U. S. DEPARTMENT OF AGRICULTURE SO1 L CONSERVATION SERVICE &INEERING DIVISION

TECHNICAL RELEASE NO. 41 GEOLOGY

MARCH 1969

U. S. Department of Agriculture Soil Conservation Service Engineering Division

GRAPHICAL SOLUTIONS

OF

GEOLOGIC PROBLEMS

D. H. Hixson Geologist

Technical Release No. 41 Geology

March 1969

GRAPHICAL SOLUTIONS OF GEOLOGIC PROBLEMS

Contents

Introduction

Scope

Orthographic Projections Depth to a Dipping Bed Determine True Dip from One Apparent Dip and the Strike Determine True Dip from Two Apparent Dip Measurements at Same Point

Three Point Problem Problems Involving Points, Lines, and Planes

Problems Involving Points and Lines Shortest Distance between Two Non-Parallel, Non-Intersecting Lines

Distance from a Point to a Plane Determine the Line of Intersection of Two Oblique Planes

Displacement of a Vertical Fault Displacement of an Inclined Fault

Stereographic Projection True Dip from Two Apparent Dips Apparent Dip from True Dip Line of Intersection of Two Oblique Planes Rotation of a Bed Rotation of a Fault Poles Rotation of a Bed Rotation of a Fault Vertical Drill Holes Inclined Drill Holes Combination Orthographic and Stereographic Technique

References

Fig. 1 Fig. 2 Fig. 3 Fig. 4 Fig. 5 Fig. 6 Fig. 7 Fig. 8

Page

Figures

Orthographic Projection Orthographic Projection True Dip from Apparent Dip and Strike True Dip from Two Apparent Dips True Dip from Two Apparent Dips True Dip from Two Apparent Dips Three Point Problem Three Point Problem

Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. F ig . Fig.

Distance from a Poin t t o a Line Shor t e s t Distance between Two Lines Distance from a Poin t t o a Plane Nomenclature of Fau l t Displacement I n t e r s e c t i o n of Two Oblique Planes Displacement of a V e r t i c a l Fau l t ( r epea t ed ) Displacement of an I n c l i n e d Faul t Wulff Net True Dip from Two Apparent Dips Apparent Dip from True Dip I n t e r s e c t i o n of Two Oblique Planes Rot a t i o n Rota t ion about an I n c l i n e d Axis Rotat ion about an I n c l i n e d Axis Rela t ionship between Planes and Poles Rota t ion us ing Poles Rota t ion about an I n c l i n e d Axis True Dip from V e r t i c a l Core Holes True Dip from I n c l i n e d Core Holes I n c l i n e d Fau l t Problem I n c l i n e d Fau l t problem

Page

17 1 9 21 23 2 5 2 7 2 9 31 3 3 3 5 37 39 41 4 3 45 4 7 4 9 5 1 5 3 5 5 57 57

GRAPHICAL SOLUTIONS OF GEOLOGIC PROBLEMS

Introduction

The geologists and engineers in the Soil Conservation Service in the course of their duties encounter problems in determining the true location, attitude, or orientation of geologic structures.

These problems can usually be solved mathematically but the mathematics is often quite involved. Graphical methods that are rapid and accurate to use will give valid results.

Scope

This technical release covers some of the basic techniques in graphical solutions of three dimensional problems involving points, lines, and planes. These techniques will give attitude, location, distance, and dimensions in the solution. The techniques of using hemispherical nets for the solution of problems are also covered. These techniques while they give attitudes and direction do not provide distance and dimension in the solutions.

The procedures presented herein are not new or original but have been used by geologists for a number of years. This is a compilation that has assembled material from various sources into one document that will be readily accessible to SCS geologists. If additional informa- tion is desired about these techniques the reader is referred to the references listed at the end of this TR.

