Euler spiral

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(Redirected from Cornu spiral)

A double-ended Euler spiral.

An Euler spiral is a curve whose curvature changes linearly with its curve length. Euler spirals are also commonly referred to as clothoids or Cornu spirals.

Euler spirals are widely used as transition curve in rail track / highway engineering for connecting and transiting the geometry between a tangent and a circular curve. They also have applications to diffraction computations.

The curvature of a circular curve is equal to the reciprocal of the radius. The principle of linear variation of the curvature of the transition curve defines the geometry of the Euler spiral: its curvature begins at zero at the straight section (the tangent), increases linearly and ends at the curvature of the horizontal circular curve where the Euler spiral meets.

[edit] Application

Main article: Track transition curve

An object traveling on a circular path experiences a centripetal acceleration. When a vehicle traveling on a straight path approaches a circular path, it experiences a sudden centripetal acceleration starting at the tangent point; and thus centripetal force acts instantly causing much discomfort.

On early railroads this instant application of lateral force was not an issue since low speeds and wide-radius curves were employed (lateral forces on the passengers and the lateral sway was small and tolerable). As speeds of rail vehicles increased over the years, it became obvious that an easement is necessary so that the centripetal acceleration increases linearly with the traveled distance. Given the expression of centripetal acceleration $V² / R$ , the obvious solution is to provide an easement curve whose curvature, $1 / R$ , increases linearly with the traveled distance. This geometry is Euler spiral.

Unaware of the solution of the geometry by Leonhard Euler, Rankine cited cubic curve (a polynomial curve of degree 3), which is an approximation of Euler spiral for small angular change as parabola is to circular curve.

Marie Alfred Cornu (and later some civil engineers) also solved the calculus of Euler spiral independently. Euler spirals are now widely used in rail and highway engineering for providing a transition or an easement between a tangent and a horizontal circular curve.

[edit] Formulation

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[edit] Symbols

$R\,$	Radius of curvature
$R_c\,$	Radius of Circular curve at the end of the spiral
$\theta\,$	Angle of curve from beginning of spiral (infinite $R$ ) to a particular point on the spiral.
	This can also be measured as the angle between the initial tangent and the tangent at the concerned point.
$\theta _s\,$	Angle of full spiral curve
$L , s\,$	Length measured along the spiral curve from its initial position
$L_s , s_o\,$	Length of spiral curve

[edit] Derivation

Imagine an Euler spiral connect between a tangent and a circular curve:

the tangent extends from –ve x direction to the origin; and
the spiral starts at the origin in the +ve x direction and turns slowly in anticlockwise direction to meet a circular curve tangentially.

Graphically, the spiral can be visualized as a small portion of the curve in the first quadrant of the above illustrated curve. The tangent lies on the -ve x axis and the circular curve connects tangentially at the end of the spiral.

From the definition of the curvature,

$\frac {1}{R} = \frac {d\theta}{dL} \propto L$

i.e.

$R L = \text{constant} = R_c L_s\,$

$\frac {d\theta}{dL} = \frac {L}{R_c L_s}$

We write in the format,

$\frac {d\theta}{dL} = 2a^2 L$

Where

$2a^2= \frac {1}{R_c L_s}$

Or

$a = \frac {1}{\sqrt {2R_c L_s} }$

Thus

$\theta = (a L)^2\,$

Now

$\begin{align} x & = \int_0^L dL \cos\theta \\ & = \int_0^L \cos (a L)^2 dL \end{align}$

If

$L' = a L \,$

Then

$dL = \frac{dL'}{a}\,$

Thus

$x = \frac{1}{a} \int_0^{L'} \cos L'^2 dL'$

$\begin{align} y & = \int_0^L dL \sin\theta \\ & = \int_0^L \sin (a L)^2 dL \\ & = \frac{1}{a} \int_0^{L'} \sin L'^2 dL' \end{align}$

[edit] Expansion of Fresnel integral

Main article: Fresnel integral

If a = 1, which is the case for normalized Euler curve, then the formula for the Cartesian coordinates are Fresnel integrals (or Euler integrals):

