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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
Azimuth, Angles, & BearingsAzimuth to Bearings
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
Azimuth, Angles, & Bearings
In the Southwest quadrant, subtract 180°from the Azimuth to get the Bearing
205°45’52” – 180° = S 25°45’52”W
Azimuth, Angles, & BearingsAzimuth to Bearings
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
Azimuth, Angles, & Bearings
In the Northern hemisphere Bearings are
measured from North towards East or West
N 47°30’46”E
N 53°26’52”W
Azimuth, Angles, & Bearings
Bearings to Azimuths
In the Southern Hemisphere the Bearings are measured from South to East or West
S 71°31’40”E
S 29°25’36”W
Azimuth, Angles, & Bearings
Bearings to Azimuths
In the Northeast quadrant, the Bearing and Azimuth are the same.
N 45°30’30”E = AZ 45°30’30”
N
Azimuth, Angles, & Bearings
Bearings to Azimuths
In the Southeast quadrant, subtract the Bearing from 180° to get the Azimuth.
180° - S 51°25’13”E = AZ 128°34’47”
Bearings to Azimuths
Azimuth, Angles, & Bearings
In the Southwest quadrant, add the Bearing to 180° to get the Azimuth.
S 46°20’30”W + 180° = AZ 226°20’30”
Azimuth, Angles, & Bearings
Bearings to Azimuths
In the Northwest quadrant, subtract the Bearing from 360° to get the Azimuth.
360° - N 51°25’41”W = AZ 308°34’19”
Azimuth, Angles, & Bearings
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:
Azimuth, Angles, & Bearings
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”
Azimuth, Angles, & Bearings
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
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’
Bearing; Bearing Intersection
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’
Bearing; Bearing Intersection
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
Bearing; Bearing Intersection
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
Bearing; Bearing Intersection
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
Bearing; Bearing Intersection
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
Bearing; Bearing Intersection
Sixth:Calculate the angles between the bearings:
S 88°46’23”E
N 0
2°0
8’1
9”E
A
B
C
Bearing; Bearing Intersection
Angle A:
S 88°46’23”E
S 43°20’24”E
45°25’59”
Bearing; Bearing Intersection
Angle B:
N 43°20’24”W
N 02°08’19”E
45°28’43”
+
Bearing; Bearing Intersection
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)
Bearing; Bearing Intersection
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.
Bearing; Bearing Intersection
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.)
Bearing; Bearing Intersection
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.
Bearing; Bearing Intersection
Now we have the bearing and
distance from a known coordinate
(the south ¼ corner) to an
unknown point (the center of
section)
Bearing; Bearing Intersection
Eight:
Use Coordinate Geometry to
calculate the coordinates of the
center of section
Bearing; Bearing Intersection
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’
Bearing; Bearing Intersection
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.
Bearing; Bearing Intersection
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
Bearing; Distance Intersection
D’
CAUTION!!
There can be two answers to this problem
Bearing; Distance Intersection
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
Bearing; Distance Intersection
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
Bearing; Distance Intersection
We now have:
C’
We need to find
Bearing; Distance Intersection
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
Bearing; Distance Intersection
We now have:
44°58’23”
All we need to find Angle D
Bearing; Distance Intersection
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’
Bearing; Distance Intersection
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
Bearing; Distance Intersection
We now have:
44°58’23”
66°37’03”
Bearing; Distance Intersection
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
Bearing; Distance Intersection
Now we have a bearing and distance
from point A, a known coordinate, to
point D
Bearing; Distance Intersection
Use coordinate geometry to
calculate the coordinates of point D
North = 309.08’
East = 45.22’
Bearing; Distance Intersection
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’
Distance, Distance Intersection
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
Distance, Distance Intersection
Needed:
The coordinate for Point “C”
Distance, Distance Intersection
Example:
Point A:
North = 10,104.94’
East = 5,910.69’
Distance, Distance Intersection
Example:
Point B:
North = 10,108.47’
East = 6,383.80’
Distance, Distance Intersection
Example:
North = 3.53’
East = 473.11’
Distance, Distance Intersection
Example:
First:
Inverse between points A and B
To find the bearing and distance
Distance, Distance Intersection
Example:
Tan-1473.11’
3.53’=89.572508828°
89.572508828° = 89°34’21”
Distance, Distance Intersection
Example:
Because point B is North and
East of Point A, the bearing
becomes:
N 89°34’21”E
Distance, Distance Intersection
Example:
3.532 + 473.112 = 473.12’
Distance, Distance Intersection
Example:
We now have:
A B
Distance, Distance Intersection
Example:
The distance from point A to point C is
192.49’
The distance from point B to point C is
339.44’
Distance, Distance Intersection
Example:
Now we have:
AB
C
Distance, Distance Intersection
Example:
We need to use the Law of Cosines to calculate one of the angles.
