theodolite traversing numericals

Theodolite Traversing Numericals

  1. Theodolite Traversing Numericals – Useful Formula  

 

  1.1. Consecutive Co-ordinates  

The latitude and departure of any station concerning the preceding station are known as consecutive coordinates. It is also known as Dependent Co-ordinates.
Consecutive co-ordinates,
Latitude = LcosΘ
Departure = LsinΘ

 

  1.2. Independent Co-ordinates  

The total latitude and departure of any point concerning common origin are known as independent coordinates.

 

  1.3. Closing Error  

In a closed traverse plot, the distance shifted of the first station due to error in-field measurement of angles and distances is called Closing Error.
In a close traverse, if there is no error, then, ƩL = 0 and ƩD = 0.
Magnitude of Closing error = √ [(ƩL)2 + (ƩD)2]
Direction of Closing error = ƩD / ƩL
Relative Precision = Error of closure / Perimeter = e / p or 1: p/e

 

  2. Balancing the linear and angular misclosure in traverse methods  

 

  2.1. Bowditch’s Method (Compass Rule) 

Assumptions:
i. Error in linear measurements is directly proportional to the square root of the length.
ii. Error in angular measurement is inversely proportional to the square root of the length.
It is employed where linear and angular measurements are closer in precision. And the total error is distributed in proportional to the length of sides.
CL = ƩL * (l/Ʃl)
CD = ƩD * (l/Ʃl)
Where,
CL = Correction to latitude of any side
CD = Correction to departure of any side
ƩL = Total error in latitude
ƩD= Total error in Departure
l = Length of any side
Ʃl = Perimeter of traverse = algebraic sum of lengths.

 

  2.1. Transit Method  

i. Employed where angular measurement is significantly more precise than linear measurement.
ii. Total error is distributed in proportion to latitudes and departures of the sides.
CL = ƩL * L/ƩLt and 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)

 

  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 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.
Screenshot 37

 

 

  4. Numericals – Theodolite Traversing Numericals  

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

LineLengthWCB
AB89.3145°10′
BC219.7672°05′
CD151.18161°52′
DE159.10228°43′
EA232.26300°42′

 

Solution,

 

By Bowditch method:

For consecutive coordinate,

Latitude = LcosΘ

Departure = LsinΘ

Perimeter = ∑l = Sum of lengths = 851.61

(Θ= WCB)

Screenshot 38

 

For correction,

Magnitude of closing error= √((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 ,correction will be negative)

Corrected departure= Departure-CD

 

By Transit method:

For consecutive coordinate,

Latitude = LcosΘ and Departure = LsinΘ

(Θ= WCB)

Screenshot 42

For correction,

CL = ƩL * L/ƩLt 
CD = ƩD * D/ƩDt
Where, CL = Correction to latitude of any side
CD = Correction to 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 ,correction will be negative)

Corrected departure= Departure-CD

 

 

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

LineLength Bearing
AB89.3145º10′
BC??
CD151.18161º52′
DE159.1228º43′
EA232.26300º42′

 

Solution,

Screenshot 43

For closed traverse,

∑L=0 (Sum of latitude is zero)

62.968+xcosy-143.672-104.971+118.579=0

xcosy=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 eqn (ii) by (i)

tany=(208.879/67.096)

∴y=72º11’31.1″

Putting value of y on any equation,

∴x=219.391

 

 

1.3. Calculate the missing data.

LineLength Bearing
AB89.3145º10′
BC219.7672º05′
CD151.18161º52′
DE?228º43′
EA232.26?

 

Solution,

Screenshot 44

 

In a closed traverse ABCDA

LineLengthBearingLatitudeDeparture
AB89.3145º10′62.96863.335
BC219.7672º05′67.606209.103
CD151.18161º52′-143.67247.052
DAxyxcosyxsiny

 

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))

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″

 

 

 

 

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