TRAVERSING AND
TRAVERSE
COMPUTATION
TRAVERSING
AND TRAVERSE COMPUTATIONS
 Interior Angle Traverse
 Deflection Angle Traverse
TRAVERSING AND TRAVERSE COMPUTATIO
INTERIOR ANGLE TRAVERSE
Used principally in land surveying
Angles formed between the adjacent sides of a closed figure are known as
interior angle. Maybe measured either clockwise or counter clockwise direction, however, the usual
practice is to measure the angles clockwise.
TRAVERSING AND TRAVERSE COMPUTATIO
TRAVERSING AND TRAVERSE COMPUTATIO
Sample Problem
The observed interior angle of a closed traverse are as follows: A, 153°30’00”; B, 58° 20’20”; C, 139°19’00”; D, 78°21’20”; E, 110°28’00”.
Determine the angular error of the same amount of each station. Tabulate values accordingly.
Solution:
a)
Determining the angular Error of Closure
n = 5 (number of interior angles in the traverse)
Sum1 =
angle A + angle B + angle C + angle D + angle E
=
153°30’00” + 58° 20’20” + 139°19’00 ” + 78°21’20” + 110°28’00”
=
539°58’40” (sum of observed interior angle)
Sum2 =
=
(n – 2) 180°
540°
=
(5 – 2) 180°
(sum of the interior angles for a five-sided closed traverse)
TRAVERSING AND TRAVERSE COMPUTATIO
Closure
=
Sum2 – Sum1 = 540° - 539°58’40”
=
+0°01’20”
(angular error for the observations made)
Closure
+0°01’20”
Correction =
n =
5
=
+16” (correction to be applied to each interior abgle)
b)
Adjusting the interior angles
Angle A’ = A + corr
=
Angle B’ = B + corr
=
Angle C’ = C + corr
=
Angle D’ = D + corr =
Angle E’ = E + corr
=
153°30’00” + 16”
58°20’20” + 16”
139°19’00” + 16”
78°21’20” + 16”
110°28’00” + 16”
= 153°30’16”
= 58°20’36”
= 139°19’16”
= 78°21’36”
= 110°28’16”
Solution check:
Sum2 = angle A + angle B + angle C + angle D + angle E
OBSERVED
INT.
CORR
ANGLE
ADJUSTED
INT. ANGLE
A
B
C
D
E
153°30’00”
58° 20’20”
139°19’00 ”
78°21’20”
110°28’00”
+16”
+16”
+16”
+16”
+16”
153°30’16”
58° 20’36”
139°19’16 ”
78°21’36”
110°28’16”
SUMS
539°58’40”
+1’20”
540°00’00”
STAT
540° = 153°30’16” + 58° 20’36” + 139°19’16 ” + 78°21’36” + 110°28’16”
540° = 540°
TRAVERSING AND TRAVERSE COMPUTATIO
DEFLECTION ANGLE TRAVERSE
Used frequently for the location survey of roads, railroads, pipelines, transmission lines,
canals, and other similar types of surveys.
Employed to a lesser extent in land surveys
TRAVERSING AND TRAVERSE COMPUTATIO
Illustrative Problem
The following are the observed deflection angles of a closed traverse:
A = 28°25’00” (L)
E = 108°13’30” (L)
B = 68°03’30” (L)
F = 16°50’00” (R)
C = 120°34’00” (L)
G = 110°00’30” (L)
D = 58°30’00” (R)
Compute the error of closure and adjust the angular values by assuming that the error is the same for each angle. Tabulate values
accordingly.
