foresight at C = Reduced level at C 141,66 – 3,97 = 137,69

C2 How find the collimation height at C (Change point has both BS and FS)

Reduced level at C + Backsight at C = Collimation height at C 137,69 + 4,16 = 141,85

D1 How to find the reduced level at D

Collimation height at C – Foresight at D = Reduced level at D 141,85 – 1,62 = 140,23

EXERCISE 3

Question 3.1

Reduce the following notes by using the collimation height methodDo the necessary checks

Point BS IS FS Collimation height Reduced levelBM 2,40 150,00A 2,00

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B 1,90C 2,80 1,40D 2,00E 1,30 2,60F 0,60 3,00G 1,70

Question 3.2


Point BS IS FS Collimation height Reduced levelBM1 4,61 884,86

A 3,54B 1,69C 1,80 3,21D 2,40E 2,89 3,62F 1,20G 2,20H 1,46

BM2 1,205,70

–1,614,09

1,61

Question 3.3


Point BS IS FS Collimation height Reduced level A 4,420B 5,500C 3,880D 3,160 1,470E 1,950F 4,550G 4,650 3,970 145,990

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H 3,900I 6,320J 2,340K 4,160 1,620L 1,620M 5,200

3.3 RECIPROCAL LEVELLING

y1 y

x x1

station 1 C

station 2 3m D

3m

Let us assume the following staff reading

Reading taken 3m from C Reading taken 3m from D C (y = 1,480m) D (x = 1,738m)

D (y1 = 1,852m ) C (x1 = 1,380)

One can see that all reading taken when staff was at C are larger than that of staff at D

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We can therefore say a point with larger reading is lower and the one with smaller readings is high. In this case we can say point D is lower and C is higher

Therefore to find the reduced level of point use the following

1. Calculate the difference from readings taken at station C (e.g between and )

2 Calculate the difference from readings taken at station D (e.g between and

3 Calculate the mean difference between the stations

4 Calculate the Reduced level required

4.1 R/L of a lower point = R/L of a higher point – mean difference

4.1 R/L of a higher point = R/L of a lower point + mean difference

EXAMPLE Using the data given let, let us say that the reduced level of D is 352,710m

And we need to calculate the reduced level of C

Solution

Difference between readings taken 3m from C = 1,852 – 1,480 = 0,372 m

Difference between readings taken 3m from D = 1,738 – 1,380 = 0,358 m

Mean difference =

= 0,365 m

Reduced level of C = 352,710 – 0,365 = 352,345

EXERCISE 4Question 4.1

4.1.1 The following readings were taken by ‘reciprocal’ levelling across a swamp. What is the correct elevation of point B if the elevation of A is 100,00 m

Reading taken 3m from A Reading taken 3m from DA 2,91 B 1,71B 2,49 A 1,39

4.1.2 Make a neat sketch showing the above mentioned set ups and readings

Question 4.2

3,35 1,95

2,85 1,55

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B 3m

A 3m

The above sketch shows the reading taken by reciprocal levelling across a river. If the elevation Of B is 139,50 Calculate the Reduced level of A

Question 4.3

4.3.1 The following readings were taken by ‘reciprocal’ levelling across a swamp. What is the correct elevation of point B if the elevation of A is 123,50 m

Reading taken 3m from A Reading taken 3m from DA 1,50 B 3,30B 1,00 A 3,00

4.3.2 Make a neat sketch showing the above mentioned set ups and readings

CHAPTER 4

TACHEOMETRIC

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LEVELLING

4. TACHEOMETRY

Use of Vertical angles

MHVD

B

θ ΔhHeight of B

IH

Height of A

Datum line

When the instrument is at a lower point

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( that means the angle given is less than 90º measured from the vertical line)

Higher point = Lower point + IH + VD – MH

This means :Reduced level B = height of A + Instrument height + vertical distance – mean height

θ IH VD

A

Δh Height of A MH

B Height of B

Datum line

When the instrument is at a higher point ( that means the angle given is more than 90º measured from the vertical line)

Lower point = Higher point + IH – VD – MH

This means :Reduced level B = height of A + Instrument height – vertical distance – mean height

FORMULAE

1. I = TH – BH

TH =Top hairBH = Bottom hair

2. VD = KI cosθ Sinθ or VD = KI sin 2θ (cosθ Sinθ = sin 2θ)

VD = Vertical distanceK = 100I = as calculated in 1θ = Slope angle

3.

