Vertical Curves-kenya-road-design-manual-for highway and expressway design

kenya-road-design-manual-for highway and expressway design-in kenyan context

1. 1

Vertical Curves
Vertical curves (VC) are used to connect intersecting gradients in the vertical plane. Vertical
curves are introduced between grade lines:
• To provide a minimum sight distance, reducing the possibility of any accident.
• To avoid discomfort to passengers (Schofield & Breach, 2007).
A vertical curve may be circular or parabolic. Parabolic curves are preferred because of the
following reasons:
1. It is flatter at the top and hence provides a longer sight distance, reducing the
possibility of any accident.
2. Rate of change of grade is uniform throughout and hence produces best riding
qualities.
3. It has simplicity in computation work.

Fig. 11.57
From intersection point:
• Grade falling as chainage increases is positive.
• Grade falling as chainage decreases is positive.
Types of Vertical Curve
There are two types of vertical curves:

2. 2

1. Summit curve 2. Sag curve
1. Summit Curve

This is provided in either of the following three cases:
(a) When an upgrade is followed by a downgrade (Fig. 11.58(a)).
(b) When a steeper upgrade is followed by a milder upgrade (Fig. 11.58(b)).
(c) When a milder downgrade is followed by a steeper downgrade (Fig. 11.58(c)).
2. Sag Curve

This is provided in either of the following three cases:
(a) When a downgrade is followed by an upgrade (Fig. 11.58(d)).
(b) When a steeper downgrade is followed by a milder downgrade (Fig. 11.58(e)).
(c) When a milder upgrade is followed by a steeper upgrade (Fig. 11.58(f)).
Total Change of Grade
It is the algebraic difference of two grades. For example, if a +g1% (upgrade) is followed by a
–g2% (downgrade) the total change of grade is [+ g1 – (–g2)]% = (g1 + g2) %.

3. 3



Fig. 11.58 Types of vertical curves
Length of Vertical Curves
The length of a vertical curve is the length from the point of commencement of the curve i.e.,
the point of rising to the point of tangency of the curve i.e., the point where curve meets
horizontal straight again.
12
total change of grade
Length of curve
permissible rate of change of grade
gg
rr
= = + ... (18.4)
As upgrades are treated positive and downgrades negative, equation (18.4) may be written as

4. 4

()
12
gg
L
r
−−
= ... (18.5)
Basic Assumptions in Curve Computations
In order to simplify vertical curve computations, the following assumptions are made;
1. The vertical curve is a parabola of equation y = Cl
2
, where y is the vertical offset from
gradient to curve, distance l from the start of the curve, and C is a constant.

Fig. 10.39 Vertical curve approximations
2. The length of curve each side of the gradient intersection is L/2, i.e., the curve is of equal
length each side of I. Thus, T1C = CT2 = T1I = IT2 = L/2
3. Horizontal, sloping, and curve distances between points on the curve are all assumed
equal, i.e., distance T1D = T1BT2 = T1CT2 = (T1I + IT2).
4. The curve bisects BI at C, thus BC = CI = Y (the mid-offset).
5. From similar triangles T1BI and T1T2J, if BI = 2Y, then T2J = 4Y. 4Y represents the vertical
divergence of the two gradients over half the curve length (L/2) and therefore equals
AL/200.

5. 5


Fig. 10.17
6. The gradients expressed in percentages indicate the change in level over a 100 m
horizontal length.
Factors Considered in Designing Vertical Curves
a) Overtaking Sight Distances
Refers to the length of road ahead that is visible to the driver. The sight distance must be
greater than the stopping distance in which the vehicle can be brought to rest.
b) Stopping Sight Distance
Stopping distance is dependent upon:
• Speed of the vehicle.
• Braking efficiency.
• Gradient of the road.
• Coefficient of friction between tyre and road.
• Road conditions e.g., wet, dry or slippery.
• Driver’s reaction time
c) Rate of change of gradient (R)
Refers to the rate at which the curve passes from one gradient (g1%) to the next (g2%) and is
similar in concept to rate of change of radial acceleration in horizontal transitions.
When linked to design speed, it should never exceed 0.3 m/s
2
.

