Transportation Route Alignments

7.1 Introduction CHAPTER 7 Transportation Route Alignments 7.1 Introduction The famous Silk Road has been one of the best known ancient trade routes. This route consisted of many sections, links, and alternates. The Silk Road was built as a caravan route connecting China and Western Europe. The route was more than 6,000 kilometers long. First Egyptians, then Romans, and other westerners established trade relationships with China. Far East products (primarily silk, paper, and spices) were transported along this route for many centuries. Caravan rest areas were built at distances 30-40 km. from one another. In this way, it was possible to travel between two caravan stations in eight to ten hours on foot. The Silk Road, as well as many other ancient and modern roads have ben highly influenced by terrain, and they followed the most convenient path, i.e. the path of least resistance. Clearly, the ideal alignment would be a great circle from the point of origin to the point of destination. The most convenient path is usually that one that follows the natural alignment of the countryside. A route alignment represents a defined three-dimensional path. It is usual and convenient to present route alignments in two dimensions: horizontal and vertical alignments. Vertical alignment (also called profile grade line) is composed of a sequence of straight line profiles connected by vertical parabolic curves. In some cases the profile grade increases from a flat alignment. In some other cases, profile grade decreases from a flat alignment. A plus grade denotes the situation when profile grade is increasing from a flat alignment (Figure 7.1). The opposite situation is known as a minus grade (Figure 7.1). 7-1

7.1 Introduction G 1 + Tangent Tangent Curve Curve -G 2 Figure 7.1 Definition of Vertical Profiles (Plus Grade G 1 and Minus Grade G 2 ). Grades are expressed in [%] or in [m/m]. Horizontal alignment represents projection of the alignment onto xy plane (Figure 7.2). z (0,0,0) y Horizontal alignment x xy Figure 7.2 Horizontal Alignment. The position of a specific point on a highway is traditionally determined using concept of stations. A datum point on a highway alignment is specified. This initial point is designated station 0 + 000.000. The positions of all other points on the highway are calculated by measuring corresponding distances on a horizontal plane along the highway from the initial point. Each station contains one kilometer of highway alignment distance. For example, the point on a highway located 2345.6 m from the previously specified point, is designated station 2 + 345.600. The station 3 + 465.800 corresponds to the highway point located 3465.8 meters from the initial point. 7-2

7.2 Vertical Alignment et us create along horizontal alignment the surface orthogonal to the xy plane. et us also denote h the distance measured along the horizontal alignment. When we stretch this surface, it becomes flat. We call it hz plane. The vertical alignment represents the projection of the alignment onto hz plane (Figure 7.3). xy hz z hz Vertical alignment h Figure 7.3 Vertical Alignment hz. The specification of a route alignment depends on demand for transportation services expected between two points along the future route and should consider construction, operating and maintenance cost as well as safety. 7.2 Vertical Alignment Vertical alignment is composed of straight sections that are connected by vertical curves. These straight sections are called grades or tangents. Once a vertical alignment is designed, the elevations of all the points along the highway are established. Vertical curve are typically classified as crest curves or sag curves (Figure 7.4). The most important points on the vertical curve are: the Vertical Point of Curvature VPC, the Vertical Point of Intersection VPI, and the Vertical Point of Tangency VPT (Figure 7.5). The VPC is the initial point of the curve, while the Vertical Point of Tangency VPT is the final point of the vertical curve. The vertical curve length is indicated in the Figure 7.5. The grades of the initial and final tangent segments are denoted by G 1, and G 2, respectively. The point VPI represents the point of intersection of the initial and final grade. In the most cases, the first half of the vertical curve length is located before the VPI, and the second half after the VPI (Figure 7.5). Such curves are called equal tangent vertical curves. 7-3

7.2 Vertical Alignment Crest +G 2 Crest -G1 Crest +G 1 - G 2 +G 1 -G 2 Sag +G 1 +G 2 Sag -G 1 +G 2 -G 1 Sag -G 2 Figure 7.4 Types of Vertical Curves (Crest and Sag Vertical Curves). x VPI parabolic curve VPC PVT +G 1 -G2 /2 /2 Figure 7.5 Elements of a Vertical Curve. 7-4

