Math For Surveyors. James A. Coan Sr. PLS

Transcription

1 Math For Surveyors James A. Coan Sr. PLS

2 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

3 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

4 The Right Triangle

5 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

6 The Right Triangle The above trigonometric formulas Can be manipulated using Algebra To find any other unknowns

7 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

8 Oblique Triangles An oblique triangle is one that does not contain a right angle

9 Oblique Triangles This type of triangle can be solved using two additional formulas

10 Oblique Triangles The Law of Sines a Sin A = b Sin B = c Sin C C b a A c B

11 Oblique Triangles The law of Cosines a 2 = b 2 + c 2-2bc Cos A C b a A c B

12 Oblique Triangles When solving this kind of triangle we can sometimes get two solutions, one solution, or no solution.

13 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

14 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

15 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

16 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

17 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

18 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

19 Azimuth Angles Bearings

20 Azimuth: Azimuth, Angles, & Bearings An Azimuth is measured clockwise from North. The Azimuth ranges from 0 to 360

21 Azimuth, Angles, & Bearings Azimuth: N

22 Azimuth, Angles, & Bearings Azimuth to Bearings In the Northeast quadrant the Azimuth and Bearing is the same. N E AZ = = N E

23 Azimuth, Angles, & Bearings Azimuth to Bearings In the Southeast quadrant, subtract the Azimuth from 180 to get the Bearing = S E

24 Azimuth, Angles, & Bearings Azimuth to Bearings In the Southwest quadrant, subtract 180 from the Azimuth to get the Bearing = S W

25 Azimuth, Angles, & Bearings Azimuth to Bearings In the Northwest quadrant, subtract the Azimuth from 360 to get the Bearing = N W

26 Azimuth, Angles, & Bearings Bearings to Azimuths In the Northern hemisphere Bearings are measured from North towards East or West N E N W

27 Azimuth, Angles, & Bearings Bearings to Azimuths In the Southern Hemisphere the Bearings are measured from South to East or West S E S W

28 Azimuth, Angles, & Bearings Bearings to Azimuths In the Northeast quadrant, the Bearing and Azimuth are the same. N N E = AZ

29 Azimuth, Angles, & Bearings Bearings to Azimuths In the Southeast quadrant, subtract the Bearing from 180 to get the Azimuth S E = AZ

30 S W = AZ Azimuth, Angles, & Bearings Bearings to Azimuths In the Southwest quadrant, add the Bearing to 180 to get the Azimuth.

31 Azimuth, Angles, & Bearings Bearings to Azimuths In the Northwest quadrant, subtract the Bearing from 360 to get the Azimuth N W = AZ

32 Angles: Azimuth, Angles, & Bearings To find the Angle between two Azimuths, subtract the smaller Azimuth from the larger Azimuth Larger Azimuth Smaller Azimuth Angle

33 Angles: Azimuth, Angles, & Bearings If both bearings are in the same quadrant, subtract the smaller bearing from the larger bearing. S E S E

34 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 E N W

35 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 (N W+S W)=

36 Coordinate Geometry COGO

37 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).

38 Coordinate Geometry The location of any point can be found with respect to the origin of the coordinate system,

39 Coordinate Geometry or with respect to some other known point on the coordinate system.

40 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.

41 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.

42 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

43 Coordinate Geometry Pythagorean Theorem 2 2 Dist = DNorth + DEast This will give you the distance from Pt.A to Pt.B

44 Coordinate Geometry Example 1 Known: The coordinates for point A The bearing and distance from point A to point B

45 Coordinate Geometry Example 1 Point A coordinates N 10, E 5, The bearing from Point A to point B N E The distance from Point A to Point B 1, feet

46 Coordinate Geometry Example 1: East B North A N 10, E 5,000.00

47 Coordinate Geometry Warning! You must convert the degrees, minutes, and seconds of your bearing to decimal degrees before you find the trig function

