Math for Surveyors
Math For Surveyors
James A. Coan Sr. PLS Topics Covered 1) 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 Covered 8) 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 B Side Opposite (a)
A C Side Adjacent (b)
a b a SineA = CosA = TanA = c c b
c c b CscA = SecA = CotA = a b a The Right Triangle
The above trigonometric formulas
Can be manipulated using Algebra
To find any other unknowns The Right Triangle Example: a a SinA = SinA· c = a = c c SinA
b b CosA = CosA· c = b = c c CosA
a a TanA = TanA·b = a = b b TanA 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 Oblique Triangles
The Law of Sines a b c = = Sin A Sin B Sin C
C
b a
A c B Oblique Triangles
The law of Cosines a2 = b2 + c2 - 2bc Cos A
C
b a
A c B 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.
C
b a
B A 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.
C
a b B A c Oblique Triangles
When angle A is acute (less than 90°) and side a is longer than side c, then there is one solution.
C
b a
A c B 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 is no solution.
h b a
A B c Oblique Triangles
When angle A is acute and side a = h, and h is less than side c there can be only one solution
C b a = h
A c B Oblique Triangles When angle A is an acute angle and h is less than side a as well as side c, there are two solutions. C
b h C’ a’ a
A c B Azimuth
Angles
Bearings Azimuth, Angles, & Bearings Azimuth:
An Azimuth is measured clockwise from North.
The Azimuth ranges from 0° to 360° Azimuth, Angles, & Bearings
Azimuth: N 360° 0°
270° 90°
180° Azimuth, Angles, & Bearings Azimuth 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, & Bearings Azimuth 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, Angles, & Bearings 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 Azimuth, Angles, & Bearings Azimuth to Bearings
In the Northwest quadrant, subtract the Azimuth from 360° to get the Bearing.
360° - 341°25’40” = N 18°34’20”W Azimuth, Angles, & Bearings
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 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
N 45°30’30”E = AZ 45°30’30” 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” Azimuth, Angles, & Bearings
Bearings to Azimuths 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 Azimuth 215°20’10” Smaller Azimuth 110°30’20” Angle Azimuth, Angles, & Bearings Angles:
If both bearings are in the same quadrant, subtract the smaller bearing from the larger bearing. S 82°35’40”E S 25°15’10”E 57°20’30” Azimuth, Angles, & Bearings
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”E N 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 Coordinate Geometry
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
C East B
DEast B& D SineA = CscA = B& D DEast
DNorth North B& D CosA = SecA = B& D DNorth DEast DNorth TanA = CotA = DNorth D A East This will give you the angle from Pt. A to Pt. B Coordinate Geometry
Pythagorean Theorem
Dist = DNorth2 + DEast 2
This will give you the distance from Pt.A to Pt.B Coordinate Geometry
Example 1
Known:
•The coordinates for point A
•The bearing and distance from point A to point B 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
Example 1: East B North
A N 10,000.00 E 5,000.00 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’
Note: Point B is Point A Northwest of N 10,000.00 Point A E 5,000.00 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 Geometry Example 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: Third Find the distance between point A and point B
Dist = D North2 + D East2
Dist = 978.692 + 1,075.292
The distance from A to B = 1,453.99’ Coordinate Geometry
Example 2:
Fourth
Find the bearing from point A to point B
1,075.29 D East Tan A = Tan A = D North 978.69 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 b c = = Sin A Sin B Sin C C
A B 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
Need to find The center quarter corner The Law of Sines
a Points a, b, c, & d Have known b coordinates d
c The Law of Sines First a Inverse between points c and d d b
c The Law of Sines
a This gives a bearing and distance from c to d d b
c The Law of Sines
Next a Inverse between a & c
d b And inverse between d & b
c The Law of Sines
After inversing a you will have a bearing and distance Dist & Bear Bear & Dist d b between a & c as well as d & b
c 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
Angle Bearing Angle d b Bearing
c 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
Angle Angle (e) Bearing d b Bearing
c Angle 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 Dist. c-e = Angle e Angle d
(Dist d-c)(Angle d)=(Dist c-e)(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 C o r.
Center ¼ corner
W ¼ C o r. E ¼ Cor.
S ¼ C o r. Bearing; Bearing Intersection Given: 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 Intersection First:
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
2,489.07 Bearing = Tan -1 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.
195.16’ Bearing = Tan-1 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.
5,289.28’ Bearing = Tan-1 113.28’’
Bearing = S 88°46’23”E Bearing; Bearing Intersection
We now have the following:
N ¼
S 88°46’23”E W ¼ E 1/4 08’19”E N02 ° S 1/4 Bearing; Bearing Intersection Sixth: Calculate the angles between the bearings:
S 88°46’23”E A C 08’19”E
B N02 ° 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: C = 180°-(A + B)
180°-(45°25’59”+45°28’43”)=89°05’18”
N 88°46’23”W Check: + S 02°08’19”W 90°54’42”
180° - 90°54’42” = 89°05’18” Bearing; Bearing Intersection
Now we have: 89°05’18” 45°25’59”
We can use the Law of Sines to solve for one of the unknown sides. 45°28’43” 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
Distance from the S ¼ cor. to the C ¼ cor.
