math for surveyors - esri · 2011. 6. 28. · topics covered. 1) the right triangle 2) oblique...

Math For Surveyors

James A. Coan Sr. PLS

Topics Covered1) The Right Triangle

2) Oblique Triangles

3) Azimuths, Angles, & Bearings

4) Coordinate geometry (COGO)

5) Law of Sines

6) Bearing, Bearing Intersections

7) Bearing, Distance Intersections

Topics Covered8) Law of Cosines

9) Distance, Distance Intersections

10) Interpolation

11) The Compass Rule

12) Horizontal Curves

13) Grades and Slopes

14) The Intersection of two grades

15) Vertical Curves

The Right

Triangle

Side Adjacent (b)

Sid

e O

ppos

ite (a

)

A

B

C

CosA bc

= TanA ab

=SineA ac

=

CscA ca

= SecA cb

= CotA ba

=

The above trigonometric formulas

Can be manipulated using Algebra

To find any other unknowns

The Right Triangle

SinA c a· = aSinA

c=

The Right Triangle

Example:

CosA c b· = bCosA

c=

TanA b a· = aTanA

b=

SinA ac

=

CosA bc

=

TanA ab

=

Oblique Triangles

An oblique triangle is one that does

not contain a right angle

Oblique Triangles

This type of triangle can be solved

using two additional formulas

a

Sin A=

b

Sin B

c

Sin C=

The Law of Sines

Oblique Triangles

A B

C

ab

c

Oblique Triangles

The law of Cosines

a2 = b2 + c2 - 2bc Cos A

A B

C

ab

c

Oblique Triangles

When solving this kind of triangle we can

sometimes get two solutions, one solution,

or no solution.

Oblique Triangles

When angle A is obtuse (more than 90°) and side a is shorter than or equal to side c, there is no solution.

A B

C

ab

c

Oblique Triangles

When angle A is obtuse and side a is greater than side c then side a can only intersect side b in one place and there is only one solution.

AB

C

ab

c

Oblique Triangles

When angle A is acute (less than 90°) and side a is longer than side c, then there is one solution.

A B

C

ab

c

Oblique Triangles

When angle A is acute, and the height is given by the formula h = c Cos A, and side a is less than h, but side c is greater than h, there isno solution.

A B

b

c

a

h

Oblique Triangles

When angle A is acute and side a = h, and h is less than side c there can be only one solution

A B

C

a = hb

c

Oblique TrianglesWhen angle A is an acute angle and h is less than side a as well as side c, there are two solutions.

A B

C

C’a

a’

b

c

h

Azimuth

Angles

Bearings

Azimuth, Angles, & BearingsAzimuth:

An Azimuth is measured clockwise from North.

The Azimuth ranges from 0° to 360°

Azimuth, Angles, & Bearings

Azimuth:0°

90°

180°

270°

360°N

Azimuth, Angles, & BearingsAzimuth to Bearings

In the Northeast quadrant the Azimuth and Bearing is the same.

N

E

AZ = 50°30’20” = N 50°30’20”E


In the Southeast quadrant, subtract the Azimuth from 180° to get the Bearing.

180° - 143°23’35” = S 36°36’25”E

Azimuth to Bearings


In the Southwest quadrant, subtract 180°from the Azimuth to get the Bearing

205°45’52” – 180° = S 25°45’52”W


In the Northwest quadrant, subtract the Azimuth from 360° to get the Bearing.

360° - 341°25’40” = N 18°34’20”W

Bearings to Azimuths


In the Northern hemisphere Bearings are

measured from North towards East or West

N 47°30’46”E

N 53°26’52”W



In the Southern Hemisphere the Bearings are measured from South to East or West

S 71°31’40”E

S 29°25’36”W



In the Northeast quadrant, the Bearing and Azimuth are the same.

N 45°30’30”E = AZ 45°30’30”

N



In the Southeast quadrant, subtract the Bearing from 180° to get the Azimuth.

