Target selection

[Jim Mosher]

The selection of Cabeus A as the impact target has been announced
here:

http://www.nasa.gov/centers/ames/news/releases/2009/09-118AR.html

and here:

http://www.nasa.gov/mission_pages/LCROSS/main/LCROSS_crater.html

both of which include some graphics with little explanation. I would
assume the target is the purple spot in the "Zoomed in image of south
pole Potential Water Concentrations" slide?

http://www.nasa.gov/centers/ames/images/content/385730main2_ARC-LCROSS-WEH-Part2_226.jpg

Has anyone found a transcript of the press conference?

-- Jim

[cano...@yahoo.com]

It is up on NASA at YouTube.

http://www.youtube.com/profile?user=NASAtelevision&view=videos

Colaprete stated the sweet spot PSR is in the upper half of small
crater on the west corner of Caebus A. See the NMSU finder image.

He also stated in response to a reporter question that - apparently
referring to integrated magnitudes - that the plume brightness would
be between 4-5 mags. Id. at minute 32:35.

- Kurt

On Sep 11, 2:01 pm, Jim Mosher <jimmos...@gmail.com> wrote:
> The selection of Cabeus A as the impact target has been announced
> here:
>
>  http://www.nasa.gov/centers/ames/news/releases/2009/09-118AR.html
>
> and here:
>
>  http://www.nasa.gov/mission_pages/LCROSS/main/LCROSS_crater.html
>
> both of which include some graphics with little explanation.  I would
> assume the target is the purple spot in the "Zoomed in image of south
> pole Potential Water Concentrations" slide?
>

>  http://www.nasa.gov/centers/ames/images/content/385730main2_ARC-LCROS...

[cano...@yahoo.com]

P.S. - He also stated that although this was the final choice, the
Team had selected this location as the primary preliminary candidate
target back on the launch date in June.

[Jim Mosher]

That's interesting since nothing in this vicinity was in the "short
list" released to this Group on July 14th:

http://groups.google.com/group/lcross_observation/msg/85b9db372ca99859

-- Jim

[Derek C Breit]

They obviously didn't want to scare anyone off, because if I had known this is the way they were leaning, I wouldn't have been here for the last two months.

There is zero chance I can image this with this being the target..



Derek C Breit
BREIT IDEAS Web Design
breit...@hotmail.com





----------------------------------------
> Date: Sat, 12 Sep 2009 11:23:05 -0700
> Subject: [LCROSS_OBS: 1059] Re: Target selection
> From: jimm...@gmail.com
> To: lcross_ob...@googlegroups.com

>
>
> On Sep 11, 1:39 pm, "canopu...@yahoo.com" wrote:
>> P.S. - He also stated that although this was the final choice, the
>> Team had selected this location as the primary preliminary candidate
>> target back on the launch date in June.
>
> That's interesting since nothing in this vicinity was in the "short
> list" released to this Group on July 14th:
>
> http://groups.google.com/group/lcross_observation/msg/85b9db372ca99859
>
> -- Jim
> >

_________________________________________________________________
Your E-mail and More On-the-Go. Get Windows Live Hotmail Free.
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[jim phillips]

Why?

Jim Phillips

[Rick Baldridge]

Good question!

 

Even though the plume will not be against a dark background, that doesn't mean it won't be observable.  The plume brightness will be added to the lit lunar landscape in the background and will still be observable.  Certainly, the ideal condition would have been seeing a V-shaped plume up above the limb or against a shadowed region, but infortunately it didn't work out that way.  Cameras and video certainly have the capability of recording the plume, and I expect even visual observers will be able see it.  Don't give up!!!

 

Others on this discussion group who are far more knowledgeable than myself will certainly be taking the plume flux calculations submitted by Diane Wooden yesterday and giving numerical estimates of the plume's brightness, contrast ratio, size, etc.  NASA/Ames will be providing info to professional and amateur astronomers in the coming few weeks to aid in exposure calculations. 

 

 

Rick B.

[Derek C Breit]

Why wouldn't I have been here for the last two months?
and Why can't I image it??

Same answer either way..

My video camera is far too sensitive for the bright limb..

And it is not even near the edge of the limb where the plume might extend into space..

All I would get would be a white screen. If I shortened the exposures to where I just began to see albedo features on the bright limb, then I couldn't see even a mag 2 point source.

I am sure everyone else can image it just fine. I am not good enough to do it.

Oh Well..

