Least Squares Fit to Main Harmonics The observed flow u’ may be represented as the sum of M harmonics: u’ = u 0 + Σ j M =1 A j sin (  j t +  j ) For M = 1 harmonic (e.g. a diurnal or semidiurnal constituent): u’ = u 0 + A 1 sin (  1 t +  1 ) With the trigonometric identity: sin (A + B) = cosBsinA + cosAsinB u’ = u 0 + a 1 sin (  1 t ) + b 1 cos (  1 t ) taking: a 1 = A 1 cos  1 b 1 = A 1 sin  1 so u’ is the ‘harmonic representation’

The squared errors between the observed current u and the harmonic representation may be expressed as  2 :  2 = Σ N [u - u’ ] 2 = u 2 - 2uu’ + u’ 2 Then:  2 = Σ N {u 2 - 2uu 0 - 2ua 1 sin (  1 t ) - 2ub 1 cos (  1 t ) + u u 0 a 1 sin (  1 t ) + 2u 0 b 1 cos (  1 t ) + 2a 1 b 1 sin (  1 t ) cos (  1 t ) + a 1 2 sin 2 (  1 t ) + b 1 2 cos 2 (  1 t ) } Using u’ = u 0 + a 1 sin (  1 t ) + b 1 cos (  1 t ) Then, to find the minimum distance between observed and theoretical values we need to minimize  2 with respect to u 0 a 1 and b 1, i.e., δ  2 / δu 0, δ  2 / δa 1, δ  2 / δb 1 : δ  2 / δ u 0 = Σ N { -2u +2u 0 + 2a 1 sin (  1 t ) + 2b 1 cos (  1 t ) } = 0 δ  2 / δ a 1 = Σ N { -2u sin (  1 t ) +2u 0 sin (  1 t ) + 2b 1 sin (  1 t ) cos (  1 t ) + 2a 1 sin 2 (  1 t ) } = 0 δ  2 / δ b 1 = Σ N {-2u cos (  1 t ) +2u 0 cos (  1 t ) + 2a 1 sin (  1 t ) cos (  1 t ) + 2b 1 cos 2 (  1 t ) } = 0

Σ N { -2u +2u 0 + 2a 1 sin (  1 t ) + 2b 1 cos (  1 t ) } = 0 Σ N {-2u sin (  1 t ) +2u 0 sin (  1 t ) + 2b 1 sin (  1 t ) cos (  1 t ) + 2a 1 sin 2 (  1 t ) } = 0 Σ N { -2u cos (  1 t ) +2u 0 cos (  1 t ) + 2a 1 sin (  1 t ) cos (  1 t ) + 2b 1 cos 2 (  1 t ) } = 0 Rearranging: Σ N { u = u 0 + a 1 sin (  1 t ) + b 1 cos (  1 t ) } Σ N { u sin (  1 t ) = u 0 sin (  1 t ) + b 1 sin (  1 t ) cos (  1 t ) + a 1 sin 2 (  1 t ) } Σ N { u cos (  1 t ) = u 0 cos (  1 t ) + a 1 sin (  1 t ) cos (  1 t ) + b 1 cos 2 (  1 t ) } And in matrix form: Σ N u cos (  1 t ) Σ N cos (  1 t ) Σ N sin (  1 t ) cos (  1 t ) Σ N cos 2 (  1 t ) b 1 Σ N u N Σ N sin (  1 t ) Σ N cos (  1 t ) u 0 Σ N u sin (  1 t ) = Σ N sin (  1 t ) Σ N sin 2 (  1 t ) Σ N sin (  1 t ) cos (  1 t ) a 1 B = A X X = A -1 B

Finally... The residual or mean is u 0 The phase of constituent 1 is:  1 = atan ( b 1 / a 1 ) The amplitude of constituent 1 is: A 1 = ( b a 1 2 ) ½ Pay attention to the arc tangent function used. For example, in IDL you should use atan (b 1,a 1 ) and in MATLAB, you should use atan2

For M = 2 harmonics (e.g. diurnal and semidiurnal constituents): u’ = u 0 + A 1 sin (  1 t +  1 ) + A 2 sin (  2 t +  2 ) Σ N cos (  1 t ) Σ N sin (  1 t ) cos (  1 t ) Σ N cos 2 (  1 t ) Σ N cos (  1 t ) sin (  2 t ) Σ N cos (  1 t ) cos (  2 t ) N Σ N sin (  1 t ) Σ N cos (  1 t ) Σ N sin (  2 t ) Σ N cos (  2 t ) Σ N sin (  1 t ) Σ N sin 2 (  1 t ) Σ N sin (  1 t ) cos (  1 t ) Σ N sin (  1 t ) sin (  2 t ) Σ N sin (  1 t ) cos (  2 t ) Matrix A is then: Σ N sin (  2 t ) Σ N sin (  1 t ) sin (  2 t ) Σ N cos (  1 t ) sin (  2 t ) Σ N sin 2 (  2 t ) Σ N sin (  2 t ) cos (  2 t ) Σ N cos (  2 t ) Σ N sin (  1 t ) cos (  2 t ) Σ N cos (  1 t ) cos (  2 t ) Σ N sin (  2 t ) cos (  2 t ) Σ N cos 2 (  2 t ) Remember that: X = A -1 B and B = Σ N u cos (  1 t ) Σ N u sin (  2 t ) Σ N u cos (  2 t ) Σ N u Σ N u sin (  1 t ) u0a1b1a2b2u0a1b1a2b2 X =

Goodness of Fit: Σ [ - u pred ] Σ [ - u obs ] 2 Root mean square error: [1/N Σ (u obs - u pred ) 2 ] ½

Fit with M 2 only

Fit with M 2, K 1

Fit with M 2, S 2, K 1 record frequency ≤ ω 1 – ω 2 Rayleigh Criterion: record frequency ≤ ω 1 – ω 2

Fit with M 2, S 2, K 1, M 4, M 6

M2S2K1M2S2K1 Tidal Ellipse Parameters Major axis: M minor axis: m ellipticity = m / M Phase Orientation

Tidal Ellipse Parameters u a, v a, u p, v p are the amplitudes and phases of the east-west and north-south components of velocity amplitude of the clockwise rotary component amplitude of the counter-clockwise rotary component phase of the clockwise rotary componentphase of the counter-clockwise rotary component The characteristics of the tidal ellipses are: Major axis = M = Q cc + Q c minor axis = m = Q cc - Q c ellipticity = m / M Phase = -0.5 (theta cc - theta c ) Orientation = 0.5 (theta cc + theta c ) Ellipse Coordinates:

M2S2K1M2S2K1

Now, let us consider the velocity of a fixed particle on a rotating body at the position The body, for example the earth, rotates at a rate, To an observer from space (us): This gives an operator that relates a fixed frame in space (inertial) to a moving object on a rotating frame on Earth (non-inertial)

This operator is used to obtain the acceleration of a particle in a reference frame on the rotating earth with respect to a fixed frame in space 0 Acceleration of a particle on a rotating Earth with respect to an observer in space Coriolis Centripetal

The equations of conservation of momentum, up to now look like this: Coriolis Acceleration CvCv ChCh

Making: f is the Coriolis parameter This can be simplified with two assumptions: 1)Weak vertical velocities in the ocean ( w << v, u ) 2)Vertical component is ~5 orders of magnitude < acceleration due to gravity

Eastward flow will be deflected to the south Northward flow will be deflected to the east f increases with latitude f is negative in the southern hemisphere