1. ACs processing TAO-cruise August 2006

2. 1. Total vs 0.2  m signal total signal 0.2  m signal The 0.2  m signal is removed from the total signal using the 0.2  m interpolated values 0.2  m interpolated values

3. 2. Spectral discontinuity correction Present also in 0.2  m measurements

4. 2. Spectral discontinuity correction Longer Portion of the Spectrum Shorter Portion of the Spectrum Q: Which is the correct one, LPS or SPS? A: For a p we don’t care because we know that a p (NIR)≈0.

5. 2. Spectral discontinuity correction Correction method a)assume that the LPS is the correct one Longer Portion of the Spectrum

6. 2. Spectral discontinuity correction Correction method a)assume that the LPS is the correct one b)using the first 2 s of the LPS, predict the value of a p at the last of the SPS

7. 2. Spectral discontinuity correction Correction method a)assume that the LPS is the correct one b)using the first 2 s of the LPS, predict the value of a p at the last of the SPS c)compute the difference between the predicted and observed values of a p at the last of the SPS d)subtract such difference from the SPS

8. 2. Spectral discontinuity correction Case of c p a)There is no at which we a-priori know the value of c p b)Arbitrarily assume that LSP is correct c)Apply correction as for a p d)Evaluate the bias introduced by such arbitrary assumption

9. 2. Spectral discontinuity correction Bias evaluation (c p ) a)NOTE: For this data set, c p (750) (minimal values) ranges from 0.02 to 0.08 m -1 b)For most of the c p data the discontinuity is not important c)When it is important, our arbitrary assumption introduces biases of the order of 0.0007/0.02=4% (conservative estimate) d)There is no preferential direction for the shift

10. 2. Spectral discontinuity correction Importance of this correction for a p a)NOTE: For this data set, a p (676) ranges from 0.001 to 0.005 m -1 b)For most of the a p data the discontinuity is not important c)When it is important, the discontinuity can be of the order of 0.0003/0.005=6% to 0.0003/0.001=30% of a p (676) d)There is no preferential direction for the shift

11. 3. Interpolation to common Because the ACs produces a p and c p spectra with different wavelength centers, we interpolated each spectrum to have equally spaced values every 2 nm from 400 to 750 nm

12. 4. Correction for residual T- dependence In theory by subtracting the 0.2  m interpolated signal from the total signal, we should not need any TS-correction of our a p spectra However, because the a p that we are measuring have really low values in the red region (0.001-0.005 m -1 ), even small  Ts between two consecutive 0.2  m filtrations could introduce relatively large biases in a p

13. What ranges of  T and  S are we observing between 1-hr distant data points? 4. Correction for residual T- dependence from Sullivan et al. 2006

14. 4. Correction for residual T- dependence What ranges of  T and  S are we observing between 1-hr distant data points? ignore  S

15. 4. Correction for residual T- dependence 1. b( ) is independent of  T (and  S) scattering correction note that part of the b-corr depends on  T 2.look at Zaneveld’s scattering correction #3:

16. 4. Correction for residual T- dependence How to get  T? a)Assume that there exists a spectral region ( ref ) where: b)Set ref =710:740 nm and NIR=730 nm c)For each spectrum, find  T that minimizes: where: d)Apply T-correction to a p using the derived  T

17. 4. Correction for residual T- dependence

18. no correction

19. 4. Correction for residual T- dependence b-correction

20. 4. Correction for residual T- dependence Tb- correction

21. 4. Correction for residual T- dependence

23. 5. Compute c p T The temperature-corrected c p is finally calculated as:

24. 6. Biofouling

26. 7. CD content processed ACs data merged with PAR, lat, long, SST and salinity data Sullivan et al. 2006 table interpolated every 2 nm (instrument specific)

27. Some highlights

29. -125ºW-140ºW

30. 125º140º

33. biofouling