Solution:
a)
Determining the Error of Closure.
n = 7 (number of deflection angles)
∑ DEFLL = A + B + C + E + G
= 28°25’00” + 68°03’30” + 120°34’00” + 108°13’30” + 110°00’30”
=435°16’30”
(sum of the left deflection angles)
TRAVERSING AND TRAVERSE COMPUTATIO
∑ DEFLR = D + F
= 58°30’00” + 16°50’00”
= 75°20’00”
DIFF1
(sum of the right deflection angles)
= ∑ DEFLL - ∑ DEFLR
= 435°16’30” - 75°20’00”
= 359°56’30”
ERROR = 360° - DIFF
= 3’30”
(difference between the two sums)
=
360° - 359°56’30”
(angular error of closure)
CORR = ERROR
=
3’30”
n
7
= 0’30” (correction to be applied to each observed deflection angle)
b) Adjusting the Deflection Angles
A’ = A + Corr = 28°25’00” (L) + 30” = 28°25’30” (L)
Note. By analyzing the observed values, it will be seen that
B’ = B + Corr = 68°03’30” (L) + 30” = 68°04’00” (L)
the left deflection angles must be increased and the sum of
the right deflection angles must be decreased.
C’ = C + Corr = 120°34’00” (L) + 30” = 120°34’30” (L)
D’ = D + Corr = 58°30’00” (R) - 30” = 58°29’30” (R)
E’ = E + Corr = 108°13’30” (L) + 30” = 108°14’00” (L)
F’ = F + Corr = 16°50’00” (R) - 30” = 16°49’30” (R)
G’ = G + Corr = 110°00’30” (L) + 30” = 110°01’00” (L)
TRAVERSING AND TRAVERSE COMPUTATIO
Solution check:
∑ Adj DEFLL = A’ + B’ + C’ + E’ + G’
= 28°25’30” + 68°04’00” + 120°34’30” + 108°14’00” + 110°01’00”
= 435°19’00”
(sum of the adjusted left deflection angles)
∑ Adj DEFLR = D’ + F’
= 58°29’30” + 16°49’30”
c) Tabulated Solution
= 75°19’00”
STA
DIFF2 = ∑ Adj DEFLL - ∑ Adj DEFLR
= 435°19’00” - 75°19’00”
= 360°00’00”
(∑ DEFLL )
(∑ DEFLL )
(∑ Adj DEFL L)
(∑ Adj DEFLR)
OBSERVED DEFL ANGLE
LEFT
A
B
C
D
F
E
G
28°25’00”
68°03’30”
120°34’00”
SUMS
435°16’30”
108°13’30”
110°00’30”
RIGHT
58°30’00”
16°50’00”
75°20’00”
CORR
+30”
+30”
+30”
-30”
+30”
-30”
+30”
ADJUSTED DEFL ANGLE
LEFT
RIGHT
28°25’30”
68°04’00”
120°34’30”
108°14’00”
110°01’00”
435°19’00”
58°29’30”
16°49’30”
75°19’00”
TRAVERSING
AND TRAVERSE
COMPUTATIONS
TRAVERSING AND TRAVERSE COMPUTATIO
ANGLE TO THE RIGHT TRAVERSE
Measured clockwise or right hand direction from the back sight on the back line to a forward line.
(n+2) 180° (n-2) 180°
TRAVERSING AND TRAVERSE COMPUTATIO
1.) ANGLE TO THE RIGHT TRAVERSE.
A five-sided closed traverse proceeds in a clockwise direction and the angle to
the right of each station were observed as follows:
αb
αa
αa = 240°30’
αb = 238°15’
αc = 289°53’
αc
αd = 220°04’
αe = 271°13’
αe
n=5
αd
SOLUTION.
a.) Determining the Error of Closure.
Sum1 = αa + αb + αc + αd + αe
= 240°30’ + 238°15’ + 289°53’ + 220°04’ + 271°13’
= 1259°55’
Sum2 = (n + 2) x 180° = (5 + 2) x 180°
= 1260°00’
Error = Sum2 - Sum1 = 1260°00’- 1259°55’
= + 0°05’
b.) Adjusting the Observed Angles.
Corr =
=
α’a = αa ± Corr = 240°30’ + 0°01’ = 240°31’
α’b = αb ± Corr = 238°15’ + 0°01’ = 238°16’
α’c = αc ± Corr = 289°53’ + 0°01’ = 289°54’
α’d = αd ± Corr =220°04’ + 0°01’ = 220°05’
α’e = αe ± Corr =271°13’ + 0°01’ = 271°14’
Solution Check:
Sum2 = α’a + α’b + α’c + α’d + α’e
1260°00’ = 240°31’ + 238°16’ + 289°54’+ 220°05’ + 271°14’
TRAVERSING AND TRAVERSE COMPUTATIO
c.) Tabulated Solution.
