Chapter 1

Problem 1.1

Geographic directions can written in two different ways, by azimuth or quadrant

Azimuth

Degrees from north normally using three digits. Examples:

        000°       North
        090°       East
        180°       South
        270°       West
        360°       North
       -090°       West
       -270°       East, but unnecessarilly confusing
       -045°       NorthWest
       -022°       North by North West

Quadrant

Degrees from a reference direction in a specified direction

        N        North
        E        East
        S        South
        W        West
        N45°W    North West at 45°
        N45°E    North East at 45°
        S45°W    South West at 45°
        S10°W    South West at 10°
        N3°E     North East at 3°

Problem 1.3

Is the line element contained within the plane? Or does the line intersect the plane in a single point or every point?

Line - Plane Intersection Reference

Use the interative plane/line visualization below to help understand the problem.

Definitions

$ \theta $ - Plane Strike
$ \delta $ - Plane Dip
$ \phi $ - Line Trend
$ \alpha $ - Line Plunge
$ \theta_\delta $ - Plane Dip Direction

Requirements

Line plunge is not steeper than the plane
- $ \alpha <= \delta $
Line plunges in the same direction as the plan
- $ | \phi - \theta_\delta | <= 90 $
Direction must satify an equation

Special Cases

Line along the Dip
- $ \phi = \theta_\delta $ and $ \alpha = \delta $
Line along the Strike
- $ \phi = \theta $ and $\alpha = 0$

Green Plane

Dip:
Strike:

Blue Line

Trend:
Plunge:
x:
y:

Problem 1.4

Definitions

$ \delta $ - Plane Dip
$ \theta $ - Plane Strike
$ \theta_\delta $ - Plane Dip Direction
$ \alpha $ - Apparent Dip
$ \theta_\alpha $ - Apparent Dip Direction
$ \beta $ - Angle between Strike and Apparent Dip Direction

Question

Find $\delta$, Given $ \theta , \alpha$ and $ \theta_\alpha $ using Orthographic Projection.

This method is done graphically using a ruler and protractor.

Example: Find $\delta$, given $ \theta = N45^\circ E , \alpha = 10^\circ$ and $ \theta_\alpha = N15^\circ E$

First we need to put down on paper the things that we know

$\theta $ - Strike (Geographic direction)
$\theta_\alpha $ Apparent Dip direct (Geographic Direction)
$\alpha $ - Apparent Dip

The two geographic directions are just two lines at the appropriate angles relative to North. The strike is in red and the apparent dip direction is in blue.

We also know the direction of the dip, just because we know the strike. $$ \theta_\delta = \theta + 90 $$

Lets draw that too, in fuchsia, just not at the intersection of the other two lines.

You will notice it is perpendicular to the line of strike $ \theta $ and intersects the apparent dip direction $ \theta_\alpha $ forming a right triangle, in green.

Focusing on the triangle in green, find the intersection of the apparent dip $\theta_\alpha $ and dip directions $ \theta_\delta $, call it point $DA$. The intersection between the strike $\theta$ and dip direction $\theta_\delta$ we will call $SD$ and the strike $\theta $ and apparent dip direction $\theta_\alpha$ intersection is $SA$.

Reality Check What we have drawn so far are only geographic direction, i.e. horizontal lines along the surface. Now we are going to draw some vertical lines and triangles, but they will be drawn as "folded up" to the horizontal plane or the surface.

At $DA$, draw a line perpendicular from $\theta_\delta$ and another from $\theta_\alpha$, away from the triangle.

At $SA$, measure an angle $\alpha$ from the apparent dip direction line away from the green surficial triangle. Lets call the intersection, point $A$.

At this point we have constructed a triangle associated with the apparent dip $\alpha$, something we were given. This triangle is formed by

The line between $SA$ to $DA$ along the surface in the direction of the apparent dip $\theta_\alpha$
The line from $DA$ to $A$ which represents a direction straight down or vertical.
The line from $SA$ to $A$ which represents a line along the plane in question. This line has a slope.

Measure the distance between the $DA$ and $A$ and transfer that along the other line. Lets call the new point $D$.

Draw a line from $A$ to $SD$. This should form another triangle associated with the dip $\delta$, the thing we are interested in. As with the apparent dip triangle it is formed by

The line between $SD$ to $DA$ along the surface in the direction of the dip $\theta_\delta$
The line from $DA$ to $D$ which represents a direction straight down or vertical.
The line from $SD$ to $D$ which represents a line along the plane in question. This line has a slope.

Measure angle formed by $D - SD - DA$ to determine the dip of the plane $\delta$

Problem 1.5

Find $\alpha$, Given $ \theta , \delta$ and $ \theta_\alpha $ using Orthographic Projection.

The procedure is the same as in previous problem by

Draw $\theta$, $\theta_\alpha$, and $\theta_\delta$ forming the triangle.
Draw two perpendicular lines at $DA $
Measure $\delta$ from $\theta_\delta$ at $SD$ making point $D$
Measure distance $DA-D$ line and transfer to other line making point $A$
Draw line from $A$ to $SA$
Measure angle between line $SA-A$ and $SA-DA$

Problem 1.8

Find $\delta$, Given $ \theta , \alpha$ and $ \theta_\alpha $
Compute $\beta = \theta - \theta_\alpha$
Make sure $0 <= \beta <= 90$
Solve for $\delta$ $$\tan \delta = \frac{\tan \alpha}{\sin \beta}$$

Problem 1.9

Find $\alpha$, Given $ \theta , \delta$ and $ \theta_\alpha $
Compute $\beta = \theta - \theta_\alpha$
Make sure $0 <= \beta <= 90$
Solve for $\alpha$ $$\tan \delta = \frac{\tan \alpha}{\sin \beta}$$ $$\tan \delta \sin \beta = \tan \alpha$$

Problem 1.12

Compute $ \beta $ - Angle between the strike and apparent dip
Make sure $0 <= \beta <= 90$
Use the alignment diagram