1. Traverse – A series of lines connecting
successive points whose lengths and directions
have been determined from field measurements.
2. Traversing – The process of measuring the
lengths and directions of lines of a traverse for
the purpose of locating the position of certain
points.
3. Traverse Stations – Any temporary or permanent
point of reference over which the instrument is
set-up. It is usually marked by a peg or a hub
driven flush with the ground with numbers or
letters.
SETTING UP THE TRANSIT
1. Positioning the tripod
2. Mounting the transit
3. Centering the transit (using a plumb
bob or the optical plummet)
4. Leveling the transit
5. Final leveling and centering
TYPES OF TRAVERSE
1. Interior angle traverse – The type of traverse
used principally in land surveys. It is based on
the interior angles measured between the
adjacent sides of a closed figure. The sum of
the interior angles in any polygon must be equal
to (n-2) times 180 degrees. Where n is the
number of sides of the polygon.
TYPES OF TRAVERSE
TYPES OF TRAVERSE
2. Deflection angle traverse – The type of traverse used
frequently for the location survey of roads, railroads,
pipelines, transmission lines, canals, and other similar
types of survey. It is based on the horizontal angle
measured clockwise or counter-clockwise from the
prolongation of the preceding line to the succeeding
line. Such angles vary from 0 to 180 degrees and must
be designated as right (R) or left (L).
For any closed traverse in which the sides do not cross
one another, the summation of the deflection angles,
considering those turned to the left as opposite in sign
to those turned to the right, should be equal to 360
degrees.
TYPES OF TRAVERSE
TYPES OF TRAVERSE
3. Angle to the right traverse – The type of traverse
employed when numerous details are to be
located from the traverse stations. Such type of
traverse is commonly used on city, tunnel, and
mine surveys, and in locating details for
topographic map. It is based on the clockwise
angle measured from the backsight on the back
line to a forward line. For a closed traverse, the
sum of the angles to the right should equal (n+2)
times 180 degrees if the traverse proceeds in a
clockwise direction, and similar to interior
angles if in a counterclockwise direction.
TYPES OF TRAVERSE
TYPES OF TRAVERSE
4. Azimuth traverse – Is by far one of the
quickest and most satisfactory method where
at one set-up of the transit or theodolite
several
angles
or
directions
can
be
determined.
Azimuths
are
measured
clockwise either from the north or south end
of a selected meridian to the line. These
angles may lie anywhere between 0 and 360
degrees.
TYPES OF TRAVERSE
TYPES OF TRAVERSE
TRAVERSE COMPUTATIONS
Computations and Adjustments required for a closed
traverse:
1. Determining latitudes and departures and their
respective algebraic sums.
2. Calculating the total error of closure and the relative
precision of the traverse
3. Computing the respective traverse correction and
balancing the traverse.
4. Determining the adjusted traverse data and adjusted
position of each traverse station.
5. Computing the area
6. Subdividing the area
TRAVERSE COMPUTATIONS
TRAVERSE COMPUTATIONS
Latitude – Is the projection of a line onto the
reference meridian or the north-south line.
Positive (+) for lines with north bearings,
and negative (-) for lines with south
bearings.
Departure – Is the projection of a line onto the
reference parallel or the east-west line.
Positive (+) for lines having easterly
bearings, and negative (-) for lines with
westerly bearings.
Linear Error of Closure (LEC) – Is a short line of
unknown length and direction connecting
the initial and final station of the traverse.
TRAVERSE COMPUTATIONS
TRAVERSE COMPUTATIONS
PROBLEM EXERCISES:
1. Given the following deflection angles of a closed
traverse. Assume bearing of line AB is S 40° E. Compute;
a. Error of closure and values of corrected deflection
angles
b. Azimuth, and Bearing of all the traverse lines
c. All Interior angles, and angles to the right
Sta.
