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AAIT, Department of Civil Engineering

- 1 - Lecture Note:- Surveying I

CHAPTER Four

TRAVERSING PRINCIPLES

4.1. Introduction

A traverse consists of a series of straight lines connecting successive points. The points

defining the ends of the traverse lines are called traverse stations or traverse points.

Distance along the line between successive traverse points is determined either by

direct measurement using a tape or electronic distance measuring (EDM) equipment, or by

indirect measurement using tachometric methods. At each point where the traverse changes

direction, an angular measurement is taken using a theodolite.

Traverse party: it usually consists of an instrument operator, a head tape man and rare tape

man.

Equipments for the traverse party:-The equipments for the traverse party are the

theodolite, tapes, hand level, leveling staff, ranging pole & plumb bobs, EDM & reflector,

stakes & hubs, tacks, marking crayon, points, walkie talkies, & hammer etc.

Purpose of traverse: It is a convenient, rapid method for establishing horizontal control

particularly when the lines of sights are short due to heavily built up areas where

triangulation and trilateration are not applicable. The purpose includes:

- Property surveys to locate or establish boundaries;

- Supplementary horizontal control for topographic mapping surveys;

- Location and construction layout surveys for high ways, railway, and other private

and public works;

- Ground control surveys for photogrammetric mapping.

CA

BD



4.2 Traversing by compass and theodolite

4.2.1. Types of traverse

1. Open traverse: It starts at a point of known position and terminates at a point of

unknown position.

- It is not possible to check the consistence of angles and distance measurement.

- To minimize errors, distances can be measured twice, angles turned by repetition,

magnetic bearings observed on all lines and astronomic observations made

periodically (not done in engineering works).

2. Closed traverse:- It originates at a point of known position and close on another point

of known horizontal position .

This type of traverse is preferable to all others since computational checks are possible

which allow detection of systematic errors in both distance and direction.

Traverses also categorized on the method of observing angles.

X

CA

BD

D

B

Closed link traverse

X

CA Closed loop traverse

X C

A B

D



4.2.2 Compass Traverse

When compasses are used to run traverses, forward and back bearings are observed from

each traverse station and distances are taped. If local attraction exists at any traverse station,

both the forward and back bearings are affected equally. Thus interior angles computed from

forward and back bearings are independent of local attraction. Since these angles are

independent of local attraction, the sum of these interior angles provides a legitimate

indication of the angular error in the traverse.

Assuming that all bearings are of equal precision and non-correlated, this error is distributed

equally among the number of interior angles. Since none of the traverse lines has an absolute

direction that is known to be correct, it is necessary to select a line affected least by local

attraction.

Exercise:

The following are bearings taken on a closed compass traverse. Compute the interior angle

and correct them.

Assuming the observed bearing of XY to be clear of local attraction and adjust the bearing

of all other sides.

Line Forward bearing Backward bearing

XY S 27030’E N 27030’W

YC S 45015’ W N 44015’ E

CD N 730 00’ W S 720 15’ W

DE N 120 45’ E S 130 15’ W

EX N 60000’ E S 590 00’ W



4.2.3 Interior angle traverse:

Interior angle traverse is the one that is employed for closed loop traverse. Successive

stations occupied and back sight is taken to the preceding station with horizontal circle set

zero. The instrument is then turned on its upper motion until the next station is

bisected/sighted and the interior angle is observed. The horizontal circle reading gives the

interior angle in the clockwise direction. Horizontal distances are determined by stadia and

angles should be observed twice by double sighting.

Azimuth of a line =360-(back azimuth of preceding line + Clockwise interior angle).

In closed figure

n

i

ni1

180)2(

n is the number of stations

The error of closure can be distributed to all angles equally assuming that all observations are

made with equal precision.

Example:

A clockwise interior angle in a closed traverse is as follows

A= 84058’, B=157038’, C=24037’

D= 1530 14’ , E=1030 54’, F= 139’ 06’ G= 2360 49’

Compute the error of closure and adjust the interior angle.

