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An equal-tangent crest curve has been designed for 60 mph to connect grades of 0.5% and -2.1% for a new vehicle that has a 3 ft driver's eye height and to avoid an object located only 1 ft high. Standard practical stopping distance design standards were used except the acceleration was assumed to be 10 ft/sec². If the PVT of the curve is at elevation 500 ft and located at station 400+50, what is the station and elevation of the high point of the curve?

An equal-tangent crest curve has been designed for 60 mph to connect grades of 0.5% and -2.1% for a new vehicle that has a 3 ft driver's eye height and to avoid an object located only 1 ft high. Standard practical stopping distance design standards were used except the acceleration was assumed to be 10 ft/sec². If the PVT of the curve is at elevation 500 ft and located at station 400+50, what is the station and elevation of the high point of the curve?

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Given: V = 65mph = 95.3fps; H₂ = 3ft; H₂ = 1ft; standard practical stoppping sight distance except a = 9ft/sec² Find: station and elevation of high point Solution: S=Vit, + S = 95.3(2.5) + V² 29(±G) Assume SL am G₂-G₂ 21 L = - -,G=0 (standard use) 95.3² 2 x 32.2(9/32.2) = 743ft AS² 200(√H, +√H₂) -0.021-0.015 2(1332) = 11.5-(-2.1)|(743²) 200(√3+√1)² L>S=881ft, assumption OK L 1332 2% K=A1.5-(-2.1)| 370 Xhi = KX |G₂| = 370 x 1.5 = 555ft station of high point = PVT-L+x=(1000 +75)-(13+32) + (5 +55) (992 +98) y = ax² +bx+c -0.0000135; b=G₁ = 0.015; G₂L G₂L c = PVC elevation = PVT elevation -- 1332 2 = 1332ft 1332 PVT 1000-75 (1000) = 1000 -0.015 x -(-0.021) x =1004ft y of high point = -0.0000135(555)² +0.015(555) + 10041008.16' -IN
Transcribed Image Text:Given: V = 65mph = 95.3fps; H₂ = 3ft; H₂ = 1ft; standard practical stoppping sight distance except a = 9ft/sec² Find: station and elevation of high point Solution: S=Vit, + S = 95.3(2.5) + V² 29(±G) Assume SL am G₂-G₂ 21 L = - -,G=0 (standard use) 95.3² 2 x 32.2(9/32.2) = 743ft AS² 200(√H, +√H₂) -0.021-0.015 2(1332) = 11.5-(-2.1)|(743²) 200(√3+√1)² L>S=881ft, assumption OK L 1332 2% K=A1.5-(-2.1)| 370 Xhi = KX |G₂| = 370 x 1.5 = 555ft station of high point = PVT-L+x=(1000 +75)-(13+32) + (5 +55) (992 +98) y = ax² +bx+c -0.0000135; b=G₁ = 0.015; G₂L G₂L c = PVC elevation = PVT elevation -- 1332 2 = 1332ft 1332 PVT 1000-75 (1000) = 1000 -0.015 x -(-0.021) x =1004ft y of high point = -0.0000135(555)² +0.015(555) + 10041008.16' -IN
An equal-tangent crest curve has been designed for 60 mph to connect grades of 0.5% and -2.1% for a new vehicle that has a 3 ft driver's eye height and to avoid an object located only 1 ft high. Standard practical stopping distance design standards were used except the acceleration was assumed to be 10 ft/sec². If the PVT of the curve is at elevation 500 ft and located at station 400+50, what is the station and elevation of the high point of the curve? Example prop below: An equal-tangent crest curve has been designed for 65 mph to connect grades of 1.5% and-2.1% for a new vehicle that has a 3 ft driver's eye height and to avoid an object located only 1 ft high. Standard practical stopping distance design standards were used except the acceleration was assumed to be 9 fftt/sseeee2. If the PVT of the curve is at elevation 1000 ft and located at station 1000+75, what is the station and elevation of the high point of the curve?
Transcribed Image Text:An equal-tangent crest curve has been designed for 60 mph to connect grades of 0.5% and -2.1% for a new vehicle that has a 3 ft driver's eye height and to avoid an object located only 1 ft high. Standard practical stopping distance design standards were used except the acceleration was assumed to be 10 ft/sec². If the PVT of the curve is at elevation 500 ft and located at station 400+50, what is the station and elevation of the high point of the curve? Example prop below: An equal-tangent crest curve has been designed for 65 mph to connect grades of 1.5% and-2.1% for a new vehicle that has a 3 ft driver's eye height and to avoid an object located only 1 ft high. Standard practical stopping distance design standards were used except the acceleration was assumed to be 9 fftt/sseeee2. If the PVT of the curve is at elevation 1000 ft and located at station 1000+75, what is the station and elevation of the high point of the curve?
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