Question

Analysis/Complexity of analysis) and WA ation and experimentation). (ivesiglU Question 1 (WA 4-CLO C3, 50%) Observations, using Theodolite, were taken in the field for an anti-clockwise polygon traverse, A, B, C, D Interior Angles 56° 18' 35 SA-740 9' 30" B-114° 6' 1" <C- 730 1' 38" <D-1070 42' 50"

Answer 1

Sol a) computation of angular error and it adjustment:

the sum of interior angle should be 360, as the close shape is quadrilateral,

Line	length	interior angle
		dergree	min	sec	in decimal	corrected
AB	18.0746	74	9	30	74.1583	71.9084
BC	12.2984	114	6	1	114.1003	111.8504
CD	18.6336	73	1	38	73.0272	70.7773
BA	12.872	107	42	50	107.7139	105.4640
Sum	61.8786				368.9997	360.0000

Sum is 368.9997,

closing error = 368.9997- 360

                     =8.9997 (degree)

This error is equally distributed in all four angle and corrected angle value is shown in figure.

Solution b )& c) . calculate the distance correction to be applied

Here we are applying Bowditch rule for distance correction value.

correction in latitude = error in latitude*length of line/ (perimeter of traverse)

in table WCB (whole circle bearing) for all four calculated using the WCB of ltine AB and interior angle B,C,D.

col1 show the latitude of AB line = length of AB*cos(WCB of line AB)

col2 show the longitude of AB line = length of AB*sin(WCB of line AB)

sum of col1 give the error in latitude which = -0.0453.

sum of col2 give the error in longitude which = -0.137.

col 3 show the correction value for latitude calculated using Bowditch rule = 0.0453*18.0746/61.8786

                                                                                                                                    = 0.0132

col 4 show the correction value for longitude calculated using Bowditch rule = 0.137*18.0746/61.8786

                                                                                                                                    = 0.0400

col 5 show the corrected latitude = col1+col3

col 6 show the corrected longitude = col 2+col 4

finally in order to calculate the Adjusted coordinate point A coordinate are assumed as (1000, 1000)

and coordinate of next points are calculated by algebrically adding the corrected latitude and longitude from respective latitude and longitude of previous point.

Line	length	interior angle					WCB				latitude	longitude	correction latitude	correction longitude	corrected latitude	corrected longitude	Adjusted coordinate
		degree	min	sec	in decimal	corrected	degree	min	sec	in decimal	col1	col2	col3	col4	col5	col6	latitude	longitude
AB	18.0746	74	9	30	74.1583	71.9084	56	18	35	56.3097	10.0335	12.8861	0.0132	0.0400	10.0468	12.9262	1000.0000	1000.0000
BC	12.2984	114	6	1	114.1003	111.8504				124.4594	-6.9475	-13.7825	0.0090	0.0272	-6.9385	-13.7553	1010.0468	1012.9262
CD	18.6336	73	1	38	73.0272	70.7773				233.6821	-11.0671	-14.0059	0.0136	0.0413	-11.0534	-13.9646	1003.1082	999.1709
BA	12.872	107	42	50	107.7139	105.4640				308.2181	7.9357	14.7652	0.0094	0.0285	7.9452	14.7937	992.0548	985.2063
Sum	61.8786				368.9997	360.0000				416.3097	-0.0453	-0.1370