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 Side Adjacent (b) C SineA = a c CosA = b TanA = a c b CscA = c SecA = c CotA = b a b a
The Right Triangle The above trigonometric formulas Can be manipulated using Algebra To find any other unknowns
The Right Triangle Example: SinA = a c SinA c = a a SinA = c CosA = b c CosA c = b b CosA = c TanA = a b TanA b = a a TanA = b
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 Sin A = b Sin B = c Sin C C b a A c B
Oblique Triangles The law of Cosines a 2 = b 2 + c 2-2bc 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 A c B
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 b a A c B
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. b h a A c B
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 C a h a A c B
Azimuth Angles Bearings
Azimuth: Azimuth, Angles, & Bearings 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
S 46 20 30 W + 180 = AZ 226 20 30 Azimuth, Angles, & Bearings Bearings to Azimuths In the Southwest quadrant, add the Bearing to 180 to get the Azimuth.
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
Angles: Azimuth, Angles, & Bearings 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
Angles: Azimuth, Angles, & Bearings 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 SineA CosA TanA East = D B& D North = D B& D East = D DNorth North A CscA = B& D DEast SecA CotA = B& D DNorth North = D DEast This will give you the angle from Pt. A to Pt. B
Coordinate Geometry Pythagorean Theorem 2 2 Dist = DNorth + DEast 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
Example 1 Coordinate Geometry Find North: Cos. Bearing x Distance = North Cos. N 36 47 16 E x 1,327.56 = 1,063.19
Example 1: Coordinate Geometry Find East Sine Bearing x Distance = 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
Example 1: Coordinate Geometry North A + North = North B East A + East = East B
Example 1: Coordinate Geometry 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 Northwest of Point A Point A N 10,000.00 E 5,000.00
Example 2: First Coordinate Geometry Find the deference in North between point A and point B Point B = 10,978.69 Point A = 10,000.00 Deference = 978.69
Example 2: Second Coordinate Geometry 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
Example 2: Third Coordinate Geometry Find the distance between point A and point B Dist = North 2 + East 2 Dist = 978.69 2 + 1,075.29 2 The distance from A to B = 1,453.99
Coordinate Geometry Example 2: Fourth Find the bearing from point A to point B East Tan A = North Tan A = 1,075.29 978.69
Example 2: Fifth Coordinate Geometry 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 C A c B
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 Points a a, b, c, & d Have known coordinates d b c
First Inverse The Law of Sines a between points c and d d b c
The Law of Sines This gives a a bearing and distance from c to d d b c
The Law of Sines Next Inverse between a & c a And inverse between d & b d b c
The Law of Sines After inversing a you will have a bearing and distance between a & c d Bear & Dist Bear & Dist b 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 d Bearing Angle b Bearing c Angle
What we now know: The Law of Sines The bearing from c to d The bearing from d to b The bearing from c to a
What we now know: The Law of Sines 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 d Bearing Angle (e) 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 Angle e = Dist. c-e Angle d (Dist d-c)(angle d)=(dist c-e)(angle e) Dist c-e = (Dist d-c)(angle d) 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. Center ¼ corner W ¼ Cor. E ¼ Cor. S ¼ Cor.
Given: Bearing; Bearing Intersection 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
First: Bearing; Bearing Intersection 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 = 2,637.64 2 +2,489.07 2 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. Bearing = Tan -1 195.16 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. 5,289.28 Bearing = Tan -1 113.28 Bearing = S 88 46 23 E
Bearing; Bearing Intersection S 1/4 We now have the following: N ¼ W ¼ S 88 46 23 E N 02 08 19 E E 1/4
Bearing; Bearing Intersection Sixth: Calculate the angles between the bearings: A S 88 46 23 E C B N 02 08 19 E
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 Check: N 88 46 23 W + S 02 08 19 W 90 54 42 180-90 54 42 = 89 05 18
Bearing; Bearing Intersection Now we have: 45 25 59 89 05 18 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 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
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 N 00 10 25 E B 205.36 Smith Property N 89 30 15 E 352.25 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 N 00 10 25 E 205.36 B N 89 30 15 E 352.25 There can be two answers to this problem A
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: We need to find D C N 00 10 25 E 205.36 B C N 89 30 15 E 352.25 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 N 00 10 25 E 205.36 B N 89 30 15 E 352.25 All we need to find Angle D A
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
Bearing; Distance Intersection We now have: D C N 00 10 25 E 205.36 B 44 58 23 N 89 30 15 E 352.25 66 37 03 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 A Distance B
The Law of Cosines a 2 = b 2 + c 2-2bc Cos A Solving for Cos A, we get Cos A = a 2 b 2 c 2-2bc
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 North East Pt B North 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 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: 3.53 2 + 473.11 2 = 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 A B
Distance, Distance Intersection Example: We need to use the Law of Cosines to calculate one of the angles. Cos A = a2 b 2 c 2-2bc
Distance, Distance Intersection Example: Cos A = 339.442 192.49 2 473.12 2-2(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: Correction = C L S C = The total error in Latitude ( North) or Departure ( East) with the sign changed. L = The total length of the Survey. S = The length of a particular course.
