Question

cleocd Question No. 3 A travorse survey was cerried peoduced these resalts as shown in the table out along the beiaris ofa building plot Line 17500 5* DA 268 12 a) Calcalate die error in traverte clonure b) Caldulate the adi sted total coordinates of points B,C and D with respect to A. point A cooedinates ae(50,50jusing WCB Question 4 Find the closed error of the closed loop traverse EPGHE, and F.G and HiThe coordinsgte of stafion B is 100m B, 1000m N and the bearing of line EF is 120 s0 00 coordlnates of Angle 13535 06 18 245.6 0p 18

Answer 1

3)

Calculation of latitudes and departures:

Latitude of a line = * cos ($\theta$)

Departure of a line = * sin ($\theta$)

Where, = Length of the line

$\theta$ = Whole circle bearing (WCB) of the line

Line AB:

Length () = 125.20 m (given)

WCB ($\theta$) = 80^$\degree$45

Therefore,

Latitude of line AB = 125.20 * cos (80^$\degree$45)

= + 20.125 m

Departure of line AB = 125.20 * sin (80^$\degree$45)

= + 123.572 m

Similarly for other lines latitude and departure can be calculated which are shown in the table below:

Line	Length (m)	WCB	Latitude (m)	Departure (m)
AB	125.2	80$\degree$ 45	20.125	123.572
BC	232.59	143$\degree$ 26	-186.808	138.567
CD	175	243$\degree$ 15	-78.762	-156.271
DA	268.9	336$\degree$ 32	246.66	-107.08

Sum of latitudes = 20.125 + (-186.808) + (-78.762) + 246.66

(〉 1)= +1.215

Similarly,

Sum of departures = 123.572 + 138.567 + (-156.271) + (-107.08)

D)--1.212

Therefore error of closure (e) is calculated using the following expression:

$\Rightarrow e = [(-\sum L)^{2} + (-\sum D)^{2}]^{0.5}$

$\Rightarrow$ e = [ 1.215² + 1.212²]^0.5

$\Rightarrow$ e = 1.716 m

Therefore error of closure of the traverse is 1.716 m

Adjusting the traverse:

Sum of latitudes is nothing but the error in latitude.

Therefore, Error in latitudes = +1.215 m

As the error is positive, correction for latitudes must be negative.

Similarly,

Sum of departures is nothing but the error in departures.

Therefore, error in departures = -1.212

As the error is negative, correction for departures must be positive.

And perimeter

Let us apply compass rule to balance the traverse.

According to compass rule,

Correction for latitude of any line = Error in latitude * (Length of the line / Perimeter of traverse)

Correction for departure of any line = Error in departure * (Length of the line / Perimeter of traverse)

Where,

Perimeter of traverse = Sum of lengths of all the sides

= 125.20 + 232.59 + 175 + 268.90

= 801.69 m

Now,

For line AB:

Correction for latitude = 1.215 * (125.20 / 801.69)

= 0.1897

Correction for departure = 1.212 * (125.20 / 801.69)

= 0.1893

Therefore,

Corrected latitude of line AB = Computed latitude + Correction for latitude

= 20.125 - 0.1897 [$\because$ Correction for latitude is negative]

= 19.9353

Corrected departure of line AB = Computed departure + Correction to departure

= 123.572 + 0.1893 [Correction for departure is positive]

= 123.7613

Similarly for other lines, Correction can be applied which are shown in the table below:

						Correction		Corrected
	Line	Length (m)	WCB	Latitude (m)	Departure (m)	Latitude	Departure	Latitude	Departure
	AB	125.2	80$\degree$ 45	20.125	123.572	-0.1897	0.1893	19.9353	123.7613
	BC	232.59	143$\degree$ 26	-186.808	138.567	-0.3525	0.3516	-187.1605	138.9186
	CD	175	243$\degree$ 15	-78.762	-156.271	-0.2652	0.2646	-79.0272	-156.0064
	DA	268.9	336$\degree$ 32	246.66	-107.08	-0.4075	0.4065	246.2525	-106.6735
Sum		801.69		1.215	-1.212			0	0

Calculation of coordinates:

Coordinate of A is given as 50, 50

i.e.,

North Coordinate of A = 50

East coordinate of A = 50

Therefore,

North coordinate of B = North coordinate of A + Latitude of AB

= 50 + 19.9353

= 69.9353

East coordinate of B = East coordinate of A + Departure of AB

= 50 + 123.7613

= 173.7613

$\therefore$ Coordinate of B is (69.9353 N, 173.7613 E)

Similarly for station C:

North coordinate of C = North coordinate of B + Latitude of BC

= 69.9353 + (-187.1605)

= -117.2252 [ Negative sigh indicates South coordinate]

East coordinate of C = East coordinate of B + Departure of BC

= 173.7613 + 138.9186

= 312.680

$\therefore$ Coordinate of C is (117.2252 S, 312.680 E)

Similarly for station D:

North coordinate of D = North coordinate of C + Latitude of CD

= -117.225 + (-79.0272)

= -196.2522 [ Negative sigh indicates South coordinate]

East coordinate of D = East coordinate of C + Departure of CD

= 312.680 + (-156.0064)

= 156.6736

$\therefore$ Coordinate of D is (196.2522 S, 156.6736 E)