Circumferences must be tangent to clothoids, and radius of curvature must be the same at each tangent point.
Euler spiral (Clothoid)
Clothoid (Euler spiral) is a curve whose curvature [color=#0000ff][i]k[/i][/color] changes linearly with its curve length (denote [color=#0000ff][i]s[/i][/color] or [color=#0000ff][i]L[/i][/color]). [br][center] [math] k=\frac{L}{A^2}[/math]. [/center] [br]Clothoids are widely used as transition curves in railroad engineering for connecting and transiting the geometry between a tangent and a circular curve.[br]Clothoid has the desirable property that the curvature [color=#0000ff][i]k[/i][/color] is linearly related to the arc length [color=#0000ff][i]L[/i][/color]. Although its defining formulas for coordinates are transcendental functions (Fresnel integrals), the important characteristics can be derived easily from equation[color=#0000ff] [i]k = L/A[sup]2[/sup] [/i][/color] where [i]A[sup]2[/sup][/i] is constant. [br]Some applications avoid using transcendental functions by proposing polynomial approximations to the clothoid, i.e.[br][center] [math]y=\frac{x^3}{6A^2}[/math]. [/center] [br][br][img]https://www.geogebra.org/resource/vvdswmpf/Yu4eL8i7AD6OlKYu/material-vvdswmpf.png[/img]
Task 1
Determine the length [color=#0000ff][i]L[/i][/color] of Euler spiral [math]k=\frac{L}{12^2}[/math] for transition between straight road and circular arc of radius [color=#0000ff][i]r = 9 m[/i][/color].[br]Solution: Curvature of a bend must be the same as curvature of a clothoid, i.e. [math]\frac{1}{9}=\frac{L}{12^2}[/math] and [color=#0000ff][i]L = 16 m[/i][/color].
Clothoid k = L/A
Task 2
Determine an angle length [color=#0000ff][i]α [/i][/color] between the tangent of Euler spiral [math]k=\frac{L}{12^2}[/math] at [color=#0000ff][i]L = 16 m[/i][/color] and x-axis. [br][br]Solution: Formula for direction part of a clothoid. [br] [math]\alpha=\frac{L^2}{2.A^2}=\frac{16^2}{2.12^2}=\frac{8}{9}[/math] and [color=#0000ff][i][color=#0000ff][i]α[/i][/color] = 50,92°[/i][/color].
Clothoid between two circumferences
Task 3: Cubic Approximation
Replace the Euler spiral [math]k=\frac{L}{15^2}[/math] with a cubic parabola.[br][br]Solution: From the Taylor expansion of degree 3, we get an approximation of the form [br][math]y=\frac{x^3}{6A^2}[/math]. [br]Substituting [i]A[/i]=15, we get the third degree algebraic transient expressed explicitly as [math]y=\frac{x^3}{6\cdot15^2}[/math].
Proof:
Slide 60, prezentation [url=https://docs.google.com/presentation/d/e/2PACX-1vQFV-RCjmsOYPsxR9mG-2VwsnOVDnTIyC-_qfNIJkPJj8pPfNgDWNYgVI2E7fu5Rarb3wYsBxH-A1ii/pub?start=false&loop=false&delayms=3000]07_Curves[/url]