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- A symmetrical parabolic curve passes through point A whose elevation is 23 23m at a distance of 54 meters from PC The elevation of the P.C. at station 4+100 is 22.56m. The grade of the back tangent is +2% and the length of the curve is 120m. Determine the grade of the forward tangent a. -2.3% b. -1.4% c. -3.3% d. 2.5%A symmetrical parabolic curve passes through point A whose elevation is 23.23m at a distance of 54 meters from P.C. The elevation of the P.C. at station 4+100 is 22.56m. The grade of the back tangent is +2% and the length of the curve is 120m. Determine the grade of the forward tangent. Determine the stationing of the highest point.A forward tangent of +6% was designed to intersect a back tangent of -3% at a proposed underpass along a street so as to maintain a minimum clearance allowed under a bridge which crosses perpendicular to the underpass. A 200 m. curve lies on the side of the back tangent while a 100 m. curve lies on the side of the forward tangent. The stationing and elevation of the grade intersection is 12+530.20 and 100 m., respectively. The center line of the bridge falls at station 12+575.20. The elevation of the underside of the bridge is 117.48 m. Determine the minimum clearance of the bridge if it has a width of 10 m.
- Two tangents intersect at v with back tangent having a grade of - 2.75q and forward tangent of 6.25%. A 195m curve lies on the side of PC AND 105 on the other side. Station Ing of V is at 10+535 having an elevation of 102 if a bridge is located above the curve, with its centerline having an elevation of 119m and stationing of 10+580, with underside of the bridge located 120cm below the given election, determine the height of the highest vehicle it can accommodate if the bridge has a width of 12mCompute the exiting grade (in % to 2 dp) of the vertical curve connecting the following profile formed by the following points (1+110, 110 m asl); (2+137, 168 m asl) and (3+115, 154 m asl)A symmetrical parabolic curve has a descending grade of -0.80% which meets ascending rate of +0.4% at station 10+020 having an elevation of 240.62m. The maximum allowable change in grade per 20m station is 0.15%. Determine the stationing of the lowest point of the curve. (Answer: 10+046.67) Determine the elevation of the lowest point of the curve. (Answer: 240.83)
- A vertical Summit parabolic curve has a vertical offset of 0.375m from the curve to thetangent grade at sta10 + 050. The curve has a slope of +4% and -2% grades intersectingat the P.I. The offset distance of the curve at P.I. is equal to 1.5m. If the stationing of theP.C. is at 10 + 000.a. Compute the required length of curve.b. Compute the horizontal distance of the vertical curve turning point from the point ofintersection of the grades.c. Compute the elevation of the vertical curve turning point if the elevation of P.T. is86.42m.A 6°simple curve connects two tangents that intersect the PC which is at station 0 + 168.150 is to be retained and a 4°curve is to connect the first curve with the second tangent that is shifted 20 meters to a parallel position outward, find the PCC and the new PT. Use chord basis.an unsymmetrical parabolic curve has a forward tangent of -8% and a back tangent of +5%. the length of the curve on the left side of the curve is 40m long while the right side is 60m long PC is at Sta. 66 + 780 and at Elev. 110m determine the height of fill at the crop (30m from PC with top elevation of 108.,40m)
- A parabolic curve JS, 480 m long is connected by tangents having an upgrade of +5.9% and a downgrade of -2.8% intersecting at Sta 20+870 at elevation 460 m. Find the distance from S to the highest point of the curve.A 6% grade meets a -4% grade at a vertex whose directly under an overpassbridge whose underside is at elevation 152. 74 m. and carries another road across the grades at right angles. a. The length of the curve on the side of the back tangent is 70 m., while onthe right side of the forward tangent is 100 m. long. Compute the verticalclearance from centreline of the bridge to the vertical curve if theelevation at PC is 142 m.b. If the underside of the bridge is level and is 10 m. wide, Find the actualclearance on the left and right edge of the bridge.A forward tangent of +6% was designed to intersect a back tangent of -3% at a proposed underpass along N. Bacalso Avenue so as to maintain a minimum clearance allowed under a bridge which crosses perpendicular to the underpass. A 200 m curve lies on the side of the back tangent while a 100 m curve lies on the side of the forward tangent. The station and elevation of the grade intersection is 12 + 530.20m and 100 m, respectively. The centerline of the bridge falls at station 12 + 575.20m. The elevation of the underside of the bridge is 117.48 m. If the bridge is 10m wide, determine the: a. Vertical Clearance on the center, left side, and right side of the bridge. b. Minimum clearance of the bridge. c. Location, station, and elevation of the drain cover grating to ensure full drainage of storm water.