CIVL 1112 Surveying - Traverse Calculations 1/13Surveying - Traverse Surveying - Traverse Introduction Almost all surveying requires some calculations to reduce measurements into a more useful form for determining distance, earthwork volumes, land areas, etc. A traverse is developed by measuring the distance and angles between points that found the boundary of a site We will learn several different techniques to compute the area inside a traverse Surveying - Traverse Distance - Traverse Methods of Computing Area A simple method that is useful for rough area estimates is a graphical method In this method, the B traverse is plotted to scale on graph paper, and the number of squares inside A the traverse are counted C D Distance - Traverse Distance - Traverse Methods of Computing Area Methods of Computing Area B B 1 1 a b Area ABC ac sin a b Area ABD ad sin 2 2 A A c 1 C d C Area BCD bc sin c 2 D Area ABCD Area ABD Area BCD Traverse Balancing Angles Balancing Angles A A surveying heuristic is that the total angle should not Error of closure vary from the correct value by more than the square root of the number of angles measured times the precision of the instrument B For example an eight-sided traverse using a 1’ transit. the surveyor may correct one N N angle by 1’ B Latitude AB If an error of 2’ is made. the Bearing A Departure AB C surveyor may correct each angle by 3’/12 or 15” Latitude CD D S S .Traverse Surveying . The closure of a traverse is checked by computing the mistakes in measuring have been made latitudes and departures of each of it sides If an error of 1’ is made.Traverse Surveying . D the maximum error is: 1' 8 2.2)(180°) e Area CDE cd sin d 2 where n is the number of sides of the traverse E To compute Area BCD more data is required Surveying . the surveyor may correct two Bearing Departure CD angles by 1’ each W E W E If an error of 3’ is made in a 12 sided traverse.CIVL 1112 Surveying .Traverse Balancing Angles Latitudes and Departures If the angles do not close by a reasonable amount. it is 1 necessary to have a closed traverse a c Area ABE ae sin 2 The interior angles of a closed traverse should total: A D 1 (n .83 ' 3' Angle containing mistake C Surveying .Traverse Surveying .Traverse Methods of Computing Area Balancing Angles B b C Before the areas of a piece of land can be computed.Traverse Calculations 2/13 Distance . B A W E E S 29° 38’ E 142.53 ft.500 for suburban land.63 ft.Traverse Latitudes and Departures Error of Closure The latitude of a line is its projection on the north–south Consider the following statement: meridian “If start at one corner of a closed traverse and walk its lines until you return to your starting point.40 ft.18’ 189.Traverse Calculations 3/13 Surveying .53 ft.Example Latitudes and Departures . you will have walked as N B The departure of a line is far north as you walked south and as far east as you have its projection on the east– walked west” Latitude AB west line W E A northeasterly bearing has: Therefore latitudes = 0 and departures = 0 Bearing A Departure AB + latitude and + departure S Surveying .Traverse Latitudes and Departures .)cos(615 ') 188.CIVL 1112 Surveying .Traverse Error of Closure Error of Closure When latitudes are added together.Traverse Surveying . D N 81° 18’ W B C S . 