Highway/rail curves – long chord of a simple circular curve: For a simple circular curve of radius R and central (deflection) angle Δ, what is the expression for the length of the long chord joining the curve’s end points?

Difficulty: Easy

Correct Answer: 2R * sin(Δ/2)

Explanation:


Introduction / Context:
In highway and railway alignment, basic elements of circular curves—tangent length, long chord, mid-ordinate—are frequently needed. The long chord connects the points of tangency and depends on the curve radius R and the central angle Δ subtended at the curve center.


Given Data / Assumptions:

  • Simple circular curve of radius R.
  • Central (deflection) angle Δ in degrees or radians (consistent units).
  • We seek the long chord length between curve end points.


Concept / Approach:
The end points and the circle center form an isosceles triangle with vertex angle Δ at the center and equal sides of length R. The half-chord is opposite the half-angle Δ/2. Using elementary trigonometry, the half-chord equals R * sin(Δ/2), so the full chord is twice that value.


Step-by-Step Solution:

Construct triangle: sides R, R, and chord C; vertex angle at center = Δ.Use sine in the right triangle formed by bisecting the isosceles triangle: (C/2) / R = sin(Δ/2).Compute C/2 = R * sin(Δ/2).Therefore C = 2R * sin(Δ/2).


Verification / Alternative check:
Check limits: as Δ → 0, sin(Δ/2) ≈ Δ/2 (in radians), so C ≈ R * Δ, which matches the arc approximation for small angles.


Why Other Options Are Wrong:

  • R * cos(Δ/2), 2R * cos(Δ/2), R * sin(Δ/2) represent other curve elements (e.g., mid-ordinates or half-chords) but not the full long chord.


Common Pitfalls:
Using degrees in calculators set to radians; confusing chord with arc length; forgetting the Δ/2 factor.


Final Answer:
2R * sin(Δ/2)

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