Horizontal Curves Horizontal Curves Curve Types Curve Types Geometric Definition Horizontal Curve Terms & Formulas Route Stationing Curve Layout Methods Horizontal Curves Curve Types Horizontal Curves Degree of Curve Horizontal Curves Horizontal Curve Formulas Horizontal Curves Horizontal Curve Formulas 1. R = 5729.58' 1746.379m = Da Da 2. L = 100' I I = 30.480m Da Da 3. T = R tan( I / 2) 4. LC = 2 R sin( I / 2) Horizontal Curves Horizontal Curve Terms Horizontal Curves Horizontal Curve Formulas 5. E = T tan( I / 4) 6. M = E cos( I / 2) 7. PC STATION = PI − T 8. PT STATION = PC + L Horizontal Curves Example: Horizontal Circular Curve Given: PI = 64+32.20 I = 24°20’ (Defl ∠) = 24.3333° Da = 4°00’ (Selected) (100’ arc for 4° ∠ Change) Find: R, L, T, LC, E, M, PC, PT Route Stationing From example: PTOLD = PI + T = 6432.20 + 308.82 = 67+41.02 PTNEW = 67+31.71 DIFF = 9.31’ (shortening of traverse-subtract from station of next tangent point) Horizontal Curves Route Stationing Preliminary - series of tangents ϖ deflections @PI’s Final - series of tangents and curves - total traverse length shortened due to insertion of curves Horizontal Curves Curve Layout Methods Radial - set up at center of curve (arc) & swing radius through deflection angle *not possible / practical on large radius curves Chord - deflection Incremental chord (Transit, theodolite, tape) Sec24-7; Fig 24-5 Total chord (Total Station) Sec 24-10 Curve Layout Methods Deflection ∠ to any point on curve dp sp = Da I = 100' L ⇒ dp = s p Da 100' = s p Da 30.480 m dp = angle subtended by arc from PC to point P δp = deflection angle from PI to P, instr @ PC sp = distance along curve in ft Da = degree of curve Horizontal Curves Curve Layout Methods Horizontal Curves Curve Layout Methods From geometry δp = dp /2 (*angle formed by tangent &chord = ½ angle included by chord) s p Da s p Da 90−δp δp ∴δ p = = 200 ' 60.960 m δp (c p / 2 ) c p sin δ p = = R 2R c p = 2 R sin δ p Curve Layout Mehods From example: δ62 δ63 General relation for chords s p Da s p Da δp = = c p = 2 R sin δ p 200 ' 60 .960 m For full station where S = 100’ δ 100 ' = D 2 c100 ' = 2 R sin( D ) 2 s62 Da 76.62(4) = = 1°31'57" 200 200 c62 = 2 R sin δ 62 = 2(1432 .39 ) sin(1°31'57" ) = 76 .61' δ 62 =