STA
CORR
A
OBSERVED ANGLE
TO THE RIGHT
240°30’
+01’
ADJUSTED ANGLE
TO THE RIGHT
240°31’
B
238°15’
+01’
238°156
C
289°53’
+01’
289°54’
D
220°04’
+01’
220°05’
E
271°13’
+01’
271°14’
SUMS
1259°55’
+05’
1260°00’
TRAVERSING AND TRAVERSE COMPUTATIO
AZIMUTH TRAVERSE
Measured clockwise either from the North or South end of a selected reference
meridian to the line.
TRAVERSING AND TRAVERSE COMPUTATIO
2.) AZIMUTH TRAVERSE.
Determine the bearing and azimuth from north of all traverse
lines, and the angle to the right of each station.
STA
OCC
STA
OBS
DISTANCE
(m)
AZIMUTH STA
FROM
OCC
SOUTH
A
E
B
210.10
90°28’
170°30’
B
A
C
155.34
350°30’
123°05’
C
B
D
206.85
303°05’
56°13’
STA
OBS
DISTANCE
(m)
AZIMUT
H FROM
SOUTH
D
C
E
174.50
236°13’
357°58’
E
D
A
330.00
177°58’
270°28’
AZIMUTH TRAVERSE.
TRAVERSING AND TRAVERSE COMPUTATIO
a.) Determining Bearing and Azimuth (from North) of all Traverse Lines.
Station A
αab = 180°00’ - 170°30’
= 9°30’ (Bearing of line AB: N 9°30’ W)
Ꝋab = 180°00’ + 170°30’
= 350°30’ (Azimuth from North of line AB)
Station B
αbc = 180°00’ - 123°05’
= 56°55’ (Bearing of line BC: N 56°55’ W)
Ꝋbc = 180°00’ + 123°05’
= 303°05’ (Azimuth from North of line BC)
Station C
α
= 56°13’ (Bearing of line CD: S 56°13’ W)
AZIMUTH TRAVERSE.
TRAVERSING AND TRAVERSE COMPUTATIO
Station D
αde = 360°00’ - 357°58’
= 2°02’ (Bearing of line DE: S 2°02’ E)
Ꝋ
de = 357°58’ + 180°00’
= 177°58’ (Azimuth from North of line DE)
Station E
αea = 360°00’ - 270°28’
= 89°32’ (Bearing of line EA: S 89°32’ E)
Ꝋea = 270°28’ + 180°00’
= 90°28’ (Azimuth from North of line EA)
AZIMUTH TRAVERSE.
TRAVERSING AND TRAVERSE COMPUTATIO
b.) Determining Angle to the Right at Each Station.
Station A
фa = 170°30’ - 90°28’
= 80°02’
Station B
фb = (360°00’ - 350°30’) + 123°05’
= 132°35’
Station C
фc = (360°00’ - 303°05’) + 56°13’
= 113°08’
AZIMUTH TRAVERSE.
TRAVERSING AND TRAVERSE COMPUTATIO
Station D
ф = 357°58’ - 236°13’
d
= 121°45’
Station E
фe = 270°28’ - 177°58’
= 92°30’
Solution Check:
фa + фb + фc + фd + фe = (n-2) 180°
80°02’ + 132°35’ + 113°08’ + 121°45’ + 92°30’ =
(5 – 2) 180°
540° = 540°
LATITUDES AND DEPARTURES
 The Latitude of a line is its projection onto the
reference meridian or a north-south line.
 Latitudes are sometimes referred to as northings
and southings bearing positive (+) as being north
and negative (-) as being south.
