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Geometric Design (Transportation Engineering)
This document provides an outline and content for a course on geometric design (CEE320). It covers key topics like vertical and horizontal alignment, including fundamentals of vertical curves, crest and sag vertical curves, horizontal curve fundamentals, superelevation, and stopping sight distance. Examples are provided to demonstrate calculations for elements like vertical curve length. Design controls and standards from AASHTO and WSDOT are presented for elements such as superelevation rates and side friction factors. The course content is intended to teach students the critical concepts and calculations for geometric roadway design.
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Geometric Design (Transportation Engineering)
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- 2. CEE320 Winter2006 Outline 1. Concepts 2. VerticalAlignment a. Fundamentals b. Crest Vertical Curves c. Sag Vertical Curves d. Examples 3. Horizontal Alignment a. Fundamentals b. Superelevation 4. Other Non-Testable Stuff
- 3. CEE320 Winter2006 Concepts • Alignment isa 3D problem broken down into two 2D problems – Horizontal Alignment (plan view) – Vertical Alignment (profile view) • Stationing – Along horizontal alignment – 12+00 = 1,200 ft. Piilani Highway on Maui
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- 7. CEE320 Winter2006 Vertical Alignment • Objective: –Determine elevation to ensure • Proper drainage • Acceptable level of safety • Primary challenge – Transition between two grades – Vertical curves G1 G2 G1 G2 Crest Vertical Curve Sag Vertical Curve
- 8. CEE320 Winter2006 Vertical Curve Fundamentals •Parabolic function – Constant rate of change of slope – Implies equal curve tangents • y is the roadway elevation x stations (or feet) from the beginning of the curve cbxaxy ++= 2
- 9. CEE320 Winter2006 Vertical Curve Fundamentals G1 G2 PVI PVT PVC L L/2 δ cbxaxy++= 2 x Choose Either: • G1, G2 in decimal form, L in feet • G1, G2 in percent, L in stations
- 10. CEE320 Winter2006 Relationships Choose Either: • G1,G2 in decimal form, L in feet • G1, G2 in percent, L in stations G1 G2 PVI PVT PVC L L/2 δ x 1and0:PVCAt the Gb dx dY x === cYx == and0:PVCAt the L GG a L GG a dx Yd 2 2:Anywhere 1212 2 2 − =⇒ − ==
- 11. CEE320 Winter2006 Example A 400 ft.equal tangent crest vertical curve has a PVC station of 100+00 at 59 ft. elevation. The initial grade is 2.0 percent and the final grade is -4.5 percent. Determine the elevation and stationing of PVI, PVT, and the high point of the curve. G1 =2.0% G 2 = - 4.5% PVI PVT PVC: STA 100+00 EL 59 ft.
- 12. G1 =2.0% G 2 = -4.5% PVI PVT PVC:STA 100+00 EL 59 ft.
