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Vertical Alignment
1
AM 2022
19-Apr-22
Vertical Alignment
 Vertical Curve Properties
 Offsets
 Some additional properties of vertical curves exist.
 Offsets, which are vertical distances from the initial tangent to
the curve, play a significant role in vertical curve design.
the curve, play a significant role in vertical curve design.
 The formula for determining offset (for an equal tangent
parabola) is listed below where Y = offset (in m) at any distance.
 Y=[A/(200L)] * x2 ……in %
 Where:
 A: The absolute difference between g2and g1, multiplied by 100 to translate
to a percentage i.e., A = |g2 – g1|
to a percentage i.e., A = |g2 – g1|
 L: Curve Length
 x: Horizontal distance from PVC along curve
 [NB: a=(G2-G1)/2L or A/2L (from previous slides)]
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Vertical Alignment
 Vertical Curve Properties
 Offsets
 Offsets are vertical distances from initial tangent to the curve
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Vertical Alignment
 Vertical Curve Properties
 Offsets cont’d
 Considering that;
 Y = offset (in m) at any distance, x, from the PVC
 Y=[A/(200L)] * x2
 It follows from the figure that,
 Ym = AL/800 at x=L/2
 Yf =AL/200 at x=L
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AM 2022
19-Apr-22
Vertical Alignment
 ‘K’ Values
 Since it has been shown that the rate of change of grade,
 d2y/dx2, equals 2a.
 The reciprocal of 2a, K, is thus the distance required to effect a unit
change of grade (1%). (2a= A/L, K=L/A )
 Thus, K = L/A, is the horizontal distance required to produce a one
 Thus, K = L/A, is the horizontal distance required to produce a one
percent change in gradient is a measure of curvature (multiplying K by
100 gives the “equivalent radius” of the vertical curve)
 Vertical curves are specified in terms of this factor, K, and their
horizontal length as shown in the relationship: L = A.K
 The K-value can be used directly to compute the high/low points for
crest/sag vertical curves (provided the high/low point is not at a curve
end i.e., PVC or PVT).
end i.e., PVC or PVT).
 xhl = K*|G1|
 Where:
 x = distance from the PVC to the high/low point
 (NB: The low point occurs when eqn dy/dx = 2ax+b = 0)
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Vertical Alignment
 ‘K’ Values cont’d
 K-values are significant in the design of vertical curves.
 Minimum K-value required must provide for the maximum
stopping sight distance encountered on the vertical curve being
designed.
The value of K is also helpful in determining desirable lengths of
 The value of K is also helpful in determining desirable lengths of
vertical curves for various design speeds.
 Once the algebraic difference between grades is known, the
designer can use K values provided to determine the desirable
length for a vertical curve.
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AM 2022
19-Apr-22
Vertical Alignment
 Stopping Sight Distance and Crest Curves
 In the design of vertical curves, its important to note that;
 Sight distance is a very important controlling factor
 Stopping sight distance should be provided as a minimum and
that
that
 Rate of change of grade should be kept within tolerable limits
 Two different factors are important for crest curves
 The driver’s eye height in vehicle, H1
 Height of a roadway obstruction object, H2
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Vertical Alignment
 Stopping Sight Distance and Crest Curves cont’d
 Curve Design
 It is necessary, when designing vertical curves, to provide adequate SSD
 Because curve construction is expensive, we want to minimize curve
length, subject to adequate SSD
 Minimum Curve Length
 By using the properties of a parabola for an equal tangent curve, it can
be shown that the minimum length of curve, Lm, for a required SSD is:
 Where:
 L = length of vertical curve (m)
 SSD = sight distance (m)
 A = algebraic difference in grades (%) (g1-g2)
 h1= height of eye above road surface (m)
 h2= height of object above road surface (m)
8
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Vertical Alignment
 Stopping Sight Distance and Crest Curves cont’d
 Minimum Curve Length cont’d
 Where:
 L = length of vertical curve (m)
 SSD = sight distance (m)
 A = algebraic difference in grades (%) (g1-g2)
 h1= height of eye above road surface (m)
h = height of object above road surface (m)
 h2= height of object above road surface (m)
9
AM 2022
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Vertical Alignment
 Stopping Sight Distance and Crest Curves cont’d
 Minimum Curve Length
 When the height of eye (h1) and the height of object (h2) are 1,080
mm and 600 mm respectively (AASHTO)and substituting these
values into previous two equations yields:
values into previous two equations yields:
 For SSD
 Compare:
 GRZ (1994) uses 1.00m and 0.100m for h1 and h2 respectively
 SATCC (1998) uses 1.05m and 0.15m for h1 and h2 respectively
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Vertical Alignment
 Stopping Sight Distance and Crest Curves cont’d
 Minimum Curve Length (Crest curve)
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Vertical Alignment
 Sag Curves
 Four criteria for establishing length of sag curves
 Headlight sight distance
 Passenger comfort
 Drainage control
 General appearance
 General appearance
 Headlight Sight Distance
 At night, the portion of highway that is visible to the driver is
dependent on the position of the headlights and the direction of
the light beam
 Headlights are assumed to be 600 mm and 1-degree upward
divergence of the light beam from the longitudinal axis of the
vehicle
vehicle
 Sag Vertical Curve Length
 The most controlling factor is headlight sight distance
 If for economic reasons such lengths cannot be provided, fixed
source lighting should be provided to assist the driver.
