Coordinate Geometry Fundamentals - WikispacesGeometry...Max. value Origin Direction Letters ......

Coordinate

Geometry

Fundamentals

2012 Edition

Professor Todd W. Horton, PE, PLS

Engineering Science & Technologies Department

Parkland College

Champaign, Illinois

[email protected]

mailto:[email protected]

Coordinate Geometry Fundamentals 2


Table of Contents

Page

Angles & Directions 4-8

Triangle Solutions 9

Inverse Computations 10-12

Traverse Computations 13-22

Sideshot Computations 23-25

Perpendicular Offset Computations 26-28

Intersection Computations

Direction – Direction 29-32

Distance – Distance 33-36

Direction – Distance 37-41

Area by Coordinates Computations 42-45

Horizontal Curves 45-51

Horizontal Curves – Tangent Offset 52-53

Horizontal Curves – Chord Offset 54-55

Vertical Curves 56-60


Angles & Direction Angle: the difference in direction of two lines.

Angular units: Degrees, minutes, and seconds:

1 degree = 1/360 of a circle

1 minute = 1/60 of a degree

1 second = 1/60 of a minute

Grads: 1 grad = 1/400 of a circle or 0º 54’ 00”

Radians: 1 radian = ½ π of a circle or 57º 17’ 44.8”

Perform as many checks on angular data as possible.

Close the horizon when turning angles in the field.

In a closed traverse, compute the sum of the interior angles.

The sum should equal (n-2)*180º, where n is the number of interior angles.

Angular adjustments:

Do not express seconds with decimal fractions unless the instrument used reads to

decimal fractions of a second.

Examine field notes for angles with poor closure and for problems with turning

the angles. Apply excess to these angles.

If unable to view field notes or if no apparent error source exists, then apply

excess of equal adjustment to angles with the shortest sides.

Bearings and azimuths:

Bearing: the acute horizontal angle between a reference meridian (north or south) and a line.

Azimuth: the horizontal angle measured from a north meridian clockwise to a line.


Max. value Origin Direction Letters

Bearing 90º North or South CW or CCW Yes

Azimuth 360º North CW No

Convert azimuths to bearings

First, determine the proper quadrant letters:

1. For 0° to 90°, use NE (quadrant 1 in most software programs).

2. For 90° to 180°, use SE (quadrant 2 in most software programs).

3. For 180° to 270°, use SW (quadrant 3 in most software programs).

4. For 270° to 360°, use NW (quadrant 4 in most software programs).

Then find the numerical value, using the following relationships:

1. NE quadrant: bearing = azimuth

2. SE quadrant: bearing = 180° - azimuth

3. SW quadrant: bearing = azimuth - 180°

4. NW quadrant: bearing = 360° - azimuth

Convert from bearing to azimuths

Convert from bearing to azimuths by using these relationships:

1. NE quadrant: azimuth = bearing

2. SE quadrant: azimuth = 180° – bearing

3. SW quadrant: azimuth = 180° + bearing

4. NW quadrant: azimuth = 360° - bearing

Reverse Directions

Back azimuth (reverse direction) = azimuth + / - 180º

Back bearing (reverse direction) = same numeric value with opposite directions


Find these directions:

Azimuth Bearing

Line 0-1

Line 0-2

Line 0-3

Line 4-0


Compute the interior angles of this closed traverse.


Traverse loop azimuth computations

1. To compute azimuths in the counterclockwise direction, add the interior angle to the

back azimuth of the previous course.

2. To compute azimuths in the clockwise direction, subtract the interior angle from the


See the Traverse Example Problem for an example of this

computation.

Azimuth computations Practice Problem


Right Triangle Solutions

For angle A, c

aA sin ,

c

bA cos ,

b

aA tan

Given Required Solutions

a, b A, B, c b

aA tan , 22 bac

a, c A, B, b Bc

aA cossin , acacb

A, a B, b, c B = 90º - A, A

ab

tan ,

A

ac

sin

A, b B, a, c B = 90º - A, Aba tan , A

bc

cos

A, c B, a, b B = 90º - A, Aca sin , Acb cos

Oblique Triangle Solutions

Sine law: c

C

b

B

a

A sinsinsin Cosine law: Abccba cos2222

Given Required Solutions

A, B, a b, c, C A

Bab

sin

sin , C = 180º - (A+B),

A

Cac

sin

sin

A, a, b B, c, C a

AbB

sinsin , C = 180º - (A+B),

A

Cac

sin

sin

a, b, C A, B, c

ba

BAba

BA

2

1tan

2

1tan

A

Cac

sin

sin , A+B = 180º - C

a, b, c A, B, C 2

cbas

,

bc

csbsA

2

1sin ,

ac

csasB

2

1sin , C = 180º - (A+B)

a, b, c area 2

cbas

, csbsassarea

A, b, c area 2

sin Abcarea

A, B, C, c area A

CBaarea

sin2

sinsin2

A

a

b

c

C

B

Right Triangle

C

B

A

c

a

b

B

C A

c

a

b

Oblique Triangles


Inverse Computations

Given known coordinates of any two points of a system, the distance and direction

between them can be determined.

1. Determine latitude (ΔN) and departure (ΔE) between the two points.

a. Subtract origin northings and eastings from destination northings and eastings.

b. Be careful to note the sign (+ or -) of each answer.

