SIMPLE CURVES 1. CE BOARD (May 1998) The perpendicular distance from a point on a simple curve to a point Q on the tangent at PC is 64 m. If the distance from PC to Q is 260 m, find the radius of the curve.
c.) If the stationing of the point of curvature is at 10 + 020, compute the stationing at a point on the curve which intersects with the line making a deflection angle of 8° with the tangent through PC.
2. CE BOARD (November 2004) A 6° simple curve has a central angle of 44°. The stationing of the point of curvature is 13 + 080.
4. CE BOARD (May 2005) The tangents of a simple curve along the North Diversion Road have bearing of N 20° E and N 80° E, respectively. The radius of the curve is 200 m.
a.) Find the distance from the midpoint of the curve to the intersection of the tangents at PC and PT. b.) Find the distance from the midpoint of the curve to the midpoint of the long chord joining the point of curvature and the point of tangency. c.) Find the stationing of a point on the curve which intersects with the line making a deflection angle of 6° with the tangent through the PC. 3. CE BOARD (November 2004) A simple curve has a central angle of 36° and a degree of curve of 6°. a.) Find the nearest distance from the midpoint of the curve to the point of intersection of the tangents. b.) Compute the distance from the midpoint of the curve to the midpoint of the long chord joining the point of tangency and point of curvature.
a.) Compute the external distance of the curve. b.) Compute the middle ordinate of the curve. c.) Compute the stationing of point A on the curve having a deflection angle of 6° from the PC which is at 1 + 200.00. 5. CE BOARD (November 1998) The deflection angles of two intermediate points A and B of a simple curve are 3° 15’ and 8° 15’, respectively, from the PC. If the chord distance between A and B is 40 m, find the length of the curve from PC to B. 6. CE BOARD (November 2008) The deflection angles from PC of two intermediate points A and B on a simple curve are 3° 15’ and 8° 15’, respectively. The length of the chord between A and B is 40 m. a.) Find the radius of the simple curve.
b.) Find the length of the curve from PC to A. c.) Find the length of the chord from PC to B. 7. The deflection angles of two intermediate points A and B of a highway curve are 6° 30’ and 12° 30’ from PC, respectively. The chord distance between two said points is 30 m. The long chord is 160 m long. a.) Compute the angle of intersection of the simple curve. b.) Compute the tangent distance of the curve. c.) Compute the external distance of the curve. 8. A simple curve has an external distance of 11.54 m and a middle ordinate of 11.10 m. The tangents intersect at V (Sta. 321 + 021). Use arc basis. a.) Find the radius of the curve.
b.) Compute the stationing of the point D on the curve along a line joining the center of the curve which makes an angle of 54° with the tangent line passing thru the PC. c.) What is the length of the line from D to the intersection of the tangent AB? 10. A 5° curve intersects a property line CD at point D. The back tangent intersects the property line at point C which is 105.27 m from the PC which is at station 2 + 040. The angle that the property line CD makes with the back tangent is 110° 50’. a.) Compute the length of curve from the PC to the point of intersection of the line from the center of the curve to point C and the curve. b.) Compute the distance CD. c.) Compute the stationing of point D on the curve.
b.) Find the length of the long chord c.) Find the stationing of PT. 9. CE BOARD (November 2004) A simple curve of the proposed extension of Mantabahadra Highway have a direction of tangent AB which is due north and tangent BC bearing N 50° E. Point A is at the PC whose stationing is 20 + 130.46. The degree of curve is 4°. a.) Compute the long chord of the curve.
COMPOUND CURVES 1. CE BOARD (November 1994) Station PT of a compound curve is at 15 + 480.14. I 1 = 30°, I 2 = 36°, D1 = 4°, and D2 = 5°. What is the stationing of PCC? Use arc basis.
b.) Determine the stationing of the PCC.
2. CE BOARD (May 1996) A compound curve has the following data I1 = 28°, I2 = 31°, D1 = 3°, and D2 = 4°. Find the stationing of PCC. Use Sta. PI – 30 + 120.50.
6. The long chord of a compound curve is 200 m and the angles it makes with the tangents of the curve are 20° and 18°, respectively. If the common tangent of the curve is parallel to the long chord.
3. The chords of a compound curve from PC to PCC and from PCC to PT are 130.60 m and 139.16 m, respectively. Its common tangent makes an angle of 20° and 36°, respectively, with the tangents at PC and PT. Determine the length of the long chord of the compound curve. 4. The length of the common tangent of a compound curve is 225 m. The first curve has a degree of curve 3° and a central angle of 24°. The second curve has a central angle of 44°. Find the degree of the second curve. Use arc basis. 5. The common tangent AB of a compound curve is 76.42 m. with an azimuth of 268° 30’. The vertex V being inaccessible. The azimuth of the tangents AV and VB was measured to be 247° 50’ and 282° 50’, respectively. If the stationing of A is 43 + 010.46 and the degree of the first curve was fixed at 4° based on the 20-m. chord. Using chord basis.
c.) Determine the stationing of the PT.
a.) Find the radius of the first curve. b.) Find the radius of the second curve. c.) Find the total length of the curve. 7. The long chord from the PC to the PT of a compound curve is 300 meters long and the angles it makes with the longer and shorter tangents are 12° and 15° respectively. If the common tangent is parallel to the long chord. a.) Find the radius of the first curve. b.) Find the radius of the second curve. c.) If stationing of PC is 10 + 204.30, find the stationing PT.
