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You can only use a planar surface so far, before the
equidistance assumption creates large errors
Distance
error
from
Kiester
to
Warroad is
greater than
two football
fields in length
So we assume a spherical Earth
P. Wormer, wikimedia commons
Longitudes are
great circles,
latitudes are
small circles
(except the
Equator, which
is a GC)
Spherical Geometry
Longitude
Spherical Geometry
Note:
The Greek letter l (lambda) is almost always used
to specify longitude
while
f, a, w and c, and other symbols are used to
specify latitude
Smithsonian
Latitude - angle to a parallel circle
Spherical
Geometry
Three
measures:
Latitude,
Longitude,
and Earth
Radius +
height above/
below the
sphere (hp)
How do we measure latitude/longitude?
Well, now, GNSS, but originally, astronomic
measurements:
latitude by north star or solar noon angles, at equinox
Longitude by the equation of time
Method 1:
The Earth rotates 360 degrees in a day, or 15 degrees
per hour. If we know the time difference between 2
points, we can determine the longitude.
Method 2:
Create a table of moon-star distances for each day/time
of the year at a reference location (Greenwich
observatory)
Measure the same moon-star distance somewhere else
at a standard or known time. The distance will be slightly
different, and we can use the difference to calculate
longitude
1 hr = 15 deg
angle from Greenwich Meridian to local point along Equator
15h 18m 55s - 12h = 3h 18min 55sec
in decimal hours, 3+18/60 + 55/3600 = 3.315277h
so angle = 3.31527 * 15 = 49.729167h
= 49 deg 0.729*60 min
= 49 deg 43.75 min
= 49 deg 43 min 0.75*60 sec
= 49 deg 43 min 45 sec
Longitude with no
clock is more
difficult - earliest
accurate method
use precalculated moonstar distances
Given date, and
time of night (to/
from midnight)
the star/moon
distance depends
on longitude, and
can be precalculated,
placed in tables
Now, international services broadcast time signals
over radio and other channels, so you can know the
exact Greenwich time instantly all over the world.
VLBI - Very Long
Baseline Interferometry
A Combination of
systems, but
based ultimately
on astronomical
measurements
Distances and Angles on a Sphere
Given L0, what is the Azimuth and
distant to L1?
Typically solve this
problem with a spherical
triangle and law of sines or
cosines, with one corner of
the triangle at the nearest
pole
Note that both
angles and
distances are
measured in
spherical units
(degrees or
radians) and not
linear units (e.g.,
miles or km)
Distances and
Angles
(Azimuths)
between points
on a Sphere
Law of Sines
sin(a)
sin(b)
——— = ———sin(A)
sin(B)
sin(c)
sin(b)
——— = ———sin(C)
sin(B)
sin(a)
sin(c)
——— = ———sin(A)
sin(C)
Law of Cosines
cos(a) = cos(b)cos(c)+sin(b)sin(c)cos(A)
or cyclically,
cos(c) = cos(b)cos(a)+sin(b)sin(a)cos(C)
Remember, A is angle, a is side
What is the great circle
distance between St. Paul,
MN
(44.9537° N, 93.0900° W)
North Pole
C
and
a
St. John
’s, Labra
dor
b
St. John’s, Labrador
(47.5605° N, 52.7128° W)
B
.P
au
l, M
N
A
St
We know a, b, and C,
so we can use the
law of cosines to
solve
c
a can be calculated from latitude of
St John’s
= 90 - 47.5605° N
= 42.4395
47
.
52 5605
.71
°
28 N,
°W
b can be calculated from latitude
of St Paul
= 90 - 44.9537° N
= 45.0463
44
93 .95
.0 37
90 °
0° N,
W
C is difference in longitudes
= 93.0900° - 52.7128°
= 40.3772
LOC, cos(c) = cos(b)cos(a)+sin(b)sin(a)cos(C)
cos(c)= cos(45.0463)cos(42.4395)+sin(45.0463)sin(42.4395)cos(40.3772)
c = 0.48385
distance = R * angle = 6,371km * 0.48385 = 3,082.6km
Cotangent formulas, derived from LOS, useful for
calculating azimuths, distances
By definition, angle A is azimuth from St Paul to St John’s
sin(C)
Tan(A) =
sin(b)
- cos(b)cos(C)
tan(a)
pgs 37 and 38 of text, similar
formula for angle B, can back
calculated for Azimuth from B to A
sin(C)
Tan(A) =
sin(b)
- cos(b)cos(C)
tan(a)
C = 40.3772
a = 42.4395
b = 45.0463
Tan(A) =
sin(40.3772)
sin(45.0463)
tan(42.4395)
Tan(A) =
- cos(45.0463)cos(40.3772)
0.64781
0.23569
A = ArcTan (2.745) = 70.00deg
Note some online calculators use an ellipsoidal calculation, so values may differ a
bit….and some are just plain wrong!
Note, your initial azimuth won’t get you to your
destination on the shortest path
(on great circle)
(rhumb line,
follows
constant
azimuth)
Due to
longitudinal
convergence, all
fixed azimuth
paths that are
not on a great
circle will spiral
to the nearest
pole
Three-dimensional Earth centered coordinate system
Why use 3-D Cartesian?
Certain common calculations
are easier, and so they’ve been
adopted as standards by most
governments
It is easy to convert from spherical coordinates to
3-D coordinates and back
(also true for ellipsoidal coordinates, more about
those later)