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IV Semester Civil CE2254-Surveying II by M.Dinagar A.P / Civil

MAHALAKSHMI ENGINEERING COLLEGE

TIRUCHIRAPALLI-621213

QUESTION BANK

CE2254 - SURVEYING - II

UNIT 1- TACHEOMETRIC SURVEYING

PART – A (2 marks)

1. Why is an anallatic lens provided in tacheometer? (AUC Apr/May 2010 & May/June 2013)

This arrangement is made to reduce the additive constant to zero and simplifies the

mathematical calculations and only multiplying constant is present. This lens reduce the

brilliance of the image.

2. What are the multiplying constant and additive constant of a tacheometer?

(AUC Apr/May 2010)

In a tacheometer the constant i

f is known as the multiplying constant or stadia interval

factor and the constant (f + d) is known as additive constant of the instrument where f is the focal length of the instrument and d is the distance of the vertical axis of the instrument from the focus O.

3. Consider the horizontal distance equation D = KS + C. what are represented by K, S and

C. (AUC Apr/May 2011)

D = KS + C

Where D = horizontal distance from instrument and levelling staff

K or i

f = multiplying constant

C or (f + d) = additive constant

S = staff intercept

f = focal length of object glass

i = length of image

d = distance between optical centre and vertical axis of instrument

4. What is parallax? How it can be eliminated? (AUC Apr/May 2011)

Parallax is a condition arising when the image formed by the objective is not in the plane

of the cross-hairs. Accurate sight is possible only when parallax is eliminated. It is eliminated by

focusing the eye piece and objective.

5. What are the different systems of tacheometric survey? (AUC May/June 2009)

1. Stadia system

i. Fixed hair method

ii. Movable hair method

2. Tangential system

6. What is a base net? (AUC May/June 2009)

Some site conditions may not be favorable to get the required length of a base line. In

such a situation a short base line is selected and the same is then extended. Such group of

triangles which are meant for extending the base is known as base net.


7. Define Stadia diagrams. (AUC Nov/Dec 2010)

The diaphragm of the tacheometer is provided with two stadia hairs (upper and lower

ones). Telescope of the tacheometer is directed towards the staff held at a point whose distance

from the instrument is to be found. The readings pertaining to stadia hair are taken. The

difference in these readings gives the staff intercepts.

8. Write any two advantages of tacheometric surveying.

(AUC Nov/Dec 2010 & May/June 2013)

The multiplying constant should have a nominal value of 100 and the error contained in

this value should not exceed 1 in 1000.

The axial horizontal line should be exactly midway between the other two lines.

The telescope should be truly anallatic.

The telescope should be powerful having a magnification of 20 to 30 diameters.

9. What is a tacheometer? (AUC May/June 2012)

The conventional transit theodolite fitted with a stadia diaphragm and an anallatic lens is

called a tacheometer.

10. Enumerate the errors caused due to manipulation and sighting in tacheometric

surveying. (AUC May/June 2012)

i. instrumental errors

ii. errors of observation

iii. errors due to natural causes

11. State the uses of tacheometry. (AUC Nov/Dec 2012)

Preparation of topographic maps which require both elevations and horizontal distances.

Survey work in difficult terrain where direct methods are inconvenient

Detail filling and Checking of already measured distances

Reconnaissance surveys for highways, railways, etc.

Hydrographic surveys and

Establishing secondary control.

12. What is subtense bar? What are its advantages? (AUC Nov/Dec 2012)

The subtense bar is an instrument used for measuring the horizontal distance between the

instrument station and a point on the ground.

Advantages:

Field work for long sights is more accurate as it is easier to intersect a fixed vane with a

movable wire than to read the staff graduations. This is very effective in setting out curves for

railways, etc. there is no staff or target rod is needed.

13. What are the three types of telescope used in stadia surveying?

External focusing telescope

External focusing anallatic telescope

Internal focusing telescope

14. List merits and demerits of movable hair method in tacheometric survey.

Merits:

Movable hair method is more accurate.

Long distances can be taken with greater accuracy than in stadia method.

Demerits:

Lacks speed in the field.

Variables m and i should be measured accurately.

Because of these limitations this method is almost obsolete.


15. Compare tangential and stadia method.

S.No Tangential method Stadia method

1 Transit theodolite is used in this system and it does not possess with a stadia hairs, the staff is provided with two vanes or targets at a known distance.

The stadia method is based on the principle

that the ratio of the perpendicular to the

base is constant in similar isosceles

triangles.

2 The horizontal and vertical distances are computed by measuring angles of elevation or depression. In all cases staff is held vertically.

The horizontal and vertical distances are computed by measuring angles of elevation or depression. In all cases staff is held vertically and inclined also.

16. What is the difference between a theodolite and tacheometer?

S.No Theodolite Tacheometer

1 They are precision instruments having telescopic sights for establishing horizontal and sometimes vertical angles.

As per this property of isosceles triangles the ratio of the distance of the base from the apex and the length of the base is always constant.

2 Theodolite is precisely leveled and it

is used from a tripod.

Tacheometer is a somewhat obsolescent term for a type of distance meter or range finder. Tacheometer is precisely leveled and it is used from a tripod.

