california coordinate system
DESCRIPTION
California Coordinate System. Capital Project Skill Development Class (CPSD) G100497. California Coordinate System. Thomas Taylor, PLS Right of Way Engineering District 04 (510) 286-5294 [email protected]. Course Outline. History Legal Basis The Conversion Triangle - PowerPoint PPT PresentationTRANSCRIPT
California Coordinate System
Capital Project Skill Development Class (CPSD)G100497
California Coordinate System
Thomas Taylor, PLSRight of Way EngineeringDistrict 04
(510) [email protected]
Course Outline History Legal Basis The Conversion Triangle Geodetic to Grid Conversion Grid to Geodetic Conversion Convergence Angle Reducing Measured Distances to Grid
Distances Zone to Zone Transformations
History
Types of Plane SystemsPlane
Ellipsoid
Tangent PlaneLocal Plane
Point of Origin
Intersecting CylinderTransverse Mercator
Axis of Ellipsoid
Ellipsoid
Axis of Cylinder
Line of intersection
Apex of Cone
Intersecting Cone2 Parallel Lambert
Axis of Cone & Ellipsoid
Ellipsoid
A number of Conformal Map Projections are used in the United States.
Universal Transverse Mercator. Transverse Mercator. Oblique Transverse Mercator. Lambert Conformal Conical.
The Transverse Mercator is used for states (or zones in states) that are long in a North-South direction.
The Lambert is used for states (or zones in states) that are long in an East-West direction.
The Oblique Mercator is used in one zone in Alaska where neither the TM or Lambert were appropriate.
What Map Projection to Use?
Characteristics of the Lambert Projection The secant cone intersects the surface of the ellipsoid at
two places. The lines joining these points of intersection are known as
standard parallels. By specifying these parallels it defines
the cone. Scale is always the same
along an East-West line. By defining the central
meridian, the cone becomes orientated with respect to the ellipsoid
Legal Basis
Public Resource Code
R0
B0
R b
E0
Nb
What are constants or given information within the Tables?
Nb is the northing of projection origin 500,000.000 meters
E0 is the easting of the central meridian 2,000,000.000 meters
R0 is the mapping radius through the projection origin
What must be calculated using the constants?
R is the radius of a circle, a function of latitude, and interpolated from the tables
u is the radial distance from the central parallel to the station, (R0 – R)
is the convergence angle, mapping angle
, mapping angle, convergence angle.
What will be given?
northing/easting
LatitudeLongitude
(N,E), (X,Y), Latitude Longitude
B0 is the central parallel of the zone
Rb is mapping radius through grid base
Ru
Geodetic to Grid Conversion
Determine the Radial Difference: u
)B(L)B(L)(L B)(L u 44
3321
0B-BB
B = north latitude of the station
B0 = latitude of the projection origin (tabled constant)
u = radial distance from the station to the central parallel
L1, L2, L3, L4 = polynomial coefficients (tabled constants)
Geodetic to Grid Conversion
Determine the Mapping Radius: R
u - RR 0
R = mapping radius of the station
R0 = mapping radius of the projection origin (tabled constant)
u = radial distance from the station to the central parallel
Geodetic to Grid Conversion
Determine the Plane Convergence: g
)L)sin(B-(L 00g = convergence angle
L = west longitude of the station
L0 = longitude of the projection and grid origin
(tabled constant)
Sin(B0) = sine of the latitude of the projection origin
(tabled constant)
Geodetic to Grid Conversion
Determine Northing of the Station
n = N0 + u + [R(sin(g))(tan(g/2))]
or
n = Rb + Nb – R(cos(g))
n = the northing of the station
N0 = northing of the projection origin (tabled constant)
Rb, Nb = tabled constants
Geodetic to Grid Conversion
Determine Easting of the Station
e = E0 + R(sin(g))
e = easting of the station
E0 = easting of the projection and grid origin
Example # 1
Compute the CCS83 Zone 6 metric coordinates of station “Class-1” from its geodetic coordinates of:
Latitude = 32° 54’ 16.987”
Longitude = 117° 00’ 01.001”
Example # 1
Determine the Radial Difference: u
34-0.4292043B
44733.3339229-904718611.32B
44733.3339229 - '16.987' 54' 32B
Example # 1
Determine the Radial Difference: u
1441-47599.846u
4)0.429204330.016171(-
).4292043345.65087(-0
).4292043348.94188(-0
4334)4(-0.42920110905.327 u
4
3
2
Example # 1
Determine the Mapping Radius: R
2349754239.92R