180° - S 51°25’13”E = AZ 128°34’47”



In the Southwest quadrant, add the Bearing to 180° to get the Azimuth.

S 46°20’30”W + 180° = AZ 226°20’30”



In the Northwest quadrant, subtract the Bearing from 360° to get the Azimuth.

360° - N 51°25’41”W = AZ 308°34’19”


Angles:

To find the Angle between two Azimuths, subtract the smaller Azimuth from the larger Azimuth.

325°50’30” Larger Azimuth215°20’10” Smaller Azimuth110°30’20” Angle

Azimuth, Angles, & BearingsAngles:

If both bearings are in the same quadrant, subtract the smaller bearing from the larger bearing.

S 82°35’40”ES 25°15’10”E

57°20’30”

Angles:


If both angles are in the same hemisphere(NE and NW) or (SE and SW), add the two bearings together to find the angle

N 30°15’26”EN 21°10’14”W

51°25’40”


Angles:

If one bearing is in the NE and the other is in the SE or (NW and SW), add the two together and subtract the sum from 180°

180°-(N15°50’25”W+S 20°10’15”W)=143°59’20”

Coordinate Geometry

COGO

The science of coordinate geometry states

that if two perpendicular directions are

known such as an X and Y plane (North

and East).

Coordinate Geometry

The location of any point can

be found with respect to the origin of

the coordinate system,

Coordinate Geometry

or with respect to some other known point on

the coordinate system.

Coordinate Geometry

This is accomplished

by finding the deference between the X

and Y coordinates (North and East) of a

known and unknown point and adding

that deference to the known point.

Coordinate Geometry

The magnitude and direction (Azimuth and

distance) can also be found between

two points if the coordinates of the two

points are known.

Coordinate Geometry

East

Nort

h

A

BC

SineA EastB D

=D

&

CosA NorthB D

=D

&

TanA EastNorth

=DD

CscA B DEast

=&

D

SecA B DNorth

=&

D

CotA NorthEast

=DD

Coordinate Geometry

This will give you the angle from Pt. A to Pt. B

Coordinate Geometry

Dist North East= +D D2 2

Pythagorean Theorem

This will give you the distance from Pt.A to Pt.B

Example 1

•The coordinates for point A

Known:

•The bearing and distance from point

A to point B

Coordinate Geometry

Coordinate Geometry

Example 1

Point A coordinates

N 10,000.00 E 5,000.00

The bearing from Point A to point B

N 36°47’16”E

The distance from Point A to Point B

1,327.56 feet

Coordinate Geometry

A

B

N 10,000.00

E 5,000.00

East

Nort

h

Example 1:

Coordinate Geometry

Warning!

You must convert the degrees,

minutes, and seconds of your bearing

to decimal degrees before you find the

trig function

Coordinate Geometry

Example 1

Find North:

Cos. Bearing x Distance = D North

Cos. N 36°47’16”E x 1,327.56’ = 1,063.19’

Coordinate Geometry

Example 1:

Find East

Sine Bearing x Distance = D East

Sine N 36°47’16”E x 1,327.56’ = 795.01’

Coordinate Geometry

Example 1:

Because point B is Northeast of point A

you must add your calculated distances

(both North and East) to the

coordinates of A to find the coordinates

of point B

Coordinate Geometry

Example 1:

North A + North = North B

East A + East = East B

Coordinate Geometry

Example 1:

N 10,000.00’ + 1,063.19’ = 11,063.19’

E 5,000.00’ + 795.01’ = 5,795.01’

Point B

North = 11,063.19’

East = 5,795.01’

Coordinate Geometry

Example 2:

Given:

Coordinates of point A

N 10,000.00 E 5,000.00

Coordinates of point B

N 10,978.69’ E 3,924.71’

Coordinate Geometry

Example 2: Point B

N 10,978.69’

E 3,924.71’

Point A

N 10,000.00

E 5,000.00

Note:

Point B is

Northwest of

Point A

Coordinate Geometry

Example 2:

First

Find the deference in North between point A and point B

Point B = 10,978.69’

Point A = 10,000.00’

Deference = 978.69’

Coordinate GeometryExample 2:

Second

Find the deference in East between point A and point B

Point A = 5,000.00’

Point B = 3,924.71’

Deference=1,075.29’

Coordinate Geometry

Example 2:

ThirdFind the distance between point A and point B

The distance from A to B = 1,453.99’

Dist = D North2 + D East2

Dist = 978.692 + 1,075.292

Coordinate Geometry

Example 2:

Fourth

Find the bearing from point A to point B

1,075.29

978.69Tan A = Tan A =

D NorthD East

Coordinate Geometry

Example 2:

Fifth

The angle from point A to point B is

47°41’34”

Because point B is Northwest of point A the bearing is N 47°41’34”W

The Law of Sines

a

Sin A=

b

Sin B

c

Sin C=

A B

C

c

The Law of Sines

The law of Sines can be used to

solve several Surveying problems,

such as finding the center of

section

The Law of Sines

Example 1:

Given

Coordinates for all 4 section quarter corners

The center quarter corner

Need to find

The Law of Sines

a

b

c

d

Points

a, b, c, & d

Have known coordinates

The Law of Sines

a

b

c

d

First

Inverse

between points

c and d

The Law of Sines

a

b

c

d

This gives a

bearing and

distance from c

to d

The Law of Sines

a

b

c

d

Next

Inverse between a & c

And inverse between d & b

The Law of Sines

a

b

c

dBear & DistB

ear

& D

ist

After inversing

you will have a

bearing and

distance

between a & c

as well as d & b

The Law of Sines

Because the bearings of all three

lines are known the angles between

them can be calculated.

The Law of Sines

a

b

c

dBearing

Bea

ring

Angle

Angle

Angle

The Law of Sines

What we now know:

The bearing from c to d

The bearing from d to b

The bearing from c to a

The Law of Sines

What we now know:

The angle at d

The angle at c

The angle at the center of section (e)

The distance from c to d

The Law of Sines

a

b

c

dBearing

Bea

ring

Angle

Angle

Angle (e)

Law of Sines

We can now solve for the following:

The distance from d to e

or

The distance from c to e

or

Both distances

By using the Law of Sines

Law of Sines

Dist. d-c

Angle e=

Dist. c-e

Angle d

Dist c-e = (Dist d-c)(Angle d)

Angle e

(Dist d-c)(Angle d)=(Dist c-e)(Angle e)

Law of Sines

At this point we have a known bearing

and distance from point c ( with known

coordinates) to point e (the center of

section)

Law of Sines

We now have all of the information

we need to calculate the coordinates

at the center of section ( the center

¼ corner)

Law of Sines

In Surveying this type of

a problem is called a

Bearing; Bearing Intersection


N ¼ Cor.

E ¼ Cor.

S ¼ Cor.

W ¼ Cor.

Center ¼ corner

Bearing; Bearing IntersectionGiven:

W ¼ Cor.; N=12,645.70, E=5,021.63

N ¼ Cor.; N=15,234.25, E=7,705.86

E ¼ Cor.; N=12,532.42, E=10,319.91

S ¼ Cor.; N=10,008.06, E=7,510.70

Bearing; Bearing IntersectionFirst:

Find the difference in North and East

from the South ¼ corner and the West

¼ corner.

North = 2,637.64’

East = 2,489.07’


Second:

Find the distance by inverse between the

South ¼ corner and the West ¼ corner.

Distance = 2,637.642+2,489.072

Distance = 3,626.65’


Third:

Find the bearing by inversing between

the South ¼ corner and West ¼ corner

Bearing = Tan-12,489.07

2,637.64

Bearing = N 43°20’24”W


Fourth:

Find the bearing between the South ¼

corner and the North ¼ corner.

Bearing = Tan-1195.16’

5,226.19’

Bearing = N 02°08’19”E


Fifth:

Find the bearing between the West ¼

corner and the East ¼ corner.

