It seems our synethtic model images of the Moon have a distribution across the lunar disc that does not match observations. When we fit such an image to the ‘band’ near the photo-equator and subtract the fitted image from the observed image we get a residuals image with ‘Mickey Mouse ears’ – i.e. the residuals in intensity are bi-polar.
We are testing a semi-empirical method where the BS of the synthetic image is replaced by the observed BS, scaled to the right value. When such an image is fitted to the observed image we get no Mickey Mouse ears problem:
Left image: Residuals after a well-fitting synthetic image has been subtracted from the observed. Note the polar nature of the residuals – the cusps are clearly too bright while the area between the cusps is too dark. Right image: residuals when a well-fitting image, based on the observed BS, is subtracted. There is very little large-scale structure now.
The two fits deliver terrestrial albedo as one of the fitting parameters.
For the left image (i.e. the entirely synethtic model image) we get A=0.37298, for the semi-empirical image we get A=0.36311, that is a difference of 2.7 %, which is considerable, given our science goals of 0.1%.
Whether the fit has been improved where it counts is a different matter. The RMSE for the two fits are identical (RMSE=0.026).
The large change in fitted albedo is a (regrettable) feature of our fitting method – change something a little bit and the fits change by several percent.
We suggest that a meaningful test of fit quality will be to fit the ensemble of images with the two methods above and see if the scatter is less in one than in the other. TBD.
