Surveying TYS Exp 250

8/17/2019 Surveying TYS Exp 250

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Civil EngineeringSurveying

WORKBOOKWORKBOOKWORKBOOKWORKBOOKWORKBOOK

2016

Detailed Explanations of

Try Yourself Questions


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Fundamental Concepts of

Surveying

1

T1 : Solution

Rod reading (m) ν (m) ν2

2.322

2.346

2.352

2.306

2.312

2.300

2.306

2.326

Mean : 2.321

0.001

0.025

0.031

0.015

0.009

0.021

0.015

0.005

0.000001

0.000625

0.000961

0.000225

0.000081

0.000441

0.000225

0.000025

= 0.002584Σ 2ν

From equation,

E s =

± = ±

−

and E m

=

= ± = ±


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Linear Measurements

2

T1 : Solution

Considered length of chain, l = 20 m

Actual length of chain, l ′ = 20 – 0.1 = 19.9 m

∴ True volume =

′ ×

l

l

= 533.9 cu.m

T2 : Solution

Let the permissible error in the angular measurement be θ∴ Displacement due to angular error = l sinθ = 15 sin θAccuracy in linear measurement is 1 in 20

∴ Displacement due to linear error =

=

Combined error on ground = θ +

Combined error on plan = Scale × Combined error on ground

=

θ +

and, combined error on plan should be less than 0.025 cm.

∴

θ + = 0.025

⇒ θ = 0°∴ Angular error is not permitted.


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4 Civil Engineering • Surveying

T3 : Solution

A normal tension of 101.76 N is applied, so, there will be no pull and sag correction.

Corrections required are slope, pull, temperature and standardisation.

Slope correction = – L (1 – cos θ)= – 29.786 (1 – cos 4°30′) = – 0.09182 m

Standardisation correction = −′ = + l l

l

Temperature correction = α (T – 20) × L= 1.12 × 10–5 (10 – 20) × 29.786 = –0.003336 m

Total correction = –0.09182 + 0.00397 – 0.003336 = – 0.09113 m

Correct horizontal distance = 29.786 – 0.09113 = 29.695 m 29.70 m

T4 : Solution

(i) Correction for pull:

C p

= ( )

− =

( )

− ×= +

× ×(ii) Correction for temperature:

C t

= α (T m

– T o ) L

= 11.5 × 10–6 × (35 – 15) × 1000

= 0.23 m (+ve)

(iii) Correction for slope:

C d = ( )

−= = × −×

(iv) Correction for mean sea level:

C R

=

1 ×= = −

×Total correction = 0.05878 + 0.23 – 2 × 10–3 – 0.15625 = 0.1305 m

Corrected length of the base line

= 1000 + 0.1305 = 1000.1305 m

T5 : Solution

Least count for an extended vernier =

⇒ 10′′ =

′

∴ n = 60For an extended vernier

‘n ’ division of vernier should be equal to ‘(2n – 1)’ divisions of main scale

∴ M = 2n – 1 = 119 and N = n = 60


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Compass Surveying, Traverse

Surveying and Theodolite

3

T1 : Solution

Line ABLine ABLine ABLine ABLine AB Latitude = 150 cos 75° 42′ = 150 × 0.2470 = + 37.04 m

Departure = 150 sin 75° 42′ = 150 × 0.9690 = + 145.35 m

Line BCLine BCLine BCLine BCLine BC Latitude = 100 cos 32° 48′ = 100 × 0.8406 = + 84.06 m

Departure = 100 sin 32° 48′ = 100 × 0.5417 = + 54.17 m

Line CDLine CDLine CDLine CDLine CD Latitude = – 300 cos 28° 54′ = – 300 × 0.8755 = – 262.63 m

Departure = 300 sin 28° 54′ = 300 × 0.4833 = + 144.98 m

Lince DELince DELince DELince DELince DE Latitude = – 800 cos 5° 36′ = – 800 × 0.9952 = – 796.18 mDeparture = 800 sin 5° 36′ = 800 × 0.0976 = + 78.06 m

Total latitude of E = – 796.18 – 262.63 + 84.06+37.04 = – 937.71 m

Total departure of E = 78.06 + 144.98 + 54.17 + 145.35 = + 422.56 m

Hence, latitude of point F = – 937.71/2 = – 468.85 m

and departure of point F = 422.52/2 = + 211.26 m

Total departure of C = 145.35 + 54.17 = 199.52 m

Total latitude of C = 37.04 + 84.96 = 122.0 m

Hence, departure of CF = 211.28 – 199.52 = 11.76 m

and latitude of CF = – 468.85 – 122.0 = – 590.85 m

Hence, length of CF = +

= + = 590.97 m

Bearing of CF, tan θ = 11.76/590.97 = 0.0199

Hence, θ = 1° 8′24′′

As the latitude of CF is negative and the departure is positive, the line CF lies in the second quadrant.

