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 coordinate.

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	LATITUDE L✖COS𝛳	DEPARTURE L✖SIN𝛳
AB	-109.570	187.884
BC	146.836	282.069
CD	297.108	-228.804
DE	257.562	-118.465
EA	❔	❔

                              

 SEE ALSO:-FOR COORDINATE CALCULATION

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


SEE ALSO:-PLOTTING
HOW TO FIND HEIGHT OF AN OBJECT
WHY USED CORRECTION FOR CURVATURE AND REFRACTION IN LEVELING