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MH = Mean height

4. HD = KI cos2 θ

HD = Horizontal distance

5. or

= gradient

Example

Given The height of point A is 24,135m, height of instrument at A is 1,60mTH = 3,01 BH = 2,41 vertical angle is 86º 30’

Required to calculate 1.1 The reduced level of B 1.2 The horizontal distance A-B

SOLUTION

1.1 θ = 90 – 86º 30’( A is lower than B) = 3º 30’

VD = KI cosθ Sinθ

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VD = 100(3,01 – 2,41) cos3º 30’ Sin 3º 30’ = 3,656m

Reduced level at B = 24,135 + 1,60 + 3,656 – 2,71 = 26,681m

1.2 HD = KI cos2 θHD = 100(3,01 – 2,41) (cos 3º 30’)2 = 59,776m

EXERCISE 5

Question 5.1

If the height of point A is 900,00m and the height of the instrument at A is 1,162m Observation to point B

TH = 3,492BH =1,250Vertical angle = 97º 24’ 00”

5.1.1 Calculate the Reduced level of point B5.1.2 Calculate the horizontal distance A-B

Answers 870,156m 220,480m

Question 5.2

A surveyor obtained the following information between two points, A and B:

Height of A = 430,17mInstrument height at A = 1,56mStadia reading 0,97 1,43 1,89Vertical angle = 82º 10’ 40”

5.2.1 Calculate the horizontal distance A-B5.2.2 Calculate the Reduced level of B

Answer : 90,296m 442,705m

Question 5.3

A theodolite is set up at point P and the height of the instrument is 1,61m. The elevation of

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Point Q is 1 718,440m

TH = 2,14BH = 0,93Vertical angle = 96º 10’

Calculate the height of point P

Answer : 1 731,288m

Question 5.4

A theodolite is set up at point A and the height of the instrument is 1,20m. When sighted to a staff held at B, the upper and lower stadia lines read 2,00 and 0,80 respectively. The angle of elevation of the instrument is 6º 20’

Calculate the Reduced level of A if the reduced level of B is 104,68m

Answer : 917,23m

Question 5.5

A theodolite is set up at point C and the instrument height is 1,58m. The elevation of C is 2 015,67m. Sighting staff held at D, the following observation are recorded

STADIA CIRCLE READINGSTH MH BH VERTICAL HORIZONTAL

2,55 1,6 0,65 98º 16’ 00” 210º 50’ 40”

Calculate:5.5.1 Vertical angle5.5.2 Vertical distance5.5.3 Horizontal distance5.5.4 Height of D5.5.6 The slope distance5.5.7 The gradient of the slope

Answer: ---- 27,034 186,0721988,616 188,0260,145

Question 5.6

6.1 If the angle of elevation is of the instrument 5º 22’ 30”. Calculate the zenith distance

6.2 If the angle of depression is of the instrument 6º 25’ 32”. Calculate the zenith distance

6.3 If the vertical circle reading 97º 30’ 35”. Calculate the vertical angle and state whether it is an angle of elevation or depression

6.4 If the vertical circle reading 82º 18’ 45”. Calculate the vertical angle and state whether it is an angle of elevation or depression

Answer not given

Question 5.7

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The following information was recorded during field survey

The slope distance from point C to D is 500mThe reduced level at point C is 1,284mThe reduced level at point D is 12,484m

Calculate the following:

5.7.1 The horizontal distance A to B5.7.2 The slope angle at A

Answer: 499,875m 1,284º

CHAPTER 5

TRAVERSING

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5.1 MEAN OBSERVATION ANGLE (Use of horizontal angles)

CIRCLE LEFT NORTHWhen this circle on the left of the telescope is turned to face leftThe readings will be between º and 180ºUsually in the vicinity of 90º

Direction of sight360º start position

0º

B 25º 40’ 05”

G 289º 34’ 00” C 80º 22’ 40”

270º A 90º

F 229º 20’ 10”

E 186º 31’ 50” D 135º 14’ 20”

180º

CIRCLE RIGHT NORTHWith the circle on the right of the telescope The readings will be between 180º and 360ºUsually in the vicinity of 270º

180º Direction of sight Start position

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B 205º 40’ 25”

G 109º 34’ 14” C 260º 22’ 40”

90º 270º

F 49º 21’ 10”