$C(L) =\int_0^L\cos L^2 \, dL$

$S(L) = \int_0^L\sin L^2 \, dL$

Expand C(L) according to power series expansion of cosine:

$\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \cdots$

$C(L) = \int_0^L \cos L^2 \, dL$

$= \int_0^L (1 - \frac{L^4}{2!} + \frac{L^8}{4!} - \frac{L^{12}}{6!} + \cdots) \, dL$

$= L - \frac{L^5}{5 \times 2!} + \frac{L^9}{9 \times 4!} - \frac{L^{13}}{13 \times 6!} +\cdots$

Expand S(L) according to power series expansion of sine:

$\sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots$

$S(L) = \int_0^L \sin L^2 \, dL$

$= \int_0^L (L^2 - \frac{L^6}{3!} + \frac{L^{10}}{5!} - \frac{L^{14}}{7!} + \cdots) \, dL$

$= \frac{L^3}{3} - \frac{L^7}{7 \times 3!} + \frac{L^{11}}{11 \times 5!} - \frac{L^{15}}{15 \times 7!} +\cdots$

[edit] Normalization and conclusion

For a given Euler curve with:

$2RL = 2R_c L_s = \frac{1}{a^2} \,$

or

$\frac{1}{R} = \frac{L}{R_c L_s} = 2a^2L \,$

then

$x=\frac{1}{a} \int_0^{L'} \cos L'^2 \, dL'$

$y=\frac{1}{a} \int_0^{L'} \sin L'^2 \, dL' \,$

where $L' = aL \,$ and $a = \frac{1}{\sqrt{2R_c L_s}}$ .

The process of obtaining solution of $(x, y)$ of an Euler spiral can thus be described as:

Map L of the original Euler spiral by multiplying with factor $a$ to $L'$ of the normalized Euler spiral;
Find $(x', y')$ from the Fresnel integrals; and
Map $(x', y')$ to $(x, y)$ by scaling up (denormalize) with factor $1 / a$ . Note that $1 / a > 1$ .

In the normalization process,

$\begin{align} R'_c & = \frac{R_c}{\sqrt{2 R_c L_s}} \\ & = \sqrt{\frac{R_c}{2L_s}} \\ \end{align}$

$\begin{align} L'_s & = \frac{L_s}{\sqrt{2R_c L_s}} \\ & = \sqrt{\frac{L_s}{2R_c}} \end{align}$

Then

$\begin{align} 2R'_c L'_s & = 2 \sqrt{\frac{R_c}{2L_s} } \sqrt{\frac{L_s}{2 R_c}} \\ & = \tfrac{2}{2} \\ & = 1 \end{align}$

Generally the normalization reduces L' to a small value (<1) and results in good converging characteristics of the Fresnel integral manageable with only a few terms.

[edit] Illustration

Given:

$\begin{align} R_c & = 300\mbox{m} \\ L_s &= 100\mbox{m} \end{align}$

Then

$\begin{align} \theta_s & = \frac{L_s} {2R_c} \\ & = \frac{100} {2 \times 300} \\ & = 0.1667 \ \mbox{radian} \\ \end{align}$

And

$2R_c L_s = 60,000 \,$

We scale down the Euler spiral by √60,000, i.e.100√6 to normalized Euler spiral that has:

$\begin{align} R'_c = \tfrac{3}{\sqrt{6}}\mbox{m} \\ L'_s = \tfrac{1}{\sqrt{6}}\mbox{m} \\ \\ \end{align}$

$\begin{align} 2R'_c L'_s & = 2 \times \tfrac{3}{\sqrt{6}} \times \tfrac{1}{\sqrt{6}} \\ & = 1 \end{align}$

And

$\begin{align} \theta_s & = \frac{L'_s}{2R'_c} \\ & = \frac{\tfrac{1}{\sqrt{6}}} {2 \times \tfrac{3}{\sqrt{6}}} \\ & = 0.1667 \ \mbox{radian} \\ \end{align}$

The two angles $\theta_s\,$ are the same. This thus confirm that the original and normalized Euler spirals are having geometric similarity. The locus of the normalized curve can be determined from Fresnel Integral, while the locus of the original Euler spiral can be obtained by scaling back / up or denormalizing.