Deflection Angles Sta. Deflection Angles
A
85° 20’ (L)
E
34° 18’ (L)
B
10° 11’ (R)
F
72° 56’ (L)
C
83° 32’ (L)
G
30° 45’ (L)
D
63° 27’ (L)
Solution:
σ 𝑫𝒆𝒇. 𝑨𝒏𝒈𝒍𝒆𝒔 = 85° 20’ - 10° 11’ + 83° 32’ + 63° 27’ + 34° 18’ +
72° 56’ +
30° 45’
= 360⁰ 07’
Error
= 360⁰ - 360⁰ 07’
=
- 0⁰ 07’
𝐄𝐫𝐫𝐨𝐫
− 0⁰ 07’
Correction =
=
𝟕
𝟕
= - 0⁰ 01’
Adjusted Def. Angle A = 85° 20’ - 0⁰ 01’
= 85° 19’ L
Do the same for the rest of the angles and arrange values in
tabulated form.
Tabulation 1
STA.
DEF.
ANGLE
ERROR
CORR.
ADJ. DEF.
ANGLE
A
+85° 20’ L
- 0° 7’
- 0° 1’
+85° 19’ L
B
-10° 11’ R
- 0° 1’
-10° 12’ R
C
+83° 32’ L
- 0° 1’
+83° 31’ L
D
+63° 27’ L
- 0° 1’
+63° 26’ L
E
+34° 18’ L
- 0° 1’
+34° 17’ L
F
+72° 56’ L
- 0° 1’
+72° 55’ L
G
+30° 45’ L
- 0° 1’
+30° 44’ L
SUM
360° 7’
- 0° 1’
360° 0’
Plotting using Deflection Angles
72⁰ 55’ L
G
30⁰ 44’ L
F
N
34⁰ 17’ L
A
85⁰ 19’ L
40⁰
E
B
10⁰ 12’ R
63⁰ 26’ L
D
C
83⁰ 31’ L
STA.
ADJ. DEF. ANGLE
A
+85° 19’ L
B
-10° 12’ R
C
+83° 31’ L
D
+63° 26’ L
E
+34° 17’ L
F
+72° 55’ L
G
+30° 44’ L
SUM
360° 0’
Interior Angles and Angles to the right
72⁰ 55’ L
F
G
30⁰ 44’ L
θG = 149⁰ 16’
A
θF = 107⁰ 05’
34⁰ 17’ L
85⁰ 19’ L
θA = 94⁰ 41’
40⁰
θB = 190⁰ 12’
E
θE = 145⁰ 43’
B
10⁰ 12’ R
63⁰ 26’ L
θD = 116⁰ 34’
θc = 96⁰ 29’
D
C
83⁰ 31’ L
N
Tabulation 2
ADJ. DEF. INTERIOR ANGLES TO
ANGLE
ANGLES THE RIGHT
STA.
DEF.
ANGLE
A
+85° 20’ L
+85° 19’ L
94⁰ 41’
94⁰ 41’
B
-10° 11’ R
-10° 12’ R
190⁰ 12’
190⁰ 12’
C
+83° 32’ L
+83° 31’ L
96⁰ 29’
96⁰ 29’
D
+63° 27’ L
+63° 26’ L
116⁰ 34’
116⁰ 34’
E
+34° 18’ L
+34° 17’ L
145⁰ 43’
145⁰ 43’
F
+72° 56’ L
+72° 55’ L
107⁰ 05’
107⁰ 05’
G
+30° 45’ L
+30° 44’ L
149⁰ 16’
149⁰ 16’
SUM
360° 7’
360° 0’
900⁰ 00’
900⁰ 00’
Bearings
72⁰ 55’ L
F
G
30⁰ 44’ L
S 76⁰ 03’ W
A
S 45⁰ 19’ W
85⁰ 19’ L
34⁰ 17’ L
N 31⁰ 02’ W
S 40⁰ E
E
B
S 29⁰ 48’ E
10⁰ 12’ R
N 03⁰ 15’ E
63⁰ 26’ L
N 66⁰ 41’ E
D
C
83⁰ 31’ L
N
Azimuths
F
G
320⁰
45⁰ 19’
A
S 45⁰ 19’ W
76⁰ 03’
S 76⁰ 03’ W
N 31⁰ 02’ W
S 40⁰ E
E
148⁰ 58’
330⁰ 12’
B
183⁰ 15’
S 29⁰ 48’ E
N 03⁰ 15’ E
N 66⁰ 41’ E
D
246⁰ 41’
C
N
Tabulation 3
LINE
BEARING
AZIMUTH
AB
S 40° 00’ E
320° 00’
BC
S 29° 48’ E
330° 12’
CD
N 66° 41’ E
246° 41’
DE
N 3° 15’ E
183° 15’
EF
N 31° 02’ W
148° 58’
FG
S 76° 03’ W
76° 03’
GA
S 45° 19’ W
45° 19’
PROBLEM EXERCISES:
2. From the given data of a closed traverse, compute the following:
a. Bearing of all sides of the traverse
b. Deflection angles in every stations
c. Interior angles.