48 Example 1 Coordinate Geometry Find North: Cos. Bearing x Distance = North Cos. N E x 1, = 1,063.19

49 Example 1: Coordinate Geometry Find East Sine Bearing x Distance = East Sine N E x 1, =

50 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

51 Example 1: Coordinate Geometry North A + North = North B East A + East = East B

52 Example 1: Coordinate Geometry N 10, , = 11, E 5, = 5, Point B North = 11, East = 5,795.01

53 Coordinate Geometry Example 2: Given: Coordinates of point A N 10, E 5, Coordinates of point B N 10, E 3,924.71

54 Coordinate Geometry Example 2: Point B N 10, E 3, Note: Point B is Northwest of Point A Point A N 10, E 5,000.00

55 Example 2: First Coordinate Geometry Find the deference in North between point A and point B Point B = 10, Point A = 10, Deference =

56 Example 2: Second Coordinate Geometry Find the deference in East between point A and point B Point A = 5, Point B = 3, Deference=1,075.29

57 Example 2: Third Coordinate Geometry Find the distance between point A and point B Dist = North 2 + East 2 Dist = , The distance from A to B = 1,453.99

58 Coordinate Geometry Example 2: Fourth Find the bearing from point A to point B East Tan A = North Tan A = 1,

59 Example 2: Fifth Coordinate Geometry The angle from point A to point B is Because point B is Northwest of point A the bearing is N W

60 The Law of Sines a Sin A = b Sin B = c Sin C C A c B

61 The Law of Sines The law of Sines can be used to solve several Surveying problems, such as finding the center of section

62 The Law of Sines Example 1: Given Coordinates for all 4 section quarter corners Need to find The center quarter corner

63 The Law of Sines Points a a, b, c, & d Have known coordinates d b c

64 First Inverse The Law of Sines a between points c and d d b c

65 The Law of Sines This gives a a bearing and distance from c to d d b c

66 The Law of Sines Next Inverse between a & c a And inverse between d & b d b c

67 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

68 The Law of Sines Because the bearings of all three lines are known the angles between them can be calculated.

69 The Law of Sines a Angle d Bearing Angle b Bearing c Angle

70 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

71 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

72 The Law of Sines a Angle d Bearing Angle (e) b Bearing c Angle

73 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

74 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

75 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)

76 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)

77 Law of Sines In Surveying this type of a problem is called a Bearing; Bearing Intersection

78 Bearing; Bearing Intersection N ¼ Cor. Center ¼ corner W ¼ Cor. E ¼ Cor. S ¼ Cor.

79 Given: Bearing; Bearing Intersection W ¼ Cor.; N=12,645.70, E=5, N ¼ Cor.; N=15,234.25, E=7, E ¼ Cor.; N=12,532.42, E=10, S ¼ Cor.; N=10,008.06, E=7,510.70

80 First: Bearing; Bearing Intersection Find the difference in North and East from the South ¼ corner and the West ¼ corner. North = 2, East = 2,489.07

81 Bearing; Bearing Intersection Second: Find the distance by inverse between the South ¼ corner and the West ¼ corner. Distance = 2, , Distance = 3,626.65

82 Bearing; Bearing Intersection Third: Find the bearing by inversing between the South ¼ corner and West ¼ corner 2, Bearing = Tan -1 2, Bearing = N W

83 Bearing; Bearing Intersection Fourth: Find the bearing between the South ¼ corner and the North ¼ corner. Bearing = Tan , Bearing = N E

84 Bearing; Bearing Intersection Fifth: Find the bearing between the West ¼ corner and the East ¼ corner. 5, Bearing = Tan Bearing = S E

85 Bearing; Bearing Intersection S 1/4 We now have the following: N ¼ W ¼ S E N E E 1/4

86 Bearing; Bearing Intersection Sixth: Calculate the angles between the bearings: A S E C B N E

87 Bearing; Bearing Intersection Angle A: S E S E

88 Bearing; Bearing Intersection Angle B: + N W N E

89 Bearing; Bearing Intersection Angle C: C = 180 -(A + B) 180 -( )= Check: N W + S W =

90 Bearing; Bearing Intersection Now we have: We can use the Law of Sines to solve for one of the unknown sides

91 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.)