- 13. CEE320 Winter2006 Other Properties G1 G2 PVI PVT PVC x Ym Yf Y 2 200 x L A Y = 800 AL Ym= 200 AL Yf = 21 GGA −= •G1, G2 in percent •L in feet
- 14. CEE320 Winter2006 Other Properties • K-Value(defines vertical curvature) – The number of horizontal feet needed for a 1% change in slope A L K = 1./ GKxptlowhigh =⇒
- 15. CEE320 Winter2006 Crest Vertical Curves G1 G2 PVI PVTPVC h2 h1 L SSD () ( )2 21 2 22100 hh SSDA L + = ( ) ( ) A hh SSDL 2 21200 2 + −= For SSD < L For SSD > L Line of Sight
- 16. CEE320 Winter2006 Crest Vertical Curves •Assumptions for design – h1 = driver’s eye height = 3.5 ft. – h2 = tail light height = 2.0 ft. • Simplified Equations ( ) 2158 2 SSDA L = ( ) A SSDL 2158 2 −= For SSD < L For SSD > L
- 17. CEE320 Winter2006 Crest Vertical Curves •Assuming L > SSD… 2158 2 SSD K =
- 18. CEE320 Winter2006 Design Controls forCrest Vertical Curves from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
- 19. CEE320 Winter2006 Design Controls forCrest Vertical Curves fromAASHTO’sAPolicyonGeometricDesignofHighwaysandStreets2001
- 20. CEE320 Winter2006 Sag Vertical Curves G1 G2 PVI PVTPVC h2=0h1 L LightBeam Distance (SSD) ( ) ( )βtan200 1 2 Sh SSDA L + = ( ) ( )( ) A SSDh SSDL βtan200 2 1 + −= For SSD < L For SSD > L headlight beam (diverging from LOS by β degrees)
- 21. CEE320 Winter2006 Sag Vertical Curves •Assumptions for design – h1 = headlight height = 2.0 ft. – β = 1 degree • Simplified Equations ( ) ( )SSD SSDA L 5.3400 2 + = ( ) ( ) + −= A SSD SSDL 5.3400 2 For SSD < L For SSD > L
- 22. CEE320 Winter2006 Sag Vertical Curves •Assuming L > SSD… SSD SSD K 5.3400 2 + =
- 23. CEE320 Winter2006 Design Controls forSag Vertical Curves from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
- 24. CEE320 Winter2006 Design Controls forSag Vertical Curves fromAASHTO’sAPolicyonGeometricDesignofHighwaysandStreets2001
- 25. CEE320 Winter2006 Example 1 A caris traveling at 30 mph in the country at night on a wet road through a 150 ft. long sag vertical curve. The entering grade is -2.4 percent and the exiting grade is 4.0 percent. A tree has fallen across the road at approximately the PVT. Assuming the driver cannot see the tree until it is lit by her headlights, is it reasonable to expect the driver to be able to stop before hitting the tree?
- 26. CEE320 Winter2006 Example 2 Similar toExample 1 but for a crest curve. A car is traveling at 30 mph in the country at night on a wet road through a 150 ft. long crest vertical curve. The entering grade is 3.0 percent and the exiting grade is -3.4 percent. A tree has fallen across the road at approximately the PVT. Is it reasonable to expect the driver to be able to stop before hitting the tree?
- 27. CEE320 Winter2006 Example 3 A roadwayis being designed using a 45 mph design speed. One section of the roadway must go up and over a small hill with an entering grade of 3.2 percent and an exiting grade of -2.0 percent. How long must the vertical curve be?
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- 29. CEE320 Winter2006 Horizontal Alignment • Objective: –Geometry of directional transition to ensure: • Safety • Comfort • Primary challenge – Transition between two directions – Horizontal curves • Fundamentals – Circular curves – Superelevation Δ
- 30. CEE320 Winter2006 Horizontal Curve Fundamentals R T PCPT PI M E R Δ Δ/2Δ/2 Δ/2 RR D π π 000,18 180 100 = = 2 tan ∆ = RT D RL ∆ =∆= 100 180 π L
- 31. CEE320 Winter2006 Horizontal Curve Fundamentals − ∆ =1 2cos 1 RE ∆ −= 2 cos1RM R T PC PT PI M E R Δ Δ/2Δ/2 Δ/2 L
- 32. CEE320 Winter2006 Example 4 A horizontalcurve is designed with a 1500 ft. radius. The tangent length is 400 ft. and the PT station is 20+00. What are the PI and PT stations?
