CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
1/14
Distance - Traverse
Introduction
Methods of Computing Area
♦ Almost all surveying requires some calculations to reduce measurements into a more useful form for determining distance, earthwork volumes, land areas, etc.
♦ A simple method that is useful for rough area estimates is a graphical method
♦ 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
♦ In this method, the traverse is plotted to scale on graph paper, and the number of squares inside the traverse are counted
B
A C D
Distance - Traverse
Distance - Traverse
Methods of Computing Area B a
Area ABC =
b α
A
c
Methods of Computing Area 1 ac sin α 2
C
B a A
Area ABD =
b
α
β
d c
C
1 ad sin α 2 1 2
Area BCD = bc sin β
D
Area ABCD = Area ABD + Area BCD
Distance - Traverse
Surveying - Traverse
Methods of Computing Area B
b
C
a A
c
α
β
e d E
D
Balancing Angles
1 Area ABE = ae sin α 2 1 2
Area CDE = cd sin β
♦ To compute Area BCD more data is required
♦ Before the areas of a piece of land can be computed, it is necessary to have a closed traverse ♦ The interior angles of a closed traverse should total:
(n - 2)(180°) where n is the number of sides of the traverse
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
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Surveying - Traverse
Balancing Angles
Balancing Angles A
Error of closure
B
D
♦ A surveying heuristic is that the total angle should not vary from the correct value by more than the square root of the number of angles measured times the precision of the instrument ♦ For example an eight-sided traverse using a 1’ transit, the maximum error is:
±1' 8 = ±2.83' = ±3' C
Angle containing mistake
Surveying - Traverse
Surveying - Traverse
Balancing Angles
Latitudes and Departures
♦ If the angles do not close by a reasonable amount, mistakes in measuring have been made
♦ The closure of a traverse is checked by computing the latitudes and departures of each of it sides
♦ If an error of 1’ is made, the surveyor may correct one angle by 1’ ♦ If an error of 2’ is made, the surveyor may correct two angles by 1’ each ♦ If an error of 3’ is made in a 12 sided traverse, the surveyor may correct each angle by 3’/12 or 15”
N
N
B
Latitude AB E
W Bearing ∠
A
Departure AB
Bearing ∠ W C Latitude CD
S
Surveying - Traverse
Surveying - Traverse Error of Closure
♦ The latitude of a line is its projection on the north– south meridian
♦ Consider the following statement:
♦ The departure of a line is its projection on the east– west line
B
Latitude AB E
W Bearing ∠
A S
Departure AB
♦ A northeasterly bearing has a + latitude and + departure
D
S
Latitudes and Departures
N
Departure CD
“If start at one corner of a closed traverse and walk its lines until you return to your starting point, you will have walked as far north as you walked south and as far east as you have walked west”
♦ Therefore --
Σ latitudes = 0
and
Σ departures = 0
E
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
3/14
Surveying - Traverse
Error of Closure
Error of Closure
♦ When latitudes are added together, the resulting error is called the error in latitudes (EL)
♦ If the measured bearings and distances are plotted on a sheet of paper, the figure will not close because of EL and ED
♦ The error resulting from adding departures together is called the error in departures (ED)
B
Error of closure
ED EL
Eclosure =
A
C
Latitudes and Departures - Example
A
N S 6° 15’ W
234.58’
Departure AB
189.53’
B E
B
C
Surveying - Traverse
−S = (189.53 ft ) cos(6o15') = −188.40 ft S
Surveying - Traverse
Latitudes and Departures - Example
Latitudes and Departures - Example Bearing
Side
N Departure BC
+E = (175.18 ft ) sin(29o38') = 86.62 ft
B
E 175.18’
S 29° 38’ E C S
Latitude AB
189.53’
175.18’
D N 81° 18’ W
W
E
S 6° 15’ W
175.18’
N 12° 24’ W
−W = (189.53 ft ) sin(6o15') = −20.63 ft
A
W
S 29° 38’ E
142.39’
Eclosure
perimeter
Surveying - Traverse
Latitudes and Departures - Example N 42° 59’ E
Precision =
2
2
♦ Typical precision: 1/5,000 for rural land, 1/7,500 for suburban land, and 1/10,000 for urban land
D
Surveying - Traverse
+ (ED )
(E L )
AB BC CD DE EA
Eclosure =
Latitude BC
−S = (175.18 ft ) cos(29 38') = −152.27 ft o
S S N N N
(E L )
Precision =
2
degree
m inutes
6 29 81 12 42
15 38 18 24 59
+ (ED ) =
Eclosure
2
perimeter
=
Length (ft)
Latitude
Departure
189.53 175.18 197.78 142.39 234.58 939.46
-188.403 -152.268 29.916 139.068 171.607 -0.079
-20.634 86.617 -195.504 -30.576 159.933 -0.163
W E W W E
2
( −0.079 )
2
+ ( −0.163)
0.181 ft 1 = 939.46 ft 5,176
= 0.182 ft
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
Surveying - Traverse
Group Example Problem 1 A
Group Example Problem 1 ♦ Compute the latitudes, departures, Eclosure, and the precision for the following traverse.