A3

12

4



Solution:

Station Observed

Interior angleCorrection

Adjusted

interior Angle

A 840 58’ -00 2’ 84056’

B 157038’ -0 02’ 157036’

C 240 37’ -0 02’ 24 35’

D 1530 14’ -0 02’ 153 12’

D 103 054’ -0 02’ 103 52’

F 139 006’ -0 02’ 139 04’

G 2360 47’ -0 02’ 236 45’

Sum 9000 14’ 00” 900000’00’

(n-2)1800 9000 00 ‘ 00”

Error of

closure 00 14’00”

Exercise:

Calculate azimuth and bearing of all lines for Azimuth of AB Az AB=315012’

4.2.4 Deflection angle traverses

This method of running traverses is widely employed than the other especially on open

traverses. It is mostly common in location of routes, canals, roads, highways, pipe lines, etc.

Successive traverse stations are occupied with a theodolite with horizontal circle set at zero

and back sight taken to the preceding station with a telescope reverse. The telescope is then

plunged and the line of sight is directed to the next station, by turning the instrument about

the vertical axis on its upper motion and the deflection angle is observed. Angles have to be

observed by double sighting.

Azimuth of line =Azimuth of preceding line + R

Azimuth of line =Azimuth of preceding line - L



In the above figure the azimuth of line AX and DY and are used to check the angular

closure for the traverse

AXA + 2 + 4-1-3-3600=ADY

2

1 1

01 360 AA

n

i

n

iLiiR

Where A1=Azimuth of starting station

A2=Azimuth of closing station

R= deflection to the right

L= deflection to the left.

The angular error of closure can be computed and the adjustment of the observed angles is

made assuming equal weights for all angles, the error of closure may be distributed equally

among the deflection angles.

Example:

The following are deflection angles observed in a closed loop traverse.

A=850 20’ L; B=100 11’ R; C= 830 32’ L; D= 630 27’ L

E= 340 18’ L; F= 720 56’ L; G= 300 45’ L

Compute the error of closure and adjust the deflection angles assuming that all observations

are made with equal precision.

4

3

2

1

Y

X

C

A

B D



Solution:

A1=A2 (for closed loop)

For closed loop traverse

0360LiRi

For the given traverse

L =370018’ R =10011”

10011’-370018’=-360007’

360007’-360000’=0.07’

Correction per angle= "00'0107

'070 00

This correction angle is added to deflection to the right and subtracted to deflection to the

left.

Station From/To Circle

observed

Deflection

angle

Correction Corrected

deflection

A G 0000’

B 85020’ 85020’L -0001’ 85019’

B A 0000’

C 10011’ 10011’R +0001’ 10012’

C B 0000’

D

D C

E

E D

F

F E

G

G F

A

0360LiRi



Exercise:

Azimuth of line AB is given as Az AB= 85024’.Calculate the azimuth and bearing of all

other lines.

4.2.5 Angle to the right traverse

This method can be used in open, closed, or closed loop traverses. Successive theodolite

stations are occupied and back sight is taken to the preceding station with the horizontal

circle set zero. Then foresight is taken on the next station using the upper motion in the

clockwise direction. The reading gives the angle to the right at the station and angles should

be observed by double sighting.

Azimuth a line= angle to the right - Back azimuth of preceding line.

Error of closure

YX AA 443211 180)14(

The condition of closure can be expressed by

A1+1+2+ -----n-(n-1)180-A2=0

Where A1 & A2 are Azimuths of the starting and closing stations.

n=no of traverse stations (exclusive of fixed stations).

Any misclosure can be distributed equally to all angles assuming equal precision.

Exercise:

Try the above example in 5.2.4 (Deflection angle traverse.)

1

4

4

3

2

X

3 1

2 Y



4.2.6. Azimuth traverse

This method is used extensively on topographic and other surveys where a large number of

details are located by angular and linear measurements from the traverse stations. Successive

stations are occupied, beginning with the line of known or assumed azimuth. At each station

the theodolite is oriented by setting the horizontal circle index to read the back azimuth (fore

ward azimuth 1800) of the preceding line, and then back sighting to the preceding traverse

station. The instrument is then turned on the upper motion, and a foresight on the following

traverse station is taken. The reading indicated by the horizontal circle on the clockwise

circle is the azimuth of the forward line.

Any angular error of closure of a traverse becomes evident by the difference between initial

and final observations taken along the first line.

