# horizontal curves. introduction use of curves, horizontal and vertical. types of horizontal curves:...

Post on 26-Mar-2015

216 views

Embed Size (px)

TRANSCRIPT

- Slide 1

Horizontal Curves Slide 2 Introduction Use 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 sections. Simple Curve: circular arc connecting two tangents. The most common Spiral Curve: a curve whose radius decreases uniformly from infinity at the tangent to that of the curve it meets. Slide 3 Compound Curve: a curve which is composed of two or more circular arcs of different radii tangent to each other, with 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. Slide 4 Slide 5 Easement 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 spiral curve, inversely proportion to the radius. When to Use What Simple 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. Slide 6 Degree of Circular Curve Curves are identified either by: Radius: 1000-m radius Degree 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 = 5729.58 R In chord definition, D = 50 R 2 sin -1 () degrees, R in ft. Pay attention to units, we will use ft for length, how about angles? Slide 7 Slide 8 Circular Curves Notations Definitions: Point of intersection (vertex) PI, back and forward tangents. Point of Curvature PC, beginning of the curve Point of Tangency PT, end of the Curve. Tangent Distance T: Distance from PC, or PT to PI Long Chord LC: the line connecting PC and PT Length of the Curve L: distance for PC to PT: measured along the curve, arc definition. measured along the 100 chords, chord definition External Distance E: The length from PI to curve midpoint. Middle ordinate M: the radial distance between the midpoints of the long chord and curve. POC: any point on the curve. POT: any point on tangent Intersection Angle I: the change of direction of the two tangents, equal to the central angle subtended by the curve. Slide 9 Slide 10 Circular Curves Formulas Remember that : R is radius, perpendicular to the tangents at PC, and PT D is the curve degree, use ARC definition. L = 100 I D ft, D and I in same units. = I D (sta)= R I I in rad R = 5729.58 D (ft) LC = 2R sin ( I 2 ) T = R tan ( I 2 ) E = R [ 1 cos (I/2) - 1 ] M = R ( 1 - cos I 2 ) E = T tan ( I 4 ) M = E cos I 2 Slide 11 Circular Curve Stationing --Route Survey are usually staked out as a series of 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 + L Station equations at PT: the route considering the curve is 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 ( I 2 ) Slide 12 Circular Curves Layout by Deflection Angles with a Total Station or an EDM All stations will be positioned from PC. Compute the chord length and the deflection angle from the direction PC-PI as follows: (see fig 25-6) Example a = S a D 200 (degrees) Where: d a = D SaSa 100 or, d a = S a D 100 Theory; the angle between the tangent and a chord is equal to half the central angle subtended by the chord, so get a Also, sin a = CaCa 2R from which C a = 2R sin a C a = 2R sin a Slide 13 In 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 ft then, = (82.92) (2)/200 = 0.8292 = 00 49 45 C= 2 (5729.58/2) sin(00 49 45) = 82.92 ft At station 64+00, Sa = 182.92 ft Then = (182.92) (2)/200 = 1.8292 = 1 49 45 C = 2 (5729.58/2) sin(1 49 45) = 182.89 ft At the PT: Sa = ?, and Ca = ? Slide 14 Slide 15 Slide 16 PI (V) PC T C Slide 17 Circular Curve Layout by Coordinates with a Total Station Given: Coordinates and station of PI, a point from which the curve could be observed, a direction (azimuth) from that point, AZ PI-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 mentioned for future reference and will not be covered.} Slide 18 Solution: - from X PI, Y PI, T, AZ PI-PC, compute X PC, Y PC - compute the length of chords and the deflection angles. - use the deflection angles and AZ PI-PC, compute the azimuth of each chord. - knowing the azimuth and the length of each chord, compute the coordinates of curve points. - for each curve point, knowing its coordinates and the total station point, compute the azimuth and the length of the line connecting them. - at the total station point, subtract the given direction from the azimuth to each curve point, get the orientation angle. Slide 19 Slide 20 Special Circular curve Problems Passing 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 the coordinates of PI, PC. In that system the coordinates of the origin O is: Xo = -T = -R tan (I/2) Yo = -R Slide 21 2- Measure the angle and the distance PI-P 3- Compute the coordinates of P: Xp = - d cos Yp = - d sin 4- Substitute in the general equation of a circle: R 2 = (X P Xo ) 2 +(Y P Yo) 2 Solve the equation to compute R: R 2 = ( X P + R tan I 2 ) 2 + (Y P + R) 2 Slide 22 Slide 23 Intersection of a circular curve and a straight line Form the line and the circle equations, solve them simultaneously to get the intersection point. Intersection of two Circular Curves simultaneously solve the two circle equations. Slide 24 Sight distance on Horizontal Curves {Required topic} What 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 less than the minimum safe sight distance: - Move the obstruction - Reduce the speed. Slide 25 Slide 26 PI (V) PC T C Slide 27 Spiral Curves {This topic will not be covered} Used to provide gradual transition in horizontal curvature, and hence superelevation. Definitions: Back and forward tangents. Entrance and exit spirals. Geometrically identical. 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 L S : the arc length of the spiral. Slide 28 Spiral Geometry Basic spiral properties: Radius changes uniformly from infinity at TS to the radius of the circular curve at the SC. So, its degree of curve D S changes uniformly from 0 o to D at the SC. Average degree of curve is D/2. In circular curves, L = (I/D) 100 ft, or I = LD stations similarly, S = L s (D/2) S and D in deg, L in stations Spiral angles at any point is proportional to the square of the distance L p from TS to the point. P = S In Fig 25-15, M is the mid point of the spiral, L p = L s /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/4 Then, M = ( L s /2) (D/4) = (L s D/8) = S /4 L P LS)2LS)2 ( Slide 29

Recommended