An equal tangent vertical curve is to be constructed between grades of -2.0% (initial) and 1.0% (final). The PVI is at station 11 + 000.000 and at elevation 420 m. Due to a street crossing the roadway, the elevation of the roadway at station 11 + 071.000 must be at 421.5 m. Design the curve.
Q: An equal tangent vertical curve is to be constructed between grades of -2.0% (initial) and 1.0%…
A: Solution:GradeG1=-2.0%Grade G2=1.0%Point of vertical intraction elevation=420 mPoint of vertical…
Q: Determine the minimum length (ft) for a vertical curve with 55mph design speed if the grades to be…
A: The length of the vertical curve is given by: L=K×G Where, K = Rate of vertical curvature G = Grade…
Q: f the vertical curve has
A: Let the length be 'L'PVCelv=PVIelv-g1(L2)=105-(-0.04)(L2)PVCelv=105+0.02LEquation of the curve…
Q: SITUATION. A -3% grade meets a +5% grade near an underpass. In order to maintain the minimum…
A: Procedure: Find the station of lowest point use the location of lowest point Find the elevation…
Q: An equal tangent vertical curve is to be constructed between grades of -6.75% (initial) and 4.75%…
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Q: A grade g1 of -2% intersects g2 of +1.6% at a vertex whose station and elevation are 87+00 and…
A: Given Data: g1=-2%g2=+1.6%station and elevation = 87+00 and 743.24L=400
Q: A vertical summit curve has its highest point of the curve at a distance of 48m from the PVT. The…
A: Given : vertical summit curve PVI has elevation 100 m at 25+160 G1 = 6%G2 = -4%
Q: It is required to construct a vertical curve with 400m from the PC at STA 3+16, elev.214m to the…
A: Solution attached below
Q: A vertical summit curve is having tangents which intersects at point Pl at Station 12+584.57 and…
A: Given Data:PVI=12+584.57Elv=139.32mgrade=-5.25%PVC=12+527.65Elv=122.32m
Q: An equal tangent vertical curve is intended to connect grades of 1.9% and -1.9%. If this curve is to…
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Q: A vertical curve was designed to connect intersecting tangents of grades of 4% ar - 6% for back and…
A: aWe knowOffset=Ax22L
Q: It is required to construct a vertical curve with 400m from the PC at STA 3+16, elev.214m to the…
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Q: PROBLEM 2 A SYMMETRICAL PARABOLIC SUMMIT CURVE CONNECTS TWO GRADES OF +4% AND -6%. IT IS TO PASS…
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Q: An equal tangent sag equal tangent verticle curve (with a negative initial and a positive final…
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Q: Q7.An equal-tangent vertical curve is to be constructed between grades of -2.0% and +1.0%. The PVI…
A: Step 1
Q: Question 3 An equal-tangent vertical curve is to be constructed between grades of -2 percent and +1…
A: Given :
Q: A symmetrical parabolic summit curve connects two grades of +5.7% and -49%. It is to pass through a…
A: Length is given by L2=PVI-PVCL=225145-24976L=338m
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Q: An upwards grade of 4.6% and a down-grade of -2.8% were connected with a vertical curve of 400m. The…
A: PVCSTATION=PVISTA-L2PVCsta=600-4002=400PVCELV=PVIELV-g1L2PVCelv=100-0.046×4002PVCelv=90.8m
Q: A 500-ft-long equal-tangent crest vertical curve connects tangents that intersect at station 340 +…
A: The distance between the initial point of vertical curve and the point of tangent intersection can…
Q: A -3% grade meets a 5% grade near an underpass. In order to maintain the minimum clearance allowed…
A:
Q: A 400-ft equal-tangent sag vertical curve has its PVC at station 78+00 and elevation 800 ft. The…
A: Length of curve (L) = 400 ftStation of PVC = 78+00Elevation of PVC = 800 ft
Q: A-3% grade meets a +5% grade at a vertex (El. 146.24m) directly under an overpass bridge having…
A: PVCelv=PVTelv-g1L2PVCelv=146.24+0.03L2
Q: A sag vertical curve is to be designed to join a -4% grade to a +2% grade. If the design speed is 45…
A:
Q: An equal tangent vertical curve is intended to connect grades of 4.8% and -3%. If this curve is to…
A: GIVEN- AN EQUAL TANGENT VERTICAL CURVE CONNECTING TWO GRADE 4.8% AND -3%, IT IS REQUIRED TO FIND…
Q: A vertical crest curve has an approach grade +4% and a departure grade -3%. The Vertical…
A: The curve in the surveying is of two type like horizontal and vertical curve. The vertical curve…
Q: SITUATION. A -3% grade meets a +5% grade near an underpass. In order to maintain the minimum…
