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Problem 6 Use the following data to calculate the layout of an equal tangent vertical curve...
Problem #5 Given the following data for a vertical curve of a road: G1 = -2.5%,...
Problem #5 Given the following data for a vertical curve of a road: G1 = -2.5%, G2 = +5.5% and station of PVI = 250+50. At station 252+50, the vertical curve must intersect an existing road at elevation 48.48 ft. The elevation on the tangent section at station 256+50 is 67.55. (a) Draw a good sketch of the vertical profile indicating all major points. (b) What is the length of the vertical curve that satisfies the above geometrics? (C) What...
(A) For an equal tangent vertical curve g1 = -4%, g2 = +3%, L = 200.00...
(A) For an equal tangent vertical curve g1 = -4%, g2 = +3%, L = 200.00 ft, BVC Station = 10+00.00, BVC Elev = 5000.00 ft. EVC Station = ? (b) For an equal tangent vertical curve g1 = -4%, g2 = +3%, L = 200.00 ft, BVC Station = 10+00.00, BVC Elev = 5000.00 ft. EVC Elevation = ? (D) For an equal tangent vertical curve g1 = -4%, g2 = +3%, L = 200.00 ft, BVC Station =...
Problem #3 (20 Points) Vertical Curve Problem (Use both Tangents) See Attached Vertical Curve Drawing Station...
Problem #3 (20 Points) Vertical Curve Problem (Use both Tangents) See Attached Vertical Curve Drawing Station Tangent Elevation Tangent Offset В VC 45+00 45+50 46+00 PVI 678.14 46+18.47 46+50 47+00 47+50 EVC I. gl=+4.3%, g2 =-3.196, L = 300, 2. Calculate stations for BVC and EVC and insert into table 3. Calculate station for high/low point and insert into table (this will be located between stations already entered in table, and not yet highlighted). 4. Calculate ALL tangent elevations, tangent...
Help please 2 (36 Pts) A 4250 ft equal tangent parabolic vertical curve has a back...
Help please
2 (36 Pts) A 4250 ft equal tangent parabolic vertical curve has a back tangent grade (gl) of + 3.6 % and a forward tangent grade (g2) of-1.8 %. If the elevation of the PVI is 4264.82 ft. at station 78+ 25.00; what is: (a) the elevation and stationing of the PVC and PVT; (b) the elevation of the curve at station 83 + 63.00; and (c) the elevation and stationing of the highest and lowest points on...
6. Based on the following Vertical Curve information (units in ft.): PVi Sta. 22+00, PVI Elev.-...
6. Based on the following Vertical Curve information (units in ft.): PVi Sta. 22+00, PVI Elev.- 1 134.50, gi-30%, g2+ +1.4%, L :300, what is the station and elevation of the BVC? a) 17+00; 1138.00 17+00; 1139.44 h9+50; 1142.00 d) 19+50; 1138.00 7. Based on the following Vertical Curve information (units in ft.): PVI Sta 1 134.50, gi .-30%, g2+ +1.4%, L . 500, what is the elevation on the curve at Sta. 23 22+00, PVI Elev. a) 1138.49 b)...
A 200m equal tangent sag vertical curve has the PVC at station 3 + 700.000 and...
A 200m equal tangent sag vertical curve has the PVC at station 3 + 700.000 and elevation 321m. The initial grade is -3.5% and the final grade is 0.5%. Determine the elevation and stationing of the PVI, PVT, and lowest point on the curve.
12. Two tangent of a vertical curve intersects at station 32+11.61 and elevation 64.18 ft. back...
12. Two tangent of a vertical curve intersects at station 32+11.61 and elevation 64.18 ft. back tangent grade is gl = +4% and forward tangent grade g2--6%, and length L = 600 ft a. b. c. d. Calculate the elevation of PVC in feet. Calculate the elevation of PVT in ft Calculate the station of the high point. Calculate the elevation of high point in ft.
Question 3 (10 marks) A 1,500ft long sag vertical curve (equal tangent) has a PC at...
