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





Bakit po inadd yong 154.40 at 6.10

Jhun Vert's picture

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.

saan po galing yung 154.40

Jhun Vert's picture

AB - T1.

yung nakuha mong 30.6 i minus mo sa 185

yup I noticed that too. siguro nagkamali lang sa solution. tayo nalang po mag adjust hehe

Jhun Vert's picture

Nakita ko na... doon pala sa next line, na add pala talaga... yeah kayo na mag adjust... di ko na i-edit... hahaha

san nakuha 28 degree

Jhun Vert's picture

The azimuths are reckoned from South.

Paano po nakuha yung 119.68, 155.95? Salamat po.

Hi! bakit po 60 yung sa v2? diba dapat po 62?