Kurt -
Thank you for posting the revised graphic at:
http://01227941410742638900-a-g.googlegroups.com/web/20091103DipstickMarkup_kf.jpg
illustrating your concerns about the Goddard-LOLA masking simulations
in LCROSS Principal Investigator Tony Colaprete's October 2, 2009
target area presentation at:
http://lcross.arc.nasa.gov/docs/LCROSS_Target_Update_100209.ppt
The second part of your graphic, using the Cornell-Smithsonian radar
image is also helpful, although the line that says "limit observable
limb" seems incorrect. The observable limb should be roughly parallel
to, but distant from this line by about as much as your present line
is distant from the impact site -- far enough to include the prominent
peak "M5" (seen beyond the dipstick in the Goddard/LOLA simulations,
but not visible in the segment of the radar image you show) on the
Earthward side of the limb. Finally, to align the images, the "to
Earth observer" arrow should be parallel to the dipstick direction.
Regarding your thesis that the LCROSS impact was hidden from Earth to
a greater extent than Tony's graphics indicate, it is certainly true
that computers and their programmers can make errors, but in the
absence of such errors, computers are usually much better at
accurately visualizing three-dimensional relationships than most of us
are, and this seems to be the case here.
I must apologize for misunderstanding your earlier message in which I
thought your concern was that the simulation showed markings on the
dipstick at elevations lower (less far from the Moon's center) than
the elevation of the ridge (M1) behind which it was seen. That is
actually no more surprising than the fact that a bit beyond the
dipstick we can see the still lower floor of crater (much of which is
in shadow) -- both simple consequences of the fact that we are looking
down on the scene from above.
Now I see your concern is with how the markings on the dipstick
compare to points of known elevation to the east and west. In
addition to some ancillary issues, those relationships may be harder
to visualize than we may at first imagine.
The first ancillary issue is how the stripes on the dipstick/candy
cane are intended to be read. You seem to be confident the even
kilometer points come at the *bottom* of the stripes (bottom of green
stripe = 2.0 km, top = 2.5 km, middle = 2.25 km, etc.). From the
measurements he cites in Slide 8, I would guess Tony takes the even
kilometer points to come in the *middle* of the stripes (middle of
green stripe = 2.0 km, bottom = 1.75 km, top = 2.25 km, etc.). Thus
masking by M1, along the centerline of the dipstick, to the from a
little above the bottom of the green stripe (in Hawaii and New Mexico,
Slides 11 and 13) to near the middle of the green stripe (from
California, Slide 9) he calls "1.8 to 2 km"; and the sunlight/shadow
line near the midpoint of the yellow stripe (Slide 15) he calls "about
1 km".
The second ancillary issue is how to read the LRO/LOLA altimeter
contours in Slide 4. Somewhere you implied that the contours have
been adjusted so that "0" matches the rim height of Cabeus. I don't
think this is true. Slide 2 says the modeled impact point elevation
of -3.82693 km is its offset from a reference surface 1737.4 km above
the Moon's center. My guess is that the "0" LOLA contour is the same
reference elevation. So when in your graphic you mention a point at
"DEM 2.5 km" one would be looking for a contour of -3.8+2.5 = -1.3 km.
The contour levels themselves are very difficult to read because no
distinction is made between up and down slopes. The levels are a bit
less ambiguous if one superimposes them on the shaded nadir view of
Slide 16, after enlarging the latter by 193% and rotating it 90° CCW.
I am guessing your reference points are at the right-hand ends of your
light green lines. I believe I may be able to see the point you refer
to as "DEM 3.8 km" (= the LOLA "0" contour), on the Earthward rim of a
little crater sandwiched between two larger ones. It looks like you
have perhaps set the top edge of the green line tangent to the crater
rim.
I am unable to locate or verify your "DEM 2.5 km" point. That light
green line appears to end on the far inner wall on a crater that is
outside the LOLA contours of Slide 4.
You seem to find it disturbing that when you extend the light green
line from the "DEM 3.8 km" point horizontally to the left, the top of
the green line intersects the dipstick at what you read as about 3.4
km and what Tony would perhaps read as 3.2 km.
