Route Surveying - CEMRWEBddean/CE305/Disk1/6... · 3/23/2010 2 Topic Outline cont. • Unit...

3/23/2010

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Route Surveying

CE 305 Intro To Geomatics

By

Darrell R. Dean, Jr., P.S., Ph.D.

Topic Outline

• Horizontal alignment• Horizontal alignment

• Types of Horizontal Curves

• Degree of Curve

• Geometric elements of curve

St ti b t ti• Station number computation

• Deflection angle per foot of curve

• Total deflection angle

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Topic Outline cont.

• Unit deflection angle• Unit deflection angle

• Chord length

• Curve stake out by deflection angles

• “Moving up” on the curve

“B ki i ” th• “Backing in” on the curve

• Tangent offset

• Radial stake out

Horizontal Alignment

• The geometric pattern of a roadway as• The geometric pattern of a roadway as represented in a horizontal plane

• Geometric Elements– Tangents - straight line sections

– Curves

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Horizontal Alignment cont.

Tangent Sections

Point of• Point of Intersection (PI)

• Each PI is identified by a station number

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Tangent Sections cont.

Station• Station Numbers– 1 station = 100 ft.

– accumulative distance alongdistance along centerline

– e.g. 5+24.03

Tangent Sections cont.

• Deflection Angle– at each PI the

deflection angle is measured from the prolongation of the previous tangent

t d b th– represented by the symbol

– includes a direction of turning , L or R

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Curves

• A Horizontal curve is• A Horizontal curve is the arc of a circle

• The radials at the beginning and end of the curve intersect the respective tangents atrespective tangents at a 90° angle.

• Horizontal curves are tangent curves

Curves cont.

• Beginning of curve is• Beginning of curve is called the Point of Curvature - PC

• The end of the curve is called the Point of Tangency - PTTangency PT

• Any point on curve is designated as a POC

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Curves cont.

• Sta Nos increase left• Sta. Nos. increase left to right - sketches

• Back tangent - name given to tangent before PC

• Forward tangent• Forward tangent -name given to tangent after PT

Curves cont.

• The central angle of• The central angle of the curve equals the deflection angle

int angles n 2 180

int angles 360

A 90 180 90 360

. ( )

.

( )

A 90 180 90 360

A 360 360

A

Central angle Deflection angle

( )

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Types of Horizontal Curves

• Simple Curve• Simple Curve

• Compound Curve

• Reverse Curve

• Broken back Curve

• Spiral Curve

Curve Types cont.

• Compound Curve• Compound Curve

– 2 or more simple curves, centers on same side of centerline

PCC Point of– PCC = Point of Compound Curve

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Curve Types cont.

• Reverse Curve• Reverse Curve– 2 simple curves,

centers on opposite sides of centerline

– PRC = Point of Re erse C r eReverse Curve

Curve Types cont.

• Broken Back Curve• Broken Back Curve– Two simple curves

connected by a relatively short tangent

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Curve Types cont.

• Spiral Curve• Spiral Curve

– 3 Components• Spiral

• Simple Circular

• Spiral

– TS = Tan to Spiral– TS = Tan. to Spiral

– SC = Spiral to Curve

– CS = Curve to Spiral

– ST = Spiral to Tan.

Degree of Curve• An index of curve

“sharpness”sharpness

• Arc definition - angle (Da ) subtended by a 100-ft. arc

• Chord definition -angle (D ) subtendedangle (Dc ) subtended by a 100-ft. chord

• Hwy. Curves based on arc definition

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Degree of Curve cont.• Relationship to radius

100 ft

D

2 R ft

360

2 R D 360 100

R360 100

2 D

5729 58

a

a

a

( ) ( )

.

R5729 58

D

or

D5729 58

R

a

a

.

.

( Da must be in decimal degrees)

Degree of Curve cont.

• Relationship to radius• Relationship to radius

SinD

2

50

Rc

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Geometric Elements of Curve

defl. or delta angle

R = radius

T = tangent distance

L = length of curve

LC l h dLC = long chord

E = external Distance

M = middle ordinate

Geometric Elements cont.

Measured or calculated

T R tan 2

D5729 58

R

L100

D 180R

LC 2 R sin 2

a

a

( / )

.

