UNIVERSITI TUN HUSSEIN ONN MALAYSIA

FINAL EXAMINATION ANSWERS SCRIPTSEMESTER I

SESI 2010/2011

COURSE NAME : ENGINEERING GEOMATIC

COURSE CODE : BFC 2103

PROGRAMME : 2 BFF

DATE : NOV 2010

DURATION : 3 HOURS

INSTRUCTION : ANSWER ONLY FOUR (4) QUESTIONS.

THIS PAPER CONSISTS OF ELEVEN (11) PAGES

BFC 2103

Q1 (a) Give the exact definition of Geomatic? Discuss the five (5) disciplines in Geomatic.

(10 marks)

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Geomatics is an “umbrella” term for a cluster of activities and technologies dealing with the locations and identities of earth features.

Geomatics has many sub-disciplines disccused as follow:(i) Land Surveying

Land surveying is a traditional name applied to measuring features directly in the field with tools such as theodolites, EDM, levels, and level staffs. A common use of surveying today is to map site in advance.

(ii) PhotogrametryPhotogrammetry is the science of measuring from photos as opposed to direct field measurement. Most is done with aerial photos taken from an airplane with a precise mapping camera.

(iii) Remote SensingRemote Sensing is a general term about gathering information about an object from a far-removed sensor. In common usage, the term remote sensing applies to acquisition and analysis of satellite imagery.

(iv) GeodesyGeodesy is the term applied to the study of the earth’s size and shape.

(v) Geographic Information System (GIS)(vi) The Global Positioning System (GPS)

(b) Briefly describe the type of errors in linear measurement and explain their source of error. What types of errors are classified as blunders and mistakes in linear measurement?

(10 marks)

The types of errors in linear measurement and their sources of error are described as follows:(i) Systematic or Cumulative Errors

These errors exist in any survey measurement and each additional measurement increases the effect of the error. Such errors which may be either +ve or –ve.

(ii) Compensating or Accidental ErrorsAlthough every precaution may be taken, certain unavoidable errors always exist in any measurement. Such errors are generally or less important than systematic errors. Sometimes +ve and sometimes –ve.

(iii) Gross ErrorsThese errors arise from mistakes, carelessness or lack of experience. They are quite random and allowance cannot be made for them.

Examples of errors are classified as blunders and mistakes in linear measurement are:(i) displacement of arrows or station mark,(ii) miscounting tape lengths,(iii) misreading the tape, and(iv) wrong booking

(c) What is the significant difference between topographic survey and engineering survey in terms of their uses and scales.

(5 marks)

Topographic SurveysThese are surveys where the physical features of the earth are measured. The uses of topographic survey are for engineering design, navigational, recreational, geological, and military. The map or plan produced in topographic survey has scales ranging from 1:25,000 to 1:1,000,000

Engineering SurveysThese embrace all survey works required before, during and after any engineering works. Some common uses of engineering surveys are for building work, location drawings, site plan, civil engineering works, and highway surveys. Typical scales range used in this type of survey are from 1:50 to 1:50,000.

Q2 (a) What are the applications of leveling in civil engineering.(4 marks)

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Typical examples of leveling applications includes:(i) establishing new vertical control (TBM),(ii) providing data for road cross-sections or volumes of earthworks,(iii) providing a level or inclined plane in the setting out of contruction works, and(iv) determining the height of discrete points.

(b) The following consecutive readings in meters were taken with a automatic level:

(0.795, 1.855, 3.190, 3.015, 0.655, 0.625, 0.955, 0.255, 1.635, 0.860, 2.375).

