---
title: "Adjust traverse A-X to E-Y bearings"
subject: "Surveying II"
chapter: 1
key: "semester_4/surveying_ii:traverse-adjust-latitude-departure-ax-ey"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/traverse-adjust-latitude-departure-ax-ey
site: BCT Engineering
---

# Adjust traverse A-X to E-Y bearings

Surveying II · Chapter 1: Traversing

- Asked in: 2070 Magh, 10 marks
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/traverse-adjust-latitude-departure-ax-ey
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1

## Question

Below given table lists measured angles to the right for the traverse. The bearings A-X and E-Y have known value of 139°05'45" and 86°20'47" respectively. Adjust this traverse for departure and latitude misclosure.

| Station | Length (m) | Measured angle |
|---|---|---|
| A | | 283°50'10" |
| | 1045.50 | |
| B | | 256°17'18" [?] |
| | 1007.38 | |
| C | | 98°12'41" |
| | 897.81 | |
| D | | 103°30'34" |

## Short answer

The final misclosure cannot be computed from the printed data because the closing line data and end coordinates are missing. Compute line bearings by adding each angle to the right, correct angular misclosure, then calculate latitudes and departures and apply Bowditch corrections proportional to line lengths.

## Answer

**Note on the data:** the table has four angles (A, B, C, D) and three lengths (AB, BC, CD). The last line of the traverse (up to the closing station and its reference line E-Y) has no length or angle, and no coordinates are given for A or for the end station. Hence the final misclosure cannot be numerically completed from the printed data. The method is shown below and every quantity that can be computed from the data is worked out.

#### Step 1: Bearings from the angles to the right
Angles are measured clockwise from the back station to the forward station, so

$$
\text{Bearing of forward line}=\text{Bearing of back line}+\text{angle to the right}\ (\pm360^\circ)
$$

With the back line A-X (139°05'45") at A:

- AB = 139°05'45" + 283°50'10" = 62°55'55"
- BC = 62°55'55" + 180° + 256°17'18" − 360° = 139°13'13"
- CD = 139°13'13" + 180° + 98°12'41" − 360° = 57°25'54"
- Line ahead of D = 57°25'54" + 180° + 103°30'34" − 360° = 340°56'28"

#### Step 2: Angular misclosure and its adjustment
The bearing of the final line (E-Y = 86°20'47") is carried through all $n$ angles of the traverse. The angular misclosure is

$$
\alpha=\text{computed bearing of E-Y}-86^\circ20'47''
$$

and each of the $n$ angles is corrected by $-\alpha/n$ (the $k$-th bearing by $-k\alpha/n$) before latitudes and departures are computed. The angle at B is marked doubtful, so it should be checked first if $\alpha$ comes out large.

#### Step 3: Latitudes and departures (bearings before angular adjustment)
$$
\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta
$$

| Line | Length (m) | Bearing | Latitude | Departure |
|---|---|---|---|---|
| AB | 1045.50 | 62°55'55" | +475.753 | +930.983 |
| BC | 1007.38 | 139°13'13" | -762.815 | +657.973 |
| CD | 897.81 | 57°25'54" | +483.296 | +756.629 |
| **Sum** | 2950.69 |  | +196.234 | +2345.585 |

#### Step 4: Misclosure in latitude and departure
After including all lines up to the closing station $Z$ with known coordinates $(N_Z, E_Z)$ and $(N_A,E_A)$ for A:

$$
e_N=(N_A+\Sigma\text{Lat})-N_Z,\qquad e_E=(E_A+\Sigma\text{Dep})-E_Z,\qquad e=\sqrt{e_N^2+e_E^2}
$$

Precision $=e/\Sigma L=1/N$.

#### Step 5: Adjustment (Bowditch)
$$
C_{lat}=-e_N\frac{L}{\Sigma L},\qquad C_{dep}=-e_E\frac{L}{\Sigma L}
$$

The corrections are added to each latitude and departure, and the adjusted values are summed from A to get the adjusted coordinates of the stations B, C, D, ... The last computed station then agrees with the known coordinates.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the bearing of the forward line computed from angles measured to the right?
   A. Bearing of back line + angle to the right (±360°), after adding 180° when the back line is the previous forward line reversed
   B. Bearing of back line - angle
   C. Angle alone
   D. Bearing of back line + 90°
   Answer: A. The forward bearing = back bearing + clockwise angle; when carrying from line to line the previous bearing is first reversed by 180°. (Key point: Bearing from angles to the right)

2. With A-X = 139°05'45" and the angle at A of 283°50'10", what is the bearing of AB?
   A. 62°55'55"
   B. 142°55'55"
   C. 422°55'55"
   D. 104°44'35"
   Answer: A. 139°05'45" + 283°50'10" = 422°55'55", minus 360° gives 62°55'55". (Key point: Bearing of AB)

3. How is the angular misclosure α corrected, with n angles?
   A. Each angle is corrected by -α/n, the k-th bearing by -kα/n
   B. Only the doubtful angle B is changed
   C. All bearings by -α
   D. Not corrected
   Answer: A. Equal distribution of the angular error, accumulating along successive bearings. (Key point: Angular misclosure correction)

4. How is the misclosure in latitude and departure found?
   A. e_N = (N_A + ΣLat) - N_Z and e_E = (E_A + ΣDep) - E_Z, then e = sqrt(e_N^2 + e_E^2)
   B. ΣLat + ΣDep
   C. ΣL minus 360°
   D. Difference of bearings
   Answer: A. Computed end coordinates are compared with the known end coordinates. (Key point: Latitude/departure misclosure)

5. How are the corrections to each line's latitude calculated?
   A. C_lat = -e_N x L/ΣL
   B. C_lat = +e_N x lat/Σlat
   C. Equal shares of e_N
   D. C_lat = e_E x L/ΣL
   Answer: A. Bowditch's corrections are proportional to line length and of opposite sign to the misclosure. (Key point: Bowditch distribution)

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