17. What is tangential tacheometry?

The tangential method of tacheometry is being used when stadia hairs are not present in

the diaphragm of the instrument or when the staff is too far to read.

18. The readings on a staff held vertically 60 m from a tacheometer were 1.460 and 2.055. The

line of sight was horizontal. The focal length of the objective lens was 24 cm and the

distance from the objective lens to the vertical axis was 15 cm. Calculate the stadia

interval.

Solution:

D = i

f x 0.595 + (f + d)

Substituting the values,

60 x 100 = 100595.024

xxi

+ (24 + 15)

600 – 39 = i

1428

i = cm55.2561

1428

Stadia interval = 2.55 cm

19. What is the difference between staff intercept and stadia intercept?

The difference between the targets is kept fixed in a staff intercept whereas in a stadia

intercept the distance between the stadia hairs is variable.


PART – B (16 marks)

1. (i) A vane 3 m above the foot of a staff was sighted at a point 1200 m away from the

instrument. The observed vertical angle was 1° 30'. The reduced level of the instrument

station was 250.50 m and the height of the instrument axis is 1.5 m. Find the reduced

level of the staff station. Apply the combined correction for curvature and refraction in

finding the R.L. of the station. (AUC Apr/May 2011)

Solution:

Combined angular correction for curvature and refraction

= sec"1sin2

)21(

R

dm

Assume m = 0.07 and R sin 1” = 30.88

Combined angular correction = "71.1688.302

1200))07.02(1(

X

X

Corrected vertical angle = 1o 30’ – 0o 0’ 16.71” = 1o 29’ 43.29”

Vertical height, V = D tan α = 1200 x tan (1o 29’ 43.29”)

V = 31.32 m

RL of staff station, Q = RL of instrument station P + HI + V – h

= 250.5 + 1.5 + 31.32 – 3

RL of staff station Q = 280.32 m

(ii) Determine the gradient from a point A to a point B from the following observations

made with a tacheometer fitted with an anallatic lens. The constant of the instrument was

100 m and the staff was held vertically.

(AUC Apr/May 2011)

Inst

station

Staff

station Bearing

Vertical

angle Staff readings

P A 134° + 10° 32' 1.360, 1.915, 2.470

B 224° + 05° 06' 1.065, 1.885, 2.705


Solution:

Let the station of observation be P. Here K = 100 and c = 0

a) Observation from P to A:

S1 = 2.470 – 1.360 = 1.11 m

Distance PA = KS1 cos2 θ1 + c cos θ1

= 100 x 1.11 x cos2 (10o 32’) + 0

D1 = 107.3 m

V1 = D1 tan θ1 = 107.3 x tan (10o 32’)

V1 = 19.95 m

Difference in elevation between P & A = 19.95 – 1.915 = 18.035 m

b) Observation from P to B:

S2 = 2.705 – 1.065 = 1.64 m

Distance PB = KS2 cos2 θ2 + c cos θ2

= 100 x 1.64 x cos2 (5o 06’) + 0

D2 = 162.7 m

V2 = D2 tan θ2 = 162.7 x tan (05o 06’)

V2 = 14.52 m

Difference in elevation between P & B = 14.52 – 1.885 = 12.635 m

Angle APB = bearing of PB – bearing of PA = 224o – 134o = 90o

Distance AB = 2222 7.1623.107PBPA

Distance AB = 194.89 m

Difference in elevation between B & A,

= 18.035 – 12.635

= 5.4 m (A is higher)

Gradient from A to B = 0277.089.194

4.5

tan ABofceDis

elevationinDifference

Gradient from A to B = 1 in 36

2. To determine the gradient between two points A and B a tacheometer was set up at

another station C and the following observations were taken, keeping the staff vertical.

Staff at Vertical angle Stadia readings (m)

A + 4O 1.300, 1.610, 1.920

B + 0O 10’ 40’’ 1.100, 1.410, 1.720

If the horizontal angle ACB is 35O 20’ 00’’. Determine the average gradient between A and

B, k = 100, c = 0. (AUC Nov/Dec 2010)


Solution:

Let the station of observation be C. Here K = 100 and c = 0

a) Observation from C to A:

S1 = 1.920 – 1.300 = 0.620 m

Distance CA = KS1 cos2 θ1 + c cos θ1

= 100 x 0.62 x cos2 (4o 20’) + 0

D1 = 61.65 m

V1 = D1 tan θ1 = 61.65 x tan (4o 20’)

V1 = 4.67 m

Difference in elevation between C & A = 4.67 – 1.610 = 3.06 m

b) Observation from P to B:

S2 = 1.720 – 1.100 = 0.62 m

Distance PB = KS2 cos2 θ2 + c cos θ2

= 100 x 0.62 x cos2 (0o 10’ 40”) + 0

D2 = 62 m

V2 = D2 tan θ2 = 62 x tan (0o 10’ 40”)

V2 = 0.19 m

Difference in elevation between C & B = 0.19 – 1.410 = - 1.22 m

Angle ACB = 35o 20’

Distance AB = ACBCBCACBCA cos.222

= )'2035(cos6265.6126265.61 22 OXXX

Distance AB = 37.52 m

Difference in elevation between A & B,

= 3.06 – (-1.22)

= 4.28 m (A is higher)

Gradient from A to B = 114.052.37

28.4

tan ABofceDis


Gradient from A to B = 1 in 9

3. In a subtense measurement of a leg of a traverse, two targets were set up at right angles

to the line of sight from the Instrument Station but on a sloping ground. From the

following data, calculate the distance of P and Q from the instrument stations.

