The other day, Chris Rowan of *Highly Allochthonous* posted some pictures (and video!) of the Teton Range in Wyoming, a normal fault-bounded block of rock that has rotated along a north-south axis, with the west side dropping down and the east side rising up relative to the floor of Jackson Hole. This is classic “Basin and Range” extension, but the great thing about the Tetons is that it is so fresh and raw. Standing in Jackson Hole, you can look up at one particular peak which allows you to calculate how much offset has occurred along the Teton fault.

This peak is Mount Moran (*slightly Photoshopified*):

Here’s how the National Park Service would annotate that view, from here:

I’m interested in other details, though (like dates and elevations), so here’s a quick sketch I worked up on my new pad of NOVA sticky notes:

Most of the mountains are Arhcean gneisses of the “basement complex.” Cross-cutting these are a series of mafic dikes, including the prominent one that pokes out of the face of Mount Moran. The diabase dike, sometimes called “The Black Dike” is a prominent feature, but to me, the really interesting tidbit is that thin little scrap on top: a bit of the Cambrian-aged Flathead Sandstone. This sedimentary stratum overlies a profound nonconformity, and that same layer is found way down beneath Jackson Hole, at a depth of about 20,000 feet (20,000′) below the surface. (As Mount Moran is 12,605′ tall, that means that at its lowest point, the nonconformity is actually close to 14,000′ below sea level!)

Well, that sandstone layer can serve as a marker bed, seeing as how it’s been broken and offset along the Teton fault. Consider the following sketch to get a sense of how the Flathead Sandstone is 6000′ above the Jackson Hole valley floor on the west (right) and 20,000′ below on the east (left):

The Teton fault is 55 km long, and it dips to the east at 45°–75°. For the sake of simplicity, I’ll use a value of 60° to make my estimate of displacement. This is in accordance with the generally high-angle nature of normal faults, in accordance with Andersonian predictions (*a topic which deserves a post of its own!*). Given the vertical offset along with this angle, with can figure out how much offset has taken place. I’ve pulled out a highlighter now to color in the Flathead Sandstone:

Of course, this requires us to employ some trigonometry. We can do this with two separate triangles, as with the example above, or we can slide that vertical bar over to the right (west), and make it into one big triangle, where we add our vertical distances above and below the valley together:

The vertical part of this triangle, 26,000′ feet tall, is the “throw” of the fault, the vertical component of the displacement vector. We can use it, plus the dip angle, to figure out what the displacement is.

The way I was taught trigonometry in school, we memorized a pseudoIndian word, “SOHCAHTOA,” as a mnemonic device. For right triangles, this meant that: this relates the angle we’re interested in (let’s call it ψ) to the lengths of the sides of the right triangle, where** S** refers to sin(ψ), **C** refers to cos(ψ), and** T** refers to tan(ψ). That’s sine, cosine, and tangent, respectively. “**O**” is the length of the side opposite the corner of the triangle with the ψ** **angle. “**A**” is the length of the non-hypotenuse side adjacent to the ψ-angled corner. “**H**” is the length of the hypotenuse itself.

So with our Mount Moran calculation, we’re interested in the length of the hypotenuse, which is the same as the offset of the Flathead Sandstone. We use the “SOH” part of “SOHCAHTOA”:

sin(60°) = O/H

sin(60°) = 26,000’/H

H*[sin(60°)] = H*(26,000’/H)

H*[sin(60°)] = 26,000′

H = (26,000′)/[sin(60°)]

Let’s pull out the old TI-83:

So the length of the hypotenuse is 30,022′ — and assuming that all the slip along the fault has been dip-slip (no strike-slip or “transform” motion), then we’ve got our answer: the Flathead Sandstone marker bed has been offset by around 30,000′ feet. Nice!

This calculation has got me in a mathy mood. Let’s check out the rate of displacement, while we’re at it. It is estimated that extension began on the Teton fault around 13 million years ago (13 Ma). If we have seen 30,000′ (9,144 m) of displacement in that time, what is the average rate of displacement?

30,000′ / 13,000,000 years

3’/1,300 years *(just lopping off four zeros from each side)*

(12 inches/foot)*3′ = 36 inches/1,300 years

0.028 inches/year

*or:* 1 inch every ~36 years.

But of course fault motion usually doesn’t proceed at a slow and steady rate; it sticks and then slips infrequently in sudden jumps that we call earthquakes. The last major earthquakes on the Teton fault were 8,000 and 4,800 years ago. Both of these saw between 4 and 10 feet of offset. Check out the map of historical seismicity in the area, from the USGS:

Notice the intense cluster of quake epicenters associated with Yellowstone National Park, and the cluster in the Gros Ventre range, active this summer. Notice also the big blue smudge of Jackson Lake, a 25,540 acre lake where the Snake River is dammed up by first a glacial moraine, then augmented by humans via a dam.

Now notice the big gaping hole in seismic data in Jackson Hole…** There has been no historical seismicity on the Teton fault.** Jackson Lake is held up by an earthen dam, and earthen dams do poorly when shaken. The town of Jackson (8,000 residents, plus tourists) is downstream of Jackson Lake.

This strikes me as worrisome.

Filed under: basin and range, cambrian, earthquake, faults, igneous, math, national parks, sandstone, structure, wyoming | 8 Comments »