175.Traverse Surveying .39’ S 6° 15’ W Latitude AB N 12° 24’ W 175.58’ 189. the figure will not close because of EL and ED The error resulting from adding departures together is Error of closure B ED called the error in departures (ED) EL ED 2 2 EL Eclosure A Eclosure C Precision perimeter Typical precision: 1/5. the resulting error is If the measured bearings and distances are plotted on a called the error in latitudes (EL) sheet of paper. 1/7.)sin(615 ') 20. and 1/10.53’ W (189.Traverse Surveying .000 for rural land.Example A N N 42° 59’ E S 6° 15’ W Departure AB 234.18’ S (189.53 ft.000 for urban land D Surveying . Traverse Surveying .933 W B E 939. 1 Precision N 64° 09’ W perimeter 939.46 ft.18 ft.Traverse Surveying .18 ft.182 ft.7 ft.)cos(2938 ') 152.576 E (175.CIVL 1112 Surveying .607 159.Traverse Latitudes and Departures .46 -0.0 ft.617 N 29° 16’ E 651.53 -188.576 B EA N 42 59 E 234.163 S 38° 43’ W EL ED 0.163 0.079 -0.079 -0.) Latitude Departure A degree m inutes S 77° 10’ E AB S 6 15 W 189.) Latitude Departure degree m inutes A popular method for balancing errors is called the AB S 77 10 E 651.Example Side Bearing Length (ft.079 0.504 DE N 12 24 W 142.268 86.5 ft.Traverse Latitudes and Departures .Traverse Calculations 4/13 Surveying . Latitude BC S 29° 38’ E S (175. 826.0 DE N 29 16 E 660.)sin(2938 ') 86. 939.916 -195.78 29.068 -30.18 -152. navigator and mathematician.176 C Surveying .Example Latitudes and Departures .Traverse Surveying . 5. CD N 81 18 W 197.78 29.7 compass or the Bowditch rule CD N 64 9 W 491.Traverse Group Example Problem 1 Balancing Latitudes and Departures Balancing the latitudes and departures of a traverse attempts to obtain more probable values for the locations of the corners of the traverse Side Bearing Length (ft.46 -0.607 159. .634 Departure BC BC S 29 38 E 175. which was posed in the American journal The Analyst in 1807.18 -152.634 BC S 29 38 E 175.068 -30.182 ft.268 86.2 BC S 38 43 W 826. as a proposed solution to the problem of compass traverse adjustment. surveyor.617 CD N 81 18 W 197.5 The “Bowditch rule” as devised by Nathaniel Bowditch. C S Surveying . 2 2 2 2 Eclosure D 491.58 171.18 ft.27 ft.62 ft.916 -195.403 -20.39 139.504 DE N 12 24 W 142.53 -188. Eclosure 0.403 -20.2 ft. EA N 42 59 E 234.) Latitude Departure N degree m inutes AB S 6 15 W 189.58 171.39 139.163 175.Example Group Example Problem 1 Side Bearing Length (ft.933 660. Traverse Balancing Latitudes and Departures Balancing Latitudes and Departures N N Latitude AB Departure AB S (189.53 ft.63 ft.58 171.46 -0.079 ft.)cos(6 15 ') 188.576 EA N 42 59 E 234.)sin(615 ') 20.53 AB S 6 15 W 189.Traverse Balancing Latitudes and Departures Balancing Latitudes and Departures The compass method assumes: A N 42° 59’ E S 6° 15’ W 1) angles and distances have same error 2) errors are accidental 234.53 ft.Traverse Surveying .016 ft.39 DE N 12 24 W 142.Example Recall the results of our example problem Recall the results of our example problem Side Bearing Length (ft) Latitude Departure Side Bearing Length (ft) Latitude Departure degree m inutes degree m inutes AB S 6 15 W 189.39 ft.933 939.53 ft.079 -0. W (189.53 ft.78 CD N 81 18 W 197.58 EA N 42 59 E 234.78 29. W A E W A E Correction in Lat AB LAB Correction in Dep AB LAB S 6° 15’ W EL perimeter S 6° 15’ W ED perimeter 189.163 ft.18 BC S 29 38 E 175.607 159. Correction in Dep AB 0. 