The level was shifted (move) after the fourth and eight readings. The first reading was taken on a benchmark whose Reduce Level (R.L) is 550.605 meters. Create a page of a level book and enter the readings. Calculate the reduced levels of a stations by the RISE and FALL Method and apply arithmetical checks.Note : Use Form Q2 to answer this question

(15 marks)

(c) In a two peg test of a automatic level, the following readings were taken:

(i) Instrument at B, midway between A and C where AB=BC Staff reading on A = 1.726 Staff reading on C = 1.262

(ii) Instrument at D where CD = AB/10

Staff reading on A = 2.245 Staff reading on C = 1.745

Determine whether or not the instrument is in adjustment?(6 marks)

Difference in height determined from A = 1.726 – 1.262 = 0.464 m (2M)

Difference in height determined from D = 2.245 – 1.745 = 0.500 m (2M)

Collimation error = 0.500 – 0.464 = 0.036 m (2M)

S3 (a) Define the following:

i. Close TraverseA closed traverse is one that begins and ends at the same point or one that begins and ends at points whose positions have been previously determined.

( 3 marks)

ii. Open TraverseAn open traverse is a series of measured straight lines (and angles) that do not geometrically close.

( 3 marks)

(b) Table Q3 shows the final bearing and distance from second class field work book.

Table Q3 : Traverse bearing and distance

Line Final Bearing Final Distance (m)

Coordinates

North East

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1234.50 6789.00

1 - 2 063°30'00" 63.2642 - 3 077°25'00" 75.1193 - 4 173°43'30" 82.1474 - 5 231°55'00" 87.2735 - 1 322°19'00" 114.829

Note: Use Form Q3 to answer this question.

Determine the following :(i) Linear Misclosure

( 5 marks)

(ii) Latitude and departure correction using Bowditch method( 5 marks)

(iii) Coordinate for every stations( 5 marks)

(iv) The traverse area using coordinate method( 4 marks)

S4 (a) Briefly explain the tacheometry systems below:

i. Optics Tacheometry

The stadia method (Optics Tacheometry) is an approximate procedure for measuring distances and elevations using a theodolite and stadia rod (leveling staff). In this method, the tacheometer is directed at the staff and the distance id indirectly computed by reading the upper and lower stadia hairs on the telescope view.

( 3 marks)

ii. EDM Tacheometry

Uses a total station which contains an EDM, able to read distance by reflecting off a prism.

( 3 marks)(b) Tacheometry survey using stadia technique was performed from station O. Table Q4(a) shows all

the observation data.

Table Q4(a) : Techeometry data

Station : OInstrument height : 1.500 meterStation reduced level : 12.635 meter

Vertical Angle StadiaNotesUpper Middle Lower

16°20'40" 2.120 1.435 0.750 To A10°32'40" 3.050 1.837 0.625 To B

Based on this data, determine :

i. Horizontal distance for each observation point when the constant values (K) = 100 and (c) = 0.HAB = 100 S Cos2 = 100 (1.370) Cos2 (16°20'40") = 126.150 meter

(2M)

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A

C1

30°

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HAC = 100 S Cos2 = 100 (2.425) Cos2 (10°32'40") = 234.379 meter(2M)

( 4 marks)

ii. Reduced level for every observation point.RLB = RLA + IH + V – h = 12.635 + 1.500 + (50 S Sin 2) – 1.435 = 12.635 + 1.500 + [50 (1.370) Sin 2(16°20'40")] – 1.435 = 12.635 + 1.500 + [36.995] – 1.435 = 49.695 meter

(3M)

RLC = RLA + IH - V – h = 12.635 + 1.500 + (50 S Sin 2) – 1.837 = 12.635 + 1.500 + [50 (2.425) Sin 2(10°32'40")] – 1.837 = 12.635 + 1.500 + [43.628] – 1.837 = 55.926 meter

(3M) ( 6 marks)

(c) Table Q4(b) shows the data from tacheometry survey using total station.

Table Q4(b) : Tacheometry observations data

Fr. To R.L. Ins. Bearing Horz. Prism Diff. NotesStn Stn Stn Height   Dist. Height Height  1 2 8.940 1.543 00° 00'            72° 05' 21.333 1.350 0.250 A        102°00' 18..490 1.350 -0.347 B        102°00' 28.897 1.350 0.634 C

Calculate :

i. Reduced level for point A, B and CRLA = RL1 + IH + V – TH = 8.940 + 1.543 + 0.250 – 1.350 = 9.383 meter

