(a)
Using the vertical curvature determine minimum length of vertical curve.
Answer to Problem 4P
Explanation of Solution
Given information:
Station
Elevation
Grades
Grades
Design speed of
Calculation:
Length of curve is given by formula,
A is the grade difference between
Rate of vertical curvature from textbook Table
Substitute in equation
Minimum length of vertical curve is
(b)
Stations and elevations of BVC and EVC.
Answer to Problem 4P
Station at BVC is
Tangent elevation at BVC is
Explanation of Solution
Given information:
Station
Elevation
Grades
Grades
Design speed of
Calculation:
Therefore station at BVC is
Therefore tangent elevation at BVC is
(c)
Elevation at each
Answer to Problem 4P
Station | BVC distance
| Tangent elevation | offset | Curve elevation(Tangent elevation) |
Explanation of Solution
Given information:
Station
Elevation
Grades
Grades
Design speed of
Calculation:
Determine the offset,
For every
Station | BVC distance
| Tangent elevation | offset | Curve elevation(Tangent elevation) |
(d)
Station and elevation of high point.
Answer to Problem 4P
Explanation of Solution
Given information:
Station
Elevation
Grades
Grades
Design speed of
Calculation:
Distance of the high point of BVC is given by,
Difference beteween both elevation of BVC and highpoint.
Conclusion:
Therefore elevation of high point is
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Chapter 15 Solutions
Traffic and Highway Engineering
- Problem 2. This is a four-part problem. A +2.5% grade intersects with a –1.5% grade at station (735 + 30.75) at an elevation of 475 ft. Part A. If the design speed is 65 mi/h, determine the minimum length, in ft, of vertical curve. Part B. If the design speed is 65 mi/h, determine the elevation, in ft, of the point of vertical tangency also known as end of vertical curve. Part C. If the design speed is 65 mi/h, determine the elevation, in ft, of the highpoint of the curve. Part D. If the design speed is 65 mi/h, determine the distance, in ft, from the point of vertical curvature (also known as the beginning of curve) to the highpoint of the curve.arrow_forwardA parabolic curve has a descending grade of -0.8% and an ascending grade of +0.4% that intersect at station 10 + 020 with an elevation of 240.600 m. If the maximum allowable rate of change of grade for this curve is 0.15% per 20 m stations, determine the following: a. The maximum probable length of the parabolic curve b. The location and elevation of the lowest point of the curve c. The elevation at station 10 + 000arrow_forwardA +8 percent grade intersects with a -2 percent grade at station (535 + 24.25) at an elevation of 300 ft. (a) If the design speed is 45 mi/h, determine the minimum length (in ft) of vertical curve using the rate of vertical curvature. (Assume the stopping sight distance is less than the length of the curve.) 610 ft (b) Using the length found in part (a), find the stations and elevations (in ft) of the BVC and EVC and the elevation (in ft) of each 100 ft station. (In the table below, the first row corresponds to the BVC and the last row corresponds to the EVC. Round your elevations to at least one decimal place.) Station Elevation (ft) 532 + 19.25 275.6 533 + 00 306.1 534 + 00 535 + 00 299.7 536 + 00 537 + 00 538 + 00 538 + 29.25 293.9 (c) Using the length found in part (a), find the station and elevation (in ft) of the highpoint. station elevation ftarrow_forward
- 12. Find the minimum length of curve for the following scenarios. Entry Grade Exit Grade Design Speed Reaction Time A 3% 8% 45 mi/hr 2.5 s В -4% 2% 65 mi/hr 2.5 s 0 % -3% 70 mi/hr 2.5 sarrow_forwardA+1 percent grade intersects with a-2 percent grade at station (535+ 24.25) at an elevation of 300 t. (a) if the design speed is 65 mi/h, determine the minimum length (in t) of vertical curve using the rate of vertical curvature. (Assume the stopping sight distance is less than the length of the curve.) it (b) Using the length found in part (a), find the stations and elevations (in R) of the BVC and EVC and the elevation (in t) of each 100 t station. (In the table below, the first row corresponds to the BVC and the last row corresponds to the EVC. Round your elevations to at least one decimal place.) Station Elevation (t) 533 + 00 534 + 00 535 + 00 536 + 00 537 + 00 538 + 00 (c) Using the length found in part (a), find the station and elevation (in t) of the highpoint. station nlevationarrow_forwardA crest vertical curve joining a + 3 percent and – 4 percent grade is to be designed for 75 mph. If the tangent intersect at station (345 + 6000) at an elevation of 250 ft, determine the stations and elevations of the PVC and PVT.arrow_forward
- A grade of -3% is followed by a grade of +2% which intersect at station 10+251 and elevation of116 m. If the maximum change in grade per 20 m station is 0.2%, determine the following:a. Elevation of PCb. Elevation of a point on the curve at station 10+251c. Length of parabolic curve that shall connect these two gradelines.arrow_forwardA grade of -3.0% is followed by a grade of +2.1% which intersects at station 10+250 and elevation of 110 m. If the maximum change in grade per 20 m station is 0.20%, determine the following A. Length of parabolic curve that shall connect these two gradelines. B. Elevation of PC. C. Elevation of a point on the curve @ Sta.10+250.arrow_forwardA grade of -3.6% is followed by a grade of +2.1% which intersects at station 10+256 and elevation of 116 m. If the maximum change in grade per 20 m station is 0.20%, determine the following A. Length of parabolic curve that shall connect these two grade lines. B. Elevation of PC. C. Elevation of a point on the curve @ Sta.10+250.arrow_forward
- Note : If you don't know the solution please leave it but don't give me wrong solution. A vertical curve joining entering grade of -4% with an exiting grade of - 7% begins at station 2+607 and is being designed with K value of 72. What is the station of the highest point on the curve . Code the station as the total distance from the origin, i.e., if the station is 1+100 code it as 1100.arrow_forwardA +3.00% grade intersects a -2.40% at station 46+70.00 and elevation 853.48 ft. A 400.00 ft curve will be used to connect the two grades. Compute Station and elevation for the curve's endpoints.arrow_forwardA 500-m long equal tangent vertical curve has a point of vertical curvature at station 2 + 600 and elevation 450 m. The initial grade is -1% and final grade is +1% b) determine the highest and the lowest points on this vertical curve by using the K-value.arrow_forward
- Traffic and Highway EngineeringCivil EngineeringISBN:9781305156241Author:Garber, Nicholas J.Publisher:Cengage Learning