0.18 ft.18 ft.504 DE N 12 24 W 142.58 ft.163 Surveying .Traverse Calculations 5/13 Surveying . Correction in Lat AB 0.403 -20.46 ft. .53 ft.53 ft.033 ft. 189.CIVL 1112 Surveying .617 CD N 81 18 W 197.Example Latitudes and Departures .916 -195.Traverse Latitudes and Departures . 189. total error in latitude (departure) as the length of the N 12° 24’ W 175. EL LAB 189.634 BC S 29 38 E 175.Traverse Surveying . ED LAB Correction in Lat AB Correction in Dep AB perimeter perimeter B B S S 0. line is the perimeter of the traverse” 175.39 139.18 -152.40 ft.068 -30. 189.Traverse Surveying . D N 81° 18’ W C Surveying .53 -188.53 ft.268 86. B The rule states: E S 29° 38’ E “The error in latitude (departure) of a line is to the 142. 939.46 ft. 939. 933 234.974 DE N 12 24 W 142.000 0.39 139.268 86.18 -152.634 0.617 0.253 86.18 ft.080 -30.634 0.Traverse Balancing Latitudes and Departures Balancing Latitudes and Departures Combining the latitude and departure calculations with Corrections Balanced Length (ft.017 0.033 -188.403 -20.18 -152.78 29. E (175.012 0.163 Corrections computed on previous slides Corrected latitudes and departures Surveying .068 -30.634 0.18 ft.403 -20.) Latitude Departure Latitude Departure Latitude Departure 189.470 AB S 6 15 W 189.000 No error in corrected latitudes and departures .068 -30.388 -20. 939.916 -195. 175.933 939.53 -188.617 0.46 ft.78 29.016 0.576 0.607 159.Traverse Balancing Latitudes and Departures Balancing Latitudes and Departures Corrections Balanced Corrections Balanced Length (ft.403 -20. 175.030 -152.163 ft.163 939.601 Side Bearing Length (ft.030 ft.53 -188.Traverse Surveying .015 0.388 -20.648 197.18 ft. S 29° 38’ E EL LBC S 29° 38’ E ED LBC Correction in LatBC Correction in DepBC perimeter perimeter C C S S 0.58 171.53 -188.016 0.39 139.18 ft.916 -195.27 ft.253 86.034 29.53 -188.933 0.CIVL 1112 Surveying . Correction in LatBC 0.18 -152.) Latitude Departure Latitude Departure Latitude Departure Length (ft.015 ft. Surveying .033 -188.)sin(2938 ') 86. 0.Traverse Balancing Latitudes and Departures Balancing Latitudes and Departures N N Latitude BC Departure BC S (175.933 -195.080 -30.46 ft.470 234.015 0.025 139.034 29.634 0.78 29. Correction in DepBC 0.58 171.62 ft.46 -0.017 0.627 159.068 -30.041 171.58 171.504 142.403 -20.39 139.Traverse Calculations 6/13 Surveying .607 159.268 86.648 degree m inutes 197.079 -0.504 0.388 -20.015 0.020 0.Traverse Surveying .020 0.916 -195.39 139.Traverse Surveying .079 ft.648 CD N 81 18 W 197.933 -195.015 0.033 189.000 0.607 159.041 171.000 EA N 42 59 E 234.) Latitude Departure Latitude Departure Latitude Departure corrections gives: Corrections Balanced 189.916 -195.18 ft.012 0.607 159.974 939.78 29.617 0. 939.268 86.601 142.163 0.268 86.46 -0.46 -0.016 0.551 BC S 29 38 E 175.504 0.601 175.18 -152. 175.030 -152.163 0.576 234.576 142. W B E W B E Correction in LatBC LBC Correction in DepBC LBC EL perimeter ED perimeter 175.079 -0.617 0.627 159.58 171.030 -152.030 175.)cos(29 38 ') 152.016 0.253 86.551 939.068 -30.) Latitude Departure Latitude Departure Latitude Departure 175.504 197.18 ft.025 139.079 -0.033 -188.46 -0.933 0.576 0.079 -0. Traverse Group Example Problem 2 Group Example Problem 3 Balance the latitudes and departures for the following In the survey of your assign site in Project #3.06 ft.46 -599.06 391.0 0.93 ft.18 ft. Reference 175.18 ft. Meridian D N 81° 18’ W C .27 meridian The meridian distance is positive (+) to the east and DA 781.53 ft.CIVL 1112 Surveying .Traverse Group Example Problem 3 Calculating Traverse Area In the survey of your assign site in Project #3. 