S 77° 10’ E
N 29° 16’ E
4/14
651.2’
B
660.5’
Side AB BC CD DE
S 38° 43’ W D
491.0’
Bearing degree
m inutes
77 38 64 29
10 43 9 16
S S N N
Length (ft)
Latitude
Departure
651.2 826.7 491.0 660.5 2629.4
-144.642 -645.031 214.084 576.190 0.601
634.933 -517.076 -441.870 322.902 -1.110
E W W E
826.7’
N 64° 09’ W C
Surveying - Traverse
Surveying - Traverse Balancing Latitudes and Departures
Group Example Problem 1 Eclosure =
(E L )
2
Side
+ (ED ) = 2
2
( 0.601 )
Bearing
AB BC CD DE
S S N N
Precision =
degree
m inutes
77 38 64 29
10 43 9 16
Eclosure
perimeter
=
E W W E
2
+ ( −1.110 )
= 1.262 ft
Length (ft)
Latitude
Departure
651.2 826.7 491.0 660.5 2629.4
-144.642 -645.031 214.084 576.190 0.601
634.933 -517.076 -441.870 322.902 -1.110
1.262 ft 1 = 4.800 × 10 −4 = 2, 629.4 ft 2,083
♦ Balancing the latitudes and departures of a traverse is to attempt to obtain more probable values for the locations of the corners of the traverse ♦ A popular method for balancing errors is called the compass or the Bowditch rule ♦ The ‘Bowditch rule’ was devised by Nathaniel Bowditch, surveyor, navigator and mathematician, as a proposed solution to the problem of compass traverse adjustment, which was posed in the American journal The Analyst in 1807
Surveying - Traverse
Surveying - Traverse
Balancing Latitudes and Departures
Balancing Latitudes and Departures A
♦ The compass method assumes: N 42° 59’ E
1) angles and distances have same error 2) errors are accidental ♦ The rule states:
“The error in latitude (departure) of a line is to the total error in latitude (departure) as the length of the line is the perimeter of the traverse”
234.58’
S 6° 15’ W 189.53’
B E
S 29° 38’ E
142.39’
N 12° 24’ W
175.18’ 175.18’
D N 81° 18’ W
C
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
5/14
Surveying - Traverse Balancing Latitudes and Departures
Latitudes and Departures - Example
N
♦ Recall the results of our example problem Side
Bearing
AB BC CD DE EA
S S N N N
degree
m inutes
6 29 81 12 42
15 38 18 24 59
W E W W E
Latitude AB
Length (ft)
Latitude
Departure
189.53 175.18 197.78 142.39 234.58 939.46
-188.403 -152.268 29.916 139.068 171.607 -0.079
-20.634 86.617 -195.504 -30.576 159.933 -0.163
−S = (189.53 ft ) cos(6o15') = −188.40 ft
A
W
E
Correction in LatAB
EL
S 6° 15’ W 189.53’
B
LAB
=
perimeter
EL ( LAB )
Correction in LatAB =
perimeter
S
Correction in LatAB =
+0.079 ft (189.53 ft )
= +0.016 ft
939.46 ft
Surveying - Traverse
Surveying - Traverse
Balancing Latitudes and Departures
Balancing Latitudes and Departures
N
N Latitude BC
Departure AB
−S = (175.18 ft ) cos(29o38') = −152.27 ft
−W = (189.53 ft ) sin(6 15') = −20.63 ft o
A
W
E
Correction in DepAB
ED
S 6° 15’ W 189.53’
B
LAB
=
B
W
perimeter
ED ( LAB )
Correction in DepAB =
Correction in LatBC
EL
175.18’
S 29° 38’ E
perimeter S
+0.163 ft (189.53 ft )
= +0.033 ft
939.46 ft
=
Correction in LatBC =
C
S
Correction in DepAB =
E
Correction in LatBC =
LBC
perimeter
EL ( LBC )
perimeter
