MMJ14203
GEOMATICS ENGINEERING
MRS SITI KAMARIAH MD SA’AT
FKTM
UniMAP
TRAVERSING
TYPES OF COORDINATE
• It is split into four quadrants with the typical mathematical convention of the axis
to the north and east being positive and to the south and west, negative.
• The x-axis is referred to as the east-axis (E) and the y-axis as the north-axis
(N), with angles (a) measured clockwise from the N-axis, (E,N)
Rectangular Coordinate
• Polar coordinate used distance and whole circle bearing of the line
• Used to define the relative position of one point to another
• (d,β)
Polar Coordinate
RECTANGULAR COORDINATES
E=EB-EA
EA
NA
(EA,NA)
EB
N
B
(EB,NB)
Point A
Point B
North
East
N=NB-NA
RECTANGULAR COORDINATES
POLAR COORDINATES

d
North
East
Point A
Point B
 ~ whole-circle bearing/Azimuth
d ~ distance
Computing the polar for a line involves calculating E and N given the
horizontal distance (d) and WCB (β) of the line.
POLAR COORDINATES
COORDINATE CONVERSIONS
2
2
1
N
E
d
N
E
tan













 






cos
d
N
sin
d
E
Rectangular to polar Polar to rectangular
d

E
N d

E
N
EXAMPLE 6.1:
 Given the coordinates of A and the distance and bearing of
AB, calculate the coordinates of B.
EA = 48 964.38 m, NA= 69 866.75 m,
WCB AB = 299◦58’46”
Horizontal distance = 1325.64 m
Solution:
 EB= EA+ ∆EAB = 48 964.38 + dsinβ
= 48 964.38 + 1325.64 sin 299◦58’46”
=
 NB= NA+ ∆NAB = 69 866.75 + dcosβ
= 69 866.75 + 1325.64 cos 299◦58’46”
=
EXAMPLE 6.2:
 Given the following coordinates for two points A and B,
compute the length and bearing of AB.
EA = 48 964.38 m, NA = 69 866.75 m
EB= 48 988.66 m , NB= 62 583.18 m
Answer:
dAB = 7283.61 m
β= 179◦48’33”
2
2
1
N
E
d
N
E
tan













 
DEFINITIONS
Traverse
• Series of straight lines connecting
survey stations (begin at known points
as baseline)
Traversing
• Determination of horizontal coordinates
by measuring horizontal angles &
distances
WHAT IS A
TRAVERSE?
 Control survey
 A series of established stations tied
together by angle and distance.
 The angles are measured using
theodolites/total station/compass, while
distances can be measured using total
stations or tapes or EDM.
WHAT IS A TRAVERSE?
 A polygon of 2D (or 3D) vectors
 Sides are expressed as either polar coordinates (,d) or as
rectangular coordinate differences (E,N)
 A traverse must either close on itself
 Or be measured between points with known rectangular
coordinates.
TYPES OF TRAVERSES
 Open Traverse: Use as pipeline, highways, railways, etc.
 Closed traverse: Use to locate lakes, land boundaries and control mapping
A closed
traverse
A traverse between
known points
OPEN TRAVERSE
CLOSED TRAVERSE
WHY TRAVERSING?
APPLICATIONS OF TRAVERSING
 Establishing coordinates for new points
(

,
d
)
(,d)
(
,d)
(E,N)new
(E,N)new
(E,N)known
(E,N)known
APPLICATIONS OF TRAVERSING
 These new points can then be used as a framework for
mapping existing features
(