92 Bearing; Bearing Intersection Distance from the S ¼ cor. to the C ¼ cor. 3, Sin = Dist. S1/4 to C ¼ Sin Dist. = (3, )(Sin ) Sin Dist = 2,584.07

93 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)

94 Bearing; Bearing Intersection Eight: Use Coordinate Geometry to calculate the coordinates of the center of section

95 Bearing; Bearing Intersection Cos. Bearing x distance = North Cos. N E x 2, = 2, Sin. Bearing x distance = East Sin. N E x 2, = 96.43

96 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.

97 Bearing; Bearing Intersection S ¼ North = 10, Delta North = C ¼ North = 12, S ¼ East = 7, Delta East = C ¼ East = 7,607.13

98 Another way the Law of Sines is used in Surveying is calculating a Bearing; Distance intersection

99 Bearing; Distance Intersection Example: D C N E B Smith Property N E A

100 Bearing; Distance Intersection Given: A = N 10, ; E 5, C = N 10, ; E 5, Bearing from C to D = N E Distance from A to D = We need to find the coordinate for point D

101 Bearing; Distance Intersection CAUTION!! D C D N E B N E There can be two answers to this problem A

102 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

103 Bearing; Distance Intersection First: Inverse between A and C A to C, North = A to C, East = Bearing, A to C = N W Distance, A to C =

104 Bearing; Distance Intersection We now have: We need to find D C N E B C N E A

105 Bearing; Distance Intersection Second: Bearing C-D = N E Bearing C-A = S E Angle C = 180 -(Bearing C-D + Bearing C-A) Angle C = 180 -( ) Angle C =

106 Bearing; Distance Intersection We now have: D C N E B N E All we need to find Angle D A

107 Bearing; Distance Intersection Third: Use the Law of Sines to Find Angle D Sin = Sin. Angle D Sin. D = (Sin )( )

108 Bearing; Distance Intersection The Sine of D = Angle D = Now we can find the Bearing from A to D

109 Bearing; Distance Intersection We now have: D C N E B N E A

110 Bearing; Distance Intersection Forth: Calculate the bearing from D to A Bearing D to C = S W Angle D = Bearing from D to A = S W

111 Bearing; Distance Intersection Now we have a bearing and distance from point A, a known coordinate, to point D

112 Bearing; Distance Intersection Use coordinate geometry to calculate the coordinates of point D North = East = 45.22

113 Bearing; Distance Intersection Finish: Northing of A = 10, North A to D = Northing of D = 10, Easting of A = 5, East A to D = Easting of D = 5,397.46

114 The last intersection problem we need to discuss is the Distance, Distance Intersection

115 In order to solve a Distance, Distance Intersection we need to use The Law of Cosines!

116 The Law of Cosines can be used when you have a Triangle with all three distances but no angles. Example: C A Distance B

117 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

118 As stated, using the Law of Cosines a surveyor can solve a Distance, Distance Intersection Problem

119 WARNING You can get two answers to this kind of a problem

120 Distance, Distance Intersection c Pt A North East Pt B North East C

121 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

122 Distance, Distance Intersection Needed: The coordinate for Point C

123 Distance, Distance Intersection Example: Point A: North = 10, East = 5,910.69

124 Distance, Distance Intersection Example: Point B: North = 10, East = 6,383.80

125 Distance, Distance Intersection Example: North = 3.53 East =

126 Distance, Distance Intersection Example: First: Inverse between points A and B To find the bearing and distance

127 Distance, Distance Intersection Example: Tan = =

128 Distance, Distance Intersection Example: Because point B is North and East of Point A, the bearing becomes: N E

129 Distance, Distance Intersection Example: =

130 Distance, Distance Intersection Example: We now have: A B

131 Distance, Distance Intersection Example: The distance from point A to point C is The distance from point B to point C is

132 Distance, Distance Intersection Example: Now we have: C A B

133 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

134 Distance, Distance Intersection Example: Cos A = (192.49)(473.12) Cos A = Angle A =

135 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

136 Distance, Distance Intersection Example: C B A

137 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

138 Distance, Distance Intersection Example: C A N 10, E 5,910.69

139 Distance, Distance Intersection Example: We can calculate the coordinates at point c by using coordinate geometry (Cogo)

140 Distance, Distance Intersection Example: North = Cos x North =

141 Distance, Distance Intersection Example: East = Sine x East =

142 Distance, Distance Intersection Example: Northing coordinate at C = North A = 10, North A to C = North C = 10,221.63

143 Distance, Distance Intersection Example: Easting coordinate at C = East A = 5, East A to C = East C = 6,063.78

144 Distance, Distance Intersection Example: Coordinates at C North = 10, East = 6,063.78

145 The Compass Rule ( Bowditch Rule)

146 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.