The Compass Rule Example: Found Record Info. A B Found C C A= N10,000.00 E 5,000.00 C = N10,199.16 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 ( North) and the Departure ( 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 -1.21 457.98 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
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 y 2 = (x 2 x 1 )(y 3 y 1 ) (x 3 x 1 ) + y 1
Example: Given x 1 = 42 31 00 y 1 = 0.9168665 (tangent of x 1 ) x 2 = 42 31 17 y 2 = Unknown (tangent of x 2 ) x 3 = 42 32 00 y 3 = 0.9174020 (tangent of x 3 )
Example: Find: the tangent of 42 31 17 by interpolation (42 31 17 42 31 00 )(0.9174020-0.9168664) y 2 = (42 32 00 42 31 00 ) + 0.9168665 Y 2 = 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
Parts of a Curve Horizontal Curve Arc PC PT RP
Horizontal Curve Parts of a Curve PI PC PT RP
Horizontal Curve Parts of a Curve PI PC Chord PT RP
Horizontal Curve Parts of a Curve PI Delta Angle PC Chord PT RP Delta Angle
Horizontal Curve Parts of a Curve PI E PC M Chord PT E = External RP CL Curve M = Middle Ordinate
Horizontal Curve Parts of a Curve PI ½ Delta PC PT ½ Delta RP CL Curve
Horizontal Curve Formulas: Length of Arc: Length of Arc (L) = 360 (2pR)
Horizontal Curve Formulas: Tangent Distance (T) Tangent (T) = Radius (Tan D/2)
Formula: Horizontal Curve Chord Distance (C) Chord Distance (C) = 2R SinD/2
Formula: Horizontal Curve Radius (R) Radius (R) = T TanD/2 OR Radius (R) = T CotD/2
Horizontal Curve Degree of Curve: NOTE: Arc distance must always be 100 100 1 Degree of Curve (D) = 5729.58 R
Formula: Horizontal Curve 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
Example: Horizontal Curve 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 External (E) = R Cos D/2 - R 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
Results: Horizontal Curve 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.C. P.R.C. P.T. Curve 1 P.I. 1 R.P. 2
Horizontal Curve Compound Curve: Curve 2 P.I. 1 Tan. P.C.C. P.I. 2 Tan. P.C. Curve 1 Rad. Rad. R.P. 1 R.P. 2 Rad. P.T. 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: Grade = Difference in Elevation Distance D Elev. Horizontal Distance
Grades Example: Grade = 16.84 352.45 = 0.0477798 ft / ft 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 Tangent = opposite = D elevation adjacent = distance Angle opposite 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 Elevation Distance
Slopes A slope is a ratio of the horizontal distance to the vertical distance. Horizontal distance Vertical distance
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 Grades The purpose of locating the intersection of two grades is to fix the point of intersection (PVI) of those grades. Station 1 Elev 1 PVI Sta.? Elev.? Station 2 Elev 2
The Intersection of two Grades Formulas: b 1 = Elev 1 - G 1 100 x Station 1 (in feet) b 2 = Elev 2 - G 2 100 x Station 2 (in feet)
The Intersection of two Grades Formulas: b 1 b 2 PVI Station = G 1 100 - G 2 100
The Intersection of two Grades Example: Station 1 = 7+00 Elevation 1 = 201.40 Grade 1 = -1.00% Station 2 = 13+00 Elevation 2 = 207.50 Grade 2 = +2.00%
The Intersection of two Grades Example: b 1 = 201.40 - -1.00 100 x 700 = 208.40 b 2 = 207.50 - +2.00 100 x 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 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 PVC Length L L / 2 L / 2 x PVI PVT Sump
Vertical Curves Formulas: Elevation = r 2 x2 + G 1 x + PVC Elevation r = G 2 G 1 L x = Distance from the PVC
Vertical Curves Sump or Peak (Low or High point) Formula: x = -G 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: G 1 = -1.5% = -0.015 ft/ft G 2 = +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 G 2 G 1 r = L r = 0.025 (-0.015) 300.00 = 0.0001333
Vertical Curves Second: Calculate elevations Elevation = r 2 x2 + G 1 x + PVC Elevation 4+00 = 0.0001333 2 50 2 +(-0.015)(50)+326.25 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 Formula: x = -G 1 r -(-0.015) x = = 112.50 0.0001333 The Sump Station is at 4+62.50
Vertical Curves Elevation at the Sump: Elevation = 4+00 = 0.0001333 2 r 2 x2 + G 1 x + PVC Elevation 112.50 2 +(-0.015)(112.50)+326.25 Elevation at Station 4+62.50 = 325.41
Vertical Curves L A L B PVC PVI A PVT A PVC B PVI B PVT L/2 A L/2 A L/2 B L/2 B Unsymmetrical Vertical Curve G 3 = G
Vertical Curves Calculate G 3 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 End