Photos from hiking up Mount Ernest Ross

Many friends and I went camping for a weekend in September near Nordegg, staying at the Kootenay Plains Cavalcade campground. Several photos were taken!

click for photo album!

Click the image!
If you want a super-high resolution JPEG, around 10MB each, note the “Name” of the photo, click [THIS LINK], enter the username and password that I should have already sent you personally on Facebook, then download the file whose named matches the “Name” of the photo. For example, the name field of photo #2 is “P1011756”, as displayed on it’s ‘detail page’, go to, wait for your browser (please not Internet Explorer) to ask for then enter your username and password, then right-click and save “P1011756.jpg” to your hard drive. I’ve tweaked almost all of the images in Aperture (a Mac OS X photo editing program), sometimes to extremes (’cause it looks neat!) — if you want the original [Olympus] RAW or untouched JPEG, it’s no problem to upload it, so just ask. If you don’t get a username and password, or have trouble downloading an image, ask and I’ll make it work.


An idea coming from the below xkcd, Depth Perception, and a few working examples[1,2], I wondered what it would take to get the parallax to look at the Moon or even the Sun in 3D as if they were the size of a bungalow as seen from the sidewalk.

xkcd 941, Depth Perception

Let’s take this bungalow to be about 5 m tall ($d_2$), 40 ft from the sidewalk, and the space between our eyes to be 7 cm ($d_1$). That’s right, I can mix units haphazardly. The angle to a point 40 ft away will differ by a small amount as viewed from each eye — I’ll call this difference $beta_0$. It looks familiar, and probably has a standard name and definition in optics.

beta_0 = dfrac{d_2-d_1} {d_2+d_1}
= dfrac{5 m – 7 cm}{5 m + 7 cm}
approx 0.972386588

The relation between object size and image size becomes this:

d_1 = d_2 dfrac{1-beta_0}{1+beta_0}
approx d_2 times 0.014

The Moon would require images 48 km apart; the Sun, 19’488 km (3x radius of Earth) apart. The Earth orbits the Sun at about 30 km/s, so those 19’488 km would conveniently go by in under 11 minutes. If there weren’t any clouds, you could have a setup that films the Sun and displays two output streams: one delayed 11 minutes behind the other. Any surface details that had changed during that time would be lost or at least blurry. As for the Moon, it would take a bit of planning with a friend that lives 50 km away and similar cameras with identical settings.

Something tells me that this only works for objects far away, as that distance isn’t in the calculations.

What’s that? I’m supposed to be doing homework? Oops.