(2M)RLB = RL1 + IH - V – TH = 8.940 + 1.543 - 0.347 – 1.350 = 8.786 meter

(2M)RLC = RL1 + IH + V – TH = 8.940 + 1.543 + 0.634 – 1.350 = 9.767 meter

(2M)( 6 marks)

ii. Horizontal distance for AC

AC2 = (1--A)2 + (1--C)2 – [2(1--A) (1--C) Cos2 30°] = 21.3332 + 28.8972 – [2(21.333)(28.897) Cos230°]AC = 19.117 meter

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( 3 marks)S5 (a) Table Q5(a) shows the area of contour lines from 100m to 140m. Based on this value determine the

volume using trapezium and simpson method.

Table Q5(a) : Contour line and area

Contour line (m) Area (m²)100 3250110 3101120 2875130 1337140 571

(10 marks)

Trapezium MethodI = d (1/2 (A1 + An) + A2 + A3 + An-1) = 10 (1/2 (3250 + 571) + 3101 + 2875 +1337) = 92235 m³

(5M)Simpson MethodI = d/3 ((A1 + An) + 4 (A2 + A4 +An-1) + 2 (A3 )) = 10/3 ((3250 + 571) + 4 (3101 + 1337) + 2 (2875)) = 91076 m³

(5M)

(b) Figure Q5 shows all point observed using the levelling equipement with grid method. The reduced level values for each point are given in Table Q5(b). Each point will be dug to same level of 10 m above datum. Determine the mean value and volume using both methods.

(i) Triangle Method(5 marks)

(ii) Square Method(5 marks)

Table Q5(b) : Reduced level for each point

Point Reduced Level (m)A 13.10B 13.48C 14.01D 13.94E 13.56F 13.87G 14.53H 14.27

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A B C D 10m 10m 10m

10m 10m 10m

10m 10m

E F G H

FIGURE Q5

Triangle Method:

Point Reduced Level

Formed Level X Num (N) N x X

A 13.10 10.00 3.10 1 3.10B 13.48 10.00 3.48 3 10.44C 14.01 10.00 4.01 3 12.03D 13.94 10.00 3.94 2 7.88E 13.56 10.00 3.56 2 7.12F 13.87 10.00 3.87 3 11.61G 14.53 10.00 4.53 3 13.59H 14.27 10.00 4.27 1 4.27

18 70.04

Average Level = (70.04/18) = 3.89 meter Volume = Area x Average Level = (30 x 10) x 3.89 = 1167 m3

(5M)

Square Method:Average Level = (46.65/12) = 3.89 meter Volume = Area x Average Level = (30 x 10) x 3.89 = 1167 m3

(5M)

Point Reduced Level

Formed Level X Num (N) N x X

A 13.10 10.00 3.10 1 3.10B 13.48 10.00 3.48 2 6.96C 14.01 10.00 4.01 2 8.02D 13.94 10.00 3.94 1 3.94E 13.56 10.00 3.56 1 3.56F 13.87 10.00 3.87 2 7.74G 14.53 10.00 4.53 2 9.06H 14.27 10.00 4.27 1 4.27

12 46.65

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Irregular Boundary

Fig. 4 : Irregular Boundary

Fig. 5: Simpson’s Rule

BFC 2103

(c) Define the algorithm to calculate the area for irregularly curved boundaries.

(5 marks)

The Trapezoidal Rule

Land parcels are not always contained by regular straight line or circular arc boundaries, especially when they front water courses or ridge lines. Methods for surveying these boundaries and computing the enclosed areas are as follows:

Area = L [ (1st ordinate + last ordinate)/2 + (sum of other ordinates) ]

where O1 .. On are ordinates; P is the uniform distance between ordinates.

The Simpson’s Rule

In Simpson’s Rule, it is assumed that the irregular boundary is comprised of parabolic arc. The assumption is that each adjacent sub-areas are a single bounded parabola rather than each sub-area being a trapezoid.

A = d/3 [(1st + last ordinates) + 2(odd ordinates) + 4(even ordinates)]

where O1 .. On are ordinates; d is the distance between ordinates.

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Form Q2

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Borang Q3

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