( / )

E R1

cos 21 R sec 2 1

M R 1 cos 2

(( / )

) ( ( / ) )

( ( / ))

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Geometric Elements cont.



T R tan 2

D5729 58

R

L100

D 180R

a

a

( / )

.

LC 2 R sin 2

E R sec 2 1

M R 1 cos 2

( / )

( ( / ) )

( ( / ))

Computing PC and PT Sta. Nos.

• Survey and/or Design Elements Required• Survey and/or Design Elements Required – PI Sta. No.

– – R or Da

• Computed Geometric Elements Requiredp q– T = tan. distance

– L = length of curve

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PC and PT Sta. Nos. cont.

• PC = PI - T• PC = PI - T

• PT = PC + L

• These equations are for the first curve on the alignment.

S b t• Subsequent curves require a different approach

Calc. Sta. Nos. - 1st Curve

Given: (1) PI at 2 + 22 80 (2) = 61°10’35” andGiven: (1) PI at 2 + 22.80, (2) = 61 10 35 , and (3) R = 133.09 ft.

Calculate: PC and PT station numbers for this curve

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Sta. Nos. - 1st Curve cont.

T R tan 2 ( / )T 133 09 tan 61 10 35 2

T 78 67 ft 0 78 67 sta

D5729 58

133 0943 03 02

L100

D

a

a

( )

. ( ' "/ )

. . . .

.

.' "

L100

43 03 0261 10 35

L 14210 ft 1 4210 sta

a

' "

' "

. . . .

Sta. Nos. - 1st Curve cont.

PI 2+22.80

- T 0+78.67

PC= 1+44.13

+ L 1+42.10

PT= 2+86.23

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Calc. Sta. Nos. for 2nd and Subsequent Curves

•First: For all curves calculate the geometric elements, i.e., T, L, etc.

Calc. Sta. Nos. for 2nd and Subsequent Curves

•Second: Calculate the tangent distance (Si-1) between the previous (PT i -1 )and the (PC i ) of interest

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Subsequent Curve Sta. Nos. cont.

•Third: Calculate the (PC i ) of interest by adding (Si-1) to the previous (PT i -1 )


•Fourth: Calculate the (PT i ) of interest by adding (Li) to the (PC i)

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•Data Required for Curve #2

T 78.67

T 165.96 tan(42

T

1

2

2

ft

49 34 2

65 08 ft

L 42 49 34 165 96

.

' "/ )

. .

' "

q

L 42 49 34180

165 96

L 124 05 ft

2

2

.

. .

Subsequent Curve Sta. Nos. cont.•Data Required for Curve #2 cont.

S PI PI T Ti 1 i i 1 i 1 i ( ) ( )

•Calculate distance - S

•E.g., for Curve #2, i=2g

S PI PI T T

S PI PI T T2 1 2 2 1 2 1 2

1 2 1 1 2

( ) ( )

( ) ( )

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•Calculation of S1 for curve # 2 cont.1

•S1 = (470.74 - 222.80) - ( 78.67 + 65.08) = 104.19 ft.

Subsequent Curve Sta. Nos. cont.•Data Required for Curve #2 cont.

T 78.67

T1

2

ft

65 08 ft

L 124 05 ft2

.

. .

. .

#2 cont.

PT1 = 2 + 86.23

S1 = 104.19 ft.

PC2 = PT1 + S1 = 286.23 + 104.19 = 390.42 = 3 + 90.42

PT2 = PC2 + L2 = 390.42 + 124.05 = 514.47 = 5 + 14.47

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Calculate Sta. No. for End of Project (E.O.P.)

• The station for the end of the line will• The station for the end of the line will change as curves are introduced to connect tangents.

Sta. No. for E.O.P. cont.•Steps to Calculate E.O.P.

1 C l l t di t (S ) f th l t P i t f•1. Calculate distance (Sn ) from the last Point of Tangency (PTn ) to the alignment terminus

• S n = (PI n + 1 - PI n ) - Tn

n = number of curves

PI n + 1 = Sta. no. of terminus after last curve and before curves are introduced

PI n = PI of last curve on the alignment

Tn = tangent dist. for the last curve

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Sta. No. for E.O.P. cont.

•Steps to Calculate E.O.P. cont.Steps to Calculate E.O.P. cont.

