Geometric Design Part 2: Horizontal Curves

Learning Objectives and Outcomes

  • Objective: Design the alignment of horizontal curves and conduct safety checks.

  • Abilities Developed:

    • Calculate Design Control Criteria (DCC) of horizontal curves.

    • Check the alignment of horizontal curves.

    • Perform safety checks on horizontal curves.

Presentation of Horizontal Curves

  • Representation: Horizontal alignment of highways is depicted in 2D drawings as plan views,

    • Drawn from a satellite perspective.

  • Characteristics of Drawings:

    • Show connected straight lines and arcs of varying radii.

    • These form the centerline of the highway.

  • Focus of Chapter:

    • Design of a curve modeled as a part of a circle with fixed radius R.

Design Control Criteria (DCC) for Horizontal Curves

  • A horizontal curve has:

    • Superelevation (e)

    • Radius (R)

    • Side Friction Coefficient (f): The friction between tires and pavement.

Forces Acting on a Vehicle on a Horizontal Curve

  • The vehicle, influenced by its weight (W), has a speed (V) which leads to:

    • Side frictional force (F)

    • Normal reaction force (N)

  • Force Balance Equation: e+f(1−ef)=V2gRe + f(1 - ef) = \frac{V^2}{gR}

    • The term ef is negligible, thus simplifying to:
      e+f=V2gRe + f = \frac{V^2}{gR}

  • From this, it can be derived:
    R≥V2g(e+f)R \geq \frac{V^2}{g(e + f)}

  • Constraints on Superelevation:

    • e≤0.12{ e \leq 0.12 }

    • Common values are $0.04 \leq e \leq 0.06$.

    • Where V is in ft/s and acceleration due to gravity (g) is 32.2 ft/s².

Example Calculations for Horizontal Curves

  • Example Scenario: A loop ramp at a cloverleaf interchange with:

    • Design Speed (V): 35 mph (51.33 ft/s)

    • Superelevation (e): 0.045

    • Friction (f): 0.09

    • Gradient (G): -3.3%

    • Time (t): 2.5 s

    • Acceleration (a): 11.2 ft²/s

  • Stopping Sight Distance (SSD) Calculation: SSD=Vt+V22a+GgSSD = Vt + \frac{V^2}{2a + Gg}

    • Plugging in values yields:
      SSD=51.33×2.5+(51.33)22(11.2−0.033×32.2)=258.28 ftSSD = 51.33 \times 2.5 + \frac{(51.33)^2}{2(11.2 - 0.033 \times 32.2)} = 258.28 \text{ ft}

  • Minimum Radius Calculation:
    R≥V2g(e+f)=(51.33)232.2(0.045+0.09)=606.11 ftR \geq \frac{V^2}{g(e + f)} = \frac{(51.33)^2}{32.2(0.045 + 0.09)} = 606.11 \text{ ft}

Additional Examples and Applications

  • Connector between Freeways with Speed 55 mph:

    • Given: $f = 0.12$

    • Suggested values for $e$ and $R$ (0.04 ≤ e ≤ 0.06):