7.2 Vertical Alignment is, Profile grade tangents are connected by the parabolic curve. the basic definition of a parabola where: y a! x 2 + b! x + c (7.1) x - horizontal distance to point on curve measured from the VPC y - elevation of point on curve located at distance x from VPC By the definition x 0 in the point of the vertical curve VPC. This means that the elevation in VPC is: y a! 0 2 + b! 0 + c c We see that the coefficient c represents elevation in PVC. et us determine coefficients a and b. The first derivative of Equation (7.1) is: dy ---- 2! a! x + b dx (7.2) At the initial point VPC, the slope equals G 1, and x 0, i.e.: dy ---- 2! a! x + b 2! a! 0 + b b dx b G 1 (7.3) equals: We conclude that the coefficient b equals to the initial slope G 1. The second derivative dy 2 ------- 2! a dx 2 (7.4) The second derivative of Equation 7.1 represents the rate of slope change, i.e.: dy 2 ------- dx 2 G 2 G ------------------ 1 (7.5) 7-5

7.2 Vertical Alignment This implies, 2! a G 2 G ------------------ 1 We calculate the coefficient a as: (7.6) a G 2 G ------------------ 1 2! (7.7) The grades in Equation 7.7 are expressed in [m/m]. et A the difference in grades ( A G 2 G 1 ). The difference in grades A is positive for sag curves and negative for crest curves (Equation 7.5). Table 7.1 shows the maximum recommended grades for urban and rural freeways. Table 7.1 Maximum Grades [%] for Urban and Rural Freeways (Source AASHTO, 1990) Design speed [m.p.h.] Design speed [m.p.h.] Design speed [m.p.h.] Type of terrain 50 60 70 evel 4 3 3 Rolling 5 4 4 Mountainous 6 6 5 Example 7.1 The length of the equal tangent vertical curve (Table 7.6 ) equals 300 [m]. The initial and the the final grades are known to be: G 1 2.5 [%] ; G 2 1.5 [%] The grades intersect at the station 3 + 650 and at an elevation of 210.500 m. (a) Determine the station and the elevation of the VPC and VPT. (b) Calculate the elevation of the point on the curve 100 meters from the VPC.(c) Determine the station and the elevation of the highest point on the curve. 7-6

7.2 Vertical Alignment VPI x VPC VPT g 1 2.5 [%] /2 /2 g 2 1.5 [%] 300[ m ]] Figure 7.6 Vertical Curve for Problem 1. Solution: (a) The grades of the initial G 1 and final tangent G 2 are: 2.5 G 1 -------- 0.025 m 100 m --- 1.5 G 2 -------- 0.015 m 100 m --- The station of the VPC and VPT. are: VPCSta 300 VPISta -- 3 + 650 -------- 3 + 650 150 3 + 500 2 2 VPTSta VPCSta + 3 + 500 + 300 3 + 800 The elevation of the VPC and VPT. are: E VPC E VPI G 1! -- " $ 2# % 210.500 0.025! 300 -------- " $ 2 # % 210.500 3.75 206.75[ m] 7-7

7.2 Vertical Alignment E VPT E VPI G 2! -- " $ 2# % 210.500 0.015! 300 -------- " $ 2 # % 210.500 2.25 208.25[ m] (b) The elevation y of any point on the curve located at distance x from VPC is described by the parabola: y a! x 2 + b! x + c (7.8) After substituting the known values for the coefficients, we get: y G 2 G ------------------ 1! x 2 + G 2! 1! x + E VPC (7.9) y ------------------------------------ 0.015 0.025! 2! 300 x 2 + 0.025! x + 206.75 y 0.000067! x 2 + 0.025! x + 206.75 For x 100 we get: y 0.000067! 100 2 + 0.025! 100 + 206.75 208.58 The elevation of the point equals 208.58 [m]. (c) Grades are opposite in sign. This means that the highest point can be estimated by setting the first derivative the equation of the parabola to zero, i.e.: dy ---- 2! a! x + b 0 dx 2! ( 0.000067)! x + 0.025 0 x 185.57[ m] The station of the highest point is: 7-8

7.2 Vertical Alignment HighSta VPCSta + 185.57 3 + 500 + 185.57 3 + 685.57 The elevation of the highest point is: y 0.000067! x 2 + 0.025! x + 206.75 For x 185.57 we get: y 0.000067! 185.57 2 + 0.025! 185.57 + 206.75 209.08[ m] Example 7.2 Offset Y of curve from initial grade line (Figure 7.7) is especially important parameter in vertical curve design. We consider again the curve from the previous example. VPI Offsets x Y VPC VPT G 1 2.5[%] /2 /2 G 2 1.5[%] 300[ m ]] Figure 7.7 Vertical Curve for Problem 7.2. The length of the equal tangent vertical curve (Figure 7.7) is 300 [m]. The initial and the final grades are: G 1 2.5 [%] ; G 2 1.5 [%] The grades intersect at station 3 + 650 and at an elevation of 210.500 m. (a) Determine the offset at distance 100 [m] from the VPC. (b) Determine the offset at the end of vertical curve. Solution: 7-9