78° 11’ 1800.Traverse Calculating Traverse Area Calculating Traverse Area N A The most westerly and easterly points of a traverse may be found using the departures of the traverse N 42° 59’ E S 6° 15’ W 234.54 -0. D Surveying . have to balance data collected in the following form: Corrections Balanced B N 69° 53’ E Length (ft) Latitude Departure Latitude Departure Latitude Departure A N 51° 23’ 713.Traverse Surveying .Traverse Surveying .) Latitude Departure Latitude Departure Latitude Departure The meridian distance of a line is the east–west degree m inutes AB N 69 53 E 713.93 distance from the midpoint of the line to the reference BC CD 606.22 781. 189. you will The best-known procedure for calculating land areas is have to balance data collected in the following form: the double meridian distance (DMD) method Corrections Balanced Side Bearing Length (ft.0 -285. 1 Precision = Surveying . 450.Traverse Surveying .0 450.00 606. Begin by establishing a arbitrary reference line and using the departure values of each point in the traverse to B determine the far westerly point E S 29° 38’ E 142.50 750.0 -164.27 ft.18 ft. 105° 39’ 600.18 negative (-) to the west Eclosure = ft.39 ft.58 ft.72 C 124° 47’ 391. you will traverse. N 12° 24’ W 175.00 339.Traverse Calculations 7/13 Surveying .00 259. 034 29.033 -188.933 -195.601 175.648 CD 29.53 ft.648 B C 142.403 -20.39 139.388 -20.551 E EA 171.253 86.Traverse Surveying .504 0.030 -152.253 86.Traverse DMD Calculations DMD Calculations Balanced Meridian distance The DMD of any side is equal to of line AB the DMD of the last side plus the Side Latitude Departure N DMD A departure of the last side plus the departure of the present side AB -188.78 29.470 B DE 139.388 -20.933 0.551 Meridian 234. 939.470 D C N 12° 24’ W 175. -30.Traverse Surveying .020 0.080 -30.634 0.58 ft.041 171.648 Reference N 42° 59’ E S 6° 15’ W 197.18 ft.268 86.016 0.576 0.Traverse Calculations 8/13 Surveying .025 139.017 0.53 -188.974 The DMD of line AB is departure of line AB .46 -0.068 -30.015 0.CIVL 1112 Surveying .601 BC -152.080 -30.627 159.974 Point E is the farthest D N 81° 18’ W E A C to the west Surveying .000 B -20.079 -0. 234.933 -195.18 ft.627 159.Traverse Surveying . 159.163 0.Traverse DMD Calculations DMD Calculations The meridian distance of The meridian distance of line AB N A line EA is: Meridian distance is equal to: of line AB N the meridian distance of EA N A + ½ the departure of line EA B + ½ departure of AB A E The DMD of line AB is twice the Reference B meridian distance of line AB Meridian E D C E DMD of line EA is the departure of line Surveying .551 E D 175. S 29° 38’ E -195.) Latitude Departure Latitude Departure Latitude Departure 189.000 0.974 189.18 -152.012 0.607 159.39 ft.601 B A E 86.601 -20.58 171.617 0.916 -195.Traverse Calculating Traverse Area Calculating Traverse Area Corrections Balanced N A Length (ft.470 142. 