A -6% grade and a +2% grade intersect at STA 12+200 whose elevation is at 25.632m. Two grades are to be connected by a parabolic curve, 160m long. A. Find the grade of the first quarter point on the curve. In decimal form B. Find the stationing of the lowest point of the curve.

A -6% grade and a +2% grade intersect at STA 12+200 whose elevation is at 25.632m. Two grades are to be connected by a parabolic curve, 160m long. A. Find the grade of the first quarter point on the curve. In decimal form B. Find the stationing of the lowest point of the curve.

Traffic and Highway Engineering

5th Edition

ISBN:9781305156241

Author:Garber, Nicholas J.

Publisher:Garber, Nicholas J.

Chapter15: Geometric Design Of Highway Facilities

Section: Chapter Questions

Problem 7P

See similar textbooks

Related questions

Q: A certain horizontal curve has a central angle of 42.2°. If STA PC is 18+363 and STA PT is at…

Q: 3. A reverse curve is to be inserted between parallel tangents 50 m apart with AD 350 m, and R2 =…

A: H=350m

Q: A circular curve of 609.6 m radius joins two points A and C which lie on the two straights AB and…

Q: A grade from A to the vertex V is -6% and from V to B is +2%. It is required to connect those grade…

A: The given data is shown below: GradeA-V=-6%GradeV-B=+2%

Q: It is required to construct a vertical curve with 400m from the PC at STA 3+16, elev.214m to the…

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: 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 downgrade of 3.2% meets a rising grade of 4.6% at station 73 + 180, where the elevation is 998 m.…

A: The expression for elevation of B is given as elevation of B=elevation of point of vertical…

Q: The change in grade of a vertical parabolic curve is 0.60%. The grade of the said curve is -4% and…

Q: A 3% ascending grade meets a descending grade at the vertex at station 2+050 and elevation 100m with…

Q: A vertical summit curve is having tangents which intersects at point Pl at Station 12+617.78 and…

A: PVCstn=PVIStn-L212506.53=12617.78-L2=222.5MPVCelv=PVIelv-g1L2123.37=134.19-g1×222.52g1=0.0972×100g1=…

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…

Q: 3. Lay out a simple curve with tangents at point A and B respectively. A line from point A to the…

Q: t is required to construct a vertical curve with 400m from the PC at STA 3+16, elev.214m to the…

Q: The length of curve at side of PC is 45 while on the PT is 50 m. A sewer line with its top having an…

Q: Compute the elevation and station of the highest point of the following vertical curve. 320-m…

A: Given, L = 320 mn1 = 2 %n2 = -1.6 %

Q: the three courses AB, BC and CD have bearing N56°30'E, N80°30'E and S32°17'E, respectively. if BC=…

A: Given Data: AB=N56°30'EBC=N80°30'ECD=S32°17'E

Q: A vertical summit curve is having tangents which intersects at point Pl at Station 12+617.78 and…

Q: A symmetrical parabolic summit curve connects two grades of +5.9% and -4.2%. It is to pass through a…

A: solution given grades g1=5.9% 0.059g2=-4.2%=-0.042PVI station = 25+145m = 25145mtsPVC station…

Q: tioning of which is 35 + 280 and the elevation is 198.13 m. If the elevation of the grade…

Q: A 3% ascending grade meets a descending grade at the vertex at station 2+050 and elevation 100m with…

Q: A +5.2% grade is followed by a -2% grade of a vertical summit parabolic curve at station 2+230 with…

A: Stations are horizontal measurements measured along the Survey Center Line of a project. With…

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: 4. The vertical parabolic curve has its P.I. at station 14 + 750 with elevation of 76.30m. The grade…

A: (a) Consider the free body diagram

Q: 26.10 A grade of +1.5% intersects a grade of -2.5% at station of 1 + 080 of elevation 68.94 m. if…

A: Given Data:g1=1.5%g2=-2.5%station=1+080Elevation=68.94m

Q: A vertical parabolic sag curve has tangent grades of -1.241% and +0.718%. If the grade changes…

Q: a vertical summit curve is having tangents which intersects at point Pl at station 12+615.61 and…

A: This question belongs to subject surveying. In this question we find out the elevation of point…

Q: Find the area enclosed by the curve r = 5cosθ

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: b. 500 m c. 360 m d. 527 m 2. Find the stationing of PC. a. 7+589

A: Length=Ar=0.03+0.020.00220m L=500m

Q: A -6% grade and a 2% grade intersect at vertex (Sta. 12+200). The two grades are to be connected by…

A: Given : -6% grade meets +2% grade 160 m long parabolic curve Elevation of PC = 14.375 m

Q: It is required to construct a vertical curve with 400m from the PC at STA 3+16, elev.214m to the…

Q: A certain horizontal curve has a central angle of 39.3°. If STA PC is 18+334 and STA PT is at…

Q: A simple curve connects two tangents AB and BC with bearings N85030’E and S68030’E, respectively.…

Q: A 175m long sag parabolic curve connects a -8.2% grade to a +2.4% grade. The grades intersect at…

Q: A certain horizontal curve has a central angle of 31.9°. If STA PC is 18+303 and STA PT is at…

A: PC=18+303PT=18+74718+747-18+303=444Rtan∆2=444R=4440.2858=1553.53

Q: A symmetrical parabolic curve passes through point A whose elevation is 23.33m at a distance of 54m…