7.2 Vertical Alignment is: The coefficient c represents elevation in VPC (Figure 7.8). The equation of the initial tangent y k! x + c (7.10) where k is the coefficient to be determined. y VPI x Y VPC VPT c x Figure 7.8 Initial Tangent and Parabola. The equation of the parabola is: y a! x 2 + b! x + c (7.11) The first derivative is: dy ---- 2! a! x + b dx (7.12) The coordinates of the VPC are ( 0, c). The first derivative in this point is calculated to be: 2! a! 0 + b b We conclude that: k b (7.13) 7-10

7.2 Vertical Alignment The equation of the initial tangent is: y b! x + c The offset Y of curve from initial grade line equals: (7.14) Y ( b! x + c) ( a! x 2 + b! x + c) (7.15) Y a! x 2 (7.16) Y G 2 G ------------------ 1! 2! x 2 G 1 G ------------------ 2! x 2 2! (7.17) m where G 1 and G 2 are expressed in ---. m When calculating offsets, grades are expressed in [%]. Equation 7.17 becomes: Y G 1 G ------------------ 2! x 2 200! (7.18) The general offset formula for both sag and crest vertical curves are: Y G 1 G --------------------- 2! x 2 200! (7.19) The offset at a distance 100 [m] from the VPC is: Y 2.5 ( 1.5) -------------------------------! 200! 300 100 2 0.67[ m] The offset at the end of vertical curve is calculated to be: Y( 300) 2.5 ( 1.5) -------------------------------! 200! 300 300 2 6[ m] 7-11

7.3 Stopping Sight Distance Considerations in Vertical Curve Design 7.3 Stopping Sight Distance Considerations in Vertical Curve Design Vertical curve design requires consideration of the average stopping sight distances. Drivers should have a clear view of the road ahead to stop before and obstacle while driving on a vertical curve. This visible length of a highway ahead is called sight distance. et us assume for this discussion that the vehicle speed is the design speed. The minimum stopping sight distance SSD is the sum of two components: (a) the distance vehicle traveled from the time point the object is observed to point in time when brakes are applied; and (b) the distance traveled by vehicle from the point in time when the brakes are applied until the vehicle is fully stopped. The American Association of State Highway and Transportation Officials (AASHTO) has recommended the minimum sight distances. These distances are shown in Table 7.2. Table 7.2 Stopping Sight Distances on Wet Pavements (Source AASHTO, 1994). Design Speed [km/h] Assumed Speed for Condition [km/h] Brake Reaction Time [s] Brake Reaction Distance [m] Coefficient of friction Braking Distance on evel [m] Computed [m] Rounded for Design [m] 30 30-30 2.5 20.8-20.8 0.40 8.8-8.8 29.6-29.6 30-30 40 40-40 2.5 27.8-27.8 0.38 16.6-16.6 44.4-44.4 50-50 50 47-50 2.5 32.6-34.7 0.35 24.8-28.1 57.4-62.8 60-70 60 55-60 2.5 38.2-41.7 0.33 36.1-42.9 74.3-84.6 80-90 70 63-70 2.5 43.7-48.6 0.31 50.4-62.2 94.1-110.8 110-120 80 70-80 2.5 48.6-55.5 0.30 64.2-83.9 112.8-139.4 120-140 90 77-90 2.5 53.5-62.5 0.30 77.7-106.2 131.2-168.7 140-170 100 85-100 2.5 59.0-69.4 0.29 98.0-135.6 157.0-205.0 160-210 110 91-110 2.5 63.2-76.4 0.28 116.3-170.0 179.5-246.4 180-250 120 98-120 2.5 68.0-83.3 0.28 134.9-202.3 202.9-285.6 210-290 The distances shown in the table are calculated under the assumption of a worse case scenario. For example, it is assumed that the driver s perception/reaction time is 2.5 seconds. It is also assumed that the brakes are applied in a situation when the pavement is wet. 7-12

7.3 Stopping Sight Distance Considerations in Vertical Curve Design 7.4.1 Crest Vertical Curve Design et us note the crest vertical curve shown in Equation 7.9. et S the available sight distance. The length of the crest vertical curve is denoted by. The height of the eye of the average driver is denoted by h 1. The h 2 denotes the height of the object perceived by the driver. The formulae for the crest vertical curve length are presented in Table 3. The formulae are derived using mathematical properties of the parabola made up by two tangent segments. S VPI h1 VPC PVT h 2 Figure 7.9 Crest Vertical Curve: Stopping Sight Distance Analysis. Table 7.3 Formulas to Estimate the ength of a Crest Vertical Curve. Relation between S and ength of the Vertical Curve S & A! S 2 --------------------------------------------- 200! ( h 1 + h 2 ) 2 S > 2! S 200! ( h 1 + h 2 ) 2 --------------------------------------------- A Older AASHTO design guidelines assumed 1.08[ m] and h 2 0.15[ m]. Newer design guidelines have modified the value of h 2 to be 0.60 meters without apparent degradation in the safety of the design (AASHTO 2004). This new design rationale produces more cost-effective vertical designs without safety degradation. h 1 7-13