+0.079 ft (175.18 ft ) 939.46 ft
Surveying - Traverse
Surveying - Traverse
Balancing Latitudes and Departures
Balancing Latitudes and Departures
= +0.015 ft
N Departure BC
Length (ft)
+E = (175.18 ft ) sin(29 38') = 86.62 ft
Latitude
Departure
Corrections Latitude Departure
Balanced Latitude Departure
o
W
B
E 175.18’
S 29° 38’ E C S
Correction in DepBC =
Correction in DepBC
ED
=
Correction in DepBC =
+0.163 ft (175.18 ft ) 939.46 ft
LBC
perimeter
ED ( LBC )
189.53 175.18 197.78 142.39 234.58 939.46
-188.403 -152.268 29.916 139.068 171.607 -0.079
-20.634 86.617 -195.504 -30.576 159.933 -0.163
0.016 0.015 0.017 0.012 0.020
perimeter Corrections computed on previous slides
= +0.030 ft
0.033 0.030 0.034 0.025 0.041
-188.388 -152.253 29.933 139.080 171.627 0.000
-20.601 86.648 -195.470 -30.551 159.974 0.000
CIVL 1112
Surveying - Traverse Calculations
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Surveying - Traverse
Surveying - Traverse
Balancing Latitudes and Departures
Balancing Latitudes and Departures
Length (ft) 189.53 175.18 197.78 142.39 234.58 939.46
Latitude
Departure
-188.403 -152.268 29.916 139.068 171.607 -0.079
Corrections Latitude Departure
-20.634 86.617 -195.504 -30.576 159.933 -0.163
0.016 0.015 0.017 0.012 0.020
Balanced Latitude Departure
0.033 0.030 0.034 0.025 0.041
-188.388 -152.253 29.933 139.080 171.627 0.000
-20.601 86.648 -195.470 -30.551 159.974 0.000
Length (ft) 189.53 175.18 197.78 142.39 234.58 939.46
Corrected latitudes and departures
Bearing
Length (ft) Latitude
minutes
6 29 81 12 42
15 38 18 24 59
W E W W E
189.53 175.18 197.78 142.39 234.58 939.46
-188.403 -152.268 29.916 139.068 171.607 -0.079
Departure -20.634 86.617 -195.504 -30.576 159.933 -0.163
Corrections Latitude Departure 0.016 0.015 0.017 0.012 0.020
0.016 0.015 0.017 0.012 0.020
0.033 0.030 0.034 0.025 0.041
-188.388 -152.253 29.933 139.080 171.627 0.000
-20.601 86.648 -195.470 -30.551 159.974 0.000
Group Example Problem 2
♦ Combining the latitude and departure calculations with corrections gives:
degree
-20.634 86.617 -195.504 -30.576 159.933 -0.163
Balanced Latitude Departure
Surveying - Traverse
Balancing Latitudes and Departures
AB S BC S CD N DE N EA N
-188.403 -152.268 29.916 139.068 171.607 -0.079
Corrections Latitude Departure
Departure
No error in corrected latitudes and departures
Surveying - Traverse
Side
Latitude
0.033 0.030 0.034 0.025 0.041
Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627 0.000
-20.601 86.648 -195.470 -30.551 159.974 0.000
Surveying - Traverse
♦ Balance the latitudes and departures for the following traverse. Corrections Balanced Length (ft) Latitude Departure Latitude Departure Latitude Departure 600.0 450.0 750.0 1800.0
450.00 -285.00 -164.46 0.54
339.00 259.50 -599.22 -0.72
-0.18 -0.13 -0.22
0.24 0.18 0.30
449.82 -285.14 -164.69 0.00
339.24 259.68 -598.92 0.00
Surveying - Traverse
Group Example Problem 2
Group Example Problem 3
♦ Balance the latitudes and departures for the following traverse.
