What is the degree of curve D?

Central angle subtended by a standard arc length (30 m)

Angle at the tangent point between tangent and long chord

Deflection angle of the whole curve

How are the circular arc length and total length of a composite curve found?

Lc = πR(Δ - 2θs)/180; total = 2Ls + Lc

Lc = πR(Δ + 2θs)/180; total = 2Ls + Lc

Lc = πR(Δ - θs)/180; total = 2Ls - Lc

Lc = πRΔ/180; total = Ls + Lc

Why does the tangent length Ts contain the term Ls/2?

Because half of the spiral lies on each side of the original tangent point

Because the deflection angle at SC is Ls/2

Because the spiral is half the length of the arc

2065 Chaitra (old course) · 7 marks

Degree of curve, elements of composite curve

Short answer

Degree of curve D is the central angle subtended by a 30 m arc; R = 1718.87/D. A composite curve has a circular arc between two transition spirals. Key elements: spiral angle θs = Ls/(2R), shift s = Ls²/(24R), tangent length Ts = (R+s)tan(Δ/2)+Ls/2, circular length Lc = πR(Δ-2θs)/180.

What is degree of curve? Find the elements of a composite curve including sketch.

Degree of curve

The degree of curve $D$ is the central angle subtended by a standard arc of length $C$ (30 m). $R = \dfrac{180\,C}{\pi D} = \dfrac{1718.87}{D}$ . A larger $D$ gives a sharper curve.

Elements of a composite curve

A composite curve is a circular arc of radius $R$ with a transition (spiral) curve of length $L_s$ at each end.

              V (IP)
             / \
            /Δ  \
           /     \
     TS  /         \ ST
        o___.___.___o
           SC     CS
         \   (R)   /
          \   |   /
           \  O  /

Element	Formula
Length of transition	$L_s = v^3/(CR)$
Spiral angle	$\theta_s = L_s/(2R)$ rad (= $90L_s/\pi R$ degrees)
Shift of the circular curve	$s = L_s^2/(24R)$
Tangent length (IP to TS or ST)	$T_s = (R+s)\tan\dfrac{\Delta}{2} + \dfrac{L_s}{2}$
Length of circular arc	$L_c = \dfrac{\pi R}{180}(\Delta - 2\theta_s)$
Total length	$2L_s + L_c$
Deflection angle at SC	$\alpha_s = \theta_s/3$
Deflection angle to a point at distance $l$	$\alpha = 573\,l^2/(RL_s)$ minutes
Offsets of SC	$x \approx L_s$ , $y = L_s^2/(6R)$
External distance	$E_s = (R+s)\sec\dfrac{\Delta}{2} - R$

Chainages: TS = IP − $T_s$ , SC = TS + $L_s$ , CS = SC + $L_c$ , ST = CS + $L_s$ . The circular arc is shifted inwards by $s$ from its original position, and each transition is placed so that about half of its length lies on each side of the original tangent point (hence the $L_s/2$ in $T_s$ ).

✓ Answer checked by Claude Opus 5.5 (high effort)

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