7.3 Stopping Sight Distance Considerations in Vertical Curve Design In order to calculate the minimum curve length we assume that the available sight distance S is the same as the stopping sight distance SSD. The minimum curve length equals: min A! SSD 2 A! SSD 2 min ------------------------------------------------------- ----------------------- SSD < 200! ( 1.08 + 0.15) 2 404 (7.20) min 404 2! SSD -------- SSD > A (7.21) If the new AASHTO design criteria is used instead (with h 2 0.60 meters), the resulting equations are: A! SSD 2 A! SSD 2 min ------------------------------------------------------- ----------------------- SSD < 200! ( 1.08 + 0.60) 2 658 min 658 2! SSD -------- SSD > A Stopping sight distances are shown in column 4 of Table 7.4. For example, for a design speed of 120 [km/h], the stopping sight distances range from[210-290] meters. The lower value (210 meters) is called minimum SSD. The higher value (290 meters) is known as desirable SSD. The usual assumption made by the highway designers is that SSD <. In this book we exclusively use the following formula to calculate : min A! SSD 2 min ----------------------- 404 (7.22) et K be the following expression: K SSD 2 ------------ 404 (7.23) 7-14

7.3 Stopping Sight Distance Considerations in Vertical Curve Design The quantity K is called the rate of vertical curvature. The K values are given in the Table 7.4. Table 7.4 Design Controls for Crest Vertical Curves Based on Stopping Sight Distance (Source AASHTO, 1994). Design Speed [km/h] Assumed Speed for Condition [km/h] Coefficient of friction Stopping Sight Distance Rounded for Design [m] Rate of Vertical Curvature K [length[m] per % of A] Computed Rate of Vertical Curvature K [length[m] per % of A] Rounded for Design 30 30-30 0.40 30-30 2.17-2.17 3-3 40 40-40 0.38 50-50 4.88-4.88 5-5 50 47-50 0.35 60-70 8.16-9.76 9-10 60 55-60 0.33 80-90 13.66-17.72 14-18 70 63-70 0.31 110-120 21.92-30.39 22-31 80 70-80 0.30 120-140 31.49-48.10 32-49 90 77-90 0.30 140-170 42.61-70.44 43-71 100 85-100 0.29 160-210 61.01-104.02 62-105 110 91-110 0.28 180-250 79.75-150.28 80-151 120 98-120 0.28 210-290 101.90-201.90 102-202 The length min can be calculated using the following equation: min K! A (7.24) The rate of vertical curvature K equals: K ---------- min A (7.25) The units for min are meters, while the units for A are %. This means that the units of K are [length[m] per % of A]. 7-15

7.3 Stopping Sight Distance Considerations in Vertical Curve Design Example 7.3 The initial and the final grade of an equal tangent vertical curve are: G 1 2.5 [%] ; G 2 1.5 [%] The highway designers are considering design speeds ranging from 100 [km/h] to 120 [km/h]. Calculate corresponding minimum length of the vertical curve that satisfies the minimum stopping sight distance. Solution: The grades are: G 1 2.5 [%] G 2 1.5 [%] The difference in grades A is: A G 2 G 1 1.5 2.5 4 A 4 4 The min is: min K! A 4! K The minimum curve length that satisfy minimum stopping sight distance are: For a design speed of 100 [km/h]: min 4! K 4! 62 248[ m] For a design speed of 110 [km/h]: min 4! K 4! 80 320[ m] 7-16