♦ In the survey of your assign site in Project #3, you will have to balance data collected in the following form:
Corrections Balanced Length (ft) Latitude Departure Latitude Departure Latitude Departure
N 600.0 450.0 750.0 1800.0
450.00 -285.00 -164.46 0.54
339.00 259.50 -599.22 -0.72
-0.18 -0.13 -0.22
0.24 0.18 0.30
449.82 -285.14 -164.69 0.00
339.24 259.68 -598.92 0.00
N 69° 53’ E
A
51° 23’
713.93’
781.18’
B 105° 39’
78° 11’ 124° 47’
D
391.27’
606.06’
C
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
7/14
Surveying - Traverse
Group Example Problem 3
Group Example Problem 3
♦ The first step is to compute the bearings for each side.
♦ Find the bearing of side DA:
N
713.93’
51° 23’
N
B
N 69° 53’ E
A
105° 39’
781.18’
606.06’
391.27’
α
S
Surveying - Traverse
DA = N 58° 44’ W
N
51° 23’
α
C
124° 47’
α = 58° 44’
B
W A
78° 11’
D
α = 180° - 69° 53’ - 51° 23’ N 69° 53’ E
D
Surveying - Traverse
Group Example Problem 3
Group Example Problem 3
♦ Find the bearing of side BC:
♦ Find the bearing of side CD:
N
N
N
α = 105° 39’ - 69° 53’ B
N 69° 53’ E
S 35° 46’ E
B
α = 35° 46’
α = 66° 03’ 35° 46’
α
W
°105 39’
A
α = 180° - 78° 11’ - 35° 46’
N
W
C
35° 46’
C
78° 11’
CD = S 66° 03’ W
α
BC = S 35° 46’ E S
D
Surveying - Traverse
S
Surveying - Traverse
Group Example Problem 3
Group Example Problem 3
♦ In the survey of your assign site in Project #3, you will have to balance data collected in the following form:
♦ In the survey of your assign site in Project #3, you will have to balance data collected in the following form:
Side
Corrections Length (ft) Latitude Departure Latitude Departure
Bearing
Balanced Latitude Departure
AB BC CD DA
N S S N
69 35 66 58
53 46 3 44
Corrections Length (ft) Latitude Departure Latitude Departure
Bearing
Side
degree m inutes
Balanced Latitude Departure
degree m inutes
E E W W
Eclosure = Precision =
713.93 606.06 391.27 781.18 2492.44
245.544 -491.760 -158.832 405.450 0.402
0.802 ft 1 3,107
670.376 354.233 -357.582 -667.722 -0.694
-0.115 -0.098 -0.063 -0.126
0.199 0.169 0.109 0.218
245.429 -491.857 -158.895 405.323 0.000
670.575 354.402 -357.473 -667.505 0.000
AB BC CD DA
N S S N
69 35 66 58
53 46 3 44
E E W W
Eclosure =
Precision =
713.93 606.06 391.27 781.18 2492.44
245.544 -491.760 -158.832 405.450 0.402
0.802 ft 1 3,107
670.376 354.233 -357.582 -667.722 -0.694
-0.115 -0.098 -0.063 -0.126
0.199 0.169 0.109 0.218
245.429 -491.857 -158.895 405.323 0.000
670.575 354.402 -357.473 -667.505 0.000
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse Calculating Traverse Area
Surveying - Traverse Calculating Traverse Area N
♦ The best–known procedure for calculating land areas is the double meridian distance (DMD) method
E
C
Surveying - Traverse Calculating Traverse Area Length (ft) 189.53 175.18 197.78 142.39 234.58 939.46
E
D
E
Surveying - Traverse Calculating Traverse Area
Latitude
Departure
-188.403 -152.268 29.916 139.068 171.607 -0.079
-20.634 86.617 -195.504 -30.576 159.933 -0.163
-20.601
-30.551
-195.470
Corrections Latitude Departure 0.016 0.015 0.017 0.012 0.020
B
A
B
159.974
N
B S 29° 38’ E
N 12° 24’ W
175.18’ 175.18’
D N 81° 18’ W
C
-20.601 86.648 -195.470 -30.551 159.974 0.000
C C
Point E is the farthest to the west
♦ The meridian distance of line EA is:
A
S 6° 15’ W 189.53’
-188.388 -152.253 29.933 139.080 171.627 0.000
86.648
A
DMD Calculations
N 42° 59’ E
0.033 0.030 0.034 0.025 0.041