Derek




________________________________
> From: thefa...@hotmail.com
> To: lcross_ob...@googlegroups.com

> Subject: [LCROSS_OBS: 1063] Re: Target selection

> Date: Sun, 13 Sep 2009 00:16:10 +0200

_________________________________________________________________
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[Derek C Breit]

Don't give up!!!
----------------


Too Late for that..
But I will go away from this group so as to not discourage anyone anymore than I already have..

Derek

[cano...@yahoo.com]

Anyone have any thoughts on the potential of using difference image
processing - something amateurs usually apply to bring out details on
comets?

Clear Skies - Kurt

[cano...@yahoo.com]

On Sep 12, 4:40 pm, Derek C Breit <breit_id...@hotmail.com> wrote:
> Don't give up!!! . . . Too Late for that.. But I will go away from this group
> so as to not discourage anyone anymore than I already have. - Derek

Derek, While I again in principle that we are near the point of
folding up our tents and moving on, I'd hope that you check back in
another week. You will certainly be welcome. Clear Skies - Kurt

[Jim Mosher]

Rick,

With respect, I don't think your conclusion that "the plume will not
be against a dark background" is quite correct. The fact that we can
see the shadow in the 20-km diameter target crater means our line of
sight from Earth is steeper than the angle of the sunlight; that is,
we are almost by definition looking down on the shadow. The rising
plume can be thought of as being like a little peak growing out of the
shadowed crater floor, and where it breaks into sunlight, the peak
will be seen surrounded by shadow on all sides. So I believe the
lowest parts of the plume (which are also -- if one has enough
resolution -- the densest, brightest, and most readily observable)
will be seen starting in (and seen against) the shadowed floor,
spreading out laterally and limbward. With imperfect resolution the
effect may be like a slight shortening or ripple in the shadow edge.
It is the extension of the sunlit plume beyond the shadow edge that
will be extremely difficult to detect: that part of the plume will be
intrinsically dimmer and seen projected first against the sunlit part
of the crater floor and then against the bright terrain beyond.

--

Regarding Dr. Wooden's graphs of "flux density", they are showing
something a physicist might call "spectral radiance". The visual
magnitudes that amateur (and many professional astronomers) are fond
of, are essentially the spectral radiance at a visual wavelength of
~0.5 microns expressed on a sign-reversed logarithmic scale with a
factor of 100 change corresponding to 2 units on the magnitude scale.
The relationship between her W m^-2 micron^-1 arc-sec^-2 at 0.545
microns (call it "SR545") and visual magnitudes per square arc-sec
(mpsas) can hence be written:

mpsas = -2.5 x log(SR545) - 18.58

where the -2.5 comes from a factor of 100 increase in "SR545" changing
the mpsas -5 units; and the constant has been evaluated based on the
Sun having a visual magnitude of -26.74 (above the Earth's atmosphere)
and a spectral radiance of 1840 W m^-2 micron^-1 at 545 nm (also above
the Earth's atmosphere). The relationship is the same whether we are
talking about the Sun's total brightness (visual magnitude) and
irradiance (the numbers just mentioned) or surface brightness (mpsas)
and radiance, since the latter numbers are just the former ones
divided by the same number of square arc-sec.

In any event, Dr. Wooden's red curve indicates a surface brightness of
9x10^-10 W m^-2 micron^-1 arc-sec^-2 at 0.545 microns for a projected
particle density of 1x10^7 particles/m^2, a density that apparently
prevails at some unspecified point near the base of the sunlit ejecta
plume. Putting that spectral radiance in the formula, one can see
it's equivalent to 4.0 mpsas (about the same as the surface brightness
of Mars). A particle density 10 times lower than that (Dr. Wooden's
green curve) gives a spectral radiance ("flux density") 10 times lower
or 4.0 + 2.5 = 6.5 mpsas, and so on. Dr. Wooden also points out that
if the same mass of material is distributed among smaller particles
(her dotted blue line), because their number increases more rapidly
(it follows the cube of the radius) than their cross-section decreases
(following the square of the radius), the result will be more
reflective in inverse proportion to the radius. Thus a given mass of
1 micron radius particles is roughly 35 times more reflective than the
same mass distributed in 35 micron radius clumps.

I don't know what the light brown "V = 5 mag AOV star" curve is meant
to represent.

