Tiêu đề 119 Spiral Curves.pdf ✅

MSTC 119: Spiral Curves 1. Spiral Transition Curve Along a horizontal curve, the vehicle experiences an outward (radial) acceleration of v 2 R , which it does not experience before entering the curve. For the comfort of the passengers, the rate of change of the radial acceleration should be constant. This is the purpose of a spiral easement curve. The locus of a point satisfying this condition forms a clothoid (or simply, spiral) shape. aR t = constant v 2 Rt = constant If t is the time taken by the vehicle to cover a length along the spiral, then v = L t or t = L v , v 2 Rt = constant v 2 R ( Ls v ) = constant v 3 RL = constant When the whole spiral is covered, R = Rc and L = Ls , v 3 RcLs = constant 1.1. Desirable Length of Spiral The acceptable rate of change of the radial acceleration is somewhere between 1 ft/s2 and 3 ft/s2 . Using the average value, 2 ft/s2 = 0.6096 m/s2 and expressing the design velocity in kph, ( K 3.6 ) 3 RLs = 0.6096 Ls = K 3 0.6096(3.6 3)R Ls = 0. 036K3 R ; K in kph,R in m The last equation gives the desirable length of the spiral.

At point SC, L = Ls , θc = L 2Rc The values computed here are in radians since they are derived from calculus. To convert to degrees, simply multiply by 180° π . To find the offset of the spiral from the original tangent, sin θ = dx dL Since these differentials concern very small angles, the small angle theory approximates that sin θ ≈ θ. θ = dx dL From earlier, θ = L 2 2RcLs L 2 2RcLs = dx dL dx = L 2dL 2RcLs To find the offset distance, simply integrate the previous equation. x = ∫ L 2dL 2RcLs x = L 3 6RcLs + C At the point of curvature x = 0 and L = 0, 0 = 0 3 6RcLs + C C = 0 Therefore, the offset distance is x = L 3 6RcLs At point SC, L = Ls , xc = L 2 6Rc
2.1. Distance along a Tangent From the figure, dy = dL − ε Using the Pythagorean theorem, dL 2 = dx2 + dy 2 dL 2 = dx2 + (dL − ε) 2 0 = dx 2 − 2dL ε (ε 2 is negligible) ε = dx 2 2dL ε = dL 2 ( dx dL) 2 From earlier, L 2 2RcLs = dx dL , ε = dL 2 ( L 2 2RcLs ) 2 ε = L 4dL 8RcLs Substituting in dy, dy = dL − L 4dL 8RcLs To find the distance along the tangent, simply integrate the previous equation. y = L − L 5 40RcLs + C At the point of curvature y = 0 and L = 0, 0 = 0 − 0 5 40RcLs + C C = 0 Therefore, the offset distance is y = L − L 5 40RcLs