d. Linear Error of Closure (LEC), direction of the side of error, and
relative precision
Line
Distance
Azimuth
1-2
189.10 m.`
151° 15’
2-3
139.80
103° 50’
3-4
186.15
36° 58’
4-5
157.05
338° 43
5-1
297.50
251° 12’
Solution: Plotting by Azimuths
3
2
36⁰ 58’
4
338⁰ 43’
251⁰ 12’
5
103⁰ 50’
151⁰ 15’
N
1
Bearings
3
S 36⁰ 58’ W
N 76⁰ 10’ W
2
36⁰ 58’
103⁰ 50’
N 28⁰ 45’ W
4
S 21⁰ 17’ E
251⁰ 12’ 5
338⁰ 43’
N 71⁰ 12’ E
151⁰ 15’
1
N
Tabulation 1
LINE
AZIMUTH
BEARING
1-2
151° 15’
N 28°45’ W
2-3
103° 50’
N 76°10’ W
3-4
36° 58’
S 36°58’ W
4-5
338° 43’
S 21°17’ E
5-1
251° 12’
N 71°12’ E
Deflection Angles
76⁰ 10’
3
66⁰ 52’
S 36⁰ 58’ W
47⁰ 25’
N 76⁰ 10’ W
2
N 28⁰ 45’ W
99⁰ 57’
N 71⁰ 12’ E
4
S 21⁰ 17’ E
1
58⁰ 15’
N 71⁰ 12’ E
5
87⁰ 31’
21⁰ 17’
N
Interior Angles
66⁰ 52’
3
47⁰ 25’
N 76⁰ 10’ W
S 36⁰ 58’ W𝜽3 = 113⁰ 08’
2
𝜽2 = 132⁰ 35’
4
S 21⁰ 17’ E
𝜽4 = 121⁰ 45’
𝜽5 = 92⁰ 29’
58⁰ 15’
N 71⁰ 12’ E
5
87⁰ 31’
N 28⁰ 45’ W
99⁰ 57’
𝜽1 = 80⁰ 03’
1
N
Tabulation 2
LINE
AZIMUTH BEARING
STA.
INTERIOR
DEF. ANGLE ANGLE
1-2
151° 15’
N28°45’ W
1
+99° 57’(L)
80° 03’
2-3
103° 50’
N76°10’ W
2
+47° 25’ (L)
132° 35’
3-4
36° 58’
S36°58’ W
3
+66° 52’ (L)
113° 08’
4-5
338° 43’
S21°17’ E
4
+58° 15’ (L)
121° 45’
5-1
251° 12’
N71°12’ E
5
+87° 31’ (L)
92° 29’
SUM
360° 00’
540° 00’
Tabulation 3
LINE
AZIMUTH BEARING LENGTH, m. LATITUDE DEPARTURE
1-2
151° 15’
N28°45’ W
189.10
+165.789
-90.955
2-3
103° 50’
N76°10’ W
139.80
+33.426
-135.745
3-4
36° 58’
S36°58’ W
186.15
-148.731
-111.941
4-5
338° 43’
S21°17’ E
157.05
-146.339
57.006
5-1
251° 12’
N71°12’ E
297.50
+95.874
281.628
969.60
0.019
- 0.007
SUM
Solution:
CL = + 0.019
CD = - 0.007
LEC =
LINE
AZIMUTH
BEARING
LENGTH,
m.