I have two problems:

1. The 1x10^7 particles/m^2 column density used for Dr. Wooden's red
curve (giving a 4.0 mpsas visual surface brightness) is 13 times
greater than what appears to be the 750,000 particles/m^2 maximum on
the scale of the Goldstein graph from which it is said to be derived:

http://lcross.arc.nasa.gov/observation.htm

Apparently some part of the cloud first reaching sunlight at 4-20
seconds is expected to be much denser than that at 25 seconds
illustrated in the poster on the above page.

2. Even with the higher estimated particle densities, I continue to be
unable to understand how such a small number of particles in terms of
fraction of space occupied by particles -- still only ~0.04 even for
1x10^7 particles/m^2 of 35 micron radius particles, and presumably
only partially sunlit -- can be reasonably expected to produce such
large reflectances: giving a result essentially as bright as the solid
lunar surface.

-- Jim


On Sep 12, 3:34 pm, "Rick Baldridge" <rickbaldri...@comcast.net>
wrote:

> Good question!
>
> Even though the plume will not be against a dark background, that doesn't mean it won't be observable.  The plume brightness will be added to the lit lunar landscape in the background and will still be observable.  Certainly, the ideal condition would have been seeing a V-shaped plume up above the limb or against a shadowed region, but infortunately it didn't work out that way.  Cameras and video certainly have the capability of recording the plume, and I expect even visual observers will be able see it.  Don't give up!!!
>
> Others on this discussion group who are far more knowledgeable than myself will certainly be taking the plume flux calculations submitted by Diane Wooden yesterday and giving numerical estimates of the plume's brightness, contrast ratio, size, etc.  NASA/Ames will be providing info to professional and amateur astronomers in the coming few weeks to aid in exposure calculations.  
>
> Rick B.
>
>
>
>   ----- Original Message -----
>   From: jim phillips
>   To: lcross_ob...@googlegroups.com
>   Sent: Saturday, September 12, 2009 3:16 PM
>   Subject: [LCROSS_OBS: 1063] Re: Target selection
>
>   Why?
>
>   Jim Phillips
>

>   > From: breit_id...@hotmail.com
>   > To: lcross_ob...@googlegroups.com
>   > Subject: [LCROSS_OBS: 1060] Re: Target selection
>   > Date: Sat, 12 Sep 2009 12:31:34 -0700
>
>   > They obviously didn't want to scare anyone off, because if I had known this is the way they were leaning, I wouldn't have been here for the last two months.
>
>   > There is zero chance I can image this with this being the target..
>
>   > Derek C Breit
>   > BREIT IDEAS Web Design

>   > breit_id...@hotmail.com

>
>   > ----------------------------------------
>   > > Date: Sat, 12 Sep 2009 11:23:05 -0700
>   > > Subject: [LCROSS_OBS: 1059] Re: Target selection

>   > > From: jimmos...@gmail.com

>   > > To: lcross_ob...@googlegroups.com
>
>   > > On Sep 11, 1:39 pm, "canopu...@yahoo.com" wrote:
>   > >> P.S. - He also stated that although this was the final choice, the
>   > >> Team had selected this location as the primary preliminary candidate
>   > >> target back on the launch date in June.
>
>   > > That's interesting since nothing in this vicinity was in the "short
>   > > list" released to this Group on July 14th:
>
>   > >http://groups.google.com/group/lcross_observation/msg/85b9db372ca99859
>
>   > > -- Jim
>
>   > _________________________________________________________________
>   > Your E-mail and More On-the-Go. Get Windows Live Hotmail Free.

>   >http://clk.atdmt.com/GBL/go/171222985/direct/01/- Hide quoted text -
>
> - Show quoted text -

[cano...@yahoo.com]

Jim wrote:

> mpsas = -2.5 x log(SR545) - 18.58

> where the -2.5 comes from a factor of 100 increase in "SR545" changing
> the mpsas -5 units; and the constant has been evaluated based on the
> Sun having a visual magnitude of -26.74 (above the Earth's atmosphere)
> and a spectral radiance of 1840 W m^-2 micron^-1 at 545 nm (also above
> the Earth's atmosphere). The relationship is the same whether we are
> talking about the Sun's total brightness (visual magnitude) and
> irradiance (the numbers just mentioned) or surface brightness (mpsas)
> and radiance, since the latter numbers are just the former ones
> divided by the same number of square arc-sec.

Does that work, Jim? The Johnson V stellar magnitudes are based on
the response across a band of spectral frequency, not just one
wavelength. Is the magnitude SR545 (for one frequency) directly
comparable to the zero point of 18.58 in the visual magnitude
system.