•2. Calculate the E.O.P. station number

•E.O.P. = PT n + S n

n = number of curves

PT n = PT of last curve on the alignmentn

Sta. No. for E.O.P. cont.•E. g., calculate the E.O.P. for the following case Given: T2 = 65 08 PT 2 = 5 + 14 47Given: T2 65.08, PT 2 5 + 14.47

S2 = (713.39 - 470.74) - 65.08 = 242.65 - 65.08 = 177.57 ft.

E.O.P. = 514.47 + 177.57 = 692.04 = 6 + 92.04

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Sta. No. for E.O.P. cont.

Deflection angle per foot of curve• = deflection angle

in minutes per foot of curve

D

2100

D

200units deg per ft

a

a

( . .)

g p

60

200D

0 3 D

units min per ft

a

a

( )

.

( . .)

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Total Deflection (dt ) Angles

• Total deflection angles• Total deflection angles are used to stake out horizontal curves.

• They are measured at the PC by backsighting the PI,backsighting the PI, setting zero, and turning the angle toward the curve

Total Deflection cont.

• Total Deflection = d• Total Deflection = dt

d 0 3 D l 60t a ( . ) /

• Where:l = arc length in ft.l = POC (sta. no.) - PC (sta. no.)Da = degree of curve

Units for dt = decimal degrees

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Total Deflection cont.

• Example Calculation• Example Calculation

• l = 635.18-517.83

• l = 117.35 ft.

• dt =

d 0 3 D l 60t a ( . ) /

(0.3)·(6.75)·117.35/60 = 3°57’38”

Unit Deflection Angle (du )

• The unit deflection• The unit deflection angle is the angle between the local tangent and the chord for any arc on the curve.

• d u = ((0.3)· Da ·l)/60

• l = POC#2 - POC#1

• l = arc length ft.

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Unit Deflection Angle (du ) cont.


• d u = ((0.3)· Da ·l)/60

• l = arc length (ft.)

• l = 750.00 - 600.00

• l = 150.00 ft.

• d u = ((0.3)·6.75·150.00)/60

• d u = 5°03’45”

Chord Length For Any Arc• Chord length equation

sin d

1

2Chord

RChord 2 R sin d

d 0 3 D l 60

u

u

u a

( )

( )

( . ) /

l arc length ft.

Chord 2 R sin(0.3 D l / 60)a

Note: May substitute dt for du

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Chord Length cont.


l = 635.18 - 517.83 = 117.35 ft. = arc length

R5729 58

6 45848 83 ft

d 0 3 6 75 117 35 60

.

'. .

/d 0 3 6 75 117 35 60

d 3 57 38

chord 2 848 83 sin 3 57 38

chord 117 26 ft

t

t

. . . /

' "

. ( ' ")

. .

Curve Stake out by the Deflection Angle Method

• The Deflection Angle gMethod is an angle-distance intersection method.

• Angles are measured after backsighting the PI and setting zero

• Chord distances are measured from the last station staked

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Deflection Angle Method cont.

• Procedures• Procedures– Office: Calculate

total deflections for full and half stations and chord lengths for all arc segments as requiredrequired

Deflection Angle Method cont.Procedures: cont.

dtl = arc length (ft.)Station0°00'00"0.00PC = 22+85.9601°32'55"14 0423+00

•sample problem total deflection angles•Da = 22°03’35” R = 259.73 ft.• = 70°00’00” L = 317.32 ft.•PC = 22+85.96 PT = 26+03.28

01°32'55"14.0423+00114.0424+00

25+0026+00PT = 26+03.28

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dtl = arc length (ft.)Station0°00'00"0.00PC = 22+85.96

•sample problem total deflection angles•Da = 22°03’35” R = 259.73 ft.• = 70°00’00” L = 317.32 ft.•PC = 22+85.96 PT = 26+03.28

01°32'55"14.0423+0012°34'42"114.0424+0023°36'30"214.0425+0034°38'17"314.0426+0035°00'00"317.32PT = 26+03.28


•sample problem chord lengths•Da = 22°03’35” R = 259.73 ft.• = 70°00’00” L = 317.32 ft.•PC = 22+85.96 PT = 26+03.28

chord (ft.)duArc (ft.)14.0401°32'55"14.04 14.0401 325514.04

501003.28

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•sample problem chord lengths•Da = 22°03’35” R = 259.73 ft.• = 70°00’00” L = 317.32 ft.•PC = 22+85.96 PT = 26+03.28

chord (ft.)duArc (ft.)14.0401°32'55"14.0449.9205°30'54"5099.3811°01'48"1003.2800°21'42"3.28


