Transition Spirals. Provide steady rate of change of curvature R at TC = R at SC = 5729.578/D Circular curve is shorter Circular curve is set back Spiral starts much further back Critical design factor – L s. Length of Spiral. Chosen based on: Speed Ability to steer Driver reactions
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Transition Spirals • Provide steady rate of change of curvature • R at TC = • R at SC = 5729.578/D • Circular curve is shorter • Circular curve is set back • Spiral starts much further back • Critical design factor – Ls
Length of Spiral • Chosen based on: • Speed • Ability to steer • Driver reactions • AASHTO Empirical Formula
Length of Spiral • Ls < 100 – no spiral necessary? • Example: • Interstate Highway: V = 75 mph • Degree of Curvature: D = 4° • Drivers capable of steering,but trucks may tip: C = 2
Spiral Example Given D = 4°, Ls = 450’, I = 40°
Spiral Staking • Establish TS, ST • TS = PT-Ts ST = PT+Ts • Establish SC • Follow tangent for L.T., establish SPI • Turn s to find tangent at SC • Measure S.T. to set SC • Establish CS • Ic = I-2 s • Transit at SC, sight SPI, plunge, turn Ic/2 to sight along L.C. • Measure 2Rsin(Ic/2) to set CS
Spiral Staking • Transit at TS,sight on PI • Determine chord lengths, L • Say TS = 78+24, Ls = 450’, SC = 82+74, s = 3° • Determine L, , for Sta 79, 80, 81, 82, 82+74