ENGI 3703 Surveying and Geomatics

Transcription

1 Horizontal Curves (Chapter 24) We ll jump ahead a little today to support the last field school activity, Lab 6 - Horizontal Curve Layout. Today we ll define i) the properties of a horizontal curve and its geometry, ii) one method by which we layout such curves in the field. Lect 4 - Sept 17/07 Slide 1 of 10

2 Need for Horizontal Curves Horizontal curves are often used in highway construction to smooth transitions between straight segments of a route or to avoid route obstacles such as waterways and excess cuts or fills. There are many types of curves. However, circular curves with a constant radius are very common. Segment 1 Horizontal Curve Segment 2 Horizontal curves help change from one tangent to another. Horizontal curves are circular to minimize steering effort. Curves need to be long enough to avoid unsafe or uncomfortable conditions (minimum radius for highway standards ~350 m or 5 o degree of curvature). Additional features can help reduce the driving effort Superelevation Transition (or spiral) curves, which slowly transition from an infinite radius (a tangent) to the radius of the circular curve. In practice, drivers define their own transitions into circular curves. Lect 4 - Sept 17/07 Slide 2 of 10

3 Types of Horizontal Curves Horizontal Curves come in a variety of flavors. Simple Curve Compound Curve r spiral everse Curve Easement or Transitional Curve spiral Lect 4 - Sept 17/07 Slide 3 of 10

4 Degree of Curvature and Curve adius (Section 24.2) Both degree of curvature and radius of curvature are equivalent and both define the severity of a curve or the angle subtended by an arc of a given length. Minimizing the the degree of curvature and maximizing the radius of a curve both produce long gradual curves. Metric Design of roadways tend to specify the curve radius while English unit design tends to specify the degree of curvature or the angle at the center of a circular arc subtending a i) chord length of 100ft or ii) arc length of 100ft. AC DEFINITION Degree of Curve Central angle subtended by a circular AC of 100 ft (highways) COD DEFINITION Degree of Curve Central angle subtended by a circular CHOD of 100 ft (railways) Note: can be directly calculated from D Eg. adius of 350m is approx. equal to a degree of curvature of 5 o ft m / ft 350m = 4 o Lect 4 - Sept 17/07 Slide 4 of 10

5 Elements of a Horizontal Curve (Section 24.3) PI = Point of Intersection of tangent lines PC = Point of Curvature also called the beginning of curve (BC) PT = Point of Tangency also called the end of curve (EC) L = Length of Curve LC = Length of Long Chord M = Middle Ordinate from LC mid-point of mid-point of curve E = External Distance from PI to mid-point of cord. T = Tangent Distance from PI to PC (or PT). I = Intersection Angle =adius D=DOC=Degree of curve POC = Point on Curve POT = Point on Tangent Lect 4 - Sept 17/07 Slide 5 of 10

6 Horizontal Curve Formula (Section 24.3) T = *tan I 2 L = I (I in rads) L E = T tan I 4 M = E cos I 2 LC = 2 *sin I 2 Problem: Given I from route alignment and from highway design criteria determine T, L, M, E PC, PT and chord lengths and deflection angles for full station POCs. Lect 4 - Sept 17/07 Slide 6 of 10

7 Derivation of M & E M = *cos I 2 L cos I = 2 E + or 1 E = cos I 1 2 Lect 4 - Sept 17/07 Slide 7 of 10

8 Stationing on Horizontal Curves (Section 24.2) (Example, every 20 m) PC T PI PC sta = PI sta T PT sta = PC sta + L L PT Note: PT may have two stations, one from curve calculation above and another if the route stationing was first completed using tangent stations (PT sta = PI stat + T) Lect 4 - Sept 17/07 Slide 8 of 10

9 Deflection Angles and Chord Lengths for Full Stations Sections ( ) Need to stake at full stations (XX+00.00) of arc length, s and at PC and PC. Set up the instrument on PC, backsight PI, turn deflection angles ( a,, b ), measure chord distance (c a, c, c b ) and total cord (c T ) from PC. Closure error (falling distance) is determined by differencing calculated c b and measured c b c a = 2 *sin d a 2 c = 2 *sin D 2 c b = 2 *sin d b 2 () c T = 2 *sin = D 2 = s I 2L Lect 4 - Sept 17/07 Slide 9 of 10

10 Example 24.3 (pp ) -in class. Lect 4 - Sept 17/07 Slide 10 of 10