Answered: A horizontal curve has a PI at…

Engineering

Civil Engineering

A horizontal curve has a PI at STA(52+40.00), I= 35 degrees, R = 725 ft. What are the station numbers of PC and PT?

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: Curves Question 9 What is the degree of the curve (in degree) for a radius of 573 m using chain of…

A: Given- Radius R=573 m chain length=20 m It is needed to find the degree of the curve.

Q: Given a simple circular curve, in feet, with the following known information = 30°45'00" o Da =…

Q: In a railroad layout, the centerline of two parallel tracks are connected with a reversed curve for…

Q: A railroad curve runs a 6º curve to the PCC, 300m long from the PC of the Compound Curve, thence…

Q: För a proposed highway curve with a radius of 500.00 meters, the point of curvature (PC) or…

Q: A circular curve of 500m radius is to be laid to connect two straights APB and CQB intersecting at…

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 horizontal curve has a PI at STA(52+40.00), I= 35 degrees, R = 725 ft. What are the station numbers of PC and PT?

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: Given

VIEW

Step 2: Calculate station number of PC

VIEW

Step 3: Calculate station number of PT

VIEW

Solution

VIEW

Step by step

Solved in 4 steps with 3 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

The PC of a vertical parabolic curve is at station 154 + 104 of elevation 80.45 m and the grade of the PC is +2.5 %. The elevation of station 154 + 184 is 81.95 m. What is the elevation of 154 + 204? (INCLUDE ILLUSTRATION/FBD)
I need help with 24.14
What is the station of the PT of a horizontal curve with the following data: A=40°50', R=600 ft and sta of PI = 20+00? O 20+00.00 O 22+04.27 O 22+23.13 O 24+27.30 O 25+04.30
1. You are traversing along a centerline, (S 45° 00’ E), at Station 1+50, you establish a BC, for a curve with R transitions into a compound curve, also concave to the north, with R = 300', Delta = 30°, which sets an EC for this configuration. Determine the T1 and T2 from the attached Exhibit A sketch. 400', Delta 40°, which is concave to the north, and then
A -1.10% grade meets a +0.60% grade at station 36+00 and elevation 800.00 ft. The +0.60 grade then joins a +2.40 grade at station 39+00. Determine the elevation at the low point of the curve. Express your answer in feet to six significant figures. 15. ΑΣΦ Elev = Submit Part B Request Answer Determine the station at the low point of the curve. 35+40.763 36+44.763 38+42.763 vec 36+42.763 D ? ft
The curve elevation at station 20+00 is VC 1, PVI=21+72.41, PVI Elevation= 430.00 ft, L= 800 ft -3.20% -4.00% +1.50% VC 2, PVI 35+00 O 432.96 ft O 434.00 ft O 435.52 ft O 437.04 ft
A horizontal curve is being designed with tangent length of 850 ft and central angle of 0.7315 radians. If the Pl is at station 215+00, determine the station of PT. A. 222+74 B. 223+50 C. 274+64 D. 271+95
Rank the curves from highest to lowest module
For the PI Inaccessible problem: Sta. X = 70+00.00, XY =800.00 ft, angle x = 35°00'00", angle y = 30°00'00", and R for the simple curve = 1,000.00 ft. The station of A (T.C.) is: O None of the above 68+04.28 65+58.65 69+32.51 O 68+69.23
Determine the distance required to transition pavement cross-slope from a normal crown section with a normal crown cross-slope of 2% to superelevation of 6%. The road is two-lane-two-way (one lane in each direction); superelevation rotation axis is the centerline of the road; lane width is 12 ft; the superelevation runnoff transition rate is 1:400. 384 ft 192 ft 288 ft 96 ft
Given broken line AB = 57.6 m., BC = 91,5 m. and CD = 91.5 m. arranged as shown. A reverse curve is to connect these three lines thus forming the center line of a new road. %3D a) Find the length of the common radius of the reverse curve, b) Find the total length of the reverse curve, CIf the P.C. is at Sta. 10+000, what are the stationing of P.R.C, and P.T. Solu tion: a) T, =Rtan 11 %3D T = R tan 32 T+T = 91.5 91.5 m R (tan 11 + tan 32 ) = 91.5 PT R = 111.688 m. 82 RI IR 7,5 7.5 64 56.7 G'16
ft Given the following vertical curve data: PVI at 7 + 25.712, L= 100, g1 = -3.2%, 82 = +1.8%, and elevation of PVI = 210,440, compute the elevations of the curve low point and even 20- stations. %3D 110 I