3,626.65’ Dist. S1/4 to C ¼ = Sin 89°05’18 Sin 45°25’59”
(3,626.65’)(Sin 45°25’59”) Dist. = Sin 89°05’18”
Dist = 2,584.07’ 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
Cos. N 02°08’19”E x 2,584.07’ = 2,582.27’
Sin. Bearing x distance = East
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 Bearing; Distance Intersection
Example: D
C
Smith Property 10’25”E 205.36’ N00 ° N 89°30’15”E 352.25’ B A Bearing; Distance Intersection Given: 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 Bearing; Distance Intersection
CAUTION!! D
C D’ 10’25”E 205.36’ N00 ° N 89°30’15”E 352.25’ B A 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’ Bearing; Distance Intersection
We now have: D We need to find
C C’ 10’25”E 205.36’ N00 ° N 89°30’15”E 352.25’ B A Bearing; Distance Intersection Second: 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” Bearing; Distance Intersection
We now have: D
C 44°58’23” 10’25”E 205.36’ N00 ° N 89°30’15”E 352.25’ B A All we need to find Angle D Bearing; Distance Intersection Third:
Use the Law of Sines to Find Angle D
312.37’ 405.67’ = Sin. 44°58’23” Sin. Angle D
(Sin 44°58’23”)(405.67’) Sin. D = 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 Bearing; Distance Intersection
We now have: D
66°37’03” C 44°58’23” 10’25”E 205.36’ N00 ° N 89°30’15”E 352.25’ B A 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 Finish:
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’ 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.
Example: C
Distance A B The Law of Cosines
a2 = b2 + c2 - 2bc Cos A
Solving for Cos A, we get
a2 – b2 – c2 Cos A = -2bc As stated, using the Law of Cosines a
surveyor can solve a Distance,
Distance Intersection Problem WARNING
You can get two answers to this kind of a problem Distance, Distance Intersection
c
Pt A Pt B North North East East
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:
473.11’ Tan -1 =89.572508828° 3.53’
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:
C
B A Distance, Distance Intersection
Example:
We need to use the Law of Cosines to calculate one of the angles.
a2 – b2 – c2 Cos A = -2bc Distance, Distance Intersection Example:
339.442 – 192.492 – 473.122 Cos A = -2(192.49)(473.12)
Cos A = 0.799791540
Angle A = 36°53’23” 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:
C
B A 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: C
A N 10,104.94’ E 5,910.69’ Distance, Distance Intersection Example:
We can calculate the coordinates at point c by using coordinate geometry (Cogo) Distance, Distance Intersection
Example:
North = Cos 52°40’58” x 192.49’
North = 116.69’ Distance, Distance Intersection
Example:
East = Sine 52°40’58” x 192.49’
East = 153.09’ 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 Land Survey System (PLSS)
It also has many other applications in Surveying. The Compass Rule
The Formula: C Correction = S L 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:
Record C Info. C’ B Found Found A
N10,000.00’ N10,199.16’ A= C’= E 5,000.00’ E 5,408.96’ 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 and field coordinates for point A
N 10,000.00’ E 5,000.00’
Record coordinates for point B
N 10,131.05’ E 5,204.85’ The Compass Rule Example
Record coordinates for point C
N 10,200.37’ E 5,408.15’
Field coordinates for point C’
N 10,199.16 E 5,408.96 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’ x 243.19’ = -0.64’ 457.98’ The Compass Rule Example
Correction of the Departure from point A to point B using the Compass Rule is
0.81’ x 243.19’ = 0.43’ 457.98’ 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 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:
Formula
(x2 – x1)(y3 – y1) y2 = + y1 (x3 – x1) 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
(42°31’17” – 42°31’00”)(0.9174020 - 0.9168664) y = + 0.9168665 2 (42°32’00” – 42°31’00”)
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 Horizontal Curve
PC PT Horizontal Curve Parts of a Curve
Arc
PC PT
RP Horizontal Curve Parts of a Curve PI
PC PT
RP Horizontal Curve Parts of a Curve PI
PC PT Chord
RP Horizontal Curve Parts of a Curve PI Delta Angle
PC PT Chord
Delta Angle RP Horizontal Curve Parts of a Curve PI
E
PC M PT Chord
CL Curve RP E = External M = Middle Ordinate Horizontal Curve Parts of a Curve PI ½ Delta
PC PT
½ Delta CL Curve RP Horizontal Curve
Formulas: Length of Arc:
Length of Arc (L) = 360° (2pR) 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 Radius (R) = TanD/2
OR
Radius (R) = T CotD/2 Horizontal Curve
Degree of Curve: 100’ NOTE: Arc distance must 1° always be 100’
Degree of Curve (D) = 5729.58’ R Horizontal Curve Formula: Delta Angle (D)
Delta Angle (D) = 180°L pR Horizontal Curve Formula: External
R External (E) = - R Cos D/2 Horizontal Curve
Formula: Middle Ordinate
Middle Ordinate (M) = (Sin D/2) T - E Horizontal Curve Example:
Given: Length of Arc (L) = 396.72’
Radius (R) = 526.54’ Horizontal Curve Example: 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 Curve
Find 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 Curve Find the Tangent (T) Tangent (T) = R Tan D/2
Tangent (T) = 526.54’ x Tan 21°35’05”
Tangent (T) = 208.31’ Horizontal Curve Chord Distance (C) = 2R SinD/2
Chord (C) = 2 x 526.54’ x Sin 21°35’05”
Chord (C) = 387.40’ Horizontal Curve Degree 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
R External (E) = - R Cos D/2
526.54’ External (E) = - 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 Curve Results: 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 Reverse Curve:
R.P. 1 P.I.2 Curve 2
P.T. P.C. P.R.C.
