Surveying Solutions To Simple Problems

CIVIL 2

1(a). ORIENTATION The bearing of a line is it’s orientation with respect to the north grid. This may be found for an initial line by starting at a station with known coordinates and orientating the Total Station with respect to another point with known National Grid coordinate. Before starting any plot, the extent of the survey should be taken into account such that the plotted survey will fall centrally on to the sheet. This implies the purpose of orientation is to ensure that the stations are plotted to the best fit on the plotting sheet. In case of an arbitrary north, in order to ensure a good fit a bearing of 900 or 2700 is assigned to the longest side. This line is then positioned parallel to the longest side of the sheet so that the survey will fit the paper properly. In conclusion we use two datum rays when opening a traverse such that the plot will fit centrally on the sheet. e.g. a datum referring to 900. This is done to save paper and for convenience.

(b) (i) TRIANGULATION: Here a network of of single or overlapping triangles is established from basic shapes of a simple triangle, braced quadrilateral, or centre point polygon. C D C E D C F B A B A B Triangle braced quadrilateral Centre point polygon The vertices of the triangles represent control points normally referred to as triangulation stations. All the angles of the triangles are measured and at least one baseline measured. If the coordinates of the A and B are known, the sine rule can be applied to baseline AB and the measured angles to get co-ordinates of the triangulation stations. i.e. sin C/AB=sin A/BC= sin B/AC

(i)RESECTION This is a method of locating a point by taking angle observation from at least 3 known stations in a network. It is known as the 3-point resection B a C γ Φ P α β

C

b

δ A PROCEDURE Setup at P and orient instrument at C. Turn it to B and measure the angle. Repeat procedure Generally the procedure is carried out in the field by observing the angles subtended at the unknown point by at least 3 unknown stations. However it should be noted that if point A,B,C and P lie in the circumference of the same circle, the procedure is indeterminate. The condition is present when α +γ+ β=180o One method of solving this involves the following function α + γ + β + Φ+ δ=180o or γ = [360- (α +γ+ β)] – Φ= R- Φ where R can be deduced. In triangles ABP and APC DAP = c.sin γ/sin α =b.sin Φ/sin β It is the deduced that sin γ/sin Φ =bsin α /csin β = constant which is the evaluated. Substituting γ =R- Φ giving a further expression for the constant. =sin(R- Φ)/sin Φ =(sinRcos Φ-cosRsin Φ)/sin Φ = sinRcot Φ-cosR Therefore cot Φ=(constant+cosR)/sinR

(c) A=(45530mE,12560.2mE) ADJUSTMENTS LINE

DISTANCE

∆E

∆N

δE

ADJUSTED

δN

∆E

COORDINATES

∆N

E

N

STTN

AB

122

55.4

108.7

0.046

-0.046

55.45

108.654

45585.4

12668.9

B

BC

155.95

73.2

-137.7

0.059

-0.059

73.26

-137.76

45658.7

12531.1

C

CD

64.95

-20.1

-61.8

0.025

-0.025

-20.08

-61.825

45638.6

12469.3

D

DE

60

-56.4

-20.5

0.023

-0.023

-56.38

-20.523

45582.2

12448.8

E

EF

123.16

-52.3

111.5

0.047

-0.047

-52.25

111.453

12560.2

A

Σ

526.06

-0.2

0.2

0.2

-0.2

0

0

Errors computed using Bowditch’s method

45530.O

δE or δN = - eE (or- eN) X Length of traverse leg/Total length of traverse.

WCB

2702’24’’

151059’46’’

197059’35’’

249059’52’’

334052’57’’

eE =-0.2, eN =0.2 e=√(0.22+(-0.2)2)=0.28 Fractional misclosure= ΣD/e =526.06/0.28.The fractional misclosure is 1 in 1800m. WHOLE CIRCLE BEARINGS B

151059’46’’

C A

197059’35’’

2702’24’’

D E

249059’52’’

334052’57’’

2.

BOOKING SHEET BS

IS

FS

HOC

1.423

124.245

0.772

SPOT HEIGHT

REMARKS

122.822

9.789

POINT 1

2.410

121.835

8.802

POINT 2

1.004

123.241

10.208

POINT 3

2.164

122.081

9.048

POINT 4

1.100

123.145

10.112

POINT 5

1.452

122.793

9.760

POINT 6

1.662

122.583

9.550

POINT 7

2.990

121.255

8.222

POINT 8

1.400

122.845

9.812

POINT 9

1.440

122.805

9.772

POINT 10

3.212

121.033

8.000

POINT 11

1.62

122.625

9.592

POINT 12

1.401

122.844

9.811

POINT 13

1.789

122.456

9.423

POINT 14

1.406

122.839

9.806

POINT 15

1.554

122.691

9.658

POINT 16

1.602

122.643

9.610

POINT 17

123.594

10.561

POINT 18

0.651 Σ=1.423

RL

Σ=0.651

Lowest reduced level =121.033m RL,Formation level = 121.033-8=113.033m

REDUCED LEVELS

SKETCH

FORMATION LEVEL

Depth above the formation level = Reduced levels- Reduced level of Formation level. Total Volume = ∆/4(Σsingle depths+2Σdouble depths+ 3Σtriple depths+ 4Σquadriple depths)+δv ∆= 20X20=400m2 Σsingle depths = 9.789+10.112+9.658+10.561+9.806 = 49.926m Σdouble depths=8.802+10.208+9.048+9.76+9.772+9.423+8.0+9.61 =74.623 Σtriple depths = 9.811 Σquadriple depths=9.55+8.222+9.812+9.592 =37.176 Therefore total volume = 400/4[49.926+2X74.623+3X9.811+4X37.176] = 37730.9m 3