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Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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CivilAdvanced Survey
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The formula for long chord length can be given as __________

A

L=2R×tan⁡∆2L = 2R \times \tan \frac{∆}{2}L=2R×tan2∆​

B

L=2R×sin⁡∆2L = 2 R \times \sin \frac{∆}{2}L=2R×sin2∆​

C

L=R(sec∆2−1)L = R ( s e c \frac{∆}{2} - 1 )L=R(sec2∆​−1)

D

L=R(1−cos⁡∆2)L = R ( 1 - \cos \frac{∆}{2} )L=R(1−cos2∆​)

Correct Answer

⚙️ TE • Technical Concept & PrincipleCivilAdvanced Survey
Option B

L=2R×sin⁡∆2L = 2 R \times \sin \frac{∆}{2}L=2R×sin2∆​

Quick Summary:

The long chord (LLL) of a simple circular curve is the straight line distance joining the point of curve (PC) to the point of tangency (PT) · It represents the longest chord formed between the initial and final tangent points of a simple circular curve · For a curve of radius RRR and deflection angle Δ\DeltaΔ, the formula for long chord length is given by L=2Rsin⁡(Δ2)L = 2R \sin\left(\frac{\Delta}{2}\right)L=2Rsin(2Δ​).

⚙️TETechnical SolutionConcept & Principle
💡 Explanation

The long chord (LLL) of a simple circular curve is the straight line distance joining the point of curve (PC) to the point of tangency (PT) · It represents the longest chord formed between the initial and final tangent points of a simple circular curve · For a curve of radius RRR and deflection angle Δ\DeltaΔ, the formula for long chord length is given by L=2Rsin⁡(Δ2)L = 2R \sin\left(\frac{\Delta}{2}\right)L=2Rsin(2Δ​).

🔢 Key Formulas

L=2Rsin⁡(Δ2)L = 2R \sin\left(\frac{\Delta}{2}\right)L=2Rsin(2Δ​) — Long chord length

T=Rtan⁡(Δ2)T = R \tan\left(\frac{\Delta}{2}\right)T=Rtan(2Δ​) — Tangent length

l = \frac{\pi R \Delta}{180° — Length of curve

M=R(1−cos⁡(Δ2))M = R \left(1 - \cos\left(\frac{\Delta}{2}\right)\right)M=R(1−cos(2Δ​)) — Mid-ordinate

E=R(sec⁡(Δ2)−1)E = R \left(\sec\left(\frac{\Delta}{2}\right) - 1\right)E=R(sec(2Δ​)−1) — Apex distance / External distance

⚙️ Working Principle

Consider the isosceles triangle formed by the center of the curve OOO, the Point of Curve T1T_1T1​, and the Point of Tangency T2T_2T2​. The angle subtended at the center by the long chord T1T2T_1T_2T1​T2​ is equal to the deflection angle Δ\DeltaΔ. Bisecting this central angle creates two congruent right-angled triangles with hypotenuse equal to radius RRR and opposite side equal to half the long chord length (L/2L/2L/2) · Applying simple trigonometry, sin⁡(Δ2)=L/2R\sin\left(\frac{\Delta}{2}\right) = \frac{L/2}{R}sin(2Δ​)=RL/2​, which yields L=2Rsin⁡(Δ2)L = 2R \sin\left(\frac{\Delta}{2}\right)L=2Rsin(2Δ​).

📌 Key Points
  • ▸

    Long chord joins the initial point of curve (PC or T1T_1T1​) and the point of tangency (PT or T2T_2T2​).

  • ▸

    It bisects the central angle subtended by the curve into two equal angles of Δ/2\Delta/2Δ/2.

  • ▸

    It forms the baseline when setting out simple circular curves using the method of offsets from the long chord.

🛠️ Applications / Uses
  • ▸

    Design and layout of horizontal simple circular curves in roads and railways.

  • ▸

    Setting out curves using the offset from long chord method.

📄 Additional Information
  • ▸

    Option A (L=2Rtan⁡(Δ2)L = 2R \tan\left(\frac{\Delta}{2}\right)L=2Rtan(2Δ​)) is incorrect as Rtan⁡(Δ2)R \tan\left(\frac{\Delta}{2}\right)Rtan(2Δ​) represents the single Tangent Length (TTT).

  • ▸

    Option C (L=R(sec⁡(Δ2)−1)L = R \left(\sec\left(\frac{\Delta}{2}\right) - 1\right)L=R(sec(2Δ​)−1)) represents the External or Apex Distance (EEE).

  • ▸

    Option D (L=R(1−cos⁡(Δ2))L = R \left(1 - \cos\left(\frac{\Delta}{2}\right)\right)L=R(1−cos(2Δ​))) represents the Mid-Ordinate (MMM).

📊 Diagram / Illustration
Long Chord Length Formula CardLong Chord Length (L)L = 2R × sinΔ2R = Radius of curveΔ = Deflection angle
✅

B is correct — The length of the long chord is given by L=2Rsin⁡(Δ2)L = 2R \sin\left(\frac{\Delta}{2}\right)L=2Rsin(2Δ​), derived from right-triangle trigonometry on the bisected central deflection angle.

Core Concepts Used
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Simple Circular Curve Geometry Deflection Angle and Central Angle Relationship Elements of Curve Surveying
💡 EXAM TIP

In competitive exam questions, carefully distinguish between Mid-Ordinate (M=R(1−cos⁡(Δ/2))M = R(1 - \cos(\Delta/2))M=R(1−cos(Δ/2))), Apex Distance (E=R(sec⁡(Δ/2)−1)E = R(\sec(\Delta/2) - 1)E=R(sec(Δ/2)−1)), Tangent Length (T=Rtan⁡(Δ/2)T = R \tan(\Delta/2)T=Rtan(Δ/2)), and Long Chord (L=2Rsin⁡(Δ/2)L = 2R \sin(\Delta/2)L=2Rsin(Δ/2)) as options frequently swap these trigonometric functions.

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