Angle of elevation to target at P = 48O 00’

Angle of elevation to target at Q = 12’ 40”

Horizontal angle at instrument subtended by PQ = 1O 40’ 20’’


Height of target above ground = 1.600 m

Slope measurement PQ = 28.0 m. (AUC Nov/Dec 2010)

Solution:

Assume the instrument station be C.

Here D’ = 28 m

The distance between instrument station and bar point,

D = )"20'401(cos

28

cos

'o

D

D = 28.012 m

From Δ CC’P,

CP = )48(cos

012.28

cos

'o

CC

CP = 41.86 m

From Δ CC’Q,

CQ = )"40'12(cos

012.28

cos

'CC

CQ = 28.01 m

4. Derive the expressions for horizontal and vertical distances by fixed hair method when

the line of sight is inclined and staff is held vertically. (AUC Apr/May 2010)

Fig. Angle of elevation


Let P = Instrument station;

Q = Staff station

M = position of instruments axis;

O = Optical centre of the objective

A, C, B = Points corresponding to the readings of the three hairs

s = AB = Staff intercept;

i = Stadia interval

θ = Inclination of the line of sight from the horizontal

L = Length MC measured along the line of sight

D = MQ’ = Horizontal distance between the instrument and the staff

V = Vertical intercept at Q, between the line of sight and the horizontal line

h = height of the instrument;

r = central hair reading

β = angle between the two extreme rays corresponding to stadia hairs.

Draw a line A’CB’ normal to the line of sight OC.

Angle AA’C = 900 + β/2, being the exterior angle of the ∆COA’.

Similarly, from ∆COB`, angle OB’C = angle BB’C = 900 – β/2.

Since β/2 is very small (its value being equal to 17’ 11” for k = 100), angle AA’C and angle

BB’C may be approximately taken equal to 900.

Angle AA’C = angle BB’C = 900

From ∆ ACA’,

A’C = AC cos Ө or A’B’ = AB cos Ө = s cos Ө ………. (a)

Since the line A’B’ is perpendicular to the line of sight OC, equation D = k s + C is directly

applicable. Hence, we have

MC = L = k x A’B’ + C = k s cosӨ + C . . . . . . . (b)

The horizontal distance

D = L cosӨ = (k s cosӨ + C) cosӨ

D = k s cos2Ө + C cosӨ . . . . . . (1)

Similarly, V = L sin Ө = (k s cosӨ + C) sinӨ = k s cosӨ x sinӨ + C sinӨ

V = k s 2

2sin+ C sinӨ . . . . . . (2)

Thus equations (1) and (2) are the distance and elevation formulae for inclined line of sight.

(a) Elevation of the staff station for angle of elevation:

If the line of sight has an angle of elevation Ө, as shown in the figure, we have

Elevation of staff station (Q) = RL of instrument station (P) + h + V – r.


(b) Elevation of the staff station for the angle of depression:

Fig. Angle of depression

If the line of sight has an angle of elevation Ө, as shown in the figure, we have

Elevation of Q = RL of P + h – V - r

5. Determine the gradient from a point P to point Q from the following observations carried

out with a tacheometer fitted with an anallatic lens.

Inst station

Staff point

Bearing Vertical angle

Staff readings

O P 340° + 17° 0.760, 1.455, 2.170

O Q 70° + 12° 0.655, 1.845, 3.150

Assume that the staff is held vertical and that the multiplying constant of the instrument

is 100. (AUC Apr/May 2010)

Solution:

Let the station of observation be O. Here K = 100 and c = 0

a) Observation from O to P:

S1 = 2.170 – 0.760 = 1.41 m

Distance OP = KS1 cos2 θ1 + c cos θ1

= 100 x 1.41 x cos2 (17o) + 0

D1 = 128.95 m

V1 = D1 tan θ1 = 128.95 x tan (17o)

V1 = 39.42 m

Difference in elevation between O & P = 39.42 – 1.455 = 37.965 m

b) Observation from O to Q:

S2 = 3.150 – 0.655 = 2.495 m

Distance OQ = KS2 cos2 θ2 + c cos θ2

= 100 x 2.495 x cos2 (12o) + 0

D2 = 238.71 m

V2 = D2 tan θ2 = 238.71 x tan (12o)

V2 = 50.74 m


Difference in elevation between O & Q = 50.74 – 1.845 = 48.895 m

Angle POQ = bearing of OP – bearing of OQ = 340o – 70o = 270o

Angle POQ = 270o – 180o = 90o

Distance PQ = 2222 71.23895.128OQOP

Distance PQ = 271.31 m

Difference in elevation between P & Q,

= 48.895 - 37.965

= 10.93 m (Q is higher)

Gradient from P to Q = 0403.031.271

93.10

tan PQofceDis


Gradient from P to Q = 1 in 25

6. Explain the objectives and theory of anallatic lens. (AUC Apr/May 2010)

A convex lens specially provided in a telescope between the object lens and eyepiece to

eliminate the additive constant (f + d) from the tacheometric distance equations is known as

Anallatic lens. It is fitted in external focusing telescopes only.