- 33. CEE320 Winter2006 Superelevation cpfp FFW=+ αααα cossincossin 22 vv s gR WV gR WV WfW = ++ α α Fcp Fcn Wp Wn Ff Ff α Fc W 1 ft e ≈ Rv
- 34. CEE320 Winter2006 Superelevation αααα cossincossin 22 vv s gR WV gR WV WfW = ++ ()αα tan1tan 2 s v s f gR V f −=+ ( )ef gR V fe s v s −=+ 1 2 ( )efg V R s v + = 2
- 35. CEE320 Winter2006 Selection of eand fs • Practical limits on superelevation (e) – Climate – Constructability – Adjacent land use • Side friction factor (fs) variations – Vehicle speed – Pavement texture – Tire condition
- 36. CEE320 Winter2006 Side Friction Factor fromAASHTO’s A Policy on Geometric Design of Highways and Streets 2004 New Graph
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- 38. CEE320 Winter2006 WSDOT Design SideFriction Factors fromthe2005WSDOTDesignManual,M22-01 New Table For Open Highways and Ramps
- 39. CEE320 Winter2006 WSDOT Design SideFriction Factors fromthe2005WSDOTDesignManual,M22-01 For Low-Speed Urban Managed Access Highways New Graph
- 40. CEE320 Winter2006 Design Superelevation Rates- AASHTO from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004 New Graph
- 41. CEE320 Winter2006 Design Superelevation Rates- WSDOT from the 2005 WSDOT Design Manual, M 22-01 emax = 8% New Graph
- 42. CEE320 Winter2006 Example 5 A sectionof SR 522 is being designed as a high-speed divided highway. The design speed is 70 mph. Using WSDOT standards, what is the minimum curve radius (as measured to the traveled vehicle path) for safe vehicle operation?
- 43. CEE320 Winter2006 Stopping Sight Distance Rv Δs Obstruction Ms() v s R SSD π 180 =∆ D RSSD s sv ∆ =∆= 100 180 π SSD −= v vs R SSD RM π 90 cos1 − = − v svv R MRR SSD 1 cos 90 π
- 44. CEE320 Winter2006 Supplemental Stuff • Crosssection • Superelevation Transition – Runoff – Tangent runout • Spiral curves • Extra width for curves FYI – NOT TESTABLE
- 45. CEE320 Winter2006 Cross Section FYI –NOT TESTABLE
- 46. CEE320 Winter2006 Superelevation Transition from the2001 Caltrans Highway Design Manual FYI – NOT TESTABLE
- 47. CEE320 Winter2006 Superelevation Transition from AASHTO’sA Policy on Geometric Design of Highways and Streets 2001 FYI – NOT TESTABLE
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- 49. CEE320 Winter2006 Superelevation Runoff -WSDOT from the 2005 WSDOT Design Manual, M 22-01 FYI – NOT TESTABLE New Graph
- 50. CEE320 Winter2006 Spiral Curves No Spiral Spiral fromAASHTO’s A Policy on Geometric Design of Highways and Streets 2001 FYI – NOT TESTABLE
- 51. CEE320 Winter2006 No Spiral FYI –NOT TESTABLE
- 52. CEE320 Winter2006 Spiral Curves • WSDOTno longer uses spiral curves • Involve complex geometry • Require more surveying • Are somewhat empirical • If used, superelevation transition should occur entirely within spiral FYI – NOT TESTABLE
- 53. CEE320 Winter2006 Desirable Spiral Lengths fromAASHTO’s A Policy on Geometric Design of Highways and Streets 2001 FYI – NOT TESTABLE
- 54. CEE320 Winter2006 Operating vs. DesignSpeed 85th Percentile Speed vs. Inferred Design Speed for 138 Rural Two-Lane Highway Horizontal Curves 85th Percentile Speed vs. Inferred Design Speed for Rural Two-Lane Highway Limited Sight Distance Crest Vertical Curves FYI – NOT TESTABLE
- 55. CEE320 Winter2006 Primary References • Mannering,F.L.; Kilareski, W.P. and Washburn, S.S. (2005). Principles of Highway Engineering and Traffic Analysis, Third Edition. Chapter 3 • American Association of State Highway and Transportation Officials (AASHTO). (2001). A Policy on Geometric Design of Highways and Streets, Fourth Edition. Washington, D.C.