Orthographic Projections

A geologic structure has a fixed position in the earth's crust. When this position has been determined by a survey method (transit, plane table and alidade, or compass bearing) the observer must consider this position as fixed. If the observer wishes to view this structure from another position, he must look directly at the position he wishes to see. In a sense, the observer must think and visualize the struc- ture in three dimensions. If he does this, he can always observe directly the view he wishes to see.

The orthographic projection is a right-angle type of projection. It uses parallel lines for projection at right angles to an image plane. The image plane is the plane on which a view is projected. A folding line is the intersection of two image planes.

Drawing equipment needed for solution of problems by orthographic pro- jection are: paper, T-square and/or triangles, scale, protractor, and drawing pencils. Dividers and a compass are useful at times, but a scale can usually be used. The lines drawn should be fine and sharp and points well defined. All angles and measurements must be laid off accurately.

The p l an view i s t h e b a s i c view. A l l o the r views must be r o t a t e d about f o l d i n g l i n e s i n t o t h e image p lane which i s t h e p lane of t h e paper . Figure 1 i s an i l l u s t r a t i o n of s e v e r a l views p ro j ec t ed by or thographic p ro j ec t ion . Basic information given i s : a stratum of rock outcrops, the width of the outcrop i s 0, and the stratum dips 45' due south. The i sometr ic ske tch (not t o s c a l e ) i n t h e lower right-hand corner is t h e block of rock we a r e cons ider ing i n t h i s f i g u r e .

From t h e given information views 1 and 2 a r e cons t ruc ted . The p lan view (view 1) i s drawn f i r s t . The view t h a t can be drawn next i s a north- south c ros s s ec t ion . This c ros s s e c t i o n i s r o t a t e d i n t o t h e p lane of t h e paper by r o t a t i o n around a f o l d i n g l i n e (FL) drawn i n a north-south d i r e c t i o n . The po in t s A l , B1, ... H 1 , a r e p ro j ec t ed perpendicular t o t h e fo ld ing l i n e t o view 2. Poin t C2F2 i s loca t ed any convenient d i s t a n c e below t h e f o l d i n g l i n e , the'45O angle l a i d o f f , and a l i n e drawn from C2F2 t o t h e i n t e r s e c t i o n of t h e l i n e p r o j e c t i n g p o i n t s A l B l from view 1 t o view 2. Line E2D2-A2G2 i s p a r a l l e l t o C2F2-B2H2 and t h e va r ious p o i n t s found at t h e i n t e r s e c t i o n of p r o j e c t i o n l i n e s from view 1.

It i s d e s i r a b l e t o l a b e l a l l p o i n t s and f o l d i n g l i n e s . I n Figure 1 t h e p o i n t s a r e a l l l a b e l e d wi th a l e t t e r and number. The l e t t e r des igna t ion remains t h e same i n a l l views while t h e number p o r t i o n changes t o t h e view number. A convenient method of l a b e l i n g f o l d i n g l i n e s i s by use of t h e symbol FL t o i n d i c a t e a fo ld ing l i n e and a two- number des igna t ion showing t h e view p ro jec t ed from and t h e view pro- j e c t e d t o . I n Figure 1 i n t h e l a b e l FL 112, FL i n d i c a t e s t h e f o l d i n g l i n e ; t h e 1 i n d i c a t e s on which s i d e of t h e f o l d i n g l i n e view 1 i s l o c a t e d and t h a t it was drawn f i r s t ; and t h e 2 i n d i c a t e s which s i d e of t h e fo ld ing l i n e view 2 i s l oca t ed and t h a t it was drawn second and by p r o j e c t i o n from view 1.

View 3 was drawn t h i r d by p r o j e c t i o n from view 1 a s i n d i c a t e d by t h e no ta t ion FL 113 on t h e fo ld ing l i n e . View 3 i s r e l a t e d t o view 1 i n t h e same manner a s view 2 i s r e l a t e d t o view 1. Therefore , a l l p o i n t s i n view 3 ( ~ 3 , B3, ... ~ 3 ) a r e l o c a t e d t h e same d i s t a n c e from FL 113 a s p o i n t s A2, B2, ... H2 a r e from FL 112 i n view 2.