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Vertical Alignment
 Sag Curves cont’d
 Like crest curves, we need expressions for determining the
minimum length of sag curve required for adequate SSD
 For sag curves the formula is:
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AM 2022
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Vertical Alignment
 Sag Curves cont’d
 Minimum length of sag curve required for adequate SSD
14
AM 2022
19-Apr-22
Vertical Alignment
 K Values -AASHTO
 Design charts or tables are used to determine minimum length of vertical curve to
provide stopping sight distance for both crest and sag vertical curves, and passing
sight distance on crests.
 Design Controls for Crest Vertical Curves Based on Stopping Sight Distance (Source: AASHTO)
 NB: Rate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A), K = L/A.
15
AM 2022
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Vertical Alignment
 K Values –AASHTO cont’d
 Design Controls for Sag Vertical Curves (Source: AASHTO)
NB: Rate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A), K = L/A.
16
AM 2022
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Vertical Alignment
 K Values for Adequate SSD (SATCC)
17
AM 2022
19-Apr-22
Vertical Alignment
 Coordination of Horizontal & Vertical Alignment
 The alignment design must ensure that all the design elements are
complementary to each other.
 A number of design situations exist that could produce
unsatisfactory combinations of elements, despite the fact that the
unsatisfactory combinations of elements, despite the fact that the
design standards have been followed for the particular class of
road in question. They are therefore comparatively unsafe.
 Avoiding such designs is more important for the higher classes of
road because of:
 Higher design speeds,
 Greater traffic volumes
 Greater traffic volumes
 Any accidents resulting from poor design are likely to be more
severe and more frequent.
18
AM 2022
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Vertical Alignment
 Coordination of Horizontal & Vertical Alignment (cont’d)
 The principles outlined below should be applied to roads of
all classes, i.e.;
 The horizontal and vertical alignment should not be
designed independently.
designed independently.
 Horizontal and vertical curvature must be kept in balance
 Use long, gentle curves and short tangents and not the
opposite (but ensure adequate SSD)
 Hazards can be concealed by inappropriate combinations
of horizontal and vertical curves and therefore such
combinations can be very dangerous.
combinations can be very dangerous.
 The next slides show some examples of good and poor
alignment combinations
19
AM 2022
19-Apr-22
Vertical Alignment
 Coordination of Horizontal & Vertical Alignment (cont’d)
 Examples of good and poor combinations of horizontal and vertical
alignment
 (Source: GRZ, RDA, Low Volume Roads Manual Vol 2, Geometric Design and Road Safety,
2019)
20
AM 2022
19-Apr-22
Vertical Alignment
 Question 1:
 A 500m equal-tangent sag vertical curve has the PVC at station
100+00 with an elevation of 1000m.
 The initial grade is -4% and the final grade is +2%.
 Determine the stationing and elevation of the PVI, the PVT,
 Determine the stationing and elevation of the PVI, the PVT,
and the lowest point on the curve.
 Solution:
 Curve length is 500m. PVT is at station 105+00 (100+00 + 5+00) and the
PVI is in the very middle at 102+50, since it is an equal tangent curve.
 From eqn: y = ax2 +bx + c,
 c =1000m, b = -0.04 and a = 0.00006
 ePVI = 993.75m (at x = 250m) and
 ePVT is 995m (at x = 500m)
 Lowest Point: dy/dx = 2ax+b = 0, x = 333.3m with an elevation of 993.3m
21
AM 2022
19-Apr-22
Vertical Alignment
 Question 2:
 Determine the minimum length of a crest vertical curve
between a +0.5% grade and a -1.0% grade for a road with a
100km/hr design speed.
 The vertical curve must provide 190m SSD.
 The vertical curve must provide 190m SSD.