4.2.7 Stadia traverse

In stadia traverse the horizontal distance between traverse station is determined by stadia

method. The stadia traverse is sufficiently accurate and considerably more rapid and

economical than corresponding surveys made with theodolite and tape. Its advantage is that

elevations can be determined concurrently with horizontal position.

L1= distance BC observed at station B

L2= distance BC observed at station C

2

21 LLLSo BC

4.2.8 Plane table and alidade

Traversing with the plane table involves the same principle as running a traverse with a

theodolite. Successive plane table and alidade stations are occupied, the table is oriented and

back sight on the preceding station is taken. A fore sight is then taken to the next station and

its location is plotted on the plane table sheet. Distances and difference in elevations are

determine by stadia using the alidade and scaled off on the paper. Check lines can also

provided for checking the consistency.



4.3 Traverse Computations

Field operation for traverses yields angles or directions and distance for a set of lines

connecting a series of traverse stations. Angles can be checked for error of closure and

corrected so that preliminary corrected values can be computed. And observed distances can

be reduced to equivalent horizontal distanced. The preliminary directions and reduced

distances are suitable for use in traverse computations, which are performed in a plane

rectangular coordinate system.

Computation with plane coordinates by considering the figure below.

Let the reduced horizontal distance of traverse lines ij and jk be dij and djk respectively, and

Ai and Aj be the azimuths of ij and jk. Let Xij and Yij be the departure & latitude.

Xij= dij sin Ai = departure

Yij = dij cos Ai =latitude

If the coordinates of i are xi and yi

So, the coordinates of j are:

Y

X

djk

i k

yij

xij

dij

Aj

Aj j

xij

xj

xk

yi

yi yk

yij



Xj=xi+xij ; yj= yi+yij

Xk=xj+xjk ; yk=yj-yjk

=xi+xij+xik ; =yi+yij-yjk

xjk =djk sin Aj yjk=djk cos Aj

Note: the signs of azimuth functions

If the coordinates for the two ends of a traverse line are given, distance between two ends

can be determined as:

dij =[(xj-xi)2+(yj-yi)2]1/2

The azimuth of line ij from north and south is

yjyi

xjxiAsij

yiyj

xixjAij

11 tantan

After coordinates for all the traverse points (all the departure and latitudes) for all lines have

been computed, a check is necessary on the accuracy of the observations and the validity of

calculations. In a closed traverse, the algebraic sum of the departures should equal the

difference between the x- coordinates at the beginning and ending stations of the traverse.

Similarly, the algebraic sum of the latitudes should equal the difference between the y

coordinates at the beginning and ending stations.

In a closed loop traverse, the algebraic sum of the latitudes and the algebraic sum of the

departures each must equal zero.

For a traverse containing n stations starting at i=1 and ending at station i=n, the foregoing

conditions can be expressed as follows.

III Sin - Cos -

II sin+ Cos -

I Sin + Cos +

IV Sin - Cos +

EW

N

S



latitudesiyiYY

departuresixiXX

n

i

n

in

n

i

n

in

1

1

1

11

1

1

1

11

1,

1,

The amounts by which the above equation will fail to be satisfied are called simply closures.

The closure correction in departure dx and dy, which are of opposite signs to errors, are:

1,

1,

1

11

1

11

iYiYYd

iXiXXd

n

iny

n

inx

And for closed loop traverse dx=- departures and dy=- latitudes.

4.4 Sources of errors and precision of traversing

When traversing, errors may arise in the following measurements.

a. Linear measurements: Errors in linear measurement are those as systematic,

and random errors. These errors in linear measurements should by corrected

considering its type.

b. Angular measurements: There are two main sources of errors in the

measurement of the traverse angle.

i) Observational errors:

Due to lateral refraction, haze and wind the line of sight may not be truly

straight. It is there fore important to keep the line of sight 1m above ground on

hot sunny days. In wind and haze no need to attempt at accurate readings.

If the signal is too large it is not possible to bisect accurately, and if the signal is

not plumbed vertically above the station mark, wrong direction will be observed.

There fore try and always observe directly on the station mark. If this is not

possible, sight to a plumb-bob or accurately plumbed target or signal; never sight

to a hand held ranging rod.

Errors can arise in mis reading and mis booking observations of the vernier or

micrometer, so always check that the reading booked appears on the instrument.



ii) Instrumental errors.