View 5 was cons t ruc ted by p r o j e c t i o n from view 4 perpendicular t o FL 415. View 4 i s r e l a t e d t o view 5 and t o view 1, t h e r e f o r e , po in t A5 i s t h e same d i s t a n c e from FL 415 a s po in t A 1 i s from FL 411.

This r e l a t i o n s h i p of views i s t h e b a s i s of or thographic p r o j e c t i o n . Enough information t o cons t ruc t 2 views must be a v a i l a b l e i f a d d i t i o n a l views a r e t o be cons t ruc ted .

FIGURE 1.- Orthographic projection.

Depth t o a Dipping Bed

Depth t o a dipping bed may be r e a d i l y determined i f t h e d i p and s t r i k e of t h e bed and su r f ace e l eva t ions a r e known.

Figure 2 i s t h e g raph ica l s o l u t i o n of a problem with t h e fol lowing d a t a known: A t point A the top of a shale bed wi th a dip of 15' t o the S45OW outcrops; a t point B 100 f ee t due eas t of A the bottom of the shale bed outcrops wi th the same dip and s t r i k e ; across a ridge 292 f ee t due west i s the la, point i n a va l l ey on the center l ine of a structure. Assume a l l three points are the same elevation. A t what depth would the top of the shale be encountered i n a t e s t hole, what thickness of shale would be penetrated by a ve r t i ca l t e s t hole, what i s the true thickness of the shale, and what i s the outcrop width of the shale?

To so lve t h i s problem, p o i n t s A and B and t h e t e s t ho le a r e l o c a t e d on a p l an view. The s t r i k e of t h e s h a l e at p o i n t s A and B i s drawn ( ~ 4 5 ~ ~ ) and t h e d i r e c t i o n of t h e d i p ind ica t ed . A fo ld ing l i n e (FL 1/2) i s drawn e a s t of p o i n t B and t h e s t r i k e of t h e two beds i s p ro j ec t ed t o t h e f o l d i n g l i n e . An angle of 15O i s l a i d o f f between t h e FL and po in t A ( o r B) and t h e t o p and bottom of t h e s h a l e bed i s drawn. The t e s t ho le i s p ro j ec t ed a t r i g h t angles t o FL 1 / 2 and t h e depth, t h i ckness , e t c . , measured from t h e drawing. If t h e e l e v a t i o n of t h e t e s t ho le i s d i f f e r e n t t han p o i n t s A and B, t h e d i f f e r e n c e i n e l e v a t i o n can be sub t r ac t ed o r added t o t h e depth (60 f e e t ) a s s ca l ed from t h e drawing.

Test hole 0

0 50 100 200

P Scale in feet

FIGURE 2.-Orthographic project ion.

Determine True Dip from One Apparent Dip and t h e S t r i k e

A bed s t r i k e s north-south and has an apparent dip of 20' t o the N65'E. What i s the true dip of the bed?

Figure 3 shows t h e s o l u t i o n of t h i s problem by two methods--orthographic p r o j e c t i o n and tangent vec to r method.

I n t h e or thographic p r o j e c t i o n ( ~ i ~ u r e 3-A) a p l an view i s drawn showing t h e s t r i k e and d i r e c t i o n of apparent d ip . The apparent d i p i's r o t a t e d i n t o t h e plane of t h e paper around FL 112 and t h e apparent d i p angle of 20° i s l a i d o f f with a p r o t r a c t o r . A t any convenient d i s t a n c e along t h e f o l d i n g l i n e , such a s po in t A , a perpendicular i s dropped from FL 112 t o t h e dipping bed and t h e d i s t a n c e D i s measured. Folding l i n e 311 i s drawn a t r i g h t angles t o t h e s t r i k e of t h e bed and po in t A i s p ro j ec t ed FL 311 at r i g h t angles t o FL 311. The d i s t a n c e D i s measured i n view 3, t h e bed drawn i n and t h e angle of t r u e d i p (22') measured wi th a p r o t r a c t o r . The d i r e c t i o n of d i p i s a t r i g h t angles t o t h e s t r i k e o r due e a s t .