,
d
)
(
,d)
(,d)
(,d) (,d)
(

,
d
)
(E,N)new
(E,N)new
(E,N)new
(E,N)new
(E,N)new
(E,N)known
(E,N)known
APPLICATIONS OF TRAVERSING
 They can also be used as a basis for setting out new work
(E,N)new
(E,N)new
(E,N)known
(E,N)known
EQUIPMENT
 Traversing requires :
 An instrument to measure angles
(theodolite) or bearings (magnetic
compass)
 An instrument to measure distances
(EDM or tape)
EQUIPMENT
 Or use of total station that can measure both distance and angles
OBSERVATION OF TRAVERSE
 The methods used in observing angles and direction:
 Interior Angles
 To reduce mistakes in reading, recording and computing, always turned clockwise from
backsight station to foresight station
 Angles to the right
 To avoid ambiguity in angles to right, forward traverse station must be established.
 Deflection Angles
 For route survey
 Azimuth
 By total station, reading azimuth at all lines and thus eliminate the need to calculate
them.
OBSERVATION OF TRAVERSE
 Observation of traverse length
 By simplest and economical ways
 High precision
 Use stationing
CHOOSING LOCATION OF
TRAVERSE STATIONS
Some practical guidelines:
1. Minimum no. of stations (each line of sight as long as
possible)
2. Ensure: adjacent stations always inter-visible
3. Avoid acute traverse angles
4. Stable & safe ground conditions for instrument
5. Marked with paint or/and nail; to survive subsequent traffic,
construction, weather conditions, etc.
CHOOSING LOCATION OF TRAVERSE
STATIONS
6. Include existing stations/reference objects for checking
with known values
7. Traverse must not cross itself
8. Network formed by stations (if any): as simple as
possible
9. Do the above without sacrificing accuracy or omitting
important details
TRAVERSE FIELD NOTES
COMPUTATION SEQUENCE
1. Compute (E, N) or (X, Y) for each traverse line
2. Calculate linear misclose
3. Calculate accuracy and precision
4. Adjust linear misclose.
COMPUTATION OF LATITUDES AND
DEPARTURES
 Latitude-north/south rectangular component of line (North +;South -)
Latitude (ΔN) = distance(H) cos α
 Departure-east/west rectangular component of line (East +;West -)
Departure (ΔE) = distance(H) sin α
Where:
α = bearing or azimuth of the traverse course
H = the horizontal distance of the traverse course
LOCATION OF A POINT
CLOSURE OF LATITUDES AND DEPARTURES
(E,N) FOR EACH LINE
 The rectangular components for each line are computed from the polar
coordinates (,d)
 Note that these formulae apply regardless of the quadrant so long as
whole circle bearings are used






cos
d
N
sin
d
E
LATITUDE / DEPARTURE COMPUTATIONS
EXAMPLE: OBSERVATION SEQUENCE
7
7
.
1
9
99.92
6
0
.
6
3
129.76
3
2
.
2
0
A
B
C
D
E
2
3
o
53
o
1
6
8
o
227
o
3
0
1
o
1
2
1
o
Start at A
EXAMPLE: VECTOR COMPONENTS
Line Bearing, β Distance, d N
(Lat=
dcosβ)
E
(Dep=
dsinβ)
AB 23o
77.19 71.05 30.16
BC 53o
99.92 60.13 79.80
CD 168o
60.63
-
59.31 12.61
DE 227o
129.76
-
88.50
-
94.90
EA 301o
32.20 16.58
-
27.60

(399.70)
(-
0.05) (0.07)
CLOSURE ERROR AND CLOSURE
CORRECTION
LINEAR MISCLOSE &
ACCURACY
 Convert the rectangular misclose components to polar coordinates
 Precision is given by
2
2
1
N
E
d
N
E
tan













 
)
misclose
linear
/
length
traverse
(
:
1
Beware of quadrant when
calculating  using tan-1
Linear misclosure, E =
Precision= 1/(total distance/misclosure)
N
E