LATITUDE
DEPARTU
RE
1-2
151° 15’
N28°45’W
189.10
+165.789
-90.955
2-3
103° 50’
N76°10’W
139.80
+33.426
-135.745
3-4
36° 58’
S36°58’ W
186.15
-148.731
-111.941
− (𝐶𝐷 )
− (𝐶𝐿 )
4-5
338° 43’
S 21°17’ E
157.05
-146.339
57.006
− (−0.007)
− (0.019)
5-1
251° 12’
N 71°12’ E
297.50
+95.874
281.628
SUM
969.60
0.019
- 0.007
(𝐶𝐿 )2 +(𝐶𝐷 )2
= (0.019)2 +(0.007)2
LEC = 0.020 m.
Tan θ =
=
= S 20⁰ 13’ 29” E
D = 969.60 m.
𝐿𝐸𝐶
RP =
𝐷
=
0.020
969.6
=
𝟏
𝟒𝟖,𝟓𝟎𝟎
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.
APPROXIMATE METHODS OF TRAVERSE ADJUSTMENT
1.
2.
3.
4.
5.
6.
7.
ARBITRARY METHOD
THE COMPASS RULE
TRANSIT RULE
LEAST SQUARE METHOD
CRANDALL METHOD
GRAPHICAL METHOD
COORDINATE METHOD
1. ARBITRARY METHOD
-The method of traverse adjustment based solely on
the estimation and personal judgment of the
surveyor. Latitudes and departures are adjusted in
a discretionary manner according to the surveyor’s
assessment of the conditions surrounding the
survey.
One traverse line for example is measured over
rugged and difficult terrain, it can be possible that
applying all or most of the correction into this line
will balance the survey satisfactorily without
adjusting the other remaining traverse lines. This
method can be as good, if not better than any of the
conventional methods of adjustment.
2. COMPASS RULE
-This
method is based on the assumption
that all lengths were measured with equal
care
and
all
angles
taken
with
approximately the same precision. It is
also assumed that the errors in the
measurement are accidental and that the
total error in any side of the traverse is
directly proportional to the total length of
the traverse.
COMPASS RULE
cl =
𝑑
CL ( )
𝐷
and
cd =
𝑑
CD ( )
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
CD = total closure in departure or the
algebraic sum of the east and west
departures
d = length of any course
D = total length or perimeter of the traverse
COMPASS RULE
All computed corrections should be added
to check whether their respective sums equal the
closure in latitude and departure. It will be observed
that during the process of adjustment a discrepancy
of 0.001 may result when rounding off computed
values. This imbalance is usually eliminated by
applying an arbitrary correction such as revising one
of the computed corrections.
To determine the adjusted latitude (or departure) of
any course, the sign of the correction for latitude (or
departure) is made opposite the sign of the closure
error in latitude (or departure) and is added to the
computed latitude (or departure) of the course.
ADJUSTED LENGTHS AND DIRECTIONS
L’ =
𝐋𝐚𝐭 ′ 𝟐
+
𝐃𝐞𝐩′ 𝟐 and
Tan α =
𝑫𝒆𝒑′
𝑳𝒂𝒕′
Where:
L’ = adjusted length of course
Lat’ = adjusted latitude of a course
Dep’ = adjusted departure of a course
α = adjusted horizontal angle between the reference
meridian and the course
Note: the adjusted length and bearing of each course
seldom differs significantly from the original
values. It may either be greater or less than the
original values.
Example Problems:
1. Given in the accompanying tabulation are the
observed data for a traverse obtained from a
transit tape survey. Determine the latitudes and
departures of each course, the linear error of
closure, bearing of the side of error and relative
precision. Adjust the traverse using compass
rule and compute also the adjusted lengths and
bearings after applying correction to the
latitudes and departures. Tabulate values
accordingly.
Traverse Data
LINE
DISTANCE, m.
AZIMUTH
AB
495.85
185° 25’
BC
847.62
226° 08’
CD
855.45
292° 21’
DE
1,020.87
347° 30’
EF
1,117.26
83° 52’
FA
660.08
124° 49’
Solution:
292° 21’
C
855.45 m.
847.62 m.
