# Geometric Design. Homework Ch 3 # 1,3,7,8,9 Geometric Design Roadway is designed using v, R, e – Need to know the required Stopping Sight Distance for.

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Geometric Design

Homework Ch 3 # 1,3,7,8,9

Geometric Design Roadway is designed using v, R, e – Need to know the required Stopping Sight Distance for the roadway – D = 1.47vt +v 2 /(30(f+g) v is in mph (1.47, 30 are conversion factors) SSD – distance required to make a full stop Decision Sight Distance – adds in a maneuver time (3.5 – 4.5s) to the reaction time based on AASHTO guidelines – D = 1.47v(tr+tm) +v 2 /(30(f+g) Gives Decision Sight Distance

Alignment 2 types of alignment – Vertical – Horizontal

Vertical Curves 2 types of curves – Crest – Sag Different criteria for SSD – crest - 6” object over hill – sag headlights

Vertical Curves Grade controls – Trucks and f control – Max grade Interstate = 9% – Max grade local street = 15% – min grade = 0.5% - drainage – Fig 3.15 – length of grade is important as well – Also look at length of curve and change in grade Don’t want cars grabbing air

Vertical Curves Truck performance Note crawl speed Fig 3.16, 3.17, 3.18

Vertical Curves

Vertical Curve Design Vertical Curves are Parabolas A = G 2 - G 1 –algebraic diff in grades (in %) E = AL/8 – L is in stations (ft/100) – gives E in ft y = [(G2-G1)/2L]*x^2 + G1x+ Y0 – Gives y in ft – x is in stations, G1, G2 in %

Vertical Curve Design Hi/Lo point occurs at dy/dx = 0 – x = -LG 1 /(G 2 -G 1 ) – Not in middle of L – All calcs in stations

Vertical Curve Design SSD for Crest curves (dead cat distance) – 2 equations - L>=SSD, – L=(!G2-G1!*(SSD^2))/(200*(h1^0.5+h2^0.5)^2 – L<=SSD – L=2*SSD - (200*(h1^0.5+h2^0.5)^2/(!G2-G1!) – These provide minimum values – only one is true – eye height h 1 = 3.5 ft – object height h 2 = 0.5 ft – 200*(h1^0.5+h2^0.5)^2 = 1329 (not what is in the book)

Given a design speed of 45 mph, approach grade of +3% and a departure grade of -4% meeting at VPI Station = 100+50, VPI Elevation = 347.85. – What is the PVC & PVT stations and elevations – Where is the high point located and what is elevation

Vertical Curve Design L for Sag curves – L>SSD L=(!G2-G1!*(SSD^2))/(200*(h+SSD*tan  ) – L { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/11/3246294/slides/slide_13.jpg", "name": "Vertical Curve Design L for Sag curves – L>SSD L=(!G2-G1!*(SSD^2))/(200*(h+SSD*tan  ) – LSSD L=(!G2-G1!*(SSD^2))/(200*(h+SSD*tan  ) – L

Given a design speed of 45 mph, approach grade of -4% and a departure grade of -1% meeting at VPI Station = 100+50, VPI Elevation = 347.85. – What is the PVC & PVT stations and elevations – Where is the high point located and what is elevation

Horizontal Alignment

3 parts to alignment – tangents - straight sections – curves – transitions between curves and tangents

Curves L=2  R  /360 D - degree of curve D different for RR and highways

Curves External Distance PI -> LE = R(sec(I/2)-1) Middle Ordinate L->LC M = R(1-cos(I/2) Tangent PC->PI T = Rtan(I/2) Curve Length PC->PT L = 100(I/D) Chord Length PC-> PT LC = 2Rsin(I/2) Degree of curve D = 5729.58/R – How many degrees to create an arc of 100’ for a given radius

Horizontal Distance Need to maintain SSD around curve Look at M and L to determine safe speed – L = SSD – I changes to I=L*D/100 – M = R(1- cos (I/2)) – If M > available M then NO GOOD

Example Given the following info for a horizontal curve, determine if a speed limit of 60 mph is safe. – I = 45 degrees, R =1500 ft, 2-12 foot lanes, 6 foot shoulder, e = 0.04, f=0.12, barn located 14 feet off centerline of inside lane.

Design Vehicles Fig 2.4 Gives tightest turn radius for a low speed turn by a vehicle 10 specific types of vehicles used – determine speeds, curb radius, pavement width

Delineation of Vehicle Paths Flow Control pavement markings, medians, islands Channelization provide improved flow through an intersection

Channelization

Lab Develop an Excel Spreadsheet which will solve horizontal and vertical curve problems

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