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Curves. Horizontal. 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). S a D. C a = 2R sin  a.  a =. (degrees).

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Horizontal

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  1. Curves Horizontal

  2. 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) Sa D Ca = 2R sin a a= (degrees) 200 Where: da = D Sa D = or, da = Sa 100 100 Theory; the angle between the tangent and a chord is equal to half the central angle subtended by the chord, so get a Ca from which Ca = 2R sin a Also, sin a = 2R

  3. Circular Curve Layout by Coordinates with a Total Station {this topic and all the following until “sight distance” is mentioned for future reference and will not be covered.} Given: Coordinates 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.

  4. 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, compute the coordinates of curve points. - for each curve point, knowing it’s 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.

  5. Special Circular curve Problems Passing a curve through a certain point: - When? -The problem: fig(25-10) 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. 2- Measure the angle and the distance PI-P 3- Compute the coordinates of P 4- Solve the equation: (XP + R tan ) 2 I R2 = + (YP+ R)2 2

  6. 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.

  7. 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 sight distance: - Move the obstruction - Reduce the speed.

  8. 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 LS: the arc length of the spiral.

  9. Spiral Geometry • Basic spiral properties: • Radius changes uniformly from infinity at TS to the radius of the circular curve at the SC. So, it’s 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 stations • similarly, S= Ls (D/2) S and D in deg, L in stations • Spiral angles at any point is proportional to the square of the distance 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/4 • Then, M = ( Ls/2) (D/4) = (Ls D/8) = S /4 LP ( LS)2

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