E 06º 31’ 20” D 315º 14’ 30” 0º

360º

ENTRY OF THE READINGS

Point Circle Left Circle Right Mean angle Correction Corrected AngleB 25º 40’ 05” 205º 40’ 20”C 80º 22’ 40” 260º 22’ 40”D 135º 14’ 20” 315º 14’ 30”E 186º 31’ 50” 06º 31’ 20”F 229º 20’ 10” 49º 21’ 10”G 289º 34’ 00” 109º 34’ 14”B 25º 40’ 05” 205º 40’ 25”

5.1.1 How to calculate the mean observation angle

5.1.1.1. Given the reading of one point ONLY

When the Circle right is Smaller than Circle left add 180º to Circle rightWhen the Circle right is larger than Circle left subtract 180º from Circle right

Example

Given Circle left Circle right 128º 45’ 12” 308º 45’ 20”

Solution (Circle right is larger 308º – 180º = 128º)

This is enter as in the following table

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Point Circle Left Circle Right Mean angle128º 45’ 12” 308º 45’ 20”128º 45’ 12” 128º 45’ 20” 128º 45’ 16”

NOTE: Mean angle = = 128º 45’ 16”

5.1.1.2 Given reading from more than one point

Example Given : Circle left Circle right A 87º 40’ 20” 267º 40’ 35”B 110º 32’ 30” 290º 32’ 45”C 184º 14’ 20” 04º 13’ 33”D 286º 25’ 30” 106º 25’ 50”A 87º 40’ 40” 267º 40’ 55”

Solution: (Use the table with the calculations below)

Circle left Circle right Mean angle Correction Corrected angle

A 87º 40’ 20” 267º 40’ 35”B 110º 32’ 30” 290º 32’ 45”

(1) 22º 52’ 10” (2) 22º 52’ 10” (3) 22º 52’ 10” 22º 52’ 05”B 110º 32’ 30” 290º 32’ 45”C 184º 14’ 20” 04º 13’ 33”

(4) 73º 41’ 50” (5) 73º 41’ 48” (6) 73º 41’ 49” 73º 41’ 44”C 184º 14’ 20” 04º 13’ 33”D 286º 25’ 30” 106º 25’ 50”

(7) 102º 11’ 10” (8) 102º 11’ 17” (9) 102º 11’ 13,5” 102º 11’ 8,5”D 286º 25’ 30” 106º 25’ 50”A 87º 40’ 40” 267º 40’ 55”

(10) 161º 15’ 10” (11) 161º 15’ 05” (12) 161º 15’ 7,5” 161º 15’ 2,5” Total 360º 00’ 20” 360º 00’ 00”

Correction = =

STEPS TO FOLLOW

NOTE: When the number subtracting is smaller add 360º to the number

B minus A

1. 110º 32’ 30” – 87º 40’ 20” = 22º 52’ 10”2. 290º 32’ 45” – 267º 40’ 35” = 22º 52’ 10”

(1) + (2) = 45º 44’ 20”

3. Mean angle = = 22º 52’ 10”

C minus B

4. 184º 14’ 20” – 110º 32’ 30” = 73º 41’ 50”5. 364º 13’ 33” – 290º 32’ 45” = 73º 41’ 48” ( add 360º to the 4º)

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(4) + (5) = 147º 23’ 38”

6. Mean angle = = 73º 41’ 49”

D minus C

7. 286º 25’ 30” – 184º 14’ 20” = 102º 11’ 10”8. 106º 25’ 50” – 04º 13’ 33” = 102º 11’ 17”

(7) + (8) = 204º 22’ 27”

9. Mean angle = = 102º 11’ 13,5”

A minus D

10 447º 40’ 40” – 286º 25’ 30” = 161º 15’ 10” ( add 360º to the 87º)11 267º 40’ 55” – 106º 25’ 50” = 161º 15’ 05”

(10) + (11) = 322º 30’ 15”

12. Mean angle = = 161º 15’ 7,5”

13. Add ALL the mean angles 14. If the answer is less than 360

Divide the difference by the number of mean angleAdd the answer to each mean angle

15. If the answer is more than 360 Divide the difference by the number of mean angleSubtract the answer from each mean angle

From the example above :In this case the total mean angles is 360º 00’ 20” which is 20” moreTherefore divide 20” by 4 and the answer is 5”We then subtract 5” from each mean angle

OTHER FORMULAE

Collimation error (E)

