Friday fold: wavelength contrast

I scored this photo off the Internet more than five years ago, the first time I taught Structural Geology at George Mason University. I failed to note the website I got it from, and now that website has apparently disappeared, at least as far as the view from Google is concerned. If anyone knows the provenance of this image, please let me know so that I can properly attribute it.

I hesitate to post something like this without knowing who took it, but I did note to myself that it came from the Point Lake Greenstone Belt in the Northwestern Territories of Canada. This image and its implications follow so nicely on to our discussion last week about fold wavelength and the Ramberg-Biot equation that I can’t resist it. Ready? Brace yourself…


I think that this is one of the coolest structural geology photos ever taken. Here it is graced with some annotations:


Maximum compressive stress was in this case from the back to the front. The same vein, oriented ~parallel to σ1, is folded in two very different ways, depending on which rock type it is cutting across. As with a week ago, we can explain this behavior using the Ramberg-Biot equation:

L = 2 π t (η / 6ηo)

where L is the wavelength of the fold (in other words, the distance from one fold hinge to the next fold hinge); t is the thickness of the folded layer; η is the viscosity (resistance to flow) of the quartz vein (or, in general, the more competent of the two layers); and ηo is the viscosity of the rock unit (sandstone or shale) that the quartz vein cuts across.

If you keep t and η constant (for say, the rightmost of the two quartz veins), then the only thing left to vary would be ηo. So sandstone will have one ηo, while shale will have another ηo. The sandstone is more resistant to flowing than the shale is. The viscosity contrast between the quartz vein and the sandstone is less (they’re both made of quartz) than the viscosity contrast between the quartz vein and the shale (which have very different material properties).

The high viscosity contrast with the shale makes for a very big number, which raised to the ⅓ power (i.e., you take the cube root) makes for a very small number. This small number, multiplied by the constants of 2, π, and t, gives you L, which will also be a small number: hence the wavelength is small, and as a result, the folds are crunkled up next to one another like sardines in a can.

On the other hand, the low contrast between the viscosities of the quartz vein and the quartz sandstone means that you get a rather small number. Say η = 3. If ηo is also about 3, then you have: (3/(6*3)), or the fraction 1/6. Expressed as a decimal instead of a fraction, this is 0.167. Take the cube root of that, and you end up with a bigger number, in this case 0.55. Multiply that by 2, π, and t, and you get your new wavelength, L. Because you have a larger number in the (η / 6ηo) part of the equation, and everything else is the same, you end up with a larger wavelength. The result is only one fold antiform in the sandstone. In the neighboring shale, ~23 antiforms are packed into the same distance along strike of the vein.

Wild stuff, right? Happy Friday. Let’s hope your weekend is of sufficiently high contrast to the sludge of the week that you get all loose and wiggly, like the top part of the photo… : )


Friday fold: multilayer buckle folding demo

Check out this video I found online whilst uploading last week’s Friday fold:

This video was produced and published on YouTube by Markus Beckers, Michael Ketterman, Dennis Laux and Janos Urai.

It’s a nice demonstration of how multiple layers of material of different properties and different thicknesses can yield up different flavors of folds. In the movie, there are two materials present: white silicone and gray foam. The silicone layers are stronger (“more competent”) than the foam. But the two silicone layers are different thicknesses. It turns out that this ends up being a decisive factor in determining the way they fold.

We can explain this behavior using the Ramberg-Biot equation:

L = 2 π t (η / 6ηo)

where L is the wavelength of the fold (in other words, the distance from one antiform fold hinge to the next antiform fold hinge); t is the thickness of the folded layer; η is the viscosity (resistance to flow) of the silicone layer (or, in general, the more competent of the two layers); and ηo is the viscosity of the foam layers.

In other words, the (η / 6ηo) part of the equation reflects the viscosity contrast between the affected layers. In the video, this viscosity contrast is a constant, since we’re looking at two layers of the same stuff surrounded by the same matrix of other stuff. The only difference is the thickness of the two silicone layers.

So as far as our video up top is concerned, pay attention to the t value and the L value: the thicker the layer is, the larger the wavelength of the resulting fold. The thin layer has a lower t value, and so it ends up with a shorter wavelength: i.e., there are more folds packed into the same amount of vertical space as its stouter neighbor. The thick layer’s higher t value means it wıll have a proportıonately higher L value. It will have a longer wavelength, and fewer undulations will fit into the available vertical space.

Happy Friday, everyone! I’m heading back to DC tomorrow (from Turkey), so more regular posting wıll resume next week.

Mount Moran

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.


Is this dike a feeder?

A new paper in the journal Geology examines an interesting question: how can you tell feeder dikes from non-feeder dikes?