7.3 Stopping Sight Distance Considerations in Vertical Curve Design For a design speed of 120 [km/h]: min 4! K 4! 102 408[ m] In some situations there is also a need to design vertical curves allowing a vehicle to overtake another one. Passing sight distance is especially important in the case of two-lane highways. The classical AASHTO design guidelines assumed 1.08[ m] and h 2 1.3[ m] (The height of the vehicle is assumed to be 1.3 meters). Recent trends in vehicle technology has dictated a reduction in the object height used for passing sight distance (PSD). The new AASHTO standard recommends 1.08[ m] instead of the original 1.3 meters recognizing that smaller vehicles h 2 populate the market today. In order to calculate the minimum curve length we assume that the available sight distance S equals the passing sight distance PSD. The minimum curve length is then: h 1 min A! PSD 2 A! PSD 2 min ---------------------------------------------------- ------------------------ PSD < 200! ( 1.08 + 1.3) 2 946 (7.26) min 946 2! PSD -------- PSD > A (7.27) If new AASHTO standards are used the following expressions are derived: A! PSD 2 A! PSD 2 min ------------------------------------------------------- ------------------------ PSD < 200! ( 1.08 + 1.08) 2 864 min 864 2! PSD -------- PSD > A Passing sight distances recommended by the AASHTO are shown in the Table 7.5. 7-17

7.3 Stopping Sight Distance Considerations in Vertical Curve Design Table 7.5 Design Controls for Crest Vertical Curves Based on Passing Sight Distance. Design Speed [km/h] Minimum Passing Sight Distance for Design [m] Rate of Vertical Curvature K [length[m] per % of A] Rounded for Design 30 217 50 40 285 90 50 345 130 60 407 180 70 482 250 80 541 310 90 605 390 100 670 480 110 728 570 120 792 670 7.5.2 Sag Vertical Curve Design Sight distance on a sag vertical curve is not limited during a day. H S VPC ' PVT VPI Figure 7.10 Sag Vertical Curve: Stopping Sight Distance Analysis 7-18

7.3 Stopping Sight Distance Considerations in Vertical Curve Design Driving conditions at night are the key factor that influences sag vertical curve design. The highway length lighted by the vehicle headlights depends on headlight height above the highway H, and on the angle ' between the headlight beam and the car horizontal plane (Equation 7.10). It has been shown that the minimum curve length could be calculated using the relations shown in the Table 7.6 9 (AASHTO, 2004). Table 7.6 ength of the Sag Vertical Curves. Relation between S and ength of the Vertical Curve S & A! S 2 ---------------------------------------------- 200! ( h + S! tan' ) S > 2! S 200! ( h + S! tan' ) ---------------------------------------------- A AASHTO design guidelines assumes h 0.6[ m] and ' 1 [ ]. In order to calculate the minimum curve length we assume that the available sight distance S equals the stopping min sight distance SSD. The minimum curve length equals: A! SSD 2 A! SSD 2 A! SSD 2 ---------------------------------------------- ------------------------------------------------------------- ---------------------------------------- SSD & 200! ( h + S! tan' ) 200! ( 0.6 + SSD! tan1 ) 120 + 3.49! SSD (7.28) 200! ( h + SSD! tan' ) 120 + 3.49! SSD 2! SSD ------------------------------------------------------ 2! SSD ---------------------------------------- SSD > A A (7.29) The usual assumption made by the highway designers is that the following formula to calculate : min SSD <. In this book we use 7-19

7.3 Stopping Sight Distance Considerations in Vertical Curve Design A! SSD 2 min ---------------------------------------- 120 + 3.49! SSD (7.30) et K be: K SSD 2 ---------------------------------------- 120 + 3.49! SSD (7.31) The K values are given in the Table 7.7. Table 7.7 Design Controls for Sag Vertical Curves Based on Stopping Sight Distance (Source AASHTO, 1994). Design Speed [km/h] Assumed Speed for Condition [km/h] Coefficient of friction Stopping Sight Distance for Design [m] Stopping Sight Distance Rounded for Design [m] Rate of Vertical Curvature [length[m] per % of A] Computed Rate of Vertical Curvature [length[m] per % of A] Rounded for Design 30 30-30 0.40 29.6-29.6 30-30 3.88-3.88 4-4 40 40-40 0.38 44.4-44.4 50-50 7.11-7.11 8-8 50 47-50 0.35 57.4-62.8 60-70 10.20-11.54 11-12 60 55-60 0.33 74.3-84.6 80-90 14.45-17.12 15-18 70 63-70 0.31 94.1-110.8 110-120 19.62-24.08 20-25 80 70-80 0.30 112.8-139.4 120-140 24.62-31.86 25-32 90 77-90 0.30 131.2-168.7 140-170 29.62-39.95 30-40 100 85-100 0.29 157.0-205.0 160-210 36.71-50.06 37-51 110 91-110 0.28 179.5-246.4 180-250 42.95-61.68 43-62 120 98-120 0.28 202.9-285.6 210-290 49.47-72.72 50-73 Example 7.4 The highway alignment should follow a crest vertical curve, then a constant grade section, and finally a sag vertical curve (Figure 7.11). The starting and ending alignment grades are 0 [%]. The elevation and the stationing at point A are: 7-20