Balanced Latitude Departure
Surveying - Traverse
A
142.39’
175.18’
D N 81° 18’ W
D
E
175.18’
N 12° 24’ W
♦ Begin by establishing a arbitrary reference line and using the departure values of each point in the traverse to determine the far westerly point
234.58’
S 29° 38’ E
142.39’
♦ The most westerly and easterly points of a traverse may be found using the departures of the traverse
Reference Meridian
189.53’
B
Reference Meridian
Calculating Traverse Area
S 6° 15’ W
234.58’
♦ The meridian distance is positive (+) to the east and negative (-) to the west
Surveying - Traverse
A N 42° 59’ E
♦ The meridian distance of a line is the east–west distance from the midpoint of the line to the reference meridian
N
8/14
Reference Meridian
N
B
E
D
A
C
E
DMD of line EA is the departure of line
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse DMD Calculations N
Meridian distance of line AB
♦ The meridian distance of line AB is equal to:
♦ The DMD of line AB is twice the meridian distance of line AB
Surveying - Traverse DMD Calculations
AB BC CD DE EA
Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627
-20.601 86.648 -195.470 -30.551 159.974
DMD -20.601
DMD Calculations
to the DMD of the last side plus the departure of the last side plus the departure of the present side
B
E
Surveying - Traverse
AB BC CD DE EA
Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627
-20.601 + 86.648 + -195.470 -30.551 159.974
DMD -20.601 45.447
♦ The DMD of line BC is DMD of line AB + departure of line AB + the departure of line BC
Surveying - Traverse DMD Calculations
Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627
♦ The DMD of any side is equal
A
Side
Surveying - Traverse
AB BC CD DE EA
Meridian distance of line AB
DMD Calculations
♦ The DMD of line AB is departure of line AB
Side
N
the meridian distance of EA + ½ the departure of line EA + ½ departure of AB
B
Side
Surveying - Traverse DMD Calculations
A
E
9/14
-20.601 86.648 + -195.470 + -30.551 159.974
Side DMD -20.601 45.447 -63.375
♦ The DMD of line CD is DMD of line BC + departure of line BC + the departure of line CD
AB BC CD DE EA
Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627
DMD -20.601 -20.601 45.447 86.648 -195.470 + -63.375 -30.551 + -289.397 159.974
♦ The DMD of line DE is DMD of line CD + departure of line CD + the departure of line DE
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse DMD Calculations Side AB BC CD DE EA
Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627
DMD -20.601 -20.601 45.447 86.648 -63.375 -195.470 -30.551 + -289.397 159.974 + -159.974
Surveying - Traverse Traverse Area - Double Area ♦ The sum of the products of each points DMD and latitude equal twice the area, or the double area
AB BC CD DE EA
Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627
-20.601 86.648 -195.470 -30.551 159.974
Surveying - Traverse Traverse Area - Double Area ♦ The sum of the products of each points DMD and latitude equal twice the area, or the double area
AB BC CD DE EA
Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627
-20.601 86.648 -195.470 -30.551 159.974
Side AB BC CD DE EA
♦ The double area for line CD equals DMD of line CD times the latitude of line CD
-188.388 -152.253 29.933 139.080 171.627
-20.601 86.648 -195.470 -30.551 159.974
DMD -20.601 45.447 -63.375 -289.397 -159.974