I find nothing disturbing about this. In the first place, there is no
significance I can think of to the horizontal direction in the
simulation. The dipstick presumably represents a line emanating
radially outward from the lunar surface at the impact point (it is
seen end on in the nadir view of Slide 16). I have no idea why it is
shown tipped at an angle, but one could imagine a second radial stick
emanating out of the "DEM 3.8 km" point on the Earthward rim of the
little crater. That stick would be tipped at an even greater angle.
This point is expected to project farther out from the disk center
than the 3.8 km mark on the dipstick simply because its base point is
farther from disk center than the base of the dipstick. Moreover, the
connecting lines needed to compare the two projections would not be
straight lines, but rather circular arcs about disk center. Such
circular arcs, if properly drawn, would intersect both sticks at right
angles. I am not completely sure what the radius of curvature in the
Goddard/LOLA simulation is, but if I try to imagine such an arc
tangent to the Earthward rim of the little crater it looks to me like
its top edge would intersect the dipstick not a little below the top
of the blue 3 km stripe (as you show), but rather a bit above the top
of the black 5 km stripe.
This does not indicate there is anything wrong with the markings on
the dipstick, but only that a "DEM 3.8 km" point, whose base is
farther from disk center (the little crater rim), projects farther
from disk center than a point whose base is closer to disk center (the
Centaur impact point); and given the ~7.3° at which we are looking
down on the scene, the magnitude of the difference seems consistent
with the little crater reference point being ~20 km farther from disk
center than the Centaur impact point.
> Would the tipping angle for Hawaii not be lower? 84.675S for
> the planned impact latitude of the Centaur less 3.6 for
> topocentric libration for Hawaii gives 81.2.
I may be misunderstanding the question. What I am calling the
"tipping angle" (90° - distance from center) becomes larger as the
feature moves towards disk center, and we look more directly down on
it. Calculating the distance from disk center by combining latitude
with libration in latitude works only for objects on the central
meridian (which Cabeus is not). A correct calculation requires
comparing the longitude and latitude of the sub-observer point with
the longitude and latitude of the point of interest, using spherical
trigonometry. The Sun Angle calculator at:
http://the-moon.wikispaces.com/Sun+Angle
implements the correct equations. For an impact point at 48.725W,
84.675S, the distances from disk center, and tip of the impact
surface, at 2009 Oct 09 11:35 UT were:
Observer Sub-Observer
Location Long Lat Long Lat Distance Tip
Hawaii -155.47 19.83 -2.165 -3.689 82.66 7.34
Sunnyvale -122.27 37.46 -2.681 -3.540 82.77 7.23
New Mexico -106.70 32.29 -2.891 -3.652 82.65 7.35
If one compares the Goddard/LOLA simulations carefully, the very
slightly greater tipping towards Earth as seen from Hawaii and New
Mexico is evident in the peak of M5 being closer to the 20 km mark on
the dipstick compared to the view from "California" (whether
"California" means Sunnyvale or Mount Wilson is unclear, but would
only slightly change the results). As expected, the horizontal
shearing due to the 0.7° difference in longitudinal libration is more
evident than the 0.1° variation in tip.
> Your computation also seems inconsistent with the LOLA model
> or the LOLA model errs, since the 1.5km marker is not visible
> on the disptick.
You are quoting a computation that does not take into account the
spherical curve of the Moon's surface, and that was based on your
estimate (of unknown origin) of the difference in elevation between
the masking ridge and impact point.
Neglecting the Moon's curvature, over a distance 40 km at a tip of
7.3° you would expect to see to a point 5.1 km below the obstruction
height. Correcting for the Moon's curve, for the same distance and
tip, you can see to something like 4.66 km below the elevation of the
obstruction.
> The 1.5km dipstick marker is 1.5km below the masking ridge.
Your meaning escapes me. By Tony's reading of his own graphic, the
mark visible on the dipstick behind the masking ridge is 1.8 to 2.0
km. That would put the 1.5 km dipstick marker about 0.3 to 0.4 km
below the ridge at that point.
> If your computation is correct, than the 1.5km marker (the
> top of the yellow band) should be visible.
Yes, the top of the yellow band is nearly visible at the western edge
of the dipstick. But a computation uncorrected for the Moon's
curvature over 40 km is not expected to be correct; nor is the data
available to me on the height of the masking ridge relative to the
impact point accurate enough for such a comparison to be meaningful.
