With the same data as in example 1.12, calculate the new rac if P.T. is unchanged and P.C. is changed. Also calculate the chainage of new P.I.. P.C. and the P.T.

 Solution. (Fig. 1.23)Δ' = Δ +θ = 90°+ 20° = 110°

 T = R  tan(Δ/2) 

= 300tan((90°)/2) = 300m 

VV'=(VT2 sinθ/ sinΔ')

 =T. (sin 20°)/(sin110°) 

= 300tan 20° = 109.2 m

Also V' T2 =VT2 . (sinbΔ)/(sin Δ') 

T' = T  (sin Δ)/(sin Δ') 

= 300 (sin 90°)/(sin 110°) 

R'(tanΔ'/2)= 300/(cos 20°) 

R' = 300 cos 20° . cot((Δ')/2) 

= (300 cot 55°)/(cos20°) = 223.5m 

Length of new tangent t = T' = R'tan((Δ')/2) 

= 223.5 tan 55° = 319.3 m 

Length of new curve (πR'Δ')/(180°) = (π(223.5)(110°))/(180°) = 429.2m