The tangent vec to r method i s used i n f i g u r e 3-B t o so lve t h e problem. I n t h i s s o l u t i o n t h e s t r i k e and d i r e c t i o n of apparent d i p a r e p l o t t e d i n t h e p l an view. A t a b l e of t r igonometr ic func t ions ( o r a s l i d e r u l e ) i s used t o o b t a i n t h e va lue of t h e tangent of 20' (0 .364) . Along t h e apparent d i p l i n e i n t h e p lan view 3.64 u n i t s ( a u n i t i s any conven- i e n t l e n g t h ) a r e l a i d o f f . A perpendicular i s dropped from t h e apparent d i p l i n e t o t h e l i n e r ep re sen t ing t h e bear ing of t r u e d i p and t h e d is - t ance measured from t h e i n t e r s e c t i o n t o po in t A (4 .0 u n i t s i n t h i s c a s e ) . The t a b l e of t r igonometr ic func t ions o r t h e s l i d e r u l e i s used t o f i n d t h e angle whose tangent i s 0.40. The t r u e d i p i s 21.8' due e a s t .

Determine True Dip from Two Apparent Dip Measurements a t Same Poin t

2bo apparent dips measured a t point A are: 30 0N40°E and 15°N150E. Find the angle and direct ion of t rue dip.

Figure 4 i s a s o l u t i o n t o t h i s problem by orthographic p ro j ec t ion . The two apparent d ips o r i g i n a t i n g a t po in t A a r e p l o t t e d i n t h e p lan view. FL 2 / 1 and FL 1 / 3 p a r a l l e l t o t h e two apparent d ips a r e drawn and t h e apparent d i p angles (15' and 30') a r e p l o t t e d .

A t any convenient po in t on FL 211 a perpendicular i s dropped t o t h e dipping bed and t h e d i s t a n c e D measured. This po in t on FL 2 / 1 i s pro- j ec t ed t o t h e bear ing of t h e apparent d i p ( p o i n t 1) i n t h e p lan view. The po in t where a perpendicular with a l eng th of D from FL 113 t o t h e dipping bed i s l oca t ed on FL 113 and p ro j ec t ed t o t h e p lan view a s po in t 2. Po in t s 1 and 2 i n t h e p lan a r e t h e l o c a t i o n of p o i n t s of t h e same e l e v a t i o n on t h e dipping bed. A l i n e connecting p o i n t s 1 and 2 i s t h e t r u e s t r i k e of t h e bed (Nh0w).

The t r u e d i p i s perpendicular t o t h e s t r i k e o r ~ 8 6 ' ~ . The amount of t r u e d i p i s found by l ay ing o f f t h e same d i s t a n c e D i n a view perpen- d i c u l a r t o and along t h e same s t r i k e l i n e def ined by p o i n t s 1 and 2. This i s shown i n view 4.

A l e s s c l u t t e r e d drawing f o r t h e s o l u t i o n of t h i s problem can be con- s t r u c t e d by us ing t h e l i n e s i n d i c a t i n g t h e d i r e c t i o n of apparent d ips and t r u e d i p a s t h e f o l d i n g l i n e s . This i s i l l u s t r a t e d i n Figure 5 .

Figure 6 i s a s o l u t i o n by t h e tangent vec to r method of t h e same problem. The p l an view i s drawn. The tangent of 15' i s .268 and t h e tangent of 30' i s .577. Therefore, 2.68 and 5.77 u n i t s a r e l a i d o f f a long t h e r e spec t ive apparent d i p bear ing l i n e s . Perpendiculars a r e drawn. A t r u e d i p bear ing l i n e i s drawn from poin t A t o t h e i n t e r - s e c t i o n of t h e two perpendiculars and t h e d i s t a n c e (8 .27 u n i t s ) measured t o g ive a t r u e d i p of 39.6ON86.5OE.