2. Determine length using 22 ENc .

3. Determine reference direction: north or south.

4. Determine local angle using

N

E1tan .

5. Determine line direction.

Northing Easting

Destination Point 2 N2 E2

- Origin Point 1 - N1 - E1

ΔN ΔE


Inverse Example Problem Inverse from Point J to Point H.

Point J N 3913.66 E 2207.65

Point H N 4692.08 E 5909.33

Step 1

Point Northing Easting

Destination H 4692.08 5909.33

- Origin J - 3913.66 - 2207.65

+ 778.42 + 3701.68

ΔN ΔE

Note that both ΔN and ΔE are both positive, thus line JH lies in the northeast quadrant.

Step 2

64.378268.370142.778 22 HD ft

Step 3

Since line JH lies in the northeast quadrant, the reference direction is North.

Step 4

"28'077842.778

68.3701tan 1

This is the local angle relative to North.

Step 5

North 0˚00’00”

+ local angle +78˚07’28”

Line direction 78˚07’28”

Solution: Line JH, azimuth 78˚07’28”, length 3782.64 ft.


Inverse Practice Problem Inverse from Point 1 to Point 3.


1 5046.79 6323.23

3 5615.27 6304.67


Traverse Computations

This computation process is dependent on these conditions.

The traverse is a closed loop traverse.

The traverse was measured in a counter-clockwise direction, allowing direct

measurement to the right of interior angles.

The traverse angular closure is within acceptable limits for the accuracy standards

required.

A starting azimuth is known or will be assumed.

A starting coordinate is known or will be assumed.

1. Find the angular closure error.

a. Sum the measured interior angles of the traverse.

b. Compute the theoretical sum of interior angles using the following equation.

180)2(nangles where n equals the number of traverse angles (sides).

c. Subtract the theoretical sum of interior angles from the measured sum of interior

angles. This difference is the angular error in the traverse.

2. Adjust the interior traverse angles.

a. Find the total angular correction. The total angular correction equals the

angular error but is opposite in sign.

b. Divide the total angular correction by the number of traverse angles. This

result will be the correction to each individual traverse loop interior angle.

Example:

For a six-sided traverse loop with a -18” error, the correction will be:

- ( -18” / 6 angles ) = + 3” correction per angle

c. Add this correction to each interior angle.

3. Compute the traverse leg azimuths.

To compute azimuths in the counterclockwise direction, add the interior angle to

the back azimuth of the previous course.


To compute azimuths in the clockwise direction, subtract the interior angle from

the back azimuth of the previous course.

4. Compute the latitude (ΔN) and departure (ΔE) for each traverse leg.

Latitude: )(cos AzHDN

Departure: )(sin AzHDE

where

ΔN equals the change in Northing (latitude)

ΔE equals the change in Easting (departure)

HD equals the measured horizontal distance along the traverse leg

Az equals the computed azimuth of the traverse leg

a. Find errors in latitude and departure, Elat and Edep.

The sum of the latitude and departure columns should be fairly close to 0.00 feet.

b. Find the linear error, Elin. This is the positional closure error of the traverse.

)(22

deplatlin EEE

c. Compute the relative error, Erel, in the traverse and check it against the

appropriate standard.

lin

rel

E

LE

1 where L equals the total length of the traverse legs

Express the Erel result as a fraction with a numerator of 1.

Round the denominator to the nearest 1000.

This error ratio is the indicator of positional error.

d. If the positional error is acceptable, continue computing the traverse. If the

error is excessive, recheck all your computations. If your computations are error

free, your field work may need to be remeasured.

e. Compute the corrections for latitude and departure.

Balancing Methods:

Least Squares: Based on the theory of probability. Linear and angular

adjustments are made simultaneously. Hand methods are long and complex thus

not often used. Computer applications are commonly used for this procedure.


Crandall Method: Used when larger random error exists in distance.

Directional adjustments from balancing are held fixed and distances are balanced

by a weighted least squares procedure.

Transit Rule: Used when larger errors occur in distance than in direction.

Seldom used today.

Compass Rule: (Bowditch Rule) Used when accuracy of angles and distances

are equal. Most commonly used method today.

For the Compass Rule, corrections are defined as follows.

latlat EL

HDC

depdep E

L

HDC

where

Clat equals the latitude correction for a traverse leg

Cdep equals the departure correction for a traverse leg

HD equals the measured horizontal distance along the traverse leg

L equals the total length of the traverse legs

The sum of all latitude corrections Clat should equal and be opposite in sign to Elat.

The sum of all departure corrections Cdep should equal and be opposite in sign to Edep.

f. For each traverse leg, add the latitude (departure) and the latitude

(departure) correction to produce the balanced latitude (departure).

g. Sum the balanced latitudes and sum balanced departures. Each sum should

equal zero since all errors have been corrected.

5. Compute traverse point coordinates.

a. Starting at a point of known or assumed coordinates, add the latitude and

departure of the next traverse leg to the starting coordinate to find the next

point coordinates.

b. Using the newly computed coordinate as a new starting point, add the next

latitude and departure to find the next point coordinates.

c. Repeat this process until all latitudes and departures have been properly

applied.

d. When all is complete, the ending coordinates should match the starting

coordinates.