Theory of an anallatic lens:

Consider the figure, in which O is the optical centre of the objective of an external

focusing telescope.

Let A, C, and B = the points cut by the three lines of sight corresponding to three wires.

b, c, and a = top, axial and bottom hairs of the diaphragm.

ab = i = interval b/w the stadia hairs (stadia interval)

AB = s = staff intercept;


f = focal length of the objective

f1 = horizontal distance of the staff from the optical centre of the objective

f2 = horizontal distance of the cross-wires from O.

d = distance of the vertical axis of the instrument from O.

D = horizontal distance of the staff from the vertical axis of the instruments.

C = centre of the instrument, corresponding to the vertical axis.

A’B’ = position of the image

K = distance between anallatic lens and object glass

F1 & F2 = conjugate focal distances of the object glass

From the laws of lenses, we know

)1......(..........111

21 FFF

)2......(..........111

21 fff

(negative sign indicates boa and B’O’A’ are same side of the anallatic lens)

Let A’B’ = i and ab = i’

Since the rays BOb and AOa pass through the optical centre, they are straight so that AOB and

aOb are similar. Hence,

From similar triangles ΔABO and A’B’O, we get

2

1

F

F

i

S…………….. (3)

From similar triangles ΔABO and A’B’O, we get

1

2

' f

f

i

i…………….. (4)

Multiply equation 3 & 4 we get,

1

2

2

1

' f

fx

F

F

i

S ………… (5)

Sub, the values of 1

2

2

1 &f

f

F

F in eqn (5)

f

ffx

F

FF

i

S 21

'………… (6)

or

f

KFfx

F

FF

i

S )(

'

21

Ff

fKFKfFF )()(1

Ff

KfFF )(1

'i

S -

Ff

fKF )(

F1 = 'i

S x

)( KfF

Ff

)(

)(

KfF

fKF

But D = F1 + d


Where d = distance of objective from the vertical axis of the theodolite

D = 'i

S x

)( KfF

Ff

)(

)(

KfF

fKF+ d ………… (7)

The condition of distance D should be proportional to the intercept S,

- )(

)(

KfF

fKF+ d = 0

)(

)(

KfF

fKF= d

K = f + dF

Fd …………….. (9)

The distance between anallatic lens and objective should be made equal to K. by adopting

suitable values of F, f, i' & K in the first term of eqn (7),

100)(' KfFi

Ff

D = 100 S ………………… (10)

From the above equation, if the theodolite is fitted with an anallatic lens, the horizontal

distance between the instrument axis and staff position is obtained by multiplying the staff

intercept by multiplying constant.

7. The following are the observation taken by a theodolite.

Inst station

Staff station

Target Vertical angle

Staff reading

Remarks

A BM Lower -12° 0.650

RL of B.M.

= 500 m

Upper -9° 2.550

A B Lower -6° 1.255

Upper +4° 3.100

Find out the observation of BM and station B and the distance between the BM and

station B. (AUC Apr/May 2010)

Solution:

a) Observation to BM:

S1 = 2.550 – 0.650 = 1.9 m

h1 = 0.650 m

D1 = )9(tan)12(tan

9.1

)(tan)(tan 00

21

1S

D1 = 35.07 m


V1 = D1 tan θ1 = 35.07 x tan (12o)

V1 = 7.45 m

RL of instrument axis, A = RL of BM + h1 + V1

= 500 + 0.65 + 7.45

RL of instrument axis A = 508.1 m

b) Observation to B:

h2 = 1.255 m; S2 = 3.100 – 1.255 = 1.845 m

D2 = )4(tan)6(tan

845.1

)(tan)(tan 00

43

2S

D2 = 10.54 m

V2 = D2 tan θ3 = 10.54 x tan (6o)

V2 = 1.108 m

RL of B = RL of instrument axis – V2 – h2

= 508.1 – 1.108 – 1.255

RL of B = 505.74 m

Distance between BM and station B = D1 + D2

= 35.07 + 10.54

Distance between BM and station B = 45.61 m

8. Explain how you would compute the horizontal and vertical distances from the

instrument station in the tangential method of tacheometry. With the help of a schematic

diagram, deduce the equations for the horizontal distance and the vertical distance when

both the vertical angles measured are angles of elevation. (AUC Apr/May 2011)


9. A theodolite was set up at a distance of 150 m from a tower. The angle of elevation to the

top of the tower was 10° 08', while the angle of depression to the foot of the tower was

03° 12'. The staff reading on the B.M. of R.L. 50.217 with the telescope horizontal was

0.880. Find the height of the tower and the reduced level of the top and foot of the tower.

(AUC Apr/May 2011)

Solution:

Here D = 150 m; θ1 = 10o 08’, θ2 = 3o 12’, HI = 0.880 m

S = D (tan θ1 + tan θ2) = 150 [tan (10o 08’) + tan (3o 12’)]

S = 35.19 m

V = D tan θ2 = 150 x tan (3o 12’)

V = 8.39 m

RL of instrument axis = RL of BM + Height of instrument (HI) = 50.217 + 0.88

RL of instrument axis = 51.097 m

RL of foot of tower = RL of instrument axis – V = 51.097 – 8.39

RL of foot of tower = 42.707 m

RL of top of tower = RL of foot of tower + S = 42.707 + 35.19

RL of top of tower = 77.897 m


10. Explain the principles and uses of a Beaman stadia arc. (AUC May/June 2009)


11. Explain different errors that may arise in stadia. (AUC May/June 2009)


12. Distinguish between vertical and normal holding a staff in tacheometry survey.

(AUC May/June 2009)


13. The following readings were taken with an anallatic tacheometer. The value of the

constant was 100 and the staff was held vertically.