b positive
add 180
o
QUADRANTS AND TAN FUNCTION

+
b negative
add 180o
+

b negative
add 360
o
+
+
b positive
okay
FOR THE EXAMPLE…
 Misclose (E, N)
 (0.07, -0.05)
 Convert to polar (,d)
  = -54.46o
(2nd
quadrant) = 125.53o
 d = 0.09 m
 Accuracy
 1:(399.70 / 0.09) = 1:4441
COMPASS RULE (BOWDITCH METHOD)
– DISTRIBUTES THE ERRORS IN LAT/DEP.
Where:
C lat AB = correction in latitude AB
∑ lat = error of closure in latitude
AB = distance AB
P = perimeter of traverse
Where:
C dep AB = correction in departure AB
∑ lat = error of closure in departure
AB = distance AB
P = perimeter of traverse
C dep AB = AB
Σ dep P
C lat AB= AB
Σ lat P
THE EXAMPLE…
 East misclose 0.07 m
 North misclose –0.05 m
 Side AB 77.19 m
 Side BC 99.92 m
 Side CD 60.63 m
 Side DE 129.76 m
 Side EA 32.20 m
 Total perimeter 399.70 m
VECTOR COMPONENTS (PRE-ADJUSTMENT)
Side N E d CN CE Nadj Eadj
EA
71.05 30.16
77.19
AB
60.13 79.80
99.92
BC
-59.31 12.61
60.63
CD
-88.50 -94.90
129.76
DE
16.58 -27.60
32.20
Misc
(-0.05) (0.07)
399.7
THE ADJUSTMENT COMPONENTS
Side N E CN CE Nadj Eadj
1A
71.05 30.16
-
0.010 0.014
AB
60.13 79.80
-
0.012 0.016
BC
-
59.31 12.61
-
0.008 0.011
CD
-
88.50
-
94.90
-
0.016 0.023
D1
16.58
-
27.60
-
0.004 0.006
Misc
(-
0.05) (0.07) (-0.050) (0.070)
C lat AB= AB X Σ lat
P
C dep AB= AB X Σ dep
P
ADJUSTED VECTOR COMPONENTS
Side N E CN CE Nadj Eadj
1A
71.05 30.16
-
0.010 0.014 71.060 30.146
AB
60.13 79.80
-
0.012 0.016 60.142 79.784
BC
-
59.31 12.61
-
0.008 0.011
-
59.302 12.599
CD
-
88.50
-
94.90
-
0.016 0.023
-
88.484
-
94.923
D1
16.58
-
27.60
-
0.004 0.006 16.584
-
27.606
Misc
(-0.05) (0.07)
-
0.050 0.070 (0.000) (0.000)
SUMMARY OF INITIAL TRAVERSE
COMPUTATION
Balance the angle
1
Compute the bearing
or azimuth
2
Compute the latitude
and departure, the
linear error of closure,
and the precision ratio
of the traverse
3
SOURCES OF ERROR IN TRAVERSING
Poor selection of station resulting in bad sighting condition cause by
• shadow,
• Line of sight passing too close to the ground
• Lines are too short
• Sighting into the sun
Error in observations of angles and distances
Failed to observed angles an equal times direct and reversed
MISTAKES IN TRAVERSING
Occupying or
sighting the
wrong stations
Incorrect
orientation
Confusing
angles to the
right and left
Mistakes in note
taking
Misidentification
of station
QUIZIZZ
Let’s know your understanding to the topics
THANK YOU