American Congress On Surveying And Mapping

Minimum Angle, Distance And Closure Requirements For Survey Measurements

Which Control Land Boundaries For ALTA/ACSM Land Title Surveys

(Note 1)

Note (1) All requirements of each class must be satisfied in order to qualify for that particular class of

survey. The use of a more precise instrument

does not change the other requirements, such as number of angles turned, etc.

Note (2) Instrument must have a direct reading of at least the amount specified (not an estimated reading),

i.e.: 20" = Micrometer reading theodolite,

<1'> = Scale reading theodolite, 10" = Electronic reading theodolite.

Note (3) Instrument must have the capability of allowing an estimated reading below the direct reading to

the specified reading.

Note (4) D & R means the Direct and Reverse positions of the instrument telescope, i.e., Urban Surveys

require that two angles in the direct and two

angles in the reverse position to be measured and meaned.

Note (5) Any angle measured that exceeds the specified amount from the mean must be rejected and the set

of angles re-measured.

Note (6) Ratio of closure after angles are balanced and closure calculated.

Note (7) All distance measurements must be made with a properly calibrated EDM or Steel tape, applying

atmospheric, temperature, sag, tension,

slope, scale factor and sea level corrections as necessary.

Note (8) EDM having an error of 5 mm, independent of distance measured (Manufacturer's specifications).

Note (9) EDM having an error of 10 mm, independent of distance measured (Manufacturer's

specifications).

Note (10) Calibrated steel tape.

Excerpted from MINIMUM STANDARD DETAIL REQUIREMENTS for ALTA/ACSM LAND

TITLE SURVEYS as adopted by American Land Title Association

American Congress on Surveying and Mapping and National Society of Professional Surveyors,1999

http://www.acsm.net/alta.html

Dir. Reading

of

Instrument

(Note 2)

Instrument

Reading

Estimated

(Note 3)

Number

of Observ-

ations Per

Station

(Note 4)

Spread From

Mean of

D&R

Not To

Exceed

(Note 5)

Angle

Closure

Where

N=No. of

Stations

Not To

Exceed

Linear

Closure

(Note 6)

Distance

Measure-

ment

(Note 7)

Minimum

Length of

Measure-

ments

(Notes 8,

9, 10)

20" <1'> 10"

5" <0.1'> N.A.

2 D&R

5"<0.1'> 5"

10" N

1:15,000

EDM or

Doubletape

with Steel

Tape

(8) 81m,

(9) 153m,

(10) 20m

http://www.acsm.net/alta.html


Traverse Example Problem Find the coordinates of points R, S, T and P.

Adjust the angles, balance the traverse, and compute coordinates for this traverse.

Vertex Angle Line Distance Direction Position

Q 75º 01’ 24" 5000.00, 5000.00

QP 1170.73 N 76º 32’ 44” E

P 41º 19’ 20"

PN 458.39

N 251º 04’ 40"

NM 339.25

M 54º 06’ 24"

ML 868.95

L 118º 27’ 52"

LQ 428.09


Adjust interior angles:

Vertex Field measured

angle adjustment

Adjusted

angle

Q 75º 01’ 24” +0º 0’ 04” 75º 01’ 28”

P 41º 19’ 20” +0º 0’ 04” 41º 19’ 24”

N 251º 04’ 40” +0º 0’ 04” 251º 04’ 44”

M 54º 06’ 24” +0º 0’ 04” 54º 06’ 28”

L 118º 27’ 52” +0º 0’ 04” 118º 27’ 56”

Sum 539º 59’ 40” 540º 00’ 00”

- 540º

error -0º 0’ 20”

-(- 20”) / 5 angles = +4” / angle

adjustment

Traverse loop azimuth computations:

3. To compute azimuths in the counterclockwise direction, add the interior angle to the


4. To compute azimuths in the clockwise direction, subtract the interior angle from the



Azimuth computations:

76º 32’ 44 “ azimuth QP This is the starting direction.

+ 180º

256º 32’ 44 “ back azimuth QP

+ 41º 19’ 24” interior angle P

297º 52’ 08” azimuth PN

- 180º

117º 52’ 08” back azimuth PN

+ 251º 04’ 44” interior angle N

368º 56’ 52”

- 360º

8º 56’ 52” azimuth NM

+ 180º

188º 56’ 52” back azimuth NM

+ 54º 06’ 28” interior angle M

243º 03’ 20” azimuth ML

- 180º

63º 03’ 20” back azimuth ML

+ 118º 27’ 56” interior angle L

181º 31’ 16” azimuth LQ This was the last remaining unknown azimuth.

- 180º Use the adjusted closing traverse angle to

check your computations.

1º 31’ 16” back azimuth LQ

+ 75º 01’ 28” interior angle Q

76º 32’ 44” azimuth QP OK: This checks with the starting value.


Traverse Practice Problem

Adjust the angles, balance the traverse, and compute coordinates for this traverse.

Vertex Angle Line Distance Direction Position

A 87º 53’ 02” N 2000.00, E 4000.00

AB 186.63 S 42º 15’ 33” E

B 189º 29’ 34”

BC 206.92

C 78º 48’ 29”

CD 198.15

D 118º 22’ 27”

DE 187.93

E 140º 41’ 31”

EF 214.57

F 104º 45’ 08”

FA 201.51


Sideshot Computations

Sideshot positions should be computed relative to balanced traverse coordinates.

1. Find the backsight azimuth (see Inverse Computations).

2. Compute the foresight azimuth.

3. Find the latitude and departure of the

foresight line.