A: Given: The grade g1 is -3%. The grade g2 is +5%. The length of one side of the curve is 200 m. The…
Q: A 4.5% grade intersects a -1% grade at STA 5+180 at an elevation of 195m. A vertical curve of length…
A: Given: The length of vertical curve is 120 m. Here: PVC Station=PVI Station- Length of vertical…
Q: An equal tangent vertical curve is intended to connect grades of 3.1% and -0.8%. If this curve is to…
A:
Q: A 600 ft equal tangent vertical The initial grade is -3.5% and the final grade is +0.5%. Determine…
A: L=600ft
Q: A crest vertical curve (equal tangent) has PVI at station 210+00 and elevation 420.75 ft. The…
A: SPVI = Stationing of PVI = 210+00EPVI = Elevation of PVI = 420.75 ftg1 = Initial grade = 2.5% =…
Q: Two tangents intersect at V with back tangent having a grade of -2.75% and forward tangent of 6.25%.…
A: PI: PI means Point of Intersection. The point origin where the back and forward tangents intersect…
Q: vertical sag curve has tangent grades of -4.3% and +4.5% meeting at point S whose elevation is 205…
A: solution given grades g1=-4.3% g2=4.5%intersection elevation at point S =205m(PVIelevation)length…
Q: A sag curve and crest curve connect -2.75% tangent section of highway (to the west) with -1.25%…
A: We knowA=G1-G2Acrest=Gc+2.75Asag=Gc+1.25Length of curveL=KSAS+KCACfrom tables for…
Q: A vertical equal-tangent, 100O ft long, connects an initial grade of -2.5% and a final grade of +2%.…
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Q: A vertical summit curve has its highest point of the curve at a distance 48m from the P.T. The back…
A: Consider the free body diagram ;
Q: A symmetrical vertical summit curve has tangents of +4% and -2%. The allowable rate of change of…
A: Given Data: g1=+4% , g2=-2% Rate of change of grade=r=0.3% Stationing of PT= 10+020 Elevation of…
Q: A 400-ft equal tangent sag vertical to curve has its PC at station 100+00 and elevation 500ft. The…
A:
Q: A symmetrical vertical summit curve has tangents of + 4% and – 2%. The allowable rate of change of…
A:
Q: A 600ft equal tangent sag vertical curve has the PVC at station 170 +00 and elevation 1000ft. The…
A: Given that, Equal tangent = 600 ft = L PVC station = 170 +00 C= 1000 ft G1 = -3.5 % G2 = +0.5%
Q: A parabolic vertical curve joins a grade of - 3% to a grade of +5.5 %. The PVI is at station 3+51…
A: Given Data: downward gradient ,g1=3%=-0.003; upward gradient g2=5.5%=0.055 Therefore, deviation…
Q: An engineer has been decided to design a vertical curve of a (- 2.5) percent grade meting a (3.2)…
A:
Q: A -3% grade meets a +5% grade near an underpass. In order to maintain the minimum clearance allowed…
A: Procedure: First, we need to find the length of the curve L. After that, we need to find the x and…
Q: Determine the minimum length of a sag vertical curve between a -1.06% grade and + 2.1% for a road…
A: Given data in question Design speed Sight distance Gradients Sag curve To find out Minimum length…
Q: A 520 ft long equal-tangent crest vertical curve connects tangents that intersect at station 340+00…
A: From the given data it is required to calculate the location and elevation of PVC, PVT and high…
Q: The data given for a vertical sag curve on a roadway plan and profile sheet are as follows: PVI…
A: Survey knowledge is useful for dividing the entire length of the curve station-wise by that it is…
Q: ith a PSD of 500
A: We Know Relationship between Length and Sight distance is L=AS22158 imperial unitsL>S
Q: A symmetrical parabolic summit curve is to connect two roads with grades of +4% and -2.5% that meet…
A:
Q: A 5.3% grade intersects a -1.9% grade at STA 5+180 at an elevation of 186m. A vertical curve of…
A: Given :- g1 = 5.3% g2 = -1.9% Sta of PVI = 5+180 Elevation of PVI = 186 m Length of vertical curve…
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- Compute and tabulate full-station elevations for an unequal-tangent vertical curve to fit the requirements shown below. A -3.40% grade meets a +2.75% grade at station 95+00 and elevation 1320.64 ft. Length of first curve 400 ft., second curve 600 ft. Match the correct elevation to the appropriate Station Value. Group of answer choices Sta. 91+00 (BVC) [ Choose ] Sta. 92+00 [ Choose ] Sta. 93+00 [ Choose ] Sta. 94+00 [ ChooseA vertical summit curve is having tangents which intersects at point Pl at Station 12+617.78 and 134.19 meters elevation. The grade for the forward tangent is -6.28%. Station PC is at station 12+506.53 and having an elevation of 123.37 meters. Determine