45 degrees is the point where they are stable - but to actually drive, it's something like 35 degrees - strongly dependant on surface.

Doug

I think as we get closer we may see sufficient evidence of dust trails on the lee side of various protrusions on the rim, just drive around some of those while the spring winds begin to pick up.

OK, showing my lack-of-cred as a space geek, since I don't even know how to search for this...

What's available in the way of other orbital photos of Victoria, correlated with time of day and incident sunlight angles? If we can see shadows of the rim wall inside the crater, can we even begin to estimate height and slope?

One thing that stands out about Victoria is that the rim seems to be fairly "sharp" in the MOC image, most of the way around. Are we looking at a real "edge" or a difference in albedo due to soil differences, compaction, etc?

Unfortunately, all the high res MOC images will have been from pretty much the same time of day (intentionally - MGS orbits at a particular 'local' time relative to the surface it's overflying) - roughly 2pm I think

Not too sure if there's anything else really to bring to the table - even Odyssey imagery's taken from pretty much the same time over day each time, 4pm - but is much lower res than the MOC imagery.

HRSC MEX imagery is lower res as well - different times of day for that, but not of much use really.

Doug

The best ones, the most recent ones that I found is located at:

http://photojournal.jpl.nasa.gov/targetFamily/Mars

2006-06-28 Mars Mars Global Surveyor (MGS)
Mars Exploration Rover (MER)
Mars Orbiter Camera
2500x2199x3
PIA08564:
'Beagle Crater' on Opportunity's Horizon (Orbital View)
Full Resolution: TIFF (16.51 MB) JPEG (633.1 kB)

Rodolfo

Well if we know the time of day and the date then we can get a solar angle at least to a reasonable approximation. There are a couple of points with clear shadows in PIA08564 that are between 7.5 and 9.5 pixels from the shadow source. For the sake of kicking this off just taking a purely random guess at a date of Dec 5 2005 @ 2PM. Noon solar elevation is 81.7deg azimuth is 180deg (southerly) though and at 2PM it's passed back north to approx 60deg. Back to our old trig friends and the height of the "cliff" edge is between 9.8 and 12.5 pixels. A rough check seems to put the scale of the image at around 1m/pixel. So we would have approximately a 13-16m "cliff".
However if the image was taken at midwinter then the solar elevation at 2 PM drops to approximately 30deg at 2.00PM. In that case the cliff could be between 4 and 5.5m.

Anyway if someone can put an exact date and time on the image we can refine this somewhat but it isn't totally useless for getting an idea of how deep the crater walls themselves are and how steep the entrance ramp features may be.

Sorry for the delayed reply, Diane. Had things to do.
It's not that easy to research crater dimensions, since there are a substantial number of variables: crater diameter, yes, but also target rock induration and layering, degree of slumping and backwash, age and erosion state etc. The theory is a work-in-progress.
Keith Holsapple has designed a nice calculation tool to draw up a crater of any desired size:
http://keith.aa.washington.edu/craterdata/scaling/index.htm
but the result is a nice fresh crater, not the old, worn, sagging kind like Vikky (We still love her; youth isn't everything.)

I would say that the rim edge is real. It should be about as sharp as the edges of the many evaporite outcrops and crater edges we've seen elsewhere. Of course the dark sand adjacent to the edge will enhance the look.

Welcome to UMSF.

Well - that image from the JPL beacon map isn't 'out' yet....

But..

http://www.msss.com/moc_gallery/r22_s04/im...2/R2200640.html

Solar longitude (Ls): 99.27°
Local True Solar Time: 14.49 decimal hours
Emission angle: 16.71°
Incidence angle: 45.18°
Phase angle: 58.47°
North azimuth: 92.96°
Sun azimuth: 41.85°
Spacecraft altitude: 384.57 km
Slant distance: 399.70 km
Get out your calculator.

Sheesh, I read this, and it comes out in Al Bundy's voice...

--Bill

Don't even need a calculator. The shadow is approximately 8 pixels across, the measurement error is +-2. The incidence angle is roughly 45deg, Tan(45) is 1. The vertical cliff is as high as the shadow, ie 8+-2m.

The inner crater floor could be a further 5-15m down but there are clearly relatively flat zones inside the initial rim that should be navigable without having to risk going into the dunefield.

That's the enjoyable kind of theory.

Vikky was young once. She probably knows how keep her surprises.

Thanks to all for the welcome!

(In response to your question which seems to have disappeared, no, I'm not Diane Lane. I did hang out in sci.space.history some years ago, so I do see some familiar names here.

From one USENET sci.space.* "veteran" ("survivor," actually) to another: You have my sympathies.

Welcome aboard the HMS Sanity (for the most part), just the same!

I'll get back to you on this, Bill, as soon as I figure out who Bundy is. I hope he's not that serial killer I read about some years back.

LOL not that Bundy!

Al Bundy is a fictional character from the U.S. Sit-Com television series Married... with Children, played by Ed O'Neill.

In comment to the original question of the thread...

If we can't drive into Victoria, perhaps we can drive into Beagle. Or the small dimple crater on Victoria's southeast corner, almost directly on her rim.

There are several other small craters in the area, but most appear to be pretty well filled in. I'd say Beagle or that unnamed dimple crater would be our best bets to find a north-facing slope on which to winter.

-the other Doug