Hence, the bearing of line CF is 180° – 1° 8′24′′ = 178° 51′36′′ = 178.86°.


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T2 : Solution

In traverse ABCDE (shown in figure), the point A, D and E are on same line.

Line ABLine ABLine ABLine ABLine AB

Length of the line AB = 110 mLatitude of line AB = +110 cos 83° 12′ = +13.02 m

Departure of the line AB = +110 sin 83° 12′ = +109.23 mLine BCLine BCLine BCLine BCLine BC

Length of the line BC = 165 m

Latitude of line AB = +165 cos 30° 42′ = +141.88 mDeparture of the line BC = +165 sin 30° 42′ = +84.24 m

° ′

° ′

° ′

° ′

α

β

γ

Now, for the traverse ABC :

Σ latitude = 0

latitude of CA = 0 – 13.03 – 141.88 = –154.91 mand Σ departure = 0

departure of CA = 0 – 109.23 – 84.24 = – 193.47 m

Since the latitude and departure are both negative, the line CA lies in the SW quandrant. Its bearing is

given by

tan θ =

=

or tan θ = S51° 18′ W

Length of CA = +

= + =Since A, D and E are on the same line so,

Bearing of AD = bearing of DE = 16° 18′From triangle ACD

α = (360° – 346° 6′) + 16° 18′= 13° 54′ + 16° 18′ = 30° 12′

β = 346° 6′ – (51° 18′ + 180°) = 114° 48′γ = 51° 18′ – 16° 18′ = 35°


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7Workbook

From the sine rule,

=

γ α

CD =

γ α

=

°

° ′

= 282.62 m


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Levelling and Contouring

4

T1 : Solution

True difference in level of A and B,

H = ( ) ( )

− + −′ ′

where, h b

= reading on staff at B when instrument at A

h a = reading on staff at A when instrument at A

′ = reading on staff at B when instrument at B

′ = reading on staff at A when instrument at B

⇒ H = 1.19 mError due to (collimation + curvature + refraction) = h b – h a – H = 0.13 m

Error due to (curvature + refraction) = 0.0673 d 2

where, d = distance between station in km

∴ Error due to (curvature + refraction) = 0.0673 (1.2)2 = 0.097 m∴ Error due to collimation = 0.13 – 0.097 = 0.033 m

∴ Error due to collimation per metre =

−= ×

T2 : Solution

There is no use of Intermediate sight

Fall in elevation = ∑ Foresight – ∑Backsight= 0.388 m

R.L. of first station – Fall in elevation = R.L. of last station

∴ R.L. of First station = 124.238 m


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9Workbook

T3 : Solution

Distance of Ship A from light house =

Distance of Ship B from Ship A =

Distance of Ship B from light house = × Let the observer at Ship B can see upto ‘h ’ m height of light house

Distance of Ship B from light house = +

⇒ × = +

⇒ 2 × 3.855 × 3 =

⇒ h = 36 m∴ The height of light house visible to observer at Ship B = (49 – 36) m

= 13 m

T4 : Solution

Starting from the point 7, the R.L. of point 6 is obtained.