Bearing = Tan-15,289.28’

113.28’’

Bearing = S 88°46’23”E


We now have the following:

S 88°46’23”E

N 0

2°0

8’1

9”E

N ¼

E 1/4

S 1/4

W ¼


Sixth:Calculate the angles between the bearings:

S 88°46’23”E

N 0

2°0

8’1

9”E

A

B

C


Angle A:

S 88°46’23”E

S 43°20’24”E

45°25’59”


Angle B:

N 43°20’24”W

N 02°08’19”E

45°28’43”

+


Angle C:

180°-(45°25’59”+45°28’43”)=89°05’18”

Check: N 88°46’23”W

S 02°08’19”W

90°54’42”

+

180° - 90°54’42” = 89°05’18”

C = 180°-(A + B)


Now we have:

45°25’59”

45°28’43”

89°05’18”

We can use the Law of Sines to solve for one of the unknown sides.


Seven:

Solve for the distance from the south

quarter corner ( S ¼) to the center of

section (C ¼ Cor.)

OR

The distance from the West quarter

corner ( W ¼) to the center of section

( C ¼ Cor.)


3,626.65’

Sin 89°05’18=

Dist. S1/4 to C ¼

Sin 45°25’59”

Dist. = (3,626.65’)(Sin 45°25’59”)

Sin 89°05’18”

Dist = 2,584.07’

Distance from the S ¼ cor. to the C ¼ cor.


Now we have the bearing and

distance from a known coordinate

(the south ¼ corner) to an

unknown point (the center of

section)


Eight:

Use Coordinate Geometry to

calculate the coordinates of the

center of section


Cos. Bearing x distance = North

Sin. Bearing x distance = East

Cos. N 02°08’19”E x 2,584.07’ = 2,582.27’

Sin. N 02°08’19”E x 2,584.07’ = 96.43’


Because the bearing from the S ¼ cor.

To the center of section is Northeast you

must add both the North and the

East to the known coordinates at the

S ¼ corner to get the coordinates of the

center of section.


S ¼ North = 10,008.06’

Delta North = 2582.27’

C ¼ North = 12,590.33’

S ¼ East = 7,510.70’

Delta East = 96.43’

C ¼ East = 7,607.13’

Another way the Law of Sines

is used in Surveying is

calculating a

Bearing; Distance intersection

N 89°30’15”E 352.25’

N 0

0°1

0’2

5”E

205.3

6’

AB

C

D

Bearing; Distance Intersection

Example:

Smith Property

Bearing; Distance IntersectionGiven:

A = N 10,003.05’ ; E 5,352.24’

C = N 10,205.36’ ; E 5,000.62’

Bearing from C to D = N 74°56’30”E

Distance from A to D = 312.37’

We need to find the coordinate for point D

N 89°30’15”E 352.25’

N 0

0°1

0’2

5”E

205.3

6’

AB

C

D


D’

CAUTION!!

There can be two answers to this problem


Because there can be two answers

to this type of problem the surveyor

must have an understanding of

what they are looking for.

There is no magic bullet


First:

Inverse between A and C

A to C, North = 202.31’

A to C, East = 351.62’

Bearing, A to C = N 60°05’07”W

Distance, A to C = 405.67’

N 89°30’15”E 352.25’

N 0

0°1

0’2

5”E

205.3

6’

AB

C

D


We now have:

C’

We need to find


Bearing C-D = N 74°56’30”E

Bearing C-A = S 60°05’07”E

Angle C’ = 180°-(Bearing C-D + Bearing C-A)

Angle C’ = 180°-(74°56’30” + 60°05’07”)

Angle C’ = 44°58’23”

Second:

N 89°30’15”E 352.25’

N 0

0°1

0’2

5”E

205.3

6’

AB

C

D


We now have:

44°58’23”

All we need to find Angle D


Third:

Use the Law of Sines to Find Angle D

312.37’

Sin. 44°58’23”=

405.67’