E =

CL = Circle leftCR = Circle right

Index error

E = ( CL + CR) – 360º

EXERCISE 6

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QUESTION 6.1

The following reading were obtained from a survey station “P”.Calculate the mean observed angles QPR , RPS, SPT, TPQCheck the calculations

Target station Circle left Circle rightQ 168 : 11 : 43 348 : 11 : 42R 17 : 11 : 50 197 : 11 : 56S 104 : 03 : 42 284 : 03 : 43T 105 : 04 : 43 285 : 04 : 43Q 168 : 11 : 45 348 : 11 : 42

QUESTION 6.2

The following reading were obtained from a survey station “A”.Calculate the mean observed angles BAC , CAD, DAE, EAF, FAG, GABCheck the calculations

Target station Circle left Circle rightB 07 : 00 : 40 187 : 01 : 00C 96 : 43 : 20 276 : 43 : 00D 187 : 22 : 10 07 : 21 : 50E 204 : 15 : 15 24 : 15 : 40F 276 : 35 : 10 96 : 35 : 50G 342 : 20 : 05 162 : 20 : 00B 07 : 01 : 00 187 : 00 : 40

Question 6.3

6.3.1 A direction on a circle left observation of a theodolite was 36º 10’ 21”. The same direction on a circle right observation was 216º 11’ 08”.Calculate the collimation error of the theodolite

6.3.2 Vertical angle on the circle left and circle right to a distant point areCL 93º 56’ 10”CR 266º 04’ 40”Calculate the index error of the theodolite

6.3.3 The following directions were observed from A to R. Reduce the mean directionCirlce left Circle right96º 43’ 20” 276º 43’ 00”

6.3.4 The following directions were observed from P toQ. Reduce the mean directionCirlce left Circle right7º 01’ 00” 187º 00’ 40”

5.2 CO – ORDINATES (JOIN CALCULATIONS)

SA co-ordinate system

–x (180º)

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((θ + 90) (θ + 180)

+y (90º) – y (270º)

θ (θ + 270)

+x (0º and 360º)

FORMULAE

Note: If given AB ( meaning from A to B) then its B minus AIf given BA ( meaning from B to A) then its A minus B

Distance or length

AB =

Direction or angle

θ =

NOTE:

2. To get the direction ( D) use the following

2.1 If is positive and is positive, we are in the 1st quadrant D =

2.2 If is positive and is negative, we are in the 2nd quadrant D = + 90

2.3 If is negative and is negative, we are in the 3rd quadrant D = + 180

2.4 If is negative and is positive, we are in the 4th quadrant D = + 270

EXAMPLE

Question

The co-ordinates of point A and B are

Y X

A – 248,17 – 58,47B – 150,27 – 260,80

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40

Calculate the orientated direction and distance between A and B (

Answer ( remember it’s A to B )

= 97,9

=

Distance from A to B

m

θ =

θ =

θ = 25º 49′ 14,24″

Since is negative and is positive, we are in the 4th quadrant D = + 270

Direction = 25º 49′ 14,24″ + 270 = 295º 49′ 14,24″

Question

The co-ordinates of point A and B are

Y X

A – 248,17 – 58,47B – 150,27 – 260,80

Calculate the orientated direction and distance between B and A (

Answer ( remember it’s B to A )

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=

= m

Distance from A to B

m

θ =

θ =

θ = 25º 49′ 14,24″

Since is positive and is negative, we are in the 2nd quadrant D = + 90

Direction = 25º 49′ 14,24″ + 90 = 115º 49′ 14,24″

EXERCISE 7

7.1 The co-ordinates of point A and B are Y X

A +310 248,17 – 1 058,47B +309 295,17 – 1 688,04

Calculate the orientated direction and distance between A and B

7.2 The co-ordinates of point A and B are Y X

A –467,89 +120,45B –120,45 +467,89

Calculate the orientated direction and distance between B and A

7.3 Study the information below showing point that formulate boundaries of 42

42

a property. Calculate the lengths of all the side of the property

SIDES (Metres) ANGLES OFDIRECTION

CO-ORDINATESY

ABBCCDDEEFFA

335 : 24 : 20 66 : 04 : 20155 : 24 : 20246 : 04 : 20155 : 24 : 20246 : 04 : 20

ABCDEF

Constant: 0,00

– 53 937,40– 53 963,29– 53 930,38– 53 918,31– 53 921,96– 53 908,23

+3 700 000,00

+ 60 210,88+ 60 267,26+ 60 281,86+ 60 255,49+ 60 253,86+ 60 223,86

* 377 P5 – 54 006,63 + 60 267,44 * 378 P5 – 53 879,14 + 60 320,80

All beacons are 12 mm iron pegs

A

723 F 726

E D

B

728

C

Scale 1/1000The figure A B C D E F represents 2100 square metres of land beingERF PORTION OF ERF WALMERSituated in the municipality and Administrative District of Port Elizabeth surveyed in June 1981This diagram is annexed to

