Transcript California Coordinate System

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
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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 Systems
Point of
Origin
Plane
Apex of
Cone
Ellipsoid
Axis of Cone
& Ellipsoid
Axis of
Ellipsoid
Tangent Plane
Local Plane
Line of
intersection
Ellipsoid
Intersecting Cone
2 Parallel Lambert
Axis of
Cylinder
Ellipsoid
Intersecting Cylinder
Transverse Mercator
What Map Projection to Use?
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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.
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Characteristics of the Lambert
Projection
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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
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Public Resource Code
g , q , mapping angle,
convergence angle.
What will be given?
(N,E), (X,Y), Latitude(F), Longitude(l)
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
Rb is mapping radius through grid base
B0 is the central parallel of the zone
northing/easting
R0 is the mapping radius through the projection origin
Latitude(F),Longitude(l)
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)
g , q is the convergence angle, mapping angle
Geodetic to Grid Conversion
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Determine the Radial Difference: u
B  B - B0

3
4
u  L1(B)  L2 ( )  L3 (B )  L4 (B )
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
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Determine the Mapping Radius: R
R  R0 - u
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
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Determine the Plane Convergence: g
g  (L0 - L)sin(B0 )
g = 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
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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
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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
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Determine the Radial Difference: u
B  32 54'16.987'' - 33.3339229
447
B  32.904718611
 - 33.3339229
447
B  -0.4292043
34
Example # 1
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Determine the Radial Difference: u
u  110905.327
4(-0.42920
4334)
 8.94188(-0
.429204334
)
2
 5.65087(-0
.429204334
)
3
 0.016171(0.42920433
4)
u  -47599.846
1441
4
Example # 1
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Determine the Mapping Radius: R
R  9706640.0762- (-47599.84
61441)
R  9754239.92
234
Example # 1
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Determine the Plane Convergence: g
g
g
g
g
 (L 0 - L)sin(B0 )
 (116 15'-117 00'01.001'' )(0.549517575763)
 (116.25 - 117.000278
056)(0.549517575763)
 -0.4122909785
or
g  -24' 44. 24752''
Example # 1
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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
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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
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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
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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
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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
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Determine latitude: B
B = 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
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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
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Determine the Longitude
L = L0 – (g/sin(B0))
L = 119° 00’ 00’’ – (0/sin(36.6258593071°))
L = 119° 00’ 00’’
Example # 2
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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
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Determine latitude: B
B = 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
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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
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Determine Grid Azimuth: t or Geodetic Azimuth: a
t=a –g +d
t = 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
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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
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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”
Combined Grid Factor
(Combined Scale Factor)
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Elevation Factors
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Before a Ground Distance can be reduced to the
Grid, it must first be reduced to the ellipsoid of
reference.
R
Ground
h
EF =
R+N+H
R
N
H
=
=
=
h
=
Radius of Curvature.
Geoidal Separation.
Mean Height above
Geoid.
Ellipsoidal Height
Ellipsoid
Geoid (MSL)
H
N
Combined Grid Factor
(Combined Scale Factor)
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A scale factor is the Ratio of a distance on the
grid projection to the corresponding distance on
the ellipse.
Zone Limit
A’
- Grid Distance A-B
is smaller than Geodetic
Distance A’-B’.
- Grid Distance C-D is
larger than Geodetic
Distance C’-D’.
C
A
B
D
C’
D’
Scale
Increases
Scale
Decreases
Scale
Decreases
Scale
Increases
Zone Limit
B’
Converting Measured Ground
Distances to Grid Distances
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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
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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
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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
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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
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Determine Grid Distance
Ggrid = cgf(Gground)
Note: Gground is a horizontal ground distance
Converting Grid Distances to
Horizontal Ground Distances
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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
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Determine Radius of Curvature of the
Ellipsoid: Ra
Ra = r0/k0
Ra = 6374328/0.999894636561
Ra = 6374999.69189
Example # 4
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Determine the Elevation Factor: re
re = Ra/(Ra + N + H)
re = 6374999.69189/(6374999.69189 – 30.5 + 3333.333)
re = 0.9994821768
Example # 4
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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
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Determine the Combined Grid Factor: cgf
cgf = re k
cgf = 0.9994821768(0.9998976162)
cgf = 0.999379846
Example # 4
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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
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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