 While the parallel or the east-west line. Departures
bearing east as positive (+) and west as negative (-)
TRAVERSING AND TRAVERSE COMPUTATI
Latitudes and Departures
FORMULAS:
Latitude:
Dep = d sinα
Departures:
Lat = d cosα
TRAVERSING AND TRAVERSE COMPUTATI
Latitudes and Departures
Example:
COURSE
DISTANCE (m)
BEARINGS
AB
550.30
N28° 10’E
BC
395.48
S69° 35’E
CD
462.70
S27° 50’ E
DE
631.22
N50° 00’E
EF
340.05
S25° 05’E
FG
275.86
DUE EAST
TRAVERSING AND TRAVERSE COMPUTATI
Latitudes and Departures
Solution:
DETERMINING LATITUDES: Lat = d cos α
DETERMINING DEPARTURES: Dep = d sin α
Latab = 550.30 cos 28°10’ = 485.13 m
Depab = 550.30 sin 28°10’ = 259.76 m
Latbc = 395.48 cos 69°35’ = - 137.96 m
Depbc = 395.48 sin 69°35’ = 370.64 m
Latcd = 462.70 cos 27°50’ = - 409.17 m
Depcd = 462.70 sin 27°50’ = 216.04 m
Latde = 631.22 cos 50°00’ = 405.74 m
Depde = 631.22 sin 50°00’ = 483.54 m
Latef = 340.05 cos 25°05’ = - 307.98m
Depef = 340.05 sin 25°05’ = 144.16 m
Latfg = 275.86 cos 90°00’ = 0.00 m
Depfg = 275.86 sin 90°00’ = 275.86 m
TRAVERSING AND TRAVERSE COMPUTATI
Latitudes and Departures
Tabulated solution:
TRAVERSING AND TRAVERSE COMPUTATI
Linear Error of Closure
 It is usually a short line of unknown length
and direction connecting the initial and final
traverse stations of the traverse.
TRAVERSING AND TRAVERSE COMPUTATI
Relative Error of Closure
 Ratio of the linear error of closure to the
perimeter or total length of the traverse.
TRAVERSING AND TRAVERSE COMPUTATI
ERROR OF CLOSURE
Example:
TRAVERSING AND TRAVERSE COMPUTATI
ERROR OF CLOSURE
Solution:
DETERMINING COURSE LATITUDES: Lat = d cosα DETERMINING COURSE DEPARTURES: Dep = d sin α
Latab = 233.10 cos 122° 30’ = -125.24 m
Depab = 233.10 sin 122° 30’ = 196.59 m
Latbc = 242.05 cos 85° 15’ = 20.04 m
Depbc = 242.05 sin 85° 15’ = 241.22 m
Latcd = 191.50 cos 20° 00’ = 179.95 m
Depcd = 191.50 sin 20° 00’ = 65.50 m
Latde = 234.46 cos 333° 35’ = 209.98 m
Depde = 234.46 sin 333° 35’ = - 104.31 m
Latef = 270.65 cos 254° 08’ = - 74.00m
Depef = 270.65 sin 254° 08’ = - 260.34 m
Latfa = 252.38 cos 213° 00’ = -211.66 m
Depfa = 252.38 sin 213° 00’ = -137.46 m
TRAVERSING AND TRAVERSE COMPUTATI
Solution:
D = dab + dbc + dcd + dde + def + dfa
= 233.10 + 242.05 + 191.50 + 234.46 + 270.65 +
= 1,424.14 m
ΣNL = Latbc + Latcd + Latde
= 20.04 + 179.95 + 209.98
= 409.97 m
ΣSL = Latab + Latef + Latfa
= - 125.24 – 74.00 – 211.66
= 410.90 m
ΣED = Depab +Depbc + Depcd
= 196.59 + 241.22 + 65.50
= 503.31 m
ΣWD = Depde + Depef + Depfa
= -104.31 – 260.34 – 137.46
= -502.11 m
252.38
CL = ΣNL + ΣSL
(total correction
in latitude)
= 409.97 + ( 410.90)
= - 0.93 m
CD = ΣED + ΣWD
(total correction
of closure)
= 503.31 + (502.11)
= 1.20 m
LEC = √CL2 + CD2
= √(-0.93)2 +
(1.20)2
= 1.52 m
Tan θ = -CD / -CL
= -(1.20) / -(-0.93)
=-1.290323
θ = 52° 13’
(therefore, the bearing of the side of
error is N52° 13’W)
RP = LEC / D
= 1.52 / 1,424.14
= 1 / 936.93 say 1 / 900 (precision of the measurements)
TRAVERSING AND TRAVERSE COMPUTATI
Tabulated solution:
Traversing and
traverse computations
TRAVERSING AND TRAVERSE COMPUTATI
Traverse Adjustment
 The procedure of computing the linear error of closure and applying
corrections to the individual latitudes and departures for the purpose of
providing a mathematically closed figure. There are different rules and
methods used in adjusting a traverse. Some are performed in graphical
method and others are in analytical method.
