# Theodolite Traversing Numericals

Table of Contents

## 1. Theodolite Traversing Numericals – Useful Formula

### 1.1. Consecutive Co-ordinates

**Latitude = LcosΘ**

**Departure = LsinΘ**

### 1.2. Independent Co-ordinates

### 1.3. Closing Error

**Magnitude of Closing error = √ [(ƩL)2 + (ƩD)2]**

**Direction of Closing error = ƩD / ƩL**

## 2. Balancing the linear and angular misclosure in traverse methods

### 2.1. Bowditch’s Method (Compass Rule)

**CL = ƩL * (l/Ʃl)**

**CD = ƩD * (l/Ʃl)**

### 2.1. Transit Method

**CL = ƩL * L/ƩLt and CD = ƩD * D/ƩDt**

## 3. Gales Table

## 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.**

Line | Length | WCB |

AB | 89.31 | 45°10′ |

BC | 219.76 | 72°05′ |

CD | 151.18 | 161°52′ |

DE | 159.10 | 228°43′ |

EA | 232.26 | 300°42′ |

**Solution**,

**By Bowditch method:**

For consecutive coordinate,

**Latitude = LcosΘ **

**Departure = LsinΘ**

Perimeter = ∑l = Sum of lengths = 851.61

(Θ= WCB)

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)

For correction,

**CL = ƩL * L/ƩLt**

**CD = ƩD * D/ƩDt**

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.**

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+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.**

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 | xcosy | xsiny |

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″