Cos A = a2 – b2 – c2
-2bc
Distance, Distance Intersection
Cos A = 339.442 – 192.492 – 473.122
-2(192.49)(473.12)
Cos A = 0.799791540
Angle A = 36°53’23”
Example:
Distance, Distance Intersection
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
Distance, Distance Intersection
Example:
A
B
C
Distance, Distance Intersection
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
Distance, Distance Intersection
Example:
A
N 10,104.94’
E 5,910.69’
C
Distance, Distance Intersection
Example:
We can calculate the coordinates at
point c by using coordinate geometry
(Cogo)
Distance, Distance Intersection
Example:
North = 116.69’
Cos 52°40’58” x 192.49’North =
Distance, Distance Intersection
Example:
East = 153.09’
Sine 52°40’58” x 192.49’East =
Distance, Distance Intersection
Example:
Northing coordinate at C =
North A = 10,104.94
North A to C = 116.69’
North C = 10,221.63’
Distance, Distance Intersection
Example:
Easting coordinate at C =
East A = 5,910.69’
East A to C = 153.09’
East C = 6,063.78’
Distance, Distance Intersection
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
The Compass Rule Example
First:
Using the record information calculate the
coordinates for points B and C
The Compass Rule Example
Second:
Calculate the Latitude (D North) and the
Departure (D East) from point C’ to point C
The Compass Rule Example
Third
Use the Compass Rule to calculate the
corrections for point B
The Compass Rule Example
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’
The Compass Rule Example
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’
The Compass Rule Example
C coordinates = N10,200.37’ E 5,408.15’
C’ coordinates= N10,199.16’ E 5,408.96’
1.21 -0.81
The Compass Rule Example
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’
The Compass Rule Example
Latitude from point A to point B = 131.05’
Departure from point A to point B = 204.85’
The Compass Rule Example
Correction of the Latitude from point A to
point B using the Compass Rule is
-1.21’457.98’
x 243.19’ = -0.64’
The Compass Rule Example
Correction of the Departure from point A to
point B using the Compass Rule is
0.81’457.98’
x 243.19’ = 0.43’
The Compass Rule Example
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
Horizontal CurveParts of a Curve
Chord
PC PT
RP
PI
Horizontal CurveParts of a Curve
Chord
Delta Angle
Delta Angle
PC PT
RP
PI
Horizontal CurveParts of a Curve
Chord
E
M
CL Curve
E = External M = Middle Ordinate
PC PT
RP
PI
Horizontal CurveParts of a Curve
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
Formulas used with grades:
Distance =D Elevation
Grade
Grades
Formulas used with 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)
The Intersection of two Grades
Formulas:
PVI Station =
b1 – b2
G1
100-
G2
100
The Intersection of two Grades
Example:
Station1 = 7+00
Elevation1 = 201.40’
Grade1 = -1.00%
Station2 = 13+00
Elevation2 = 207.50’
Grade2 = +2.00%
The Intersection of two Grades
b1 = 201.40’ - -1.00100
x 700’ = 208.40
Example:
b2 = 207.50’ -+2.00
100x 1300’ = 181.50
The Intersection of two Grades
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
r2 x2 + G1x + PVC ElevationElevation =
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:
r2 x2 + G1x + PVC ElevationElevation =
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