Sin. Angle D

Sin. D = (Sin 44°58’23”)(405.67’)

312.37’


The Sine of D = 0.917876488…

Angle D = 66°37’03”

Now we can find the Bearing from A to D

N 89°30’15”E 352.25’

N 0

0°1

0’2

5”E

205.3

6’

AB

C

D


We now have:

44°58’23”

66°37’03”


Forth:

Calculate the bearing from D to A

Bearing D to C = S 74°56’30”W

Angle D = 66°37’03”

Bearing from D to A = S 08°19’27”W


Now we have a bearing and distance

from point A, a known coordinate, to

point D


Use coordinate geometry to

calculate the coordinates of point D

North = 309.08’

East = 45.22’


Northing of A = 10,003.05

North A to D = 309.08’

Northing of D = 10,312.13’

Easting of A = 5,352.24’

East A to D = 45.22’

Easting of D = 5,397.46’

Finish:

The last intersection problem we

need to discuss is the

Distance, Distance

Intersection

In order to solve a Distance, Distance

Intersection we need to use

The Law of Cosines!

The Law of Cosines can be used when

you have a Triangle with all three

distances but no angles.

A

C

BDistance

Example:

The Law of Cosines

a2 = b2 + c2 - 2bc Cos A

Solving for Cos A, we get

a2 – b2 – c2

-2bcCos A =

As stated, using the Law of Cosines a

surveyor can solve a Distance,

Distance Intersection Problem

WARNINGYou can get two answers to this kind

of a problem

Distance, Distance Intersection

Pt A North East

Pt B North East

c

C’


Problem:

Find the coordinates for Point C

Given:

Coordinates for points A and B

Distance from point A to point C

Distance from point B to point C


Needed:

The coordinate for Point “C”


Example:

Point A:

North = 10,104.94’

East = 5,910.69’


Example:

Point B:

North = 10,108.47’

East = 6,383.80’


Example:

North = 3.53’

East = 473.11’


Example:

First:

Inverse between points A and B

To find the bearing and distance


Example:

Tan-1473.11’

3.53’=89.572508828°

89.572508828° = 89°34’21”


Example:

Because point B is North and

East of Point A, the bearing

becomes:

N 89°34’21”E


Example:

3.532 + 473.112 = 473.12’


Example:

We now have:

A B


Example:

The distance from point A to point C is

192.49’

The distance from point B to point C is

339.44’


Example:

Now we have:

AB

C


Example:

We need to use the Law of Cosines to calculate one of the angles.

Cos A = a2 – b2 – c2

-2bc


Cos A = 339.442 – 192.492 – 473.122

-2(192.49)(473.12)

Cos A = 0.799791540

Angle A = 36°53’23”

Example:


Example:

Now we have

1) The bearing from Pt. A to Pt. B

2) The angle at Point A

We can calculate a bearing from Pt. A

to Pt. C


Example:

A

B

C


Example:

We now have:

1) A coordinate at point A

2) A bearing from point a to point C

3) A distance from point A to point C


Example:

A

N 10,104.94’

E 5,910.69’

C


Example:

We can calculate the coordinates at

point c by using coordinate geometry

(Cogo)


Example:

North = 116.69’

Cos 52°40’58” x 192.49’North =


Example:

East = 153.09’

Sine 52°40’58” x 192.49’East =


Example:

Northing coordinate at C =

North A = 10,104.94

North A to C = 116.69’

North C = 10,221.63’


Example:

Easting coordinate at C =

East A = 5,910.69’

East A to C = 153.09’

East C = 6,063.78’


Example:

Coordinates at C

North = 10,221.63’

East = 6,063.78’

The Compass Rule

( Bowditch Rule)

The Compass Rule

Mainly used for:

1) Traverse closure computations

2) Used throughout the Public LandSurvey System (PLSS)

It also has many other applications in Surveying.

The Compass Rule

The Formula:

Correction = CL

S

C = The total error in Latitude (D North) or Departure (D East) with the sign changed.

