1. Transition Spirals Provide steady rate of change of curvature
R at TC = 
R at SC = /D
Circular curve is shorter
Circular curve is set back
Spiral starts much further back
Critical design factor – Ls

2. Length of Spiral Chosen based on: AASHTO Empirical Formula Speed
Ability to steer
Driver reactions
AASHTO Empirical Formula

3. 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

4. Spiral Geometry

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

6. 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

7. 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