Error in the adjustment of the theodolite. Always observe on both faces of the

theodolite when measuring horizontal and vertical angles.

The theodolite must be properly leveled before observations are made. So that

ensure the plate bubble remains in the same position in its tube when the

theodolite is rotated through 360.

Ensure that the theodolite is stable with the legs firmly planted in solid ground

and that the tripod adjusting screws are properly tightened.

The theodolite must be properly centered over the station mark with an optical

plummet or plumbing rod.

If the horizontal circle is moved between observations the reduced angles will be

in error. This can occur for any of the following reasons.

i) Screwing the theodolite too loosely to the tripod head.

ii) Omitting to secure the movable head.

iii) Omitting to clamp the lower plate.

iv) Using the lower tangent screw instead of the upper tangent screw.

v) Moving the orientation screw on single-axis theodolites.

4.5 Checking and adjusting traverse.

Traverse adjustment should be applied before the results of the traverse are usable for

determining areas or coordinates for publishing the data, or for computing lines to be

located from the traverse stations, to make the traverse mathematically consistent. The

closure in latitudes and departures must be adjusted out.

4.5.1 The compass rule.

Consider a traverse station i,

xi =correction to Xi

yi = correction to Yi

Xt =total closure correction of the traverse in the X coordinate.



Yt =total closure correction of the traverse in the Y coordinate.

Li =distance from station i to the next station.

L =total length of traverse

dxt= (Xn-X1)- departure

dyt = (Yn-Y1)- latitude

Then the corrections are

tt dYL

LiyianddX

L

Lixi

.

Alternatively, corrections may be applied to the departure and latitudes prior to calculating

coordinates

tt dyL

dijyijanddx

L

dijxij

xij and yij are respective corrections to the departure and latitude of line ij which has a

length of dij.

Example:

In a closed traverse the distance between traverse stations and the deflection angle are as

hereunder. Compute the error of closure and adjust the traverse using compass rule.

Line Distance (m) Deflection angle Azimuth

AB 225.94 A 10234’L 67050’

BC 143.39 B 85055’R 153045’

CD 188.47 C 150 47’L 2058’

=557.80 D 78020’R

AXA=170024’ ADY=80038’



Solution:

Using the formula

A1+ R-L-3600=A2

170024’+164015’-253021’-3600=81018’80038’

Error of closure 81018’-80038’=0040’00’

Correction per angle= "00'1004

"00'400 00

Station Correction Correct deflection angle Azimuth

A +0010’ 102044’ 67040’

B -0010’ 85045’ 153025’

C +0010’ 150057’ 2028’

D -0010’ 78010’

Traverse Computation

Station

Distance

AzimuthDeparture Latitude Coordinates ,m

X Y

X

Y

A 100 100

B 225.94 670 40’ 208.992 85.856 308.992 185.856

C 143.39 153025’ 64.167 -128.231 373.159 57.625

D 188.47 2028’ 8.111 188.295 381.27 245.920

=281.27 =145.92 [382.0] [245.0]

92.0

0.145,

73.0

0.282

ADAD YYXX



Adjustment of coordinates by the compass rule

Station Distance Corrections Adjusted

coordinates,m

xi yi X Y

A 100 100

B 225.94 0.296 -0.373 309.288 185.483

C 143.39 0.483 -0.609 373.642 56.016

D 188.47 0.730 -0.920 382.0 245.0

Adjusted distance and Azimuth

DAB=[(XB-XA)2+(YB-YA)2]1/2

DAB=[(309.288-100)2+(185.483-100)2]1/2

=226. 073m

AAB=tan-1

AB

AB

YY

XX

=tan-1

100483.185

100288.309

=67046’57.36’’

4.6 Computation of Area

Area computation is one of the primary objective of land survey. A closed traverse is run, in

which the lines of the traverse are made to coincide with property lines as possible. The

length and bearings of all straight boundary lines are determined either directly or by

computation.

In ordinary land surveying, the area of a tract of land is taken as its projection up on a

horizontal plane, and it is not the actual area of the surface of land. For precise

determination of the area of a large tract, such as state or nation, the area is taken as the

projection of the tract up on the earth’s spheroidal surface to mean sea level.