Inst station

Height of axis

Staff station

Vertical angle

Staff reading Remarks

A 1.46 B.M -5O 30’ 0.92, 1.76, 2.55 RL of BM

= 209.05 m

A 1.46 B +3O 24’ 0.96, 1.70, 2.45

B 1.40 C +6O 12’ 0.90, 1.97, 3.04

Determine the horizontal distances between A, B and C and also the elevations of the

three stations. (AUC May/June 2009)


Solution:

a) From the observation A to BM:

S1 = 2.55 – 0.92 = 1.63 m; θ1 = - 5O 30’; k = 100; C = 0

V1 = 2

)'3052(sin63.1100

2

2sin 11

oXXXSk

V1 = 15.55 m

b) From the observation A to B:

S2 = 2.45 – 0.96 = 1.49 m; θ2 = 3O 24’

AB = D1 = KS2 cos2 θ2

= 100 x 1.49 x cos2 (3O 24’)

AB = D1 = 148.476 m

V2 = 2

)'2432(sin49.1100

2

2sin 22

oXXXSk

V2 = 8.82 m

c) From the observation B to C:

S3 = 3.04 – 0.90 = 2.14 m; θ3 = 6O 12’

BC = D2 = KS3 cos2 θ3

= 100 x 2.14 x cos2 (6O 12’)

BC = D2 = 211.50 m

V3 = 2

)'1262(sin14.2100

2

2sin 33

oXXXSk

V3 = 22.98 m

d) Reduced levels of stations:

RL of instrument axis at A = RL of BM + Axial hair reading + V1

= 209.05 + 1.76 + 15.55

= 226.36 m

RL of station A = RL of instrument axis at A – (HI)A

= 226.36 – 1.46

= 224.9 m

RL of station B = RL of station A + (HI)A + V2 – Axial hair reading

= 224.9 + 1.46 + 8.82 – 1.7

= 233.48 m

RL of station C = RL of station A + (HI)A + V3 – Axial hair reading

= 224.9 + 1.46 + 22.98 – 1.97

= 247.37 m


14. You are given a theodolite fitted with stadia hairs, the object glass of telescope being

known to have a focal length of 230 mm and to be at a distance of 138 mm from the

trunnion axis. You are told that the multiplying constant for the instrument is believed to

be 180. The following tacheometric readings are then taken from an instrument station A,

the reduced level of which is 15.05 m.

Inst at

H.I Sight

to Vertical angle

Stadia readings Remarks

A 1.380 m B +30O 1.225, 1.422, 1.620 Staff held vertical RL

of B = 40.940 m

A 1.380 m C +45O 1.032, 1.181, 1.330 Staff held normal to

line of sight

Find the distance AB, AC and reduced level of C. (AUC May/June 2012)

Solution:

Focal length, f = 230 mm = 0.23m

Distance of optical centre and trunnion axis, d = 138 mm = 0.138 m

Additive constant of the instrument, (f + d) or C = 0.23 + 0.138 = 0.368 m

Multiplying constant of the instrument, K = 180

RL of instrument station, A = 15.05 m

RL of plane of collimation = 15.05 + 1.38 = 16.43 m

i) Staff held vertically:

S1 = 1.620 – 1.225 = 0.395 m

Distance, AB = KS1 cos2 θ1 + C cos θ1

= (180 x 0.395 x cos2 (30O)) + (0.368 x cos (30O)

= 53.64 m

ii) Staff held normal:

S2 = 1.330 – 1.032 = 0.298 m

Distance, AC = KS2 cos θ2 + C cos θ2 + h sin θ2

= (180 x 0.298 x cos (45O)) + (0.368 x cos (45O)) + (1.181 x sin (45O))

= 39.02 m

iii) RL of C:

V = KS2 sin θ2 + C sin θ2 = (180 x 0.298 x sin (45O)) + (0.368 x sin (45O))

= 38.18 m

RL of C = RL of instrument axis + V – h cos θ

= 16.43 + 38.18 – (1.181 x cos (45O))

= 53.33 m

15. The vertical angles to vanes fixed at 1 m and 3 m above the foot of staff held vertically at

a station A were 03o 10’ and 050 24’ respectively. Find the horizontal distance and the

reduced level of A if the height of the instrument axis is 138.556 m above datum.