Curve 1 R.P. 2 P.I.1 Horizontal Curve Compound Curve: Curve 2
Tan. P.C.C. P.I.1 P.I.2 Rad. Tan. R.P. 2 P.T. P.C. Rad. Rad. R.P. 1 Curve 1 Tan. 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 - Grades Example:
Difference in Elevation Grade = Distance Elev. D
Horizontal Distance Grades Example:
16.84’ Grade = = 0.0477798 ft / ft 352.45’ 16.84’
352.45’ Grades
Grades can also be expressed as a Percent (%) by multiplying the grade times 100
0.0478 ft / ft x 100 = 4.78 % Grades
A grade is also the tangent of an angle
opposite = D elevation Tangent = adjacent = distance
Angle elev. D opposite
adjacent, distance Grades
Formulas used with grades:
Grade x distance = D Elevation Grades
Formulas used with grades:
D Elevation Distance = Grade Grades
Formulas used with grades:
D Elevation Grade = Distance Slopes
A slope is a ratio of the horizontal distance to the vertical distance.
Horizontal distance Vertical Vertical distance Slopes
Example:
2.0’ A 2:1 slope down = 1.0’
3.0’ A 3:1 slope up = 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 Grades The purpose of locating the intersection of two grades is to fix the point of intersection (PVI) of those grades. 1 2 1 2 Station Elev Station Elev PVISta.? ? Elev. The Intersection of two Grades
Formulas:
G1 b1 = Elev1 - x Station1 (in feet) 100
G2 b2 = Elev2 - x Station2 (in feet) 100 The Intersection of two Grades
Formulas:
b1 – b2
PVI Station = G1 G2 - 100 100 The Intersection of two Grades
Example:
Station1 = 7+00 Station2 = 13+00
Elevation1 = 201.40’ Elevation2 = 207.50’
Grade1 = -1.00% Grade2 = +2.00% The Intersection of two Grades
Example:
-1.00 b = 201.40’ - x 700’ = 208.40 1 100
+2.00 b = 207.50’ - x 1300’ = 181.50 2 100 The Intersection of two Grades
Example:
208.40 – 181.50 PVI Station = = -896.67 -1.00% - +2.00% 100 100
Use the absolute value: -896.67 = 8+96.67 The Intersection of two Grades Example:
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
Length “L”
L / 2 L / 2 PVT PVC x PVI
Sump Vertical Curves
Formulas:
r Elevation = x2 + G x + PVC Elevation 2 1
G – G r = 2 1 L
x = Distance from the PVC Vertical Curves
Sump or Peak (Low or High point)
Formula: -G x = 1 r
The high or low point is Always on the lesser grade side (absolute value)
x = Distance from the PVC 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 r = L
0.025 – (-0.015) r = = 0.0001333… 300.00’ Vertical Curves
Second: Calculate elevations
r Elevation = x2 + G x + PVC Elevation 2 1
0.0001333 4+00 = 502+(-0.015)(50)+326.25’ 2
Elevation at Station 4+00 = 325.67’ Vertical Curves Station 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 -G Formula: x = 1 r
-(-0.015) x = = 112.50’ 0.0001333…
The Sump Station is at 4+62.50 Vertical Curves
Elevation at the Sump:
r Elevation = x2 + G x + PVC Elevation 2 1
0.0001333 4+00 = 112.502+(-0.015)(112.50)+326.25’ 2
Elevation at Station 4+62.50 = 325.41’ Vertical Curves
LA LB B B A A I V PVI PVT PVT PVC P PVC
L/2A L/2A L/2B L/2B Unsymmetrical Vertical Curve
G3 = G’ Vertical Curves
Calculate G3 from the center of the first curve to the center of the second curve
Calculate each part of the curve as if it was a
regular vertical Curve The
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