You are the expert, Just asking. - Kurt

[Jim Mosher]

Kurt,

I am *no* expert on this, but yes, unless there was something
extremely unusual about the shape of the spectrum presented,
information about the expected LCROSS spectral radiance at 545 nm
(0.545 microns) -- which I have quite arbitrarily called "SR545" in
the formula -- is sufficient to estimate an astronomer's "visual
magnitude" with high precision. And yes, the "Johnson V" photometric
system is based (as I understand it) on the number of photo-electrons
detected through a filter with a relatively broad bandpass intended to
vaguely approximate the response of the human eye. An article about
"Standard Photometric Systems" by the Australian National University's
Michael Bessell in Volume 43 of the /Annual Reviews of Astrophysics/
(2005):

www.mso.anu.edu.au/~bessell/araapaper.pdf

depicts it (Figure 1) as having a vaguely sawtooth-shaped bandpass
starting at roughly 490 nm, peaking at 530 nm, and ending at roughly
660 nm. In Table 1 he describes it as having an "effective
wavelength" of 544.8 nm and a "width" of 84 nm. In theory, one needs
to add up the responses at each individual wavelength weighted by this
curve, a process mathematicians call "integration"; but if we know
that the Sun, with SR545 = 1840 W m^-2 micron^-1 gives visual
magnitude = -26.74 in this system, then any other light source with a
similar spectrum, differing only in overall intensity, is going to
give the same result corrected only for the difference in intensity at
any one wavelength (if the spectra are the same, the ratio will be the
same at all wavelengths, but SR545 is the one most characteristic of
the V-band). I did not check Dr. Wooden's curves in detail, but since
the LCROSS signal is reflected sunlight, and the Moon is not strongly
colored, I would assume they are very Sun-like spectra, at least
between 490 and 660 nm, and probably out to where she shows "thermal"
emission beginning to overtake "scattering" (~3.7 microns).

--

Incidentally, there has been, I believe, some confusion on this forum
about the use of the word "integrated magnitude".

To the best of my admittedly limited knowledge this phrase is normally
used to mean adding up elements of varying surface brightness over a
two-dimensional area to obtain an overall magnitude of that object
treated as an unresolved point source. This is again a process of
mathematical integration, but it is unrelated to integration over
wavelength (something that is necessary only when trying to guess the
photometric response of a V-filter to a light source with a very non-
Sun-like spectrum). Surface brightness can be expressed in "W m^-2
micron^-1 arcsec^-2" (as Dr. Wooden has done), in "W m^-2 micron^-1
sr^-1" (as physicists normally do, 1 steradian = 4.25x10^10 arcsec^2),
or can be converted to mpsas using the formula you mention. Since Dr.
Wooden's "flux densities" are for 1 square arc-sec, they convert
directly to magnitudes for 1 square arc-sec (what we have been calling
"mpsas").

A very important property of surface brightnesses, known at least
subconsciously to all photographers, is that (in whatever system they
are expressed) the surface brightness of an object does *not* vary
with viewing distance. For example, if Mars has a surface brightness
of 3.9 mpsas, (a value that does not change much because its distance
from the light source -- the Sun -- is always about the same) each
square arc-sec will have that same surface brightness (and require the
same photographic exposure) whether we are viewing from a distance
where the planet as a whole subtends an angle of 4 arc-sec (near
conjunction with the Sun) or an angle of 20 arc-sec (near opposition
to the Sun). On the other hand, the surface brightness of Mars would
be expected to be lower than that of the Moon (at most phases) because
it is farther from the light source, and the planets still farther
from the Sun have still lower intrinsic surface brightnesses.

An example of integrating mpsas over area to obtain an "integrated
magnitude" would be an idealized Mars with a uniform surface
brightness of 3.9 mpsas and a diameter of 20 arc-sec. The projected
area is pi * (10)^2 = 314 sqr-arcsec, so the idealized planet as a
whole is 314 times brighter than a single sqr-arcsec. This changes
the magnitude from 3.9 (for a single sqr-arcsec) to 3.9 - 2.5*log
(314) = -2.3 (for 314 sqr-arcsec). -2.3 is the magnitude of the
object as a whole, or its "integrated magnitude", which means, if
unresolved and treated as a point source, it would be
indistinguishable (in terms of counts through the V-band detector)
from a star of that magnitude. If the calculation were repeated with
an assumed diameter of 4 arc-sec, the result would be based on an area
25x smaller, giving an integrated magnitude of 3.9 - 2.5*log(314/25) =
+1.2 near conjunction.