Surveying - Traverse Traverse Area - Double Area ♦ The sum of the products of each points DMD and latitude equal twice the area, or the double area
AB BC CD DE EA
Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627
-20.601 86.648 -195.470 -30.551 159.974
DMD Double Areas -20.601 3,881 -6,919 45.447 -63.375 -289.397 -159.974
♦ The double area for line BC equals DMD of line BC times the latitude of line BC
Surveying - Traverse Traverse Area - Double Area ♦ The sum of the products of each points DMD and latitude equal twice the area, or the double area Side
DMD Double Areas -20.601 3,881 -6,919 45.447 -1,897 -63.375 -289.397 -159.974
Balanced Latitude Departure
♦ Notice that the DMD values can be positive or negative
Side DMD Double Areas -20.601 3,881 45.447 -63.375 -289.397 -159.974
♦ The double area for line AB equals DMD of line AB times the latitude of line AB
Side
Surveying - Traverse DMD Calculations
♦ The DMD of line EA is DMD of line DE + departure of line DE + the departure of line EA
Side
10/14
AB BC CD DE EA
Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627
-20.601 86.648 -195.470 -30.551 159.974
DMD Double Areas -20.601 3,881 -6,919 45.447 -1,897 -63.375 -40,249 -289.397 -159.974
♦ The double area for line DE equals DMD of line DE times the latitude of line DE
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
Surveying - Traverse
Traverse Area - Double Area
Traverse Area - Double Area
♦ The sum of the products of each points DMD and latitude equal twice the area, or the double area Side AB BC CD DE EA
♦ The sum of the products of each points DMD and latitude equal twice the area, or the double area
Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627
-20.601 86.648 -195.470 -30.551 159.974
Side DMD Double Areas -20.601 3,881 -6,919 45.447 -1,897 -63.375 -40,249 -289.397 -27,456 -159.974
AB BC CD DE EA
♦ The double area for line EA equals DMD of line EA times the latitude of line EA
1 acre = 43,560 ft2
DMD Double Areas -20.601 3,881 45.447 -6,919 -63.375 -1,897 -289.397 -40,249 -159.974 -27,456 2 Area = -72,641
36,320 ft2 0.834 acre
Area =
♦ The acre was selected as approximately the amount of land tillable by one man behind an ox in one day.
2
Area =
DMD Double Areas -20.601 3,881 45.447 -6,919 -63.375 -1,897 -289.397 -40,249 -159.974 -27,456 2 Area = -72,641
♦ The word "acre" is derived from Old English æcer (originally meaning "open field", cognate to Swedish "åker", German Acker, Latin ager and Greek αγρος (agros).
Balanced Latitude Departure -20.601 86.648 -195.470 -30.551 159.974
-20.601 86.648 -195.470 -30.551 159.974
Traverse Area - Double Area
♦ The sum of the products of each points DMD and latitude equal twice the area, or the double area
-188.388 -152.253 29.933 139.080 171.627
-188.388 -152.253 29.933 139.080 171.627
Surveying - Traverse
Traverse Area - Double Area
AB BC CD DE EA
Balanced Latitude Departure
1 acre = 43,560 ft2
Surveying - Traverse
Side
11/14
36,320 ft 0.834 acre
Surveying - Traverse Traverse Area - Double Area ♦ The word "acre" is derived from Old English æcer (originally meaning "open field", cognate to Swedish "åker", German Acker, Latin ager and Greek αγρος (agros). ♦ A long narrow strip of land is more efficient to plough than a square plot, since the plough does not have to be turned so often. ♦ The word "furlong" itself derives from the fact that it is one furrow long.
