Geographic Information Processing
Computational Basics
Coordinate Systems
Coordinate transformations
Projections
UTM calculations
Calculations
Wikipedia
3/22/2016
Distances and areas
Spherical Trigonometry
© 2009 Raymond P. Jefferis III
Lect 02 - 1
Coordinate Systems
• Earth ellipsoid
• The Datum
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 2
The Earth Ellipsoid
• An ellipsoid represents shape of the earth
more accurately than a sphere
• It requires more specification parameters
than a spherical model (which would only
require a radius)
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 3
Ellipsoid Measures
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 4
Earth Dimensions
• Equatorial radius 6,378,206 m
• Polar semi-axis:
6,356,583 m
• Radius of sphere: 6,370,997 m
(equal area)
• Radius of geoid:
6,378,200 m
(gravitational equipotential surface approximately at the mean sea level)
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 5
Parameters of Earth Ellipsoid
Reference Ellipsoidal parameters:
Name
Major axis, “a” [km]Flattening, “f”
WGS84
6378.13700
1/298.257223563
GRS80/NAD83 6378.13700
1/298.257222101
WGS66
6378.145
1/298.25
GRS67/IAU68 6378.16000
1/298.2472
WGS72
6378.135
1/298.26
NAD27
6378.2064
1/294.9786982138
Note: polar radius = equatorial radius * ( 1 - f ) or b = a * ( 1 - f )
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 6
Differences
• WGS84
b = 6356.7523142 [km]
• NAD83
b = 6356.7523141 [km]
The difference is 0.1 mm (100 µ) !
(Note: this is about the thickness of a human hair.)
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 7
Radius Approximation
For the purposes of this course, a spherical
earth having a radius of 6, 378, 200 meters
(The geoid radius) can be used.
Note: Using such a spherical model results in
significant errors over large distances.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 8
The Datum
• A mathematical model of the earth
– A starting point (Latitude, Longitude) on earth
– Parameters for the spheroidal model used
– Result: Origin separation between spheroid and
earth models
• Examples:
– NAD27 (North American Datum of 1927)
– NAD83 (North American Datum of 1983)
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 9
Geodetic Datum
• A point on the surface of the Earth is used
to pin down the position and orientation of
the ellipsoid model relative to the earth
itself. This is the Datum.
• This is not the only method, but it was used
for many years.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 10
NAD27
The North American Datum of 1927 is a
based on the Clarke 1866 spheroid, anchored
at "Meades Ranch" in Kansas. Each point in
the model is referred to this point on the
earth.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 11
Meades Ranch, KS
North American Datum of 1927 (NAD 27) at 39 ｡13′26.686″N, 98｡32′30.506″W, Wikipedia
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 12
NAD83
The North American Datum of 1983 is based
on the GRS 1980 ellipsoid
This is nearly identical to the similarly
geocentric WGS 1984 ellipsoid.
It is earth-centered based on the GRS80
spheroid.
There is no initial point.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 13
Coordinate Transformations
• Conversion of Datum
Model equations are used for converting
from one datum to another.
• Software - GEOTRANS
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 14
Conversion of Datum Model
Molodensky formula
See URL:
http://www.kartografie.nl/
geometrics/Coordinate%20
transformations/coordtrans.html
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 15
GEOTRANS
Converts between MANY datum models:
http://earth-info.nga.mil/GandG/geotrans/
Uses the Molodensky formulas
http://earth-info.nga.mil/GandG/coordsys/datums/standardmolodensky.html
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 16
Molodensky Transformation
• Transfers local coordinates to WGS84
WGS 84   Local  
WGS 84   Local  
hWGS 84  hLocal  h
• Increments are calculated from the subject
formulas (See previous reference)
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 17
Projections
Projections represent the surface of the earth
projected onto another surface, allowing the
viewer access to more or less rectangular
coordinates - albeit with scale changes over
the viewed surface.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 18
Projected Earth Topography
GTOPO30 global digital elevation model (DEM), U.S. Geological Survey
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 19
Web References on projections
http://en.wikipedia.org/wiki/Map_projection
http://members.shaw.ca/quadibloc/maps/mapint.htm
http://members.shaw.ca/quadibloc/maps/mcyl01.htm
http://www.geometrie.tuwien.ac.at/karto/
http://www.kartografie.nl/geometrics/Coordinate%2
0transformations/coordtrans.html
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 20
Transverse Mercator Projection
• The meridians and
parallels form a nearly
rectangular net around
the center of the map.
• The map will be used only
in this regular area, which
is centered on a given
longitude and latitude.
