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Using the same method, determine the slope for the remainder The shortening is the difference between the original length

of the earthquake position and depth values in the table. (L O ) and the folded length (L F ):
Results are shown below:
L O = 30 m shortening = L O − L F
__
4
−2.06 × 1 0 m
slope = L F = 20 m = 30 m − 20 m

2
54.2 km
shortening = ? = 10 m
m _
= −380

km 19.2. The amount of shortening is determined by subtracting the
4
__ folded length of the rock from the original length of the rock
−3.13 × 1 0 m

slope =
3
87.5 km layers in their originally horizontal position.

m _
= −358 The folded length of the rock is sum of the diameters of
km each semicircular anticline or syncline. Since each fold has the
__ same radius of curvature, the diameter for one fold can be
4
−4.86 × 1 0 m
slope =

4
115.6 km determined and multiplied by the number of folds to obtain the
m _
= −420 folded length:
km
r = 54 m D = 2r(number of folds)
__
4
−6.02 × 1 0 m

slope = number of folds = 3 = 2(54 m)(3)
5
139.9 km
D = ? = 320 m
m _
= −430
km

The original length of the rock is the circumference of the circle
Average the results. divided by 2 to obtain the length of the arc of each fold. Since
m _
m _
m _
( −410 + −380 + −385 each fold has the same radius of curvature, the arc length can
) (
)
) (

km
km
km
be multiplied by the number of folds to obtain the original
m _
m _
) (
)
(
+ −420 + −430 length:
km
km
slope average =
5 r = 54 m C = (number of folds)
π2r
_

m _ number of folds = 3 2
= −405
km C = ? π2(54 m)
_

=
2 (3)
18.15. Calculate the future position from the velocity, and state the
= 510 m
position relative to its current position.
_ d _ The shortening is the difference between the original length
mm
v = 45 yr v = ∴ d = vt (L O ) and the folded length (L F ):

t
mm
km
6
_
t = 20 × 10 yr Convert velocity from _ to :

yr
yr
d = ? mm _ L O = 510 m shortening = L O − L F
km
yr (

45 _ 6 ) L F = 320 m = 510 m − 320 m
1 × 10 mm
shortening = ? = 190 m
−5 km
_
4.5 × 10 19.3.

yr
_
−5 km 6

d = 4.5 × 10 (20 × 10 yr) West Fault West Fault

yr
2
Δy
Δy
_
= 9.0 × 10 km northwest of its current position Δy = 105 m slope = ∴ Δx = _

m _ Δx slope
slope = 2.1
m
105 m
Δx = _
m _
Δx = ?
CHAPTER 19 2.1
m
105 m _
19.1. The amount of shortening is determined by subtracting the = _

folded length of the rock from the original length of the rock 2.1 m _
m
layers in their originally horizontal position. = 500 m
The folded length of the rock is the diameter of the
semicircular arch: East Fault East Fault
Δy = 105 m _
105 m
r = 10 m D = 2r Δx = m _
m
D = ? = 2(10 m) slope = 2.9 m _ 2.9
m
105 m _
= 20 m = _

Δx = ? 2.9 m _

The original length of the rock is the circumference of the circle m

= 360 m
divided by 2 to obtain the length of the semicircular arch:
The extension is the sum of the horizontal displacement (Δx) of the east
r = 10 m _
π2r
C = and west faults:
C = ? 2
_
π2(10 m)

= Δx w = 500 m extension = Δx w + Δx e
2
Δx e = 360 m = 500 m + 360 m
= 30 m
extension = ? = 860 m
E-41 APPENDIX E Solutions for Group A Parallel Exercises 683