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1、HorizontalCurvesIntroductionUse of curves, horizontal and vertical.Types of horizontal curves: Circular and spiral. We will cover circular curves only, spiral curves are given for future reference.Definitions:Horizontal Curves: curves used in horizontal planes to connect two straight tangent section
2、s.Simple Curve: circular arc connecting two tangents. The most commonSpiral Curve: a curve whose radius decreases uniformly from infinity at the tangent to that of the curve it meets.Compound Curve: a curve which is composed of two or more circular arcs of different radii tangent to each other, with
3、 centers on the same side of the alignment.Broken-Back Curve: the combination of short length of tangent (less than 100 ft) connecting two circular arcs that have centers on the same side.Reverse Curve: Two circular arcs tangent to each other, with their centers on opposite sides of the alignment.Ea
4、sement Curves: curves used to lessen the effect of the sudden change in curvature at the junction of either a tangent and a curve, or of two curves.Super elevation: a difference of elevation between the edges of the cross section, to overcome the effect of centrifugal force. Changes gradually in a s
5、piral curve, inversely proportion to the radius.When to Use WhatSimple circular curves are the most common type.Spirals are used at highway exits, sometimes, and all the times in railroad curves.The rest of curves are used when the designer has to. Degree of Circular CurveCurves are identified eithe
6、r by:Radius: 1000-m radiusDegree of Curve (D): central angle subtended by a 100ft: circular arc (in arc definition), or chord in chord definition.The arc definition is commonly used in highways, chord definition is commonly used in railroads.In arc definition, D =5729.58RIn chord definition, D =50R
7、2 sin-1() degrees, R in ft.Pay attention to units, we will use ft for length, how about angles?Circular Curves NotationsDefinitions: Point of intersection (vertex) PI, back and forward tangents.Point of Curvature PC, beginning of the curvePoint of Tangency PT, end of the Curve.Tangent Distance T: Di
8、stance from PC, or PT to PILong Chord LC: the line connecting PC and PTLength of the Curve L: distance for PC to PT:measured along the curve, arc definition.measured along the 100 chords, chord definitionExternal Distance E: The length from PI to curve midpoint.Middle ordinate M: the radial distance
9、 between the midpoints of the long chord and curve.POC: any point on the curve.POT: any point on tangentIntersection Angle I: the change of direction of the two tangents, equal to the central angle subtended by the curve. Circular Curves FormulasRemember that : R is radius, perpendicular to the tang
10、ents at PC, and PT D is the curve degree, use ARC definition.L = 100 IDft, D and I in same units. = ID(sta) = R I I in radR = 5729.58D(ft)LC = 2R sin(I2)T = R tan(I2)E = R1cos (I/2)- 1M = R(1 - cosI2)E = T tan(I4)M = E cos I2Circular Curve Stationing-Route Survey are usually staked out as a series o
11、f tangents having continuous stations. The station of the PI and the value of (I) are determined.Stations of PC, and PT are computed from PI, R is given by designers: Compute Station of PC = station of PI - T Station of PT = station of PC + LStation equations at PT: the route considering the curve i
12、s shorter than it was computed considering the tangents.= (station of PI + T) - (station of PC + L)This amount should be subtracted from stations of all the points after PT. T = R tan(I2)Circular Curves Layout by Deflection Angles with a Total Station or an EDMAll stations will be positioned from PC
13、. Compute the chord length and the deflection angle from the direction PC-PI as follows: (see fig 25-6)Exampleda= Sa D200(degrees)Where: da = DSa100or, da= Sa D100Theory; the angle between the tangent and a chord is equal to half the central angle subtended by the chord, so get daAlso, sinda= Ca2Rfr
