Problem Statement
A chain line PQR crosses a river, Q and R being on the near and distant banks respectively. A perpendicular QS, 90 m long, is set out at Q on the left of the chain line. The respective bearings of R and P taken at S are 77° 30′ 20′′ and 167° 30′ 20′′. Find the chainage of R given that PQ is 45 m and the chainage of Q is 650 m.
Step-by-Step Solution
Key Information
- Chain line PQR crosses river (Q near bank, R distant bank).
- Perpendicular distance QS = 90 metres (set out at Q, left of PQR). Thus ∠PQS = 90° and ∠RQS = 90°.
- Bearing of R from S (Bearing SR) = 77° 30′ 20′′.
- Bearing of P from S (Bearing SP) = 167° 30′ 20′′.
- Length PQ = 45 metres.
- Chainage of Q = 650 metres.
- Goal: Find the chainage of R.
Step 1: Calculate Angle RSP
The angle RSP at point S is the difference between the bearings of SP and SR, measured from S.
∠RSP = Bearing of SP − Bearing of SR
∠RSP = 167° 30′ 20′′ − 77° 30′ 20′′
∠RSP = 90°
This indicates that triangle RSP is a right-angled triangle, with the right angle at S.
Step 2: Identify Similar Triangles & Calculate RQ
We know QS is perpendicular to the chain line PQR at Q, so ∠PQS = 90° and ∠RQS = 90°. Triangles PSQ and RQS are right-angled at Q.
Additionally, we found ∠RSP = 90°.
Consider the angles within the right triangles:
In ΔPSQ: ∠SPQ + ∠PSQ = 90°
In ΔRQS: ∠SRQ + ∠RSQ = 90°
Also: ∠PSQ + ∠RSQ = ∠RSP = 90°
From these relationships, it can be shown that ∠SPQ = ∠RSQ and ∠PSQ = ∠SRQ. Therefore, triangles PSQ and RQS are similar (AAA similarity).
For similar triangles, the ratio of corresponding sides is equal:
PQ / QS = QS / RQ
Rearranging to find RQ (the river width along the chain line):
RQ = QS² / PQ
RQ = (90 m)² / 45 m
RQ = 8100 m² / 45 m
RQ = 180 metres
Step 3: Calculate Chainage of R
The chainage of R is the chainage of Q plus the distance RQ along the chain line.
Chainage of R = Chainage of Q + Length RQ
Chainage of R = 650 m + 180 m
Chainage of R = 830 metres
Final Result
Conceptual Explanation & Applications
Core Concepts:
- Chainage: Measurement of distance along a primary survey line (PQR).
- Bearings: Clockwise angles from North defining line directions (SR, SP), used here to calculate the angle (∠RSP) at the observation point S.
- Perpendicular Offset: Setting out QS at 90° to the chain line at Q, creating right-angled triangles (ΔPQS, ΔRQS).
- Right Triangle Geometry: Utilizing properties of right-angled triangles, including the fact that angles sum to 180°.
- Similar Triangles: Recognizing that ΔPSQ and ΔRQS are similar due to their angles. This allows the use of ratios of corresponding sides (PQ/QS = QS/RQ) to find unknown lengths. This is related to the Geometric Mean Theorem when a right triangle (ΔRSP) has an altitude (QS) to its hypotenuse (PR).
Real-World Applications:
- Route Surveying (Roads, Rail, Pipelines): Establishing precise locations (chainages) for points along a route, especially when crossing obstacles like rivers or valleys.
- Bridge Design & Construction: Determining the exact span required and locating abutments based on chainage.
- Cadastral Surveying: Defining property boundaries that cross inaccessible features.
- Hydrographic Surveying: Mapping river widths and cross-sections relative to a baseline.
Why It Works:
This problem showcases how geometric principles allow surveyors to find unknown distances and chainages indirectly. By establishing a perpendicular line QS of known length from the known point Q, and measuring bearings from S to points P and R, we create a geometric figure (triangle RSP) whose internal angle at S can be calculated (∠RSP = 90°).
The perpendicular QS divides the triangle RSP into two smaller triangles (PSQ and RQS) which are both right-angled at Q. The specific geometry (right angle at Q and right angle at S) leads to these two triangles being similar. The property of similar triangles allows us to set up a proportion involving the known lengths (PQ, QS) and the unknown length (RQ).
Solving this proportion (RQ = QS²/PQ) gives the distance across the river along the chain line. Adding this distance to the known chainage of Q yields the chainage of the point R on the distant bank. This avoids the need for direct measurement across the river.