We have used B and V images of the moon from JD2455859 to make a map of the B-V colour of the reflected light from a thin crescent moon.
We use the photometric transformations derived from our open cluster observations: http://earthshine.thejll.com/#post229
The B band image of the moon was shifted to the position of the V band image, and the colour computed from the fluxes in B and V directly from the pixel fluxes.
The map below shows the result:
The scale at bottom runs from B-V=0.1 to B-V=1.5.
The darkside (DS) has a B-V colour in the range 0.6 to 0.9 — whereas the crescent has B-V ~ 1.1. The measured colour of the BS is B-V ~ 0.9 (Van den Bergh, 1962, ApJ, 67, 147), so we may be a little too red. More work needed on this.
We are seeing roughly the colour difference expected between ES and the BS — ES should be a few tenths bluer in B-V.
There is a clear residual halo in the colour map, indicating that the scattered light around the crescent falls off differently in B and V filters (as we expect).
Very interestingly, the crater pattern becomes much harder to see in the colour map. The V and B images look like this (after registration) V on the left, B on the right:
The lowlands appear to be slightly redder than the uplands in the colour map, by a few tenths of a magnitude in B-V.
More work needed on this — this is just a progress report!
Technical info:
a) Images used were:
2455859.1313477MOON_V_AIR_DCR.fits and 2455859.1365256MOON_B_AIR_DCR.fits
b) The transformations used were (airmass is 2.6 for V and 2.7 for B, extinction coefficients for V and B are 0.10 and 0.15 respectively).
Vinst = -2.5*log10(sumv/exptimev) – 2.6*0.10 ! airmass = 2.6
Binst = -2.5*log10(sumb/exptimeb) – 2.7*0.15
V = Vinst + 15.07 – 0.05*bmv
B = Binst + 14.75 + 0.21*bmv
which implies that B-V = 1.35 * (Vinst-Binst) – 0.43
(the airmasses above were incorrect in the first version of this entry, they have been corrected now)
I think you use airmass=3.5, yes? It is nearer 2.5. This would give is 1*0.05 mags or the 0,5 mag difference I noted. If this is so we have a B-V that is substantially different from van den Berghs B-V=0.9.Are we suer taht the images used from DJ2456034 are on our ‘GOOD list’? Inside the tunnel, and all that?
That’s easy — the equations in that particular form can be solved to yield B-V directly — I’ve added particular details about on this to the end of the posting…
I’m confused on how you get B and V without knowing B and V already – do you iterate, starting from guesses?
I’ll work on a version using 100 summed frames for each filter, instead of just the single frames used here…
Very nicemap! There are published values for the colour of highlands v. Mare – i’ll dig these out! Swell!