the elevation of the highest point in meters. CS Scanned with CamScanner2. Compute by hand and tabulate station elevations and the 1st and 2nd differences for an equal tangent parabolic (vertical) curve for the following data: A +3.50% (g1) grade meets a -4.00% (g2) grade at Station 12+65 (the PVI). The PVI Elevation = 300.00'. Length of Curve = 750'. Calculated elevations for full stations between the BVC and the EVC. Also, calculate the station and elevation of the curve for the PVI, BVC, EVC and the High Point. Place all the information in station order in a properly labeled and organized table using the exact format shown in the last slide of the Application of Vertical Curves Video. Show all elevations to 3 decimal places. Provide a page sized detailed and labeled sketch of the profile of the curve. Deliverables: Your calculations, sketches and tables should be submitted by the syllabus deadline as a properly named single PDF document. The order for the information in the submission PDF is to be as follows: Problem 1 Hand Calculations Problem 1…
- A +1.512% grade meets a -1.785% grade at PVI Station 31+50, elevation 562.00. The Equal Tangent Vertical curve = 700 feet. Calculate the elevations on the vertical curve at full stationsAn equal-tangent vertical curve connects a +3.2 % and a -1.1 % grade. The PVI is at station 100+20 and elevation 950 ft and the PVT is at station 105+20. Answer the following 5 questions Question 3 to Question 7. The station of the PVC is at: sta 93+20 sta 95+20 sta 97+20 sta 99+20 sta 105+20+2.00% intersects grade g2 =-1.90% at a vertex whose station and elevation are 78+60 and 7855.35 ft, respectively. An equal-tangent A grade gi = parabolic curve 420 ft long has been selected to join the two tangents. Part A Compute station elevations for the curve, for stakeout at full stations. Enter the elevations in the following order: BVC, full stations in order of ascending station coordinate, and, finally, EVC. Express your answers, separated by commas, to six significant figures. nνα ΑΣφ vec ft, ft, ft, ft, ft, ft
- A +2.5% grade meets a -1.75% grade at the station 44+25 and elevation 386.96 ft, the length of curve is 800 feet (8 stations). (1) Provide the stakeout for full stations using parabolic method STATION X g1 X (g2-g1) X2 2L Y (Elevation) PVT = PVC = (2) Determine the station and elevation at the high point of the curveA vertical summit curve is having tangents which intersects at point Pl at Station 12+618.37 and 133.75 meters elevation. The grade for the forward tangent is -513%. Station PC is at station 12+515.99 and having an elevation of 116 26 meters. What is the elevation of point A along the curve with Station 12+548.27? cs Seanned with CamScannerAn equal tangent vertical curve is to be constructed between grades of -6.75% (initial) and 4.75% (final). The PVI is at station 12+000.000 and at elevation 319 m. Due to a street crossing the roadway, the elevation of the roadway at station 12+071.00 must be at 324 m. Which of the following most nearly gives the elevation of the PVC in meters? * 335.0 O 330.4 339.0 330.0 348.1
- A Highway curve will pass through a railway at grade. The crossing must be at Sta. 4+310 and at elevation 220.82 m. the initial grade is +2% and meets a -3% grade at Sta. 4+235 at elevation of 223.38 m. The rate of change must not exceed at 2%. Compute the length of curve, station and elevation of highest point also check if the condition of rate of change if meet. 3 decimal places. ASAAP 30 MINS THANK YOUPoint “P” is the location of the center line of an existing highway. An underpass is to be designed perpendicular to the existing highway with a vertical parabolic curve such that its lowest point is directly below point “P” with a vertical clearance of 5.5 meters. Stationing of the point of intersection is 5+800 and has an elevation of 105 meters. The slope of the tangent passing thru the Pc is -4% and that of the PT is 3%. With illustrations.a) Determine the length of the curve. b) Determine the stationing of the point “P” on the right side of the curve if it has an elevation of 120 meters. c) Determine the elevation of the PT of the curve.An equal tangent vertical curve is to be constructed between grades of -2.0% (initial) and 1.0% (final). The PVI is at station 11 + 000.000 and at elevation 420 m. Due to a street crossing the roadway, the elevation of the roadway at station 11 + 071.000 must be at 421.5 m. Design the curve. with drawing