PART VICOMPOUND CURVESAny simple circular curve used in a highway, railway or canal presentspro bl ems.At the P. C. there is an instantaneous change of direction and themagnitude of this change increases as the radius decreases. In the case of ahighway or railway, a vehicle or a train cannot change direction instantaneously.In the case of the vehicle, it does not follow a truly circular pathin the first part of the curve. In the case of the train it imposes forceson the rails which may cause them to shift. The magnitude of the forces willincrease as R decreases assuming that the velocity remains constant.There is a1 so the problem of super-elevating or " bankingu a simp1 e circularcurve. The greater the design speed the greater will be the need for superelevation.Theoretically, there should be no superelevation on one side of theP. C. and there should be the required superel evation on the other, with notransition in between. Because such a situation is impractical we have to compromiseand perhaps begin our superelevation before the P.C. is reached so thatthe first part of the curve has, at least, part of the required superelevation.The situation becomes worse as R is decreased, for a constant design speed.Obviously what has been said in the preceding two paragraphs applies atthe P.T. as well.On canals, too rapid a change in direction could lead to erosion problems.Because the magnitude of the problems of change of direction and superelevationincrease as R decreases we can, at least partially, relieve the problemby combining arcs of simple curves of different radius as shown.The curve shown in Ffg . I consi sts of two outer arcs, both having the samelong radius and the same central angle. The central portion has s short radius.In this case the curve is symmetrical about the dotted Q and is called aSymmetrical ly Compounded Curve.The long radius portion between the P.C. and the P.C.C. (Point of CompoundCurvature) a1 lows for a more gradual change in direction and a more gradual(and smoother) appl icat ion of superel evation so that the short radius portionwill be safe. The second long radius portion (from the second P.C.C. to theP.T.) allows for a gradual change back to straight-line conditions beyond theP. T. Figure 1 We can have a multi-centred curve as illustrated below*This is a multi-centeredIn terns of riding and drivingqualities it would be animprovement over thesymnetri call y-compoundedcurve in Figure 1 .Later we will discuss a bettersolution to the problem-theso-called transition spiralFigure 2There are many applications of mu1 ti-centered curves which are notsymmetrical-particularly in various types of grade-separation structures. Calculate the sub-tangentdistance, T, for thesymnetrically-compoundedcurve shown above.In this case we can make use of symnetry to set up a simple solution.We can treat A00 and B0C each as separate self-contained curves. ThroughB we can draw a 1 ine V, BV2, I to BO, which is tangent to both curves.Figure 4 AV, = VIB = TIBVp = V2C = TZVIVp = TI + T2180-A - 600 Angle AVV2 = - - 2Angle VV V- A1 = 15' 1 2-By deduction, Angle V, V2V = 180' - 60'-15' = 105'T, = 500 x Tan 7'30' = 65.82We can use the sin law to solve for V1 Y in the triangle VVIVZWe can now try an example where the curve is not symmetrical. 2. Example TwoFigure 5Figure 6 AVl = VIB = TI = 400 Tan 10' = 70.53BV2 = V2C = T2 = 300 Tan lo0= 52.89CV3 = VD= T = 200 Tan lo0= 35.27 3 3In Figure 6AngleWe must now calculate the 1engthsAV and VD by the method of latitudesand departures.The approach to all such problems will be the same. As the number ofdifferent arcs is increased the only change will be the length of thelatitudes and departures problem.4'5;VV1V1 V2vzv3V3V1Traverse DataLenqth Bearing .- Ass. West123.43 ~70'~88.16 ~50'~114.19 ~30'~0Functions Latitudes-I-98.880DeparturesCos-34202.64279.86603-42.2156.6798.88+115.9867.54183.52Sin.93969.76604.30000I126.43(4)57.09(3) 1. c Southings = 98.88 = c NorthingsAll Northings are in V3V = 5' 98.882. V3V x Cos 30' = 98.8898.88 - 98.88 =V3v = pp Cos 300 * .866033. Dep. in V3V = 114.19 x .50000 =4. c Eastings = 183.52 = c WestingsBalance of Westings is VV1VV,Back=Tangent183.52=-AV57.09= VV1=v+ T1Forward Tangent = VD = VV3 + T3, = 11419 + 35.27VD = 149.45,Note that, in any compound curve, the curve must be run in sections,the instrument occupying the beginning or end of each section ofdifferent radius.It should also be noted that it is necessary to put in controlindependent of the curve in order to isolate any errors that may bemade. In the case of the assymetrical compound curve in Figures 5 and 6it is highly desirable (in fact, essential) that points V1, B, V2, C andV3 be 1 ocated before one begins running the curve.3. Example ThreeTwo tangents intersect at n = 70, P. I. = Stn 32 + 51 -82. Setup the field notes for the following compound curve (symmetrical).1st curve A = 20, D = 8'~ (20 rn chords)2nd curve A = 300, D = 120R (10 m chords)3rd curve A = 200, D = 80R (20 m chords) Solution1st curve R1 =2 08 = 143.36 m2 sin (Z-)R1 Tan A TI =2 0 1 = (143.36) tan (-T) = 25.28 m22nd curve R2 ='to2 = 47.83 m2 sin (1) 2R2Tan A2T2 = 30 - 2 = (47.83) tan - 12.82 rn3rd curve = same as 1st curve Figure 7 Figure 8AV, TI = 25.28mIn A V1V2VVIV2 = TI + T2 25.28 + 12.82using sin lawV,V = 44.93 m. . Back tangent = AV1 + VIV = 25.28 + 44.93Since curve is symmetrical the forward tangent = back tangent = 70.21 rn P.I. = Stn 32 t 51.82-B.T. 00 + 70.21P.C. 31 + 81.61CheckP.C. 31 + 81.51+ LCC1 + 25.00P,T, 33 + 06.61P.T. 33 + 06.61Figure 9Def 1Station33 + 6.61 P.T.33 + 0032 + 8032 + 6032 + 56.61 P.C .C. 232 + 5032 + 40-32 + 31.61 P.C.CmlNOTE: The checks of P.C.Cm1, P.C.C. and P.T. are essential! 2 'Angler 32 + 0031 + 81.61 P.C.-32 + 20 7'40 ' 40 "35~~'~~"33'40 ' 40"29'40 ' 40"25'40 ' 40 "2500'001121 '02 '00"I 5'02' 00"10~00'00"3'40 ' 40 "oOOO' 00"Curve #1 D = 8' .'. d = 4'= (18.39)(8)(40 = 340140~d2 = (11.61)(8)(40 = 219'20"=bT/2 curve 82 D = 12' .*.d = 6 0dl = (8.39) (12)/20 = 502t00"d2 = (6.61 )~12)/20 = 3'58'00"= ("1+*~~/2 V' curve #3 D = 8' . ' .d=4'dl = (3.39) (8)/40? 0~40'40"d2 = (6.61)(8)/40= 119120"= *1/2 4 Field procedure for compound curves is essentially the same as forsimple curves. The transit has to 6e set up at P.C.C.1, and P.C.C.2 inorder to run in the curves.PROBLEMS7. Determine the radius of the central curve of a symmetrical compoundcurve which passes through point "Au, and determine the stationingof point "A".