933 -195.388 -20.601 BC -152.933 -195.933 -195.648 45.447 CD 29.470 -63.Double Area The sum of the products of each points DMD and latitude Balanced equal twice the area.974 The DMD of line BC is DMD of line AB + departure of line AB The DMD of line CD is DMD of line BC + departure of line + the departure of line BC BC + the departure of line CD Surveying .397 EA 171.648 45.551 + -289.881 CD 29.627 159.Traverse DMD Calculations DMD Calculations Balanced Balanced Side Latitude Departure Side Latitude Departure DMD DMD AB -188.601 DMD Double Areas BC -152.375 BC -152.551 DE 139.470 + -63.080 -30.470 -63.601 3.933 -195.397 DE 139.080 -30.601 BC -152.253 86.601 -20.551 EA 171.601 AB -188.CIVL 1112 Surveying .601 AB -188.388 -20.470 + -63.648 + 45.388 -20.253 86.080 -30.974 The DMD of line DE is DMD of line CD + departure of line The DMD of line EA is DMD of line DE + departure of line DE CD + the departure of line DE + the departure of line EA Surveying .388 -20.397 EA 171.974 EA 171. or the double area Side Latitude Departure DMD Balanced Side Latitude Departure AB -188.601 -20.601 -20.447 CD 29.Traverse Surveying .627 159.253 86.375 DE 139.974 Notice that the DMD values can be positive or negative The double area for line AB equals DMD of line AB times the latitude of line AB .Traverse DMD Calculations DMD Calculations Balanced Balanced Side Latitude Departure Side Latitude Departure DMD DMD AB -188.447 139.080 -30.447 BC -152.627 159.551 -289.Traverse Surveying .627 159.933 -195.601 -20.601 + -20.974 EA 171.648 45.447 AB -188.388 -20.974 -159.253 86.Traverse Surveying .253 86.648 + 45.253 86.080 -30.447 BC -152.648 45.388 -20.933 -195.974 EA 171.551 + -289.Traverse DMD Calculations Traverse Area .375 CD 29.470 CD 29.397 CD 29.627 159.375 DE DE 139.470 -63.Traverse Calculations 9/13 Surveying .551 -289.627 159.375 DE 139.601 -20.974 + -159.080 -30.974 -159. 470 -63. or the double area equal twice the area.470 -63.627 159.601 3.2 1 acre = 43.375 -1.648 45.560 ft.253 86.648 45.974 -27.601 -20.551 -289.933 -195.974 -159.253 86.320 ft.974 EA 171.080 -30.Double Area Traverse Area .397 -40.919 CD 29.881 BC -152.627 159.388 -20.834 acre .919 CD 29.648 45.974 -159.Traverse Surveying .249 DE 139.919 BC -152.2 36.641 2 Area = -72.456 EA 171.388 -20.080 -30.919 BC -152.601 -20.080 -30.551 -289.470 -63.Traverse Traverse Area .974 -159.470 -63. or the double area equal twice the area.2 Area = 0.447 -6.388 -20.253 86.560 ft.447 -6.933 -195.601 3.080 -30.375 CD 29.388 -20.933 -195.456 The double area for line DE equals DMD of line DE times The double area for line EA equals DMD of line EA times the latitude of line DE the latitude of line EA Surveying .601 3.897 CD 29.388 -20.974 -159.Double Area The sum of the products of each points DMD and latitude The sum of the products of each points DMD and latitude equal twice the area.Traverse Calculations 10/13 Surveying .933 -195.080 -30.Traverse Traverse Area . or the double area Balanced Balanced Side Latitude Departure Side Latitude Departure DMD Double Areas DMD Double Areas AB -188.601 -20.551 -289.881 AB -188.897 DE 139.897 CD 29.397 -40.627 159.Double Area The sum of the products of each points DMD and latitude The sum of the products of each points DMD and latitude equal twice the area.919 CD 29.253 86.320 ft.974 -27.470 -63.249 DE 139.447 -6.601 -20.834 acre 0.249 EA 171.551 -289.080 -30.CIVL 1112 Surveying .397 -40.456 2 Area = -72.933 -195.627 159.Traverse Surveying .648 45.375 -1.447 -6.Double Area Traverse Area .897 DE 139.253 86.627 159.397 DE 139.601 3.447 -6.601 3.551 -289.Double Area The sum of the products of each points DMD and latitude The sum of the products of each points DMD and latitude equal twice the area.974 -159.375 -1.388 -20.397 EA 171.627 159.881 