Vertical Parabolic Curves

VERTICAL PARABOLIC CURVES A. SYMMETRICAL PARABOLIC CURVES In highway practice, abrupt change in the vertical direction of moving vehicles should be avoided. In order to provide gradual change in its vertical direction, a parabolic curve is adopted on account of its slope which varies at a constant rate with respect to horizontal distances. Properties of Vertical Parabolic Curve: D g1L/2

P.I. X2

B

X1 g1

Y1

(g1-g2)L/2 x3

H

Y2

E

Y3

P.C . A

-g2L/2

g2

H F P.T. L/

C

L/

1. The vertical offsets from the2 tangent to the curve are 2proportional to the squares of the distances from the point of tangency. y1 = H y3 = H (x1)2 (L/2)2 (x3)2 (L/2)2 y2 = (x2)2

H (L/2)2

y1 (x2)2

=

y2 = (x2)2

y3 (x3)2

2. The curve bisects the distance between the vertex and the midpoint of the long chord. From similar triangles: BF = CD (L/2) L BF = CD/2

BE

BE

=

=

CDL2

From the first property of curve: But CD/2 = BF (L/2) 2 L2

CD BE BE H

= =

= 1 CD/4 CD/4

(CD/2) 4L 2

BE

2 =

BF/2

3. If the algebraic difference in the rate of grade of the two slopes is positive, that is (g 1-g2), we have a “summit curve”, but if it is negative, we have a sag curve. 4. The length of curve of a parabolic vertical curve is measured not along the curve but along the horizontal line. 5. The stationing of the vertical parabolic curve is measured along the curve but along a horizontal line. 6. For a symmetrical parabolic curve, the number of stations to the left must be equal to the number of stations to the right, of the intersection of the slopes or forward and backward tangent. 7. The slope of the parabola varies uniformly along the curve, as shown by differentiating the equation of the parabolic curve. y dy dx

= =

kx2 2kx

The second derivative is d2y = 2k , where d2y = rate of change of grade or the slope. dx dx Therefore, the rate of change is constant and equal to: r = 2k or r = g2 1 g n n = no. of station of curve 8. The maximum offset H = 1/8 the product of the algebraic difference between the two rates of grade and the length of the curve. From the figure: CD

=

H

= BE (g1 – g2) L

=

CD/4

2 H

=

1L (g1 – g2) 4 2

Therefore,

H

=

1 (g1 – g2)L 8

LOCATION OF THE HIGHEST OR LOWEST POINT ON THE CURVE: a. From P.C. S

S1 1

=

L - S1

g1L P.I..

g1 – g2 P.C.

P.T..

L

b. From P.T. L – S2

S2

=

S2

g2L

g2 – g1

P.I. . P.C .

P.T .. L

B. UNSYMMETRICAL PARABOLIC CURVES A vertical highway curve is at times designed to include a particular elevation at a certain station where the grades of the forward and backward tangents have already been established. It is therefore necessary to use a curve with unequal tangents or a compound curve which is usually called “unsymmetrical” or asymmetrical parabolic curve where one parabola extendsfrom the P.C. to a point directly below the vertex and a second parabola which extends from this point to the P.I.. In order to make the entire curve smooth and continuous, the two parabolas are so constructed so that they will have a common tangent at the point where they joined, which is at a point directly below the vertex.

D g1L2 P.I. X1

B

(g1-g2)L2 X2

H Y1

g1 P.C .

E

Y2

g2L2

g2

H A F

P.T. L1

C

L2

Applying the squared property of parabola in solving for the vertical offsets of the parabola, we use the following: y1

=

H

y2

=

H

(x1)2

L 12

H is computed as:

(x2)2 H

=

L 22 (g1 – g2)L1 L2

2 (L1 + L2)

In solving for L1, we use the following formula:

L1

=

2HL2

L2 (g1 – g2) – 2H

LOCATION OF THE HIGHEST OR LOWEST POINT ON THE CURVE: a. From P.C. when L1g1< 2 S

1

=

H

b.

From P.T. when L1g1>

H 2

g1L12

2H

S2

=

g2L22 2H

Sample Problem No. 1: A symmetrical vertical summit curve has tangents of +4% and -2%.The allowable rate of change of grade is 0.3% per meter station. Stationing and elevation of PT is at 10+020 and 142.63 m respectively. a.

Compute the length of curve.

b.

Compute the distance of the highest point of curve from PC.

c.

Compute the elevation of highest point of curve.

Sample Problem 2: A vertical symmetrical sag curve has a descending grade of -4.2% and ascending grade of +3% intersecting at station 10+020, whose elevation is 100-m. The two grade lines are connected by a 260-m vertical parabolic sag curve. a. At what distance from the PC is the lowest point of the curve located? b. What is the vertical offset of the parabolic curve to the point of intersection of the tangent grades to midpoint of the curve? c. What is the elevation of the lowest point on your curve? d. If a 1-m diameter culvert is placed at the lowest point of the curve with top of the culvert buried 0.60-m below subgrade, what will be the elevation of the invert of the culvert?

Sample Problem 3: A symmetrical parabolic summit curve connects two grades of 6% and -4% it is to pass through a point “P” the stationing of which is 35+280 with elevation of 198.13 and the elevation of the grade intersection is 200-m with stationing 35+300. a. Determine the length of curve. b. Determine the stationing and elevation of PC. c. Determine the elevation and stationing of PT.

Sample Problem No. 4: A forward tangent having a slope of -4% intersects the back tangent having a slope of +7% at point V at stations 6+300 having an elevation of 230. It is 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 60m on the side of the back tangent.. a. Determine the length of the curve on the side of forward tangent. b. Determine the stationing of the highest point on the curve. c. Determine the elevation of the highest point of the curve.