H.I. at point 6 = 194.830 + 2.990 = 197.820 m

R.I. of point 6 = 197.820 – 4.330 = 193.490 m

H.I. of point 3 = B.M. + 5.390

= 191.620 + 5.390 = 197.010 m

R.L. of point 3 = 197.010 – 3.910 = 193.100 m

R.L. of point 4 = 197.010 – 4.730 = 192.280 m

I.S. at point 5 = 197.010 – 203.300 = – 6.290 m

F.S. at point 6 = 197.010 – 193.490 = 3.520 m

H.I. at point 1 = 193.100 + 6.520 = 199.620 m

I.S. at point 2 = 199.620 – 192.00 = 7.620 m

BS IS FS HI RL Remarks

4.390

—

3.910

—

—

—

—

4.330

—

12.630Σ

—

7.620

—

5.390

4.730

–6.290

—

—

—

—

—

6.520

—

—

—

—

3.520

2.990

13.030Σ

199.620

—

197.010

—

—

—

—

197.820

—

195.230

192.000

193.100

191.620

192.280

203.300

—

193.490

194.830

Point 1

Point 2

Point 3

BM

Point 4

Point 5

Staff inverted

Point 6

Point 7

Aritmetic Check

∑ B.S. – ∑ F.S. = Last R.L. – First R.L.12.630 – 13.030 = 194.830 – 195.230 = – 0.40 (O.K)


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T5 : Solution

As the ends of the bubble lie on the opposite ends of the zero of graduations

∴ Movement of centre, n = ( ) ( )

− + −l l

where, l 1 = left end of the bubble when the position of bubble in extreme left

r 1 = right end of the bubble when the position of bubble in extreme left

l 2

= left end of the bubble when the position of bubble in extreme right

r 2 = right end of the bubble when the position of bubble in extreme right

⇒ n =

− + −=

Radius of curvature, R =

where, n = movement of centre = 9

L = distance between staff and instrument = 150 m

d = value per division = 2 mm

s = difference between staff readings

= 1.452 – 1.37 = 0.082 m

∴ R =

× ×=

T6 : Solution

Sensitivity, φ =

××

;

where s, n and L are combined curvature and refraction correction, no. of division of the level tube and

distance of point under observation respectively

⇒ 30′′ =

××

⇒ s =

.....(i)

Correction due to combined curvature and refraction

C = – 0.0673 L2 , where L is in km .....(ii)

From equation (i) and (ii),

∴

= 0.0673 × L2 × 10–6

⇒ L =

−× × = 2161.13 m


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Plane Table Surveying,

Calculation of Area and Volume

5

T1 : Solution

By trapezoidal formula.

V =

+ + + +

=

+ + + + = 532621 m3

By primoidal formula

= ( )

+ + + +

= ( )

+ + + × +

= 458880.67 m3

So difference in capacity is

= 532621 – 458880.67

= 73,740.33 m3


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Tacheometric, Curve and

Hydrographic Surveying

6

T1 : Solution

Additive constant, C = f + d

= 25 + 15 = 40 cm = 0.4 m

Staff intercept, s = 1.790 – 1.275

= 0.515 m

Distance of object from telescope,

D = + +i

where, i = stadia interval

⇒ 50 =

× +i

∴ i = 2.596 mm 2.60 mm

T2 : Solution

Instrument P at station A and staff held vertical at B:Instrument P at station A and staff held vertical at B:Instrument P at station A and staff held vertical at B:Instrument P at station A and staff held vertical at B:Instrument P at station A and staff held vertical at B:

s = 1.795 – 1.090 = 0.705 m,

θ = 5° 44′,2θ = 11° 28′

AB = Ks cos2 θ + C cos θ= 100 × 0.705 cos2 5° 44′ + 0.3 cos 5° 44′ = 70.095 m

V =

θ+ θ

=

×° ′ + ° ′ =


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13Workbook

R.L of B = R.L. of A + H.I + V – h

= 100.00 + 1.400 + 7.038 – 1.440 = 106.998 m

Instrument Q at station A and staff held normal at B:Instrument Q at station A and staff held normal at B:Instrument Q at station A and staff held normal at B:Instrument Q at station A and staff held normal at B:Instrument Q at station A and staff held normal at B:

AB = (Ks + C ) cosθ + h sinθ

= (95s + 0.45) cos 5°44′ + h sin 5°44′

= (95s + 0.45) × 0.995 + h × 0.0999

or 70.095 = (95 s + 0.45) × 0.995 + 0.0999h

or 946.1962s + h = 697.167

or h = 697.167 – 946.1962s ...(i)

