Vertical Alignment

CTC 440

Objectives

Understand the basics of profiles Understand the basics of vertical

curves

Grades and Elevations

Grade-A change in elevation per unit horizontal length (+ or - % in direction of stationing)

Elevations are usually computed to the nearest one-hundredth of a foot or to the nearest one-thousandth of a meter (43.32 ft; 21.262m)

Examples Computing a grade between two

stations with known elevations Computing an unknown elevation at a

station given the known elevation at another station and given the grade between the stations

Compute a grade between two pts w/ known elevations

Sta 1+00; Elev.=198.30’ Sta 2+00; Elev.=203.80’ Grade=change in elev/change in

length Answer: Grade=+5.50%

Compute an unknown elevation at a station

Sta 5+30; Elev.=270.00’ Grade between 5+30 & 7+70=-

2.5% Calculate elev at 7+70 Answer: 264.00 feet

Vertical Curves-Definitions

Vertical curves are parabolic curves instead of circular curves

Crest Curves (3 types) Sag Curves (3 types)

Other Vertical Curve Parameters

PVI-Point of Vertical Intersection-intersection of the two grades

PVC-Point of Vertical Curvature-beginning of the vertical curve

PVT-Point of Vertical Tangency-end of the vertical curve

Other Vertical Curve Parameters M,middle ordinate-A computed

correction, which must be applied to the PVI to determine the elevation at the midpoint of the vertical curve

G1=grade before the PVC G2=grade after the PVT L=length of the vertical curve

Vertical Curve Elevations

Need G1, G2, and L A=G2-G1 (%) r=A/100*L Elevx=(r/2)x2+g1x+ElevPVC

g1 is in decimal form X is distance measured from PVC

Calculating M and min/max elevations

M=A*L/800

Max. or min. elev. Occurs @ x=-g1/r g1 is in decimal form To get the actual elevation substitute

x into the elevation equation: Elevx=(r/2)x2+g1x+ElevPVC

Vertical Curve Examples

English (crest) ---see below Metric (sag) ---on board if needed Comprehensive Curve (sag)---see

below

English-Crest

A vertical crest curve with a length of 400’ is to connect grades of +1% and -1.75%. The PVI is located at station 35 and has an elevation of 549.20’.

What are the elevations of the PVC, PVT and at all full stations on the curve?

English-Crest Elev at PVC=549.2-(1%*200’)=547.20’ Elev at PVT=549.2-(1.75%*200’)=545.70’

Find A=-1.75-1=-2.75 (in %) Find r=A/(100*L)=-.0000688 Find r/2=-.0000344 Find Elevations at even stations: Elev(x)=r/2x2 + g1*x + Elev(PVC)

English-Crest

X Sta r/2* X^2+ G1* X+PVC

elev=Elev

0 33+00

-.0000344

0 +.01 0547.2

0547.20

10034+0

0-.000034

4100^2 +.01 100

547.20

547.86

20035+0

0-.000034

4200^2 +.01 200

547.20

547.82

30036+0

0-.000034

4300^2 +.01 300

547.20

547.11

40037+0

0-.000034

4400^2 +.01 400

547.20

545.70(Check)

English-Crest

Highest Elev occurs @ x=-g1/r=145.35’

Plug x in elevation equation to get:

Highest elevation is 547.93’ at Sta 34+45.35

Comprehensive Curve Example

Finding a PVI Fitting a curve Finding elevations on the curve Finding M Finding low point

Fitting a Curve Between Two Set Tangents (1/6)

Given:

G1=-2%; G2=+3%Two Fixed Points (sta/elev is set):Sta 1+00; Elev=450.00Sta 7+00; Elev=460.00

Fitting a Curve Between Two Set Tangents (2/6)

Find PVI Station and Elevation:450-.02*L1=460-.03*L2 L1+L2=600’Solve for L1 and L2L1=160 ftL2=440 ftPVI Sta=2+60 (Sta 1+00 + 160’)PVI Elev=446.80’

Fitting a Curve Between Two Set Tangents (3/6)

If we’re constrained at Sta 1 and 7 then the maximum vertical curve length we can fit is 160’+160’=320’L=320’A=5%r=A/100L=.0001563r/2=.0000781

Fitting a Curve Between Two Set Tangents (4/6)

Elev (x)=r/2*X2+g1X+El PVC

X STA r/2 X2 G1 X Elev PVC

Elev

0 1+00 .0000781 0 -.02 0 450.00

450.00

100

2+00 .0000781 100^2

-.02 100 450.00

448.78

160

2+60 .0000781 160^2

-.02 160 450.00

448.80

200

3+00 .0000781 200^2

-.02 200 450.00

449.12

300

4+00 .0000781 300^2

-.02 300 450.00

451.03

320

4+20 .0000781 320^2

-.02 320 450.00

451.60

Fitting a Curve Between Two Set Tangents (5/6)

Determine Curve Elevation @ PVIM=AL/800=2’

PVI STA 2+60; Elev 446.8’Curve Elev @ PVI=446.80+2’=448.80’

Fitting a Curve Between Two Set Tangents (6/6)

Determine Low PointX=-g1/r=.02/0001563=127.96Sta=2+27.96

Elev @ x=127.96 =448.72’

General Rules for Establishing Vertical Alignment Goal-Provide a uniform, comfortable ride and

safe vehicle operation Balance cut/fills Grades>=0.5% to prevent drainage

problems Check SSD/HSD Check driveway and intersecting road tie-ins Keep simple (few curves, flat curves, gradual

grades) Check clearances (over/under bridge, over

culverts)

Next lecture

Checking sight distances on horizontal curves

Checking sight distances on vertical curves Crest-Stopping sight distance Sag-Headlight sight distance