CHAPTER 12 12.1.

EQUATION OF AN EQUAL TANGENT VERTICAL

PARABOLIC CURVE
VERTICAL CURVES
In Figure 11.2., shows a vertical parabolic curve that joins two intersection
Vertical curves are used to connect intersecting gradients in the vertical of a grade line. In the figure, BVC denotes the beginning of vertical curve,
plane. Thus, in route design they are provided at all changes of gradient. They sometimes called VPC (vertical point of curvature); V is the vertex, often called the
should be sufficiently large curvature to provide comfort to the driver, that is, they VPI (vertical point of intersection); and EVC denotes the end of vertical curve,
should have a low ‘rate of change of grade’. In addition, they should afford adequate interchangeably called the VPT (vertical point of tangency). The percent grade of
sight distances for safe stopping at a given design speed. the back tangent is 𝑔1 , that of the forward tangent is 𝑔2 . The curve length 𝐿 is the
Two basic types of vertical curves exist, crest and sag. These are illustrated horizontal distance from the BVC to the EVC.
in Figure 11.1. Curve a is a crest type, which by definition undergoes a negative
change in grade; that is, the curve turns downward. Curve b is a sag type, in which
the change in grade is positive and the curve turns upward.

Figure 11.1. Grade Line and Ground Profile of a Proposed Highway Section

There are several factors that must be considered when designing a grade
line of tangents and curves on any highway or railroad project. They include (1)
providing a good fit with the existing ground profile, thereby minimizing the depths
of cuts and fills, (2) balancing the volume of cut material against fill, (3) maintaining Figure 11.2. Vertical Parabolic Curve
adequate drainage, (4) not exceeding maximum specified grades, and (5) meeting
fixed elevations such as intersections with other roads. On the XY plane in Figure 11.3., X values are horizontal distances measured
from the BVC, and Y values are elevations measured from the vertical datum of
reference. By substituting this surveying terminology in equation, the parabola can
be expressed as

𝑌 = 𝑌𝐵𝑉𝐶 + 𝑔1 𝑋 + 𝑐𝑋 2 (Eq’n. 11.1)

12.1.1. Symmetrical Parabolic Curve 𝑆 = (𝑔
𝑔1 𝐿
(Eq’n. 11.4)
1 −𝑔2 )
In Figure 11.3., shows a symmetrical parabolic curve. A vertical curve is
symmetrical if the horizontal length is equal at both sides of the vertex. Since the
curve is parabolic, its geometric properties should conform to that of a parabola. INSTRUCTIVE PROBLEMS:
12.1. A symmetrical parabolic curve passes through point A whose elevation is
23.23 m at a distance of 54 m from the PVC. The elevation of the PVC at
station 1+000 is 23.50 m. The grade of the back tangent is +2% and the length
of curve is 100 m.
a. Compute the grade of the forward tangent.
b. Compute the stationing of the highest point of the curve.
c. Compute the elevation of the highest point of the curve.
Solution:

Figure 11.3. Symmetrical Parabolic Curve

H is the vertical offset of the curve below the vertex which is express by
the formula,
𝐿
𝐻 = (𝑔1 − 𝑔2 ) (Eq’n. 11.2)
8

Consider an arbitrary point A at ‘x’ horizontal distance from PVC. Since

the curve is parabolic, the following relations shall be true,
𝑥2 (𝐿/2)2
𝑦
= 𝐻
(Eq’n. 11.3)

Since this is a summit curve and the forward and back tangents have
opposite signs, a point with the highest elevation should exist. To locate this point,
we simply set the grade at the highest point to zero. Let ‘S’ be the location of the
summit from PVC. Assuming S<L/2. The location of the highest point of the curve
from PVC is given by,
12.2. A descending grade of 4% intersects an ascending grade of 3% at Station 12.4. The length of a summit curve is 200 m. Tangent grades for the vertical summit
1+325 at elevation 15.0 m. These two grades are to be connected by a 250 m curves are +3% and -2%. Find the radius of summit curve.
vertical parabolic curve. A reinforced concrete culvert pipe with an overall
Solution:
diameter of 105 cm is to be constructed with its top 30 cm below the subgrade.
What will be the invert elevation of the culvert?
Solution:

12.1.2. Unsymmetrical Parabolic Curve

A curve is said to be unsymmetrical if the length of curve at each side of
the vertex is not equal. Consider the unsymmetrical parabolic summit curve shown
12.3. A -6% and a +2% grade intersect at station 12+200 whose elevation is at
in Figure 11.4.
25.632 m. The two grades are to be connected by a parabolic curve, 160 m
long. Find the elevation of the first quarter point of the curve.
Solution:

Figure 11.4. Unsymmetrical Parabolic Curve

 This intermediate grade g3 will act as a common tangent, dividing the
unsymmetrical curve into two symmetrical curves of lengths L1 and L2. Each
symmetrical curve can now be analyzed using the methods of symmetrical parabolic
curve.
INSTRUCTIVE PROBLEMS:
12.5. A -2% grade meets a +8% grade near an underpass. In order to maintain the
minimum clearance allowed under the bridge and at the same time introduce a
vertical transition curve in the grade line, it is necessary to use a curve that lies
400-m on one side of the vertex of the straight grade and 240 m on the other.
The station of the beginning of the curve (400 m side) is 10+000 and its
elevation is 50 m.
a. Determine the stationing of the lowest point on the curve.
b. Determine the elevation at station 10+080.
c. Determine the elevation of PVT.
d. If the uphill edge of the underside of the bridge is at station 10+440 and at
elevation 53.41 m, what is the vertical clearance under the bridge at this
point?
Solution:
SOLUTIONS AND ANSWERS TO INSTRUCTIVE PROBLEMS: 120 120
𝐻= |2% − 𝑔2 | = |2% − (−1.36%)| = 0.506 𝑚
8 8
12.1. Given 𝐸𝑙𝑒𝑣 𝐴 = 23.23 𝑚, 𝑥 = 54 𝑚, 𝑆𝑡𝑎. 𝑃𝑉𝐶 = 4 + 100, 𝑔1 𝐿 0.02(120)
𝑆= = = 71.122 𝑚
(𝑔1 −𝑔2 ) 0.02−(−0.136)
𝐸𝑙𝑒𝑣. 𝑃𝑉𝐶 = 22.56 𝑚, 𝑔1 = +2%, 𝐿 = 120 𝑚
𝑆𝑡𝑎. 𝑆𝑢𝑚𝑚𝑖𝑡 = 𝑆𝑡𝑎. 𝑃𝑉𝐶 + 𝑆
𝑆𝑡𝑎. 𝑆𝑢𝑚𝑚𝑖𝑡 = 4,100 + 71.122
𝑺𝒕𝒂. 𝑺𝒖𝒎𝒎𝒊𝒕 = 𝟒 + 𝟏𝟕𝟏. 𝟏𝟐𝟐
c. Elevation of the highest point
𝐿 2
𝑆2 ( )
2
=
𝑦𝑠 𝐻

(71.122)2 (60)2
=
𝑦𝑠 0.506

𝑦𝑠 = 0.711 𝑚
a. Grade of the forward tangent 𝐸𝑙𝑒𝑣. 𝑆𝑢𝑚𝑚𝑖𝑡 = 22.56 + 0.02(71.122) − 0.711
𝐿 120 𝑬𝒍𝒆𝒗. 𝑺𝒖𝒎𝒎𝒊𝒕 = 𝟐𝟑. 𝟐𝟕𝟏 𝒎
𝐻 = (𝑔1 − 𝑔2 ) = |2% − 𝑔2 |
8 8

𝐿 2
12.2. Given 𝑔1 = −4%, 𝑔2 = +3%, 𝑆𝑡𝑎. 𝑉 = 1 + 325, 𝐸𝑙𝑒𝑣. 𝑉 = 15.0 𝑚,
𝑥2 ( )
2
𝑦
= 𝐻 𝐿 = 250 𝑚, 𝐷𝑐𝑢𝑙𝑣𝑒𝑟𝑡 = 105 𝑐𝑚, ℎ = 30 𝑐𝑚
(54)2 (60)2
𝑦
= 120
|2%−𝑔2 |
8

