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?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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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
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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