 Least squares method – provides the most
rigorous adjustment .
 Arbitrary method, compass rule, transit
rule, and the Crandall method – they are
all approximate methods of traverse
adjustment.
TRAVERSING
AND TRAVERSE COMPUTATIONS




Traverse Adjustment
Arbitrary method
Compass rule
Transit rule
TRAVERSING AND TRAVERSE COMPUTATI
Arbitrary Method
 The latitudes and departures are adjusted in a discretionary
manner according to the surveyor’s assessment of the
conditions surrounding the survey.
 It is the simplest to perform.
 This method does not conform to established rules or
mathematical equations since the surveyor simply relies on
his own estimation and personal judgment.
TRAVERSING AND TRAVERSE COMPUTATI
Compass Rule
 Also called Bowditch Rule.
 Named after the distinguished American Navigator Nathaniel
Bowditch (1773-1838)
 Based on the assumption that:
All lengths are measured with equal care and all angles are taken with approximately
the same precision. The errors in the measurement is accidental and that the total error in
any side of the traverse is directly proportional to the total length of the traverse. The
correction to be applied to the latitude (or departure) of any course is equal to the total
closure in latitude (or departure) multiplied by the ratio of the length of the course to the
total length or perimeter of the traverse.
TRAVERSING AND TRAVERSE COMPUTATI
Compass Rule
Cl= CL(d/D)
and
Cd=CD(d/D)
where
Cl = correction to be applied to the latitude of any course.
Cd = correction to be applied to the departure of any course.
CL = total closure in latitude or the algebraic sum of the north
and south latitudes ( ΣNL + ΣSL ).
CD = total closure in departure or the algebraic sum of the
east and west departure ( ΣED + ΣWD ).
d = length of any course.
D = total length or perimeter of the traverse.
Note:
If the sum of the north
latitudes exceeds the sum of
the south latitudes, latitudes
corrections are subtracted from
north latitudes and added to
corresponding south latitudes.
However, if the sum of the
south latitudes, exceeds the
sum of the north latitudes, the
corrections are applied in the
opposite manner. A similar
procedure is used when
adjusting the departure.