347⁰ 30’
D
N
226⁰ 08’
LINE
DISTANCE, m.
AZIMUTH
AB
495.85
185° 25’
BC
847.62
226° 08’
185⁰ 25’
CD
855.45
292° 21’
A
DE
1,020.87
347° 30’
EF
1,117.26
83° 52’
FA
660.08
124° 49’
B
495.85 m.
1,020.87 m.
660.80 m.
124⁰ 49’
E
1,117.26 m.
F
83⁰ 52’
Solution:
Line
Distance, m.
Azimuth Bearing
AB
495.85
185° 25’
N05° 25’E +493.636 + 46.807
BC
847.62
226° 08’
N46° 08’E +587.386 +611.095
CD
855.45
292° 21’
S67° 39’E -325.296
+791.187
DE 1,020.87
347° 30’
S12° 30’E -996.671
+220.957
EF 1,117.26
83° 52’
S83° 52’W -119.371 -1,110.865
FA
124° 49’
N55° 11’W +376.874
660.08
σ = 4,997.13 m.
σ=
Latitude Departure
-541.915
+ 16.558 + 17.266
Solution cont’d:
CL = + 16.558
CD = + 17.266
LEC =
(𝐶𝐿 )2 +(𝐶𝐷 )2
= (16.558)2 +(17.266)2
LEC = 23.922 m.
Tan θ =
θ=
− (𝐶𝐷 )
− (𝐶𝐿 )
Tan-1
− (17.266)
(
)
− (16.558)
= S 46⁰ 11’ 57” W
D = 4,997.13 m.
𝐿𝐸𝐶
RP =
𝐷
=
23.922
4997.13
=
𝟏
𝟐𝟎𝟎
Solution cont’d:
Correction for latitude, cl =
𝒅
𝟒𝟗𝟓.𝟖𝟓
CL ( ) = +16.558 (
)
𝑫
𝟒,𝟗𝟗𝟕.𝟏𝟑
Correction for latitude, cl = 1.643
Correction for departure, cd =
𝒅
𝟒𝟗𝟓.𝟖𝟓
CD ( ) = +17.266 (
)
𝑫
𝟒,𝟗𝟗𝟕.𝟏𝟑
Correction for departure cd = 1.713
Adjusted latitude, Lat’ = Lat ± cl = +493.636 – 1.643
Adjusted latitude, Lat’ =
+491.993
Adjusted departure, Dep’ = Dep ± cl = +46.807 – 1.713
Adjusted departure, Dep’ =
+45.094
Solution cont’d:
Line Distance
Correction
Correction
Adjusted
for Latitude for Departure Latitude
Adjusted
Departure
AB
495.85
-1.643
-1.713
+491.993 + 45.094
BC
847.62
-2.809
-2.929
+584.577 +608.166
CD
855.45
-2.834
-2.956
- 328.130 +788.231
DE
1,020.87
-3.383
-3.527
-1,000.054 +217.430
EF
1,117.26
-3.702
-3.860
- 123.073 -1,114.725
FA
660.08
-2.187
-2.281
+374.687
-544.196
σ = 4,997.13 -16.558
-17.266
0
0
Solution cont’d:
Line
Distance,m. Adjusted
Adjusted
Adjusted
Adjusted
Latitude
Departure
Length, m. Bearing
AB
495.85
+491.993
+ 45.094
494.055 N05°14’ 13”E
BC
847.62
+584.577
+608.166
843.561 N46° 07’ 59”E
CD
855.45
- 328.130
+788.231
853.802 S67° 23’ 55”E
DE
1,020.87 -1,000.054
EF
1,117.26
- 123.073
-1,114.725 1,121.498 S83°41’ 59”W
FA
660.08
+374.687
-544.196 660.711 N55° 27’07”W
σ=
4,997.13
0
+217.430 1,023.418 S12° 15’ 58”E
0
3. TRANSIT RULE
-This method is based on the assumption that the
total error in any side of the traverse is directly
proportional to the length of the latitude and
departure of the course.
The transit rule may be stated as follows: 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
arithmetic sum (absolute values) of all the latitudes
(or departures) of the traverse.
