WPC #27369 07/09

AMRC 2012 MODULE 8

Horizontal Alignment

CONTENTS Overview ......................................................................................... 8-1 Objectives ........................................................................................ 8-1 Procedures ....................................................................................... 8-1 8.1 Design Considerations and Circular Curves ......................... 8-3 8.2 Superelevation and Transitional Spiral ................................. 8-5 8.3 Geometry of Circular Curves ................................................ 8-7

8.3.1 Circular Arc Geometrics .......................................... 8-7 8.3.2 Arc Length ............................................................... 8-9 8.3.3 BC (Beginning of Curve) and EC (End of Curve)

Stationing ............................................................... 8-10 8.4 Circular Curves and Stopping Sight Distances ................... 8-15

8.4.1 External Distance ................................................... 8-15 8.5 Self-Test .............................................................................. 8-19

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Module 8

Horizontal Alignment

Overview This module explains the technical details of horizontal curves and

how to apply them in design.

Objectives Upon completion of this module, the student will be able to: describe the horizontal alignment as a design consideration describe the elements of a circular curve describe the elements of superelevation.

Procedures Study the module materials and make notes as required. Perform the self-test on these principles and review the course

materials in such a manner as to be able to successfully complete similar questions upon examination.

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SECTION 8.1

Design Considerations and Circular Curves

Horizontal curves are used to provide a travel path between two intersecting tangents. When a vehicle moves along a circular path, it creates a centrifugal force causing the vehicle to slide radially outward. The centrifugal force is counteracted by the vehicle mass acting in consort with the friction between the tire and the pavement. When the friction force is insufficient to hold the vehicle to the pavement, the road needs to be banked to further counteract the centrifugal force. The banking of the road is called superelevation. The basic relationship between the radius of the roadway, the speed of the vehicle, the friction factor, and the superelevation is given by the following equation:

2V

R127 (e f )

Where: R = Radius of the curve linking two tangents, m V = Speed of vehicle, km/h e = Rate of superelevation, m/m f = Lateral friction of pavement at the travelling speed Section 330.01 of the Ministry of Transportation’s supplement to the TAC Manual provides a more comprehensive formula for rural and high speed urban roadways. The minimum radius of a curve linking two tangents can, therefore, be developed, given the design speed, V, the corresponding lateral friction, and the maximum superelevation. The minimum radius is used to provide driver comfort and safety against side skid of the vehicle.

Figure 8.1

Typical Horizontal Tangents and Curves

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The TAC design manual, Section 2.1.2, and the horizontal section of the AASHTO design policy provide detailed explanation on the equation and on the concept of minimum curve radius.

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SECTION 8.2

Superelevation and Transitional Spiral

Superelevation is the banking of the road cross section to counteract the centrifugal force along a curve. The minimum curve radius can be established given a maximum superelevation. A maximum superelevation is required because the maximum lateral friction is assumed to hold the vehicle to the pavement without side skid. The lateral friction is affected by the frequency of snow and icing on the road surface. The frequency of slow moving vehicles is also a factor since they may slide sideways due to icy road conditions. Other factors include the terrain and whether the highway is located in rural or urban areas.

Figure 8.2

Typical Superelevated Road Cross Section The Ministry of Transportation has established a maximum rate of superelevation of 0.08 m/m. The cross section of the road provides a normal crown for drainage. A normal crown exists when the highest point of the pavement cross section is at the centre and slopes to both sides, usually at 2 percent. To achieve superelevation, the outside pavement edge of a circular curve will be raised from a minus 2 percent to the desired elevation. The inside pavement edge will also be lowered from its +2 percent normal crown to the desired elevation. The section of the roadway for the adjustment of the cross section from an adverse crown to the design superelevation is called a transitional spiral. The rate of this change is called rate of superelevation. The length of this transition is called a spiral curve. Section 2.1.2 of the TAC Manual provides typical superelevation development with tangent runoff options.

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Figure 8.3 Typical Normal Crown to Full Superelevation Transition

Note: If the road enters the curve with “normal crown,” the

outside lanes are sloping the wrong way. This is known as “adverse crown” and should be avoided in most cases.

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SECTION 8.3

Geometry of Circular Curves

The following are some terminologies and calculation procedures of circular curves. Though detailed calculations are given in the following sections, they will not form part of the final exam and are included for information only.

8.3.1 Circular Arc Geometrics Prior to computation of the curve data required for layout of the

curve, two variables must be known. They are: radius (R) which is selected by the designer and is based

partially on the previously mentioned safety rules intersection angle (Δ) external angle between tangents

(difference between bearings). Once these have been established, all other curve data can be computed. The essential additional data required for layout is as follows: curve tangent length (T) arc length (L) external distance (E) station at BC station at EC. These are illustrated in Figure 8.4.