7.3 Stopping Sight Distance Considerations in Vertical Curve Design E A 225[ m] ; ASta 3 + 120.000 The constant grade is G 1.5 [%]. The length of the constant grade section is planned to be 150 [m]. The design speed is 110 [km/h]. Determine the stationing and the elevations at points B, C, and D in Figure 7.11. A B Constant grade section c C D s Figure 7.11 Vertical Alignment Geometry for Problem 7.4. Solution: The grades are: Constant grade: G 1.5 [%] Crest vertical curve (index c): G 1 c 0 G 2 c G 1.5 [%] A c 1.5 0 1.5 [%] G 2 c G 1 c 7-21

7.3 Stopping Sight Distance Considerations in Vertical Curve Design Sag vertical curve (index s): G 1 s G 1.5 [%] G 2 s 0 A s min 0 ( 1.5) 1.5 [%] is: G 2 s G 1 s min K! A know: The rates of the vertical curvature are shown in Table 7.7 For a design speed of 110 [km/h] we K c 80 ; K s 43 The minimum curve length that satisfies the minimum stopping sight distances are: min c! 1.5! 80 120[ m] A c K c min s! 1.5! 43 64.5[ m] A c K s In other words, the length of the vertical curves are: c 120[ m] s 64.5[ m] The stationing at point B is : BSta ASta + c 3 + 120.000 + 120 3 + 240.000 The offset Y of curve from initial grade line is: 7-22

7.3 Stopping Sight Distance Considerations in Vertical Curve Design Y G 1 G --------------------- 2! x 2 200! (7.32) In the case of the crest vertical curve, the offset in the point B : G Y 1c G 2c 2 G B --------------------------! 1c G 2c 1.5 200! c --------------------------! c 200 c --------! 120 0.9[ m] 200 The elevation at point B is: E B E B E A Y B 225 0.9 224.1[ m] The stationing of point C is: CSta BSta + 3 + 240.000 + 150 3 + 390.000 The constant grade G : E B E ------------------ C G (7.33) 224.1 E ------------------------- C 0.015 m 150 m --- The elevation of the point C: E C 221.85[ m] The stationing at point D is: DSta CSta + s 3 + 390.000 + 64.5 3 + 454.500 In the case of the sag vertical curve, the offset in the point C: G Y 1s G 2s 2 G C -------------------------! 1s G 2s 1.5 200! s -------------------------! s 200 s --------! 64.5 0.48375[ m] 200 The elevation at point D: 7-23

7.3 Stopping Sight Distance Considerations in Vertical Curve Design E D E C Y C 221.85 0.48375 221.36625[ m] Example 7.5 The highway alignment should follow crest vertical curve, constant grade section, and finally a sag vertical curve (Figure 7.12). The starting and ending alignment grades are 0 [%]. The elevation and the stationing of the points A and D are known to be: E A 240[ m] ; ASta 3 + 120.000 E D 225[ m] ; DSta 3 + 620.000 Points should be connected with crest and sag equal tangent vertical curves with a constant grade section in between (Figure 7.12). The design speed equals 70 [km/h]. Determine the grade of the constant section, and the lengths of the crest vertical curve, sag vertical curve, and the constant grade section. A Y B B C Y E A E D Y C c s D Figure 7.12 Geometry for Problem 7.5. Solution: The grades are: Constant grade: G [%] 7-24

7.3 Stopping Sight Distance Considerations in Vertical Curve Design Crest vertical curve (index c): G 1 c 0 G 2 c G [%] A c G 0 G G 2 c G 1 c Sag vertical curve (index s): [%] G 1 s G [%] G 2 s 0 A s 0 ( G) G G 2 s G 1 s [%] The difference in the elevation E A E D between points A and D (Figure 7.12) are: Y B + Y + Y C E A E D (7.34) Since: i.e.: Y -- G Y -- [%] G -------- m 100 m --- and: Y B A c A c 2 ------------------! 200! c --------! c 200 c G --------! 200 c (7.35) Y C A s A s 2 ------------------! 200! s --------! s 200 s G --------! 200 s (7.36) we get: 7-25