Traversing lecture notes for geomatics engineering

  • 1.
    MMJ14203 GEOMATICS ENGINEERING MRS SITIKAMARIAH MD SA’AT FKTM UniMAP
  • 2.
  • 3.
    TYPES OF COORDINATE •It is split into four quadrants with the typical mathematical convention of the axis to the north and east being positive and to the south and west, negative. • The x-axis is referred to as the east-axis (E) and the y-axis as the north-axis (N), with angles (a) measured clockwise from the N-axis, (E,N) Rectangular Coordinate • Polar coordinate used distance and whole circle bearing of the line • Used to define the relative position of one point to another • (d,β) Polar Coordinate
  • 4.
  • 5.
  • 6.
    POLAR COORDINATES  d North East Point A PointB  ~ whole-circle bearing/Azimuth d ~ distance Computing the polar for a line involves calculating E and N given the horizontal distance (d) and WCB (β) of the line.
  • 7.
  • 8.
  • 9.
    EXAMPLE 6.1:  Giventhe coordinates of A and the distance and bearing of AB, calculate the coordinates of B. EA = 48 964.38 m, NA= 69 866.75 m, WCB AB = 299◦58’46” Horizontal distance = 1325.64 m Solution:  EB= EA+ ∆EAB = 48 964.38 + dsinβ = 48 964.38 + 1325.64 sin 299◦58’46” =  NB= NA+ ∆NAB = 69 866.75 + dcosβ = 69 866.75 + 1325.64 cos 299◦58’46” =
  • 10.
    EXAMPLE 6.2:  Giventhe following coordinates for two points A and B, compute the length and bearing of AB. EA = 48 964.38 m, NA = 69 866.75 m EB= 48 988.66 m , NB= 62 583.18 m Answer: dAB = 7283.61 m β= 179◦48’33” 2 2 1 N E d N E tan               
  • 11.
    DEFINITIONS Traverse • Series ofstraight lines connecting survey stations (begin at known points as baseline) Traversing • Determination of horizontal coordinates by measuring horizontal angles & distances
  • 12.
    WHAT IS A TRAVERSE? Control survey  A series of established stations tied together by angle and distance.  The angles are measured using theodolites/total station/compass, while distances can be measured using total stations or tapes or EDM.
  • 13.
    WHAT IS ATRAVERSE?  A polygon of 2D (or 3D) vectors  Sides are expressed as either polar coordinates (,d) or as rectangular coordinate differences (E,N)  A traverse must either close on itself  Or be measured between points with known rectangular coordinates.
  • 14.
    TYPES OF TRAVERSES Open Traverse: Use as pipeline, highways, railways, etc.  Closed traverse: Use to locate lakes, land boundaries and control mapping A closed traverse A traverse between known points
  • 15.
  • 16.
  • 17.
  • 18.
    APPLICATIONS OF TRAVERSING Establishing coordinates for new points (  , d ) (,d) ( ,d) (E,N)new (E,N)new (E,N)known (E,N)known
  • 19.
    APPLICATIONS OF TRAVERSING These new points can then be used as a framework for mapping existing features (  , d ) ( ,d) (,d) (,d) (,d) (  , d ) (E,N)new (E,N)new (E,N)new (E,N)new (E,N)new (E,N)known (E,N)known
  • 20.
    APPLICATIONS OF TRAVERSING They can also be used as a basis for setting out new work (E,N)new (E,N)new (E,N)known (E,N)known
  • 21.
    EQUIPMENT  Traversing requires:  An instrument to measure angles (theodolite) or bearings (magnetic compass)  An instrument to measure distances (EDM or tape)
  • 22.
    EQUIPMENT  Or useof total station that can measure both distance and angles
  • 23.
    OBSERVATION OF TRAVERSE The methods used in observing angles and direction:  Interior Angles  To reduce mistakes in reading, recording and computing, always turned clockwise from backsight station to foresight station  Angles to the right  To avoid ambiguity in angles to right, forward traverse station must be established.  Deflection Angles  For route survey  Azimuth  By total station, reading azimuth at all lines and thus eliminate the need to calculate them.
  • 24.
    OBSERVATION OF TRAVERSE Observation of traverse length  By simplest and economical ways  High precision  Use stationing
  • 25.
    CHOOSING LOCATION OF TRAVERSESTATIONS Some practical guidelines: 1. Minimum no. of stations (each line of sight as long as possible) 2. Ensure: adjacent stations always inter-visible 3. Avoid acute traverse angles 4. Stable & safe ground conditions for instrument 5. Marked with paint or/and nail; to survive subsequent traffic, construction, weather conditions, etc.
  • 26.