NoDatedi.f.o Registrar of Deeds

The original diagram is

No 7494/1981 annexed to Transfer/GrantNo

File No s/7902/94

S.R No

Comp. BO-8CC/x43

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CHAPTER 6

SETTING OUT

6.1 How a traveller is used with profiles to control excavation and foundation levels

a) Two profiles ate place at each end of the excavation.

b) A small cross bar is fixed at each profile at a level equal to the invert level of the excavation plus the traveller (follower)

c) The depth if the excavation is therefore controlled by dipping the traveller suchthat the line of sight between the two bars of the profiles is in line with that of thecross bar of the traveller

6.2 How to set up a rectangular site along a road and use the road as a reference on the site plan

a) Offsets must be taken from the road reserve to the boundary of the site that is required

b) A baseline parallel to the road drawn.

c) The corresponding chainage of the site boundary are then identified

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d) the required site shape is then set out from the baseline

6.3 On site, how to set out two points A and B stationed on a third point known co-ordinates

a) Set up the instrument and calculate the direction between the point on which the instrumentis set up and point A and the distance between the two points

b) Also calculate the direction between the point on which the instrument is set up and point Band the distance between the two points

c) Orientate direction from instrument to point A and measure the distance and put a peg

d) Swing the instrument to point B using the calculated direction of point B and measure the distance from the instrument to point B and put a peg

6.4 How to set out a rectangular proposed building using simple surveying instrument forearthworks

Equipment: 100m tape, Four steel pegs or droppers 2m long, Lime, Fish line, Levelling instrument. Traveller 2m high

Measure the distance of the proposed structure from all the four corners and make it 1m lessFrom each side of the building.

Put the steel pegs or droppers on these new found points and mark them for a 2m traveller taking the 150m depth of the top soil into consideration

The area to be removed top soil is the one that is indicated by ABCD

A B

C D

6.5 CUT AND FILL

Ground level

Depth of excavation at A A

Depth of excavation at BInvert level at A Δh

B

Invert level at B

Steps to follow:

1. Calculating the Fall ( difference in height)

Δh = gradient x distance from first to the last point

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2. Calculating the invert level of the first point

Invert level at A = ground level – excavation

3. Calculating the invert level of the last point

Invert level at B = invert level at A – Δh

4 Calculating invert level at other points

e.g. chainage is 00 20 40 55 60

invert level at 00 is the invert level of the first point

invert level at 60 is the invert level of the last point

Invert level at 20 = invert level at A –

Invert level at 40 = invert level at A –

Invert level at 55 = invert level at 40 –

5. How to determine Cut or Fill

Ground level of a point – invert level of that point = Cut or Fill

If the answer is positive then it is a CutIf the answer is negative then it is a Fill

EXAMPLE

Given :

Chainage Ground level A 0 50,76

20 50,0040 45,7545 47,28

B 60 49 20

The information above refers to a drain between point A and B. The depth of excavation at A is 1,96mand the fall from A to B is 1:40. Calculate the Cut and Fill in metres

1. Δh = gradient x distanceΔh = NOTE: Δh = 1,5m

2. Invert level at A = 50,76 – 1,96 = 48,8 m (write this figure in the invert level column as shown in table 1)

3. Invert level at B = 48,8 – 1,5 = 47,3 m (write this figure in the invert level column as shown in table 1)

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4. Invert level at 20 = 48,8 – 0,025(20) = 48,3 m (write this figure in the invert level column as shown in table 1)



calculating the cut and fill

At 00 chainage 50,76 – 48,8 = 1,96 CutAt 20 chainage 50,00 – 48,3 = 1,70 CutAt 40 chainage 45,75 – 47,8 = –2,05 FillAt 45 chainage 47,28 – 47,675 = –0,395 FillAt 60 chainage 50,76 – 48,8 = 1,96 Cut

