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Appendix D:Road layout: traditional method: setting-out horizontal curvesThis example is based on one of the curves of the estate road layout given previously for which the coordinates andbearings of the skeleton have been worked out.Given, in addition to the tangent bearings:R = 54.567 m, tangent length = 32.44 mDeflection angle of point A and bearing of tangentsArc length 34/1a to A = CH 10.00 − CH 1.322 = 8.678 m = R × 2θ (θ in radians)Deflection angle θ (deg) = 04° 33' 22"Whole circle bearing of A from 34/1a= 173° 25' 48" − 04° 33' 22" = 168° 52' 26"Chord length 34/1a to A = 2R sinθ = 8.669 mRepeat for subsequent points at 10 m chainage intervals (arc length) and for point 34/1b, as shown in the table.Using the proforma shown opposite, calculate all the deflection angles at all the chosen points along the arc(column 4). The sum of these deflections should equal α/2. The whole circle bearing is then worked out to each ofthe chainages.Assuming instrument is set up at 34/1a, final bearing to 34/1b =142° 42' 15"To check:142° 42' 15" = bearing of the far tangent point from 34/1aIP 34/1 to IP 34/2 = 291° 58' 42" − 180°= 111° 58' 42"Final bearing = 142° 42' 15" − sum of deflections= 142° 42' 15" − 30° 43' 33"= 111° 58’ 42"Finally calculate chord lengths (column 6)The instrument remains at 34/1a, the chord distances are from intermediate point to intermediate point rather thancalculating and taping the long chords from 34/1a to each intermediate point. On a long curve or if the sight line isobstructed it may be necessary to move the instrument to one of the intermediate points, setting it to the backbearing of point 34/1a and then proceeding as before.CIRIA C709129