Station Adjusted

Distance Azimuth

A

B 226.073 67046’52.36”

C 143.684 153023’31.1”

D 188.170 2032’44.75”



Methods of determining area:

1. The area of the tract may be obtained by use of the planimeter from a map or plan. It

may also be calculated by dividing the tract in to triangles and rectangles, scaling the

dimensions of these figures, and computing their areas mathematically.

2. Area by triangles.

It is computation of areas individually mathematically by dividing the track in to

triangles.

If length of two sides and included angle of any triangle are known,

cabarea sin2

1

If lengths of the three sides of any triangle are given,

cbaS

csbsassarea

2

1

))((

3. Area by coordinates:

When the points defining the corners of a tract of land are coordinated with respect to

some arbitrarily chosen coordinate axes or are given in a regional system, these

coordinates are useful not only in finding the lengths and bearings of the boundaries

but also in calculating the area of the tract. The calculation involves finding the areas

of trapezoids formed by projecting the line up on a parallel at right angle to this.

Considering the figure under here

c

a

b

B

CA

c



Area 12345= area 23cb+area 34dc-area 45fd-area 15fa-area 12bc

Area= 455443433232 2

1_

2

1

2

1YYXXYYXXYYXX

12125151 2

1

2

1YYXXYYXX

Rearranging the above eqn.

4153542431325212 xxyxxyxxyxxyxxyarea

The general formula for any polygon having n stations

)(.......2 112113221 nnnnnn xxyxxyxxyxxyarea or

112113221 .......2 nnnnnn yyxyyxyyxyyxarea

The area of tract can also be computed by using

1

1

3

3

2

2

1

1 .....................x

y

x

y

x

y

x

y

x

y

n

n

1332211332212 yxyxyxyxxyxyxyxyarea nnnn

4. Area by double meridian distance and latitudes:

After the adjustment of latitude and departure of traverse lines in a given traverse it is

possible to compute the area enclosed by the traverse. The method by which the adjusted

departures and latitudes are used to compute the area is called double meridian distance

method.

X5

a 1

d

5

2

e

c 3

Y2

b

4Y1

X4Y4

X2

Y5Y3

X3

X1

a’ b’ e’ d’ c’



A reference meridian is assumed to pass through the most westerly point of the survey; the

double meridian distance of the lines are computed; and double the areas of the trapezoids

or triangles formed by orthographically projecting the several traverse lines up on the

meridian are computed. The algebraic sum of these double areas is double the area within

the traverse. The meridian distance of a point is the total departure or perpendicular distance

from the reference meridian and the meridian distance of a straight line is the meridian

distance of its mid point. The double meridian distance of a straight line is the sum of the

meridian distances of the two extremities.

N

b B

A

C

c

The length of the orthographic projection of a line up on the meridian is the latitude of the

line.

The double area of the triangle or trapezoidal formed by projecting a given line up on the

meridian is:

Double area=DMD* latitude

In computing double area algebraic signs should be taken in to account.

Example:

For a traverse 123456 the adjusted distance and azimuths are given as below. Coordinate of

1(0.0, 0.0)

Meridian distance of B M.D.B =bB Meridian distance of C M.D.C =cC Double meridian distance of AB=0+ bB Double meridian distance of BC= MD@B+MD@C =bB+cC DMD of line =DMD of preceding line + departure of

Preceding line + departure of the line.

d D



Adjusted Adjusted

Line Distance Azimuth Departure Latitudes

12 405.18 106019’45” +388.84 -113.92

23 336.59 57054’01” +285.13 +178.86

34 325.18 335028’43” -134.96 +295.85

45 212.92 219028’33” -135.41 -164.31

56 252.21 266055’30” -251.85 -13.53

61 237.69 219040’28” -151.75 -182.95

Compute the area in the traverse by using all methods.