The answer is, normally you can’t. Normally, there’s no way to tell for sure whether a given dike actually funneled magma to the paleo-surface, or whether it never reached the paleo-surface. The reason for this is that usually, the paleo-surface is gone by the time the dike is exposed at the modern surface to your scrutiny. In the new paper, a team of Japanese researchers examined the plumbing of Miyakejima Volcano, which collapsed during an eruption in the year 2000. The collapse opened up a view into the volcano’s “guts,” which showed the anatomical details of many dikes.

Here’s Figure 2B from the paper (reproduced, as with Figure 3 below, with permission of the publishers of Geology), showing the extraordinary exposures on this volcano. The authors report that they were able to trace an individual dike for more than 350 meters. In this example, you can follow a feeder dike up 150 m to find where it erupted at the paleosurface in a cinder cone!

So, given such an extraordinary exposure, how do you go about assessing the geometries of the dikes? The research team used photography as their tool. Hopefully it will be obvious that examining the dikes in person would be difficult and dangerous on a subvertical cliff many hundreds of meters tall — and on an unstable and crumbly volcano, to boot! So they took photos, and then did their measurements based on the photos. They claim a resolution of about 3 cm per pixel at a distance of about 1 km.

Filtering your data through a medium like photography is a good way to introduce error and bias to your study, and the authors took some steps to avoid that. They used good zoom lenses, aimed at the outcrop face from the safety of the opposite side of the caldera, and aimed them straight on to the dike outcrops (i.e., within 10° of the strike of the dikes, not necessarily orthogonal to the cliff face, since there is no guarantee the dike would intersect the cliff face at a right angle): so the apparent thickness was as close as reasonably possible to the true thickness. For each photo, they cut off a 20% margins on each side of the image (total cropped area: -40%), as a guard against the effects of lens distortion. Finally, they double-checked their accuracy by comparing in-person measurements of objects of known size on the caldera rim to their photo-measurements of those same objects.

They defined feeder dikes as those (as in the image above) which were observed to connect directly to the bottoms of spatter cones and diatremes.  They defined non-feeder dikes as those which terminated “either by tapering away inside layers or ending bluntly at layer contacts,” where the ‘layers’ being referred to are pyroclastics and lava flows within this stratovolcano. In total, they tallied up 165 dikes, 93% of which were “non-feeders.” Of these, they selected the 27 best-exposed (21 non-feeders and 6 feeders) for their analysis.

What did they find? To quote from their abstract:

A typical feeder thickness reaches a maximum of 2–4 m at the surface, decreases rapidly to ~1 m at a depth of 20–40 m, and then remains constant to the bottom of the exposure. By contrast, a typical non-feeder thickness reaches a maximum of 1.5–2 m at 15–45 m below the tip, and then decreases slowly with depth to 0.5–1 m at the bottom of the exposure.

Width vs. depth data from five representative non-feeder dikes are plotted in their Figure 3, top row, and three representative feeder dikes in the second row of Figure 3. Check it out:

Feeder dikes open up (get wider) at the surface, but the non-feeder dikes first get wider (gradually positive trend to these plots), and then abruptly pinch out up towards the tip (sudden leftward cant at the top of the plot). The authors ponder these dramatically different profiles, and offer an explanation.

They offer two equations which describe these dike profiles pretty accurately. If you’re not mathematically inclined, take a deep breath. We’ll translate in a few column-inches! The first equation is:

b = (2Po(1-v2)L)/E

where b is the thickness of the dike, Po is the magmatic overpressure (the pressure in excess of the normal stress on the dike at the point of measurement), v is Poisson’s ratio, a measure of how much volume is conserved during strain for the host rock. In other words, when a material is compressed in one direction, how much do the other directions pooch outward? Call it ‘poochiness.’  E is Young’s modulus, a measure of the elasticity of the host rock. L is the “dike-controlling dimension,” that is whichever of the dimensions of the dike (either the dip-dimension or the strike-dimension) is smaller. So, to translate this equation into “English” enough that even Rick Sanchez could understand it, equation #1 says, ” The thickness of a dike of a given height depends on how much pressure the magma opening and filling the dike is under, along with how ‘poochy’ and elastic the host rock is.”

The second equation is:

Po = (ρrρm)gh + ρc + σd

where ρr is the density of the host rock, ρm is the density of the magma, and g is the acceleration due to gravity. The variable h is the dip dimension (height) of the dike (measured upward from the source magma chamber), ρc is the excess magmatic pressure in the source chamber before rupture (dike injection), and  σd is the difference between the maximum and minimum principal stresses. Let’s translate this one, too: “The pressure exerted by the magma filling a growing dike depends on the difference between the density of the magma and the host rock it’s intruding into, as well as the force exerted on the magma by gravity.  Another important factor is whether there are significant tectonic stresses impinging on the dike as it forms.”