BC -152.447 -6.974 The double area for line BC equals DMD of line BC times The double area for line CD equals DMD of line CD times the latitude of line BC the latitude of line CD Surveying .2 Area = 1 acre = 43.648 45.881 AB -188.397 -40.974 -159.470 -63.Traverse Surveying .249 EA 171.375 -1.253 86.897 DE 139.601 -20.974 -27.Traverse Traverse Area .919 BC -152. or the double area Balanced Balanced Side Latitude Departure Side Latitude Departure DMD Double Areas DMD Double Areas AB -188.601 -20.Double Area Traverse Area .974 EA 171. or the double area Balanced Balanced Side Latitude Departure Side Latitude Departure DMD Double Areas DMD Double Areas AB -188.601 3.641 36.375 -1. or the double area equal twice the area.881 BC -152.933 -195.551 -289.648 45.881 AB -188. Traverse Surveying . Side Latitude Departure DMD Double Areas AB 600.Traverse DPD Calculations Rectangular Coordinates The same procedure used for DMD can be used the Rectangular coordinates are the convenient method double parallel distances (DPD) are multiplied by the available for describing the horizontal position of survey balanced departures points The parallel distance of a line is the distance from the With the application of computers.Traverse Traverse Area .Traverse Surveying . a square plot.).Double Area The word "acre" is derived from Old English æcer The word "acre" is derived from Old English æcer (originally meaning "open field".Traverse Traverse Area . Surveying . cognate to Swedish "åker".0 400. Latin ager and Greek αγρος (agros).560 ft. cognate to Swedish "åker". furrow long. The acre was selected as approximately the amount of A long narrow strip of land is more efficient to plough than land tillable by one man behind an ox in one day.Traverse Calculations 11/13 Surveying .Double Area Traverse Area . cognate to Swedish (originally meaning "open field". This explains one definition as the area of a rectangle with sides of length one chain (66 ft. 2 1 acre = 43.0 BC 100. German acker.Double Area Traverse Area – Example 4 The word "acre" is derived from Old English æcer Find the area enclosed by the following traverse (originally meaning "open field". rectangular midpoint of the line to the reference parallel or east–west line coordinates are used frequently in engineering projects In the US.0 200. German acker. Latin ager and Greek αγρος "åker". German acker.0 2 Area = ft.Traverse Surveying .0 EA -300.) and one furlong (ten The word "furlong" itself derives from the fact that it is one chains or 660 ft.0 -400.2 Area = acre Surveying . since the plough does not have to be turned so often.0 -300.0 DE -400. (agros). Latin ager and Greek αγρος Balanced (agros). the x–axis corresponds to the east–west direction and the y–axis to the north–south direction .0 CD 0.CIVL 1112 Surveying .0 100. 974 BC -152.Traverse Rectangular Coordinates Example Rectangular Coordinates Example y A x coordinates Consider our previous example.373 ft.648 D C x CD 29.183 ft. Coordinates of Point B (320.021 ft.974 = 159.667 ft.080 -30.253) Balance d (0. it is assumed that the coordinates of points shown in the figure below. 340.627 159. A (100.551 C (226.080 -30. -100) Surveying .933 -195.974 Surveying .Traverse Surveying . 300) Latitude AB = -400 ft. sin(4230’) = 202.933) D CD 29. B (139.470 E = D – 30.627 159.933 ft.0.974 ft.388 -20.CIVL 1112 Surveying .252 = 0 ft.640 ft.388 -20. N 42 30’ E Departure AB =300 ft.020.183 ft.080 = 169.640) E B C = 0 ft.974 . 