Solution:

Computation of area

1. Area by triangle

m

YYXXD

93.379

)92.113()95.182(84.38875.151

)()(22

2

26

2

2662

m

D

74.310

)92.113(48.19684.3884.403 22

52

m

D

85.300

94.6448.19697.6734.403 2253

Using the formula. csbsassArea

2

cbaS

Station Coordinate, m

X Y

1 0 0

2 388.84 -113.92

3 673.97 64.94

4 539.01 306.79

5 403.4 196.48

6 151.75 182.95

1 0 0

1

2

6 5

4

3

1 2

3 4



For triangle 1, 4.5112

379.73237.69405.18

S

220.44212

)93.3794.511)(69.2374.511)(18.4054.511(4.511126

m

area


74.31021.25293.379

S

271.38985

)21.25244.471)74.31044.471)(93.37944.471(44.471256

m

area


59.33685.30074.310

S

222.42950

59.33609.474)(85.30009.474)(74.31009.479(09.479235

m

area


92.21218.32585.300

S

278.31133

92.21248.419)(18.32548.419)(85.30048.419(48.419345

m

area

Total area of traverse =a1+a2+a3+a4

=44212.20+38985.71+42950.22+31133.78

= 157,281.91m2

2. Area by coordinate method.

Using the general formula

2 area=Y1(X2-Xn)+Y2(X3-X1)+….+Yn-1(Xn-Xn-2)+Yn(X1-Xn-1).

=0 (388.84-151.75) +-113.92(673.97-0) +64.94(539.01-388.84)

+360.79(403.4-673.97) +196.48(151.75-539.01) +182.95(0-403.4).

2 area =-314536.45

Area= 157268.22m2



3. Area by DMD and latitude

Assume the meridian passes through the westerly corner of the traverse i.e station

1.

DMD of line=DMD of preceding line+ departure of preceding line + departure

of the line.

Line Departure DMD Latitudes Double Area

+388.84 388.84 -113.92 -44296.65

23 +285.13 1062.81 +178.86 +190094.20

34 -134.96 1212.98 +295.85 +358860.13

45 -135.41 942.61 -164.31 -154880.25

56 251.85 555.35 -13.53 -7513.89

61 -151.75 151.75 -182.95 -27762.66

Double area =548954.33

-234453.45

=314500.88

Area of tract =314500.88/2 =157250.44m2

a 1

d

5

2

e

c 3

b

4

f

6



Area of tract with irregular or curved boundaries

When the boundary of a tract of land follows some irregular or curved line, such as a

stream or road, it is necessary to run a traverse in some convenient location near the

boundary and to locate the boundary by offsets from the traverse line. Offsets are taken

at changing points of an irregular boundary from the near by traverse line, and when the

boundary is a gradual curve, offsets are taken at regular intervals.

If the offsets are taken sufficiently close together, the error involved in considering the

boundary as straight between offsets is small as compared with the inaccuracies of the

measured offsets. So, the areas between offsets are of trapezoidal shapes and irregular

areas can be calculated by the trapezoidal rule.

**Area by trapezoidal rule

If offsets are taken at equal interval

1321

1133221

...221

2

.....

1

nn

nnn

hhhhh

n

D

hhhhhhhh

n

D

12...

121213221

n

Dhh

n

Dhh

n

DhhArea nn

BDA

h1 h7

h6

h5h3 h4h2



If

dn

D

1interval, n=-number of offsets.

132

1 ...22 n

n hhhhh

dArea

**Area by Simpson’s one-third rule.

Simpson’s one-third rule is applied directly for odd number of offsets.

Area between line AB and curved line DFC

321

312

31

43

3

22

22

2

hhhd

dhh

hdhh

Area for the next two intervals

543 43

hhhd

Area

The summation of these partial areas for (n-1) intervals, n being on odd number

representing the number of offsets, is

)...(4)...(23 1422531 nnn hhhhhhhhd

area

Exercise:

The following offsets are taken at equal interval of 5 m, and offsets between the traverse

line and an irregular boundary is as below. Calculate the area bounded between the traverse

line and an irregular boundary by using both methods.

Distance,m 0 8 16 24 32 40 48 56 64

Offsets,m 2.9 3.8 17.9 12.7 20.2 11.4 25.7 23.3 20.9



**Area by coordinate rule

When offsets are taken at irregular intervals, the area of each figure between pairs of

adjacent offsets may be computed and the values are added. The coordinate rule for

irregular spacing of offsets can be applied by the relation; twice the area can be calculated

if each offset is multiplied by the distance to the preceding offset plus the distance to the

following offset.