(AUC May/June 2012)

Solution:

S = 3 – 1 = 2 m

D = )'103(tan)'245(tan

2

)(tan)(tan 00

21

S

D = 51.02 m


V = D tan θ2 = 51.02 x tan (03O 10’)

V = 2.82 m

RL of A = RL of instrument axis + V – h

= 138.556 + 2.82 – 1

= 140.376 m

16. A tacheometer was set up at station A and the following readings were obtained on a

vertically held staff.

Inst station

Staff station

Vertical angle

Stadia hair readings (m)

Remarks

A B.M. -02O 18’ 3.225, 3.550, 3.875 RL of B.M = 425.515 m

A B +08O 36’ 1.650, 2.515, 3.380

Find the distance between A and B, R.L of B. (AUC May/June 2012)

Solution:

Assume K = 100 and C = 0


S1 = 3.875 – 3.225 = 0.65 m; θ1 = - 2O 18’; k = 100; C = 0

V1 = 0)2

)'1822(sin65.0100(sin

2

2sin1

11

oXXXCSk

V1 = 2.61 m


S2 = 3.380 – 1.650 = 1.73 m; θ2 = 8O 36’

AB = D = KS2 cos2 θ2 + C cos θ2

= (100 x 1.73 x cos2 (8O 36’)) + 0

AB = D = 169.13 m

V2 = 02

)'3682(sin73.1100sin

2

2sin2

22

oXXXCSk

V2 = 25.58 m

c) Reduced levels of stations:

RL of station B = RL of BM + V1 + Axial hair reading (h1) + V2 – h2

= 425.515 + 2.61 + 3.55 + 25.58 – 2.515

= 454.74 m


17. Calculate the tacheometric constants from the following readings taken with a

tacheometer on to a vertical staff. (AUC Nov/Dec 2012)

Horizontal distance b/n

inst. and staff (m) Staff reading (m)

66.3 0.77, 1.10, 1.43

75.3 1.68, 2.055, 2.43

Solution:

Here D1 = 66.3 m; D2 = 75.3 m; θ = 0O;

S1 = 1.43 – 0.77 = 0.66 m;

S2 = 2.43 – 1.68 = 0.75 m;

D1 = KS1 + C

66.3 = 0.66 K + C …………. (1)

D2 = KS2 + C

75.3 = 0.75 K + C …………. (2)

By solving above two equations we get

K = 100 and C = 0.3

18. A staff held vertically at a distance of 50 m and 100 m from a transit fitted with stadia

hairs, the staff intervals with the telescope normal were 0.494 m and 0.994 m

respectively. The instrument was then set up near a B.M of R.L 1500 m and the readings

on the staff held on the B.M was 1.495 m. The staff readings at the station A with staff

held vertically and the line of sight horizontal were 1.00, 1.85, and 2.70. What is the

horizontal distance between the B.M and A and R.L of A. (AUC Nov/Dec 2012)

Solution:

a) Find the value of K and C:

Here D1 = 50 m; D2 = 100 m; S1 = 0.494 m; S2 = 0.994 m; θ = 0O;

D1 = KS1 + C

50 = 0.494 K + C …………. (1)

D2 = KS2 + C

100 = 0.994 K + C …………. (2)

By solving above two equations we get

K = 100 and C = 0.6

b) Distance b/n BM and A and RL of A:

Here RL of BM = 1500 m; θ = 0O;

S = 2.70 – 1.00 = 1.7 m

D = KS cos2 θ + C cos θ

= (100 x 1.7 x 1) + (0.6 x 1)

D = 170.6 m


RL of A = RL of BM + HI – h1

= 1500 + 1.495 – 1.85

= 1499.645 m

19. During the course of tacheometric traversing from A to D, the following observations

were made with a theodolite fitted with an anallatic lens.

Line Bearing Vertical

angle Staff reading (m)

AB 33o 35’ +5o 45’ 1.050, 1.950, 2.850

BC 115o 50’ +6o 30’ 1.300, 2.165, 3.030

CD 202o 32’ -2o 55’ 1.385, 2.250, 3.115

Assuming the staff was held vertical and the multiplying constant of the instrument as

100, calculate the distance of D from A along the traverse line. Also determine the

reduced level of B, C and D if the reduced level of A is 215.5 m and height of the

instrument axis at A, B and C are respectively 1.45 m, 1.4 m and 1.55 m.

(AUC Nov/Dec 2012)

Solution:

K = 100; C = 0; RL of A = 215.5 m; HIA = 1.45 m; HIB = 1.4 m; HIC = 1.55 m

a) From the observation A to B:

S1 = 2.85 – 1.05 = 1.8 m; θ1 = 5O 45’;


= 100 x 1.8 x cos2 (5O 45’)

BC = D2 = 178.19 m

V1 = 2

)'4552(sin8.1100

2

2sin 11

oXXXSk

V1 = 17.94 m

b) From the observation B to C:

S2 = 3.030 – 1.300 = 1.73 m; θ2 = 6O 30’


= 100 x 1.73 x cos2 (6O 30’)

BC = D2 = 170.78 m


V2 = 2

)'3062(sin73.1100

2

2sin 22

oXXXSk

V2 = 19.45 m

c) From the observation C to D:

S3 = 3.115 – 1.385 = 1.73 m; θ3 = 2O 55’

CD = D3 = KS3 cos2 θ3

= 100 x 1.73 x cos2 (2O 55’)