Question 3 (10 marks) A 1,500ft long sag vertical curve (equal tangent) has a PC at stenine 1,500f. The initial grade is -3% and the fina on 3+70 and elevation I grade is +6%. Determine the elevation and station 3+70 and a PVC at stationing of the low point, PVI, and PVT
The data given for a vertical sag curve on a roadway plan and profile sheet are...
The data given for a vertical sag curve on a roadway plan and profile sheet are as follows: PVI station 32+11.61, PVI elevation 54.18 ft, back tangent gradient gi4.00 percent, forward tangent gradient g, 7.00 percent, and length of curve L- 600 ft. Determine the curve elevations at half-station intervals along the curve, and compute the sta elevation of the lowest point. tion and い600 ft PVT 35+11.61 Elev. 75.18 35 34 PVC 29 11.61 Elev 66.18 Low point 31...
A 1080 ft long tangent crest vertical curve connects tangents ( G1= 4% , G2 =...
A 1080 ft long tangent crest vertical curve connects tangents ( G1= 4% , G2 = -2.8%). The two stations intersect at ghost station 640+00 where elevation is 1325.000 ft. Determine the stationing and elevations of the following points. 1. PVC 2. PVT 3. High point 4. 640+20
Problem #5 Given the following data for a vertical curve of a road: G1 = -2.5%, G2 = +5.5% and station of PVI = 250+50. At station 252+50, the vertical curve must intersect an existing road at elevation 48.48 ft. The elevation on the tangent section at station 256+50 is 67.55. (a) Draw a good sketch of the vertical profile indicating all major points. (b) What is the length of the vertical curve that satisfies the above geometrics? (C) What...
Problem #3 (20 Points) Vertical Curve Problem (Use both Tangents) See Attached Vertical Curve Drawing Station Tangent Elevation Tangent Offset В VC 45+00 45+50 46+00 PVI 678.14 46+18.47 46+50 47+00 47+50 EVC I. gl=+4.3%, g2 =-3.196, L = 300, 2. Calculate stations for BVC and EVC and insert into table 3. Calculate station for high/low point and insert into table (this will be located between stations already entered in table, and not yet highlighted). 4. Calculate ALL tangent elevations, tangent...
Help please
2 (36 Pts) A 4250 ft equal tangent parabolic vertical curve has a back tangent grade (gl) of + 3.6 % and a forward tangent grade (g2) of-1.8 %. If the elevation of the PVI is 4264.82 ft. at station 78+ 25.00; what is: (a) the elevation and stationing of the PVC and PVT; (b) the elevation of the curve at station 83 + 63.00; and (c) the elevation and stationing of the highest and lowest points on...
6. Based on the following Vertical Curve information (units in ft.): PVi Sta. 22+00, PVI Elev.- 1 134.50, gi-30%, g2+ +1.4%, L :300, what is the station and elevation of the BVC? a) 17+00; 1138.00 17+00; 1139.44 h9+50; 1142.00 d) 19+50; 1138.00 7. Based on the following Vertical Curve information (units in ft.): PVI Sta 1 134.50, gi .-30%, g2+ +1.4%, L . 500, what is the elevation on the curve at Sta. 23 22+00, PVI Elev. a) 1138.49 b)...
12. Two tangent of a vertical curve intersects at station 32+11.61 and elevation 64.18 ft. back tangent grade is gl = +4% and forward tangent grade g2--6%, and length L = 600 ft a. b. c. d. Calculate the elevation of PVC in feet. Calculate the elevation of PVT in ft Calculate the station of the high point. Calculate the elevation of high point in ft.
Question 3 (10 marks) A 1,500ft long sag vertical curve (equal tangent) has a PC at stenine 1,500f. The initial grade is -3% and the fina on 3+70 and elevation I grade is +6%. Determine the elevation and station 3+70 and a PVC at stationing of the low point, PVI, and PVT
The data given for a vertical sag curve on a roadway plan and profile sheet are as follows: PVI station 32+11.61, PVI elevation 54.18 ft, back tangent gradient gi4.00 percent, forward tangent gradient g, 7.00 percent, and length of curve L- 600 ft. Determine the curve elevations at half-station intervals along the curve, and compute the sta elevation of the lowest point. tion and い600 ft PVT 35+11.61 Elev. 75.18 35 34 PVC 29 11.61 Elev 66.18 Low point 31...
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