147 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.

148 The Compass Rule Example: Found Record Info. A B Found C C A= N10, E 5, C = N10, E 5,408.96

149 The Compass Rule Example Need to find: The corrected coordinates for point B

150 The Compass Rule Example First: Using the record information calculate the coordinates for points B and C

151 The Compass Rule Example Second: Calculate the Latitude ( North) and the Departure ( East) from point C to point C

152 The Compass Rule Example Third Use the Compass Rule to calculate the corrections for point B

153 The Compass Rule Example Record and field coordinates for point A N 10, E 5, Record coordinates for point B N 10, E 5,204.85

154 The Compass Rule Example Record coordinates for point C N 10, E 5, Field coordinates for point C N 10, E 5,408.96

155 The Compass Rule Example C coordinates = N10, E 5, C coordinates= N10, E 5,

156 The Compass Rule Example Total length of the survey = Length from point A to point B = Total error in Latitude with the sign changed = Total error in Departure with the sign changed = 0.81

157 The Compass Rule Example Latitude from point A to point B = Departure from point A to point B =

158 The Compass Rule Example Correction of the Latitude from point A to point B using the Compass Rule is x = -0.64

159 The Compass Rule Example Correction of the Departure from point A to point B using the Compass Rule is x = 0.43

160 The Compass Rule Example Corrected Latitude= (-0.64 ) = Corrected Departure = = Corrected coordinates for point B N 10, = 10, E 5, = 5, q.e.d.

161 Interpolation

162 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)

163 Interpolation: Formula y 2 = (x 2 x 1 )(y 3 y 1 ) (x 3 x 1 ) + y 1

164 Example: Given x 1 = y 1 = (tangent of x 1 ) x 2 = y 2 = Unknown (tangent of x 2 ) x 3 = y 3 = (tangent of x 3 )

165 Example: Find: the tangent of by interpolation ( )( ) y 2 = ( ) Y 2 =

166 Interpolation: What did we do?

167 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

168 Example = x 17/60 = =

169 Horizontal Curves

170 Horizontal Curve PC PT

171 Parts of a Curve Horizontal Curve Arc PC PT RP

172 Horizontal Curve Parts of a Curve PI PC PT RP

173 Horizontal Curve Parts of a Curve PI PC Chord PT RP

174 Horizontal Curve Parts of a Curve PI Delta Angle PC Chord PT RP Delta Angle

175 Horizontal Curve Parts of a Curve PI E PC M Chord PT E = External RP CL Curve M = Middle Ordinate

176 Horizontal Curve Parts of a Curve PI ½ Delta PC PT ½ Delta RP CL Curve

177 Horizontal Curve Formulas: Length of Arc: Length of Arc (L) = 360 (2pR)

178 Horizontal Curve Formulas: Tangent Distance (T) Tangent (T) = Radius (Tan D/2)

179 Formula: Horizontal Curve Chord Distance (C) Chord Distance (C) = 2R SinD/2

180 Formula: Horizontal Curve Radius (R) Radius (R) = T TanD/2 OR Radius (R) = T CotD/2

181 Horizontal Curve Degree of Curve: NOTE: Arc distance must always be Degree of Curve (D) = R

182 Formula: Horizontal Curve Delta Angle (D) Delta Angle (D) = 180 L pr

183 Horizontal Curve Formula: External R External (E) = - R Cos D/2

184 Horizontal Curve Formula: Middle Ordinate Middle Ordinate (M) = (Sin D/2) T - E

185 Horizontal Curve Example: Given: Length of Arc (L) = Radius (R) =

186 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)