Table 1Chainage Ground level Invert levels Cut Fill

A 0 50,76 48,8 1,9620 50,00 48,3 1,7040 45,75 47,8 2,0545 47,28 47,675 0,395

B 60 49,20 47,3 1,90

EXERCISE 8

Quesrtion 8.1

A drain is laid between A and B. The excavation at A is 1,58m and the fall from A to B is 1:75Calculate the cut and fill in metres. Use the data given below

Chainage Ground levelA 0 250,15

20 251,2640 247,9960 252,5665 253,0180 247,323

100 250,08B 120 249,67

Question 8.2

A drain is laid between a House (H) and the manhole (M). the depth of the excavation at House 1,368m and the fall from the house to the manhole is 0,76%. Use the information given below to calculate the cut and fill in metres

Chainange Ground levelHouse 0 322,80

25 326,4240 327,4260 320,42

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75 319,4680 324,22

100 325,44105 326,47

Manhole 120 322,10

Question 8.3

The information below refers to the details of a drain which is to be laid between P and Q. the depthof the excavation at Q is 2,4m the rise from Q to P is 0,82%. Calculate the invert level at the differentpoints and also the Cut and Fill

Chainange Ground levelP 0 420,90

15 424,4025 425,4340 420,4260 419,4675 424,2280 425,44

100 426,47Q 120 420,20

Question 8.4

A drain is laid between A and B. The Reduced level at B is 47,3m and the fall from A to B is 1:80Calculate the cut and fill in metres. Use the data given below

Chainage Ground level Invert levelA 0 50,76

20 50,0040 47,1660 49,5780 49,00

100 48,80B 120 48,50 47,30

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6.6 LENGTH AND THE SLOPE OF DRAINAGE PIPE]

1. Slope =

Δh = difference between invert levelsΔchainage (distance) = difference between chainage

2. Length of the pipe =

NOTE: Length of pipe is the same as the length of the slope

3. Slope as a percentage of the pipe =

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EXAMPLE

QUESTION A B

C

DATUM MSL 300

GROUND LEVEL

INVERT LEVEL

DISTANCE ( CHAINAGE)

4600

,00

30

9,27

8

31

0,80

4660

,00

31

1,10

0

31

2,40

4736

,60

30

7,90

0

The diagram below shows a longitudinal section of a pipeline

1. Calculate the total length of the piping required from A to C2. Calculate the slope as a percentage of the pipeline3. Calculate the slope as a ratio from A to B and B to C

SOLUTION

1.1

m

50

50

m

Total length AC = 60,028 + 76,667= 136,695 m

2. slope A to B

= 3,037 %

slope B to C

= 4,178 %

3 slope A to B

=

=

Ratio = 1 : 32,931

slope B to C

=

=

Ratio = 1 : 23,938

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EXERCISE 9

Question 9.1

The diagram below is a longitudinal section of a pipeline Carefully study the drawing and calculate the following:

9.1.1 the length of the required piping between point A and B and B and C and C and D9.1.2 The slope as a percentage of the pipeline between A and B and B and C and C and D9.1.3 The slope as a ratio of the pipeline between A and B and B and C

B C

A

D E

200DATUM MSL

GROUND LEVEL

INVERT LEVEL

DISTANCE 3800

,0

209

,489

2

10,9

0

3860

,0

211

,200

2

12,5

0

3923

,0

211

,02

212,

20

3936

,6

208

,00

2

09,3

2

3967

,5

207

,90

2

09,4

8

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Question 9.2

A sewer internal reticulation longitudinal section MHO20 – MHO26 is shown below

9.2.1 Calculate the length on the pipeline between MHO23 and MHO259.2.2 Calculate the slope of the terrain between MHO24 and MHO25

PIPE

NGL

MHO20

MHO21 MHO22

MHO23

MHO24 MHO25

MHO26

Datum 55m

INVERT LEVEL

GROUND LEVEL

CHAINAGE (m)

00,0

00

67,

309

65

,900

18,1

94

65,5

97

45,9

56

66,

581

65,1

34

63,9

14

54,8

90

91,9

12

65,

894

64,5

10

150,

912

63

592

62,3

37

208,

912

61

,850

60,3

95

226,

912

60,1

53

277,

981

60

,423

Question 9.353

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From ANNEXURE C on REESTON INTERNAL SERVICES AREA C’-SEWERLAYOUT details junction C1 – C7