243
𝑦= |2% − 𝑔2 |
20

22.56 + 2%(54) − 𝑦 = 23.23

𝑦 = 0.41
243
0.41 = |2% − 𝑔2 |
20

𝒈𝟐 = −𝟎. 𝟎𝟏𝟑𝟔 𝒐𝒓 − 𝟏. 𝟑𝟔%

b. Stationing of the highest point of the curve 𝐸𝑙𝑒𝑣. 𝑃𝑉𝐶 = 15.0 + 0.04(125) = 20 𝑚
 𝐸𝑙𝑒𝑣. 𝑃𝑉𝑇 = 15.0 + 0.03(125) = 18.75 𝑚 (40)2 (80)2
𝑦
= 1.60
𝑔2 𝐿 0.03(250)
𝑆= = |−0.04−0.03|
= 107.14 𝑚 𝑦 = 0.40 𝑚
(𝑔1 −𝑔2 )

𝐿 250 𝐸𝑙𝑒𝑣. 𝑄 = 25.632 + 0.06(40) + 0.40

𝐻 = 8 |𝑔1 − 𝑔2 | = 8
|−0.04 − 0.03| = 2.188 𝑚

𝐿 2
𝑬𝒍𝒆𝒗. 𝑸 = 𝟐𝟖. 𝟒𝟑𝟐 𝒎
𝑆2 ( )
2
𝑦𝑠
= 𝐻 12.4. Given: 𝐿 = 200 𝑚, 𝑔1 = +3%, 𝑔2 = −2%
(107.14)2 (125)2 𝐿𝑐 = 𝑅𝐼
𝑦𝑠
= 2.188
𝐿 ≈ 𝑅(∆𝑔)
𝑦𝑠 = 1.607 𝑚
200 ≈ 𝑅[0.03 − (−0.02)]
𝐸𝑙𝑒𝑣. 𝑃 = 18.75 − 0.03(107.14) + 1.607 = 17.143 𝑚
𝑅 ≈ 4000 𝑚 (approximately)
𝐸𝑙𝑒𝑣. 𝑜𝑓 𝐼𝑛𝑣𝑒𝑟𝑡 𝐶𝑢𝑙𝑣𝑒𝑟𝑡 = 17.143 − 0.30 − 1.05
𝑬𝒍𝒆𝒗. 𝒐𝒇 𝑰𝒏𝒗𝒆𝒓𝒕 𝑪𝒖𝒍𝒗𝒆𝒓𝒕 = 𝟏𝟓. 𝟕𝟗𝟑 𝒎
12.3. Given: 𝑔1 = −6%, 𝑔2 = +2%, 𝑆𝑡𝑎. 𝑉 = 12 + 200, 𝐸𝑙𝑒𝑣. 𝑉 = 25.632 𝑚
𝐿 = 160 𝑚

In highway design, curves are provided to minimize the force acting on the
vehicle due to acceleration which come from sudden change in alignment. The
change in the grade from PVC to PVT is often small enough such that the curve,
although parabolic, can be approximated as circular. The smaller this change in
grade is, the more accurate the results.
𝐿 160
𝐻 = 8 |𝑔1 − 𝑔2 | = |−0.06 − 0.02| = 1.60 𝑚 12.5. Given: 𝑔1 = −2%, 𝑔2 = +8%, 𝐿1 = 400 𝑚, 𝐿2 = 240 𝑚,
8

