Law of Vs Example
Determining the orientation of the formations in the map requires using the "Law of Vs"
Review
Given 3 points with known depths, or (\(x,y,z)\), to a thing of interest (water, a bed, some formation) find its orientation
Note We assume the thing is planar, if it is not, say it is folded, we cannot use this method properly.
Lets start by determining the Strike \(\theta\). Draw a line from the largest value \(C\)to the smallest value \(A\).
Step 1: Draw a line from the smallest value to the largest value
Reference This line \(A-C\) lies along the plane of interest and it changes its depth from the largest to smallest values linearly.
Step 2. Determine where along the line \(A-C\) the value of the middle point is.
The form this takes is $$ \Delta_{map} (D-A) = \frac{\Delta_{map}(C-A)}{C_z - A_z} (B_z - A_z)$$ where \(\Delta_{map}()\) is distance in map units, measured with a ruler, and \(A_z, B_z\) and \(C_z\) are the provided depths. This is linear function turning the distance along the map to the depth to the plane; equivalently: $$ y - y_0= m (x -x_0)\\ $$ You can also write this as a proportion $$ \frac{\Delta_{map} (D-A)}{B_z - A_z} = \frac{\Delta_{map}(C-A)}{C_z - A_z} $$
Step 3.Call this new point \(D\) and draw the line \(B-D\).
Reference The depth along the line \(B-D\) is same and lies in the plane of interest. An official name is a structural contour. The direction of this line is the strike \(\theta\).
Step 4. The orientation of the line \(B-D\) from North.
Warning: This is usually the place where most people get confused because you have been asked to measure an angle from the surface with only looking at the surface.
To determine the dip we will need the following equation $$\tan \delta = \frac{\Delta z}{\Delta x}$$ where \(\delta\) is the dip, \(\Delta x\) is the horizontal distance and \(\Delta z\) is the vertical distance between two points. But which two points?
Remember the dip is:
We have a strike line, \(B-D\), so lets draw a line perpendicular to it (this is the direction of the dip). But where do we draw the line? The line should go through a known point, we have \(A\) and \(C\), pick one[1].
So lets draw a line from point \(A\) to line \(B-D\) that is perpendicular to \(B-D\). This magically creates two points: \(E\) and \(E'\) where \(E_z = A_z\) and \(E'_z = B_z\)[2].
Step 5. Draw a line from a known point (either highest or lowest value \(A\) or \(C\)) to the line of known strike (here, \(B-D\)) perpendicular to the line of strike (again, \(B-D\)). Call the point on the line \(B-D\), point \(E\) or \(E'\)
Reference Point \(E\) is not on the plane, it is the same depth as \(A\). Point \(E'\) is on the plane as it has the same depth of \(B\) and is on the structure contour line \(B-D\).
Step 6. Determine the values for \(\Delta z\) and \(\Delta x\)
Compute the real-world vertical distance from \(E\) to \(E'\), or \(\Delta z = E'_z - E_z \).
Measure the distance in map units, \(\Delta_{map}(A-E)\), and convert it to real-world horizontal distance via $$\Delta x = \Delta_{map}(A-E) \frac{\Delta_{real}(\text{scalebar})}{\Delta_{map}(\text{scalebar})}$$
Reminder: \(\Delta_{real}(\text{scalebar})\) is that little number over the scale bar and \(\Delta_{map}(\text{scalebar})\) is the length of the scalebar in map units, say cm. This equation is basic unit conversion you probably learned in Introductory Chemistry.
Step 7. Compute the dip using \(\Delta x\) and \(\Delta z\)
Now that we have \(\Delta x \) and \(\Delta z\) we can rearrange the \(\tan \delta = \frac{\Delta x }{ \Delta z}\) equation to determine the dip. $$ \delta = \tan^{-1}{\frac{\Delta z}{\Delta x} } = \text{atan}{\frac{\Delta z}{\Delta x} } $$