    • If $e = 0.04$: R≥1363.13 ftR \geq 1363.13 \text{ ft}

    • If $e = 0.06: R \geq 1122.78 \text{ ft} </p></li></ul></li></ul><h3id="6953e96e−e8e4−4919−bcb4−e7c7597cbb18"data−toc−id="6953e96e−e8e4−4919−bcb4−e7c7597cbb18"collapsed="false"seolevelmigrated="true">AlignmentDesignPrinciples</h3><ul><li><p><strong>PlanViewCharacteristics:</strong></p><ul><li><p>Startsat<strong>PointofCurvature(PC)</strong>andendsat<strong>PointofTangent(PT)</strong>.</p></li><li><p>Intersectsat<strong>PointofIntersection(PI)</strong>withanangle(I).</p></li></ul></li><li><p><strong>TangentsonCurves:</strong>ThelengthofprojectionTisknownastheTangentoftheCurve.</p></li><li><p><strong>CurveSketching:</strong>TolocatePC,PI,andPT,drawstraightlinesbasedonthoseprojections.</p></li></ul><h3id="d2b0f183−7444−4b0a−907a−5a0f6a931edd"data−toc−id="d2b0f183−7444−4b0a−907a−5a0f6a931edd"collapsed="false"seolevelmigrated="true">AlignmentDesignEquations</h3><ul><li><p><strong>KeyEquations:</strong></p><ul><li><p></p></li></ul></li></ul><h3 id="6953e96e-e8e4-4919-bcb4-e7c7597cbb18" data-toc-id="6953e96e-e8e4-4919-bcb4-e7c7597cbb18" collapsed="false" seolevelmigrated="true">Alignment Design Principles</h3><ul><li><p><strong>Plan View Characteristics:</strong></p><ul><li><p>Starts at <strong>Point of Curvature (PC)</strong> and ends at <strong>Point of Tangent (PT)</strong>.</p></li><li><p>Intersects at <strong>Point of Intersection (PI)</strong> with an angle (I).</p></li></ul></li><li><p><strong>Tangents on Curves:</strong> The length of projection T is known as the Tangent of the Curve.</p></li><li><p><strong>Curve Sketching:</strong> To locate PC, PI, and PT, draw straight lines based on those projections.</p></li></ul><h3 id="d2b0f183-7444-4b0a-907a-5a0f6a931edd" data-toc-id="d2b0f183-7444-4b0a-907a-5a0f6a931edd" collapsed="false" seolevelmigrated="true">Alignment Design Equations</h3><ul><li><p><strong>Key Equations:</strong></p><ul><li><p> R = \frac{V^2}{g(e + f)} </p></li><li><p></p></li><li><p> T = R \tan \left( \frac{I}{2} \right) </p></li><li><p></p></li><li><p> M = R[1 - \cos \left( \frac{I}{2} \right)] </p></li><li><p></p></li><li><p> E = R\left(1 - \cos \left( \frac{I}{2} \right) \right) </p></li><li><p></p></li><li><p> D = \frac{18000}{\pi R} \text{ (degree)}oror D = \frac{100}{R} \text{ (radian)} </p></li><li><p></p></li><li><p> L = \frac{R I \pi}{180} </p></li><li><p></p></li><li><p> stn PT = stn PC + L

Example Problem - Curve Length Calculation

  • Given: A horizontal curve with PI at STA(180 + 00), radius $R = 403.15$ ft and angle $I = 30°$: Calculations for PC and PT Stations and Length:

    • T = R \tan \left( \frac{I}{2} \right) = 403.15 \tan(15°) = 108.02 ext{ ft} </p></li><li><p></p></li><li><p> L = \frac{R I \pi}{180} = \frac{403.15 \times 30 \pi}{180} = 211.10 ext{ ft} </p></li><li><p></p></li><li><p> PC = PI - T = 18000.00 - 108.02 = 17891.98 ext{ (STA 178 + 91.98)} </p></li><li><p></p></li><li><p> PT = PC + L = 17891.98 + 211.10 = 18103.08 ext{ (STA 181 + 03.08)} </p></li></ul></li></ul><h3id="d4446356−b95d−4cd0−a417−cf52f7c04a05"data−toc−id="d4446356−b95d−4cd0−a417−cf52f7c04a05"collapsed="false"seolevelmigrated="true">TypeofProblemsinSafetyChecks</h3><ul><li><p><strong>SafetyCheckObjectives:</strong>Checkthesightdistance(S)againstDCC(SSD),wheresafetyrequires:<br></p></li></ul></li></ul><h3 id="d4446356-b95d-4cd0-a417-cf52f7c04a05" data-toc-id="d4446356-b95d-4cd0-a417-cf52f7c04a05" collapsed="false" seolevelmigrated="true">Type of Problems in Safety Checks</h3><ul><li><p><strong>Safety Check Objectives:</strong> Check the sight distance (S) against DCC (SSD), where safety requires:<br> S \geq SSD $$

    • Sight Distance Factors:

      • Factors include the width of lanes, vertical curvature, and obstacles.

    Example: Safety Check Calculation

    • Scenario: A sound wall near a curve blocks sightlines, informing us to calculate perceived sight distance (S).

      • Using measurements, determine R, M, and assess if S ≥ SSD for preventing potential collisions.

    Type 1 Problem Sample Solution

    • Calculating Visibility with Safety Structures such as Sound Walls:

      • Given: Inside lane width = 12 ft, shoulder = 4.5 ft, radius = 148 ft.

      • Calculate new R and reference dimensions to find S.

    Type 2 Problem Approach

    • When given a required SSD calculate R and M:

      • Example: Given a design speed necessitating SSD, find R and the distance of a wall from the centerline.

    In-Class Exercises

    • Exercises aim to reinforce taught principles regarding pelting curves aligned with varying scenarios and given data.

    • Students are tasked with employing alignment design equations under mixed circumstances to derive requisite components of horizontal curves and their safety checks.