TRANSIT RULE
cl =
𝐿𝑎𝑡(𝐶𝐿)
σ 𝐿𝑎𝑡
and
cd =
𝐷𝑒𝑝(𝐶𝐷)
σ 𝐷𝑒𝑝
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
CD = total closure in departure or the algebraic
sum of the east and west departures
σ 𝐿𝑎𝑡 = summation of north and south latitude
σ 𝐷𝑒𝑝= summation of east and west departure
Example Problem:
1. Given in the accompanying tabulation are the
observed data for a traverse obtained from a
transit tape survey. Determine the latitudes and
departures of each course, the linear error of
closure, bearing of the side of error and relative
precision. Adjust the traverse using transit rule
and compute also the adjusted lengths and
bearings after applying correction to the
latitudes and departures. Tabulate values
accordingly.
Traverse Data
LINE
DISTANCE, m.
AZIMUTH
AB
495.85
185° 25’
BC
847.62
226° 08’
CD
855.45
292° 21’
DE
1,020.87
347° 30’
EF
1,117.26
83° 52’
FA
660.08
124° 49’
Solution:
Line
Distance, m.
Azimuth Bearing
AB
495.85
185° 25’
N05° 25’E +493.636 + 46.807
BC
847.62
226° 08’
N46° 08’E +587.386 +611.095
CD
855.45
292° 21’
S67° 39’E -325.296
+791.187
DE 1,020.87
347° 30’
S12° 30’E -996.671
+220.957
EF 1,117.26
83° 52’
S83° 52’W -119.371 -1,110.865
FA
124° 49’
N55° 11’W +376.874
660.08
σ = 4,997.13 m.
Latitude Departure
σ=
σ=
-541.915
+ 16.558 + 17.266
2,899.234 3,322.826
Solution cont’d:
CL = + 16.558
CD = + 17.266
LEC =
(𝐶𝐿 )2 +(𝐶𝐷 )2
= (16.558)2 +(17.266)2
LEC = 23.922 m.
Tan θ =
− (𝐶𝐷 )
− (𝐶𝐿 )
− (17.266)
)
− (16.558)
θ = Tan-1 (
= S 46⁰ 11’ 57” W
D = 4,997.13 m.
𝐿𝐸𝐶
RP =
𝐷
=
23.922
4997.13
=
𝟏
𝟐𝟎𝟎
Solution cont’d:
Correction for latitude,cl =
𝑳𝒂𝒕
493.636
CL (σ ) = +16.558 (2,899.234
)
𝑳𝒂𝒕
Correction for latitude, cl = 2.819
Correction for departure,
cd =
𝑫𝒆𝒑
𝟒𝟔.𝟖𝟎𝟕
CD (
) = +17.266 (𝟑,𝟑𝟐𝟐.𝟖𝟐𝟔)
σ 𝑫𝒆𝒑
Correction for departure, cd = 0.243
Adjusted latitude, Lat’ = Lat ± cl = +493.636 – 2.819
Adjusted latitude, Lat’ =
+490.817
Adjusted departure, Dep’ = Dep ± cl = +46.807 – 0.243
Adjusted departure, Dep’ =
+46.564
Solution cont’d:
Line Distance
Correction
Correction
Adjusted
for Latitude for Departure Latitude
Adjusted
Departure
AB
495.85
-2.819
- 0.243
+ 490.817 + 46.564
BC
847.62
-3.355
- 3.175
+ 584.031 + 607.920
CD
855.45
-1.858
- 4.111
- 327.154 + 787.076
DE
1,020.87
-5.692
- 1.148
- 1,002.363 + 219.809
EF
1,117.26
-0.682
- 5.773
- 120.053 -1,116.638
FA
660.08
-2.152
- 2.816
+374.722 - 544.731
σ = 4,997.13 -16.558
-17.266
0
0
Solution cont’d:
Line Adjusted
Adjusted
Adjusted
Adjusted
Distance
Bearing
Latitude
Departure
AB
493.020
N 05° 25’ 10”E
+ 490.817 + 46.564
BC
843.006
N 46° 08’ 43”E
+ 584.031 + 607.920
CD
852.360
S 67° 25’ 46”E
- 327.154 + 787.076
DE
1,026.181 S 12° 22’ 07”E
- 1,002.363 + 219.809
EF
1,123.073 S 83° 51’ 49”W
- 120.053 -1,116.638
FA
661.172
+374.722 - 544.731
σ = 4,997.13
N 55° 28’ 33”W
0
0
Assignment:
1. Given in the accompanying tabulation are the
observed data of a closed traverse. If the azimuth
from north of line 3-4 is 205⁰ 54’, compute the
following:
a. Interior angle, and deflection angle at every
station;
b. bearing and azimuth from north of all the
traverse lines;
c. latitude and departure of each course;
d. linear error of closure, bearing of the side of
error and relative precision;
e. Adjust the traverse using transit rule and
compute also the adjusted lengths and
bearings after applying the correction to the
latitude and departure.