The Goddard/LOLA graphics show the projected height of the ridge
varies by very nearly 1 km over the 3.5 km width of the stick.
--
In summary, while I can't vouch for the accuracy of the LRO/LOLA
Digital Elevation Model, the Goddard Space Flight Center renderings
look like an accurate depiction of it to me.
--
> Even the ultra low level of light from Milky Way over a
> million years can evaporate surface deposited hydrogen in a
> PSR. However, I do not recall any of the journal articles
> that I read including the effect of reflected interior crater
> surface light, e.g. that reflected light which undoubteldy
> occurs at the wide and flat Faustini PSR or at the Cabeus
> LCROSS impact site.
Again, I may have missed Clif's point, but I thought he was suggesting
that the floors of deep craters near the pole (like Faustini, or even
better, Shackleton) are always in shadow which, to layman, seems a
much colder and more conducive environment to finding ice than a
larger crater farther from the pole, much of whose floor is bathed in
sunlight. As a crude analogy, choosing Cabeus over Shackleton or
Faustini seems somewhat like an explorer in a broad sunlit valley of
the American Southwest seeking shelter in the shaded slope at the foot
of a north-facing hill (where, looking to the north one still sees a
vast expanse of bright sunlit landscape) as opposed to looking in a
deep pit from whose bottom one can see nothing but sky. On the Moon,
the sky is very dark, so the difference between a dark pit from which
one can only see sky and a shaded slope from which a sunlit landscape
is visible is even more stark.
I know nothing about the evaporation of ice by light from Milky Way,
but I would certainly think the amount of sunlight scattered off
crater inner walls (and floors, in the case of Cabeus) would be much
more significant and must be included in models of temperatures
expected in such environments.
Although we think of the shadowed areas of craters as being very dark,
I don't think this is correct. In human terms, we know that on Earth
moonlight (at Full Moon) is about 400,000 times weaker than sunlight,
yet it is strong enough to read the headlines in a newspaper. This
comes from having a source in the sky covering a 0.5° diameter circle
(0.2 square degrees solid angle) with the brightness of lunar surface
reflections. For sources of similar brightness, the intensity detected
is proportional to the solid angle filled by the source. From shaded
parts of the Moon's surface one sees a source (or sources) of similar
brightness, but covering a different area of the sky.
The intensity in a "very dark" crater like Shackleton can be crudely
estimated by assuming it is a truncated cone about 20 km in diameter
and 4 km deep, with a floor of some small extent, say 10 km in
diameter. If the Sun peers over the rim at an angle of 2° (about the
steepest it can get this close to the pole), we would expect about 0.7
km of inner wall (on the opposite side) to be in sunlight. Looking up
at this from a point on the floor 5 km from the centerline, one would
see the sunlit inner wall about 7° away from edge-on, which would
foreshorten it to an effective width across the line of sight of about
0.1 km. At a distance of 15 km, this corresponds to a vertical extent
of about 0.5°. To the sides, the ribbon of light would extend over
roughly 180°, but getting thinner towards the edges; so the total
solid angle of sunlit inner wall seen in the sky would be about (1/2)x
(0.5°)x180° = 45 square-degrees. This is ~200 times more area (of
similar brightness) to what the Full Moon presents in our nighttime
sky, so the intensity of light falling on this point on the floor of
Shackleton is going to be ~200 times more than we are accustomed to on
a moonlight night on Earth. This is still ~2000 times weaker than the
same scene illuminated by direct sunlight; but just as lunar craters
are much shallower than we imagine, it would seem the shadows are much
less dark -- at least at times when a piece of sunlit landscape is
visible from the shadowed point.
I find it much harder to visualize what the landscape looks like from
the LCROSS impact point on the floor of Cabeus -- I would imagine that
much of the month you see sunlit floor to the south and east, probably
some of the inner rim, and possibly the top of M5 off in the distance
to the south? Intuitively I would think it adds up to many more square
degrees of illuminated area than the thin ribbon of sunlit inner wall
sometimes visible from the floor of Shackleton, so I would think the
ambient light level would be correspondingly higher (at least when it
is local noon there). I would also expect the ambient light level in
the shaded areas of Cabeus to be much higher than in Faustini,
Shoemaker, and other craters closer to the pole whose floors are
always in shadow.
-- Jim