 Assume minimum curve length for h1=1.07m and h2 =0.15m;
 Solution
 From previous equations;
 Assuming S ≤ L then L= 134.0m
But 134.0m is less than 190m SSD,
 But 134.0m is less than 190m SSD,
 [i.e., against the assumption of S ≤ L ]
 Therefore S > L
 For S > L, L = 110.5 m ok.
22
AM 2022
19-Apr-22
Vertical Alignment
 Question 3:
 A crest vertical curve is used to join two tangent sections of a
trunk road.
 Assuming g1 is +2.7% and g2 is –2.3% and a design speed of
120km/hr determine the desirable length that should be used for
120km/hr determine the desirable length that should be used for
this curve.
 First determine A: A = +2.7% – (-2.3%) = 5.0%
 Solution
 From the table under SATCC conditions, the desirable K value for
a crest vertical curve is 110
 Now solve for L, the desirable length of the curve:
 Now solve for L, the desirable length of the curve:
 L = K * A = 110*5 = 550 meters
 [NB: Meets the minimum criteria for Lmin but must check for SSD
compliance as well.]
23
AM 2022
19-Apr-22
Vertical Alignment
 Question 4:
 A vertical alignment for a single carriageway road consists of a parabolic
crest curve connecting a straight-line uphill gradient of +4% with a
straight line downhill gradient of -3%.
 i) Calculate the vertical offset at the point of intersection of the two
i) Calculate the vertical offset at the point of intersection of the two
tangents at PI
 ii) Calculate the vertical offset and horizontal distance for the
highest point on the curve.
 Assume a design speed of 85 km/hr and use the absolute minimum K
value for crest curves. Try based on both AASHTO and SATCC
 Solution
 Solution
 K from tables (using above speed) and then Lmin
 Using equations for offsets Ym can be determined
 For x for highest point is K*G1 , Then Y at highest can be computed
24
AM 2022
19-Apr-22
Vertical Alignment
 Homework:
 Given:
 g1 =- 3.629%, g2 = 0.151%
 PVI station = 5+265.000 m
 ePVI = 350.520 m
 ePVI = 350.520 m
 L = 240 m
 Find the stationing and elevation of PVC and PVT.
 What is the elevations at the stations:
 5+160,
 5+200,
5+240,
 5+240,
 5+280,
 5+320 and
 5+360
25
AM 2022
19-Apr-22

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6. f Geometric Vertical 2.pdf

  • 2. Vertical Alignment  Vertical Curve Properties  Offsets  Some additional properties of vertical curves exist.  Offsets, which are vertical distances from the initial tangent to the curve, play a significant role in vertical curve design. the curve, play a significant role in vertical curve design.  The formula for determining offset (for an equal tangent parabola) is listed below where Y = offset (in m) at any distance.  Y=[A/(200L)] * x2 ……in %  Where:  A: The absolute difference between g2and g1, multiplied by 100 to translate to a percentage i.e., A = |g2 – g1| to a percentage i.e., A = |g2 – g1|  L: Curve Length  x: Horizontal distance from PVC along curve  [NB: a=(G2-G1)/2L or A/2L (from previous slides)] 2 AM 2022 19-Apr-22
  • 3. Vertical Alignment  Vertical Curve Properties  Offsets  Offsets are vertical distances from initial tangent to the curve 3 AM 2022 19-Apr-22
  • 4. Vertical Alignment  Vertical Curve Properties  Offsets cont’d  Considering that;  Y = offset (in m) at any distance, x, from the PVC  Y=[A/(200L)] * x2  It follows from the figure that,  Ym = AL/800 at x=L/2  Yf =AL/200 at x=L 4 AM 2022 19-Apr-22
  • 5. Vertical Alignment  ‘K’ Values  Since it has been shown that the rate of change of grade,  d2y/dx2, equals 2a.  The reciprocal of 2a, K, is thus the distance required to effect a unit change of grade (1%). (2a= A/L, K=L/A )  Thus, K = L/A, is the horizontal distance required to produce a one  Thus, K = L/A, is the horizontal distance required to produce a one percent change in gradient is a measure of curvature (multiplying K by 100 gives the “equivalent radius” of the vertical curve)  Vertical curves are specified in terms of this factor, K, and their horizontal length as shown in the relationship: L = A.K  The K-value can be used directly to compute the high/low points for crest/sag vertical curves (provided the high/low point is not at a curve end i.e., PVC or PVT). end i.e., PVC or PVT).  xhl = K*|G1|  Where:  x = distance from the PVC to the high/low point  (NB: The low point occurs when eqn dy/dx = 2ax+b = 0) 5 AM 2022 19-Apr-22