6. 6

General Equation of an Equal Tangent Vertical Parabolic
Curve
The general equation of a parabolic curve with a vertical axis is given by (Allan, 1993)
y = ax
2
+ bx + c

Figure 11.17
The slope of the curve
2
dy
ax b
dx
=+
The rate of change of grade or slope
2
2
2
dy
aK
dx
== (11.33)
Equation (11.33) represents the fact that the rate of change of gradient is uniform throughout,
and hence produces a smooth riding condition.
The gradient at any point on the curve can be found from
2
2
dy d y
dx Kdx Kx A
dx dx
= = = +
If point T1 is taken as the origin for the X axis, then at T1, x = 0, and dy/dx = g1, whence A=
g1, and equation of the curve becomes

7. 7

1
dy
Kx g
dx
=+
or ( )
21
1
ggdy
g
dx L
−
=+
Note that when
x = L, 21
g KL g=+ , i.e., 21
gg
K
L
−
=
The level of any point on the curve can be found from
( )
2
11
1
2
dy
y dx Kx g dx Kx g x B
dx
= = + = + +
If T1 is taken as the origin for the Y axis, then at T1, x = 0, y = 0, and so B = 0. It is usually
more convenient to take the level datum as the origin for the Y axis, so that when x = 0, y = B
= reduced level of T1 = H say.
In this case. the equation becomes
( )
2
21
1
2
g g x
y g x H
L
−
= + + … (25.3)
where:
y = vertical offset from gradient to curve
H = reduced level of T1
g1x = change in elevation along the back tangent with increasing X
L = horizontal distance from T1 to T2
The rate of change of grade, r, for an equal-tangent parabolic curve equals: 21
gg
r
L
−
=
… (25.4)
If the grades are in %:
( )
2
21 1
200 100
g g x gx
yH
L
−
= + +
Passing a Curve Through a Point of Known Level
In order to ensure sufficient clearance at a specific point along the curve it may be necessary
to pass the curve through a point of known level., for example under or over a bridge.
Example 1
A downgrade of 4% is followed by a rising gradient of 5%. At the start of the curve the level
is 123.06 m at chainage 3420 m, whilst at chainage 3620 m there is an overpass with an

8. 8

underside level of 127.06 m. If the designed curve is to afford a clearance of 5 m at this point,
calculate the required length.

Fig. 10.45 Clearance at a given point
Solution
Fall in elevation from T1 to E = 1
gL =4
200 8 m
100
=
∴ RL at E = 123.06 − 8 = 115.06 m
RL at point C on the curve directly below the overpass = 127.06 − 5 = 122.06 m
∴ y-offset CE at C, yc = RLC – RLE = 122.06 − 115.06 = 7 m
( )
2
21
200
c
C
g g x
y
L
−
=
()( )
2
54
7
200
c
x
L
−−
=
2
1400 9
c
Lx=
But xc = 3620 – 3420 = 200
∴ L = 257 m
Example 2
A falling gradient of 1.8% is followed by a rising gradient of 2.5%, their intersection
occurring at chainage 8 + 73.2 and RL 72.56. In order to provide sufficient clearance under
an existing bridge, the RL at chainage 7 + 95.8 on the curve must not be higher than 74.20.
Find the greatest length of simple parabolic vertical curve that can be used

9. 9


Figure 11.23
Solution
Let P be the control point under the bridge. Distance PI is given by
PI = (8 + 73.2) − (7 + 95.8) = 77.4
RL of T11.8
72.56
200
L
H= + =
Equation of curve is
( )
2
2.5 1.81.8
100 200
xx
yH
L
+
= − +
RL of P = 74.20
( )
( )
2
0.5 77.4
72.56 +0.009 0.018 0.5 77.4 0.0215
L
LL
L
−
= − − +
whence
L
2
− 355.5L + 23 963 = 0
from which
L = 265.1 or 90.4
Since 0.5 × 90.4 is less than 77.4, the second solution is impracticable.
Chainage of Highest or Lowest Point on a Vertical Curve
The position and level of the highest or lowest point on the curve is frequently required for
drainage design.