14、om which Ca = 2R sin daCa = 2R sin daIn a curve whose I = 8 24, station of PC is 62+ 17.08, D = 2 00, calculate the necessary information to stake out points at stations 63+00, 64+00, and at the PT.Answer: . a= Sa D/200 deg, and Ca = 2R sin a . At station 63+00, Sa = 6300 6217.08 = 82.92 ftthen, = (
15、82.92) (2)/200 = 0.8292 = 00 49 45”C= 2 (5729.58/2) sin(00 49 45”) = 82.92 ftAt station 64+00, Sa = 182.92 ftThen = (182.92) (2)/200 = 1.8292 = 1 49 45” C = 2 (5729.58/2) sin(1 49 45”) = 182.89 ftAt the PT: Sa = ?, and Ca = ?PI (V)PCTCCircular Curve Layout by Coordinates with a Total StationGiven: C
16、oordinates and station of PI, a point from which the curve could be observed, a direction (azimuth) from that point, AZPI-PC , and curve info.Required: coordinates of curve points (stations or parts of stations)and the data to lay them out.this topic and all the following until “sight distance” is m
17、entioned for future reference and will not be covered.Solution: - from XPI, YPI, T, AZPI-PC, compute XPC, YPC - compute the length of chords and the deflection angles. - use the deflection angles and AZPI-PC, compute the azimuth of each chord. - knowing the azimuth and the length of each chord, comp
18、utethe coordinates of curve points. - for each curve point, knowing its coordinates and the totalstation point, compute the azimuth and the length of the line connecting them. - at the total station point, subtract the given direction from theazimuth to each curve point, get the orientation angle.Sp
19、ecial Circular curve ProblemsPassing a curve through a certain point:- When?-The problem: fig(24-15) Given PI, point (P) that should be on the curve, and the tangents. Required: R. Solution: 1-Establish an arbitrary coordinate system, origin is at PI, X axis is the line PC-PI. In that system we know
20、 the coordinates of PI, PC. In that system the coordinates of the origin O is:Xo = -T = -R tan (I/2) Yo = -R 2- Measure the angle and the distance PI-P3- Compute the coordinates of P: Xp = - d cos Yp = - d sin 4- Substitute in the general equation of a circle: R2 = (XP Xo )2 +(YP Yo)2Solve the equat
21、ion to compute R:R2 = (XP + R tanI2)2+ (YP+ R)2Intersection of a circular curve and a straight lineForm the line and the circle equations, solve them simultaneously to get the intersection point.Intersection of two Circular Curvessimultaneously solve the two circle equations.Sight distance on Horizo
22、ntal CurvesRequired topicWhat is the problem?Stopping distance depends on: speed, perception and reaction time, coefficient of friction, and pavement condition.Available sight distance = C =Where m is the distance from the obstruction to the center of the road, along a radius.Two solutions if C is l
23、ess than the minimum safe sight distance:- Move the obstruction- Reduce the speed.PI (V)PCTCSpiral CurvesThis topic will not be coveredUsed to provide gradual transition in horizontal curvature, and hence superelevation.Definitions:Back and forward tangents.Entrance and exit spirals. Geometrically i
24、dentical.TS, SC, CS, ST. What is in between?SPI: the angle beteen the tangents at TS and SC.Spiral Angle S: the angle between the two tangents. Spiral Length LS: the arc length of the spiral.Spiral GeometryBasic spiral properties:Radius changes uniformly from infinity at TS to the radius of the circ
25、ular curve at the SC. So, its degree of curve DS changes uniformly from 0o to D at the SC. Average degree of curve is D/2.In circular curves, L = (I/D) 100 ft, or I = LD stationssimilarly, S= Ls (D/2) S and D in deg, L in stationsSpiral angles at any point is proportional to the square of the distan
26、ce Lp from TS to the point. P = S In Fig 25-15, M is the mid point of the spiral, Lp = Ls/2 but M is not = (S /2).Since D changes uniformly, degree of the curve = D/2 at M. But D changes uniformly, so the average degree of curvature between TS and M is (D/2)/2 = D/4Then, M = ( Ls/2) (D/4) = (Ls D/8) = S /4 LPLS)2(
