Compound and Reserve Curves

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Compound and Reserve Curves

  1. 1. Highway Curves 1 20/01/2013
  2. 2. A compound curve consists of two or more circular arcs betweentwo main tangents turning in the same direction and joining at common tangentpoints. Rarely used to provide gradually transition from straight lines to curves.Mainly are found in the design of interchange loops and ramps. 2 20/01/2013
  3. 3. O1 Compound Circular Curve 1 R1 – R2 180 - 2 O2 290 90BC t1 EC t1 PCC t2 2 V2 V1 1 180 -  t2 V1V2 = t1 + t2  V 3 20/01/2013
  4. 4. Compound Circular Curves Refer to previous Figure;• PCC, Point of Compound Curve.• Curve 1, is of the lower station.• Curve 2, is of the higher station.• Knowing 4 of these parameters; R1, R2, 1, 2 (1 + 2 = ), T1 andT2, the others will be known.• In general, 1 and 2 or  are measured in the field.• The minimum values for R1 and R2 are governed by designspeed. The larger radius  1(1/3) times (the smaller radius). Thisratio increases to 1(1/2) when dealing with interchange curves. 4 20/01/2013
  5. 5. Compound Circular CurvesAll Problems of Compound Curves Can be Solved by:• The sine law.• The cosine law.• A five-sided traverse with sides R1, T1, T2, R2 and (R1 – R2) andwith angles 90, 180 - , 90, 180 - 2 and 1.• For simplification, set the assumed azimuth, the direction of R1 to(0 00 00). 5 20/01/2013
  6. 6. Solution of Compound Circular Curves V  V1 1 C t1 2 V2 t2 B E 1 2 R1  R2 & T1  T2 O2  = 1 + 2 ( is an interior angle of triangle V1VV2) 6 20/01/2013
  7. 7. Solution of Compound Circular CurvesLengths of t1 and t2 :V1B = V1C = t1 = R1 tan (1/2)V2E = CV2 = t2 = R2 tan (2/2)V1V2 = t1 + t2 = R1 tan (1/2) + R2 tan (2/2) 7 20/01/2013
  8. 8. Solution of Compound Circular CurvesLengths of T1 and T2 : Applying the sine law in triangle V1VV2V1V = V1V2(sin2/sin) = (t1 + t2)(sin2/sin)V2V = V1V2(sin1/sin) = (t1 + t2)(sin1/sin)T1 = V1B + V1V = t1 + (t1 + t2)(sin2/sin)T1 = R1 tan(1/2) + [R1 tan (1/2) + R2 tan (2/2)](sin2/sin)T2 = V2E + V2V = t2 + (t1 + t2)(sin1/sin)T2 = R2 tan(2/2) + [R1 tan (1/2) + R2 tan (2/2)](sin1/sin) 8 20/01/2013
  9. 9. Problem B T1 V(PI)  Refer to the opposite Figure, T1 = 380 ft, R1 = 500 ft, C 1 = 2115,  = 3950 and the PI is at station 136 + 42.85. Using the chord definition for degree of curve,R1 1 compute the T2, R2 and 2 and the stationing of the C and the E. E 2 The Compound-curve equations can be developed by considering the polygon O1-B-V-E-O2 as a five-sided traverse. Since the traverse is closed, the algebraic sums of the latitudes and departures must equal zero. The values for this traverse are shown in the following Table: 9 20/01/2013
  10. 10. Basis of Equations for Compound CurvesAzimuths and Lengths of Polygon Sides Departures Latitudes Side Azimuth Length E W N S O1B 0 R1 0 R1 BV 90 T1 T1 0 VE 90 +  T2 T2 cos  T2 sin  EO2 180 +  R2 R2 sin  R2 cos  O2O1 1 R2 – R1 (R2 – R1) sin 1 (R2 – R1) cos 1 From the departures: T1 + T2 cos  - R2 sin  + (R2 – R1) sin 1 = 0 …………………………………(1) From the latitudes: T1 - T2 sin  - R2 cos  + (R2 – R1) cos 1 = 0 …………………………………..(2)  = 1 + 2 …………………………………………………………………………….(3) From these three equations, T2 , R2 and 2 can be found. 10 20/01/2013
  11. 11. Reverse curves are seldom used in highway or railway alignmentdue to their instantaneous change in direction at PRC. Pleasing to eye and areused for park roads, formal paths, waterway channels…..etc. • Like compound curves, reverse curves have six independent parameters; (R1, 1, T1, R2, 2, T2). • The solution of reverse curves problems are the same as the solution applied to compound curves problems. 11 20/01/2013
  12. 12. Reverse CurvesBack Tangent BC PI1 R2 a) Nonparallel tangents 1 PRC, Point of Reversed Curve 2 PRC R1 EC PI2 2 1 R1 R2 2 1 R2 Back Tangent PI1 PRC Forward Tangent BC 2 EC PI2 R1 R1 b) parallel tangents 1 Often, R1 = R2 12 20/01/2013
  13. 13. Problem O2 Refer to the opposite Figure, 1 and 2 are two parallel lines, s = 250 m. We 1 B T1 V1 would like to connect the two lines by reversed curves consisted of two simple 1 R2 circular curves, R1 = 200 m and R2 = 300 m. Compute the main elements of F G the reversed curves, T1 and T2, 1,2 R1 Cs and , t1 and t2. 2 O1 2 V2 T2 E Solution: T1 =  & T2 =   = 0 & 1 = 2 13 20/01/2013
  14. 14. SolutionFrom the intersection point of joint tangent, C, establish a line FG parallel tolines 1 and 2;s = BF + EGBF = R1(1 – cos 1)EG = R2(1 – cos 1)s = (R1 + R2) – (R1 + R2) cos 1cos 1 = [(R1 + R2) – s]/(R1 + R2)1 = 2 = 60 00 00t1 = R1 tan (1/2) = 115.47 mt2 = R2 tan (2/2) = 173.21 m 14 20/01/2013

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