10. 10


Fig. 10.45 Clearance at a given point
At the low or high point, a tangent to the curve will be horizontal, and its grade equal to zero.
Based on this fact, by taking the derivative of Equation (25.3) (i.e., rate of change of
grade), and setting it equal to zero, the following formula is readily derived
( )
2
21
1
2
g g x
y g x H
L
−
= + + … (25.3)
( )
21
1
g g xdy
g
dx L
−
=+
( )
21
1
0
g g x
g
L
−
+=
( )
2 1 1
g g x g L− = − … (25.4)
Simplifying equation (25.4);
1
12
gL
x
gg
=
− … (25.6)
where x is the distance from T1 to the high or low point of the curve, g1 the tangent grade
through the I, g2 the tangent grade through T2, and L the curve length.
Computation of Setting Out Data
A ‘2nd difference’ (δ2y/δl2) arithmetical check on the offset computation should
automatically be applied.
Example 1
Given the following information about a vertical curve;
Uphill gradient = 1 in 100
Downhill gradient = 0.45 in 100
Chainage of intersection point = 122.88 m

11. 11

Reduced level of intersection point = 126 m
Rate of change of gradient = 1.18 × 10
−4

Calculate:
i) The length of the curve
ii) The chainage and reduced level of the first tangent point (10 marks)
Solution
i) The length of the curve
21
gg
L
r
−
=
4
0.0045 0.01
122.881 m
1.18 10
L
−
−−
==

ii) The chainage and reduced level of the first tangent point
Chainage T1 = Chainage I – L/2 = 122.881
122.88
2
− = 61.44 m
Since this is a summit curve;
Fall in elevation from intersection point to T1 = 1
1 122.881
0.6149 m
2 100 2
L
g =  =
RLT1 = RLI − Fall = 126 – 0.6149 = 125.386 m
Example 2
A falling gradient 4% meets a rising gradient of 5% at chainage 2500 m and at level of 216
m. At chainage 2400 m the underside of a bridge has level 235 m. The two gradients are to be
joined by a parabolic vertical curve giving 14 m clearance under the bridge. Given that
setting out interval is 50 m. Compute the setting out levels along the curve giving the answer
in a tabular form.
Solution
( 2




2

2




2

2

12. 12

Rise from I to P = 1
2
L
g =4
100 4 m
100
=
RL at P = 216 + 4 = 220 m
RL at P' = 235 – 14 = 221 m
y offset at P yP = 221− 220 = 1 m
i) Computation of curve length ( )
2
21
2
n
n
g g x
y
L
−
=
()( )
2
54
1
200
P
x
L
−−
=

∴ 2
200 9
P
Lx=
But xP = 100
2
L
−
∴ 2
200 9 100
2
L
L

=−


2
200 9 100 10,000
4
L
LL

= − +

29
200 900 90,000
4
L L L= − +
29
1100 90,000 0
4
LL− + =
Solving using quadratic formula;
2
4
2
b b ac
L
a
−  −
= , where b = −1100, a = 9/4 and c = 90,000
L = 385 or 104
But L can’t be since 100
2
L
−

 here gives a negative value.
∴ L = 385 m
Distance T1P is therefore = 385
100 92.5 m
2

−=


ii) Computation of y- offsets from back tangent to curve

13. 13

( )
2
21
2
n
n
g g x
y
L
−
= , where xn = horizontal distance from T1
yn = offset from back tangent to curve
L = always assumed to be equal to curve length
General formula for y offsets ( )
2
2
54 9
2 385 77,000
n
nn
x
yx
−−
==