TRAVERSING AND TRAVERSE COMPUTATI
Example
From the field notes of a closed traverse shown below, adjust the traverse
a) Using compass rule
b) Compute the linear error of closure
c)Compute the relative error or precision
Course
Bearings
Distance
AB
Due North
400.00 m
BC
N 45°E
800.00 m
CD
S 60° E
700.00 m
DE
S 20° W
600.00 m
EA
S 86°59' W
966.34 m
TRAVERSING AND TRAVERSE COMPUTATI
Latitude = d cos α
LatAB = 400.00 cos(0) = +400.00
Course
Bearings
Distance
Latitude
(+N)(-S)
AB
Due North
400.00 m
+400.00
BC
N 45°E
800.00 m
+565.69
CD
S 60° E
700.00 m
-350.00
DE
S 20° W
600.00 m
-563.82
EA
S 86°59' W
966.34 m
-50.86
Departure
(+E)(-W)
TRAVERSING AND TRAVERSE COMPUTATI
Departure = d sin α
DepAB = 400.00 sin(0) = 0
Course
Bearings
Distance
Latitude
(+N)(-S)
Departure
(+E)(-W)
AB
Due North
400.00 m
+400.00
0
BC
N 45°E
800.00 m
+565.69
+565.69
CD
S 60° E
700.00 m
-350.00
+606.22
DE
S 20° W
600.00 m
-563.82
-205.21
EA
S 86°59' W
966.34 m
-50.86
-965.00
Total Closure in Latitude and Departure
ΣNL = 400.00 + 565.69 = 965.69
ΣSL = - 350.00 - 563.82 - 50.86 = -964.68
ΣED = 565.69 + 606.22 = 1171.91
ΣWD = -205.21- 965.00 = -1170.21
CL = ΣNL + ΣSL
CD = ΣED + ΣWD
= 965.69 + (-964.68)
CL = 1.01
= 1171.91 + (-1170.21)
CD = 1.70
D = d1 + d2 + d3 + d4 + d5
= 400.00 m + 800.00 m + 700.00 m + 600.00 m + 966.34 m D= 3466.34
Corrections for Latitude:
Cl= CL(d/D)
K1 = CL/D
Cl = dK1
K1 = 1.01/3466.34 = 0.000291374
CAB = 0.000291374 x 400.00 = 0.12
CBC = 0.000291374 x 800.00 = 0.24
CCD = 0.000291374 x 700.00 = 0.20
CDE = 0.000291374 x 600.00 = 0.17
CEA = 0.000291374 x 966.34 = 0.28
Corrections for Departure:
Cd=CD(d/D)
K2 = CD/D
Cd = dK2
K2 = 1.70/3466.34 = 0.00049043
CAB = 0.00049043 x 400.00 = 0.20
CBC = 0.00049043 x 800.00 = 0.40
CCD = 0.00049043 x 700.00 = 0.34
CDE = 0.00049043 x 600.00 = 0.29
CEA = 0.00049043 x 966.34 = 0.47
Adjusting Latitudes and Departures
AdjLat= Computed Latitude ± Cl
Cd
AdjDep= Computed Departure ±
AdjLatAB = +(400.00 – 0.12) = 399.98 AdjDepAB = +(0 – 0.20) = -0.20
AdjLatBC = +(565.69 – 0.24) = 565.45 AdjDepBC = +(565.69 – 0.40) = 565.29
AdjLatCD = -(350.00 + 0.20) = -350.20 AdjDepCD = +(606.22 – 0.34) = 605.88
AdjLatDE = -(563.82 + 0.17) = -563.99 AdjDepDE = -(205.21 + 0.29) = -205.50
AdjLatEA = -(50.86 + 0.28) = -51.14
AdjDepEA = -(965.00 + 0.47) = -965.47
Course
Latitude
Correction
Departure
Correction
Adjusted Latitude
(+N) (-S)
Adjusted
Departure
(+E) (-W)
AB
0.12
0.20
+399.98
-0.20
BC
0.24
0.40
+565.45
+565.29
CD
0.20
0.34
-350.20
+605.88
DE
0.17
0.29
-563.99
-205.50
EA
0.28
0.47
-51.14
-965.47
Linear Error of
Closure
L'= √(CL)²+(CD)²
= √(1.01)²+(1.70)²
= 1.97740
Relative Error
Tan α= (-CD/-CL)
= -(+1.70)/-(1.01)
= +1.68
LEC/D = 1.97740/3466.34
RE =1/3466.34/1.97740
= 1/1753
Meaning only 1 meter error is allowed for every 1753m
distance measured.
TRAVERSING AND TRAVERSE COMPUTATI
Transit Rule
 No sound theoretical foundation since it is purely empirical.
Not commonly used but best suited for surveys where
traverse sides are measured by stadia or subtense-bar method.
 Based on the assumption that:

The angular measurements are more precise than the linear measurements and
that the errors in traversing are accidental.
 The correction to be applied to the latitude (or departure) of any course is equal to
the latitude (or departure) of the course multiplied by the ratio of the total closure
in latitude ( or departure ) to the arithmetical sum of all the latitudes ( or
departure ) of the traverse.