    CHOOSING LOCATION OFTRAVERSE STATIONS 6. Include existing stations/reference objects for checking with known values 7. Traverse must not cross itself 8. Network formed by stations (if any): as simple as possible 9. Do the above without sacrificing accuracy or omitting important details
  • 27.
  • 28.
    COMPUTATION SEQUENCE 1. Compute(E, N) or (X, Y) for each traverse line 2. Calculate linear misclose 3. Calculate accuracy and precision 4. Adjust linear misclose.
  • 29.
    COMPUTATION OF LATITUDESAND DEPARTURES  Latitude-north/south rectangular component of line (North +;South -) Latitude (ΔN) = distance(H) cos α  Departure-east/west rectangular component of line (East +;West -) Departure (ΔE) = distance(H) sin α Where: α = bearing or azimuth of the traverse course H = the horizontal distance of the traverse course
  • 30.
  • 31.
    CLOSURE OF LATITUDESAND DEPARTURES
  • 32.
    (E,N) FOR EACHLINE  The rectangular components for each line are computed from the polar coordinates (,d)  Note that these formulae apply regardless of the quadrant so long as whole circle bearings are used       cos d N sin d E
  • 33.
  • 34.
  • 35.
    EXAMPLE: VECTOR COMPONENTS LineBearing, β Distance, d N (Lat= dcosβ) E (Dep= dsinβ) AB 23o 77.19 71.05 30.16 BC 53o 99.92 60.13 79.80 CD 168o 60.63 - 59.31 12.61 DE 227o 129.76 - 88.50 - 94.90 EA 301o 32.20 16.58 - 27.60  (399.70) (- 0.05) (0.07)
  • 36.
    CLOSURE ERROR ANDCLOSURE CORRECTION
  • 37.
    LINEAR MISCLOSE & ACCURACY Convert the rectangular misclose components to polar coordinates  Precision is given by 2 2 1 N E d N E tan                ) misclose linear / length traverse ( : 1 Beware of quadrant when calculating  using tan-1 Linear misclosure, E = Precision= 1/(total distance/misclosure)
  • 38.
    N E   b positive add 180 o QUADRANTSAND TAN FUNCTION  + b negative add 180o +  b negative add 360 o + + b positive okay
  • 39.
    FOR THE EXAMPLE… Misclose (E, N)  (0.07, -0.05)  Convert to polar (,d)   = -54.46o (2nd quadrant) = 125.53o  d = 0.09 m  Accuracy  1:(399.70 / 0.09) = 1:4441
  • 40.
    COMPASS RULE (BOWDITCHMETHOD) – DISTRIBUTES THE ERRORS IN LAT/DEP. Where: C lat AB = correction in latitude AB ∑ lat = error of closure in latitude AB = distance AB P = perimeter of traverse Where: C dep AB = correction in departure AB ∑ lat = error of closure in departure AB = distance AB P = perimeter of traverse C dep AB = AB Σ dep P C lat AB= AB Σ lat P
  • 41.
    THE EXAMPLE…  Eastmisclose 0.07 m  North misclose –0.05 m  Side AB 77.19 m  Side BC 99.92 m  Side CD 60.63 m  Side DE 129.76 m  Side EA 32.20 m  Total perimeter 399.70 m
  • 42.
    VECTOR COMPONENTS (PRE-ADJUSTMENT) SideN E d CN CE Nadj Eadj EA 71.05 30.16 77.19 AB 60.13 79.80 99.92 BC -59.31 12.61 60.63 CD -88.50 -94.90 129.76 DE 16.58 -27.60 32.20 Misc (-0.05) (0.07) 399.7
  • 43.
    THE ADJUSTMENT COMPONENTS SideN E CN CE Nadj Eadj 1A 71.05 30.16 - 0.010 0.014 AB 60.13 79.80 - 0.012 0.016 BC - 59.31 12.61 - 0.008 0.011 CD - 88.50 - 94.90 - 0.016 0.023 D1 16.58 - 27.60 - 0.004 0.006 Misc (- 0.05) (0.07) (-0.050) (0.070) C lat AB= AB X Σ lat P C dep AB= AB X Σ dep P
  • 44.
    ADJUSTED VECTOR COMPONENTS SideN E CN CE Nadj Eadj 1A 71.05 30.16 - 0.010 0.014 71.060 30.146 AB 60.13 79.80 - 0.012 0.016 60.142 79.784 BC - 59.31 12.61 - 0.008 0.011 - 59.302 12.599 CD - 88.50 - 94.90 - 0.016 0.023 - 88.484 - 94.923 D1 16.58 - 27.60 - 0.004 0.006 16.584 - 27.606 Misc (-0.05) (0.07) - 0.050 0.070 (0.000) (0.000)
  • 45.
    SUMMARY OF INITIALTRAVERSE COMPUTATION Balance the angle 1 Compute the bearing or azimuth 2 Compute the latitude and departure, the linear error of closure, and the precision ratio of the traverse 3
  • 46.
    SOURCES OF ERRORIN TRAVERSING Poor selection of station resulting in bad sighting condition cause by • shadow, • Line of sight passing too close to the ground • Lines are too short • Sighting into the sun Error in observations of angles and distances Failed to observed angles an equal times direct and reversed
  • 47.
    MISTAKES IN TRAVERSING Occupyingor sighting the wrong stations Incorrect orientation Confusing angles to the right and left Mistakes in note taking Misidentification of station
  • 48.
    QUIZIZZ Let’s know yourunderstanding to the topics
  • 49.