Tabulate all values accordingly.
Traverse Data
Line Distance, m. Azimuth(North)
Angle to the right
1-2
380.50
85° 11’
2-3
519.60
115° 16’
3-4
693.85
4-5
418.00
109° 26’
5-6
403.95
78° 07’
6-1
249.10
242° 12’
205° 54’
89° 48’
4. GRAPHICAL METHOD
-This method is the application of compass rule in
graphical form. It provides a simple graphical
means of making traverse adjustments. In this
method each traverse point is moved in a
direction parallel to the error of closure by an
amount proportional to the distance along the
traverse from the initial point to the given point.
5. RECTANGULAR COORDINATES
-The two horizontal and vertical distances measured
to a point from a pair of mutually perpendicular axes
are referred to as the rectangular coordinates of a
point. All coordinate values are computed from an
origin fixed by the intersection of an x-axis and a yaxis. The x-axis is a reference line which runs along
an east-west direction and the y-axis runs along a
north-south direction. If latitudes and departures have
been computed and adjusted, and if the coordinates
of one point are known, the coordinates of all other
points can be determined by adding successive
departures to the previous X coordinates and
successive latitudes to the previous Y coordinates.
RECTANGULAR COORDINATES
-The two horizontal and vertical distances measured to a
point from a pair of mutually perpendicular axes are referred
to as the rectangular coordinates of a point. All coordinate
values are computed from an origin fixed by the intersection
of an x-axis and a y-axis. The x-axis is a reference line which
runs along an east-west direction and the y-axis runs along a
north-south direction. If latitudes and departures have been
computed and adjusted, and if the coordinates of one point
are known, the coordinates of all other points can be
determined by adding successive departures to the previous
X coordinates and successive latitudes to the previous Y
coordinates.
Figure:
Y – Axis (North – South Line)
X3
C
B
Lat B-C
N
X2
Y3 Lat A-B
X1
Y2
A
Y1
X – Axis (East – West Line)
ORIGIN
Dep A-B
Dep B-C
Example Problem:
1. Given in the accompanying tabulation are the
observed data for a traverse obtained from a transit
tape survey. Determine the latitudes and departures
of each course, the linear error of closure, bearing
of the side of error and relative precision. Adjust the
traverse using compass rule and compute also the
adjusted lengths and bearings after applying
correction to the latitudes and departures. If
coordinates are XA = 6,000.0 m. and YA = 7,000.0 m.,
calculate the coordinates of all the stations.
Tabulate values accordingly.
Traverse Data
LINE
DISTANCE, m.
AZIMUTH
AB
495.85
185° 25’
BC
847.62
226° 08’
CD
855.45
292° 21’
DE
1,020.87
347° 30’
EF
1,117.26
83° 52’
FA
660.08
124° 49’
Plotting:
292° 21’
C
855.45 m.
847.62 m.
347⁰ 30’
D
N
226⁰ 08’
B
LINE
DISTANCE, m.
AZIMUTH
AB
495.85
185° 25’
BC
847.62
226° 08’
185⁰ 25’
CD
855.45
292° 21’
A
DE
1,020.87
347° 30’
EF
1,117.26
83° 52’
FA
660.08
124° 49’
495.85 m.
