Highway transition curve (lemniscate) in polar form If α is the angle between the polar ray and the tangent at the point of commencement of a lemniscate transition curve, which polar equation correctly represents the lemniscate used in geometric design?

Difficulty: Medium

Correct Answer: l = k √(cos 2α)

Explanation:


Introduction / Context:
Besides the clothoid (Euler spiral), the lemniscate is another transition curve used in geometric design for smooth variation of curvature. In polar coordinates, a classic lemniscate (of Bernoulli) provides a convenient mathematical description for layout and setting-out calculations in some highway/rail problems.


Given Data / Assumptions:

  • α is the angle between the initial tangent and the polar ray to a generic point on the curve.
  • k is a constant related to the chosen size/scale of the lemniscate.
  • We use the standard polar form consistent with surveying texts.


Concept / Approach:

The Bernoulli lemniscate has the polar equation r^2 = a^2 cos 2θ. In highway literature the same is often written as l^2 = k^2 cos 2α, where l is the radius vector from the origin on the initial tangent. Taking the positive square root for the relevant branch yields l = k √(cos 2α). This expresses how l varies with α and furnishes deflection relationships for setting out by angular methods.


Step-by-Step Solution:

Start from canonical form: r^2 = a^2 cos 2θ.Map symbols: r → l, a → k, θ → α.Hence l^2 = k^2 cos 2α → l = k √(cos 2α) for the applicable branch.


Verification / Alternative check:

At α = 0, cos 2α = 1 → l = k (finite, maximum). At α = 45°, cos 90° = 0 → l = 0, consistent with the lemniscate reaching the origin at 45° from the tangent.


Why Other Options Are Wrong:

Linear sine/tangent or simple cosine forms without the square root do not match the defining equation r^2 ∝ cos 2θ and would yield incorrect geometry and stationing.


Common Pitfalls:

Forgetting the squared relation; confusing lemniscate with Euler spiral (where curvature varies linearly with arc length); using degrees vs radians incorrectly in computation.


Final Answer:

l = k √(cos 2α)

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