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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 external length can be given as __________

A

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

B

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

C

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

D

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

Correct Answer

⚙️ TE • Technical Concept & PrincipleCivilAdvanced Survey
Option C

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

Quick Summary:

In highway and railway curve surveying, the external distance (or external secant length) EEE is the distance from the point of intersection (PI) to the apex of the simple circular curve. It represents the maximum distance from the tangents to the curve measured along the bisector of the deflection angle Δ\DeltaΔ. The correct expression is derived geometrically as E=R(sec⁡Δ2−1)E = R(\sec\frac{\Delta}{2} - 1)E=R(sec2Δ​−1).

⚙️TETechnical SolutionConcept & Principle
💡 Explanation

In highway and railway curve surveying, the external distance (or external secant length) EEE is the distance from the point of intersection (PI) to the apex of the simple circular curve. It represents the maximum distance from the tangents to the curve measured along the bisector of the deflection angle Δ\DeltaΔ. The correct expression is derived geometrically as E=R(sec⁡Δ2−1)E = R(\sec\frac{\Delta}{2} - 1)E=R(sec2Δ​−1).

🔢 Key Formulas

E=R(sec⁡Δ2−1)E = R\left(\sec\frac{\Delta}{2} - 1\right)E=R(sec2Δ​−1) — External distance (Apex distance)

M=R(1−cos⁡Δ2)M = R\left(1 - \cos\frac{\Delta}{2}\right)M=R(1−cos2Δ​) — Mid-ordinate (Versed sine)

T=Rtan⁡Δ2T = R \tan\frac{\Delta}{2}T=Rtan2Δ​ — Tangent length

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

⚙️ Working Principle

In a simple circular curve of radius RRR and deflection angle Δ\DeltaΔ, the distance from the center of the curve OOO to the Point of Intersection VVV is OV=Rsec⁡Δ2OV = R \sec\frac{\Delta}{2}OV=Rsec2Δ​. Since the distance from OOO to the curve apex CCC is the radius RRR, the external length E=VC=OV−OC=Rsec⁡Δ2−R=R(sec⁡Δ2−1)E = VC = OV - OC = R \sec\frac{\Delta}{2} - R = R(\sec\frac{\Delta}{2} - 1)E=VC=OV−OC=Rsec2Δ​−R=R(sec2Δ​−1).

📌 Key Points
  • ▸

    External distance EEE is the distance from the Point of Intersection (PI) to the midpoint/apex of the curve.

  • ▸

    It is also called the secant distance because it is derived using the secant trigonometric ratio in the right-angled triangle formed by the center, tangent point, and PI.

  • ▸

    As deflection angle Δ\DeltaΔ increases, external distance EEE increases exponentially.

🛠️ Applications / Uses
  • ▸

    Determining clear spacing requirements between the intersection point and the highway/railway centerline.

  • ▸

    Designing circular horizontal curves to ensure proper clearance around obstacles at hill crests or intersections.

📄 Additional Information
  • ▸

    Option A (2Rtan⁡Δ22R \tan\frac{\Delta}{2}2Rtan2Δ​) is incorrect; the tangent length is T=Rtan⁡Δ2T = R \tan\frac{\Delta}{2}T=Rtan2Δ​.

  • ▸

    Option B (2Rsin⁡Δ22R \sin\frac{\Delta}{2}2Rsin2Δ​) is incorrect; the long chord length is L=2Rsin⁡Δ2L = 2R \sin\frac{\Delta}{2}L=2Rsin2Δ​.

  • ▸

    Option D (R(1−cos⁡Δ2)R(1 - \cos\frac{\Delta}{2})R(1−cos2Δ​)) represents the Mid-ordinate (MMM), not the external distance.

📊 Diagram / Illustration
Simple Circular Curve GeometryV (PI)C (Apex)O (Center)T1T2ERR
✅

C is correct — The external distance E of a simple circular curve is given by E=R(sec⁡Δ2−1)E = R(\sec\frac{\Delta}{2} - 1)E=R(sec2Δ​−1).

Core Concepts Used
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Simple Circular Curve External Distance Elements of Circular Curve
💡 EXAM TIP

Remember the relation between External Distance EEE and Mid-Ordinate MMM: E=R(sec⁡Δ2−1)E = R\left(\sec\frac{\Delta}{2}-1\right)E=R(sec2Δ​−1) while M=R(1−cos⁡Δ2)M = R\left(1-\cos\frac{\Delta}{2}\right)M=R(1−cos2Δ​). Notice how sec⁡\secsec is used for EEE (outside the arc) and cos⁡\coscos for MMM (inside the arc).

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