A survey line CDE crosses a river, D being on the near bank, and E on the opposite bank. A perpendicular DF = 150 metres is ranged at D on the left. From F bearings of E and C are observed to be 25° and 115° respectively. If the chainage of C is 1250 meters and that of D is 1620 metres, find the chainage of E.

Problem Statement

A survey line CDE crosses a river, D being on the near bank, and E on the opposite bank. A perpendicular DF = 150 metres is ranged at D on the left. From F bearings of E and C are observed to be 25° and 115° respectively. If the chainage of C is 1250 metres and that of D is 1620 metres, find the chainage of E.

Survey diagram for river crossing chainage

Step-by-Step Solution

Key Information

  • Survey line CDE crosses river (D near bank, E far bank).
  • Perpendicular distance DF = 150 metres (ranged at D).
  • Bearing of FE = 25°.
  • Bearing of FC = 115°.
  • Chainage of C = 1250 metres.
  • Chainage of D = 1620 metres.
  • Goal: Find the chainage of E.

Step 1: Calculate Length CD

Chainage represents the distance along the survey line from a starting point. The length of segment CD is the difference between the chainages of D and C.

Length CD = Chainage of D − Chainage of C

Length CD = 1620 m − 1250 m

Length CD = 370 m

Step 2: Calculate Angle EFC

The angle EFC at point F can be found by the difference between the bearings of FC and FE, measured from F.

∠EFC = Bearing of FC − Bearing of FE

∠EFC = 115° − 25°

∠EFC = 90°

This means triangle EFC is a right-angled triangle, with the right angle at F.

Step 3: Calculate Distance DE (River Width)

We are given that DF is perpendicular to the survey line CDE at D. This means ∠FDC = 90° and ∠FDE = 90°.

Since ∠FDC = 90°, triangle FDC is right-angled at D.

Since ∠FDE = 90°, triangle FDE is right-angled at D.

From Step 2, we know triangle EFC is right-angled at F.

In a right-angled triangle EFC, if DF is the altitude to the hypotenuse CE (which holds true given DF⊥CE and ∠EFC=90°), a geometric property states: FD² = ED × DC.

Rearranging to find ED:

ED = FD² / DC

ED = (150 m)² / 370 m

ED = 22500 m² / 370 m

ED ≈ 60.81 metres

(This value represents the width of the river segment DE along the survey line).

Step 4: Calculate Chainage of E

The chainage of E is found by adding the distance DE to the chainage of D.

Chainage of E = Chainage of D + Length DE

Chainage of E = 1620 m + 60.81 m

Chainage of E = 1680.81 m

Final Result

The chainage of E is 1680.81 metres.

Conceptual Explanation & Applications

Core Concepts:

  • Chainage: A measurement of distance along a survey line, route, or alignment, typically starting from zero. Used extensively in road, rail, and pipeline projects.
  • Bearings & Angles: Using bearings (angles from North) to determine the angles between survey lines (like ∠EFC).
  • Perpendicular Offsets: Establishing a line (DF) perpendicular to the main survey line (CDE) to create right-angled triangles for calculation.
  • Right Triangle Geometry: Applying properties of right triangles, including trigonometric relationships and geometric theorems (like the altitude theorem where FD² = ED × DC in this specific configuration).

Real-World Applications:

  • Route Surveying: Determining exact locations (chainages) of points along linear projects like roads, railways, canals, or pipelines, especially when crossing obstacles like rivers.
  • Bridge & Culvert Placement: Accurately locating start and end points for structures crossing obstacles.
  • Construction Layout: Setting out points based on calculated chainages for construction activities.
  • Topographic Mapping: Mapping features relative to established survey lines and chainages.

Why It Works:
This problem combines the concepts of chainage and indirect measurement using bearings and perpendiculars. By knowing the chainages of C and D, we establish the length of the segment CD. Setting up a perpendicular DF allows us to use point F as a reference. Measuring bearings from F to C and E enables the calculation of angle EFC.

The crucial step involves recognizing the geometric relationships formed. The perpendicular DF creates right triangles FDC and FDE. The calculated angle EFC being 90° confirms triangle EFC is also a right triangle. This specific geometry allows the use of the relationship FD² = ED × DC to find the unknown river width DE along the survey line.

Finally, adding the calculated distance DE to the known chainage of D gives the required chainage of point E on the far bank. This method is practical for determining precise locations along a route when direct measurement across obstacles is not feasible.

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