Great pans and analglyphs. Thanks for the hard work. It may be a perspective thing, but it looks like the cape on the left which rises up slightly (is this the beacon people have been mentioning?) is way way above the plain to the south. Kind of like Pride Rock. That would be a most righteous spot for a giant pan. It is remarkably unflat. But the terrain is just gorgeous to look at, especially in these near UV wavelengths.
Given there'll be a drop of approximately 70m over a horizontal distance of 125-175m going in to Vic, the average slope of any bay will be from ~22 to ~30 degrees. Because of the concavity of the path to the bottom, the slopes will be significantly greater near the rim. Many of these look like 45 degree slopes over the first/uppermost few metres. It'll be a challenge to find a route out. Lots of switchbacks could perhaps get her out, but the lip (where slope should be steepest) will be a tricky obstacle.
Should be some big excitement tomorrow. Perhaps we'll see the south edge of the dunefield and fuller expanse of the far rim.
Some quick and dirty trig helps to figure out how much of the outcrop we see across the way (old news to the veterans here no doubt). You can calculate the height of the outcrop if you know the distance to it and the angle it subtends (ie. covers) in the pancam image. Each pixel of a pancam image subtends 0.0164 degrees. Count the pixel height of any feature you know the distance to, multiply by 0.0164, take the sin of that, then multiply by distance and you have the height of the feature. I make the current outcrop we see to be about 45 pixels high and (based on the consensus) 800m distant so:
45 pixels x 0.0164 degree/pixel = 0.738 deg.
sin(0.738) = 0.01289
800m x 0.01289 = 10.3m high
Here is the Excel spreadsheet that does the math for you:
Click to view attachment<Had to edit this due to fup in wording. doh.>