1. 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’

2. 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

3. 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

4. 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

5. 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 =

6. 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

7. Fit with M2 only

8. Fit with M2, K1

9. Tidal Ellipse ParametersK1
Major axis: M
minor axis: m
ellipticity = m / M
Phase
Orientation

10. 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:

11. M2 K1

12. Study Area

13. Study Area
Transect Sampled
June 12, 2012
Semidiurnal Cycle

14. Trajectories Sampled

15. East Component
North Component
Raw Velocity Components (all data)

16. -12.5º rotation
(u, v)
cm/s
(East, North)
cm/s

20. TIDE
ADCP
depth measurements

22. FLOOD u 