(NB: There are five sections of the pipeline to be considered)

Calculate the total length of pipework C1 to C7

Question 9.4

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ANNEXURE C

REESTON INTERNAL SERVICESAREA C – AS-BUILT SEWER LEVELS

The information below is for a layout plan of a pipeline from MHO10 – MHO134

Calculate the pipe length required from

9.4.1 MHO10 to MHO169.4.2 MHO16 to MHO159.4.3 MHO15 to MHO149.4.4 MHO14 to MHO13

6.7 Calculating the staff reading and height of sight rails (SR)

Collimation line

Staff staff reading at A 55

55

Staff reading at B

datum A

Traveller sight rail

Invert level A B

Δh

Invert level B

NOTE: This is base on the assumption that we are given the Depth of excavation And the ground level at A

Step 1 Calculate the collimation height

Collimation height = BS + BM

Step 2 Calculate the fall (difference in height)

Δh = gradient x distance

Step 3 Calculate the Invert level at A (if not given invert level of A)

Invert level at A = Ground level – Excavation

Step 4 Calculate the Invert level at B

Invert level at B = Invert level at A – Δh

Step 5 Calculate the staff reading at AStaff reading at A = Collimation height – (Invert level at A + height of the traveller)

Step 6 Calculate the staff reading at BStaff reading at B = Collimation height – (Invert level at B + height of the traveller)

EXAMPLE

Question

A drain PQ is to be set out using a tilting level, given the following information

Length of the drain 225mGradient PQ falling from P to Q at 0,80%Ground level at P is 229,38mDepth of excavation at P 0,90m

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Length of the traveller (Boning rod) is 1,90m

A backsight of 1,50m is taken on a benchmark with an elevation of 230,31m

Calculate the staff reading necessary to locate sight rails over P and Q

SOLUTION

Collimation height = 1,5 + 230,31 = 231,81

Falling = = 1,6

Invert level at P = 229,38 – 0,9 = 228,48

Invert level at Q = 228,48 – 1,6 = 226, 88

Staff reading at P = = 1,43

Staff reading at Q = = 3,03

EXERCISE 10

Question 10.1

A drain AB is to be set out using a tilting level, given the following information

Length of the drain 200mGradient AB falling from A to B at 1 : 125Ground level at A is 207,49mDepth of excavation at A 1,00mLength of the traveller (Boning rod) is 2,00m


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Calculate the staff reading necessary to locate sight rails over A and B

Question 10.2

A drain PQ is to be set out using a tilting level, given the following information

Length of the drain 225mGradient PQ falling from P to Q at 0,80%Ground level at P is 229,38mDepth of excavation at P 0,90mLength of the traveller (Boning rod) is 1,90m


Calculate the staff reading necessary to locate sight rails over P and Q

Question 10.3

A drain AB is to be set out using an automatic levelling instrument from the following information

Length of the drain pipe 155mGradient AB falling from A to B at 1 : 180Invert level of pipe at A 100,400mThickness of the pipe 30mmBedding depth 50mmLength of the traveller (Boning rod) is 1,800m



Question 10.4

Calculate the staff readings necessary to locate sight rails over N and M. Given the following information

Length of the drain pipe 165mGradient NM rising from N to M at 1 : 100Invert level of pipe at N is 280,400mLength of the traveller (Boning rod) is 2,00m


Calculate the staff reading necessary to locate sight rails over N and M

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Question 10.5

Calculate the staff readings necessary to locate sight rails over A and B Given the following information

Length of the drain pipe 180mGradient AB rising from A to B at 0,75%Ground level at P is 312,50mDepth of excavation at P is 1,00mLength of the traveller (Boning rod) is 2,00m



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CHAPTER 7

AREAS AND VOLUMES

7.1 AREA

Difference between the ordinate and mid-ordinateGiven the following Sketch

5m 6m 7m

3m

5m and 7m are ordinate

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m, ( 6m is in the middle of 5 and 7)

mid-ordinate = sum of two ordinate divide by two

FORMULAE

7.1.1 Mid ordinate rule

a) If given mid-ordinates

Area = width x sum of the mid ordinates

Example Given: mid-ordinate 6, 8, 7, 4 all in m and the width between the ordinate is 10m

