3.5 An equal-tangent crest curve has been designed for 70 mi/h to connect a +2% initial grade and a - 1% final
grade for a new vehicle that has a 3 ft driver’s eye height; the curve was designed to avoid an object that is 1
ft high. Standard practical stopping distance design was used but, unlike current design standards, the vehicle
was assumed to make a 0.5 g stop, although driver reactions are assumed to be the same as in current highway
design standards. If the PVC of the curve is at elevation 848 ft and station 43 + 48, what is the station and
elevation of the high point of the curve?
1- Determine the stopping sight distance (SSD) for the vertical curve:
𝑉1 2
𝑆𝑆𝐷 =
+ (𝑉1 × 𝑡𝑟 )
𝑎
2g ( ± 𝐺)
g
Here, 𝑆𝑆𝐷 is the stopping sight distance for the vertical in feet, 𝑉1 is the initial vehicle speed in ft/s, g is
the gravitational constant, 𝑎 is the deceleration rate of the vehicle, 𝐺 is the roadway grade in percent per
100, and 𝑡𝑟 is the perception or reaction time in seconds.
Assume roadway grade as zero.
Substitute 11.2 ft/s2 for 𝑎 , 2.5 s for 𝑡𝑟 , 0 for 𝐺, 16.10 ft/s2 for g , and 70 mi/h for 𝑉1 :
𝑆𝑆𝐷 =
(70
mi 2
)
h
ft
ft 11.2 𝑠2
2 × 32.2 2 (
± 0)
ft
s
32.2 2
s
+ (70
mi
× 2.5 𝑠 )
h
mi
ft 2
× 1.467 )
mi
ft
h
s
=
+ ((70
× 1.467 ) × 2.5 𝑠 ) = 470.77 ft + 256.73 ft
ft
h
s
ft 11.2 𝑠2
2 × 32.2 2 (
± 0)
ft
s
32.2 2
s
(70
= 727.5 ft
2- Determine the horizontal distance required to effect a 1 % change in the slope of the vertical curve:
𝐿𝑚 = 𝐾𝐴
𝐿𝑚
𝐾=
… … … … … (1)
𝐴
Here, K is the horizontal distance required to effect a 1 % change in the slope of the vertical curve.
3- Determine the minimum length of the vertical curve assuming that the length of the vertical curve is
greater than the SSD:
𝐿𝑚 =
𝐴 × 𝑆𝑆𝐷 2
200(√𝐻1 + √𝐻2 )
2 … … ….
(2)
Here, 𝐿𝑚 is the minimum length of the vertical curve in feet, 𝐴 is the absolute value of the difference in
grades, expressed as a percentage, 𝐻1 is the height of the driver eye, and 𝐻2 is the height of the object to
be avoided by stopping before a collision.
Substitute Equation (2) in Equation (1):
𝐴 × 𝑆𝑆𝐷 2
200(√𝐻1 + √𝐻2 )
𝐾=
𝐴
2
𝐴 × 𝑆𝑆𝐷 2
=
𝐴 × 200(√𝐻1 + √𝐻2 )
=
2
𝑆𝑆𝐷 2
2
200(√𝐻1 + √𝐻2 )
Substitute 727.5 ft for SSD, 3 ft for 𝐻1 , and 1 ft for 𝐻2 :
𝐾=
727.52
200(√3 + √1)
2
= 354.5
4- Determine the horizontal distance required to effect a 1 % change in the slope of the vertical curve:
𝐿𝑚 = 𝐾𝐴 = 𝐾 × (|𝐺1 − 𝐺2 |)
Substitute 354.5 for 𝐾, + 2 % for 𝐺1 , and -1 % for 𝐺2 :
𝐿𝑚 = 354.5 × (|2.00 − (−1.00)|) = 1063.5 ft
5- Determine the stationing of the high point:
𝑥ℎ𝑙 = 𝐾 × |𝐺1 |
Here, 𝑥ℎ𝑙 is the distance from the point of the vertical curve (PVC) to the high or low point in feet.
Substitute 354.5 for 𝐾 and + 2.0 for 𝐺1 :
𝑥ℎ𝑙 = 354.5 × |+2.00| = 709 ft
The high point is located at 7+09 stations away from the PVC.
The stationing of high point is 43 + 48 plus 7 + 09, which is 50 + 57
6- Determine the stationing of the high point on the vertical curve.
Show the equation of the vertical curve:
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 … … … (1)
Here, 𝑦 is the roadway elevation at distance 𝑥 from the beginning of the vertical curve in stations or feet,
𝑥 is the distance from the beginning of the vertical curve in stations or feet, 𝑎 and 𝑏 are the coefficients,
and 𝑐 is the elevation of the PVC.
7- Determine the value of the coefficient 𝑏 using the relation:
𝑏 = 𝐺𝑖
Substitute + 2.0 % for 𝐺𝑖 :
𝑏 = +2.0
8- Determine the value of the coefficient 𝑎 using the relation:
𝐺𝑓 −𝐺𝑖
𝑎=
2𝐿
Substitute + 2.0 % for 𝐺𝑖 , -1.0 % for 𝐺𝑓 , and 1063.5 ft for 𝐿:
−1.0 − (+2.0)
𝑎=
= −0.14
2 × 10.64
Substitute -0.14 for 𝑎, + 2.0 for 𝑏 , 848 ft for 𝑐 , and 7.09 for 𝑥 in Equation (1)
𝑦 = (−0.14 × 7.09)2 + (+2 × 7.09) + 848 = 863.17 ft
Thus,
The stationing of high point is 50 + 57
The elevation of the high point is 863.17 ft