2 Area=h1d1+h2 (d1+d2)+h3(d2+d3)+h4(d3+d4)+h5(d4+d5)+h6(d5+d6)+h7d6

The general formula

2 Area=h1x d1+h2 (d1+d2)+…+hn-1(dn-2+dn-1)+hn x dn-1.

Exercise:

Try the above exercise!

Coordinates of unoccupied point

Some times it is not possible to set the instrument up over points like top of building etc.

So, in order to determine the coordinates of points, spur lines or ties lines must be run

from the traverse to these points. In the figure below, the boundaries if a parcel of land

are defined by an iron pipe, a nail in a tree and two fence posts. The only corner that can

be occupied is the one marked by the iron pipe. The problem is to determine the lengths

and bearings of the property lines.

It will be assumed that the coordinates of the iron pipe at point A and the bearing of the

line AB are known. The procedure is then as follows:

1. Measure the lengths of the traverse sides AB, BC, CD and DA, and also the lengths

of the three tie lines BP, CR, and DS.

2. Measure the angles in the traverse ABCD, and also the angles between traverse lines

and the tie lines to P, R, and S, a shown in the figure.



3. Adjust the angles in the traverse ABCD, and compute the bearings of lies BC, CD

and DA based on the known bearing of the one AB.

4. Balance the traverse ABCD.

5. Compute the coordinates of points B, C, and D.

6. Determine the bearings of the lines BC, CD, and DA based on the coordinates

computed in (5).

7. Compute the bearing of the tie lies from the bearings computed in (5) and the

measured angles.

8. Compute the latitudes and departures of the tie lines.

9. Compute the coordinates of the unoccupied stations from the coordinates of

traverse stations B, C, and D and the latitudes and departures of the lines BP, CR,

and DS, respectively.

When the coordinates of P, R, and S have been computed, the bearings and

lengths of the boundary lines may be computed.

Example:

The following adjusted coordinates of stations A, B, C and D of traverse are given.

Additionally, unadjusted deflection angles and unadjusted lengths of supr lines are recorded

determine the lengths and bearings of the boundary lines AP, PR, RS and SA.

Station Adjusted Coordinate

Y X

A 1000.00 1000.00

B 1279.68 1018.13

C 1197.70 735.31

D 1015.45 713.45

A 1000.00 1000.00



Station From To Def-angle Line Length, m

A - - -

B A P 69008’L BP 40.15

C B R 128002’R CR 64.24

D C S 124030’L DS 24.80

Solution:

From the adjusted coordinate of the traverse the azimuth of line AB, BC, CD, DA can be

computed

"25.32'423100068.1279

100013.1018tan 0

ABN AAABlineFor

"7.5'5025368.1127970.1197

13.101831.735tan 0

BCN AABClineFor

"22'5018670.119745.1015

310.73545.713tan 0

CDN AACDlineFor

For line DA tan AN= 5.10'59345.10151000

45.7131000 0

DAA

Computation of Azimuth of tie lines

Line Azimuth Line Azimuth Line Azimuth

AB 3042’32.25’’ BC 253050’5. 7’’ CD 186050’22.8’’

363042’32.25’’ +<C +12802’00’’ -<D -124030’00’’

-<B -6908’00’’ CR 381052’5.7’’

BP 294034’32.25’’ CR 21052’5.7’’ DS 62020’22.8’’

BP N65025’27.75’’W CR N21052’5.7’’E DS N62020’22.8’’E



Computation of coordinates of boundary corners.

Station Length Bearing Latitude DepartureCoordinates

X Y

B 40.15 N65025’27.75’’W 16.70 -36.51

1018.13 1279.68

P *981.62 1296.38

C 64.24 N21052’5.7’’E 59.62 23.93

735.31 1197.70

R *759.24 1257.32

D 29.80 N62020’22.8’’E 13.83 26.39

713.45 1015.45

S *739.84 1029.28

From adjusted coordinates azimuth and distance of property lines can be computed.

For line AP tan AAP =100038.1296

100061.981

=-3032’55.15’’+360000’00’’ =356027’4.85’’

Distance AP=[(981.62-1000)2+(1296.38-1000)2]1/2 =296.95

Adjusted distance and azimuth

Line Distance, m Azimuth

AP 296.95 365027’4.85’’

PR 225.78 260002’20’’

RS 228.86 184051’50’’

SA 261.80 96025’20’’

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