Area = 10( 6 + 8 + 7 + 4)Area = 250 m2

b) If given ordinates

Area = width

Example Given: ordinate 8, 7, 9, 6, 5, 3, 0 all in m and the width between the ordinate is 15m

Area = 15

Area = 15 x 34Area = 510 m2

7.1.2 Trapezoidal rule (ONLY when given ordinates)

Area =

Example Given: ordinate 8, 7, 9, 6, 5, 3, 0 all in m and the width between the ordinate is 10m

Area =

Area = 7,5 x 68Area = 510 m2

7.1.3 Simpson’s rule (ONLY when given ordinates)

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Area =

ExampleGiven: ordinate 8, 7, 9, 6, 5, 3, 0 all in m and the width between the ordinate is 15m

Area =

Area = 5 x 100Area = 500 m2

Exercise 11

Question 11.1

From a straight line AB Offsets to the bank of a river are

A BDistance (m) 00 50 100 150 2000ffsets (m) 21 26 22 23 21

Calculate the area between AB and the river using the following methods11.1.1 Trapezoidal rule11.1.2 Mid-ordinate rule11.1.3 Simpson’s rule

Question 11.2

The following offset were taken from a baseline to the shoreline of a site along the coast at 10 mIntervals for a distance of 100 m. The offsets are 75m, 85m, 95m, 105m, 110m, 125m, 110m, 100m,98m and 80m

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Calculate the area of the site using the following methods11.2.1 Simpson’s rule11.2.2 Trapezoidal rule11.2.3 Mid-ordinate rule

Question 11.3

Calculate the area of an irregular piece of ground measurements were taken at intervals of 10mThe mid-ordinates are 42m, 46m, 44m, 48m, 52m, 50m

Question 11.4

A plot has the following ordinates 65m, 68m, 70m, 73m, 75m, 81m, 76m, 72m, 63m which were taken atIntervals of 25m

Calculate the area of the site using the following methods

11.4.1 Simpson’s rule11.4.2 Trapezoidal rule

ANSWERS

Question 1.1

CD = = 32,248 m

Question 1.2

1.2.1

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63

CD = = 93,732 m

1.2.1

m

CD = = 93,702 m

QUESTION 1.3

A-B = = 90,095B-C = = 72,180CD = = 47,184

= 209,459 m

Question 4

1.4.1

m

Slope distance = 1000 + 0,05 = 1000,050 m

OR

Slope =

= 1000,050 m

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1.4.2

CD = m

Question 1.5

Angle of the slope

Zenith distance = = 76º 51′ 9,2″

Question 1.6

= +0,0158 = – 0,0084

= 0.0177 = 0,0122

= 0,544 = 0,0791

= 93,464 = 83,0403

176,504 m

Question 1.7

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ºC

Question 2.1

Point BS IS FS Rise Fall Reduced levelA 2,40 150,00B 2,00 0,40 150,40C 1,90 0,10 150,50D 2,80 1,40 0,50 151,00E 2,00 0,8 151,80F 1,30 2,60 0,60 151,20G 0,60 3,00 1,7 149,50H 1,70 1,1 148,40

7,10–8,70–1,6

8,70 1,8 –3,4 –1,6

3,4 148,40 –150,00 –1,6

Question 2.2

Point BS IS FS Rise Fall Reduced level RemarksA 0,49 1032,40

0,27 3,29 2,80 1029,600,39 3,77 3,50 1026,10

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3,72 3,59 3,20 1022,90–1,11 4,83 1024,73

3,56 0,82 1,93 1022,80BM 3,89 1,36 2,20 1025,00 BM 1 025,00

3,72 0,99 2,90 1027,903,69 1,02 2,70 1030,603,86 1,31 2,38 1032,983,90 1,56 2,30 1035,28

B 2,40 1,50 1036,7823,93

–19,294,64

19,29

Question 2.3

Point BS IS FS Rise Fall Reduced levelBM1 3,141 401,261

A 0,086 3,055 404,316

B 2,111 4,283 4,197 400,119

C 1,406 3,517 403,636

D 2,032 0,626 404,262

E 3,108 3,638 5,67 398,592

F 1,823 2,110 5,218 403,810

G 3,111 4,934 398,876

H 4,123 2,109 5,220 404,096

I 3,281 7,404 396,692

BM2 3,083 0,198 396,890 2,414–6,785 –4,371

6,785 17,834 –6,785

–4,371

22,205 396,890 –401,261

–4,371

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SURVEY NOTES N5