What do SC and CS denote?

Spiral-to-circular and circular-to-spiral junction points

Starting chainage and curve start

Start and end of the whole composite curve

What is the shift s of the circular curve?

s = Ls^2/(24 R), the inward movement of the circular arc from the tangent

What is the spiral angle thetas at SC and the deflection angle alpha_s at SC from TS?

thetas = Ls/(2R); alpha_s = thetas/3

thetas = 2Ls/R; alpha_s = thetas/3

thetas = Ls/R; alpha_s = thetas/2

thetas = Ls/(2R); alpha_s = thetas

How is the length of circular arc Lc found?

Lc = R(Delta - 2 thetas), with angles in radians

2070 Magh · 6 marks

Elements of transition curve

Short answer

Key elements are transition length Ls = v³/(CR), shift s = Ls²/(24R), spiral angle θs = Ls/(2R), and tangent length Ts = (R+s)tan(Δ/2)+Ls/2. Chainages: TS, SC, CS, ST; deflection angle from TS is α = 573l²/(RLs) minutes; SC coordinates x≈Ls, y=Ls²/(6R).

Describe elements of transition curve.

A combined (composite) curve consists of a circular arc with a transition curve at each end. The elements are:

              V (IP)
             / \
            /   \
           /     \
     TS  /         \ ST
        o___.___.___o
           SC     CS

Element	Meaning / formula
$L_s$	Length of transition curve, $L_s = v^3/(CR)$
TS, ST	Tangent–spiral and spiral–tangent points (start and end of the composite curve)
SC, CS	Spiral–circular and circular–spiral junction points
$\theta_s$	Spiral (tangent) angle at SC, $\theta_s = L_s/(2R)$ rad
$s$	Shift of the circular curve, $s = L_s^2/(24R)$
$T_s$	Total tangent length, $T_s = (R+s)\tan(\Delta/2) + L_s/2$
$\Delta$	Total deflection angle between the straights
$L_c$	Length of circular arc, $L_c = R(\Delta - 2\theta_s)$ (θ in radians)
$\alpha$	Deflection angle at TS to a point at distance $l$ : $\alpha = 573l^2/(RL_s)$ min
$\alpha_s$	Deflection angle at SC from TS, $\alpha_s = \theta_s/3$
$x, y$	Coordinates of SC: $x \approx L_s$ , $y = L_s^2/(6R)$
Total length	$2L_s + L_c$

Chainages: TS = IP − $T_s$ ; SC = TS + $L_s$ ; CS = SC + $L_c$ ; ST = CS + $L_s$ . The shift $s$ is the distance by which the circular curve is moved inward from the original tangents so that the transition can fit; it equals the perpendicular distance between the tangent and the shifted circular arc.

The rate of change of radial acceleration $C$ and the design speed $v$ decide $L_s$ ; the super-elevation is attained over $L_s$ .

✓ Answer checked by Claude Opus 5.5 (high effort)

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