𝐿 2 𝑆𝑡𝑎. 𝑃𝑉𝐶 = 10 + 000, 𝐸𝑙𝑒𝑣. 𝑃𝑉𝐶 = 50 𝑚

𝑥2 ( )
2
𝑦
= 𝐻
 𝑬𝒍𝒆𝒗. @𝟏𝟎 + 𝟎𝟖𝟎 = 𝟒𝟖. 𝟕𝟎 𝒎
c. Elevation of PVT
𝐸𝑙𝑒𝑣. 𝑃𝑉𝑇 = 50 − 0.02(400) + 0.08(240)
𝑬𝒍𝒆𝒗. 𝑷𝑽𝑻 = 𝟔𝟏. 𝟐𝟎 𝒎
d. Vertical clearance under the bridge.
𝑥2 (𝐿2 )2
=
𝑦 𝐻
a. Stationing of the lowest point on the curve
(200)2 (240)2
2𝐻𝐿2 𝑦
= 7.5
𝐿1 = 𝐿
2 |𝑔1 −𝑔2 |−2𝐻
𝑦 = 5.21𝑚
2𝐻(240)
400 =
240|−0.02−0.08|−2𝐻 𝐸𝑙𝑒𝑣. 10 + 440 = 61.20 − 0.08(200) + 5.21
𝐻 = 7.5 𝑚 𝐸𝑙𝑒𝑣. 10 + 440 = 50.41 𝑚
𝐿1 𝑔1 400(0.02)
2
= 2
=4 𝐶𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 = 53.41 − 50.41

𝐿1 𝑔1 𝑔1 𝐿1 2 𝑪𝒍𝒆𝒂𝒓𝒂𝒏𝒄𝒆 = 𝟑. 𝟎𝟎 𝒎
Since 2
= 4 < 𝐻, use 𝑠1 = 2𝐻

𝑔1 𝐿1 2 (0.02)(400)2
𝑠1 = = = 213.33 𝑚
2𝐻 2(7.5)