Editor's Notes
- #5 Therefore, roads will almost always be a bit longer than their stationing because of the vertical alignment Draw in stationing on each of these curves and explain it
- #6 Typical set of road plans – one page only
- #13 400 ft. vertical curve, therefore: PVI is at STA 102+00 and PVT is at STA 104+00 Elevation of the PVI is 59’ + 0.02(200) = 63 ft. Elevation of the PVT is 63’ – 0.045(200) = 54 ft. High point elevation requires figuring out the equation for a vertical curve At x = 0, y = c =&gt; c=59 ft. At x = 0, dY/dx = b = G1 = +2.0% a = (G2 – G1)/2L = (-4.5 – 2)/(2(4)) = - 0.8125 y = -0.8125x2 + 2x + 59 High point is where dy/dx = 0 dy/dx = -1.625x + 2 = 0 x = 1.23 stations Find elevation at x = 1.23 stations y = -0.8125(1.23)2 + 2(1.23) + 59 y = 60.23 ft
- #14 Last slide we found x = 1.23 stations
- #15 G is in percent, x is in feet G is in decimal, x is in stations
- #17 Minimum lengths are about 100 to 300 ft. Another way to get min length is 3 x (design speed in mph)
- #22 What can you do if you need a shorter sag vertical curve than calculated? Provide fixed-source street lighting Minimum lengths are about 100 to 300 ft. Another way to get min length is 3 x design speed in mph
- #26 Assume that S&gt;L (it usually is not but for example we’ll do it this way), therefore S = 146.23 ft. which is less than L Must use S&lt;L equation, it’s a quadratic with roots of 146.17 ft and -64.14 ft. The driver will see the tree when it is 146.17 feet in front of her. Available SSD is 146.17 ft. Required SSD = (1.47 x 30)2/2(32.2)(0.35 + 0) + 2.5(1.47 x 30) = 196.53 ft. Therefore, she’s not going to stop in time. OR L/A = K = 150/6.4 = 23.43, which is less than the required K of 37 for a 30 mph design speed Stopping sight distance on level ground at 30 mph is approximately 200 ft.
- #27 Assume that S&gt;L (it usually is), therefore SSD = 243.59 ft. which is greater than L The driver will see the tree when it is 243.59 feet in front of her. Available SSD = 243.59 ft. Required SSD = (1.47 x 30)2/2(32.2)(0.35 + 0) + 2.5(1.47 x 30) = 196.53 ft. Therefore, she will be able to stop in time. OR L/A = K = 150/6.4 = 23.43, which is greater than the required K of 19 for a 30 mph design speed on a crest vertical curve Stopping sight distance on level ground at 30 mph is approximately 200 ft.
- #28 For 45 mph we get K=61, therefore L = KA = (61)(5.2) = 317.2 ft.
- #29 Trinity Road between Sonoma and Napa valleys
- #31 D = degree of curvature (angle subtended by a 100’ arc)
- #33 Since we know R and T we can use T = Rtan(delta/2) to get delta = 29.86 degrees D = 5729.6/R. Therefore D = 3.82 L = 100(delta)/D = 100(29.86)/3.82 = 781 ft. PC = PT – PI = 2000 – 781 = 12+18.2 PI = PC +T = 12+18.2 + 400 = 16+18.2. Note: cannot find PI by subtracting T from PT!
- #35 Divide both sides by Wcos(α) Assume fse is small and can be neglected – it is the normal component of centripetal acceleration
- #36 The maximum side friction factor is the point at which the tires begin to skid Design values of fs are chosen somewhat below this maximum value so there is a margin of safety
- #41 There is a different curve for each superelevation rate, this one is for 8%
- #43 For the minimum curve radius we want the maximum superelevation rate. WSDOT max e = 0.10 For 70 mph, WSDOT f = 0.10 Rv = V2/g(fs+e) = (70 x 1.47)2/32.2(0.10 + 0.10) = 1644.16 ft. This is the radius to the center of the inside lane. Depending upon the number of lanes and perhaps a center divider, the actual centerline radius will be different.
- #44 Basically it’s figuring out L and M from the normal equations
- #48 Can also rotate about inside/outside axis
- #51 Ease driver into the curve Think of how the steering wheel works, it’s a change from zero angle to the angle of the turn in a finite amount of time This can result in lane wander Often make lanes bigger in turns to accommodate for this
- #52 I-90 past North Bend looking east
- #55 According to NCHRP Report