So where does that leave us in interpreting Figure 3, showing those different profiles for feeder dikes versus non-feeder dikes? Equation #2 says that the magmatic overpressure in a dike (Po) will increase as the dike propagates upwards (gets taller, in other words: h goes up). And equation #1 says, if the magmatic overpressure increases, then the dike will get thicker. That’s why the non-feeder dikes get thicker and thicker in a nice gradual way as you trace them upwards.

An additional factor is related to the density. You can lower the density of a magma if you allow the gases in it to expand under lower pressure regimes (i.e., at shallower depths). The basaltic lava from this volcano has been previously measured to have about 2% water by weight. As this water exsolves from the magma at shallow depths (lower pressures), it will make bubbles that expand, and lower the density of the magma. However, at shallower depths, the rock surrounding the dike is under less pressure too, so they both decrease their densities in tandem.

Deviations from the expected dike geometries can be observed in some of the field measurements. For instance, in the lower-right-hand corner of Figure 3, dike “110-01” flares out to a wider thickness right as it crosses a stratum of “poorly consolidated scoriaceous tuff” within the volcano. The authors suggest that this rock type has a lower Young’s modulus. Because it’s poorly consolidated, it’s less elastic. A lower E value in equation #1 results in a larger b value, the thickness of the dike. Cool!

Now, the feeder dikes have a constant thickness all the way up. To the authors of the paper, this suggests that in the course of the eruption, these dikes reached a stress equilibrium with the surrounding host rock. Magma, being fluid, flowed away from highly-pressurized zones, and the dike thickness “evened out.” And why do the feeder dikes abruptly get wider at the top? The authors postulate a couple of possible reasons: First to consider is the elastic free-surface effect, which is essentially saying that as a dike approaches the surface of the Earth, half of the surrounding rock elasticity is lost (replaced by air), and so that control “hemming in” the dike is lost, and the dike expands. Second, erosion is probably an important factor, as the flowing magma churns away at the wall rock, breaking it down thermally as well as dynamically. In other words, some of the rock that used to be there at the edge of the fissure has been abraded or melted away as a consequence of all that lava flowing out of the dike and away over the surface.

Take home message? To quote the authors, “Feeders propagate and grow as non-feeders before they reach the surface. Therefore, the geometric difference between these types of dike… is primarily a reflection of the feeders reaching the surface.”

I’m interested in feeder dikes because Neoproterozoic feeder dikes of the Catoctin Formation are a significant piece of the geologic story of Virginia’s Shenandoah National Park:

…But these are interpreted as feeder dikes. To my knowledge, no one has claimed any particular outcrop in the Blue Ridge province as a spot where you can actually see the dike flare out and transition into a Neoproterozoic spatter cone. I picked up the Geshi, et al. paper in the first place because I wanted to know whether there was some measurable aspect of the Shenandoah dikes’ geometries that could tell me if indeed they were feeder dikes. The problem is that the exposure in Virginia (especially vertically) isn’t quite as good as the exposure on the inside of Miyakejima’s caldera. We’re lucky if we get 10 meters of vertical exposure, and there’s no suggestion from the Miyakejima data that that 10 m is sufficient to “profile” the dike sufficiently precisely to say whether it’s got a feeder geometry or not, especially if you don’t know where in the dike’s profile that 10 m vertical segment lies. So maybe all we Virginians can do is just interpret: we’ve got a bunch of Neoproterozoic dikes cutting basement rock, and atop the basement rock a bunch of Neoproterozoic lava flows, therefore some of those dikes are likely to be feeders.


Geshi, N., Kusumoto, S., & Gudmundsson, A. (2010). Geometric difference between non-feeder and feeder dikes Geology, 38 (3), 195-198 DOI: 10.1130/G30350.1

Salamander shear

Whilst discussing how to quantify strain with my GMU structural geology students recently, I hit upon a cool analogy. In order for you to understand the analogy (assuming you’re not a structural geologist), I’ll have to review some background information first. Stick with it, and I promise you a salamander at the end.

Structural geologists are interested in how rocks deform. If we have some idea of the original shape of a rock element (say, an oncolite, or a cooling column, or a fossil, or more prosaically, a sedimentary layer), and we find a deformed version of it, then we can use that object as a strain marker. By measuring its features, we can determine how parts of the strain marker have elongated, shortened, or rotated.

Elongation is pretty easy to calculate. You compare the final dimensions to the original dimensions via the equation e = (lf-lo)/lo where “lo” is the original length and “lf” is the final length.

If the strain marker has been elongated, then the value of e will be positive. If it has been shortened, then the value of e will be negative. Another value, S (for ‘stretch’) is also a useful parameter to calculate. S = lf/lo, which is the same as saying: S = e + 1

As a quick example, a line which has been deformed from an original length of 20 cm to a final length of 30 cm has an elongation of 0.5. It has a stretch of 1.5. Put another way, the line has been increased in length by 50% (of its original length), and it is now 150% as long as it originally was.

