(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 = 10+00.00, BVC Elev = 5000.00 ft. Lowest Point on the Vertical Curve Station = ?
(E) For an equal tangent vertical curve g1 = -4%, g2 = +3%, L = 200.00 ft, BVC Station = 10+00.00, BVC Elev = 5000.00 ft. Lowest Point on the Vertical Curve Elevation = ?
Homework Answers
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(A) For an equal tangent vertical curve g1 = -4%, g2 = +3%, L =
200.00...
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)...
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...
A crest vertical curve connects a +2% grade and a -2.05% grade. The PVI is at...
A crest vertical curve connects a +2% grade and a -2.05% grade. The PVI is at station 40+00.00 at an elevation of 100.00 ft. The design speed is 70 mi/h. If the length is 1084 ft, determine: (b) The station of the BVC (3 pts (c) The elevation of the BVC (3 pts (d) The station of the EVC (3 pts) (e) The elevation of the EVC (3 pts) (1) The station of the highest point on the curve (4...
Find L please! -3 % high elevation on curve is 206.00 Find L. 2. PC station...
Find L please!
-3 % high elevation on curve is 206.00 Find L. 2. PC station = 10+00.00 PC elev = 200.00 G1=+2% G2=-3% at station 15+00 elevation must be 205.00 Find L 3. PI elev 185.00 GI=+2% G2--3% high elevation on curve is 176.00 Find L 4. P1 station = 16+25.00 P1 elev = 185.00 G1 +2% G2--3% at station 15+00 elevation must be 175.00 Find L
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
Question 1 For each of the four vertical curves show in the attachment (click here for attachment...
ATTACHMENTS:
Question 1 For each of the four vertical curves show in the attachment (click here for attachment) calculate the following: 1. K and r 2. station of BVC and EVC 3. elevation of point at a distance, L/4, from BVC and EVC 4. station of turning point 5. elevation of turning point 6. elevation of mid-point of each curve Curve 1 L=200 ft Pl: Station 12+00, Elevation: 520 ft Curve 2 L = 300 ft Pl: Station 15+00, Elevation:...
12-5. What is the sta. and elev. for both the BVC and the EVC of vertical curve that joins a back tangent of +0.50% and tangent of +3.65%. The sta, and elev. of the PVI are 21 + 73.50 8 a forward...
12-5. What is the sta. and elev. for both the BVC and the EVC of vertical curve that joins a back tangent of +0.50% and tangent of +3.65%. The sta, and elev. of the PVI are 21 + 73.50 8 a forward Ans.: (a.) BVC: sta. 15+85; elev. 76.50 EVC: sta. 26+85; elev. 93.58 (b.y BVC: sta. 15+85; elev.- 70.75 EVC: sta.- 26+85; elev. 93.58 (c.) BVC: sta. 15+85; elev. - 76.50 EVC: sta.- 21+45; elev. 53.42 (d.) BVC: sta....
13. A 400-ft long equal tangent sag vertical curve has its PVC at station 100+00 and o b elevation 500 ft, the init...
13. A 400-ft long equal tangent sag vertical curve has its PVC at station 100+00 and o b elevation 500 ft, the initial grade g,--4% and the final grade g2 elevation of the lowest point of the curve. +2.5%. Determine the
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...
Problem 4 A 500 ft long sag vertical curve passes under a bridge at station 82+45....
Problem 4 A 500 ft long sag vertical curve passes under a bridge at station 82+45. The beginning of vertical curve (BVC) is at station 81+00. A-3.6% curve meets a +4.4% curve at the point of vertical intersection (PVI), which is at elevation 425.38 ft. What is the elevation of the point on the curve under the bridge? Problem 5 The grade into a vertical sag curve is -2%. The curve length is 1,400 ft. The grade out of the...
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)...
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...
A crest vertical curve connects a +2% grade and a -2.05% grade. The PVI is at station 40+00.00 at an elevation of 100.00 ft. The design speed is 70 mi/h. If the length is 1084 ft, determine: (b) The station of the BVC (3 pts (c) The elevation of the BVC (3 pts (d) The station of the EVC (3 pts) (e) The elevation of the EVC (3 pts) (1) The station of the highest point on the curve (4...
Find L please!
-3 % high elevation on curve is 206.00 Find L. 2. PC station = 10+00.00 PC elev = 200.00 G1=+2% G2=-3% at station 15+00 elevation must be 205.00 Find L 3. PI elev 185.00 GI=+2% G2--3% high elevation on curve is 176.00 Find L 4. P1 station = 16+25.00 P1 elev = 185.00 G1 +2% G2--3% at station 15+00 elevation must be 175.00 Find L
ATTACHMENTS:
Question 1 For each of the four vertical curves show in the attachment (click here for attachment) calculate the following: 1. K and r 2. station of BVC and EVC 3. elevation of point at a distance, L/4, from BVC and EVC 4. station of turning point 5. elevation of turning point 6. elevation of mid-point of each curve Curve 1 L=200 ft Pl: Station 12+00, Elevation: 520 ft Curve 2 L = 300 ft Pl: Station 15+00, Elevation:...
12-5. What is the sta. and elev. for both the BVC and the EVC of vertical curve that joins a back tangent of +0.50% and tangent of +3.65%. The sta, and elev. of the PVI are 21 + 73.50 8 a forward Ans.: (a.) BVC: sta. 15+85; elev. 76.50 EVC: sta. 26+85; elev. 93.58 (b.y BVC: sta. 15+85; elev.- 70.75 EVC: sta.- 26+85; elev. 93.58 (c.) BVC: sta. 15+85; elev. - 76.50 EVC: sta.- 21+45; elev. 53.42 (d.) BVC: sta....
13. A 400-ft long equal tangent sag vertical curve has its PVC at station 100+00 and o b elevation 500 ft, the initial grade g,--4% and the final grade g2 elevation of the lowest point of the curve. +2.5%. Determine the
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...
Problem 4 A 500 ft long sag vertical curve passes under a bridge at station 82+45. The beginning of vertical curve (BVC) is at station 81+00. A-3.6% curve meets a +4.4% curve at the point of vertical intersection (PVI), which is at elevation 425.38 ft. What is the elevation of the point on the curve under the bridge? Problem 5 The grade into a vertical sag curve is -2%. The curve length is 1,400 ft. The grade out of the...
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