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: AASHTO Empirical Formula 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 Geometry

Spiral Example Given D = 4°, Ls = 450’, I = 40°

Spiral Staking Establish TS, ST Establish SC Establish CS 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