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- 1. 20/01/2013 1
- 2. 20/01/2013 2
- 3. Figure in slide 4, equal spiral curves join a circular curve to the main tangents; • T.S. tangent to spiral point. • LS the length of the spiral curve. • S.C. spiral to circular curve point. • C.S. curve to spiral point. • S.T. spiral to tangent point. • TS tangent distance from the T.S. or the S.T to the P.I. • R the radius of the circular curve.20/01/2013 3
- 4. P.I. TS TS Circular curve T.S. S.T. C.S. S.C. LS LS R R 1st spiral curve 2nd spiral curve20/01/2013 4
- 5. S.P.I. S.C. Circular curve T.S. REnlargement of Spiral Curve 20/01/2013 5
- 6. Figure in slide 5, the spiral curve on the left side redrawn to a much larger scale. • S.P.I. the spiral point of intersection. • S the spiral angle. • S.T. and L.T. are the short and long tangents of the spiral curve. • c the degrees of curvature of the circular curve. • XS the distance measured from the T.S. along the main tangent to a point where a perpendicular line to the tangent hits the S.C. • YS the distance measured perpendicular from the XS coordinate to the S.C. • L.C. the long chord from the T.S. to the S.C.20/01/2013 6
- 7. • P.C. the point at which the circular curve becomes parallel to the spiral. The curves will be a distance p apart. • S the deflection angle from the T.S. to the S.C. • CS the correction factor, negligible when 15.20/01/2013 7
- 8. Spiral Curve Equations 5729.58 R LS C S 200 S 2 S 4 X S L S 1 10 216 S S3 YS LS 3 42 L.C. X S YS 2 2 YS S .T . sin S L.T . X S S .T . cos S S S 3 p YS R (1 cos S ) TS X S R sin S ( R p ) tan 220/01/2013 8
- 9. Steps of Laying Out A Spiral Curve • LS is selected considering: traffic design, speed, No. of lanes, c and the length needed for super-elevation. • The values for R, S, XS, YS, L.C., S.T., L.T., S, p and TS are computed. • The chord lengths are assumed and the deflection angle L is determined from ( L ) 2 S S • The curve will be staked out in identical manner used for circular curves.20/01/2013 9
- 10. Approximate Solution for Spiral Problems LS S.C. (2/3) S T.S. S = (1/3) SBasic Assumption: LS Long Chord, LS L.C.20/01/2013 10
- 11. Approximate Equations Y LS sin S X LS Y 2 2 1 q X 2 1 p Y 420/01/2013 11
- 12. Approximate Equations Using the sine law, we obtain the following: L.T . LS 2 sin (180 S ) sin ( S ) 3 2 LS L.T . sin ( S ) 3 sin S Using the sine law, we obtain: S .T . LS S sin S sin 3 S LS S .T . sin 3 sin S20/01/2013 12
- 13. An Example of Compared Values Referring to previous solved problem: Precise Methods Approximate Methods Parameters (ft) Parameters (ft) Y 10.46 Y 10.47 X 299.67 X 299.82 q 149.94 q 149.91 p 2.61 p 2.62 L.T. 200.15 L.T. 200.20 S.T. 100.10 S.T. 100.1620/01/2013 13

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