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CE 323: ENGINEERING SURVEYS LECTURE SPIRAL CURVE
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Spirals are used to overcome the abrupt change in curvature and super elevation that occurs between tangent and circular curve. The spiral curve is used to gradually change the curvature and super elevation of the road, thus called transition curve.
X Xc Y Yc Es θ θs i
is D Dc
= length of throw or the distance from tangent line that the circular curve has been offset = offset distance (right angle distance) from tangent to any point on the spiral = offset distance (right angle distance) from tangent to SC = distance along tangent to any point on spiral = distance along tangent from TS to point at right angle to SC = external distance of the simple curve = spiral angle from tangent to any point on the spiral = spiral angle from tangent to SC = deflection angle from TS to any point on the spiral, it is proportional to the square of its distance = deflection angle from TS to SC = degree of spiral curve at any point = degree of simple curve
Formulas for Spiral Curves Distance along tangent to any point on spiral: 𝑳𝟓 𝒀=𝑳− 𝟒𝟎𝑹𝟐 𝑳𝒔 𝟐 At L = Ls, Y = Yc, thus, 𝑳𝒔 𝟑 𝒀𝒄 = 𝑳𝒔 − 𝟒𝟎𝑹𝟐 Offset distance from tangent to any point on the spiral: 𝑳𝟑 𝑿= 𝟔𝑹𝑳𝒔 At L = Ls, Y = Yc, thus, 𝑳𝒔 𝟐 𝑿𝒄 = 𝟔𝑹 Length of throw: Elements of Spiral Curve TS = tangent to spiral SC = spiral to curve CS = curve to spiral ST = spiral to tangent LT = long tangent ST = short tangent R = radius of simple curve Ts = spiral tangent distance Tc = circular curve tangent L = length of spiral from TS to any point along the spiral Ls = length of spiral PI = point of intersection I = angle of intersection Ic = angle of intersection of the simple curve
ENGR. S.R. DOMINGUEZ
𝒑=
𝟏 𝑳𝒔 𝟐 𝑿𝒄 = 𝟒 𝟐𝟒𝑹
Spiral angle from tangent to any point on the spiral (in radians): 𝑳𝟐 𝜽= 𝟐𝑹𝑳𝒔 At L = Ls, θ = θs, thus, 𝑳𝒔 𝜽𝒔 = 𝟐𝑹 Deflection angle from TS to any point on the spiral: 𝟏 𝑳𝟐 𝒊= 𝜽= 𝟑 𝟔𝑹𝑳𝒔 This angle is proportional to the square of its distance: 𝒊 𝑳𝟐 = 𝟐 𝒊 𝒔 𝑳𝒔
CE 323: ENGINEERING SURVEYS LECTURE Tangent distance:
Degree of spiral curve: 𝑫 𝑳 = 𝑫 𝒄 𝑳𝒔 Super elevation: 𝒆=
𝟎. 𝟎𝟎𝟕𝟗𝑲𝟐 𝑹
Considering 75% of K to counteract the super elevation: 𝟎. 𝟎𝟎𝟒𝑲𝟐 𝒆= 𝑹 Desirable length of spiral: 𝑳𝒔 =
𝟎. 𝟎𝟑𝟔𝑲𝟑 𝑹
Illustrative Examples: 1.
2.
A spiral 80m long connects a tangent with a 6°30’ circular curve. If the stationing of the TS is 10 + 000, and the gauge of the tract on the curve is 1.50m, a. Determine the elevation of the outer rail at the midpoint, if the velocity of the fastest train to pass over the curve is 60kph. b. Determine the spiral angle at the first quarter point. c. Determine the deflection angle at the end point. d. Determine the offset from the tangent at the second quarter point. A simple curve having a radius of 280m connects two tangents intersecting at an angle of 50°. It is to be replaced by another curve having 80m spirals at its ends such that the point of tangency shall be the same. a. Determine the radius of the new circular curve. b. Determine the distance that the curve will nearer the vertex.
ENGR. S.R. DOMINGUEZ
3.
Determine the central angle of the circular curve. CE Board May 2010. The tangents of a spiral curve form an angle of intersection of 25° at station 2 + 058. Design speed is 80kph. For a radius of central curve of 300m and a length of spiral of 52.10m, a. Find the stationing at the point where the spiral starts. b. Find the stationing of the start of central curve. c. Find the length of the central curve.