61441)(-47599.84-0762.9706640R
Example # 1
Determine the Plane Convergence: g
'24752' .44 '24-
or
785-0.4122909
575763))(0.549517056117.000278-25.116(
575763))(0.549517'01.001' 00' 15'-117 (116
)L)sin(B-(L 00
Example # 1
Determine Northing of the Station
n = Rb + Nb – R(cos(g))
n = 9836091.7896 + 500000.000
– 9754239.92234(cos(-0.4122909785))
n = 582104.404
Example # 1
Determine Easting of the Station
e = E0 + R(sin(g))
e = 2000000.000 + 9754239.92234(sin(-0.4122909785))
e = 1929810.704
Problem # 1
Compute the CCS83 Zone 3 metric coordinates of station “SOL1” from its geodetic coordinates of:
Latitude = 38° 03’ 59.234”
Longitude = 122° 13’ 28.397”
Solution to Problem # 1
EB = 0.315384453°
u = 35003.7159064
R = 8211926.65249
g = -1° 03’ 20.97955” (HMS) 0r -1.05582765°
n = 675242.779
e = 1848681.899
Grid to Geodetic Conversion
Determine the Plane Convergence: g
g = arctan[(e - E0)/(Rb – n + Nb)]
g = convergence angle at the station
e = easting of station
E0 = easting of the projection origin (tabled constant)
Rb = mapping radius of the grid base (tabled constant)
n = northing of the station
Nb = northing of the grid base (tabled constant)
Grid to Geodetic Conversion
Determine the Longitude
L = L0 – (g/sin(B0))
L = west longitude of the station
L0 = longitude of the projection origin (tabled constant)
sin(B0) = sine of the latitude of the projection origin
(tabled constant)
Grid to Geodetic Conversion
Determine the radial difference: u
u = n – N0 – [(e – E0)tan(g/2)]g = convergence angle at the station
e = easting of the station
E0 = easting of the projection origin (tabled constant)
n = northing of the station
N0 = northing of the projection origin
u = radial distance from the station to the central parallel
Grid to Geodetic Conversion
Determine latitude: BB = B0 + G1u + G2u2 + G3u3 + G4u4
B = north latitude of the station
B0 = latitude of the projection origin (tabled constant)
u = radial distance from the station to the central parallel
G1, G2, G3, G4 = polynomial coefficients (tabled constants)
Example # 2
Compute the Geodetic Coordinate of station “Class-2” from its CCS83 Zone 4 Metric Coordinates of:
n = 654048.453
e = 2000000.000
Example # 2
Determine the Plane Convergence: g
g = arctan[(e - E0)/(Rb – n + Nb)]
g = arctan[(2000000.000 – 2000000.000)/
(8733227.3793 – 654048.453 + 500000.000)]
g = arctan(0)
g = 0
Example # 2
Determine the Longitude
L = L0 – (g/sin(B0))
L = 119° 00’ 00’’ – (0/sin(36.6258593071°))
L = 119° 00’ 00’’
Example # 2
Determine the radial difference: u
u = n – N0 – [(e – E0)tan(g/2)]
u = 654048.453 – 643420.4858
- [(2000000.000 – 2000000.000)(tan(0/2)]
u = 10627.967
Example # 2
Determine latitude: BB = B0 + G1u + G2u2 + G3u3 + G4u4
B = 36.6258593071° + 9.011926076E-06(10627.967)
+ -6.83121E-15(10627.967)2
+ -3.72043E-20(10627.967)3
+ -9.4223E-28(10627.967)4
B = 36° 43’ 17.893’’
Problem # 2
Compute the Geodetic Coordinate of station “CC7” from its CCS83 Zone 3 Metric Coordinates of:
n = 674010.835
e = 1848139.628
Solution to Problem # 2
g = -1° 03’ 34.026” or -1.0594517°
L = 122° 13’ 49.706”
u = 33761.9722245
B = 38° 03’ 18.958”
Convergence Angle Determining the Plane Convergence Angle and
the Geodetic Azimuth or the Grid Azimuth
g = arctan[(e – E0)/(Rb – n + Nb)]
or
g = (L0 – L)sin(B0)
Convergence Angle
Determine Grid Azimuth: t or Geodetic Azimuth: a
t = a – g + dt = grid azimuth
a = geodetic azimuth
g = convergence angle (mapping angle)
d = arc to chord correction, known as the second order term (ignore this term for lines less than 5 miles long)
Example # 3
Station “Class-3” has CCS83 Zone 1 Coordinates of n = 593305.300 and e = 2082990.092, and a grid azimuth to a natural sight of 320° 37’ 22.890”. Compute the geodetic azimuth from Class-3 to the same natural sight.
Example # 3 Determining the Plane Convergence Angle and
the Geodetic Azimuth or the Grid Azimuth
g = arctan[(e – E0)/(Rb – n + Nb)]
g = arctan[(2082990.092 – 2000000.000)/
(7556554.6408 – 593305.300 + 500000.000)]
g = arctan[0.0111198338]
g = 0° 38’ 13.536’’
Example # 3
Determine Grid Azimuth: t or Geodetic Azimuth: a
t = a – g
a = t + g
a = 320° 37’ 22.890’’ + 0° 38’ 13.536’’
a = 321° 15’ 36.426’’
Problem # 3
Station “D7” has CCS83 Zone 6 Coordinates of n = 489321.123 and e = 2160002.987, and a grid azimuth to a natural sight of 45° 25’ 00.000”. Compute the geodetic azimuth from D7 to the same natural sight.