7.3 Stopping Sight Distance Considerations in Vertical Curve Design G --------! 200 c G! G + -------------- + --------! 100 200 s E A E D (7.37) Since: c K c! A c and s K s! A s, we have: G --------! K 200 c! A c G --------! K 200 c! G G! G + -------------- + --------! K 100 200 s! A s E A E D G! G + -------------- + --------! K 100 200 s! G E A E D (7.38) (7.39) The rates of the vertical curvature (Tables 7.6 and 7.7) for a design speed of 100[km/h] are: K c 22 ; K s 20 Our equation reads: 22 --------! G 2 200 G! 20 + -------------- + --------! G 2 240 225 100 200 22! G 2 + 2! G! + 20! G 2 3000 42! G 2 + 2! G! 3000 0 The total length of the curve is: c + + s DSta ASta ( DSta ASta) c s ( DSta ASta) ( K c! G ) ( K s! G ) [( 3 + 620.000) ( 3 + 120.000) ] ( 22! G ) ( 20! G ) 500 42! G Further substitution yields: 7-26

7.3 Stopping Sight Distance Considerations in Vertical Curve Design 42! G 2 + 2! G! ( 500 42! G ) 3000 0 A quadratic equation is: 42! G 2 + 1000! G 3000 0 The solutions are: G 1 3.52 and G 2 20.29 We must reject the solution curves would be: G 2 20.29. If we accepted this solution the length of the c K c! G 22! 20.29 446.38[ m] s K s! G 20! 20.29 405.8[ m] The total length of the vertical curves would be greater than the difference in stationing between points D and A, i.e.: c + 446.38 + 405.8 852.18 > 500 DSta ASta s The final solution is: G G 1 3.52 [%] The length of the curves are: c K c! G 22! 3.52 77.44[ m] s K s! G 20! 3.52 70.4[ m] 500 77.44 70.4 352.16[ m] 7-27

7.6 Horizontal Alignment and Horizontal Curves 7.6 Horizontal Alignment and Horizontal Curves Consider the simple horizontal curve shown in Figure 7.13. We denote PC as the point of the curve. This is the point where the horizontal curve starts. The curve ends at point PT that is called point of tangency. The central angle of the curve ( is equal to the deflection angle between the curve tangents (Figure 7.13). Point PI is called point of tangent intersection. et R the radius of the horizontal curve, the length of curve (from PC to PT ), T tangent length, E the external distance, and M the middle ordinate. T PI E ( Arc ength PC M PT ( -- 2 R ( -- 2 ( -- 2 R Figure 7.13 Simple horizontal Alignment Circular Curve. The following circular curve equations are obtained (Figure 7.13): T -- R ( tan-- 2 (7.40) i.e., the tangent length is calculated to be: T R (! tan-- 2 (7.41) The length of curve is: 7-28

7.6 Horizontal Alignment and Horizontal Curves 2! R! )! ( -------------------------- 360 R! )! ( ------------------ 180 (7.42) The quantities E and M can be calculated as: E ------------ R ( R R ------------ 1 * * $ ( 1 + + %! cos-- " cos-- # 2 2 (7.43) ( ( M R R! cos-- R! 1 cos-- 2 " $ 2# % (7.44) Example 7.6 The stationing of PC is 5 + 210.000. The curve radius is 650 [m]. The central angle is 30 [ ]. Determine the stationing of PI and PT. T PI Arc ength PC PT R 650 ( -- 15 2 Figure 7.14 Horizontal curve with the radius R 650 and the central angle ( 30 Solution: The tangent length is: 7-29

7.6 Horizontal Alignment and Horizontal Curves T ( 30 R! tan-- 650! tan----- 174.075 2 2 The stationing of PI is: PISta PCSta + T 5 + 210.000 + 0 + 174.075 5 + 384.075 The length of curve is: R! )! ( 650! )! 30 650! 3.14 ------------------ -------------------------- ----------------------- 340.167 180 180 6 The stationing of PT is: PTSta PCSta + 5 + 210.000 + 0 + 340.167 5 + 550.167 Different forces act on the vehicle as vehicle travels around a horizontal curve of constant radius. Basic analysis of the forces acting in a vehicle negotiating a horizontal curve shows the following relationship: R v 2 ---------------------- g! ( e + f) where: R - curve radius v - vehicle speed [m/s] g - gravitational constant e - rate of superelevation (the vertical rise - in meters - for every 100 meters of horizontal distance) f - coeeficient of side friction (7.45) AASHTO recommendations for minimum radii for limiting values of e and f are given in the Table 7.8. 7-30

