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

Great and Small CirclesGreat circles splits the earth into equal halves

Small circles splits the earth into unequal halves

wikipedia geography.name

All lines of equal longitude are great circles

Equator is the only line of equal latitude that’s a great circle

Surface distance measurements should all be along a great circle

Caliper Corporation

Smithsonian

Note that great circle distances appear curved on projected or

flat maps

In GIS Fundamentals book, Chapter 2

Latitude - angle to a parallel circle, a small circle parallel to the Equatorial great 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 MeasurementMethod 1: The Earth rotates 360 degrees in a day, or 15 degrees per hour. If we know the time difference between 2 points, and the longitude of the first point (Greenwich), we can determine the longitude at our current location.

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

Longitude by Hour Angle

Clock set to Greenwich time - if accurate enough, calc longitude by time difference

1 hr = 15 deg longitude on Earth, and measured time to get angle from Greenwich Meridian to local point along Equator

12 h at sailor’s meridian,

15h, 18m 55s at Greenwich

1 hr = 15 deg longitude on Earth, and measured time to get angle from Greenwich Meridian to local point along Equator

12 h at sailor’s meridian, 15h, 18m 55s at Greenwich

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 pre-calculated moon-star distances

Given date, and time of night (to/from midnight) the star/moon distance depends on longitude, and can be pre-calculated, 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

C

North Pole

St. Pa

ul, M

N

St. J

ohn’s

, Lab

rado

r

Ac

a

B

b

What is the great circle distance between St. Paul, MN (44.9537° N, 93.0900° W)

and

St. John’s, Labrador (47.5605° N, 52.7128° W)

We know a, b, and C, so we can use the law of cosines to solve

C is difference in longitudes = 93.0900° - 52.7128° = 40.3772

b can be calculated from latitude of St Paul = 90 - 44.9537° N = 45.0463a can be calculated from latitude of St John’s = 90 - 47.5605° N = 42.4395LOC, 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

44.95

37° N

,

93.09

00° W

47.5

605°

N,

52.7

128°

W

Cotangent formulas, derived from LOS, useful for calculating azimuths, distances

Tan(A) = sin(b)

sin(C)

tan(a) - cos(b)cos(C)

pgs 37 and 38 of text, similar formula for angle B, can back calculated for Azimuth from B to A

By definition, angle A is azimuth from St Paul to St John’s

Tan(A) =

C = 40.3772 b = 45.0463a = 42.4395

sin(b) sin(C)

Tan(A) = sin(40.3772)

tan(a) - cos(b)cos(C)

cos(45.0463)cos(40.3772)sin(45.0463)tan(42.4395) -

Tan(A) = 0.647810.23569

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)