CURVES Lecture – 06

Problem: Two lines BA & AC are intersected by the third line EF. The angle AEF and AFE are 28o 12' and 32o 24' respectively. And EF = 599.7 ft. Find the radius of the simple curve which is tangential to BA and AC. Also find the chainage of beginning and end of the curve? Given that chainage of A is 5625 ft.

Solution: Given Data: Angle AEF = α = 28o 12' Angle AFE = β = 32o 24' EF =599.7 ft Calculations: In Triangle AEF: Angle EAF = 180o - α – β = 180o - 28o 12' – 32o 26‘ γ = 119o 24'

Solution AE / sinβ = EF / sinγ AE = (EF x sinβ) / sinγ = [599.7 x sin (32o 24')] / sin (119o 24') AE = 368.83 ft Tangent Length = LT 1 = AE = Rtan (α / 2) R = AE/tan(α / 2) R = 383.83/tan(28o 12‘/2) R = 1468.4 ft

Solution Chainage of T1 = Chainage of A– AE = 5625 – 368.83 = 5256.17 ft Chainage of T2 = Chainage of T1 + Arc length (T1PT2) = 5256.17 + [π x R x (α + β)] / 180o = 5750.36 ft

Problem: The wire is tangential to each of the following line. Find the radius of the curve and also the lengths? Line W.C.B Length AB 00o 00' ---- BC 270o 00' 550 ft CD 220o 00'

Solution: Now, I2 = 90o Ф2 = 90o I1 = 220o – 90o I1 = 130o

Solution CT2 = R tan(Ф/2) & T2B = R tan(Ф/2) = R tan (25o) = R tan (45o) BC = CT2 + T2B = R tan (25o) + R tan (45o) 550 = R [tan(25o) + tan(45o)] R = 375.09 ft  So; Tangents length: T1C = CT2 = R tan(25o) = 174.9 ft T3B = T2B = R tan(45o) = 375.1 ft

CURVE RANGING BY TANGENTIAL ANGLE METHOD: (by Theodolite & Tape) In this method curve is set out by the tangential angles (often called deflection angles) with a theodolite and a chain or a tape. The designation of formula for calculating the deflection angles is as follows: Let AB be rear tangent to the curve, T1 and T2 are the tangent points, D, E & F etc. are successive points on the curve δ1, δ2, δ3 are the tangential angles, which each of the successive chord T1D, DE, EF etc makes with respective tangents at T1 , D and E etc. Δ1, Δ2, Δ3 etc = the total tangential or deflection angles (between the rear tangent AB and each of the lines (T1D, T1E, T1F etc).

Tangential Angle Method C1, C2, C3 etc = length of chords T1D, DE, EF etc. R = radius of the curve. Chord T1D = Arc T1D (very nearly) = C1 BT1D = δ1 = ½ x Angle T1OD. Angle T1OD = 2δ1 Now: C1 = R x T1OD x π/180o as S=Rθ T1OD = (180oxC1)/πR 2δ1 = (180o x C1) / (πR) Then; δ1 = (90o x C1) / (πR) degrees = (90o x 60 x C1)/πR minutes. δ1 = (1718.9 x C1) / R minutes

Tangential Angle Method Similarly; δ2 = (1718.9 x C2) / R minutes . δn = (1718.9 x Cn) / R Total tangential angle: Δ1 = δ1 Δ2 = Δ1 + δ2 = δ1 + δ2 Δ3 = Δ2 + δ3 = δ1 + δ2 + δ3 . . Δn = Δn-1 + δ n = δ1 + δ2 +…….+ δn-1 + δn

Tangential Angle Method If C1 = C2 =C3 = ………= Cn δ1 = δ2 = δ3 = …….= δn Δ1 = δ1 Δ2 = 2δ1 Δ3 = 3 δ1 . Δn = n δ1

Problem: Two tangents intersect at chainage 39 + 60, the deflection angle being 36o. Calculate all the data necessary for setting out a 6o curve by deflection angle method. Peg interval being the 50 ft.

Solution: Since R = 5730 / D = 5730 / 6 = 955 ft Length of the arc = πRФ / 180o = (π x 955 x 36) / 180o = 600.04 ft δ1 = (1718.9 x C) / R = (1718.9 x 50) / 955 = 1o 30' 00'‘ Δ1 = δ1 = 01o 30' Δ2 = 2δ1 = 3o 00' Δ3 = 3 δ1 = 4o 30' Δ4 = 4 δ1 = 6o 00' Δ5 = 5 δ1 = 7o 30'

Solution Δ6 = 6 δ1 = 9o 00' Δ7 = 7 δ1 = 10o 30' Δ8 = 8 δ1 = 12o 00' Δ9 = 9 δ1 = 13o30' Δ10 = 10 δ1 = 15o 00' Δ11 = 11 δ1 = 16o 30' Δ 12 = 12 δ1 = 18o 00'

Procedure: To set out a curve: Set a theodolite over first tangent point (T1) and level it. With both plates changed at zero, direct the telescope the ranging rod at the point of intersection B and bisect it. Release the vernier plates and set the vernier A at the first deflection angle Δ1, the telescope being thus deflected along T1D. Pin down the zero end of the tape at T1 and holding the arrows at a distance equal to the length of first chord, swing down the chain around T1 until the arrow is bisected by the cross-hairs, thus fixing the first point D on the curve.

Procedure Unclamp upper plate. Set the vernier to 2nd deflection angle Δ2. The line of sight being now directed along T1E. Hold the zero of the chain at D and swing the other and is bisected by the line of sight, thus locating the 2nd point of the curve. Repeat the process until the end of the curve is reached.

TWO THEODOLITE METHOD: This method is used when ground is not favorable for accurate chaning e.g., rough ground. It is based on the fact that the angle between the tangent and chord is equal to angle which long chord subtends in the opposite segment. Let D, E etc be the floats on the curve the angle (Δ1) between T1B and the chord T1D = BT1D = T1T2D. Similarly, the angle BT1E = Δ2 = T1T2E The total tangent or the deflected angles Δ1, Δ2, Δ3 being calculated as in the 1st method.

Procedure: To set out the curve, procedure as follows: Set up the theodolite over T1 and another at T2. Set the vernier of each of the instrument to zero. Direct the instrument at T1 to the ranging rod at the point of intersection (B) and bisect it. Direct the instrument at T2 to the 1st tangent point T1 and bisect it. Set the vernier of each of the instrument to read the 1st deflection angle Δ1. Thus the line of sight at T1 is directed Along T2D. This point of intersection gives the required point D on the curve.

Procedure Move the ranging rod until it is bisected by the cross-hair of both the instrument, thus locating the point, the point (D), on the curve. To obtain the 2nd point on the curve, set the vernier of each of the instrument to the 2nd deflection angle Δ2 and proceed as before. Continue the above process setting angle Δ3, Δ4,…..,Δn until the lst end T2 of the curve is reached.