L = The total length of the Survey.

S = The length of a particular course.

The Compass Rule Example:

A

C

B

C’

N10,000.00’E 5,000.00’A= N10,199.16’

E 5,408.96’C’=

RecordInfo.

FoundFound

The Compass Rule Example

Need to find:

The corrected coordinates for point B


First:

Using the record information calculate the

coordinates for points B and C


Second:

Calculate the Latitude (D North) and the

Departure (D East) from point C’ to point C


Third

Use the Compass Rule to calculate the

corrections for point B


Record coordinates for point B

N 10,131.05’ E 5,204.85’

Record and field coordinates for point A

N 10,000.00’ E 5,000.00’


Field coordinates for point C’

N 10,199.16 E 5,408.96

Record coordinates for point C

N 10,200.37’ E 5,408.15’


C coordinates = N10,200.37’ E 5,408.15’

C’ coordinates= N10,199.16’ E 5,408.96’

1.21 -0.81


Total length of the survey = 457.98’

Length from point A to point B = 243.19’

Total error in Latitude with the sign changed = -1.21’

Total error in Departure with the sign changed = 0.81’


Latitude from point A to point B = 131.05’

Departure from point A to point B = 204.85’


Correction of the Latitude from point A to

point B using the Compass Rule is

-1.21’457.98’

x 243.19’ = -0.64’


Correction of the Departure from point A to

point B using the Compass Rule is

0.81’457.98’

x 243.19’ = 0.43’


Corrected Latitude=131.05’ + (-0.64’) = 130.41’

Corrected Departure = 204.85’ + 0.43’ = 205.28’

Corrected coordinates for point B

N 10,000.00’ + 130.41’ = 10,130.41’

E 5,000.00’ + 205.28’ = 5,205.28’

q.e.d.

Interpolation

Determination of an intermediate value

between fixed values from some known or

assumed rate or system of change.

(Definitions of Surveying and Associated

Terms – American Congress on Surveying and Mapping)

Interpolation:

Interpolation:

y2 = (x2 – x1)(y3 – y1)

(x3 – x1)+ y1

Formula

Example:

Given

x1 = 42°31’00”

y1 = 0.9168665 (tangent of x1)

x2 = 42°31’17”

y2 = Unknown (tangent of x2)

x3 = 42°32’00”

y3 = 0.9174020 (tangent of x3)

Example:

Find: the tangent of 42°31’17” by interpolation

y2 =(42°31’17” – 42°31’00”)(0.9174020 - 0.9168664)

(42°32’00” – 42°31’00”)+ 0.9168665

Y2 = 0.9170182

Interpolation:

What did we do?

Interpolation:

You can quickly see that we have calculated

17/60 of the difference between the two given

tangents then added this number to the

tangent of 42°31’00”

Example 2

0.9174020 – 0.9168665 = 0.0005355

0.0005355 x 17/60 = 0.0001517

0.9168665 + 0.0001517 = 0.9170182

Horizontal

Curves

PC PT

Horizontal Curve

Horizontal Curve

PC PT

Parts of a Curve

RP

Arc

PC PT

RP

PI

Horizontal CurveParts of a Curve

PC PT

RP

PI


Chord

PC PT

RP

PI


Chord

Delta Angle

Delta Angle

PC PT

RP

PI


Chord

E

M

CL Curve

E = External M = Middle Ordinate

PC PT

RP

PI


CL Curve½ Delta

½ Delta

Horizontal Curve

Formulas: Length of Arc:

360° (2pR)Length of Arc (L) =

Horizontal Curve

Formulas: Tangent Distance (T)

Tangent (T) = Radius (Tan D/2)

Horizontal Curve

Formula: Chord Distance (C)

Chord Distance (C) = 2R SinD/2

Horizontal Curve

Formula: Radius (R)

T

TanD/2Radius (R) =

OR

Radius (R) = T CotD/2

Horizontal Curve

Degree of Curve:

100’