0. determine the x and y coordinates of all the points E B E = 0 ft.601 Balance d BC -152.388 = 152.627 = 340.551. D C x A Side Latitude De parture B = A – 20. D C x E = D + 139.Traverse Rectangular Coordinates Example Rectangular Coordinates Example y A y coordinates y A (159.080 -30.648 (30.677 ft.373. AB -188.0) x EA 171.Traverse Surveying . and bearing is In this example.648 = 226.667 = 402. EA 171.601 B = A –188. 300) y B = 300 + 221. 29.470 C = B –152.253 86. Coordinates of Point A x B x (200.013 ft.551 ft. D = C + 29. Determine the coordinates of A and B are know and we want to calculate the latitude and point B departure for line AB y Latitude AB =300 ft.252 ft. y Balance d A = E + 159.183 = 521. the length of AB is 300 ft.470 = 30.Traverse Calculations 12/13 Surveying . E B CD 29. Departure AB = x B – x A A x B = 200 + 202. cos(4230’) y Coordinates of Point A Latitude AB = y B – y A B = 221.933 -195.601 D = C – 195.470 DE 139.551 = 0 ft. DE 139.253 86. BC -152.974.Traverse Surveying . 152. Departure AB = 220 ft.388 -20.551 AB -188. DE 139.013) E Side Latitude De parture A = E + 171.551 EA 171.253 86.627 159.Traverse Rectangular Coordinates Example Rectangular Coordinates Example In this example.933 -195. AB -188.601 = 139.648 Side Latitude De parture C = B + 86. 169. 373(0.Traverse End of Surveying .974 – 30.974.373) y1 y2 y3 y4 y5 y1 B (139.974) C (226.253) + 152.933(0.020.551.x2y1 – x3y2 – x4y3 – x5y4 – x1y5 Area = 0.373.403 -20. 340.253(226.53 -188.0) x Twice the area equals: + 169. 169. 169.634 -188.640) B (139.) Latitude Departure Latitude Departure Points x y degree m inutes (using a sign convention of + for next side and .373.2 ft.Traverse Group Example Problem 5 Area Computed by Coordinates Compute the x and y coordinates from the following The area of a traverse can be computed by taking each y balanced. coordinate multiplied by the difference in the two adjacent x Balanced Coordinates coordinates Side Bearing Length (ft.Traverse Area Computed by Coordinates Any Questions? There is a simple variation of the coordinate method for area computation y A (159.320.373 – 0.020(29.551(169.470 C DE N 12 24 W 142.013) E x1 x2 x3 x4 x5 x1 + 0.Traverse Calculations 13/13 Surveying .640(139.933 -195.0) + 226.Traverse Surveying .551. 0.013) E .0) method for area computation y A (159.551) – 29.253) 0.163 0.0) x = 72.78 29.373.000 0.000 100.18 -152.576 139.933 171.58 171.020 – 159.020) (0.640(139.068 -30.46 -0.433 ft.0 – 226.551 D EA N 42 59 E 234.0(30.916 -195. 169.320 ft.for last AB S 6 15 W 189.020.0(340.000 side) BC S 29 38 E 175.640 ft.607 159.253) + 139.933) + 30. 340.0.079 -0. 0.020) .2 . 152.601 A 100.013(159.640) Twice the area equals: 159.2 Surveying .0) (30. 340.551.268 86.000 Surveying .974.080 -30.020.253(226.2 Area = 36. 152.551 – 139.CIVL 1112 Surveying .0) x = -72. 29. 29.648 B CD N 81 18 W 197.013(159.0.0.Traverse Area Computed by Coordinates Area Computed by Coordinates y A (159.551) (30.933) D + 29.853 acre Area = 36.39 139.2 .933) D – 169.617 -152. 29. 0.0.933) D = x1y2 + x2y3 + x3y4 + x4y5 + x5y1 C (226.0(30.974) (0.253 86.013) E C (226.974.974(152.253) (30.373) – 152.Traverse Surveying .627 159.974 E 939.013) + B (139.504 29.640) (0.640) Twice the area equals: There is a simple variation of the coordinate = 340.933(0.340.388 -20.640. 152.