A: The given data is shown below: Elevation from PC=23.33 mDistance from PC=54 mElevation of PC at…

Q: A vertical curve is to be laid out. The curve is 900.00ft long, g1= -2.5%, g2 = +3.5%, STA@BVC (VPC)…

A: L = Length of curve = 900 ftg1 = Intial grade = - 2.5%g2 = Final grade = 3.5 %BVCSTA = 55 +…

Q: A vertical summit symmetrical parabolic curve has a length of 141m with tangent grades of -1.5% and…

A: Solution: Given Data: n1=-1.5%n2=2.5%Elevation=240 mStation=12+640.22

Q: a downgrade of 3.20 meets an upgrade of 4.6 at Sta. 73+180 where the elevation is 998m. A parabolic…

Q: The centerline of a new railway line is to be set out along a valley. The first straight AI bears…

A: The given problem is based on the compound circular curve problem. This can be solved as follows:

Q: A parabolic curve JS, 340 m long is connected by tangents having an upgrade of +5.7% and a…

Q: 3. A symmetrical parabolic curve is designed to pass through a point A whose elevation is 22.995m at…

A: In the above discussed problem the curve is parabolic curve, the general equation for this curve is…

Q: 1. A symmetrical parabolic curve has descending grade of 6% and an ascending grade of 2% and…

A: In creating a curve, the tangent distance must often be limited. Some of them are crossings of…

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

Q: A vertical summit curve is having tangents which intersects at point Pl at Station 12+618.37 and…

A: Solution Grade, g2 = -5.13% (given) = -0.0513 Find Length of curve, L L2 = Station PI - Station…

Q: An ascending grade of 4.9% intersects another grade at STA 12+598 whose elevation is 500m. The…

Q: A -6% grade and a +2% intersect at Sta 12 + 200 whose elevation is at 14.375 m. The two grades are…

Q: A 3% ascending grade meets a descending grade at the vertex at station 2+050 and elevation 100m with…

A: Stationing of highest point is also, written as Sta HP It is expressed as ; Station of highest point…

Concept explainers

Electronic Spreadsheet

Tabular listing has many advantages ranging from time management to efficient scheduling of tasks. Electronic spreadsheets allow the user to tabulate and analyze data, using its easy-to-use interface. An electronic spreadsheet is a staple application used among accountants, management staff, and IT professionals. It is used in various sectors worldwide for data management. An electronic spreadsheet is a system and computer software that could be used to do numerical computations automatically. Spreadsheet application programs are often organized as a table with rows and columns. Each row and column intersect to form a cell, which may be used to store data. These data can take the form of a written label, a number, or a formula that combines information from other cells.

Question

A -6% grade and a +2% grade intersect at STA 12+200 whose elevation is at 25.632m. Two grades are to be connected by a parabolic curve, 160m long.

A. Find the grade of the first quarter point on the curve. In decimal form

B. Find the stationing of the lowest point of the curve.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Step by step

Solved in 3 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

Compute 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)
Compute the entering grade (in % to 2 dp) of the vertical curve connecting the following profile formed by the following points (1+130, 170 m asl); (2+20, 120 m asl) and (3+10, 100 m asl)
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%
3. A grade descending at the rate of -4% intersects another grade ascending at the rate of +8% at station 2+000, elevation 100m. A vertical curve is to connect the two such that the curve will clear a boulder located at station 1+980, elevation 101.34. Determine the necessary length of the curve. Sketch the curve.
. A grade line AB having a slope of +5% intersect another grade line BC having a slope of –3% atB. The elevations of pointsA,BandCare 95 m, 100 m and 97 m respectively. Determine the elevation of the summit of the 100 m parabolic vertical curve to connect the grade lines. Resolve using equations for a parabola. (y=Ax^2+Bx+C)
The grade from point A to vertex V is -6% and from V to point B is +3%. It is desired to connect these grades with a vertical symmetrical parabolic curve that shall pass 1.8 m. directly above V. Determine the length of the curve in meter. (Consider 2-decimal places) What is the difference in elevation between A and B in meter? (Consider 2-decimal places) How far is the lowest point of the curve from B in meter? (Consider 2-decimal places)
The grade from point A to vertex V is -6% and from V to point B is +3%. It is desired to connect these grades with a vertical symmetrical parabolic curve shall pass 1.8 m directly above V. Determine the difference in elevation between A and B
A 160m long parabolic curve connects a -8% grade to a +2% grade.These grades intersect at Sta. 1+240 with elevation of 14.87m. Find the elevation and station of the lowest/highest point on the curveand at the first quarter point of the curve. (DRAW CLEAR FIGURE)
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.
2. A grade line AB having a slope of +5% intersect another grade line BC having a slope of –3% at B. The elevations of points A, B and C are 95 m, 100 m and 97 m respectively. Determine the elevation of the summit of the 100 m parabolic vertical curve to connect the grade lines. PLEASE SOLVE IT USING THE EQUATIONS OF A PARABOLA. (y=Ax2+Bx+C) DO NOT SOLVE THE REQUIRED USING THE GRADE DIAGRAM.
A 160m long sag parabolic curve connects a -8% grade to a +2% grade. The grades intersect at station 1 + 240 with elevation of 14.87m. Find the elevation of the first quarter point of the curve
A symmetrical parabolic curve has an ascending grade of 3.2% and a descending grade of 2.8%. If the PC is at elevation 68.72m, what is the elevation at the PI, length of curve is 50m.