FIGURE 4.-True dip from two apparent dips.

FIGURE 5.-True dip f rom two apparent dips.

F I G U R E 6 -Tangent vector method

Three Poin t Problem

Any t h r e e p o i n t s on a p lane de f ine t h e l o c a t i o n of t h a t p lane i f t hey a r e not i n a s t r a i g h t l i n e . Therefore, t h e d ip and s t r i k e of a t r u e plane su r f ace can be determined from t h r e e po in t s . An example of t h e s o l u t i o n of t h i s type of problem fo l lows .

In a t e s t well a t point A a key marker bed i s encountered a t an eleva- t i on of 850 feet. Point B, the second t e s t well , i s 1000 feet due west of point A and the marker bed i s encountered a t an elevation of 620 feet . A t point C, 800 feet S 2 S 0 E from point B, a third t e s t we22 encounters t h i s marker bed a t elevation 720 feet. What i s the true dip and s t r ike of t h i s marker bed?

The l o c a t i o n of t h e t h r e e p o i n t s a r e p l o t t e d a t a convenient s c a l e i n t h e p lan view ( ~ i g u r e 7-A). Line AE i s drawn above AB (h ighes t and lowest po in t s ) and d i s t ances equiva len t t o t h e d i f f e r e n c e i n e l e v a t i o n between po in t s A and B l a i d o f f at a convenient s c a l e . A l i n e i s drawn from t h e 230 mark on AE t o po in t B ( d i f f e r e n c e i n e l e v a t i o n between A and B ) and a p a r a l l e l l i n e drawn from t h e 130 po in t t o l i n e AB. The i n t e r s e c t i o n on AB i s a t an e l eva t ion of 720, t h e same a s po in t C , and a l i n e connecting t h i s i n t e r s e c t i o n and po in t C i s t h e s t r i k e of t h e bed. FL 112 i s drawn at r i g h t angles t o t h e s t r i k e and p o i n t s A , C , and B a r e p ro j ec t ed perpendicular t o t h e FL. Poin t A i s on t h e FL (h ighes t p o i n t ) , po in t C i s 130 f e e t below, and po in t B i s 230 f e e t below. A l i n e drawn through t h e s e 3 p o i n t s de f ines t h e t r u e angle of d i p i n view 2.

Figure 7B i s an a l t e r n a t e s o l u t i o n of t h e same problem. Sec t ions from t h e h ighes t po in t ( A ) t o t h e o t h e r two p o i n t s ( B and C ) can be consi- dered a s two apparent d i p s and t h e problem solved a s descr ibed previous ly under two apparent d ips from t h e same po in t . I n t h e s o l u t i o n FL 112 and FL 311 a r e drawn and 130 f e e t l a i d o f f perpendicular t o FL 112 a t C and 230 f e e t perpendicular t o FL 311 at B. Poin t D i s t h e p r o j e c t i o n i n t h e p l an view where t h e bed i s 130 f e e t below po in t A i n view 3. Poin t C and po in t D a r e both 130 f e e t below A i n t h e p lan view and a l i n e connecting t h e s e p o i n t s de f ines t h e s t r i k e . FL 411 i s drawn through po in t A and perpendicular t o t h e s t r i k e and 130 f e e t s ca l ed o f f i n view 4 perpendicular t o FL 411 along t h e s t r i k e l i n e def ined i n t h e p lan view. This i s t h e angle of t r u e d i p ; t h e bear ing of t r u e d i p i s a t r i g h t angles t o t h e s t r i k e .

I n problems s i m i l a r t o t h e above i f t h e d i p angles a r e very sma l l , it i s d i f f i c u l t t o measure them wi th a p r o t r a c t o r . I f t h e s e angles a r e converted t o l i n e a r measurements ( f e e t l m i l e , f e e t l f e e t , e t c . ) t h e v e r t i c a l s c a l e can be exaggerated (10 , 100, e t c . , t imes ) t o provide a workable drawing. It i s important t o remember i f an exaggerated v e r t i c a l s c a l e i s used t h e d i p angles cannot be measured wi th a p r o t r a c t o r .