𝑆𝑡𝑎. 𝑜𝑓 𝐿𝑜𝑤𝑒𝑠𝑡 𝑃𝑜𝑖𝑛𝑡 = 10,000 + 213.33

𝑺𝒕𝒂. 𝒐𝒇 𝑳𝒐𝒘𝒆𝒔𝒕 𝑷𝒐𝒊𝒏𝒕 = 𝟏𝟎 + 𝟐𝟏𝟑. 𝟑𝟑
b. Elevation at Sta. 10+080
𝑥2 (𝐿1 )2
𝑦
= 𝐻

(80)2 (400)2
𝑦
= 7.5

𝑦 = 0.30𝑚
𝐸𝑙𝑒𝑣. @10 + 080 = 50 − 0.02(80) + 0.30
REVIEW EXERCISES: 12.6. What length of curve will meet these conditions?
12.7. What is the stationing and elevation of the highest point of the curve?
12.1. On a railroad a -8% grade meters and +4% grade station 2+700 whose
12.6. A symmetrical parabolic summit curve connects two grades of +6% and -4%.
elevation of 30 meters. The maximum allowable change in grade per station
It is to pass through a point p the stationing of which is 35+280 and the
having a length of 20 m is 1.5%. It is desired to place a culvert to drain the
elevation is 198.13 m. If the elevation of the grade intersection is 200 m with
flood waters during heavy downpour. Where must this culvert be located? At
stationing 35+300, determine:
what elevation must the invert of the culvert be set if the pipe has a diameter
a. The length of the curve.
of 0.9 m and the backfill is 0.3 m high? Neglect the thickness of the pipe.
b. Stationing and elevation of PC and PT.
12.2. The grade of a symmetrical parabolic curve from station 9+000 to the vertex
c. The location of the highest point on the curve.
V at Sta. 9+100 is minus 6% and from Sta. 9+100 to 9+200 is minus 2%. The
d. Elevation of station 35+260 on the curve.
elevation at the vertex is 100 m. H is required to connect these grade lines with
12.7. An underpass crossing a reinforced concrete bridge along the Shaw Blvd. has
a vertical parabolic curve that shall pass 0.80 m above the vertex. Determine
a downward grade of -4% meeting an upward grade of +8% at the vertex V
the amount of cut or fill. Locate also the stationing of the PC and the PT. The
(elevation 70.00 m) at station 7+700, exactly underneath the center line of the
following field notes shows the ground elevations at their respective stations.
bridge having a width of 10 m. If the required minimum clearance under the
Station Ground Elev. Station Ground Elev. bridge is 5 m and the elevation of the bottom of the bridge is 78.10 m,
9+000 106.20 9+120 99.65
determine the following:
9+020 105.37 9+140 98.36
a. Length of the vertical parabolic curve that shall connect the two tangents.
9+040 102.49 9+160 99.00
b. Stationing and elevation where a catch basin will be placed.
9+060 102.00 9+180 98.00
12.8. Point P is the location of the center line of an existing highway. An underpass
9+080 102.18 9+200 98.00
9+100 101.80 is to be designed perpendicular to the existing highway with a vertical
12.3. A descending grade of 4.2% grade intersect an ascending grade of 3% at parabolic curve such that is lowest point is directly below P with a vertical
station 11+488 of elevations 20.80 meters. These two center gradelines are to clearance of 5.5 m. Stationing of the PI is 5+800 and has an elevation of 105
be connected by a 260 meter vertical parabolic curve. At what station is the m. The slope of the tangent passing thru the PC is -4% and that of the PT is
cross-drainage pipes be situated. If the overall outside dimensions of the +3%. Determine the length of the vertical parabolic curve, the location of P
reinforced concrete pipe to be installed is 95 cm and the top of the culvert is being on the right side of the curve and compute also the stationing of the PC
30 cm below the subgrade, what will be the invert elevation at the center? if the elevation of point P is 120 m.
12.4. A horizontally laid circular pipe culvert having an elevation of its top to be 26 12.9. An unsymmetrical parabolic curve has a forward tangent of -8% and a
m crosses at right angles under a proposed 120 m highway parabolic curve. backward tangent of +5%. The length of the curve on the left side of the curve
The point of intersection of the grade lines is at station 5+216 and its elevation is 40 m long while that of the right side is 60 m long. At station 66+780 and at
is 27.0 m while the culvert is located at station 5+228. The backward tangent elevation 110.
has a grade of 3% and the grade of the forward tangent is -1.6%. Under this a. Determine the height of fill at the outcrop at Sta. 66+760 and elevation of
condition, what will be the depth of cover over the pipe? 108.40.
12.5. A vertical highway curve is to pass through a railroad at grade. The crossing b. Determine the elevation of curve at station 66+820.
must be at station 4+210 and at elevation 220.82 m. The initial grade of the 12.10. A forward tangent having a slope of -4% intersects the back tangent having
highway is +2% and meets a -3% grade at station 4+135 at an elevation of a slope +7% at point V at stations 6+300 having an elevation of 230 m. It is
223.38 m. The rate of change must not exceed 2%. required to connect the two tangents with an unsymmetrical parabolic curve
 that shall pass through point A on the curve having an elevation of 227.57 m
at station 6+270. The length of curve is 60 m on the side of the back tangent.
a. It is required to determine the length of the curve on the side of the forward
tangent.
b. Determine the stationing and elevation of the highest point of the curve.
12.11. In a certain road construction undertaken by the DPWH it was decided to
connect a forward tangent of 3% and a back tangent of -5% by a 200 meters
symmetrical parabolic curve. It was discovered that the grade intersection at
station 10+100, whose elevation is 100 m fall on a rocky section with the
exposed boulder at elevation 102.67 m. To avoid rock excavation, the project
engineer decided to adjust the vertical parabolic curve in such a way that the
curve will just clear the rock without altering the position of PC and the grade
of the tangents. Determine the stationing and elevation of the new PT.
12.12. A forward tangent of +6% was designed to intersect a back tangent of -3%
at a proposed underpass along EDSA so as to maintain a minimum clearance
allowed under a bridge which crosses perpendicular to the underpass. A 200
m curve lies on the side of the back tangent while a 100 m curve lies on the
side of the forward tangent. The stationing and elevation of the grade
intersection is 12+530.20 and 100 m respectively. The center line of the bridge
falls at station 12+575.20. The elevation of the underside of the bridge is
117.48 m. Determine the minimum clearance of the bridge if it has a width of
10 m.