Question 1 Report
(a) Describe, with the aid of a suitable diagram, an open traverse
(b) A compass traverse round a school compound produced the following field record:
| Line | Length (mm) | Bearing |
| AB | 370 | 260º |
| BC | 450 | 198° |
| CD | 265 | 110° |
| DE | 560 | 35° |
| EA | 140 | 330° |
Use the data above to:
(i) plot the area of the compound on a scale of 1cm to 50m
(ii) find the direct distance between D and A, in metres.
(a) Open traverse
An open traverse is a series of connected survey lines whose lengths and bearings are measured, beginning at one station and ending at another station. It does not return to, or close at, the starting point. It is suitable for surveying linear features such as roads, railway lines, rivers and pipelines.
(b) Compass traverse of the school compound
The given field lengths are taken as metres. At a scale of 1 cm to 50 m, the representative fraction is:
\[1:5\,000\]
| Line | Ground length (m) | Bearing | Length on plan (cm) |
|---|---|---|---|
| AB | 370 | 260° | 7.4 |
| BC | 450 | 198° | 9.0 |
| CD | 265 | 110° | 5.3 |
| DE | 560 | 35° | 11.2 |
| EA | 140 | 330° | 2.8 |
(i) Plot of the compound
Using a protractor, set off each bearing clockwise from north at the appropriate station and draw the corresponding scaled length. The completed traverse is shown below.
(ii) Direct distance DA
From the plotted plan, the straight-line distance from D to A measures approximately \(12.7\text{ cm}\).
\[DA=12.7\times 50\text{ m}=635\text{ m}\]
Therefore, the direct distance between D and A is approximately 635 m.
Check by computation from the first three traverse lines:
| Line | Easting, \(L\sin\theta\) (m) | Northing, \(L\cos\theta\) (m) |
|---|---|---|
| AB | \(-364.4\) | \(-64.2\) |
| BC | \(-139.1\) | \(-428.0\) |
| CD | \(+249.0\) | \(-90.6\) |
| D relative to A | \(-254.5\) | \(-582.8\) |
\[DA=\sqrt{(254.5)^2+(582.8)^2}\approx636\text{ m}\]
This agrees with the scaled answer of 635 m, allowing for plotting and measurement.
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