REVERSED CURVES

A reversed curve is formed by two circular simple curves having a common tangent but lies on opposite sides.

Elements of Reversed Curve

PC

T1

T1

R2

PPRCT2R1

I2

I2T2

PTI1

R1 and R2=radii of curvaturePT=point of tangency=angle between converging tangentsPRC=point of reversed curvature=I2 + I1PC=point of curvatureP=distance between parallel tangents

Four types of reversed curve problems:

1. Reversed curve with equal radii and parallel tangents.2. Reversed curve with unequal radii and parallel tangents.

I/21I/21

I2

PCT1I2PCT1

I2I2

aR2R1T1T1

PPRCCR1PRCCR1

T1T21P

bI2I1

T2I2

PTT22PTI1

I/21I/21Long Chord from PC to PTLong Chord from PC to PT

3. Reversed curve with equal radii and converging tangents.4. Reversed curve with unequal radii and converging tangents.

I2PCPRCPCI1PT

R2T1T22T22I2PRCI1R1T2T1PTI2I2R1R1T2T1T1

Sample Problem No. 1: Two parallel tangents 10-m apart are connected by a reversed curve. The chord length from the PC to the PT is equal to 120-m.a.Compute the length of the tangent with common direction.b.Determine the equal radius of the reversed curve.c.Compute the stationing of the PRC. If the stationing of A at the beginning of the tangent with common direction is 3+420.

Sample Problem No. 2:In a railroad layout, the centerline of two parallel tracks is connected with a reversed curve of unequal radii. The central angle of the first and second curve is 16o and the distance between parallel tracks is 27.6 m. Stationing of the PC is 15+420 and the radius of the second curve is 290.a.Compute the length of the long chord from PC to PT.b.Compute the radius of the first curve.c. Compute the stationing of PT.

Sample Problem No. 3:Two tangents converge at an angle of 30, the direction of the second tangent is due east. The vertical distance of the PC from the second tangent is 116.50. The bearing of the common tangent is S40E.a. Compute the central angle of the first curve.b. If a reverse curve is to connect these two tangents, determine the common radius of the curve.c. Compute the stationing of the PT if PC is located at 10 + 620.

Sample Problem No. 4:A reversed curve connects two converging tangents intersecting at an angle of 30. The distance of this intersection from the PI of the second curve is 150-m. The deflection angle of the common tangent from the back tangent of the first curve is 20R. The degree of curve of the second simple curve is 6 and the stationing of the point of intersection of the first curve is 4+450. a. Determine the radius of the first curve. b. Determine the stationing of PRC. c. Determine the station of PT.

Plate No.3: Reversed Curve

The first curve of a reversed curve has a radius of 200-m. If the horizontal distance between PC and PT is 110-m. Determine the following:a. Radius of the second curve, so that the curve can connect two parallel tangents 18-m apart.b. The station of point of reversed curvature, if PI of the first curve is located at 10+095.c. The station of PT.

Determine the following if the radius of the second curve is 750-m and the vertical distance between the point of intersection of the two curve is 685.676-m.a. Radius of the first curve.b. Length of the reverse curve.c. Station of PT if PI of the first curve is located at 8+000.

Two parallel railway tracks have a vertical distance of 60-m. They are connected by a reversed curve, each section having the same radius. And the maximum horizontal distance between PC and PT is 220-m.a. Calculate the radius.b. Determine the common tangent.c. Determine the long chord.d. Determine the length of the reverse curve.