T1 = 0 m
2
1
9
50
77,000
y= = 0.2922 m
2
2
9
100
77,000
y= = 1.1688 m
2
3
9
150
77,000
y= = 2.6299 m
2
4
9
200
77,000
y= = 4.6753 m
2
5
9
250
77,000
y= = 7.3052 m
2
6
9
300
77,000
y= = 10.5195 m
2
7
9
350
77,000
y= = 14.3182 m
2
8
9
385
77,000
y= = 17.325 m
RLT1 = 4 385
216 223.7 m
100 2

 + =


Change in elevation over 50 m = ()
4
50 2 m
100
fall

 = −


Change in elevation between y7 and y8 = ( ) ()
4
385 350 1.4 m
100
fall − = −
Curve RL = Tangent R.L. + yn
Chainage
(m)
Offset (m) Tangent
level (m)
Curve
level (m)
1
st
diff.
(m)
2
nd
diff.
(m)
Remarks

14. 14

0 0 223.7 223.7 T1 – Start
1.708
50 0.2922 221.7 221.992 0.585
1.123
100 1.1688 219.7 220.869 0.584
0.539
150 2.6299 217.7 220.33 0.584
−0.045
200 4.6753 215.7 220.375 0.585
−0.63
250 7.3052 213.7 221.005 0.585
−1.215
300 10.5195 211.7 222.22 0.583
−1.798
350 14.3182 209.7 224.018

385 17.325 208.3 225.625 T2 – End

Example 3
An existing length of road consists of rising gradient of 1 in 20 followed by a vertical
parabolic crest curve of 100 m and a falling gradient. The reduced level of the highest point
on the curve is 173.07 m asl. Compute the levels at 20 m intervals showing 2
nd
differences.
Solution

i) Computation of y- offsets from back tangent to curve














2





2

2

15. 15

( )
2
21
2
n
n
g g x
y
L
−
= , where xn = horizontal distance from T1
yn = offset from back tangent to curve
L = always assumed to be equal to curve length
g1 = 1
0.05
20
=+
Assuming the two grades are equal, g2 = 1
0.05
20
=−
Thus, yn can be re-written as, ( )
2
0.05 0.05
200
n
n
x
y
−−
=
∴ 2
0.0005
nn
yx=− 2
0.0005
n
x=
(distance is never negative)
Offset (m) 1
st
difference 2
nd
difference
T1 = 0 m _______________ ____________
0.2
y1 = 0.0005 × 20
2
= 0.2 m 0.4
0.6
y2 = 0.0005 × 40
2
= 0.8 m 0.4
1.0
y3 = 0.0005 × 60
2
= 1.8 m 0.4
1.4
y4 = 0.0005 × 80
2
= 3.2 m 0.4
1.8
4Y = y5 = 0.0005 × 40
2
= 0.8 m

ii) Computation of distance x of summit of curve from T1
( )
1
12
0.05 100
50 m
0.05 0.05
gL
x
gg

= = =
− − −
Elevation of summit above msl, ( )
2
21
1
2
msl o
g g x
y g x H
L
−
= + +
, where Ho = RL at T1
( )
( )
2
0.05 0.05 50
173.07 50 0.05
200
o
H
−−
= +  +
Ho = 173.07 – 1.25 = 171.82 m

16. 16

iii) Computation of tangent RLs
Rise in grade along back tangent over 20 m = 0.05 × 20 = 1 m
∴ 1 m is added to every chainage starting from T1
iv) Computation of curve RLs at 20 m intervals
Curve levels are computed by subtracting y-offsets from tangent RLs
Chainage
(m)
Offset (m) Tangent
level (m)
Curve
level (m)
1
st
diff.
(m)
2
nd
diff.
(m)
Remarks
0 0 171.82 171.82 T1
0.8
20 0.2 172.82 172.62 0.4
0.4
40 0.8 173.82 173.02 0.4
0 Center of curve
60 1.8 174.82 173.02 0.4
−0.4
80 3.2 175.82 172.62 0.4
−0.8
100 5 176.82 171.82 T2