1°

Degree of Curve (D) = 5729.58’R

NOTE:

Arc distance must always be 100’

Horizontal Curve

Formula: Delta Angle (D)

Delta Angle (D) = 180°LpR

Horizontal CurveFormula: External

R

Cos D/2External (E) = - R

Horizontal Curve

Formula: Middle Ordinate

Middle Ordinate (M) = (Sin D/2) T - E

Horizontal CurveExample:

Given:

Length of Arc (L) = 396.72’

Radius (R) = 526.54’

Horizontal CurveExample:

Find:

Tangent Distance (T)

Length of Chord (C)

Radius (R)

Degree of Curve (D)

The Delta Angle (D)

The External (E)

The Middle Ordinate (M)

Horizontal CurveFind the Delta Angle (D) Delta (D) = 180°L

p R

Delta (D) = 180° x 396.72’3.1415927 x 526.54’

Delta (D) = 43.169334995° = 43°10’10”

Half Delta (D/2) = 21°35’05”

Horizontal CurveFind the Tangent (T) Tangent (T) = R Tan D/2

Tangent (T) = 526.54’ x Tan 21°35’05”

Tangent (T) = 208.31’

Horizontal CurveChord Distance (C) = 2R SinD/2

Chord (C) = 2 x 526.54’ x Sin 21°35’05”

Chord (C) = 387.40’

Horizontal CurveDegree of Curve (D) = 5729.58’

R

Degree of Curve (D) = 5729.58’526.54’

Degree of Curve (D) = 10.88156645°

Degree of Curve (D) = 10°52’54”

Horizontal Curve

External (E) = - R RCos D/2

External (E) = - 526.54’526.54’Cos 21°35’05”

External (E) = 39.71’

Horizontal Curve

Middle Ordinate (M) = (Sin D/2) T - E

(M) = Sin 21°35’05” x 208.31’ – 39.71’

(M) = 36.92’

Horizontal CurveResults: Length of Arc (L) = 396.72’ (given)

Tangent Distance (T) = 208.31’

Length of Chord (C) = 387.40’

Radius (R) = 526.54’ (given)

Degree of Curve (D) = 10°52’54”

The Delta Angle (D) = 43°10’10”

The External (E) = 39.71’

The Middle Ordinate (M) = 36.92’

Horizontal Curve

P.C.

R.P.1

P.I.1

P.R.C.

R.P.2

P.I.2

P.T.

Reverse Curve:

Curve 1

Curve 2

Horizontal CurveCompound Curve:

Tan.

Tan.

Tan.Rad.

Rad

.Rad.P.C.

P.I.1P.C.C. P.I.2

P.T.

R.P.1

R.P.2

Curve 1

Curve 2

Grades

&

Slopes

Grades

A grade is expressed as a calculation of how

steep a slope is either going up or down.

If the slope is going up, the grade is +

If the slope is going down, the grade is -

GradesExample:

Horizontal Distance

Grade = Difference in Elevation

Distance

DE

lev.

GradesExample:

352.45’

Grade = 16.84’352.45’

16.8

4’

= 0.0477798 ft / ft

Grades

Grades can also be expressed as a

Percent (%) by multiplying the grade times 100

0.0478 ft / ft x 100 = 4.78 %

GradesA grade is also the tangent of an angle

Tangent = opposite = D elevation

adjacent = distance

Angle

oppo

site

D

elev

.

adjacent, distance

Grades

Formulas used with grades:

Grade x distance = D Elevation

Grades


Distance =D Elevation

Grade

Grades


Grade =D ElevationDistance

Slopes

A slope is a ratio of the horizontal distance to the vertical distance.

Horizontal distance

Verti

cal

dist

ance

Slopes

Example:

A 2:1 slope down =2.0’

1.0’

A 3:1 slope up = 3.0’ 1.0’

Slopes and Grades

Slopes are expressed as a ratio;

2:1, 5:1, 0.25:1, 8:3, etc

Grades are expressed as ft /’ ft; 0.025 ft/ft

Or as a present ;

2.0%, 10.34%, 7.62%, etc

Locating the

Intersection of Two

Grades

The Intersection of two GradesThe purpose of locating the intersection of

two grades is to fix the point of intersection

(PVI) of those grades.

