improving the spatial thickness distribution of modelled arctic sea ice
DESCRIPTION
Improving the Spatial Thickness Distribution of Modelled Arctic Sea Ice. UK Sea Ice Meeting, 8-9 th Sept 2005. Paul Miller, Seymour Laxon, Daniel Feltham Centre for Polar Observation and Modelling University College London. Motivation. - PowerPoint PPT PresentationTRANSCRIPT
Improving the Spatial Thickness Distribution of Modelled Arctic Sea Ice
Paul Miller, Seymour Laxon, Daniel FelthamCentre for Polar Observation and Modelling
University College London
UK Sea Ice Meeting, 8-9th Sept 2005
Motivation• Rothrock et al., (2003)
showed Arctic sea ice thickness as predicted by various models
• Significant differences, both in the mean and anomaly of ice thickness during the last 50 years
• Causes of differences not well understood, but there is both parameter and forcing uncertainty
• How can we reduce this uncertainty and increase our confidence in conjectures based on model output?
Reducing Parameter Uncertainty in Sea Ice Models
• Use one of the best available sea ice models (CICE) and force it with the best available fields (ERA-40 & POLES)
• Optimise and validate the model using a comprehensive range of sea ice observations:– Sea ice velocity, 1994-2001 (SSM/I
+ buoy + AVHRR, Fowler, 2003, NSIDC)
– Sea ice extent, 1994-2001 (SSM/I, NSIDC)
– Sea ice thickness, 1993-2001 (ERS radar altimeter, Laxon et al., 2003)
We used this model and forcing to reduce uncertainty surrounding sea ice model parameters
Parameter Space
Ice strength, P*
Albedo, ice
Air drag coefficient, Cair
• Our parameter space has three dimensions
• Uncertainty surrounds correct values to use
• Space includes commonly-used values
• 168 model runs needed to optimise model
0.62
0.54
2.5 (kPa)100
0.0003
0.0016
We explored the model’s multi-dimensional parameter space to find the ‘best’ fit to the observational data
Arctic Basin Ice Thickness(<81.5oN)
{ice, Cair, P*} = {0.56, 0.0006, 5 kPa}
Miller et al 2005a
Validation Using ULS Data from Submarine Cruises
• We consider data from 9 submarine cruises between 1987 and 1997
• Rothrock et al. (2003) used this data to test their coupled ice-ocean model
• Modelled cruise means of ice draft are in good agreement with ULS observations
R = 0.98RMS difference = 0.28m
Rothrock et al., 2003, JGR, 108(C3), 3083
Spatial Draft Discrepancy
Rothrock et al., 2003, JGR, 108(C3), 3083 Optimised CICE Model
Sea Ice Rheology
• CICE sea ice rheology is plastic
• CICE has an elliptical yield curve, with ratio of major to minor axes, e (Hibler, 1979)
• Maximum shear strength determined by P*, thickness, concentration and e
• Decreasing e reduces ice thickness in the western Arctic and increases it near the Pole
2
1
e=2
e=√.5
C
S
S
P/2
Spatial Draft Discrepancy
e=2 e=√.5
Improvements Due to Increased Shear Strength
Improved Zonal Averages
Improved Cruise Averagese = 2 e = √.5
Model vs ERS Mean Winter Ice Thickness (<81.5oN)
Model-Satellite Thickness (m)
e = 2 e = √.5
Truncated Yield Curve2
1
e=2
e=√.5
S
Truncated Yield Curve2
1
e=2
e=√.5
S
Truncated
Truncated Yield Curve2
1
e=2
e=√.5
S
Truncated < 80% IceConcentration
Arctic Basin Ice Thickness Since 1980
e = 2e = √.5
e = √.5(Truncated for IC < 80%)
Conclusions
• Initial work reduced parameter uncertainty in a stand-alone sea ice model
• Observations of thickness/draft from submarine cruises were used to independently test the optimised model
• By increasing the shear strength (by changing e from 2 to .5), we reproduced the observed spatial distribution of ice draft
• Found that some tensile strength is necessary
• These results are in press, Miller et al 2005b
Paul Miller, Seymour Laxon, Daniel FelthamCentre for Polar Observation and Modelling, UCL
Slide 5
Melt Season LengthIce ConcentrationIce Motion
•Extend observational validation back to mid-1980’s using intermittent submarine data•Use optimised model to examine changes in radiative, thermal, and mechanical forcing to determine primary mechanisms in ice mass change from 1948 - present
ERS-derived Mean Winter Ice Thickness
Model-Observed Thickness (m)
e = 2 e = √.5