4. Compute the foresight point coordinates.

Adjusted traverse courses

To find the distance and azimuth between the adjusted

points, you must inverse between them.

Notice how these values differ from the corrected values

in the computation sheet above.

Line Distance Azimuth

QP 1170.67 76°32'48"

PN 458.40 297°51'57"

NM 339.24 8°56'44"

ML 869.00 243°03'18"

LQ 428.10 181°31'25"


Sideshot Example Problem Find the coordinates of point 1 given the field measurements shown here.

Occ

Pt

BS

Pt

FS

Pt Horiz Angle Horiz Distance Comment

P Q 1 244º 08’ 38” 291.53 ft IR FND IN O&C SURF

Step 1

Backsight direction:

Local angle:

"47'327636.272

54.1138tan 1

180˚00’00” + 76˚32’47” = 256˚32’47”

backsight azimuth

Step 2

Step 3

Latitude:

57.225)"25'41140(cos53.291 N

Departure:

69.184)"25'41140(sin53.291 E

Step 4

Northing Easting

P 5272.36 6138.54

Lat / Dep +(-225.57) +184.69

FS Pt 1 5046.79 6323.23

Northing Easting

Destination Q 5000.00 5000.00

- Origin P - 5272.36 - 6138.54

- 272.36 - 1138.54

ΔN ΔE

Backsight azimuth 256˚32’47”

Angle RT to FS Pt + 244º 08’ 38”

500º 41’ 25”

- 360º 00’ 00”

Foresight azimuth 140º 41’ 25”

Solution: Point 1 = N 5046.79, E 6323.23


Sideshot Practice Problem Find the coordinates of point 3 given the field measurements shown here.

Occ

Pt

BS

Pt

FS

Pt Horiz Angle Horiz Distance Comment

P Q 1 244º 08’ 38” 291.53 ft IR FND IN O&C SURF

P Q 2 240º 42’ 36” 258.67 ft S. FACE WOOD FENCE POST

M N 3 282º 45’ 42” 558.20 ft IR/CAP FND 6” DEEP, ILS 2006

M N 4 283º 07’ 40” 569.98 ft N. FACE WOOD CORNER POST

L M 5 285º 37’ 47” 143.35 ft IR/CAP FND 12” DEEP, ILS 2006

L M 6 282º 18’ 38” 165.57 ft W. FACE WOOD FENCE POST


Perpendicular Offset

Computations

An application of inverse calculations

Given a Point 3 of known coordinates (N3, E3)and a line 1-2 whose endpoint coordinates

(N1, E1 and N2, E2) are known, the perpendicular offset of the point from the line can be

determined. Stationing along the line to the point can also be found.

1. Inverse between the end points of the line 1-2.

2. Inverse between point 1 and the offset point 3.

3. Find the interior angle, Φ, between lines 1-3 and 1-2.

4. Find the perpendicular offset and direction. sin31HDoffset

5. Find the station from Point 1 along the line. cos31HDstation


Perpendicular Offset Example Problem

What is the perpendicular offset and station of point 76 relative to line 80 - 83?

Point 76 Point 80 Point 83

N 3256.82 N 3534.01 N 3144.89

E 2296.66 E 1709.58 E 2523.41

Step 1

Line 80-83

inverse

Since ΔN is negative and ΔE is positive, line 80-83 lies in the southeast quadrant.

Reference direction is South (azimuth=180˚00’00”).

07.90283.81312.389 22 HD ft "45'266412.389

83.813tan 1

Step 2

Line 80-76

inverse

23.64908.58719.277 22 HD ft "32'436419.277

08.587tan 1


Destination 83 3144.89 2523.41

- Origin 80 - 3534.01 - 1709.58

- 389.12 + 813.83

ΔN ΔE

South 180˚00’00”

+ local angle +(-64˚26’45”)



Destination 76 3256.82 2296.66

- Origin 80 - 3534.01 - 1709.58

- 277.19 + 587.08

ΔN ΔE

South 180˚00’00”

+ local angle +(-64˚43’32”)


Line 80-83,

azimuth 115˚33’15”,

length 902.07 ft.

Line 80-76,

azimuth 115˚16’28”,

length 649.23 ft.


Step 3

Close inspection of these directions shows that

Point 76 falls to the left of line 80-83.

Step 4 17.316'47”)0(sin23.649sin31 HDoffset ft

Step 5 22.649)"47'160(cos23.649cos31 HDstation ft

Perpendicular Offset Practice Problem

Find the perpendicular offset and station of Point K relative to line JL.

Point J Point K Point L

N 2537.19 N 2423.58 N 2399.34

E 1774.94 E 2223.41 E 2445.15

Line 80-83 115˚33’15”

- Line 80-76 -115˚16’28”

Interior angle Φ 0˚16’47”

Solution:

Point 76 falls 3.17 ft LEFT of line 80-83, 649.22 ft along the line from Point 80.


Intersection Computations

3 types of intersections: Solution method:

1. Direction – Direction Sine law

2. Distance – Distance Cosine law

3. Direction – Distance Sine law

Sine law: c

C

b

B

a

A sinsinsin Cosine law: Abccba cos2222

Direction – Direction Intersection

Find the coordinates of Point C given known coordinates at Points A & B and two lines

of known direction.