CD = D3 = 172.55 m

V3 = 2

)'5522(sin73.1100

2

2sin 33

oXXXSk

V3 = 8.79 m


RL of station B = RL of A – (HI)A + V1 – h1

= 215.5 – 1.45 + 17.94 – 1.950

= 230.04 m

RL of station C = RL of B + (HI)B + V2 – h2

= 230.04 + 1.4 + 19.45 – 2.165

= 248.725 m

RL of station D = RL of C + (HI)C - V3 – h3

= 248.725 + 1.55 – 8.79 – 2.550

= 238.935 m

20. Explain the different between tangential and stadia tacheometry. (AUC May/June 2013)

Stadia tacheometry:

In this type of system the diaphragm of the tacheometer is provided with two stadia hairs

(upper and lower ones). Telescope of the tacheometer is directed towards the staff held at a

point whose distance from the instrument is to be found. The readings pertaining to stadia hair

are taken. The difference in these readings gives the staff intercept. The horizontal distance is

obtained by multiplying constant.

There are two kinds of stadia systems,

i) Fixed hair method

ii) Movable hair method.

Tangential tacheometry:

In this method the stadia hairs are not used. Only the single horizontal hair is used to take

the reading. The staff consists of two vanes or targets kept apart at a known distance. To

measure the staff intercepts two points are required. The angles of elevations or depressions

are measured. Their tangents are used for finding the horizontal distances and elevations. This

method is generally not adopted as two vertical angles are required to be measured for one

single observation.


21. In tacheometer survey made with an instrument whose constants are 100 and 0.5 the

staff was inclined so as to be normal to the line of sight for each reading. Two sets of

readings were taken as given below. Calculate the gradient between the staff stations P

and Q, the RL of station R is 41.800 m. (AUC May/June 2013)

Inst at

Height of instrument

axis

Staff station

Bearing

Vertical angle

Hair reading

R

1.600 P

85O

4º 30’

1.000, 1.417, 1.838

Q 135O - 4º 00’ 1.000, 1.657, 2.313

Solution:

Here K = 100; C = 0.5; RL of R = 41.800 m

a) From the observation R to P:

S1 = 1.838 – 1.000 = 0.838 m; θ1 = 4O 30’;

D1 = KS1 cos θ1 + C cos θ1 + h1 sin θ1

= (100 x 0.838 x cos (4O 30’) + (0.5 x cos (4O 30’)) + (1.417 x sin (4O 30’))

D1 = 84.15 m

V1 = ))'304(sin5.0())'304(sin838.0100(sinsin 111

oo XXXCSk

V1 = 6.61 m

b) From the observation R to Q:

S2 = 2.313 – 1.000 = 1.313 m; θ2 = - 4O 00’ (depression)

D2 = KS2 cos θ2 + C cos θ2 – h2 sin θ2

= (100 x 1.313 x cos (4O 00’) + (0.5 x cos (4O 00’)) - (1.657 x sin (4O 00’))

D2 = 131.36 m

V2 = ))'004(sin5.0())'004(sin313.1100(sinsin 222

oo XXXCSk

V2 = 9.19 m

c) Reduced levels of stations:

RL of station P = RL of station R + (HI)R + V1 - h1 cos θ

= 41.8 + 1.6 + 6.61 – (1.417 x cos (4O 30’))

= 48.59 m


RL of station Q = RL of station R + (HI)R - V2 – h2 cos θ

= 41.8 + 1.6 + 9.19 – (1.657 x cos (4O 00’))

= 50.94 m

22. A tacheometer was set up at station A and the following readings were obtained on a

vertically held staff.

Inst at

Staff station

Vertical angle

Hair reading

Remarks

A

B.M

-2 º 18’

3.225, 3.550, 3.875 R.L. of B.M. is 437.655 m

B +8 º 36’ 1.650, 2.515, 3.380

Calculate the horizontal distance from A to B and the R.L. of B, if the constants of the

instrument were 100 and 0.4.

Solution:

d) From the observation A to BM:

S1 = 3.875 – 3.225 = 0.65 m; θ1 = - 2O 18’; k = 100; C = 0.4

V1 = ))'182sin(4.0()2

)'1822(sin65.0100(sin

2

2sin1

11

oo

XX

XXCSk

V1 = 2.62 m

e) From the observation A to B:

S2 = 3.380 – 1.650 = 1.73 m; θ2 = 8O 36’

AB = D = KS2 cos2 θ2 + C cos θ2

= (100 x 1.73 x cos2 (8O 36’)) + (0.4 x cos (8O 36’))

AB = D = 169.53 m

V2 = ))'368sin(4.0(2

)'3682(sin73.1100sin

2

2sin2

22

oo

XX

XXCSk

V2 = 25.64 m

f) Reduced levels of stations:

RL of station B = RL of BM + V1 + Axial hair reading (h1) + V2 – h2

= 437.655 + 2.62 + 3.55 + 25.64 – 2.515

= 466.95 m


23. Explain how a subtense bar is used with a theodolite to determine the horizontal distance

between two points.

The subtense bar is an instrument used for measuring the horizontal distance between the

instrument station and a point on the ground.

Procedure:

The transit theodolite is set up over the instrument station.