• Field Procedures• Field Procedures– Occupy PI set PC and

PT on alignment

– Occupy PC, Backsight the PI, set zero

– Layoff the total deflection angle for the first station and set hub at chord distance from PC

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• Field Procedures cont• Field Procedures cont.– In turn lay off the total

deflection angle for each station and set a hub for the station at the intersection of the line of sight and theline of sight and the chord distance from the last station staked.


• Field Procedures cont• Field Procedures cont.– Checks:

• Measure distance between PTs as set from PI and PC, call it the L.E.C.

– R.E.C =

L.E.C / ( 2T + L )

• Set midpoint on curve and measure distance to PI: Compare to E

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“Moving up” on the Curve

• Obstructions sometimes prohibits the• Obstructions sometimes prohibits the measuring of total deflection angles to all stations on the curve.

• A station on the curve may be occupied and the same total deflection angles computed for using at the PC may be used to complete the layout.

“Moving up” on the Curve cont.

• The last visible station on the curve is• The last visible station on the curve is staked and then occupied, i.e., you “move up” on the curve.

• The first time you “move up” you backsight the PC.

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• Orientation is always such that the circle• Orientation is always such that the circle reading, as you sight along the projection of the chord from the last station occupied through the station you are occupying, is set to equal the total deflection to the last station occupied. In the case of backsighting the PC the setting should be zero as you sight along the projection


• Stake out of stations proceeds as if the• Stake out of stations proceeds as if the procedure was taking place at the PC.

• That is, the total deflections calculated to be measured from the PC may now be used at the station occupied on the curve.

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“Moving Up” on the Curve cont.

“Backing” the Curve in from PT

• The same total deflection angles calculated• The same total deflection angles calculated for staking stations on the curve from the PC may also be used to stake the same stations from the PT.

• The PT is occupied and rather than backsighting the PI, the PC is the backsight point.

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“Backing” the Curve in from PT

• The PC is backsighted and the circle• The PC is backsighted and the circle reading is set to zero.

• Stations are located with the same angle-distance intersection technique as previously described.

• The total deflection angle to each station is as calculated for layout from the PC.

“Backing” in the Curve cont.

• Consider Triangle• Consider Triangle AOB

2

d 1802

180

d

t

t

•Therefore, curve stakeout from the PC or PT with the same set of total deflection angles is possible.

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Tangent Offset Method

• Procedures• Procedures– measure distance (x)

along the tangents from PC or PT

– measure perpendicular offset (y) to curve

• Requirements– total deflection angle

– chord distance

Tangent Offset Method cont.• Note

– for first half of curve arc length calculated from PC

– for second half of curve arc length calculated from PT

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Tangent Offset Method cont.

•Offset Equations

x chord cos d

y chord sin dt

t

( )

( )

Offset Equations

d 0 3 D l 60

l l th ftt a . /

( )l arclength ft

chord 2 R sin dt

( .)

( )


•Find x and the offset, y, for this example:

x chord cos d

y chord sin dt

t

( )

( )

Find x and the offset, y, for this example:

d 0 3 D l 60

l l th ftt a . /

( )l arclength ft

chord 2 R sin dt

( .)

( )

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•Answers:l = 54.63 ft.dt = 6°33’01”chord = 54.51 ft.x = 54.15 ft.y = 6.22 ft.

Radial Stake out

• Calculate coordinates of PC• Calculate coordinates of PC

• Azimuth from the PC to any station is the azimuth of PC to PI plus the total deflection angle to the station. Remember left deflections are negative

• Distance from the PC to any station is the chord distance.

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Radial Stake Out cont.

• With the coordinates of the PC known and• With the coordinates of the PC known and the direction and distance to each station known, calculate the station coordinates.

• Radial stake out from control points can then be achieved.

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Route Surveying - CEMRWEBddean/CE305/Disk1/6... · 3/23/2010 2 Topic Outline cont. • Unit...

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