1,020.87 m.
660.80 m.
124⁰ 49’
E
1,117.26 m.
F
83⁰ 52’
Solution:
Line
Distance, m.
Azimuth Bearing
AB
495.85
185° 25’
N05° 25’E +493.636 + 46.807
BC
847.62
226° 08’
N46° 08’E +587.386 +611.095
CD
855.45
292° 21’
S67° 39’E -325.296
+791.187
DE 1,020.87
347° 30’
S12° 30’E -996.671
+220.957
EF 1,117.26
83° 52’
S83° 52’W -119.371 -1,110.865
FA
124° 49’
N55° 11’W +376.874
660.08
σ = 4,997.13 m.
Latitude Departure
σ=
σ=
-541.915
+ 16.558 + 17.266
2,899.234 3,322.826
Solution:
CL = + 16.558
CD = + 17.266
LEC =
(𝐶𝐿 )2 +(𝐶𝐷 )2
= (16.558)2 +(17.266)2
LEC = 23.922 m.
Tan θ =
− (𝐶𝐷 )
− (𝐶𝐿 )
− (17.266)
)
− (16.558)
θ = Tan-1 (
= S 46⁰ 11’ 57” W
D = 4,997.13 m.
𝐿𝐸𝐶
RP =
𝐷
=
23.922
4997.13
=
𝟏
𝟐𝟎𝟎
Solution:
Correction for latitude,cl =
𝑳𝒂𝒕
493.636
CL (σ ) = +16.558 (2,899.234
)
𝑳𝒂𝒕
Correction for latitude, cl = 2.819
Correction for departure,
cd =
𝑫𝒆𝒑
𝟒𝟔.𝟖𝟎𝟕
CD (
) = +17.266 (𝟑,𝟑𝟐𝟐.𝟖𝟐𝟔)
σ 𝑫𝒆𝒑
Correction for departure, cd = 0.243
Adjusted latitude, Lat’ = Lat ± cl = +493.636 – 2.819
Adjusted latitude, Lat’ =
+490.817
Adjusted departure, Dep’ = Dep ± cl = +46.807 – 0.243
Adjusted departure, Dep’ =
+46.564
Solution:
Line Distance
Correction
Correction
Adjusted
for Latitude for Departure Latitude
Adjusted
Departure
AB
495.85
-2.819
- 0.243
+ 490.817 + 46.564
BC
847.62
-3.355
- 3.175
+ 584.031 + 607.920
CD
855.45
-1.858
- 4.111
- 327.154 + 787.076
DE
1,020.87
-5.692
- 1.148
- 1,002.363 + 219.809
EF
1,117.26
-0.682
- 5.773
- 120.053 -1,116.638
FA
660.08
-2.152
- 2.816
+374.722 - 544.731
σ = 4,997.13 -16.558
-17.266
0
0
Solution:
Line Adjusted
Adjusted
Adjusted
Adjusted
Distance
Bearing
Latitude
Departure
AB
493.020
N 05° 25’ 10”E
+ 490.817 + 46.564
BC
843.006
N 46° 08’ 43”E
+ 584.031 + 607.920
CD
852.360
S 67° 25’ 46”E
- 327.154 + 787.076
DE
1,026.181 S 12° 22’ 07”E
- 1,002.363 + 219.809
EF
1,123.073 S 83° 51’ 49”W
- 120.053 -1,116.638
FA
661.172
+374.722 - 544.731
σ = 4,997.13
N 55° 28’ 33”W
0
0
Solution:
Line
Adjusted
Adjusted
Latitude
Departure
Sta.
X
Y
Coordinates Coordinates
AB
+ 490.817
+ 46.564
A
6,000.000
7,000.000
BC
+ 584.031
+ 607.920
B
6,046.564
7,490.817
CD
- 327.154
+ 787.076
C
6,654.484
8,074.848
DE
- 1,002.363 + 219.809
D
7,441.560
7,747.694
EF
- 120.053
-1,116.638
E
7,661.369
6,745.331
FA
+374.722
- 544.731
F
6,544.731
6,625.278
0
0
A
6,000.000
7,000.000
σ=