TRAVERSING AND TRAVERSE COMPUTATI
Transit Rule
ΣNL = summation of
north latitudes
ΣSL = summation of
south latitudes
ΣED = summation of east
departures
ΣWD = summation of
west departures
Note:
Latitude and departure
corrections are applied
in a manner similar to
that described for the
compass rule.
TRAVERSING AND TRAVERSE COMPUTATI
Adjusted Lengths and Directions
 After the latitudes and departures of the course of a closed
traverse have so adjusted, the bearings ( or azimuth ) of the
course and their lengths should also be adjusted to correspond
to the adjusted latitudes and departures.
where:
L’ = adjusted length of a
course
Lat’= adjusted latitude of a
course.
Dep’= adjusted departure of
a course.
α’= adjusted horizontal angle
between the reference
meridian and a course.
Least square method
The best way of adjusting survey data particularly to a precise measurements. The method
of least-squares adjustment of a travers is suitable for use on electronic
computers. It is useful for determination of the best value and for estimating the
relative worth of different determination.
However, it is somewhat complex, laborious to perform, and requires lengthy
computations.
Crandall method
Application of the theory of least-squares, which can be used to compute corrections to the
measured distances to make the traverse close mathematically. Crandall rule adjustment
assumes that there are error in distance measurements and distribute an angular error
throughout the traverse.
Example
Line
AB
BC
CD
DE
EF
FA
Solve the following:
a. Latitude and departure
b. Transit rule
c. Adjusted latitude and departure
d. Adjusted lengths and bearing
Bearing
N 05° 30’ E
N 46° 02’ E
S 67° 38’ E
S 12° 25’ E
S 83° 44’ W
N 55° 09’ W
Distance
495.85 m
850.62 m
855.45 m
1020.87 m
1117.26 m
660.08 m
a. Latitude and Departure
Line
Bearing
Distance
Computed latitude
Computed departure
+N
+E
-S
AB
N 05° 30’ E
495.85 m
493.57
47.52
BC
N 46° 02’ E
850.62 m
590.53
612.23
CD
S 67° 38’ E
855.45 m
-323.53
791.09
DE
S 12° 25’ E
1020.87 m
-996.99
219.51
EF
S 83° 44’ W
1117.26 m
-121.96
FA
N 55° 09’ W
660.08 m
Sum
-1110.58
377.19
ƩNL=1469.21
-W
-541.70
ƩSL=-1444.35
ƩED=1670.35
ƩWD=-1652.28
a.
Line
Correction
LAT
DEP
Adjusted Latitude
Adjusted Departure
+N
+E
-S
AB
493.57
47.52
BC
590.53
612.23
CD
-323.53
791.09
DE
-996.99
219.51
EF
-121.96
FA
377.19
Sum
ƩNL=1469.21
-W
-1110.58
-541.70
ƩSL=-1444.35
ƩED=1670.35
ƩWD=-1652.28
TRAVERSING
AND TRAVERSE COMPUTATIONS
 Graphical Method
 Rectangular Coordinates
 Coordinate Method
TRAVERSING AND TRAVERSE COMPUTATI
Rectangular Coordinates
 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 the US, the x–axis corresponds to the east–west
direction and the y–axis to the north–south direction
TRAVERSING AND TRAVERSE COMPUTATI
Rectangular Coordinates Example
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.
Departure AB =300 ft. sin(4230’)
= 202.677 ft.
x B = 200 + 202.667 = 402.667 ft.
y B = 300 + 221.183 = 521.183 ft.
TRAVERSING AND TRAVERSE COMPUTATI
Rectangular Coordinates Example
Given in the accompanying tabulation are the adjusted latitudes and departures of a closed
traverse. Calculate the coordinates of each station along the traverse if the coordinates of
station A are X= 3,000.00m and Y= 4,000.00m. Tabulate values accordingly.
ADJUSTED
LATITUDES
ADJUSTED
DEPARTURES
LINE
+N
AB
405.50
BC
218.13
CD
DE
-S
+E
-W
202.25
175.64
71.08
325.67
415.36
355.62
EF
389.70
85.51
FA
488.52
739.08
SUMS
949.30
949.30
973.23
973.23
TRAVERSING AND TRAVERSE COMPUTATI
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
LatitudeAB= yB – yA
Latitude AB = -400 ft
Departure AB = xB – xA
Departure AB = 220 ft.