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Figure 8.4

Simple Circular Arc Where: Δ intersection angle PI point of intersection of tangents of road CL) BC beginning of curve EC end of curve L length of curve (arc length) T tangent length E external distance From Figure 8.4, the tangent length T is the distance from the BC to the PI and also the distance from the EC to the PI. Other terms used for this length are subtangent length and apex distance. Using Figure 8.5 it can be seen that a line from the PI to 0 (the centre of the radius) bisects the angle Δ at 0. Consequently, Tan (Δ/2) = T/R Therefore: T = R tan (Δ/2)

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Figure 8.5

8.3.2 Arc Length As shown in Figure 8.6, the circumference of a circle of radius, R, is 2πR and 360° is swept by that circumference. Therefore, the arc length:

L 1 circumference360

R

L 2 R360 180

R

L180

Where: Δ is in degrees. For Δ in radians L = RΔ

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Figure 8.6

Figure 8.7 Simple Circular Arc

8.3.3 BC (Beginning of Curve) and EC (End of Curve) Stationing In order to establish the station of the BC of the curve, it is necessary

to know the distance of the BC from a reference point before the curve. This reference point is very often the EC or the PI of the previous curve. Using the following example where the coordinates of PI #1 and PI #2 are known, you can calculate the curve data for curve #2.

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Figure 8.8 Given: Tangent length, T1 = 104.648 m Tangent length, T2 = 88.641 m Length of curve #1 = 553.095 m Length of curve #2 = 140.000 m Calculate the distance D between PI #1 and PI #2.

Figure 8.9

Distance D = 2 2(1210 1010) (980 810)

= 262.488 m Distance 1 = D −T2 −T1

= 262.488 − 88.641 − 104.648 = 69.199 Chainage of EC Curve #1 = 4 + 879.722

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Distance from EC1 to PI #2 = D − T1 = 262.488 − 104.648 = 157.840 Chainage of PI #2 = (4 + 879.722) + 157.840 = 5 + 037.562 Chainage of BC Curve #2 = (5 + 037.562) − T2 = (5 + 037.562) − 88.641 = 44948.921 Chainage of EC Curve #2 = (4 + 948.921) + Length of Curve = (4 + 948.921) + 140 = 5 + 088.921 Note: The chainage given to the PI is derived as you travel forward (i.e., increasing chainage) along the alignment and once you have passed through the curve the PI chainage no longer has any use in your design. It is important that you realize that you do not calculate the chainage through the curve to the EC point, travel back along the target at EC to the PI, and then expect to get the same PI chainage as had previously been calculated. Figure 8.10 emphasizes this point.

Figure 8.10 From the previous, it can be seen that: BC chainage & distance T ≠ BC chainage + distance L − distance T.

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The following example of curve data computation may serve to explain this point fully.

Figure 8.11 Intersection Angle: Δ = 106.7 − 63.4 = 43.3° 1. Curve Data:

Tangent Length = R tan (Δ/2) = 400 tan 21.65 = 158.775 m

Arc Length = 43.3 400

180

= 302.291 m

180

Note: L < 2T BC Chainage = PI Station − Distance T = [1 + 000] − 158.775 = 0 + 841.225 EC Chainage = BC Station + Distance L = [0 + 841.225] + 302.291 = 1 + 143.516 PI Station (if req’d.) = EC Station − Distance T = [1 + 143.516] − 158.775 = 984.741 m Hence, the two chainages do not equal each other. Therefore, only use the forward chainage.

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Note: At this point, we could introduce spiral curves. These are used to ease the transition from a straight to a curve and are used where cars are expected to be travelling at higher speeds. Their provision improves driver safety and comfort; however, they are not in the scope of this course.

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SECTION 8.4

Circular Curves and Stopping Sight Distances

8.4.1 External Distance The external distance is the distance between the PI and the centre of

the curve. This distance may be required where right-of-way restrictions occur or where the PI is inaccessible during setting out. Using Figure 8.12, it can be seen that:

R

cos2 R E

Therefore: R

R Ecos

2

Therefore: R

E Rcos

2

Or:

1 cos2

E Rcos

2

There are various forms of this equation but the above is the easiest for use with a calculator. Note: The terms station and chainage are interchangeable; each means the distance along the alignment from a point of known chainage.

