Alternate coordinate systems:
Solutions for large-scale maps
+
X axis
Y axis
-
(0,0)
-
• Universal Transverse Mercator (UTM)
• State Plane Coordinate System (SPCS)
• US Public Land Survey (USPLS)
+
The Geographic Coordinate System:


Strengths:
–
One set of coordinates can describe an exact location
anywhere on the globe
–
Based on the earth’s spherical shape
Issues:


Subdivision of degrees:
 1° = 60’
 1’ = 60”
1 degree = 60 minutes
1 minute = 60 seconds
The length of 1 ° longitude varies
UTM Grid System:
Why it was developed
• UTM = Universal Transverse Mercator
• Developed shortly after World War II by US
Army
• Now, UTM system used widely for precise
positioning
Mercator
Transverse Mercator
1 UTM zone:
6 degrees wide
• Is divided into north-south columns known as zones
(between 80°S and 84°N)
• Zones are numbered 1 to 60 eastward from the Int’l Date
Line
UTM Zones for the United States
Logic of the UTM Grid
Zone 12
Origin
500,000 m
Reading the UTM Grid
Easting is measured
from the central
meridian + 500,000 m
Northing is measured
from the equator
Austin is:
621,161 m E,
3,349,894 m N,
Zone 14 N,
NAD 83
Some additional facts about UTM

Accuracy level of UTM -Maximum error is:
1 m / 1000 m
(5 feet in a mile)
Designating location: GCS
• Geographic Coordinate System
33° 40’ 12” North
Deg. Min. Sec. Hemisph.
111° 55’ 30” West
Deg.
Min. Sec. Hemisph.
• Universal Transverse Mercator
621,100 m E, 3349,800 m N, Zone 14 N, NAD 83
Easting,
Northing,
Or, on topo map:
621100
m E, 3349800 m N
Zone,
Datum
Comparing coordinate systems
+
X axis
Y axis
-
(0,0)
-
• The Geographic Coordinate System
• Universal Transverse Mercator (UTM)
• State Plane Coordinate System (SPCS)
• US Public Land Survey (USPLS)
+
The Geographic Coordinate System:
Longitude and latitude
Latitude
33° 30’ 00” North,
Degrees, Minutes, Seconds N,
Longitude
112° 00’ 00” West
Degrees, Minutes, Seconds W
The United States Public Land Survey:
The National Picture
The United States Public Land Survey:
Townships & Ranges
T 2 S,
Township,
R 3 W,
Range,
Gila and Salt River Meridian
Principal Meridian
Townships
Survey
Township
6 mi.
Ranges
Choosing a Coordinate System

Cartesian coordinate systems:
–
–

When precise measurements of distance & area
are needed
Choose a system based on size & position

USPLS

UTM
Geographic Coordinate system:
–
Thematic maps – coordinates give general idea of
location on globe
Use of Known Features

Size of known objects may be visible on the
map—particularly true with aerial photographs
Strategy:
1. Create a fraction:
In real life:
Athletic field is
130 yards
Map distance / real distance
2. Simplify:
•
Numerator & denominator
are in same units
•
numerator is “1”
On the photo:
Athletic field 7/8
inch
7 inches
0.875
0
.
875
inches
0.875
8
0.875  1



130 yards 130 * 36 inches
4680 4680
5349
0.875
Photo scale =
Approx 1:5,350
Use of Latitude and Longitude
Strategy:
In real life:
1. Create a fraction:
Map distance / real distance
1° latitude = 111.13 km
2. Simplify:
• Numerator & denominator are in
same units
• numerator is “1”
On the map:
10° latitude = 1,111.3 km
10° latitude = 8.8 cm
Map scale =
approx
1:12,628,750
8.8
8.8 cm
8.8 cm
8.8


1111.33 km 1111.33 *100,000 cm 111,133,000

8.8
1
12,628,750
Using longitude instead:
If using longitude, remember
they converge at the poles.
Calculate the distance with the
following formula:
Distance = cos(latitude) x 111.113km
Map Comparison

Use a map with a known scale and
compare features on the map with unknown
scale
1. Measure the
distance between
same pair features
on both maps
6 inches
5 inches
1:5200
2. Calculate the real- world
distance on the map with
the known scale
6” = x” on ground?
6” = 5200” x 6
6” = 31,200” on ground
6 inches
5 inches
1:5200
3. Calculate the ratio – as
before – of the map with
unknown scale
6” = x” on ground?
5” = 31,200” on ground
6” = 5200” x 6
1” = 31,200” / 5
6” = 31,200” on ground
1” = 6240” on ground
6 inches
5 inches
1:5200
1:6240
Scale and area
9 square inches, 1:5200
map:
9 sq in (map) = 46,800 sq in (ground) ?

RFs are LINEAR
scales, and can be
used to measure
lines but not areas
(at least not
directly)
NO!
That would imply that, from
McAllister to Palm Walk is 207 inches
on the ground!
1
2
3
4
5
6
7
8
9
1:5200
Scale and area
3 inches on each (linear) side
3”map = 5200 x 3”ground
3”map = 15,600” ground
• To calculate AREA
from a linear scale …
15,600” / 63,360in/mi = .246 mi
.246 mi x .246 mi =.064 mi2
Strategy:
1.
Measure width & height of area (on
map):
2.
Use RF to calculate the area’s
width & height in real world
•
Convert to more convenient
real-world units
Multiply real-world width x height
3.
3”
3”
1:5200
New points about scale …
Two common RF’s:

–
–
–

1:63,360 1 inch represents 1 mile
1:100,000 1 cm represents 1 km
Use these two RFs to be able to visualize the meaning of
other RF’s
Strategies for solving scale problems:
–
To calculate RF for a map of unknown scale
1. Create a fraction:
Map distance / real distance
2. Simplify:


–
Numerator & denominator are in same units
numerator is “1”
To calculate land area:
1. Find height and width in real-world units
2. Multiply these 2 values