Solution to Problem # 3
g = 0° 55” 51.361’ (0.9309335°)
Geodetic Azimuth = 46 20’ 51.361”
Elevation Factors Before a Ground Distance can be reduced to the
Grid, it must first be reduced to the ellipsoid of reference.
Geoid (MSL)
Ellipsoid
Groundh H
NR
adius of the Ellipsoid
REF =
R + N + H
R = Radius of Curvature.N = Geoidal Separation.H = Mean Height above
Geoid.h = Ellipsoidal Height
Combined Grid Factor (Combined Scale Factor)
Scale Decreases
Zon
e Li
mit
Zon
e Li
mit
Scale Increases
A scale factor is the Ratio of a distance on the grid projection to the corresponding distance on the ellipse.
Scale Increases
Scale Decreases
A B D
CA’
B’
D’
C’
- Grid Distance A-B is smaller than Geodetic Distance A’-B’.- Grid Distance C-D is larger than Geodetic Distance C’-D’.
Combined Grid Factor (Combined Scale Factor)
Converting Measured Ground Distances to Grid Distances
Determine Radius of Curvature of the Ellipsoid: Ra
Ra = r0/k0
Ra = geometric mean radius of curvature of the ellipsoid at the projection origin
r0 = geometric mean radius of the ellipsoid at the projection origin, scaled to grid (tabled constant)
k0 = grid scale factor of the central parallel (tabled constant)
Converting Measured Ground Distances to Grid Distances
Determine the Elevation Factor: re
re = Ra/(Ra + N + H)
re = elevation factor
Ra = radius of curvature of the ellipsoid
N = geoid separation
H = elevation
Converting Measured Ground Distances to Grid Distances
Determine the Point Scale Factor: k
k = F1 + F2u2 + F3u3
k = point scale factor
u = radial difference
F1, F2, F3 = polynomial coefficients (tabled constants)
Converting Measured Ground Distances to Grid Distances
Determine the Combined Grid Factor: cgf
cgf = re k
cgf = combined grid factor
re = elevation factor
k = point scale factor
Converting Measured Ground Distances to Grid Distances
Determine Grid Distance
Ggrid = cgf(Gground)
Note: Gground is a horizontal ground distance
Converting Grid Distances to Horizontal Ground Distances
Determine Ground Distance
Gground = Ggrid/cgf
Example # 4
In CCS83 Zone 1 from station “Me” to station “You” you have a measured horizontal ground distance of 909.909m. Stations Me and You have elevations of 3333.333m and a geoid separation 0f -30.5m. Compute the horizontal grid distance from Me to You. (To calculate the point scale factor assume u = 15555.000)
Example # 4 Determine Radius of Curvature of
the Ellipsoid: Ra
Ra = r0/k0
Ra = 6374328/0.999894636561
Ra = 6374999.69189
Example # 4
Determine the Elevation Factor: re
re = Ra/(Ra + N + H)
re = 6374999.69189/(6374999.69189 – 30.5 + 3333.333)
re = 0.9994821768
Example # 4
Determine the Point Scale Factor: k
k = F1 + F2u2 + F3u3
k = 0.999894636561 + 1.23062E-14(15555)2
+ 5.47E-22(15555)3
k = 0.9998976162
Example # 4
Determine the Combined Grid Factor: cgf
cgf = re k
cgf = 0.9994821768(0.9998976162)
cgf = 0.999379846
Example # 4
Determine Grid Distance
Ggrid = cgf(Gground)
Ggrid = 0.999379846(909.909)
Ggrid = 909.3447
Problem # 4
In CCS83 Zone 4 from station “here” to station “there” you have a measured horizontal ground distance of 1234.567m. Station here and there have elevations of 2222.222m and a geoid separation 0f -30.5m. Compute the horizontal grid distance from here to there. (To calculate the point scale factor assume u = 35000)
Solution to Problem # 4
Ra = 6371934.463
re = 0.999656153
k = 0.999955870
cgf = 0.999612038
Ggrid = 1234.088m
Converting a Coordinate from one Zone to another Zone Firstly, convert the grid coordinate from
the original zone to a GRS80 geodetic latitude and longitude using the appropriate zone constants
Then, convert the geodetic latitude and longitude to the grid coordinates using the appropriate zone constants
Problem # 5
CC7 has a metric CCS Zone 3 coordinate of n = 674010.835 and e = 1848139.628. Compute a CCS Zone 2 coordinate for CC7.
Solution to Problem # 5
n = 543163.942
e = 1979770.624