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Figure 8.12 Review Figures 2.1.2.11, 2.1.2.12, and 2.1.2.13 in the TAC Manual. The term stopping sight distance and the factors relative to its determination based on speed, reaction, friction, etc. were detailed in earlier modules. The provision of a minimum clear line of sight, for the driver, “around” a horizontal curve is essential in order to avoid accidents. Obstacles (e.g., rocks, deer, etc.) which may be on the road ahead must be visible in time for the driver to execute a braking stop without collision occurring. The greater this forward line of sight, the safer the curve is to negotiate. The diagram in Figures 2.1.2.11 and 2.1.2.12 in the TAC Manual show this line of sight to be established from the centreline of the inside lane inbound to the curve to a point on the centreline of the inside lane outbound from the curve. The critical distance is “C,” the offset distance from the centreline of the inside lane to the line of sight. To provide visual safety, the determination of clearing boundaries, rock removal, cut slope variations, and building relocation can be determined at the preliminary design stage using the nomograph provided in Figures 2.1.2.11 and 2.1.2.12. For a selected curve radii, R, and a known design speed, V, the value of C, the clearance, can be determined from the nomograph. The typical road section must be taken into account in the preliminary calculation of clearing boundaries or the checking for obstructions to visibility.

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This is achieved by using the value of E, the external distance for a simple circular curve. This is the distance from the PI of a curve to the centreline of the highway. However, the value C, given in Figures 2.1.2.11 and 2.1.2.12, is the distance from the centre of the inside lane and it leaves a minimum half lane width for a two-lane road or one and a half lanes for a four-lane road. These must be added together. Required distance from PI to obstacle for two-lane highway = E + C + 1/2 lane width. This is illustrated in Figure 8.13.

Figure 8.13

Required distance from PI to sight obstacle for a four-lane highway = E + C + 1 1/2 lane widths (plus 1/2 median width where applicable). Note: Where it is not possible to achieve the required clearance, the design speed should be lowered to match the available value of C. Using the following as an example:

R 300 m

kmV 80 desirable

h

kmC 8 m for 80

h

Say we have an actual C value of 4 m and the obstacle cannot be moved, a posted speed reduction warning would be required. The design speed for a C value of 4 m is approximately 65 km/h.

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SECTION 8.5

Self-Test

Provided below is a self-test for this module. Perform the test and compare your answers with those given in your student guide. If you have difficulty with the self-test, review the course materials and make notes in order to be able to successfully complete similar questions upon examination. 1. Explain the terms of the equation used in calculating minimum radius.

2. Why do we need superelevation on a horizontal curve?

3. What is the purpose of a transitional spiral?

4. What is the distance from the PI of a horizontal curve to the BC or EC of the curve called?

5. To find the station of the EC, which of the following would you do?

A. Add the tangent length to the PI station. B. Add the curve length to the BC station. C. Add the spiral length to the BC station. D. None of the above. 6. Which of the following is likely to have the largest value?

A. curve tangent length B. arc length C. external length 7. Minimum curve radius is calculated for which of the following?

A. dry road B. wet road C. icy road D. snow

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8. Which of the following situations might adversely affect stopping sight distance around the inside of a curve?

A. building B. trees C. rock cut D. widened shoulder 9. What is the minimum acceptable radius for each of the following road classifications?

A. RLU 80 B. RAD 100 C. UAU 60 D. UFD 100

The following questions are not typical exam questions; they are included for information only. 10. Two horizontal centreline tangents on a proposed UFD 110 highway intersect at 15°. A 700 m

radius curve has been proposed between the two tangents. Is this radius acceptable? Calculate the curve data required to lay out this curve using a PI chainage of 0+946.627. 11. Two horizontal centreline tangents of a four-lane RAU 110 road, with bearings of 118.3° and

88.3° respectively, intersect at a PI with a chainage of 48+791.066. A 400 m curve has been recommended.

A. Is the suggested radius satisfactory? B. Calculate the curve data for this curve (using the suggested radius if it was found to be

satisfactory, or using the minimum acceptable radius if the suggested radius was not found to be adequate).

C. Find the minimum clearance required from the centre of the nearside lane around the

inside of the curve. 12. The sketch on the page following shows the horizontal alignment of the centreline of an existing

two-lane highway that is to be upgraded to a four-lane RAD 100. From the information given: A. Calculate the existing radius. B. Calculate the station of the EC. C. Is the existing alignment satisfactory for the upgraded classification? If not, what speed

restriction do you suggest might be posted? D. How far back will the existing bush have to be cleared to provide adequate stopping sight

distance around the inside of the curve for a driver travelling at 95 km/h? (Use 3.75 m lane width.)

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MODULE 8 Horizontal Alignment

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