Transition curve for 300 m radius, 7 m pavement

Reading: the design speed is read as 90 km/h (the digit in the scan is doubtful).

Data and assumptions

$R=300$ m, $W=7$ m, wheel base $l=6.1$ m. Rotation of the pavement about the centre line, desirable rate of introduction of superelevation 1 in $N=150$ (IRC, plain and rolling terrain).

Superelevation and widening

• $e=\dfrac{V^2}{225R}=\dfrac{90^2}{225(300)}=0.120>0.07$, so adopt e = 0.07. Friction needed $=\dfrac{V^2}{127R}-e=0.143<0.15$ (safe).
• Mechanical widening $W_m=\dfrac{nl^2}{2R}=\dfrac{2(6.1)^2}{2(300)}=0.124$ m; psychological $W_{ps}=\dfrac{V}{9.5\sqrt R}=0.547$ m. $W_e=0.67\approx0.7$ m.

(a) Length for comfort

$C=\dfrac{80}{75+90}=0.485$ m/s³, below the lower limit 0.5, so take $C=0.5$ m/s³. $V=25$ m/s:

$$L_s=\frac{V^3}{CR}=\frac{25^3}{0.5(300)}=104.2\ \text{m}$$

(b) Length for introducing superelevation

$$L_s=\frac{N\,e\,(W+W_e)}{2}=\frac{150(0.07)(7.0+0.7)}{2}=40.4\ \text{m}$$

(The IRC empirical value $2.7V^2/R=72.9$ m is also smaller than the comfort value.)

Design length

The greater value, comfort, governs: L_s = 105 m (calculated 104.2 m). The shift would be $s=\dfrac{105^2}{24(300)}=1.53$ m.