Equatorial projection: J. B. Calvert, U. Denver, 2002.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 21
Mercator Projection Formulas
x  tanh 1 B
tan 
cos(   0 )
  0
y  tan 

sin D
  sin 1  cosh

x
1
x
  0  tan 1  sinh

cos D 
where
B  cos  sin(  0 )
D  y  0
x is equatorial
projection
y is projection
on longitude λ0
λ is longitude
ϕ is latitude
From Eric Weisstein,MathWorld,
Wolfram Research
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 22
UTM Conversion Example
• Use UTM conversion spreadsheet in the
Models file:
–
–
–
–
–
Lat: 27.15˚
Lon: -94.625˚
Datum: NAD27
xsw = Easting: 339117.761 [meters]
ysw = Northing; 3015001.964 [meters]
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 23
Conversion Formulas
http://www.uwgb.edu/dutchs/UsefulData/UTMFormulas.HTM
Students should be able to program their own conversion
programs from the clear formulas given on this site.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 24
References [See note]
• Snyder, J. P., 1987; Map Projections - A Working Manual. U.S.
Geological Survey Professional Paper 1395, 383 p.
• Army, Department of, 1973; Universal Transverse Mercator Grid, U.
S. Army Technical Manual TM 5-241-8, 64 p.
• NIMA Technical Report 8350.2, "Department of Defense World
Geodetic System 1984, Its Definition and Relationships with Local
Geodetic Systems," Second Edition, 1 September 1991 and its
supplements. [ U.S. Geological Survey, Box 25286, Denver Federal
Center, Denver, Colorado 80225 or by phone at 1-800-USA-MAPS.]
Note: These references are from the excellent paper (referenced in previous
slide) of Steven Dutch, Natural and Applied Sciences, University of Wisconsin
- Green Bay.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 25
Calculations
• Area approximations
• Distances
• Spherical trigonometry
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 26
Area Approximations
Since the circumference of the earth varies
with latitude, areas of rectangles 1˚ on a side
are different at various latitudes.
The calculation can be done as follows:
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 27
Distance Formulas
a = 6378.137000 km
[WGS84 earth radius]
f = 1/298.257223563
[WGS84 flattening]
e2 = f*(2-f)
[Eccentricity]
R1 = a(1-e2)/(1-e2*(Sin(lat0))2)3/2
R2 = a/Sqrt(1-e2*(Sin(lat0))2)
North_Distance = R1*dlat
East_Distance = R2*Cos(lat0)*dlon
From: Aviation Formulary V1.43, Ed Williams
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 28
®
Mathematica
Notebook
lat0 = Degree 40.0;
lon0 = Degree 76.0
dlat = Degree 1.0;
dlon = Degree 1.0;
a = 6378.137000;
f = 1.0/298.257223563;
esq = f*(2.0 - f);
r1 = a*(1 - esq)/(1.0 esq*(Sin[lat0])^2)^(3/2)
r2 = a/(Sqrt[1 - esq*(Sin[lat0])^2])
distN = r1*dlat
distE = r2*Cos[lat0]*dlon
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 29
Example Calculation 1
At 40˚ North latitude, 76˚ West longitude, a square
of 1˚ will have the planar dimensions:
North_Distance = 111.035 km
East_Distance = 85.3939 km
This will apply to the data included on your data CD.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 30
1˚x 1˚ Areas at Various Latitudes
Lat.
0
10
20
30
40
50
60
70
80
90
3/22/2016
∆ Lon.
110, 567 m
110, 598 m
110, 692 m
110, 840 m
111, 023 m
111, 220 m
111, 406 m
111, 560 m
111, 661 m
111, 699 m
© 2009 Raymond P. Jefferis III
∆ Lat.
111, 321 m
109, 641 m
104, 649 m
96, 448 m
85, 396 m
71, 698 m
55, 802 m
38, 188 m
9, 735 m
0
m
Lect 02 - 31
Example Calculation 2
At 40˚ North latitude, 76˚ West longitude, a square of 7.5´
(standard quadrangle) will have the planar dimensions:
North_Distance = 13.8793 km
(about 463 30 meter cells)
East_Distance = 10.6742 km
(about 356 30 meter cells)
The Malvern quadrangle used in many of the lecture
examples is of this size.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 32
Note on Spherical Model
NOTE: Although spherical trigonometry will be used for
calculations in this course, the earth is really an
ellipsoid. Calculations based on the spherical model,
acceptable for the purposes of this course, will be
slightly in error.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 33
Spherical Trigonometry
• Solves triangles formed by three points on a
sphere
• If two of these points lie in a plane that
includes the center of the sphere, the arc
between them is a Great Circle.
