Spiral Calculations

Spiral elements are readily computed from the formulas given on pages 26 and 27. To use these formulas, certain data must be known. These data are normally obtained from location plans or by field measurements.
The following computations are for a spiral when D, V, PI station, and I are known.
D = 4°
I = 24°10’
PI station = 42 + 61.70
V = 60 mph

Determining Ls

1. Assuming that this is a highway spiral, use either the equation on page 21 or table 1.
2. From table 1, when D = 4° and V = 60 mph, the value for Ls is 250 feet.

Determining Δ

1. Δ = DL_s / 200
2. Δ = 4 (250) / 200 = 50

Determining o

1. o = Y – (R Vers Δ)
2. R =50 ft / Sin ½ D˚
   R = 50 ft / .0348994
   R = 1,432.69 ft
   Using Δ = 5˚, we find (see table A-9), Y = 0.029073 x Ls
   Y = 0.029073 x 250
   Y = 7.27 ft
3. o = Y – (R Vers Δ)
   (Vers Δ = 1 – Cos Δ)
   o = 7.27 – (1,432.69 x 0.00381)
   o = 1.81 ft

Determining Z

1. Z = X – (R Sin A)
2. From table A-9 we see that
   X = .999243 x L_s
   X = .999243 x 250
   X = 249.81 ft
   R = 1,432.69 ft
   Sin 5° = 0.08716
3. Z = 249.81- (1,432.69X 0.08716)
   Z = 124.94 ft

Determining T_s

1. T_s= (R + o) Tan (½ I) + Z
2. From the previous steps, R = 1,432.69 feet, o = 1.81 feet, and Z = 124.94 feet.
3. Tan 1/2 . (Tan 24º 10′) / 2 = Tan 12º 05′ = 0.21408
4. T_s = (1,432.69 + 1.81) (0.21408) + 124.94
   T_s = 432.04 ft

Determining Length of the Circular Arc (L_a)

1. L_a = (I – 2Δ / D) x 100
2. I = 24° 10’= 24.16667°
   A=5”
   D=4°
3. L, = (24.16667- 10 / 4) x 100 = 354.17 ft

Determining Chord Length

1. Chord length = Ls / 10
2. Chord length = (length = 250 ft) / 10

Determining Station Values

With the data above, the curve points are calculated as follows:

Station PI = 42 + 61.70
Station TS = -4 + 32.04 = T_s
Station TS = 38+ 29.66 +2 + 50.()() = L,
Station SC = 40 + 79.66 +3 + 54.17 = L_a
Station CS = 44+ 33.83 +2 + 50.()() = L_s
Station ST = 46+ 83.83

Determining Deflection Angles

One of the principal characteristics of the spiral is that the deflection angles vary as the square of the distance along the curve.

a / A :: L² / L_s²

From this equation, the following relationships are obtained:

a₁= (I)² / (10)² A, a₂ = 4a₁, a₃ = 9a, = 16a₁, …a₉ = 81a₁, and a₁₀ = 100a₁ = A.

The deflection angles to the various points on the spiral from the TS or ST are a1, a2, a3 . . . a9 and a10. Using these relationships, the deflection angles for the spirals and the circular arc are computed for the example spiral curve.

D = kL_s / 100

Hence, k = (D (100)) / Ls = (D (100)) / 250 = 1.6