Examination Form

Do not forget to fill your Examination form on 30th Oct (Saturday) between 10:00am - 12:30pm in Room No. 311 and 314 (SET)

Surveying Lab schedule

Below is the schedule for the Surveying Lab. Don't miss any of the labs, as nothing shall be repeated. We might also use some of the Tutorial classes to complete your course.


28th Oct and 29th Oct: Plane Table Surveying (G2 and G1 resp)
11th Nov and 12th Nov: Compass Survey (G2 and G1 resp)
18th and 19th Nov: Theodolite Survey (G2 and G1 resp)
25th and 26th Nov: Levelling Survey (G2 and G1 resp)


2nd and 3rd Dec: Revision of practicals


Your theory exams start from 8th Dec

Schedule of lectures for Term III

Below please find the schedule of lectures for Surveying course in Term III.

Since we are left only with one month of teaching, we will get a maximum of 20 lectures before your end term starts from 8th Dec.

Please attend the classes regularly to cover the entire course. No classes shall be repeated.

If there are no students in the class, the scheduled lecture shall be considered as complete. As was the case today (27th Oct) in the Tut of G1, since no one turned up for the classes, I am assuming the topic scheduled to be covered as complete.

Section III: Plane Table Surveying (4)

S.No.	Title	Number of Lectures	Tentative Date of Lecture
1	Basic principles, Equipments, Orientation and resection, Errors, Advantages and Disadvantages	2	27-28 Oct
2	Two point and three point problem and solution	2	1-2 Nov
	Total Lectures	4

Section IV: Angular Measurements (8)

S.No.	Title	Number of Lectures	Tentative Date of Lecture
1	Bearing and Angles, Reference direction, Magnetic bearing, magnetic declination, local attraction errors and adjustment	1	8 Nov
2	Theodolite surveying (types, uses, methods, adjustments , measurement of horizontal and vertical angles)	2	9-10 Nov
3	Traversing (traversing by compass and theodolite, computation of co-ordinates, sources of errors)	3	11, 15, 16 Nov
4	Triangulation (Network, strength of figures, selection of stations, measurement and computation)	2	18, 22 Nov
	Total Lectures	8

Section V: Leveling and Contouring (5)

S.No.	Title	Number of Lectures	Tentative Date of Lecture
1	Definitions, Methods of determining elevation, Classification of levels	1	23 Nov
2	Temporary and permanent adjustment of levels, method of reduction of levels, Sources of errors and precision, , Methods of representation	2	24-25 Nov
3	Definition and characteristics of contours, Methods of contouring and its usage	2	29-30 Nov
	Total Lectures	5

Section VI: Modern Surveying Techniques (2)

S.No.	Title	Number of Lectures	Tentative Date of Lecture
	Basics of Global Positioning System (GPS) and Laser Scanners	1	1 Dec
	Geographic Information System and Remote Sensing	1	2 Dec
	Total Lectures	2

Answers to the Mid term question paper

The evaluation of the mid term questions have been completed by me.
I am really disappointed by your marks. I was expecting much more than what you all have got. I can't revel the marks at this point of the time, but I am really really disappointed.
Some observations,
* Many of you have not written which question you are attempting. I had to find the question number and then specify
* Many of you have tried to attempt three from each of the sections, where as the only two was allowed from Sec-A and three from Sec-B
* None of you have actually done the numerical which was directly from Bhavakatti and Basak.
* None of you actually did the correct procedure for Question number 6
* None of you could even draw the correct diagram for Question number 7
* Only few of you have correctly done the Question 8
* I tried confusing you in Question 4 (both parts) and you got confused
* First three questions were directly from my slides, only few of you could score full marks in those
* Most surprising was your understanding of Significant Figures. Only a few were able to solve Question 5 (Part-1)

Below is the answers to all the questions. Evaluate yourselves, how much you should have got in this paper.

I expect a better performance from you in the end term.

All the best.

Solution:

1. What do you mean by Surveying?                                                                                       (0.5)

Science and art of determining the relative position of points above, on, or beneath the surface of the earth

or establishing such points with all kinds of appropriate approaches and instruments, which aims to make the natural appearances/ features known to the people

Whether it is Science and Arts? Explain.                                                                                (1)

Both.

Science

Because it uses mathematical techniques to analyze field data

Accuracy and reliability depends on understanding scientific principles underlying and affecting survey measurement

Art

Because only a surveyor who has full understanding of surveying techniques will be able to determine the most efficient methods required to obtain optimal results over a wide variety of surveying problems

List different types of Surveying based on “Type of measurement” and “Platform used”.                 (2)

Type: Planar and Geodetic

Platform used:

Land Survey: Measurements performed on the surface of the earth

Aerial survey: Measurements done from the aerial platform, such as aircraft, balloon

Satellite Survey: Data captured using sensors on-board satellites

What is the end product of Surveying?                                                                                               (0.5)

Plan/ Map

2. What is/ are the quantity measured using the following instruments, Laser Range Finder; Dumpy Level; Chain, Total Station, compass and GPS.                                                                             (3)

Laser Range Finders: Distance

Dumpy Level: Difference in elevations

Chain: Distance

Total Station: Angle and Distance

Compass: Angle

GPS: Co-ordinates

Why surveying in important to Civil Engineers?                                                                                  (1)

Civil Engineering has been associated with planning and construction of infrastructure

Construction is based on precise measurements

Measurement is achieved using “Surveying”

Surveying is the basic requirement of “Civil Engineers”

3. List the basic methods used in surveying. Explain them in brief.                                        (2)

Measurement of Horizontal distances, Vertical distances, Horizontal angles and Vertical angles

Explain the following terms with appropriate examples:

Representative factor

Scale

Accuracy

Precision

(2)

