A grade of +5% grade intersects a grade of -3.0% P.C. has a of 10 + 110.00 which has an elevation of 30MASL. These two center gradelines are to be connected by a 240 meter symetrical vertical parabolic curve. a. What is the station of the highest point of the vertical curve. b. What will be the elevation of the curve at station 10+210. c. What will be the elevation of the curve at station 10+250

Q: A 160m long parabolic curve connects a -8% grade to a +2% grade. These grades intersect at Sta.…

A:

Q: A symmetrical parabolic curve passes through point A whose elevation is 23.23m at a distance meters…

A: Using equation y=g2-g12Lx2+g1x+PVCelvg1=back tangentg2=Forward tangentL=length

Q: The tangents of a sag curve connects at the point of intersection having Station 18+407.60 and…

A: 1. Given : At point of intersection (PI) ⇒ Station = 18 + 407.60…

Q: A +3 % grade @ PC intersects with a -4.2 % grade at PI forming a symmetrical vertical curve with…

A:

Q: A vertical parabolic sag curve of Laon Laan underpass has a grade of -4% followed by a grade of +2%…

A: We know that ; r = g2 - g1L L = g2 - g1r where ; g2 = 2 % = 0.02 g1 = - 4 % = - 0.04 r = 0.6 % =…

Q: A - 6% grade and a + 2% grade intersect at sta. 12 + 200 whose elevation is at 45.673 m. The two…

A: PVCELV=PVIELV-g1L2PVCelv=45.673+0.06×1802PVCelv=51.073m

Q: A grade of -5% grade intersects a grade of 0%. P.C. has a STATION of 10 + 110.00 which has an…

A: Given that   Sag vertical curve   g1 = -5%   g2 = 0   A = G1-G2   A = -5   PC station = (10+110)…

Q: A forward tangent of +6% was designed to intersect a back tangent of  -3% at a proposed underpass…

A: GIVEN DATA:- L1 = 200m L2 = 100m g1 = -3% g2= +6% VPI 12+ 530.20 Bridge Elevation = 117.48m VPI…

Q: Q6. A -2.5% grade is connected to a +1.0% grade by means of a 180 m vertical curve. The P.I. station…

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Q: A downgrade of 4.6° meets an upgrade of 3.20% at Sta . 73+108 where the elevation is 998 m. A…

A: Given; downgrade = 4.6°  Sta = 73+108 elevation = 998 m. parabolic curve AB=440 m long

Q: A parabolic vertical curve is to be set out connecting two uniform grades of +1.8 and +3.9%. The…

A: Length=Ar=3.9-1.80.0520m=840mPVCsta=PVIsta-L2PVCsta=2705-8402=2285PVCELV=PVIELV-g1L2PVCelv=238.755-0…

Q: curve has a length of 141m with tangent grades of -1.5% and 2.5% intersecting at station 12+640.22…

A: PVCSTATION=PVISTA-L2PVCsta=12640.22-1412=12+569.72PVCELV=PVIELV-g1L2PVCelv=240-0.025×1412PVCelv=238.…

Q: evel and the through chainage of the intersection point of the two gradients are 93.60m and 66…

A: PVCSTATION=PVISTA-L2PVCsta=6671.34-2002=6571.34PVCELV=PVIELV-g1L2PVCelv=93.6-0.01×2002PVCelv=92.6

Q: Identify the type of vertical curve (sag or crest) by the following rates of change in grade. -0.5%…

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Q: A grade descending at the rate of -4% intersects another grade ascending at the rate of +8% at…

A: We have to determine the value of necessary length of the curve as per the given data in the…

Q: A-4% grade line meets another line of +3.2% grade at station 11+ 40. If a 225-ft parabolic curve…

A: This question is related to Simple curve from Surveying Engineering. In this question we calculate…

Q: A 6.2% grade is followed by -2.8% grade; the grades intersecting at sta. 10+800 of elevation 1060 m…

A: The grades are g1=6.2% and g2=-2.8%. The station of grade intersection is PVI=10+800. The elevation…

Q: A 1200 feet parabolic vertical curve has an entering grade of –2.6% and a departing grade of +3.4%.…

A: PVC Station…

Q: A vertical crest curve has an approach grade +3% and a departure grade –3%. The VIP level is 120 m…

A: To calculate 'H' based on the given data we have  2HL2= (g1 -g2)L2L g1 = 3% = 0.03 g2 = -3% = - 0.03…

Q: as PVC at Sta 19+274 and at elevation 136m. The initial grade is 6.14% and the final grade is…

A: Given data PVC at Sta 19+274 elevation 136m  initial grade is 6.14% final grade is -5.58%

Q: A parabolic curve has a descending grade of -0.85% which meets an ascending grade of 0.35% at sta.…

A: Given Data: A parabolic curve is given as: Descending grade(g1)=-0.85%Ascending…

Q: A parabolic curve has a descending grade of -0.6% which meets an ascending grade of 0.3% at sta. 10…

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Q: A -2.6% grade meets a +4.8% grade near an underpass. To maintain the minimum clearance allowed under…

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Q: A -4% grade at Sta. 5 + 700 and elev. 110.2 is to connect to a +6% grade at Sta. 6 + 100 and elev.…

A: Determine the elevation of PCC and location of PVI.

Q: Two tangents intersect at V with back tangent having a grade of -4% and forward tangent of 5%. A 205…

A:

Q: A +3.2% grade Line meets another Line of -4% grade at station 11 + 40. If a 225-ft parabolie curve…

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Q: A-3.3% grade meets a +7.4% grade at a vertex (EI. 139.57m) directly under an overpass bridge having…

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Q: Q3:- A parabolic vertical curve joins a grade of -3% to a grade of +5.5 %. The PVI is at station…

A: The downward gradient (g1) = -3% = -0.03   The upward gradient (g2) = +5.5% = 0.055The station of…

Q: An underpass crossing a reinforced concrete bridge has a downward grade of -4% meeting an upward…

A:

Q: joining two gradients AB and BC. Hesign the curve at 20 metres intervals. Table Q2 (Units are in 9

A: Length…

Drainage of Pavements

Pavement Drainage holds ample significance, as it prevents the accumulation of water or any liquid on the surface of pavements and ensures proper drainage system. With proper design and gradual slopes provided, the water gets easily removed from the surface into the drains, which are essential during heavy rains and unusual weather conditions keeping it less prone to accidents and increasing the durability of the pavement structure.