Least Squares Fit to Main Harmonics The observed flow u’ may be represented as the sum of M harmonics: u’ = u0 + ΣjM=1 Aj sin (j t + j) For M = 1 harmonic (e.g. a diurnal or semidiurnal constituent): u’ = u0 + A1 sin (1t + 1) With the trigonometric identity: sin (A + B) = cosBsinA + cosAsinB u’ = u0 + a1 sin (1t ) + b1 cos (1t ) taking: a1 = A1 cos 1 b1 = A1 sin 1 so u’ is the ‘harmonic representation’

2u0b1 cos (1t ) + 2a1 b1 sin (1t ) cos (1t ) + a12 sin2 (1t ) + The squared errors between the observed current u and the harmonic representation may be expressed as 2 : 2 = ΣN [u - u’ ]2 = u 2 - 2uu’ + u’ 2 Then: 2 = ΣN {u 2 - 2uu0 - 2ua1 sin (1t ) - 2ub1 cos (1t ) + u02 + 2u0a1 sin (1t ) + 2u0b1 cos (1t ) + 2a1 b1 sin (1t ) cos (1t ) + a12 sin2 (1t ) + b12 cos2 (1t ) } Using u’ = u0 + a1 sin (1t ) + b1 cos (1t ) Then, to find the minimum distance between observed and theoretical values we need to minimize 2 with respect to u0 a1 and b1, i.e., δ 2/ δu0 , δ 2/ δa1 , δ 2/ δb1 : δ2/ δu0 = ΣN { -2u +2u0 + 2a1 sin (1t ) + 2b1 cos (1t ) } = 0 δ2/ δa1 = ΣN { -2u sin (1t ) +2u0 sin (1t ) + 2b1 sin (1t ) cos (1t ) + 2a1 sin2(1t ) } = 0 δ2/ δb1 = ΣN {-2u cos (1t ) +2u0 cos (1t ) + 2a1 sin (1t ) cos (1t ) + 2b1 cos2(1t ) } = 0

ΣN { -2u +2u0 + 2a1 sin (1t ) + 2b1 cos (1t ) } = 0 ΣN {-2u sin (1t ) +2u0 sin (1t ) + 2b1 sin (1t ) cos (1t ) + 2a1 sin2(1t ) } = 0 ΣN { -2u cos (1t ) +2u0 cos (1t ) + 2a1 sin (1t ) cos (1t ) + 2b1 cos2(1t ) } = 0 Rearranging: ΣN { u = u0 + a1 sin (1t ) + b1 cos (1t ) } ΣN { u sin (1t ) = u0 sin (1t ) + b1 sin (1t ) cos (1t ) + a1 sin2(1t ) } ΣN { u cos (1t ) = u0 cos (1t ) + a1 sin (1t ) cos (1t ) + b1 cos2(1t ) } And in matrix form: ΣN u cos (1t ) ΣN cos (1t ) ΣN sin (1t ) cos (1t ) ΣN cos2(1t ) b1 ΣN u N ΣN sin (1t ) Σ N cos (1t ) u0 ΣN u sin (1t ) = ΣN sin (1t ) ΣN sin2(1t ) ΣN sin (1t ) cos (1t ) a1 X = A-1 B B = A X

Finally... The residual or mean is u0 The phase of constituent 1 is: 1 = atan ( b1 / a1 ) The amplitude of constituent 1 is: A1 = ( b12 + a12 )½ Pay attention to the arc tangent function used. For example, in IDL you should use atan (b1,a1) and in MATLAB, you should use atan2

For M = 2 harmonics (e.g. diurnal and semidiurnal constituents): u’ = u0 + A1 sin (1t + 1) + A2 sin (2t + 2) ΣN cos (1t ) ΣN sin (1t ) cos (1t ) ΣN cos2(1t ) ΣN cos (1t ) sin (2t ) ΣN cos (1t ) cos (2t ) N ΣN sin (1t ) Σ N cos (1t ) ΣN sin (2t ) Σ N cos (2t ) ΣN sin (1t ) ΣN sin2(1t ) ΣN sin (1t ) cos (1t ) ΣN sin (1t ) sin (2t ) ΣN sin (1t ) cos (2t ) Matrix A is then: ΣN sin (2t ) ΣN sin (1t ) sin (2t ) ΣN cos (1t ) sin (2t ) ΣN sin2(2t ) ΣN sin (2t ) cos (2t ) ΣN cos (2t ) ΣN sin (1t ) cos (2t ) ΣN cos (1t ) cos (2t ) ΣN sin (2t ) cos (2t ) ΣN cos2 (2t ) Remember that: X = A-1 B and B = ΣN u cos (1t ) ΣN u sin (2t ) ΣN u cos (2t ) ΣN u ΣN u sin (1t ) u0 a1 b1 a2 b2 X =

Σ [< uobs > - upred] 2 Σ [<uobs > - uobs] 2 Goodness of Fit: Σ [< uobs > - upred] 2 ------------------------------------- Σ [<uobs > - uobs] 2 Root mean square error: [1/N Σ (uobs - upred) 2] ½

Fit with M2 only

Fit with M2, K1

Tidal Ellipse Parameters K1 Major axis: M minor axis: m ellipticity = m / M Phase Orientation

Tidal Ellipse Parameters ua, va, up, vp 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 component phase of the counter-clockwise rotary component The characteristics of the tidal ellipses are: Major axis = M = Qcc + Qc minor axis = m = Qcc - Qc ellipticity = m / M Phase = -0.5 (thetacc - thetac) Orientation = 0.5 (thetacc + thetac) Ellipse Coordinates:

M2 K1

Study Area

Study Area Transect Sampled June 12, 2012 Semidiurnal Cycle

Trajectories Sampled

East Component North Component Raw Velocity Components (all data)

-12.5º rotation (u, v) cm/s (East, North) cm/s

105 26 18

TIDE ADCP depth measurements

1.35 0.19 0.12

FLOOD u 