Sta

tion 1

Ele

v 1

Sta

tion 2

Ele

v 2

PV

I Sta

.?

Ele

v. ?

The Intersection of two Grades

Formulas:

b1 = Elev1 - G1

100x Station1 (in feet)

b2 = Elev2 - G2

100x Station2 (in feet)


Formulas:

PVI Station =

b1 – b2

G1

100-

G2

100


Example:

Station1 = 7+00

Elevation1 = 201.40’

Grade1 = -1.00%

Station2 = 13+00

Elevation2 = 207.50’

Grade2 = +2.00%


b1 = 201.40’ - -1.00100

x 700’ = 208.40

Example:

b2 = 207.50’ -+2.00

100x 1300’ = 181.50


Example:

PVI Station = 208.40 – 181.50

-1.00%100

- +2.00%100

= -896.67

Use the absolute value: -896.67 = 8+96.67

The Intersection of two GradesExample:

Grade x distance = difference in elevation

-0.01 x 196.67’ = -1.97’

Elevation at PVI = 201.40’ – 1.97 = 199.43’

Vertical

(Parabolic)

Curves

Vertical Curves

Vertical curves are used as a

transition from one grade to another

Vertical Curves

Vertical curves are needed in six separate cases. They Are:

+

++

++

+--

--

-

-

Vertical Curves

PV

C

PV

I

PV

T

Length “L”

L / 2 L / 2x

Sump

Vertical Curves

Formulas:

r2 x2 + G1x + PVC ElevationElevation =

G2 – G1

Lr =

x = Distance from the PVC

Vertical Curves

Sump or Peak (Low or High point)

Formula:

-G1rx =

x = Distance from the PVC

The high or low point is Always on the lesser grade side (absolute value)

Vertical Curves

Example:

Given:

G1 = -1.5% = -0.015 ft/ft

G2 = +2.5% = +0.025 ft/ft

Length = 300.00 ft

Vertical Curves

Example:

Given:

PVC Station = 3+50.00; Elevation = 326.25 ft

PVI Station = 5+00.00; Elevation = 324.00 ft

PVT Station = 6+50.00; Elevation = 327.75 ft

Vertical Curves

Need to find:

Elevations at each 50 ft station along the vertical curve.

The Sump (low point) station and elevation

Vertical Curves

First: Calculate “r” G2 – G1

Lr =

0.025 – (-0.015)300.00’

r = = 0.0001333…

Vertical Curves

Second: Calculate elevations


4+00 = 0.00013332

502+(-0.015)(50)+326.25’

Elevation at Station 4+00 = 325.67’

Vertical CurvesStation X Elevation

3+50 PVC 0 326.25’

4+00 50 325.67’

4+50 100 325.42’

5+00 PVI 150 325.50’

5+50 200 325.92’

6+00 250 326.67’

6+50 PVT 300 327.75’ (Chk)

Vertical Curves

Third: Calculate the sump distance

Formula:-G1rx =

-(-0.015)0.0001333…

x = = 112.50’

The Sump Station is at 4+62.50

Vertical Curves

Elevation at the Sump:


4+00 =0.0001333

2112.502+(-0.015)(112.50)+326.25’

Elevation at Station 4+62.50 = 325.41’

Vertical Curves

PVT A

PVC

B

PVI

A

PVI B

PVT

PVC

LA

L/2A L/2A

LB

L/2B L/2B

Unsymmetrical Vertical CurveG3 = G’

Vertical Curves

Calculate each part of the curve as if it was a

regular vertical Curve

Calculate G3 from the center of the first curve

to the center of the second curve

The

End

math for surveyors - esri · 2011. 6. 28. · topics covered. 1) the right triangle 2) oblique...

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