  • 6. Vertical Alignment  ‘K’ Values cont’d  K-values are significant in the design of vertical curves.  Minimum K-value required must provide for the maximum stopping sight distance encountered on the vertical curve being designed. The value of K is also helpful in determining desirable lengths of  The value of K is also helpful in determining desirable lengths of vertical curves for various design speeds.  Once the algebraic difference between grades is known, the designer can use K values provided to determine the desirable length for a vertical curve. 6 AM 2022 19-Apr-22
  • 7. Vertical Alignment  Stopping Sight Distance and Crest Curves  In the design of vertical curves, its important to note that;  Sight distance is a very important controlling factor  Stopping sight distance should be provided as a minimum and that that  Rate of change of grade should be kept within tolerable limits  Two different factors are important for crest curves  The driver’s eye height in vehicle, H1  Height of a roadway obstruction object, H2 7 AM 2022 19-Apr-22
  • 8. Vertical Alignment  Stopping Sight Distance and Crest Curves cont’d  Curve Design  It is necessary, when designing vertical curves, to provide adequate SSD  Because curve construction is expensive, we want to minimize curve length, subject to adequate SSD  Minimum Curve Length  By using the properties of a parabola for an equal tangent curve, it can be shown that the minimum length of curve, Lm, for a required SSD is:  Where:  L = length of vertical curve (m)  SSD = sight distance (m)  A = algebraic difference in grades (%) (g1-g2)  h1= height of eye above road surface (m)  h2= height of object above road surface (m) 8 AM 2022 19-Apr-22
  • 9. Vertical Alignment  Stopping Sight Distance and Crest Curves cont’d  Minimum Curve Length cont’d  Where:  L = length of vertical curve (m)  SSD = sight distance (m)  A = algebraic difference in grades (%) (g1-g2)  h1= height of eye above road surface (m) h = height of object above road surface (m)  h2= height of object above road surface (m) 9 AM 2022 19-Apr-22
  • 10. Vertical Alignment  Stopping Sight Distance and Crest Curves cont’d  Minimum Curve Length  When the height of eye (h1) and the height of object (h2) are 1,080 mm and 600 mm respectively (AASHTO)and substituting these values into previous two equations yields: values into previous two equations yields:  For SSD  Compare:  GRZ (1994) uses 1.00m and 0.100m for h1 and h2 respectively  SATCC (1998) uses 1.05m and 0.15m for h1 and h2 respectively 10 AM 2022 19-Apr-22
  • 11. Vertical Alignment  Stopping Sight Distance and Crest Curves cont’d  Minimum Curve Length (Crest curve) 11 AM 2022 19-Apr-22
  • 12. Vertical Alignment  Sag Curves  Four criteria for establishing length of sag curves  Headlight sight distance  Passenger comfort  Drainage control  General appearance  General appearance  Headlight Sight Distance  At night, the portion of highway that is visible to the driver is dependent on the position of the headlights and the direction of the light beam  Headlights are assumed to be 600 mm and 1-degree upward divergence of the light beam from the longitudinal axis of the vehicle vehicle  Sag Vertical Curve Length  The most controlling factor is headlight sight distance  If for economic reasons such lengths cannot be provided, fixed source lighting should be provided to assist the driver. 12 AM 2022 19-Apr-22
  • 13. Vertical Alignment  Sag Curves cont’d  Like crest curves, we need expressions for determining the minimum length of sag curve required for adequate SSD  For sag curves the formula is: 13 AM 2022 19-Apr-22
  • 14. Vertical Alignment  Sag Curves cont’d  Minimum length of sag curve required for adequate SSD 14 AM 2022 19-Apr-22
  • 15. Vertical Alignment  K Values -AASHTO  Design charts or tables are used to determine minimum length of vertical curve to provide stopping sight distance for both crest and sag vertical curves, and passing sight distance on crests.  Design Controls for Crest Vertical Curves Based on Stopping Sight Distance (Source: AASHTO)  NB: Rate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A), K = L/A. 15 AM 2022 19-Apr-22
  • 16. Vertical Alignment  K Values –AASHTO cont’d  Design Controls for Sag Vertical Curves (Source: AASHTO) NB: Rate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A), K = L/A. 16 AM 2022 19-Apr-22
  • 17. Vertical Alignment  K Values for Adequate SSD (SATCC) 17 AM 2022 19-Apr-22