V = (Ks + C ) sinθ

= (95 s + 0.45) sin 5°44′

= (95 s + 0.45) × 0.0999

R.L of B = R.L of A + H.I + V – h cos θ

= 100.00 + 1.450 + (95 s + 0.45) × 0.0999 – h (0.995)

or 106.998 = 100.00 + 1.45 + (95 s + 0.45) × 0.0999 – h × 0.0995

or 9.5382 s – h = 5.5307

h = 9.5382s – 5.5307 ...(ii)

From equation (i) and (ii)

697.167 – 946.1962 s = 9.5382 s – 5.5307

955.7344 s = 702.698

or s =

=

∴ h = 9.5382 × 0.735244 – 5.631 = 1.3819 mUpper stadia wire reading

=

− =

T3 : Solution

Let the point of intersection of the two grades be C

Now, chainage of point C = 435 m

RL of point C = 251.48 m

Let the fixed point on the curve be P

Now, chainage of point P = 460 m

RL of point P = 260 m

It is obvious from the given data that point P lies towards right

of point C .

Let the length of the curve be L and the horizontal distance

between P and C be x . Also R and S be two points on tangents

T 1C and T

2C respectively and V be a point just below S on

tangent T 1C .


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From the figure above we get

Distance x = chainage at P – chainage at C = 460 – 435 =25 m

∴ RL of point V = RL of C –

× 25 = 251.48 – 0.75 = 250.73 m

also, RL of point R = RL of Point S = RL of C +

× 25 = 251.48 + 1.00 = 252.48 m

Now PS = RL of P – RL of S =260 – 252.48 = 7.52 m

and PV = RL of P – RL of V =260 – 250.73 = 9.27 m

Since the vertical curve to be provided is parabolic, therefore its equation will be of the type

y = k x 2

∴ PS = k

x and PV = k

+ x

⇒

=

+

x

x

⇒

=

+

⇒ 22.5 + 0.45 L = 0.5 L – 25 ⇒ L = 950 mHence the length of the curve will be 950 m.

T4 : Solution

Observation from A to P:Observation from A to P:Observation from A to P:Observation from A to P:Observation from A to P:

s = 2.880 – 1.230 = 1.65 m

D = ks cos2θ = 100 × 1.65 cos2 2° 24′ = 164.7 m

V =

θ ′= × × ° =

R.L. of P = 77.750 + 1.420 + 6.903 – 2.055 = 84.018 m

T5 : Solution

Horizontal distance, D = Ks cos2θ + C cos θ

=

θ + θ

i

= i

⇒ δD =

− θ δ +ii

⇒ δD = − θ δii

=

− ° × = –0.97s

= = i

T6 : Solution

In triangle BCD , cosθ =

+ − + −=

× × × ×


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15Workbook

⇒ cosθ = 0.4609

In triangle ABC , cos θ =

+ −× ×

(Cosine rule)

⇒ 0.4609 =

−

∴ AB = 89.44 m

T7 : SolutionC

B

A

River

2 0 0 m

150m P 215° 30′

True North

305° 30′

tan ∠PAB =

⇒ ∠PAB = 36.87°∠APC = 305° 30′– 215° 30′ = 90°

∴ ∠ACP = 180°– ∠PAB – ∠APC

= 53.13° = ∠BCP

∴ BC =

=∠

T8 : Solution

Let P and Q be the instrument stations and F be the top of tower

In ∆APQ ∠ APQ = 60° and ∠ AQP = 50°∠ PAQ = 180° – (50 + 60) = 70°

Applying sine rule

°=

=

° °

PA =

°× = °

QA =

°× = °

h 1 = PA tan α = 40.76 tan 30° = 23.53 m

h 2

= QA tan β = 46.08 tan 29° = 25.54 mhence, reduced level of tower top = R.L of line of collimation + h 1 (or h 2)

R.L. of F from observation at P = 22.5 + 23.533 = 46.033 m

R.L. of F from observation at Q = 20.5 + 25.54 = 46.043 m

hence elevation of F =

+=

Pond A

θ

C

D

B

F

A

29°

50°30°

60°

P

h2

h1

ChimneyQ


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Tides, Triangulation,

Field Astronomy

7

T1 : Solution

P

L o n g

. 5 0 ° 1 2

′

Long. 54°0 E′ A

B

Lat. 15°

Lat. 12°6′

Equator

In the spherical triangle ABP ,

b = 90° – lattitude of A = 90° – 15° 0′ = 75°a = 90° – latitude of B = 90° – 12°6′ = 77° 54′P = difference of longitude = 54° 0′ – 50°12′ = 3° 48′