1. Inverse line AB.

2. Compute the interior angles at A, B, and C.

3. Compute the length of line AC (or line BC) using the Sine Law.

4. Find the latitude and departure of line AC (or line BC).

5. Compute the coordinates of Point C.

C

B

A

c

a

b



Example Problem

Find the coordinates of Point C given

known coordinates at Points A & B and

two lines of known direction.

Step 1

Line AB

inverse

Since both ΔN and ΔE are negative, line AB lies in the southwest quadrant. Reference

direction is South (azimuth=180˚00’00”).

33.99248.63030.766 22 HD ft "46'263930.766

48.630tan 1


Destination B 5724.36 6198.05

- Origin A - 6490.66 - 6828.53

- 766.30 - 630.48

ΔN ΔE

South 180˚00’00”



Line AB,

azimuth 219˚26’46”,

length 992.33 ft.


Step 2

Step 3

"18'3251sin"41'5787sin

33.992

BC

BC = 777.51 ft

Step 4 Latitude: 38.777)"45'56358(cos51.777 N ft

Departure: 30.14)"45'56358(sin51.777 E ft

Step 5

Side AC 270˚59’04”

Side AB - 219˚26’46”

Angle A 51˚32’18”

Side BC 358˚56’45”

Side AC - 270˚59’04”

Angle C 87˚57’41”

Side AB 219˚26’46”

+ 180˚00’00”

Side BC - 358˚56’45”

Angle B 40˚30’01”

Angle A 51˚32’18”

Angle C + 87˚57’41”

Angle B + 40˚30’01”

Check = OK 180˚00’00”

Northing Easting

Point B 5724.36 6198.05

Lat / Dep +777.38 +(-14.30)

Point C 6501.74 6183.75

Solution: Point C = N 6501.74, E 6183.75



Practice Problem

Find the coordinates of Point C given

known coordinates at Points A & B and

two lines of known direction.


Distance – Distance Intersection


of known length.

1. Inverse line AB.

2. Compute the interior angle A using the Law of Cosines.

3. Compute the azimuth of line AC on the appropriate side (left or right) of line AB.

4. Find the latitude and departure of line AC.

5. Compute the coordinates of Point C.



Example Problem


of known length.

Step 1

Line AB

inverse

Since both ΔN and ΔE are positive, line AB lies in the northeast quadrant. Local angle

equals azimuth in the northeast quadrant.

82.79381.79390.3 22 HD ft "07'438990.3

81.793tan 1


Destination B 6338.33 8704.38

- Origin A - 6334.43 - 7910.57

+ 3.90 + 793.81

ΔN ΔE

North 0˚00’00”



Line AB,

azimuth 89˚43’07”,

length 793.82 ft.


Step 2

Abccba cos2222

bc

cbaA

2cos

2221

"24'246682.793)30.742(2

82.79330.74230.842cos

2221

A

Step 3

Step 4 Latitude: 70.681)"43'1823(cos30.742 N ft


Step 5

Side AB 89˚43’07”

Angle A - 66˚24’24”

Side AC 23˚18’43”

Northing Easting

Point A 6334.43 7910.57

Lat / Dep +681.70 +293.76

Point C 7016.13 8204.33




Practice Problem


of known length.


Direction – Distance Intersection

Find the coordinates of Point C given known coordinates at Points A & B and one line of

known length R and one line of known direction.

1. Inverse line AB.

2. Compute the interior angle B.

3. Using Angle B, distance AC,

and distance AB, find Angle C

using the Sine Law.

4. Using Angles B & C,

compute Angle A and the

azimuth of line AC.

5. Find the latitude and

departure of line AC.

6. Compute the coordinates of

Point C.

Direction – Distance Intersection Conditions

Evaluate R and Angle B to determine possible number of solutions.

Angle B acute Angle B right Angle B obtuse

R < AB 0-2 solutions

(see note next page) 0 solutions 0 solutions

R = AB 1 solution

(isosceles) 0 solutions 0 solutions

R > AB 1 solution 1 solution 1 solution


Note:

For acute Angle B and R < AB, three conditions may occur.

O solutions:

R is less than the minimum

(perpendicular) distance to line BC.

1 solution:

R equals the minimum (perpendicular)

distance to line BC.

2 solutions:

R is greater than the minimum

(perpendicular) distance to line BC.



Example Problem



Step 1

Line BA

inverse

Since ΔN is negative and ΔE is positive, line BA lies in the southeast quadrant.

Reference direction is South (azimuth=180˚00’00”).

12.106163.100016.353 22 HD ft "36'337016.353

63.1000tan 1


Destination A 7386.35 4810.28

- Origin B - 7739.51 - 3809.65

- 353.16 + 1000.63

ΔN ΔE

South 180˚00’00”

+ local angle -70˚33’36”


Line BA,

azimuth 109˚26’24”,

length 1061.12 ft.


Step 2

Step 3

Csin

12.1061

"34'4796sin

86.1377

C = 49˚52’54”

Step 4

Step 5

Latitude: 01.1097)"56'45322(cos86.1377 N ft


Step 6

Side BC 192˚38’50”

- 180˚00’00”

Side CB 12˚38’50”

Side BA 109˚26’24”

Side BC - 12˚38’50”

Angle B 96˚47’34”

Sum of angles 180˚00’00” Side BA 109˚26’24”

Angle B - 96˚47’34” + 180˚00’00”

Angle C - 49˚52’54” Angle A + 33˚19’32”

Angle A 33˚19’32” Side AC 322˚45’56”

Northing Easting

Point A 7386.35 4810.28

Lat / Dep +1097.01 +(-833.71)

Point C 8483.36 3976.57




Practice Problem




Area by Coordinates

Computations

Given a closed figure defined by points of known coordinates (Nx, Ex), the figure area

can be determined by cross-multiplication of the coordinate pairs.

