I find it difficult to teach college students about using the triangular facet on remote sensing images to judge the dip direction of sedimentary formations. Is there any intuitive way that can help?


This is a tale of two three dimensional coordinate systems. You may need to take a bit of a side track if your students are not already familiar with the concepts of three dimensional coordinate systems, three dimensional vectors, the three dimensional dot and cross products, and transformations between coordinate systems. The image below portrays two such coordinate systems, with axes in blue and red, Two sets of three dimensional orthogonal axes

I'll denote one of those coordinate systems as the East, North, Up coordinate system (ENU for short), with unit vectors $\hat E$ (East), $\hat N$ (North), and $\hat U$ (Up), and the other as the dip, strike and outward normal coordinate system (dsn for short), with unit vectors $\hat d$ (dip), $\hat s$ (strike), and $\hat n$ ((outward) normal). In terms of the above diagram, imagine the trio of blue vectors corresponding to East, North, and Up while the trio of red vectors corresponding to dip, strike, and (outward) normal. As an instructor, you might well want to create a better diagram, with labels.

There are singularity issues with both of these sets of coordinate systems. East/North/Up doesn't make sense at the Earth's poles as all horizontal directions are south at the North Pole and north at the South Pole. The dip/strike/normal doesn't make sense wherever the surface is flat. I'll ignore those issues. (Bonus question for your students: Address those singularities.)

One way to define local ENU coordinate system for a point on the surface of the Earth is to use the Earth's reference ellipsoid. The "up" unit vector is normal to the ellipsoid, with "up" pointing outward. The east unit vector is orthogonal to the $\hat U$ unit vector and the Earth's rotation axis. Use the cross product and divide by the norm. The north unit vector is the cross product of the up and east unit vectors.

Regarding the dip/strike/normal coordinate system, one direction, the outward normal, can be determined from the coordinates of the three points in a triangle in a vector-based triangular network digital elevation model. The coordinates of the three points in a triangle ($\vec A$, $\vec B$, $\vec C$) are supposed to be arranged such that $(\vec B - \vec A)\times (\vec C - \vec B)$ points outward. This points in the direction of $\hat n$: $\hat n = \frac{(\vec B - \vec A)\times (\vec C - \vec B)}{||(\vec B - \vec A)\times (\vec C - \vec B)||}$.

Aside: This is not always the case; there are DEMs that do not follow this rule. Those are DEMs one should toss in the nearest digital trashcan.)There are also some DEMs that do follow this rule but occasionally have triangles whose outward normal has a downward rather than upward component. An example is an extremely detailed DEM that models not only the topside but also the underside of an overhang. Satellite-based DEMS typically don't do this as satellites cannot see the underside of an overhang. (Extra credit: Have your students address such cases!)

We're looking for $\hat s$ and $\hat d$, the unit vectors that point in the strike and dip direction. Both of these vectors lie in the plane of the triangle formed by $\vec A$, $\vec B$, and $\vec C$. This means that both $\hat s$ and $\hat d$ are normal to the outward normal $\hat o$. The strike vector is horizontal, which means that it is also normal to the unit upward normal. To find $\hat s$, compute the cross product of $\hat U$ and $\hat n$ and normalize. (Note: Normalization is impossible if $\hat U \times \hat n \equiv 0$. This only happens if $\hat n$ is parallel or antiparallel to $\hat U$, in which case strike and dip are not well-defined.) Finally, the dip unit vector, which is what we're after, is the cross product of the strike and outward normal unit vectors: $\hat d = \vec s \times \vec n$. Bonus: Ask your students to show that if $\hat d$ exists, it always has a downward component.


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