Example 4
A grade g1 of +3.00% intersects grade g2 of −2.40% at a vertex whose station and elevation
are 46 + 70 and 853.48 m, respectively. An equal-tangent parabolic curve 600 m long has
been selected to join the two tangents. Compute and tabulate the curve for stakeout at full
stations. (Figure 25.4 shows the curve.)
Solution
The rate of change of grade: 21
2.40 3.00
0.90%
6
gg
r
L
− −−
= = = −

Stationing:
I = 46 + 70
− L/2 = 3 + 00
T1 = 43 + 70
+ L = 6 + 00
T2 = 49 + 70
Elev T1 = 1
R.L. of
100 2
gL
I

= − •



17. 17

= 3
853.48 300
100

−

 = 844.48 m

Figure 25.4 Crest curve of Example 25.1
Station X
(sta)
g1x ( )−
2
21
2
g g x
L

Curve
Elevation
1
st

Difference
2
nd

Difference
49 + 70
(T2)
600 18.00 − 16.20 846.28

49 + 00 530 15.90 −12.64 847.74 −1.32 −0.90
48 + 00 430 12.90 − 8.32 849.06 −0.42 −0.90
47 + 00 330 9.90 −4.90 849.48 0.48 −0.90
46 + 00 230 6.90 −2.38 849.00 1.38 −0.90
45 + 00 130 3.90 −0.76 847.62 2.28
44 + 00 30 0.90 −0.04 845.34
43 + 70
(T1)
0.0 0.00 −0.00 844.48
Check: T2 = ( )
2
Elev 853.48 2.40 3 846.28
2
I
L
g

− = −  =


A check on curve elevations is obtained by computing the first and second differences
between the elevations of full stations. All second differences (rate of change) should be
equal.
Example 18.4.
Calculate the reduced levels of the various station pegs on a parabolic vertical curve which is
to be set to connect two uniform grades .6% and − .7%. The chainage and required level of
the point of intersection are 2525 m and 335.65 m respectively. Assume the rate of change of
grade to be 0.05 per 20 m chain.
Solution

18. 18


Fig. 18.9
Here g1 = 0.6%; g2 − .7%: r = 0.05
The total length of the curve
()
12
0.6 0.7
26 chains
0.05
gg
L
r
−−−
= = =
The length of the curve on either side
= 13 chains = 13 × 20 = 260 m.
Chainage at point of intersection I = 2525 (given)
Chainage at point of commencement
T1 2 2 − 26 226 m
Chainage at point of tangency
T2 = 2525 + 260 = 2785 m
R.L. of point of intersection I = 335.65 m (given)
R.L. of point of commencement
T1 = 0.6
335.65 260 334.09 m
100

−  =


R.L. of point of tangency
T2 = 0.7
335.65 260 333.83 m
100

−  =


R.L. of the point
E = 12
R.L. of R.L. of 334.09 333.83
333.96 m
22
TT+ +
==
R.L. of the point
F = R.L. of R.L. of 333.96 335.65
334.805 m
22
EI++
==

19. 19

Stn.
Chainage
(m)
X g1X ( )−
2
21
2
g g X
L
R.L.
(m)
1
st

Difference
2
nd

Difference
T1 2265 0 0 −0 334.090
1 2285 20 0.12 −0.005 334.205
2 2305 40 0.24 −0.02 334.310
3 2325 60 0.36 −0.045 334.405 0.095
4 2345 80 0.48 −0.08 334.490 0.085
5 2565 100 0.60 −0.125 334.565 0.075 −0.01
6 2385 120 0.72 −0.18 334.630 0.065 −0.01
7 2405 140 0.84 −0.245 334.685
8 2425 160 0.96 −0.32 334.730
9 2445 334.765
10 2465 334.790
11 2485 334.805
12 2505 334.810 Apex
F 2525 334.805
14 2545 334.790
15 2565 334.765
16 2585 334.730
17 2605 334.685
18 2625 334.630
19 2645 334.565
20 2665 334.490
21 2685 335.405
22 2705 334.310
23 2725 334.205 −0.105
24 2745 334.090 −0.115
25 2765 333.965 −0.125 −0.01
T2 2785 520 333.830