Here’s a little diagram to summarize:

Okay — so far, so good. We can now calculate length changes. But many of the structural elements in a strain marker not only change their length, but also their orientation. That is to say, they rotate. In order to quantify that, we need to thinking about shear strain. Shear strain (γ), is calculated from angular shear (ψ), a quantity that is measurable in a rock element (again, provided you have a clear idea of what it looked like before deformation).

For any given line in a strain marker, angular shear is defined as the deviation (in degrees) from perpendicular for a line which was originally perpendicular to the line we care about. That is a ridiculous definition at first glance, and it seems to many students quite non-intuitive why one would bother looking for a line originally perpendicular to the line on which we are actually trying to measure the angular shear! I mean, is that abstractified or what?

Before I reveal my stunning new analogy for angular shear, let me diagram the definition for you, and relate it to shear strain.

The line we care about (blue) looks quite the same after deformation as it looked before deformation. So we can’t actually measure anything about it directly that has changed. But when we look at a line which we know was originally perpendicular to it (gray/red), we can measure how it has changed.

The angle between the old perpendicular-line and the new ‘perpendicular’- line is ψ, the angular shear. By convention, clockwise rotations of the perpendicular-line are given positive values, while counter-clockwise rotations are assigned negative values. In the above example, the rotation is clockwise by 35°, so we note it as “ψ = +35°”.

Another way to think about this situation is to consider any point on our perpendicular-line. Here, let’s consider point A:

A was originally at the end of the old perpendicular-line (gray), in position Ao. After deformation, it is now at the end of the new ‘perpendicular-line’ (red), in position Af. You can see that this outlines a right triangle (yellow):

Right triangles are lots of fun, because they allow us to practice trigonometry. You may remember the mnemonic phrase “SOHCAHTOA” from your high-school trigonometry class. Basically, this relates the angle 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 the change in the position of point A, which could be expressed as Δ(A), is equal to the length of the ‘opposite’ side relative to the length of the ‘adjacent’ side of the triangle. The length of the adjacent side isn’t changing through this deformation, but the length of the opposite side is changing. With a little deformation, it changes a little. With a lot of deformation, the length of that O side increases dramatically.

The formula for calculating shear strain from angular shear is: γ = tan(ψ)

So, with a large value of ψ, you get a large value for tan(ψ). The shear strain (γ) increases along the line we care about (blue) with increasing rotation of the ‘perpendicular’-line (red). The maximum value of ψ would be approaching 90°, which means that your γ would be close to infinity. The minimum angular shear, 0°, would yield a shear strain of 0. Here’s a table of other possible values.

Okay, got it? …(Whew!)

Now for the salamander!

In order to express to my structure students why we bother to pay attention to that ‘perpendicular’-line, I wanted to get them into the perspective of the line we care about: I wanted to get them into its mindset. But of course, lines are geometric entities, and don’t have minds. They are hard to identify with. So I opted for a salamander, because they are cute. Furthermore, salamanders are roughly linear organisms, and have legs poking out of them at roughly right-angles to their spinal column. Here’s my blackboard sketch:

So if you think about what the salamander experiences as it gets sheared (right-lateral), you can imagine what shear strain feels like. The legs go from an original orientation (purple chalk) to a deformed orientation (blue chalk), rotating by about 45°. Along the length of the salamander’s body (orange chalk), the angular shear is 45°, and the tangent of 45° is 1. So the shear strain along the length of this poor salamander is 1. What this means is that the tips of the salamander’s toes (the equivalent of point A in the earlier diagrams) have moved one whole leg-length to the right of their original positions. You can put some numbers to this if you want: Say the leg length is 1.5 cm, and the toes are displaced 1.5 cm. Remember SOHCAHTOA… γ = tan(ψ) = opp/adj = 1.5/1.5 = 1.

There’s something else on the diagram, too: a yellow line. This makes the point that if the leg-line of the salamander is the “line you care about,” and not the spine-line, then the spine-line can be “the line originally perpendicular to the line you care about.” If you’re measuring the angular shear along the leg lines, you would want to measure how the spine-line has rotated relative to its original position. As a result of the shearing deformation in this example, the spine-line has rotated by 45° counter-clockwise. So we’d say: “ψ = -45°”. The tangent of -45° is -1. The shear strain on the leg-line is therefore exactly the opposite of the shear strain on the spine-line.

But the key point of this analogy is to put your mind in the salamander’s perspective, and “feel” the shear strain along the length of your body as deformation proceeds. I hope that intuiting shear strain from that perspective will be useful to structure students learning these concepts.