Relating Dr. Wooden's "flux densities" to visual mpsas is easy since
that is simply a matter of plugging them into the formula. Converting
them to "integrated magnitudes" of the plume as a whole is much more
complex since one cannot add up the intensities without knowing how
many sqr-arcsec there are at each mpsas. And here graphs do not tell
us what the surface brightnesses will be at individual points --
instead we are being instructed (I believe) to look at the grain
density charts, and where those charts say, for example, 1E7 35-micron
radius(?) particles m^-2 to assume the red curve will apply, where
they say 1E6 35-micron particles m^-2 to assume the green curve
applies, and so on.

If you asking about magnitudes in other photometric bands, such as one
of the many in the infrared, then you would need to find out what the
magnitude of the Sun (or some other reference star) is in that system,
what spectral radiance (W m^-2 micron^-1) it produces at that
wavelength, and re-determine the constant in the conversion equation
based on those two numbers.

-- Jim

P.S.: the present "flux density" curves are similar to the ones you
have found in various LCROSS presentations, except that those were
plotted in "W m^-2 micron^-1 sr^-1". Since 1 steradian = 4.25x10^10
arcsec^2, the surface brightnesses in that system are 4.25x10^10
times greater than those in "W m^-2 micron^-1 arcsec^-2" .


On Sep 12, 11:05 pm, "canopu...@yahoo.com" <canopu...@yahoo.com>
wrote:

[cano...@yahoo.com]

Jim wrote:
> I don't know what the light brown "V = 5 mag AOV star" curve is
> meant to represent.

I believe that they have simply plotted the spectral irradiance curve
for a stellar mag 5 spectral classification A0V star over the top of
the plume brightness drawing. Therefore, concluding that the plume
will have the same apparent brightness as a stellar magnitude 5
star.

- Kurt

[Jim Mosher]

Maybe, but the vertical scale of the chart is in units of surface
brightness, and most stars are point sources (as seen from Earth).
The surface brightness of a point source is infinite (making it
unplottable on this scale) unless you arbitrarily give it some finite
projected surface area (square arc-sec); such as, for example,
smearing the emission out over a seeing disk. I don't see any mention
of an assumed size that would make a comparison with surface
brightnesses meaningful. An extended source with a surface brightness
of 5 mpsas is a completely different animal from a magnitude 5 star.

-- Jim

[XB70man]

Jim,


As always, you are quite correct. (I need to learn to be more precise
in my statements.) I was focusing on the extended plume 60 to 180
seconds after impact and did not make the distinction . . .

While the plume will not extend above the lunar limb as viewed from
Earth on impact day, Jim is quite correct in that the initial phases
of the plume (at its brightest!) will indeed appear projected against
the darkend interior of Cabeus A1. It is my guess (and only a guess
-- not yet supported by NASA/Ames / LCROSS project scientists) that
the impact plume will be MOST APPARENT to the amateur community as it
BISECTS the darkened crater interior within the first 30 seconds after
the Centaur impact, causing the crater to "split in two" for a minute
or so. This event should be quite apparent (in my estimation,) and
easily recordable with video, CCD and DSLR cameras using adequate
image scale (focal length / projection).

Excellent discussions on this site! I am in no way an expert at
this, and maybe none of us are, but there has been some very
interesting calculations and conjecture of how this impact will
appear. In fact, as I've stated before, advanced notice of this type
of event is VERY unique, and we must expect the unexpected. EVERY
amateur that is capable of imaging the Moon on impact day SHOULD make
the attempt. There are so many camera, telescope, exposure, filter
and weather (seeing) combinations possible that it is almost certain a
very few, good observations of this impact will be recorded by
amateurs. The LCROSS Science Team has been most impressed by the
nature of the discussions here. While this site was not intended to
be a 50-50 two-way communication between the public and LCROSS
Scientists, I can tell you that much of what has appeared here has
been taken into consideration planning the Earth-based and even Earth-
orbit observation campaign. Please keep the discussion going. Be
critical when you must. Everyone makes mistakes and this has been a
good forum to express those thoughts. We are not and are not going to
be the sole resource for gathering data to determine if there is water
on the Moon, but I think we all are very excited in trying to further
the science as best we can by observing this very exciting event and
sending in what data we can.

Keep up the good work!


Rick Baldridge
Campbell, CA

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