Question

1. Given the following horizontal curve and spiral data: Total tangent deflection angle (A) = 30° Rt. D=50 Design Speed = 50 mph Two lane road in Southern Climate PI Station @ 64+20.00 Find and e from Table 13.8 or 13.9 (see handouts) Compute: R P= Compute the stations of: T.S. = S.C. = C.S. S.T. = Compute the Long Tangent and the Short Tangent

Answer 1

Given

Design speed, V= 50 mph, $D=5^{^{\circ}}$ , Two lane road in Southern climate

Then from Table 13.9, Using above terms

superelevation, e = 0.083

and length of superelevation runoff or spiral curve for 2 lane, L = 200 ft

As we know, $e=\frac{V^{2}}{225R}$ , where R is raduis of curve

and e is restricted to 7%, is that 0.07

Then,  $R=\frac{V^{2}}{225e}$

502 R= 225 X 0.083

$R=133.87$, calculated as per question requied

P =?, is not clarified in question. what does this term means?

Procedding further,

Now R is converted into chains from metres, where length of chain is equal to 20 metres

Then $R=\frac{133.87}{20}=6.69$

Calculating,Tangent length, $T=Rtan\frac{\Delta }{2}$ , where $\Delta =30^{\circ}$   given in question

$T=6.69tan\frac{30 }{2}= 1.792 chains=35.85 m$

we calculated, length of curve, L = 200ft= 60.96 m = 3.048 chains

Chainage of P.I. = 64 + 0.20 (Chains) = $(64\times 20) +(20\times 0.20) = 1284$ m

Deduct Tangent length, T,= 1 + 0.792 (Chains) = 35.85 m

Chainage of T.S.= 63 + 0.59 (Chains) = 1271.8 m

Add length of curve, L = 3 +0.048 (Chains) =60.96 m

Chainage of S.T. = 69 +0.63 (Chains) = 69.63 chains = 1392.6 m