The subtense bar is set up and leveled at P which is the position of the alidade.

The targets B and C are located at S (3m) apart. The horizontal distance D is required to

be found.

Using the alidade the line of sight of the telescope of the theodolite is made

perpendicular to the axis of the bar.

The horizontal angle BAC is measured by the method of repetition. Let the angle be θ.

AP is perpendicular to BC and bisect P. from Δ ABC,

tan 2

2

S

D or

D tan 2 2

S

D = 1

cot22

S ……………… (1)

Thus horizontal distance can be computed.


If θ is small, then

tan 1

2 2 where θ in radians.

1

,2 206265

X if θ in seconds.

(1 radian = 206265 sec)

From equation (1) D is inversely proportional to angle θ. Thus a negative error in the

measurement of angle θ will produce positive error in D. Then if an error of δθ (-ve) will

cause an error of δD (+ve).

S = D θ D D

D D

D

D D D

D

.................................(2)D

D

Similarly if δθ is +ve; δD is negative,

.................................(3)D

D

If , ,then

.................................(4)D

D

24. A theodolite has a tacheometric multiplying constant of 100 and an additive constant of

zero. The centre reading on a vertical staff held at point B was 2.292 m when sighted

from A. If the vertical angle was +25º and the horizontal distance AB 190.326 m, calculate

the other staff readings and show that the two intercept intervals are not equal. Using

these values, calculate the level of B if A is 37.950 m angle of depression and the height

of the instrument is 1.35 m.

Solution:


D = AB = 190.326 m; h = 2.292 m; k = 100; c = 0; θ = 25O

HI = 1.35 m; RL of A = 37.95 m

V = D tan θ = 190.326 x tan (25O)

= 88.75 m

D = KS cos2 θ

190.326 = 100 x S x cos2 (25O)

S = 2.317 m

RL of station B = RL of station A + HI +V – h

= 37.95 + 1.35 + 88.75 – 2.292

= 125.758 m

25. How will you determine the stadia constants?

The stadia constants i

f and (f + d) may be determined by two measurements.

i. Laboratory measurement

ii. Field measurement

i) Laboratory measurement:

The three components contributing for the constants are f, i and d. they are determined

in the laboratory as explained below.

The focal length f of the lens can be determined by using an optical bench as per the

formula

vuf

111

The stadia intercept i can be measured from the diaphragm with the help of a vernier

caliper.

The distance d is also obtained by measuring the distance between the optical

centre and the vertical axis of the instrument.

Based on these measured values the multiplying constant i

f and additive constant

(f + d) can be determined.

ii) Field measurement:

The following procedure is followed:

A tacheometer is set over a station on a fairly level ground.

Wooden pegs are driven at distances D1, D2 and D3 from the vertical axis of the

instrument at A, B and C.

The staff intercepts i.e., the stadia hair readings, are noted at each of the pegs.

Let the intercept be S1, S2 and S3.

By substituting the values of D1, D2 …….. and S1, S2 ……. in general equation,

then

D = i

f S + (f + d)

The equations for different distances are


D1 = i

f S1 + (f + d)

and D2 = i

f S2 + (f + d) and so on

By solving the equations in pairs, several values of i

f and (f + d) can be

obtained. Average values can be obtained.

26. A line was leveled tacheometrically with a tacheometer fitted with an anallatic lens, the

value of the constant being 100. The following observations were made, the staff having

been held vertically :

Inst. Station

Ht. of axis (m)

Staff at

Vertical angle

Staff readings

Remarks

A 1.38 B.M. -1 º 54’ 1.02, 1.720, 2.420 R.L=

638.55 m

A 1.38 B +2 º 36’ 1.220, 1.825, 2.430

C 1.40 B +3 º 6’ 0.785, 1.610, 2.435

Compute the elevation of A, B and C.

Solution:


S1 = 2.42 – 1.02 = 1.4 m; θ1 = - 1O 54’; k = 100; C = 0

V1 = 2

)'5412(sin4.1100

2

2sin 11

oXXXSk

V1 = 4.64 m


S2 = 2.43 – 1.22 = 1.21 m; θ2 = 2O 36’


= 100 x 1.21 x cos2 (2O 36’)


AB = D1 = 120.75 m

V2 = 2

)'3622(sin21.1100

2

2sin 22

oXXXSk

V2 = 5.48 m

c) From the observation C to B:

S3 = 2.435 – 0.785 = 1.65 m; θ3 = 3O 6’


= 100 x 1.65 x cos2 (3O 6’)

BC = D2 = 164.52 m

V3 = 2

)'632(sin65.1100

2

2sin 33

oXXXSk

V3 = 8.91 m


RL of instrument axis at A = RL of BM + Axial hair reading + V1

= 638.55 + 1.72 + 4.64

= 644.91 m

RL of station A = RL of instrument axis at A – (HI)A

= 644.91 – 1.38

= 643.53 m

RL of station B = RL of station A + (HI)A + V2 – Axial hair reading

= 643.91 + 1.38 + 5.48 – 1.825

= 648.945 m

RL of station C = RL of station A + (HI)A + V3 – Axial hair reading

= 643.91 + 1.38 + 8.91 – 1.61

= 652.59 m

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