  • 18. Vertical Alignment  Coordination of Horizontal & Vertical Alignment  The alignment design must ensure that all the design elements are complementary to each other.  A number of design situations exist that could produce unsatisfactory combinations of elements, despite the fact that the unsatisfactory combinations of elements, despite the fact that the design standards have been followed for the particular class of road in question. They are therefore comparatively unsafe.  Avoiding such designs is more important for the higher classes of road because of:  Higher design speeds,  Greater traffic volumes  Greater traffic volumes  Any accidents resulting from poor design are likely to be more severe and more frequent. 18 AM 2022 19-Apr-22
  • 19. Vertical Alignment  Coordination of Horizontal & Vertical Alignment (cont’d)  The principles outlined below should be applied to roads of all classes, i.e.;  The horizontal and vertical alignment should not be designed independently. designed independently.  Horizontal and vertical curvature must be kept in balance  Use long, gentle curves and short tangents and not the opposite (but ensure adequate SSD)  Hazards can be concealed by inappropriate combinations of horizontal and vertical curves and therefore such combinations can be very dangerous. combinations can be very dangerous.  The next slides show some examples of good and poor alignment combinations 19 AM 2022 19-Apr-22
  • 20. Vertical Alignment  Coordination of Horizontal & Vertical Alignment (cont’d)  Examples of good and poor combinations of horizontal and vertical alignment  (Source: GRZ, RDA, Low Volume Roads Manual Vol 2, Geometric Design and Road Safety, 2019) 20 AM 2022 19-Apr-22
  • 21. Vertical Alignment  Question 1:  A 500m equal-tangent sag vertical curve has the PVC at station 100+00 with an elevation of 1000m.  The initial grade is -4% and the final grade is +2%.  Determine the stationing and elevation of the PVI, the PVT,  Determine the stationing and elevation of the PVI, the PVT, and the lowest point on the curve.  Solution:  Curve length is 500m. PVT is at station 105+00 (100+00 + 5+00) and the PVI is in the very middle at 102+50, since it is an equal tangent curve.  From eqn: y = ax2 +bx + c,  c =1000m, b = -0.04 and a = 0.00006  ePVI = 993.75m (at x = 250m) and  ePVT is 995m (at x = 500m)  Lowest Point: dy/dx = 2ax+b = 0, x = 333.3m with an elevation of 993.3m 21 AM 2022 19-Apr-22
  • 22. Vertical Alignment  Question 2:  Determine the minimum length of a crest vertical curve between a +0.5% grade and a -1.0% grade for a road with a 100km/hr design speed.  The vertical curve must provide 190m SSD.  The vertical curve must provide 190m SSD.  Assume minimum curve length for h1=1.07m and h2 =0.15m;  Solution  From previous equations;  Assuming S ≤ L then L= 134.0m But 134.0m is less than 190m SSD,  But 134.0m is less than 190m SSD,  [i.e., against the assumption of S ≤ L ]  Therefore S > L  For S > L, L = 110.5 m ok. 22 AM 2022 19-Apr-22
  • 23. Vertical Alignment  Question 3:  A crest vertical curve is used to join two tangent sections of a trunk road.  Assuming g1 is +2.7% and g2 is –2.3% and a design speed of 120km/hr determine the desirable length that should be used for 120km/hr determine the desirable length that should be used for this curve.  First determine A: A = +2.7% – (-2.3%) = 5.0%  Solution  From the table under SATCC conditions, the desirable K value for a crest vertical curve is 110  Now solve for L, the desirable length of the curve:  Now solve for L, the desirable length of the curve:  L = K * A = 110*5 = 550 meters  [NB: Meets the minimum criteria for Lmin but must check for SSD compliance as well.] 23 AM 2022 19-Apr-22
  • 24. Vertical Alignment  Question 4:  A vertical alignment for a single carriageway road consists of a parabolic crest curve connecting a straight-line uphill gradient of +4% with a straight line downhill gradient of -3%.  i) Calculate the vertical offset at the point of intersection of the two i) Calculate the vertical offset at the point of intersection of the two tangents at PI  ii) Calculate the vertical offset and horizontal distance for the highest point on the curve.  Assume a design speed of 85 km/hr and use the absolute minimum K value for crest curves. Try based on both AASHTO and SATCC  Solution  Solution  K from tables (using above speed) and then Lmin  Using equations for offsets Ym can be determined  For x for highest point is K*G1 , Then Y at highest can be computed 24 AM 2022 19-Apr-22
  • 25. Vertical Alignment  Homework:  Given:  g1 =- 3.629%, g2 = 0.151%  PVI station = 5+265.000 m  ePVI = 350.520 m  ePVI = 350.520 m  L = 240 m  Find the stationing and elevation of PVC and PVT.  What is the elevations at the stations:  5+160,  5+200, 5+240,  5+240,  5+280,  5+320 and  5+360 25 AM 2022 19-Apr-22