TRAVERSING AND TRAVERSE COMPUTATI
Rectangular Coordinates Example
Consider our previous example, determine the x and y
coordinates of all the points
TRAVERSING AND TRAVERSE COMPUTATI
Rectangular Coordinates Example
x coordinates
E = 0 ft.
A = E + 159.974 = 159.974 ft.
B = A – 20.601 = 139.373 ft.
C = B + 86.648 = 226.021 ft.
D = C – 195.470 = 30.551 ft.
E = D – 30.551 = 0 ft.
y coordinates
C = 0 ft.
D = C + 29.933 ft.
E = D + 139.080 = 169.013 ft.
A = E + 171.627 = 340.640 ft.
B = A –188.388 = 152.252 ft.
C = B –152.252 = 0 ft.
TRAVERSING AND TRAVERSE COMPUTATI
Rectangular Coordinates Example
Compute the x and y coordinates from the following balanced.
Side
Bearing
Length
(ft.)
Latitude
Departure
Balanced
Points
Latitude
Departure
Coordinates
X
AB
S 6 15 W
189.53
-188.403
-20.634
-188.388
-20.601
A
BC
S 29 38 E
175.18
-152.268
86.617
-152.253
86.648
B
CD
N 81 18 W
197.78
29.916
-195.504
29.933
-195.470
C
DE
N 12 24 W
142.39
139.068
-30.576
139.080
-30.551
D
EA
N 42 59 E
234.58
171.607
159.933
171.627
159.974
E
939.46
-0.079
-0.163
0.000
0.000
Y
100. 100.
TRAVERSING AND TRAVERSE COMPUTATI
COORDINATE METHOD
 When a survey is run and made to close on a distant terminal
station or point of known coordinates, it will be expected that
the computed coordinates of the terminal will be an error of
closure along x-axis and also along y-axis.
 This is due to inherent errors in both angular and linear
measurements during traversing.
TRAVERSING AND TRAVERSE COMPUTATI
COORDINATE METHOD
FORMULAS:
d=
Cx= Xk-Xc
Cy= Yk – Yc
RP=
x= d(
y= d(
X’= X ± x
Y’= Y ± y
Where:
d= distance between any two stations whose x and y coordinates are
known
X2= coordinate along the x-axis of a succeeding station
Y2= coordinate along the y-axis of a succeeding station
X1= coordinate along the x-axis of a preceding station
Y1= coordinate along the x-axis of a preceding station
Cx= error of closure along the x-axis
Xk= known, coordinate along the x-axis of the distant terminal station
Xc= computed coordinate along the x-axis of the distant terminal station
Yk= known, coordinate along the y-axis of the distant terminal station
Yc= computed coordinate along the y-axis of the distant terminal station
RP = relative precision of closure
D = total length or perimeter of the traverse from the initial station to the
distant terminal station
x = coordinate correction along the x-axis
y = coordinate correction along the y-axis
X’ = adjusted X coordinate of a station
Y’ = adjusted Y coordinate of a station
TRAVERSING AND TRAVERSE COMPUTATI
Coordinate Method Example
Given in the accompanying tabulation are the known and computed coordinates of stations along a
traverse. The traverse originates on station Baguio whose known coordinates are X= 6,2018.67 and
Y= 8,601.44, and closes on a station Acupan whose known coordinates are X= 5,226.10 and Y=
5,782.62. Adjust the coordinates of the traverse station and tabulate values accordingly
STATION
Computed Coordinates
X
Y
Baguio
6,208.67
8,601.44
A
7, 030.45
9,299.54
B
6,984.53
7,698.69
C
7,001.14
7,260.00
D
7,112.99
6,774.08
E
6,586.70
5,941.82
F
G
Acupan
6,147.28
5,467.06
5,226.18
6,058.24
6,066.64
5,782.98