1. List point coordinates in sequence around the area to be calculated.

2. Cross-multiply coordinate pairs to find ΣNorthings.

NorthingsENENENEN x )*...()*()*(* 1433221

3. Cross-multiply coordinate pairs to find ΣEastings.

EastingsNENENENE x )*...()*()*(* 1433221

4. Calculate the area.

AreaEastingsNorthings

2


Area by Coordinates

Example Problem

Find the area of the figure defined by Points 1 – 6.

Step 1


1 10000.0000 5000.0000

2 10326.7981 5356.3614

3 9938.7277 5298.7122

4 9448.9156 4560.3990

5 9854.7405 4760.8417

6 10070.8565 4583.9559

1 10000.0000 5000.0000

Step 2

NorthingsENENENEN x )*...()*()*(* 1433221 = 294,119,678.8 ft2

Step 3

EastingsNENENENE x )*...()*()*(* 1433221 = 293,663,353.6 ft2

Step 4

2

2.325,456

2

2ftArea

EastingsNorthings

= 228,162.6 ft

2

24.5/560,43

6.162,2282

2

acreft

ft acres

Solution: Area = 5.24 acres


Area by Coordinates

Practice Problem

Find the area of the figure defined by Points 1 – 6.


Horizontal Curves

Δ = Central Angle R

D577951.5729

D

L

100

R = Radius

T = Tangent Distance 100

LD

3602

RL

D = Degree of Curvature

E = External Distance 2

tan

RT 2

sin2

RC

M = Middle Ordinate

C = Chord Length 2

cos2

TC 4

tan2

CM

L = Curve (arc) length

PC = Point of Curvature 2cos1 RM

1

2cos

1RE

PI = Point of Intersection

PT = Point of Tangency

Any 2 known parts will

completely describe a

curve.


Circular Curve Example Problem

Given: PI Station 107+67.90 Δ = 11º 00’ 00” D = 2º 30’ 00”

Calculate all other curve parameters.

Radius Deflection for 100 ft arc

R=5729.58 / D 100 ft arc = D / 2

R=5729.58 / 2º 30’ 00” 100 ft arc = 2º 30’ 00” / 2

R=2291.83 ft 100 ft arc = 1º 15’ 00”

Tangent Distance Deflection for 50 ft arc

T=R (tan Δ/2) 50 ft arc = D / 4

T=2291.83 (tan 11º 00’ 00”/2) 50 ft arc = 2º 30’ 00” / 4

T=220.68 ft 50 ft arc = 0º 37’ 30”

Length of Curve Deflection for 25 ft arc

L=100 (Δ/D) 25 ft arc = D / 8

L=100 (11º 00’ 00”/2º 30’ 00”) 25 ft arc = 2º 30’ 00” / 8

L=440.00 ft 25 ft arc = 0º 18’ 45”

External Distance Deflection for 1 ft arc

E=T (tan Δ/4) 1 ft arc = D / 200

E=220.68 (tan 11º 00’ 00”/4) 1 ft arc = 2º 30’ 00” / 200

E=10.60 ft 1 ft arc = 0º 00’ 45”

PC Station Chord Length 100 ft arc

PC = PI Station – Tangent Distance 100’ arc = 2R (sin deflection angle)

PC = 107+67.90 – 220.68 100’ arc = 2 (2291.83) sin 1º 15’ 00”

PC = 105+47.22 100’ arc = 99.99 ft

PT Station Chord Length 50 ft arc

PT = PC Station + Curve Length 50’ arc = 2R (sin deflection angle)

PC = 105+47.22 + 440.00 50’ arc = 2 (2291.83) sin 0º 37’ 30”

PC = 109+87.22 50’ arc = 50.00 ft

Calculate the deflection for the first station from P.C. or any odd station along the curve.

1. Take the distance from the last point with a known deflection to the station you are

calculating.

2. Multiply this distance by the deflection of a 1 foot arc (D/200); this will give you the

deflection between these two points.

Example: Find the deflection angle at Sta 108+55.

(108+55 – 105+47.22) = 307.78’ 307.78’ * (0º 0’ 45”) = 3º 50’ 50”


Field Book for Circular Curve

Sta.

Dist.

Chord

Dist

Defl.

Angle

Total

Defl.

105+00

105+47.22 0 0 0 0 P.C.

105+50 2.78 2.78 0º 02’ 05” 0º 02’ 05”

106+00 50 50.00 0º 37’ 30” 0º 39’ 35”

106+50 50 50.00 0º 37’ 30” 1º 17’ 05”

107+00 50 50.00 0º 37’ 30” 1º 54’ 35”

107+50 50 50.00 0º 37’ 30” 2º 32’ 05”

108+00 50 50.00 0º 37’ 30” 3º 09’ 35”

108+50 50 50.00 0º 37’ 30” 3º 47’ 05”

109+00 50 50.00 0º 37’ 30” 4º 24’ 35”

109+50 50 50.00 0º 37’ 30” 5º 02’ 05”

109+87.22 37.22 37.22 0º 27’ 55” 5º 30’ 00” P.T.