RF: Ratio of distance on the map to the corresponding distance on the ground

Scale represents the “number of times” the actual area is reduced

Accuracy of  measurement system is the degree of closeness of measurements of a quantity to its actual (true) value

Precision of a measurement system, also called reproducibility or repeatability, is the degree to which repeated measurements under unchanged conditions show the same results

4. Solve the following,

i) A rectangular plot in plan is 20cm x 40 cm, drawn to scale of 1cm = 500 m. If the same plot is drawn on a toposheet of 1:50,000, what would be the area on the toposheet.                                               (2)

Actual size = 20x500 by 40x500 sqm => 10000m x 20000m => 200000000 sqm => 2x10⁸ sqm

4.ii) A 30 m tape used for measuring a line was found to be 30.01 m at the beginning and 30.03 at the end of the work. The area of the plan drawn to a scale of 1:1000 was found to be 5146 mm2. Compute the correct area of the field.                                                                                                             (2)

Method1:

Avg length of the tape = (30.01+30.03)/2 = 30.02m

Correct length of the tape = 30.0m (l)

Measured area = 5146 sqm

Correct area = (l’/l)² x measured area = (30.02/30.0)² x 5146 = 5152.86 sqm

Method2:

Avg length of the tape = (30.01+30.03)/2 = 30.02m = l’

Correct length of the tape = 30.0m (l)

e1 = 1-(l’/l) = 1-(30.02/30.0) = -6.67x10^-4

Correct area = (1-2e1)x computed area = (1+2x6.67x10^-4)x5146 = 5152.86 sqm

Section B

5. Solve the following,                                                                                                             (2)

i) a) ; = 0.305

b) ; = 15.2

c) 3.22+5.1021+103.313+7.1; = 118.7

d) What would be the number 165.1, if a precision of 0.001 is used and 0.1 is used.                          

165.100; 165.1

ii) State the “Principle of Surveying”.                                                                                                 (1)

Working from whole to parts

Tie the whole of the area first

Divide the area into smaller parts

Survey the smaller parts using the reference points from whole

Work from “Whole to Part”

Choose the method which is most suitable (Operations Research!)

Make provisions of adequate checks

Record field data carefully

What is degree of accuracy and state the different order of accuracy?                                              (1)

How much close you are to the reference value.

Represented by

1 in xxxx

ppm (parts per million)

The minimum degree of accuracy allowed for a particular survey and the range of the allowed degrees of accuracy is known as the order of accuracy.

Order of accuracies: 1^st order, 2^nd order, 3^rd order

6. A steel tape was exactly 20 m long at 200 C when supported throughout its length under a pull of 5 kg. A line measured with this tape under a pull of 16 kg and at a mean temperature of 320C, was found to be 680 m long. Assuming the tape is supported at every 20 m; find the true length of the line. Given that (a) Cross-sectional area of the tape= 0.03 cm2, (b) Young’s Modulus=2.1x106 kg/cm2, (c) Thermal coefficient of expansion= 11 x 10-6 per 0C and (d) weight of tape = 10 g/cc.                                                    (4)

Given:

L = 20m

T₀ = 20⁰C

P₀ = 5kg

T_m = 32⁰C

P_m = 16kg

ML = 680kg

n = 1

A = 0.03 cm²

a = 11x10^-6 per ⁰C

E = 2.1x10⁶ cm²

Given weight = 10gm/cc

Total weight of the tape = 0.03 x 20 x 100 x 10 = 600gm = 0.6kg

Corrections per tape length:

Temp correction = = 11x10^-6 (32-30) x 20 = 0.00264m (+)

Pull correction = = = 0.00349m (+)

Sag correction = =0.00117 (-)

Total correction = 0.00264 + 0.00349 – 0.00117 = 0.00496m

Hence actual length of the tape = 20 + 0.00496 = 20.00496m

True length of the line = = 680.169m

7. Answer the following,

i) Explain the term hypotenuse allowance.                                                                             (1)

Used when intermediate points are to be located for taking offsets

Length of  a slope is always more than the corresponding horizontal distance. Thus, it is possible to measure a larger distance along a slope such that the equivalent horizontal distance is known

ii) There is an obstacle in the form of a pond on the main chain line AB. Two points C and D were taken on the opposite sides of the pond. A line CE of length 100m was drawn on the left side of line CD and another line CF of length 80m was drawn on the right side of CD (opposite direction of CE), such that E, D and F are in a straight line. Determine the obstruction length CD, where ED=60m and DF=56m.                                                                                                                                                   (3)

 

For CEF,

CE² = CF² + EF² – 2.CF.EF.Cos q

100² = 80² + 116² – 2x80x116.Cos q

10000 = 6400 + 13456 – 18560 Cos q

Cos q = (10000-6400-13456)/18560 => Cos q = -0.53

For CDF,

CD² = CF² + DF² – 2.CF.DF.Cos q

CD² = 80² + 56² – 2x80x56.(-0.53)

CD² = 6400+3136+4748.8 => CD = sqrt (14284.8) => CD = 119.52

8. Determine the values of Stadia constants (additive constant and multiplying constant) from the following observations,                                                                                                   (4)

Instrument Station	Staff reading on	Distance	Stadia Readings
			Lower	Upper
O	A	150	1.255	2.750
	B	200	1.000	3.000
	C	250	0.750	3.255

D = , where S: Stadia reading difference; (f+d): Additive constant = X; (f/i): Multiplicative constant = Y

Hence D = Y.S + X

For A, 150 = Y (2.750 – 1.255) + X => 150 = 1.495 Y + X

For B, 200 = Y (3.000 – 1.000) + X => 200 = 2.000 Y + X

For C, 250 = Y (3.255 – 0.750) + X => 250 = 2.505 Y + X

Solving, Y = 99.0; X = 2