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Computations

The document summarizes the steps to compute basic measurements for two horizontal curves on a road. It calculates the radius, degree of curve, length of curve, stationing of points of curvature, intersection, and tangency, as well as superelevation rates and transition lengths for both curves based on their intersection angles, tangent distances, and design speeds. Computations are shown for Curve 1 with an intersection angle of 56 degrees and Curve 2 with an intersection angle of 8 degrees.

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Arshid Kalid Magandian
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40 views2 pages

Computations

The document summarizes the steps to compute basic measurements for two horizontal curves on a road. It calculates the radius, degree of curve, length of curve, stationing of points of curvature, intersection, and tangency, as well as superelevation rates and transition lengths for both curves based on their intersection angles, tangent distances, and design speeds. Computations are shown for Curve 1 with an intersection angle of 56 degrees and Curve 2 with an intersection angle of 8 degrees.

Uploaded by

Arshid Kalid Magandian
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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COMPUTATIONS

Basic Computations of Simple Horizontal Curve

1. The designer assumes tangent distances and design speeds of the two horizontal curves.
Curve 1: T= 150.00 m V= 50 kphs
Curve 2: T= 10.00 m V= 30 kph

2. The intersection angles, I, are measured from the contour map using straight edge and
protractor.
Curve 1: I= 56°
Curve 2: I= 8°

3. Solving the radius of each horizontal curve


𝑇 150
Curve 1: 𝑅= 𝐼 = 𝑡𝑎𝑛28° = 282.11 𝑚
𝑡𝑎𝑛
2

𝑇 10
Curve 2: 𝑅= 𝐼 = 𝑡𝑎𝑛4° = 143.01 𝑚
𝑡𝑎𝑛
2
4. Solving the External Distance, E
𝐼
Curve 1: 𝐸 = 𝑅 (𝑠𝑒𝑐 2 − 1) = 282.11 (𝑠𝑒𝑐28° − 1) = 37.40𝑚

𝐼
Curve 2: 𝐸 = 𝑅 (𝑠𝑒𝑐 2 − 1) = 143.01 (𝑠𝑒𝑐4° − 1) = 0.35𝑚

5. Solving the Degree of Curve, D


1145.916 1145.916
Curve 1: 𝐷= = = 4°3′ 43.01"
𝑅 282.11

1145.916 1145.916
Curve 2: 𝐷= = = 8°0′ 46.22"
𝑅 143.01

6. Solving the Length of Curve, LC


20𝐼 20 (56°)
Curve 1: 𝐿𝐶 = = 4°3′ 43.01" = 275.73 m
𝐷

20𝐼 20 (8°)
Curve 2: 𝐿𝐶 = = 8°0′ 46.22" = 19.97 m
𝐷

7. Stationing of Point of Curvature, PC, is measured using the tangent distance.


Curve 1: Sta. PC = 0 + 725.00
Curve 2: Sta. PC = 1 + 879.75
8. Stationing of Point of Intersection, PI
Curve 1: Sta. PI = Sta.PC + T = (0 + 725.00) + 150.00 = 0 + 875.00
Curve 2: Sta. PI = Sta.PC + T = (1 + 879.75) + 10.00 = 1 + 889.75

9. Stationing of Point of Tangency, PT


Curve 1: Sta. PT = Sta. PC + LC = (0 + 875.00) + 275.73 = 1 + 253.84
Curve 2: Sta. PT = Sta. PC + LC = (1 + 879.75) + 19.97 = 1 + 899.72

10. From the Table of Super elevation and Widening


Curve 1: W=0
Curve 2: W=0

Superelevation Diagram

1. Determining e from the Table of Minimum Radii for Design Superelevation Rates,
Design Speeds, emax = 8%
Curve 1: e = 4.40 %
Curve 2: e = 3.80 %

2. Determining Runoff Length, Lr from the Table of Superelevation Runoff Lr for


Horizontal Curves
Curve 1: Lr= 24.00 m
Curve 2: Lr= 18.00 m

3. Calculating Runout Length, Lo


𝐿𝑅 (𝑁𝐶) 24 (2%)
Curve 1: 𝐿𝑜 = = = 10.91 𝑚
𝑒 4.4%
𝐿𝑅 (𝑁𝐶) 18(2%)
Curve 2: 𝐿𝑜 = = = 9.47 𝑚
𝑒 3.8%

4. Calculating distance from the zero crown to Sta. PC/ Sta. PT to zero crown
Curve 1: 70% (Lr)= 0.70(24) = 16.80 m
Curve 2: 70% (Lr)= 0.70(18) = 12.60 m

5. Calculating distance from Sta. PC to Full Superelevation / Full Superelevation to Sta. PT


Curve 1: 30% (Lr)= 0.30(24) = 7.20 m
Curve 2: 30% (Lr)= 0.30(18) = 5.40 m

Curve 1: Total Transition = Lr+ Lo = 24 + 10.91 = 41.02 m


Curve 2: Total Transition = Lr+ Lo = 18 + 9.47 = 27.47 m

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