Calc by KAB 7-20-93 Check by AN 7-21-93

Δ = 11º 00’ 00” D = 2º 30’ 00”

Deflection Angles Chord Length

100 ft arc = D / 2 = 2º 30’ 00” / 2 100’ arc = 2R (sin deflection angle)

= 1º 15’ 00” = 2 (2291.83) sin 1º 15’ 00”

50 ft arc = D / 4 = 2º 30’ 00” / 4 = 99.99 ft

= 0º 37’ 30” 50’ arc = 2R (sin deflection angle)

1 ft arc = D / 200 = 2º 30’ 00” / 200 = 2 (2291.83) sin 0º 37’ 30”

= 0º 00’ 45” = 50.00 ft

P.T. (Note: Total deflection should equal Δ/2)


Circular Curve Practice Problem Given a circular horizontal curve with a central angle of 29º 42’ 00” and a radius of

700 feet, find the tangent length and the arc length.

Solution:

Delta Angle = 29°42'00"

Degree of Curvature = 8°11'06"

Radius = 700.00 ft

Circular Curve Length = 362.85 ft

Tangent Distance = 185.60 ft

Circular Curve Long Chord = 358.81 ft

Middle Ordinate = 23.38 ft

External = 24.19 ft


Circular Curve Practice Problem Given a curve to the right with PI at 10+71.78, T = 375.60 ft, R = 1150.00 ft, compute chord

and deflection data for all even 100 ft stations within the curve.


Solutions:

Delta Angle = 36°10'30"

Degree of Curvature = 4°58'56"

Radius = 1,150.00 ft

Circular Curve Length = 726.08 ft

Tangent Distance = 375.60 ft

Circular Curve Long Chord = 714.08 ft

Middle Ordinate = 56.83 ft

External = 59.78 ft

PI Stationing = 10+71.78

Incremental chord solution:

Station Chord Deflection

Increment

Deflection

Angle

14+22.26 PT 22.26 0°33'16" 18°05'15"

14+00.00 99.97 2°29'28" 17°31'59"

13+00.00 99.97 2°29'28" 15°02'31"

12+00.00 99.97 2°29'28" 12°33'03"

11+00.00 99.97 2°29'28" 10°03'35"

10+00.00 99.97 2°29'28" 7°34'07"

9+00.00 99.97 2°29'28" 5°04'39"

8+00.00 99.97 2°29'28" 2°35'11"

7+00.00 3.82 0°05'43" 0°05'43"

6+96.18 PC

Total chord solution:

Station Chord Deflection

Increment

Deflection

Angle

14+22.26 PT 714.08 0°33'16" 18°05'15"

14+00.00 692.89 2°29'28" 17°31'59"

13+00.00 596.91 2°29'28" 15°02'31"

12+00.00 499.80 2°29'28" 12°33'03"

11+00.00 401.75 2°29'28" 10°03'35"

10+00.00 302.94 2°29'28" 7°34'07"

9+00.00 203.55 2°29'28" 5°04'39"

8+00.00 103.79 2°29'28" 2°35'11"

7+00.00 3.82 0°05'43" 0°05'43"

6+96.18 PC


Circular Curve Practice Problem This curve cannot be staked entirely from the PC. While occupying the PC, you can only

stake up through Station 46+00.00 due to an obstruction that prevents you from seeing

the remainder of the points. What can you do?

Δ = 38°23'06"

D = 4°25'40"

R = 1,294.00 ft

L = 866.91 ft

T = 450.43 ft

LC = 850.79 ft

M = 71.92 ft

E = 76.15 ft

PI Stationing = 47+16.26

Station Chord Defl. Increment Defl. Angle

51+32.74 PT 32.74 0°43'29" 19°11'33"

51+00.00 99.98 2°12'50" 18°28'04"

50+00.00 99.98 2°12'50" 16°15'14"

49+00.00 99.98 2°12'50" 14°02'24"

48+00.00 99.98 2°12'50" 11°49'34"

47+00.00 99.98 2°12'50" 9°36'44"

46+00.00 99.98 2°12'50" 7°23'54"

45+00.00 99.98 2°12'50" 5°11'03"

44+00.00 99.98 2°12'50" 2°58'13"

43+00.00 34.17 0°45'23" 0°45'23"

42+65.83 PC Incremental chord solution


Horizontal Curve Layout – Tangent Offset

22 XRRY

R = curve radius

X = Distance along tangent to set out point

Y = Offset from tangent

Tangent Offset Example Problem

Given R = 40 feet, find offsets from the tangent to the curve at 2 foot increments along

the tangent.

22 XRRY

X (feet) Y (feet) X (feet) Y (feet)

0 0

2 0.05 22 6.59

4 0.20 24 8.00

6 0.45 26 9.60

8 0.81 28 11.43

10 1.27 30 13.54

12 1.84 32 16.00

14 2.53 34 18.93

16 3.34 36 22.56

18 4.28 38 27.51

20 5.36 40 40.00


Tangent Offset Practice Problem Given R = 75 feet, find offset Y at 5 foot increments along the tangent. In addition, find

the offset at X = 67.5 feet and at X = 72.5 feet.


