# HOW TO CALCULATE LENGTH AND BEARING OF OMITTING MEASUREMENT IN TRAVERSING

UPENDRA KUMAR

In the traverse, the survey omits the measurement of length and bearing of a line. The omitted
length and bearing of easily calculate with the use of

CASE I:-  If the length and bearing of one line is omitted. In the above picture, ABCD is the traverse. During traversing the length and bearing of the EA

is omitted. The length and bearing of the traverse shown below.

 LINE LENGTH BEARING AB 217.5 120º15’ BC 318.0 62º30’ CD 375.0 322º24’ DE 283.5 335º18’ EA ❔ ❔

First calculate the latitude and departure of each line

 LINE LATITUDEL✖COS𝛳 DEPARTUREL✖SIN𝛳 AB -109.570 187.884 BC 146.836 282.069 CD 297.108 -228.804 DE 257.562 -118.465 EA ❔ ❔

Find the algebraic sum of the latitudes and algebraic sum departures.
⇒-109.570+146.836+197.108+257.562=491.936 OR
∑L=491.936
⇒187.884+282.069-228.804-118.465=122.684
∑D=122.684

Subtract the algebraic sum of latitude and departure from zero.
∑L=0-491.936 = ∑L-491.936
∑D=0-122.684= ∑D-122.684
Then TAN𝛳= departure÷latitude=122.684÷491.936=
𝛳=14º0’12” OR S14º0’12”W

LENGTH OF EA= LAT÷COS𝛳=491.936÷COS14º0’12” =507.00mt
OR
LENGTH OF EA=√(491.936)²+(122.684)²
=507.00mt

CASE II:- In the traversing length of one side and bearing of another side missing

 LINE LENGTH BEARING LATITUDE DEPARTURE AB 217.5 S59º45’E -109.570 187.872 BC 318.0 ❔ CD ❔ N37º36’E DE 283.5 S55º18’W -161.390 -233.077 EA 173.15 S2º40’W -172.962 -8.055 In the table and figure the length of CD and bearing of BC is missing.

For this type of problem ignore the affected side BC and CD . Closed the polygon, draw a line
B TO D, and made a complete polygon ABDEA.

Compute the length and bearing of the closing line BD.

∑L=443.922
∑D=53.26

Then TAN𝛳= departure÷latitude=53.26÷443.922=6º50’29”
R.B OF BD=N6º50’29”E
Length of BD=LAT÷COS𝛳=443.922÷COS6º50’29” =447.10m.

In the triangle BCD , angle CDB= 37º36+6º50’29=44º26’29”

Here BD=447.10m, and BC=318.0m

∠C= BD/BC✖SinD
∠C= 447.10/318.0✖Sin44º26’29”
∠𝛳2=79º52’39”

Then ∠𝛳1=180º-79º52’39”+44º26’29”
∠𝛳1=55º40’52”

R.B OF BC=6º50’29”+55º40’52”

R.B OF BC=N6º231’21”E

Length of CD= BD✖Sin𝛳1÷Sin𝛳2
Length of CD=447.10✖Sin55º40’52”÷Sin79º52’39”

Length of CD=375.10m

CASE III:-  When two sides are omitted.

In the above figure, ABCDA is a traverse, The length of BC and CD IS not given.

 No. LINE LENGTH Bearing Latitude Departure 1 AB 217.50 S59º45’E -109.570 187.884 2 BC ❔ N6230’E ❔ ❔ 3 CD ❔ N3736’W ❔ ❔ 4 DE 283.50 S55º18’W -161.390 -233.077 5 EA 173.50 S2º40’W -173.312 -8.072

Compute the length and bearing of the closing line BD.

The latitude of closing line BD=444.272
The departure of closing lineBD=53.258

Then reduced bearing of BD= Tan𝛳=53.258÷444.272
= N 6º50’8”E
Length of BD=  444.272÷COS 6º50’8”

= 447.452mt.

Solve the triangle BCD.

In the triangle BCD

∠B=55º39’52”

∠C=79º54’

∠D=44º26’08”

In the triangle BCD

BC/SinD=BD/SinC
BC=BD/SinC*SinD
BC=447.452/Sin79º54’*Sin44º26’08”
BC= 318.195mt

CD=BD/SinC*SinB

CD=447.452/Sin79º54’*Sin55º39’52”
CD=375.29mt

CASE IV:-  When two angles are omitted.

Solve the problem as case iii

HOW TO FIND HEIGHT OF AN OBJECT