Question

A horizontal curve is being designed through mountainous terrain for a four-lane road with lanes that are 10 ft wide. The central angle (I) is known to be 40 degrees, the tangent is 520 ft, and the stationing of the tangent intersection (PI) is 2600 + 00. Under specified conditions and vehicle speed, the roadway surface is determined to have a coefficient of side friction of 0.08, and the curve’s superelevation is 0.09 ft/ft. What is the stationing of the PC and PT and what is the safe vehicle speed?

Answer 1

Answer:

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Explanation:

Answer 2

Following are the calculation on the vehicle speed:

Calculating the PC of stationing:  

$\to PC = PI-T$

          $= 260000 - 520.00\\\\ = 2594 +80$

Calculating the horizontal curve of the radius:  

$\to T = R \tan \frac{\Delta }{2} \\\\\to 520 = R \tan \frac{40}{2}\\\\ \to R=1428.69\ ft$

Calculating the horizontal curve of the length:

$\to L = 0.0174533 \ R\Delta$

        $= 0.0174533 \times 1428.69 \times 40^{\circ} \\\\ = 997.413 \ ft$

Calculating the PT of stationing:

$\to PT = PC +L$

          $= 259480+997.413 \\\\= 2604 + 774$

Calculating the horizontal curve of the safe vehicle speed:  

Using the equation:  

$\to e+ f_{side}=\frac{v^2}{g \ R_{v}}$

If superelevation is $e$, side friction factor is $f_{side}$, gravitational acceleration is $g$, and curve radius is $R$.

$\to 0.09 +0.08= \frac{V^2}{32.2\times 1428.69}\\\\ \to V = \sqrt{(0.09+0.08) \times (32.2 \times 1428.69)}$

       $= \sqrt{(0.17) \times (46003. 818)} \\\\= \sqrt{7820.64906} \\\\=88.434$

$\to V=88.434 \ \frac{ft}{s} = 60.29\ mph$

Therefore, the safe vehicle speed is "$60.29\ mph$".

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