Horizontal Curve Layout – Chord Offset

2

2222 CRXRY X = Distance from chord midpoint to set out point

Y = Offset from chord R = curve radius C = Chord length

Chord Offset Example Problem

Given R = 636.62 feet and C = 100.00 feet, find offsets from the chord to the curve at

5 foot increments along the tangent.

2

2222 CRXRY

X (feet) Y (feet) X (feet) Y (feet)

0 1.97

5 1.95 30 1.26

10 1.89 35 1.00

15 1.79 40 0.71

20 1.65 45 0.37

25 1.48 50 0.00


Chord Offset Practice Problem Given R = 20 feet and C = 28.28 feet (Δ = 90º), find offsets at increments of C/8.


Vertical Curves

Two major methods of computation: Tangent offset & Equation of Parabola.

Information Needed: Grade or slope on each side of curve.

Elevation and station of PVI.

Curve length (Horizontal distance PVC - PVT)

Typical Vertical Curve Diagram:

Tangent Offset Method:

Procedure:

1. Compute the elevation of the PVC and PVT.

2. Compute elevation of Chord midpoint.

3. Compute offset to curve at midpoint.

4. Determine total number of stations covered.

5. Determine tangent elevations at stations.

6. Compute curve offset at stations.

7. Combine data and determine vertical curve elevations.

Equation Of Parabola Method:

Equation: r = g2 – g1 / L r = change in grade per station

g1 = initial grade

g2 = final grade

L = length of curve in stations

Procedure:

1. Compute PVC and PVT elevations.

2. Calculate total change in grade/station.

3. Insert data to chart and compute final curve elevations.


Tangent Offset Vertical Curve Example Problem

PVC Station & Elevation 50+00 – 400’ = 46+00 Station

550.97 + (400’ * 6.0%) = 574.97 elevation

PVT Station & Elevation 50+00 + 400’ = 54+00 Station

550.97 + (400’ * 2.0%) = 558.97 elevation

Elevation of chord at midpoint: (574.97 – 558.97) / 2 = 566.97

Offset to curve at midpoint: (566.97 – 550.97) / 2 = 8.00’

4 stations each side 46+00 – 50+00: Subtract 6.00’ / station

50+00 – 54+00: Add 2.00’ / station

Determine Curve Offset:

47+00 & 53+00: (1/4)2 * 8.00’ = 0.50’

48+00 & 52+00: (1/2)2 * 8.00’ = 2.00’

49+00 & 51+00: (3/4)2 * 8.00’ = 4.50’

Compute tangent elevations & vertical curve elevations:

47+00: 574.97 – 6.00 = 568.97 + 0.50 = 569.47 VC elev.

48+00: 568.97 – 6.00 = 562.97 + 2.00 = 564.97 VC elev.

49+00: 562.97 – 6.00 = 556.97 + 4.50 = 561.47 VC elev.

50+00: 556.97 – 6.00 = 550.97 + 8.00 = 558.97 VC elev.

51+00: 550.97 + 2.00 = 552.97 + 4.50 = 557.47 VC elev.

52+00: 552.97 + 2.00 = 554.97 + 2.00 = 556.97 VC elev.

53+00: 554.97 + 2.00 = 556.97 + 0.50 = 557.47 VC elev.

54+00: 556.97 + 2.00 = 558.97 PVT elev.


Equation Of Parabola Vertical Curve

Example Problem

PVC Station and Elevation: 50+00 – 400 = 46+00 PVC Station

550.97 + (400 X 6.0%) = 574.97 elevation

PVT Station and Elevation: 50+00 + 400’ = 54+00 PVT Station

550.97 + (400 X 2.0%) = 558.97 elevation

Total change in grade / station (r): r = 2.0 – (-6.0) / 8 = 1.00%

VC Elev. = PVC Elev. + g1X + r/2 X2

Station X X2 r/2X

2 g1X PVC Elev. VC Elev.

PVC 46+00 0 0 0 0 574.97 574.97

47+00 1 1 0.5 -6.0 574.97 569.47

48+00 2 4 2.0 -12.0 574.97 564.97

49+00 3 9 4.5 -18.0 574.97 561.47

PVI 50+00 4 16 8.0 -24.0 574.97 558.97

51+00 5 25 12.5 -30.0 574.97 557.47

52+00 6 36 18.0 -36.0 574.97 556.97

53+00 7 49 24.5 -42.0 574.97 557.47

PVT 54+00 8 64 32.0 -48.0 574.97 558.97


Station and Elevation of Low/High Point

(Based on the Equation of Parabola Method)

The lowest point on a sag vertical curve or the highest point on a crest vertical curve lies at a

distance X stations (X * 100 ft) from the PVC of the curve.

r

gX 1

Substitute this value of X into the equation below to find the elevation of the high point or low

point.


Low/High Point Example Problem

r

gX 1 6

0.1

0.6X Distance = X*100 ft = 6 * 100 ft = 600 ft

46+00 + 600ft = 52+00 low point station


574.97 + (-6.0)*6 + (1.0/2)*62 = 556.97 low point elevation


Vertical Curve Practice Problem

Given the following vertical curve data, compute the elevations of the curve summit and

even full stations (that is, 100 ft stations).

PVI at 19+00, elevation 723.86

LVC = 500 ft

g1 = +2.5%

g2 = -1.0%

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