Theodolite Traversing Numericals – Formula and Full Solution

THEODOLITE MAY BE DEFINED AS THE OPTICAL SURVEY INSTRUMENT THAT IS USED TO MEASURE THE ANGLES BETWEEN THE SPECIFIED POINTS BOTH IN THE VERTICAL AND HORIZONTAL PLANE.

 c. Closing Error  

b. Transit Method  

  3. Gales Table  

Traverse computations are usually done in a tabular form, a more common form is Gales Traverse Table.

For complete traverse computations, the following steps are usually necessary.

i. Adjust the interior angles to satisfy the geometrical condition, i.e., the sum of interior angles to be equal to (2n-4)*90º and exterior angles (2n+4)*90º.

ii. Starting with the observed bearing of one line, calculate the whole circle bearings of all other lines.

iii. Calculate the consecutive coordinates (i.e., latitudes and departures).

iv. Calculate ƩL and ƩD.

v. Apply necessary corrections to the latitudes and departures of the line so that ƩL = 0 and ƩD = 0. The correction may be applied to either the transit rule or compass rule depending upon the type of traverse.

vi. Using corrected consecutive coordinates, calculate the independent coordinates to the points so that they are all positive, the whole of the traverse thus lying in the North East quadrant.

  4. Theodolite Traversing Numericals  

1.1. An abstract from a traverse sheet for a closed traverse is given below. Balance the traverse by the Bowditch method and transit method.

Solution,

By the Bowditch method,

For consecutive coordination,

Latitude = LcosΘ

Departure = LsinΘ

Perimeter = ∑l = Sum of lengths = 851.61

(Θ= WCB)

For correction,

Magnitude of closing error= √¹

or, Cos43º37’51.42″=(319.758²+ED²-232.26²)/2×319.758×ED

ED=158.982

Again,

((Sin∠D)/232.26)=((Sin∠A)/ED)

Sin∠A=0.472

∠A=28º11’4.21″

Then,

Bearing of DE=228º43′

Bearing of ED=228º43′-180º=48º43′

Bearing of EA=360º-71º47’48.37″+48º43’= 336º55’11.63″

Read Also: Correction of Length & Area Due To Incorrect Scale or Chain | 7 Numericals

1. 0.51)²+(0.224)²)=0.557m

   Direction of closing error= tan^-1(0.224/0.51)=23º42’42.6″

   CL =∑L×(l/∑l)

   CL of line AB= 0.51×(89.31/851.61)=0.053

   CL of line BC= 0.51×(219.76/851.61)=0.132

   CL of line CD= 0.091

   CL of line DE=0.095

   CL of line EA= 0.061

   similarly,

   CD = ƩD * (l/Ʃl)

   CD of line AB=0.224×(89.31/851.61)=0.023

   CD of line BC=0.224×(219.76/851.61)=0.058

   CD of line CD=0.224×(151.18/851.61)=0.040

   CD of line DE=0.224×(159.10/851.61)=0.042

   CD of line EA=0.224×(232.26/851.61)=0.061

    

   For corrected coordinates,

   Corrected Latitude= latitude – CL (since (∑L) is positive, the correction will be negative)

   Corrected departure= Departure – CD

    

   By Transit method,

   For consecutive coordination,

   Latitude = LcosΘ and Departure = LsinΘ

   (Θ= WCB)

    

   

   For correction,

   CL = ƩL * L/ƩLt 

   CD = ƩD * D/ƩDt

   Where, CL = Correction to the latitude of any side

   CD = Correction to the departure of any side

   ƩL = Total error in latitude (Sign Consideration)

   ƩD= Total error in Departure (Sign Consideration)

   L, D = Latitude, and Departure of any side

   ∑Dt = Arithmetic sum of Departures (No sign Consideration)

   ∑Lt = Arithmetic sum of Latitudes (No sign Consideration)

    

   Then,

   CL of line AB= 0.51×(62.968/497.796)= 0.065

   CL of line BC= 0.51×(67.606/497.796)= 0.069

   CL of line CD= 0.51×(143.672/497.796)= 0.147

   CL of line DE= 0.51×(104.971/497.796)= 0.108

   CL of line EA= 0.51×(118.579/497.796)= 0.121

    

   Similarly,

   CD of line AB =0.224×(63.335/638.756)= 0.022

   CD of line BC =0.224×(209.103/638.756)= 0.073

   CD of line CD =0.224×(47.052/638.756)= 0.017

   CD of line DE =0.224×(119.557/638.756)= 0.042

   CD of line EA = 0.224×(199.709/638.756)= 0.070

    

   For corrected coordinates,

   Corrected Latitude= latitude – CL (since (∑L) is positive, the correction will be negative)

   Corrected departure= Departure – CD

    

    

   ---

    

    

   1.2. Calculate the length and bearing of the same traverse leg is omitted.

   Line	Length	Bearing
   AB	89.31	45º10′
   BC	?	?
   CD	151.18	161º52′
   DE	159.1	228º43′
   EA	232.26	300º42′

    

   Solution,

    

   

   For closed traverse,

   ∑L=0 (Sum of latitude is zero)

   62.968+x cosy-143.672-104.971+118.579=0

   x cosy=67.096 ……(i)

   ∑D=0 (Sum of departure is zero)

   63.335+xsiny+47.052-119.557-199.709=0

   xsiny=208.879 …..(ii)

   Dividing equation (ii) by (i)

   tany=(208.879/67.096)

   ∴y=72º11’31.1″

   Putting the value of y on any equation,

   ∴x= 219.391

    

   ---

    

   1.3. Calculate the missing data.

   Line	Length	Bearing
   AB	89.31	45º10′
   BC	219.76	72º05′
   CD	151.18	161º52′
   DE	?	228º43′
   EA	232.26	?

    

    

   Solution,

    

   In a closed traverse ABCDA

   Line	Length	Bearing	Latitude	Departure
   AB	89.31	45º10′	62.968	63.335
   BC	219.76	72º05′	67.606	209.103
   CD	151.18	161º52′	-143.672	47.052
   DA	x	y	x cosy	x siny

    

   For closed traverse,

   ∑L=0 (Sum of latitude is zero)

   62.968+67.606+xcosy-143.672=0

   xcosy=13.098 ……(i)

   ∑D=0 (Sum of departure is zero)

   63.335+209.103+xsiny+47.052=0

   xsiny=-319.49 …..(ii)

   From eqn (i) and (ii)

   tany=-24.392

   ∴y= -87º39’8.58″

   Also, y=360º-87º39’8.58’=272º20’57.42″

   ∴x= 319.758

   Again,

   using sine law, In traingle AED

   {(Sin∠D) / 232.26} = {(Sin∠A) / ED} = {(Sin∠E) / 319.758}

   Then,

   (∠D =272º20’51.42″ – 228º43′  = 43º37’51.42″)

   ((Sin43º37’51.42″)  / 232.26 )= ( (Sin∠E) / 319.758)

   Sin∠E=0.958

   ∠E=71º47’48.37″

   Again,

   Using cosine law,

   CosD=((AD²+ED²-AE²)/2(AD)(ED[↩]