Reversed Curve to Connect Three Traversed Lines

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:

Type of Pavement = Item 311 (Portland Cement Concrete Pavement)
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.

  1. Find the radius of the first curve.
      A.   123 m
      B.   156 m
      C.   182 m
      D.   143 m
  2. Find the length of road from A to D. Use arc basis.
      A.   552 m
      B.   637 m
      C.   574 m
      D.   468 m
  3. Find the cost of the concrete pavement from A to D.
      A.   P2.81M
      B.   P5.54M
      C.   P3.42M
      D.   P4.89M




In reply to by Anonymous (not verified)

Wala akong nakita sa solution na inadd yang dalawa. Do you mean minultiply? If yes, rectangular kasi ang shape sa protion na yan kaya minultiply para makuha ang area.

In reply to by Cestudent (not verified)

I know late na po itong reply ko, pero sasagutin ko parin para dun sa mga makakakita nito in the future tas may gantong tanong din. Ima-minus niyo po yung width ng pavement sa radius na nakuha mo kaya naging ganan po ang sagoot

In reply to by Anonymous (not verified)

kaya po 60 gawa ng yung equivalent na bearing ng azimuth na 302 is 58 and then imaminus mo siya sa 90 tas ang makukuha mo dun is ipaplus mo sa 28 which is yung bearing nung sa azimuth na 242. Yun po sana na gets