• The length of a Great Circle arc [radians] is
found by the Cosine Formula.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 34
Cosine Formula
cos c  cos a  cosb  sin a  sinb  cosC
Wikipedia
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 35
Example
Find the Great Circle distance (c) between:
London
(Lat. 51.501080, Lon. - 0.126693)
Philadelphia (Lat. 39.952712, Lon. - 75.163524)
Answer:
5706.41 km
[Calculations follow, Mathematica® Notebook in
Models file “Distance” in course Web site]
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 36
Calculations
Differential longitude:
[Angle C]
-75.0368
-1.30964
degrees
radians
Polar angle [a] to B:
50.0473
0.87349
38.4989
0.671933
0.625774
5706.41
degrees
radians
degrees
radians
radians
km
Polar angle [b] to A:
Cosine of arc c:
Length of arc c:
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 37
Mathematica® Notebook
(* Lattitude of Point A [London] *)
lat1 = 51.501080;
(* Longitude of Point A [London] *)
lon1 = -0.126693;
(* Lattitude of Point B
[Philadelphia] *)
lat2 = 39.952712;
(* Longitude of Point B
[Philadelphia] *)
lon2 = -75.163524;
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 38
Computations (continued)
(* Included angle *)
dlon = lon2 - lon1;
rlon = Degree dlon;
(* POLAR ANGLE TO B *)
dp2 = 90 - lat2;
rp2 = Degree dp2;
(* POLAR ANGLE TO A *)
dp1 = 90 - lat1;
rp1 = Degree dp1;
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 39
Computations (continued)
(* Cosine of arc *)
ca = Cos[rp2]*Cos[rp1] +
Sin[rp2]*Sin[rp1]*Cos[rlon]
(* Spherical radius of Earth *)
r = 6378200.0;
(* Length of arc {km} *)
cd = ArcCos[ca]*r/1000
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 40
Bearings
• For finding bearings, the Sine Formula may
be useful.
• Calculations involving small angles may
require extended precision.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 41
Sine Formula
sin a / sin A  sinb / sin B  sin c / sinC
Wikipedia
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 42
Example - Downingtown Quadrangle
Find the distance [a] and bearing [C] between:
E. Nantmeal (Lat. 40.12246, Lon. -75.72445) (Point B)
Eagle
(Lat. 40.07907, Lon. -75.68838) (Point C)
Answer:
5.724 km
329.266 degrees
[Calculations follow. Mathematica® Notebook in Models
file “Bearing” in course Web site]
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 43
Calculations
Differential longitude:
[Angle A]
Polar angle [c] to B:
Length of arc a:
-0.03607
-0.00062954
49.8875
0.870527
49.9209
0.871285
5.724
degrees
radians
degrees
radians
degrees
radians
km
Bearing to B:
329.266
degrees
Polar angle [b] to C:
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 44
Mathematica® Notebook
(* Lattitude of Point C *)
lat1 = 40.07907;
(* Longitude of Point C *)
lon1 = -75.68838;
(* Lattitude of Point B *)
lat2 = 40.12246;
(* Longitude of Point B *)
lon2 = -75.72445;
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 45
Procedure
1. Calculate arc CB
2. Calculate angle C
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 46
Computations 1
(* Included angle *)
angA = lonB - lonC;
rA = Degree angA;
(* POLAR ANGLE TO B *)
dpB = 90 - latB;
rpB = Degree dpB;
(* POLAR ANGLE TO C *)
dpC = 90 - latC;
rpC = Degree dpC;
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 47
Computations 1 (continued)
(* Cosine of arc *)
ca =
SetPrecision[Cos[rpB]*Cos[rpC] +
Sin[rpB]*Sin[rpC]*Cos[rA], 10]
(* Spherical radius of Earth *)
r = 6378200;
(* Length of arc {km} *)
cd = ArcCos[ca]*r/1000
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 48
Results
ca = 0.9999995973
Note: extra precision (10 places) was used to
avoid roundoff to unity.
ra = 5.724
Note: result is in kilometers
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 49
Computations 2
(* By the Sine Formula:
Sin[C]=Sin[A]*Sin[c]/Sin[a] *)
snc = Sin[rA]* sc/Sin[ra];
AngleC = snc/Degree
BearingC = 360 -AngleC
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 50
Results
• AngleC = 30.7339 degrees
• Bearing = 329.266 degrees
• Slide to follow will show the path
superimposed on a topographical map.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 51
Display of Calculated Path
Topographic map of
Downingtown quadrangle
showing calculated path.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 52
Homework Problem Notes
(1) Use distance example above
(2) First point on Great Circle through origin.
Calculate distance in radians.
(3) Subsequent points at 30 degree intervals.
Use Cosine Rule.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 53
Workshop 1
Using a spreadsheet on a laboratory computer
please complete ALL of the calculations
requested on the distributed Workshop 1
handout.
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 54
Questions ?
3/22/2016
© 2009 Raymond P. Jefferis III
Lect 02 - 55