187 Horizontal Curve Find the Delta Angle (D) Delta (D) = 180 L p R Delta (D) = 180 x x Delta (D) = = Half Delta (D/2) =

188 Horizontal Curve Find the Tangent (T) Tangent (T) = R Tan D/2 Tangent (T) = x Tan Tangent (T) =

189 Horizontal Curve Chord Distance (C) = 2R SinD/2 Chord (C) = 2 x x Sin Chord (C) =

190 Horizontal Curve Degree of Curve (D) = R Degree of Curve (D) = Degree of Curve (D) = Degree of Curve (D) =

191 Horizontal Curve External (E) = R Cos D/2 - R External (E) = Cos External (E) = 39.71

192 Horizontal Curve Middle Ordinate (M) = (Sin D/2) T - E (M) = Sin x (M) = 36.92

193 Results: Horizontal Curve Length of Arc (L) = (given) Tangent Distance (T) = Length of Chord (C) = Radius (R) = (given) Degree of Curve (D) = The Delta Angle (D) = The External (E) = The Middle Ordinate (M) = 36.92

194 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

195 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.

196 Grades & Slopes

197 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 -

198 Grades Example: Grade = Difference in Elevation Distance D Elev. Horizontal Distance

199 Grades Example: Grade = = ft / ft

200 Grades Grades can also be expressed as a Percent (%) by multiplying the grade times ft / ft x 100 = 4.78 %

201 Grades A grade is also the tangent of an angle Tangent = opposite = D elevation adjacent = distance Angle opposite D elev. adjacent, distance

202 Grades Formulas used with grades: Grade x distance = D Elevation

203 Grades Formulas used with grades: Distance = D Elevation Grade

204 Grades Formulas used with grades: Grade = D Elevation Distance

205 Slopes A slope is a ratio of the horizontal distance to the vertical distance. Horizontal distance Vertical distance

206 Slopes Example: A 2:1 slope down = A 3:1 slope up =

207 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; ft/ft Or as a present ; 2.0%, 10.34%, 7.62%, etc

208 Locating the Intersection of Two Grades

209 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

210 The Intersection of two Grades Formulas: b 1 = Elev 1 - G x Station 1 (in feet) b 2 = Elev 2 - G x Station 2 (in feet)

211 The Intersection of two Grades Formulas: b 1 b 2 PVI Station = G G 2 100

212 The Intersection of two Grades Example: Station 1 = 7+00 Elevation 1 = Grade 1 = -1.00% Station 2 = Elevation 2 = Grade 2 = +2.00%

213 The Intersection of two Grades Example: b 1 = x 700 = b 2 = x 1300 =

214 The Intersection of two Grades Example: PVI Station = % % 100 = Use the absolute value: =

215 The Intersection of two Grades Example: Grade x distance = difference in elevation x = Elevation at PVI = =

216 Vertical (Parabolic) Curves

217 Vertical Curves Vertical curves are used as a transition from one grade to another

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

219 Vertical Curves PVC Length L L / 2 L / 2 x PVI PVT Sump

220 Vertical Curves Formulas: Elevation = r 2 x2 + G 1 x + PVC Elevation r = G 2 G 1 L x = Distance from the PVC

221 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

222 Vertical Curves Example: Given: G 1 = -1.5% = ft/ft G 2 = +2.5% = ft/ft Length = ft

223 Vertical Curves Example: Given: PVC Station = ; Elevation = ft PVI Station = ; Elevation = ft PVT Station = ; Elevation = ft

224 Vertical Curves Need to find: Elevations at each 50 ft station along the vertical curve. The Sump (low point) station and elevation

225 Vertical Curves First: Calculate r G 2 G 1 r = L r = (-0.015) =

226 Vertical Curves Second: Calculate elevations Elevation = r 2 x2 + G 1 x + PVC Elevation 4+00 = (-0.015)(50) Elevation at Station 4+00 =

227 Vertical Curves Station X Elevation 3+50 PVC PVI PVT (Chk)

228 Vertical Curves Third: Calculate the sump distance Formula: x = -G 1 r -(-0.015) x = = The Sump Station is at

229 Vertical Curves Elevation at the Sump: Elevation = 4+00 = r 2 x2 + G 1 x + PVC Elevation (-0.015)(112.50) Elevation at Station =

230 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

231 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

232 The End