Problem 4.1

That symbol that looks like a strike-dip symbol with the curly thing is an overturned bed

Use the dips contained within the bed to define the dips of the bed in the cross section

But dont make them too long because we will need to connect them.

Now, you need to make sure which side is "up".

The overturned bed has the upfacing or younging direction pointing down and a normal bed has the younging direction pointing up.

I like to use little arrows to indicate the younging direction

The beds can connect either 1) above or 2) below the surface

How do you pick which one? By using the younging direction

If you "unfold" the bed, the correct direction should be where the younging direction points up. *Provide the entire structure has not been overturned.*

Apparent Dip Calculator

Problem 4.2

Start by putting in the fault, it has a dip iof 82 degrees towards the east, but the strike of the fault is not perpendicular to the cross section, so an apparent dip needs to be computed. I will be nice and do this one as an example.

The strike of the fault is approximately N15W.

The direction in which we want the apparent dip is along the cross section, which could be West, or N90W, or it could be East, N90E. Thats not helpful. But the fault dips towards the East, so we will use the East direction for the apparent dip. This means that beta (the difference between the strike and the apparnet dip is 75 degrees)

Using the apparent dip equation as in the previous chapters

$$\tan \delta = \tan \alpha / \sin \beta$$

we can solve for the apparnet dip \(\alpha\).

$$\tan \alpha = \tan 82^\circ \sin 75^\circ$$

$$\alpha = 81.72^\circ$$

The apparent dip is not much different than the true dip. This is because the orientation, or stirke, of the bed is alomost 90 degrees to the cross section.

So we will use the value of 82 degrees and not quibble about 0.8 degrees.

You can now draw in the western block and Fm. K using the dips as they are, but if you feel like reducing the dip just a little bit, thats ok.

The eastern most arm of the western fault block has its dip on the other side of the fault. It is ok to use this.

Again, remember to transfer the Fm K outcrops along A-A' onto your Cross section

For the eastern fault block, the western most arm can probably be put in with a apparent dip close to the actual dip of 42.

The eastern arm of Fm K in the eastern fault block is not perpendicular to the cross section. It is more like 45 degrees. Therefore, we need to use the apparent dip equation from above.

Compute it, draw in the bed and connect it to the other arm of Fm K.

Almost done

Put in the Dike. You will need to use the apparent dip formula

Put in the Quartz diorite. This is an igneous intrusion. So, uh, I will let you figure that one out

Define the limits of the vertical scale. Determine the minimum and maximum topographic elevations. The minimum elevation will likely need to be much greater than what is on the map. For this cross section it will likely extend to either 0 feet or -2,000 feet.

Place the topography onto your cross section

Insert the faults

Work on the fault blocks one at a time

Place the unfolded sedimentary beds. Unfoled Sedimentary beds are assumed to not be folded and as such should have flat, linear bottoms. They will be very think and it might be a challenge to insert them in properly.

Now work on the deeper folded beds.

Use the apparent dip formula $$\tan \delta = \tan \alpha / \sin \beta$$ to determine the dip of each bed.

Take an average of the bed dip and strike, taking care to stay away from the hinge area of the fold where most of the curvature in the fold is. Take the Measurements within the straight-ish limbs.

If the beds are overturned, this will need to be taken into account

Methodoically insert each bed contact until the surface structue within a single fault block is complete.