Earth – Maps and Navigation size/shape/rotation of Earth is fundamental to ocean dynamics - interesting to know history of Earth’s geometry - need to know how to describe location and size of features - bit of navigational history (can use oceans/celestial) Geography “Physical Geography of the World’s Oceans”

Greek mathematician, poet, athlete, geographer and astronomer Eratosthenes of Cyrene (c ): Interesting to know

800 km Alexandria & Syene (Aswan)

Eratosthenes’ estimate of earth’s diameter 360º x 800 km 7.2º C = C = earth’s circumference = 40,000 km  = 7.2°

Eratosthenes’ estimate of earth’s diameter º 360º = 800 km C C = earth’s circumference = 40,000 km (stadia)  = 7.2° Modern estimate of Earth’s radius, r e = 6371 km C=2  r e = 40,030 km

Eratosthenes

polar radius = 6357 km equatorial radius = 6378 = radius of sphere with Earth’s volume is 6371 km non-spherical shape important for satellite orbits

Latitude and Longitude – Mollweide projection latitude line (equator only great circle latitude) longitude line (great circle) zonal meridional

Latitude and Longitude – Mollweide projection latitude line longitude line 360 deg/1 day = 15 deg/hr

special latitude & longitude lines axial tilt of the Earth with respect to the sun is 23° 26′ 21.41″

circle – subtends angle of 360° 1° = 60’ (minutes) 1’ = 60” (seconds) nautical mile = distance of 1° latitude distance / degree latitude is constant not true for longitude r e = 6371 km circumference = 2πr e = 40,030 km r e = 6371 km c lat 1° / 360° = C lat / 40,030 km C lat = 40,030 km / 360° = 111 km/°lat = 60 nm/°lat = 69 mi/°lat distance per degree of latitude

distance / degree not constant for longitude r e = 6371 km Φ = latitude circumference = 2πr e cosΦ = 40,030km cosΦ r e = 6371 km 1° / 360° = C lon / 40,030km cosΦ C lon = 111 km cosΦ km/°lon distance per degree of longitude ϕ r e cos Φ

A B dd xx yy - can use Pythagorean Theorem to obtain accurate distance estimate between two points - accurate for distances around 100 km or less

34° 20’ 34° 10’ 120° 10’120° A B  A = 34° 20’ = 34° + 20/60° = °  B = 34° 10’ = 34° + 10/60° = °  A -  B = 0.166° A =A = B =B = yy

34° 20’ 34° 10’ 120° 10’120° A B  y = (111 km/ ° lat)(0.166 ° lat) = 18.3 km A =A = B =B = yy

120° 10’120° A B =  A B =B = xx  x = (111*cos(34+15/60) km/°lon)(0.167 °lon) = 15.2 km 34° 15’

A B dd xx yy  d = (  x 2 +  y 2 ) 1/2 = ( ) 1/2 = 23.9 km

early Pacific navigation

early Atlantic navigation

Latitude determination

antique navigational instruments

horizon North Star latitude using a cross staff

sextant

using a sextant

Gemma Frisius 1508 – 1555 mathematician, cartographer, instrument maker, first to describe how an accurate clock could be used to determine longitude

Longitude determination rotation rate = 360 deg in 24 hr = 15 deg per hr 15°

Set clock on boat before departing from Greenwich. Longitude determination

What is longitude if clock on boat reads 16:18 at local noon? Longitude determination

Local noon is behind clock => boat is west of Greenwich Longitude determination

time difference = 4+ hr = 4.3 hr longitude difference = 4.3 hr x 15°/hr = 64.5° W Longitude determination

John Harrison ( ) Clocks for measuring longitude

global sea surface temperature from satellite observations

Readings for next time (Seafloor): -Read Chapter 4 “Seafloor Features”; article “Emergence of Complex Societies After Sea Level Stabilized”; article “Risk of Rising Sea Level to Population and Land Area”