♦ This explains one definition as the area of a rectangle with sides of length one chain and one furlong.
Surveying - Traverse Traverse Area – Example 4 ♦ Find the area enclosed by the following traverse
Side
Balanced Latitude Departure DMD
AB BC CD DE EA
600.0 100.0 0.0 -400.0 -300.0
Double Areas
200.0 400.0 100.0 -300.0 -400.0 2 Area =
1 acre = 43,560 ft2
Area =
ft 2 acre
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
Surveying - Traverse
Traverse Area – Example 4
DPD Calculations
♦ Find the area enclosed by the following traverse
Side AB BC CD DE EA
♦ The same procedure used for DMD can be used the double parallel distances (DPD) are multiplied by the balanced departures
Balanced Latitude Departure 600.0 100.0 0.0 -400.0 -300.0
200.0 400.0 100.0 -300.0 -400.0
1 acre = 43,560 ft2
12/14
♦ The parallel distance of a line is the distance from the midpoint of the line to the reference parallel or east–west line
DMD Double Areas 200.0 120,000 800.0 80,000 1300.0 0 1100.0 -440,000 400.0 -120,000 2 Area = -360,000
180,000 ft2 4.132 acre
Area =
Surveying - Traverse
Surveying - Traverse
Rectangular Coordinates
Rectangular Coordinates Example
♦ Rectangular coordinates are the convenient method available for describing the horizontal position of survey points ♦ With the application of computers, rectangular coordinates are used frequently in engineering projects
♦ In this example, the length of AB is 300 ft and bearing is shown in the figure below. Determine the coordinates of point B Latitude AB =300 ft cos(42°30’) = 221.183 ft
y B
A x
Coordinates of Point A (200, 300)
Surveying - Traverse
Latitude AB = y B – y A
Coordinates of Point A (100, 300)
Departure B
x
Coordinates of Point B (320, -100)
y B = 300 + 221.183 = 521.183 ft
♦ Consider our previous example, determine the x and y coordinates of all the points y
A
Latitude AB = -400 ft
Departure
x B = 200 + 202.667 = 402.667 ft
Rectangular Coordinates Example
♦ In this example, it is assumed that the coordinates of points A and B are know and we want to calculate the latitude and departure for line AB A
=300 ft sin(42°30’) = 202.677 ft
AB
Surveying - Traverse
Rectangular Coordinates Example
y
Departure
N 42° 30’ E
♦ In the US, the x–axis corresponds to the east–west direction and the y–axis to the north–south direction
AB
= xB – xA
AB
= 220 ft
Side
B
E D
C
x
AB BC CD DE EA
Balanced Latitude Depa rture -188.388 -152.253 29.933 139.080 171.627
-20.601 86.648 -195.470 -30.551 159.974
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse
Surveying - Traverse
Rectangular Coordinates Example y
A
E D
AB BC CD DE EA
y
E = 0 ft
-188.388 -152.253 29.933 139.080 171.627
A
E D
B = A – 20.601 = 139.373 ft
-20.601 86.648 -195.470 -30.551 159.974
C = B + 86.648 = 226.021 ft
AB BC CD DE EA
E = D – 30.551 = 0 ft
Surveying - Traverse
D = C + 29.933 ft
x
C
E = D + 139.080 = 169.013 ft
-188.388 -152.253 29.933 139.080 171.627
-20.601 86.648 -195.470 -30.551 159.974
A = E + 171.627 = 340.640 ft B = A –188.388 = 152.252 ft C = B –152.252 = 0 ft
Surveying - Traverse
Rectangular Coordinates Example y
C = 0 ft
Balanced Latitude Depa rture
Side
D = C – 195.470 = 30.551 ft
♦ y coordinates
B
A = E + 159.974 = 159.974 ft
x
C
Balanced Latitude Depa rture
Side
Rectangular Coordinates Example
♦ x coordinates
B
13/14
Group Example Problem 5 ♦ Compute the x and y coordinates from the following balanced.
