You Can Only Use a Planar Surface So Far, Before the Equidistance Assumption Creates Large Errors

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 Circles Great circles splits the Small circles splits the earth into equal halves 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 Note that great circle distances appear curved on projected or ﬂat maps

Smithsonian 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 Measurement Method 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 What is the great circle distance between St. Paul, North Pole MN C (44.9537° N, 93.0900° W) and a St. John’s, Labrador b (47.5605° N, 52.7128° W) B We know a, b, and C, A so we can use the c law of cosines to solve

St. John’s, Labrador

St. Paul, MN C is difference in longitudes = 93.0900° - 52.7128° = 40.3772 b can be calculated from latitude of St Paul = 90 - 44.9537° N

= 45.0463 a can be calculated from latitude of 44.9537° N, St John’s 93.0900° W 47.5605° N, = 90 - 47.5605° N 52.7128° W = 42.4395 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 b = 45.0463 a = 42.4395

Tan(A) = sin(40.3772)

sin(45.0463) - cos(45.0463)cos(40.3772) tan(42.4395) 0.64781 Tan(A) = A = ArcTan (2.745) = 70.00deg 0.23569

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)