The shortest distance between two points is the distance along the great circle passing through the two

points.

cos P =

−

⇒ cos p = cos P sina sinb + cosa cosb = cos3° 48 sin 77° 54′ sin 75° + cos 77° 54′ cos 75°= 0.94236 + 0.05425 = 0.99661

∴ p = 4° 40′ = 4°·7

Distance AB, arc ≈ radius × central angle =

× ° × π=

°


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17Workbook

T2 : Solution

P

a

B

C

Parallelof Lat.45°

A

(Lat. 45°)

Great circle

Equator

b

90°

AB is straight line portion of a great circle; since its length is 300 nautical miles, it subtends 300 minutes

(= 5°) at the centre of the earth. AP is the meridian through A. Since AB is due east of the meridian, ∠PAB = 90°. Similarly, BP is the meridian through B , and meets the parallel to latitude through A (45° N) in C . PAB

is, therefore, an astronomical triangle in which sidePA = b = co-latitude of A = 90° – 45°; side AB = p = 5° ∠A = 90°

The side PB = a can be calculated by Napier’s rule. Thus, sine of middle part = product of cosines of

opposite parts.

∴ sin (90° – a ) = cos b cos p ⇒ cos a = cos 45° cos 5°

⇒ log cos 45° =

⇒ log cos 5° =

⇒ log cos a =

∴ a = PB = 45° 13′·108BC = PB – PC

= 45° 13′·108 – 45° = 13′·108Hence, distance BC = 13.108 nautical miles = 13.108 × 1.852 = 24.275 km.

The angle at B can be found by the application of the sine formula,

i.e.

=

⇒

°=

°° ′

log sin 45° =

log sin 45° 13′·108 = (subtract)

log sin B = ′ ′′

T3 : Solution

Let M be the star having A and B as the upper and lower transits. Since the upper culmination is at the

zenith, Z and A coincide.


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Hence, zenith distance of star = zero and Polar distance of the star = AP = ZP = co-latitude of place

∴ 90° – δ = 90° – θ or θ = δ = 56° 10′At the lowest transit of the star at B , its zenith distance = ZB = ZP + PB

= (90° + θ) + (90° – δ)

= 180° – θ – δ = 180° – 2δ= 180° – 112° 20′ = 67°40′

∴ Altitude of the star at lower transit = 90° – 67° 40 ′ = 22° 20′

T4 : Solution

Since the azimuth of the star is 140° E, it is in eastern hemi-sphere.

In the astronomical triangle, ZPM , we have

ZM = 90° – α = 90° – 21°30′ = 68° 30′;ZP = 90° – θ = 90° – 48° = 42°;A = 140°

Knowing the two sides and the included angle, the third side can becalculated by the cosine rule.

Thus cos PM = cos ZM cos ZP + sin ZM sin ZP

cos A

= cos 68° 30′ cos 42° + sin 68° 30′ sin 42′ cos 140°= 0.27236 – 0.47691 = – 0.20455

∴ cos (180° – PM ) = 0.20455⇒ 180° – PM = 78°12′⇒ PM = 101°48′∴ Declination of the star = 90° – 101° 48′ = –11° 48′ = –11° 48′ SAgain, knowing all the three sides, the angle H 1 can be calculated from the cosine formula, (Eq. 13.2). Thus

cos H 1 =

−

=

° − ° °′ ′° ° ′

=

+=

∴ H 1

= 37°40′But H 1 is the angle measured in the eastward direction.

∴Hour angle of the star = 360° – H

1 = 360° – 37° 40

′ = 322°20

′

A

Z

H 1 P

N

Horizon

Equator

E

M S


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Photogrammetric Surveying

8

T1 : Solution

At h = 80 m, S 80

=

=

−As a representative fraction, the scale is

R 90 =

= =

×−

Similarly, at h = 300 m,

S 300

=

= =

−

1 cm = 60 m

As a representative fraction, the scale is

R 300 =

=

−

T2 : Solution

−=

Here, h ab =

+ =

∴

−=


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∴ H approx.– 400 =

×

or H approx.