A (159.974, 340.640)
Side
Balanced Length (ft) Latitude Departure Latitude Departure
Bearing
Points
Coordinates x y
degree m inutes
AB BC CD DE EA
B (139.373, 152.253) (0.0, 169.013) E
(30.551, 29.933)
Balanced Length (ft) Latitude Departure Latitude Departure
Points
Coordinates x y
degree m inutes
6 29 81 12 42
15 38 18 24 59
W E W W E
189.53 175.18 197.78 142.39 234.58 939.46
189.53 175.18 197.78 142.39 234.58 939.46
-188.403 -152.268 29.916 139.068 171.607 -0.079
-20.634 86.617 -195.504 -30.576 159.933 -0.163
-188.388 -152.253 29.933 139.080 171.627 0.000
-20.601 86.648 -195.470 -30.551 159.974 0.000
A B C D E
100.000 79.399 166.047 -29.423 -59.974
100.000 -88.388 -240.641 -210.708 -71.627
Area Computed by Coordinates
♦ Compute the x and y coordinates from the following balanced.
S S N N N
W E W W E
Surveying - Traverse
Group Example Problem 5
AB BC CD DE EA
15 38 18 24 59
x
Surveying - Traverse
Bearing
6 29 81 12 42
D C (226.020, 0.0)
Side
S S N N N
-188.403 -152.268 29.916 139.068 171.607 -0.079
-20.634 86.617 -195.504 -30.576 159.933 -0.163
-188.388 -152.253 29.933 139.080 171.627 0.000
-20.601 86.648 -195.470 -30.551 159.974 0.000
A B C D E
100.000 79.399 166.047 -29.423 -59.974
100.000 -88.388 -240.641 -210.708 -71.627
♦ The area of a traverse can be computed by taking each y coordinate multiplied by the difference in the two adjacent x coordinates (using a sign convention of + for next side and - for last side)
CIVL 1112
Surveying - Traverse Calculations
Surveying - Traverse Area Computed by Coordinates y
A (159.974, 340.640)
Twice the area equals: = 340.640(139.373 – 0.0)
B (139.373, 152.253) (0.0, 169.013) E
+ 152.253(226.020 – 159.974)
C (226.020, 0.0) x
+ 29.933(0.0 – 226.020) + 169.013(159.974 – 30.551) = 72,640.433 ft2
Area = 0.853 acre
Area Computed by Coordinates ♦ There is a simple variation of the coordinate method for area computation y
A (159.974, 340.640)
x1 y1
Area Computed by Coordinates ♦ There is a simple variation of the coordinate method for area computation A (159.974, 340.640)
B (139.373, 152.253) (0.0, 169.013) E
Twice the area equals: 159.974(152.253) + 139.373(0.0) + 226.020(29.933) + 30.551(169.013) + 0.0(340.640) - 340.640(139.373) – 152.253(226.020) - 0.0(30.551) – 29.933(0.0) – 169.013(159.974)
(30.551, 29.933) D C (226.020, 0.0) x
= -72,640 ft2
Area = 36,320 ft2
x2 y2
x3 y3
x4 y4
B (139.373, 152.253) (0.0, 169.013) E
x5 y5
x1 y1
Twice the area equals:
(30.551, 29.933) D C (226.020, 0.0) x
Area = 36,320.2 ft2
Surveying - Traverse
y
Surveying - Traverse
+ 0.0(30.551 – 139.373)
(30.551, 29.933) D
14/14
= x1y2 + x2y3 + x3y4 + x4y5 + x5y1 - x2y1 – x3y2 – x4y3 – x5y4 – x1y5
End of Surveying - Traverse Any Questions?