7.6 Horizontal Alignment and Horizontal Curves Table 7.8 Minimum Radius for imiting Values of e and f (Source AASHTO, 1994) Design Speed [km/h] Maximum Maximum Total e f ( e + f) Calculated Radius [m] Rounded Radius [m] 30 0.04 0.17 0.21 33.7 35 40 0.04 0.17 0.21 60.0 60 50 0.04 0.16 0.20 98.4 100 60 0.04 0.15 0.19 149.2 150 70 0.04 0.14 0.18 214.3 215 80 0.04 0.14 0.18 280.0 280 90 0.04 0.13 0.17 375.2 375 100 0.04 0.12 0.16 492.1 490 110 0.04 0.11 0.15 635.2 635 120 0.04 0.09 0.13 872.2 870 30 0.06 0.17 0.23 30.8 30 40 0.06 0.17 0.23 54.8 55 50 0.06 0.16 0.22 89.5 90 60 0.06 0.15 0.21 135.0 135 70 0.06 0.14 0.20 192.9 195 80 0.06 0.14 0.20 252.0 250 90 0.06 0.13 0.19 335.7 335 100 0.06 0.12 0.18 437.4 435 110 0.06 0.11 0.17 560.4 560 120 0.06 0.09 0.15 755.9 755 30 0.08 0.17 0.25 28.3 30 40 0.08 0.17 0.25 50.4 50 50 0.08 0.16 0.24 82.0 80 60 0.08 0.15 0.23 123.2 125 7-31

7.6 Horizontal Alignment and Horizontal Curves Table 7.8 Minimum Radius for imiting Values of e and f (Source AASHTO, 1994) Design Speed [km/h] Maximum Maximum Total e f ( e + f) Calculated Radius [m] Rounded Radius [m] 70 0.08 0.14 0.22 175.4 175 80 0.08 0.14 0.22 229.1 230 90 0.08 0.13 0.21 303.7 305 100 0.08 0.12 0.20 393.7 395 110 0.08 0.11 0.19 501.5 500 120 0.08 0.09 0.17 667.0 665 30 0.10 0.17 0.27 26.2 25 40 0.10 0.17 0.27 46.7 45 50 0.10 0.16 0.26 75.7 75 60 0.10 0.15 0.25 113.4 115 70 0.10 0.14 0.24 160.8 160 80 0.10 0.14 0.24 210.0 210 90 0.10 0.13 0.23 277.3 275 100 0.10 0.12 0.22 357.9 360 110 0.10 0.11 0.21 453.7 455 120 0.10 0.09 0.19 596.8 595 Example 7.7 Calculate minimum radius for a design speed of 120 [km/h]. The superelevation rate is limited to 0.06 due to icing effects in winter. Solution: The minimum radius formula is: R v 2 ---------------------- g! ( e + f) The speed expressed in [m/s] is: 7-32

7.6 Horizontal Alignment and Horizontal Curves 120 km 1000 m ------ 120! -------------------- [ ] 33.33 m h 3600[ s] --- s At 120 km/hr AASHTO recommends using a side friction coefficient of 0.09. This value is read from Table 7.8. Therefore, the minimum radius is: R ( 33.33) 2 9.81 ----------------------------------------------! ( 0.06 + 0.09) 755.1 [ m ] Depending on a terrain, the length of horizontal curves need to be adjusted to allow drivers be able to see ahead and potentially avoid obstacles. In the context of horizontal curves, sight distance depends on the presence of obstruction (Figure 7.15). The stopping sight distance SSD is measured along the horizontal curve. SSD Object Eye M s Highway centerline ( s ine of sight R R v Figure 7.15 Horizontal Curve: Analysis of the Stopping Sight Distance and Middle Ordinate. The radius of the vehicle traveled path is denoted by curve, we can express the stopping sight distance SSD as: R v. Using formulae for the length of the 7-33

7.6 Horizontal Alignment and Horizontal Curves SSD R v! )! ( ----------------------- s 180 (7.46) where ( s is the angle that corresponds to the arc that is equal to the SSD length (Figure 7.15). (7.47) This angle is: 180! SSD ( s ----------------------- R v! ) (7.48) The value of the midle ordinate that can satisfy SSD is: M s ( M s R v! 1 cos-- " $ 2# % (7.49) After substitution from equation (7.48) into equation (7.49), we get: 180! SSD ----------------------- ( M s R v! 1 cos-- " $ 2# % R R v 1 v! ) * * $ cos------------------------ + + %! * 2 + " # (7.50) 90! SSD M s R v! 1 " $ cos-------------------- R v! ) # % (7.51) M s 90! SSD ------ 1 cos -------------------- " $ R v! ) # % R v (7.52) M s 90! SSD cos -------------------- " $ R v! ) # % 1 ------ R v (7.53) M s 90! SSD -------------------- acos 1 ------ R v! ) " $ # % R v (7.54) 7-34

7.6 Horizontal Alignment and Horizontal Curves The stopping sight distance SSD is then, SSD R v! ) M ------------! acos 1 ------ s 90 " $ # % R v (7.55) 7-35

7.6 Horizontal Alignment and Horizontal Curves 7-36