= 400 + 2132.2 = 2532.2 m

The actual ground length is 545 m. The second approximate height is calculated as follows:

−

−=

∴

−−

=

Using this value of H to calculate the co-ordinates, we get

X a

=

−× = +

Y a =

−

× = +

X b =

−× − = −

Y b =

−× = +

∴ L = + + − =

This agrees with the measured length. Hence height of lens = 2500 m.

T3 : Solution

(i)(i)(i)(i)(i) Flying heightFlying heightFlying heightFlying heightFlying height

We have,

=

=

H = 12000 × 0.3 = 3600 m above ground

∴ Height above datum = 3600 + 400 = 4000 m(ii)(ii)(ii)(ii)(ii) Theoretical ground spacing of flight linesTheoretical ground spacing of flight linesTheoretical ground spacing of flight linesTheoretical ground spacing of flight linesTheoretical ground spacing of flight lines

The ground width covered by each photograph, with 30% side lap is given by

W = (1 – P w ) sw

where, w = width of photograph = 20 cm;

s =

= = =

i.e. 1 cm = 120 m;

P w = 0.30

W = (1 – 0.3) 120 × 20 = 1680 m

(iii)(iii)(iii)(iii)(iii) Number of flight lines requiredNumber of flight lines requiredNumber of flight lines requiredNumber of flight lines requiredNumber of flight lines required

The number of flight lines is given by Eq. 14.31 (b), i.e.

N 2 =

+

−


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21Workbook

=

+ = + = + ≈

(iv)(iv)(iv)(iv)(iv) Actual spacing of flight lines: Since the number of flight lines is to be an integral numberActual spacing of flight lines: Since the number of flight lines is to be an integral numberActual spacing of flight lines: Since the number of flight lines is to be an integral numberActual spacing of flight lines: Since the number of flight lines is to be an integral numberActual spacing of flight lines: Since the number of flight lines is to be an integral number, the, the, the, the, the

actual flight lines = 16 and the number of flight strips or spacings = 15. Hence, the actualactual flight lines = 16 and the number of flight strips or spacings = 15. Hence, the actualactual flight lines = 16 and the number of flight strips or spacings = 15. Hence, the actualactual flight lines = 16 and the number of flight strips or spacings = 15. Hence, the actualactual flight lines = 16 and the number of flight strips or spacings = 15. Hence, the actual

spacing is given byspacing is given byspacing is given byspacing is given byspacing is given by

W =

=

(v)(v)(v)(v)(v) Spacing flight lines on flight mapSpacing flight lines on flight mapSpacing flight lines on flight mapSpacing flight lines on flight mapSpacing flight lines on flight map

Flight map is on a scale of 1:60000 or 1 cm = 600 m. Hence the distance on the flight map corresponding

to a ground distance =

=

(vi)(vi)(vi)(vi)(vi) Ground distance between exposuresGround distance between exposuresGround distance between exposuresGround distance between exposuresGround distance between exposures

The ground length covered by each photograph in the direction of flight with an overlap of 60% is

given by ( ) − = − × × =l l

(vii)(vii)(vii)(vii)(vii) Exposure intervalExposure intervalExposure intervalExposure intervalExposure interval

The time interval between exposures is usually the integral number of seconds.

V =

××

(viii)(viii)(viii)(viii)(viii) Adjusted ground distance between exposuresAdjusted ground distance between exposuresAdjusted ground distance between exposuresAdjusted ground distance between exposuresAdjusted ground distance between exposures

Keeping the exposure interval as an integral number of seconds the adjusted ground distance covered

by each photograph is given by

L = V × T = 55.56 (m/sec) × 17.0 (sec) = 945 m

(ix)(ix)(ix)(ix)(ix) Number of photographs per flight lineNumber of photographs per flight lineNumber of photographs per flight lineNumber of photographs per flight lineNumber of photographs per flight line

The number of photographs per flights line given by

N 1 = ( )

+

− l l

+ =

+ = 31.6 + 1 ≈ 33

(x)(x)(x)(x)(x) TTTTTotal number of photograph rotal number of photograph rotal number of photograph rotal number of photograph rotal number of photograph requirequirequirequirequired ised ised ised ised is

N = N 1 × N 2 = 33 × 16 = 528

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Surveying TYS Exp 250