# Reversed Curve to Connect Three Traversed Lines

**Situation**

A reversed curve with diverging tangent is to be designed to connect to three traversed lines for the portion of the proposed highway. The lines *AB* is 185 m, *BC* is 122.40 m, and *CD* is 285 m. The azimuth are Due East, 242°, and 302° respectively. The following are the cost index and specification:

Number of Lanes = Two Lanes

Width of Pavement = 3.05 m per lane

Thickness of Pavement = 280 mm

Unit Cost = P1,800 per square meter

It is necessary that the *PRC* (Point of Reversed Curvature) must be one-fourth the distance *BC* from *B*.

- Find the radius of the first curve.

A. 123 m

B. 156 m

C. 182 m

D. 143 m - Find the length of road from
*A*to*D*. Use arc basis.

A. 552 m

B. 637 m

C. 574 m

D. 468 m - Find the cost of the concrete pavement from
*A*to*D*.

A. P2.81M

B. P5.54M

C. P3.42M

D. P4.89M

**Solution**

$T_2 = 122.4 - 30.6 = 91.8 ~ \text{m}$

$R_1 = \dfrac{T_1}{\tan 14^\circ} = \dfrac{30.6}{\tan 14^\circ} = 122.73 ~ \text{m}$ ← [ A ] *answer for part 1*

$R_2 = \dfrac{T_2}{\tan 30^\circ} = \dfrac{91.8}{\tan 30^\circ} = 159.00 ~ \text{m}$

Total length of road from *A* to *D*

$L = 154.40 + \dfrac{2\pi(122.73)(28^\circ)}{180^\circ} + \dfrac{\pi(159)(60^\circ)}{180^\circ} + 193.20$

$L = 154.40 + 59.98 + 166.50 + 193.20$

$L = 574.08 ~ \text{m}$ ← [ C ] *answer for part 2*

Area of road pavement

$A = 154.40(6.10) + \dfrac{\pi \left[ (119.68 + 6.10)^2 - 119.68^2 \right](28^\circ)}{360^\circ} \\ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ + \dfrac{\pi \left[ (155.95 + 6.10)^2 - 155.95^2 \right](60^\circ)}{360^\circ} \\ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ + 193.20(6.10) + 193.20(6.10)$

$A = 160.50 + 365.86 + 1015.68 + 1178.52$

$A = 2720.56 ~ \text{m}^2$

Cost of concrete pavement

$C = 1,800(2720.56) = \text{P}4,897,008$ ← [ D ] *answer for part 3*