1. Islamic University of Gaza Civil Engineering Department Surveying II ECIV 2332 By Belal Almassri

2. Chapter 9 Route Surveying – Part 2 - Quiz 1 (First Sample Solution). - Quiz 2 (Second Sample Solution). - Degree of curvature. - Chainage. - Example 9.1

3. Quiz 1 (First Sample) St.DistanceInternal Angle A238º 40′ 24.93 B65º 30′ 37.56 C82º 30′ 48.42 D91º 45′ 35.26 E61º 10′ 25.77 A The following table shows measurements taken for a closed loop traverse ABCDEA. Determine: The adjusted Azimuth of each side if α AB = 210º 40′. 5 points ε E, ε N and the linear errror of misclosure ε. 5 points

4. Solution !

5. Quiz 1 (Second Sample) St.DistanceCorected Azimuth A 29.92220º 40′ B 45.07106º 15′ C 58.10008º 50′ D 42.31280º 40′ E 30.92161º 55′ A The following table shows data for a closed loop traverse ABCDEA. Determine: ε E, ε N and the linear errror of misclosure. 5 points the corrected departure  y and latitude  x. 5 points

6. Solution !

7. Degree of Curvature - It defines the sharpness or flatness of the curve. - There are two types of the degree of curvatures. - A) Arc Definition. D Highways metric systems - B) Chord Definition. D c English systems Railways

8. The arc definition states that the degree of curve is the central angle formed by two radii that extend from the centre of a circle to the ends of an arc measuring 100 ft or 30 meters long. Example: If you have a distance of 100 ft or 30 meters along the arc and the central angle of this arc is 12° then the degree of curvature for this arc equals 12°. Note: Preferred to be less than 4 degrees in modern highways.

10. The chord definition states that the degree of curve is the central angle formed by two radii drawn from the centre of the circle to the ends of a chord of 100 feet (or 30 meters) long. Note: Both types of the degree of the curvature are in terms of R, and they are inversely related to each other. (R and D)

11. Chainage/Station: The total distance on the centre line of a road between the first point of the project until reached. Ch of PI = Ch of PC + T Ch of PT= Ch of PC + L

12. Example 9.1: A tangent with a bearing of N 56° 48’ 20” E meets another tangent with a bearing of N 40° 10’ 20” E at PI STA 6 + 26.57. A horizontal curve with radius = 100 m will be used to connect the two tangents. Compute the degree of curvature, tangent distance, length of curve, chord distance, middle ordinate, external distance, PC and PT Stations.

13. Solution !