0 100 200 301 - Scale !,I feel

Problems Involving P o i n t s , L ines , and Planes

I n working problems involv ing p o i n t s , l i n e s , and p l anes , a l i n e i s assumed t o be s t r a i g h t throughout i t s course and a p lane i s assumed t o be a t r u e plane.

A l i n e , t o be shown i n i t s t r u e l eng th i n a view, must be p ro j ec t ed t o t h a t view by l i n e s of s i g h t t h a t a r e a t r i g h t angles t o t h e l i n e i n t h e f i r s t view. S t a t e d i n another way, t o p r o j e c t a l i n e t o a view where it w i l l be shown i n i t s t r u e l e n g t h , t h e f o l d i n g l i n e (FL) i s p a r a l l e l t o t h e l i n e i n t h e f i r s t view and t h e l i n e s of s i g h t a r e perpendicular t o t h e FL.

The t r u e s lope of a l i n e can be seen only i n an e l eva t ion view which shows t h e l i n e i n i t s t r u e l eng th . The t r u e s lope can only be p ro j ec t ed from t h e p lan view. The t r u e l eng th of a l i n e can be p ro j ec t ed from views o the r than t h e p lan view.

A l i n e w i l l appear a s a po in t i n a view taken a t r i g h t angles t o t h e l i n e shown i n i t s t r u e length .

A plane w i l l appear a s a l i n e i n t h e view i n which any l i n e i n t h e plane appears a s a po in t . Therefore, t h e t r u e d i r e c t i o n and angle of d i p of a p lane w i l l be shown i n t h e e l eva t ion view at r i g h t angles t o t h e s t r i k e .

Problems Involving Po in t s and Lines A Zine dips 20°iV200E and outcrops a t point A. Point B l i e s 1000 fee t IV6O0W from point A and i s 200 fee t lower. What i s the distance and

. slope i n a due eas t d i rec t ion from point B t o the l ine? What i s the shortes t distance, d irect ion, and slope from point B t o the Zine?

Figure 8 i s t h e s o l u t i o n f o r t h e d i s t a n c e and s lope i n a due e a s t d i r e c t i o n . The p l an view (view 1) i s drawn from t h e given da t a . Note t h a t l i n e B1-C1 i s i n a due e a s t d i r e c t i o n . FL 112 is p a r a l l e l t o dipping l i n e o r i g i n a t i n g s,t A and t h e angle of d i p (20') i s l a i d o f f . Poin t B2 i n view 2 i s 200 f e e t below A2 and po in t C2 i s on t h e dipping l i n e a s p ro j ec t ed from view 1. FL 311 i s drawn p a r a l l e l t o B1-C1 and t h e s e po in t s p ro j ec t ed t o view 3. The t r u e l eng th (1050 f e e t ) and t r u e s lope (+0.5') of t h e l i n e i s found i n view 3. Note: t h e d i s t ances from FL 311 t o po in t s B3 and C3 a r e equal t o t h e d is - tances from FL 112 t o po in t B2 and C2.

200 OL_____. Scale 111 feet

Figure 9 is the solution for the shortest distance from point B to the sloping line. The plan view is drawn except for point C1 which is unknown. View 2 is drawn. The shortest distance from point B to the sloping line is perpendicular to the sloping line in view 2. Perpendicular B2-C2 is drawn and point C projected to the plan view (view 1). Line B1-C1 is the true bearing of the shortest line, the true length and slope is found by projection to view 3.

Scale in feet

FIGURE 9.-Distance from a p ~ n t to a line.

Shortest Distance Between Two Non-ParallelYgNon-Intersecting Lines An incl ined Zine outcrops a t point A and dips 20' t rue north. Another l i ne outcrops a t point 150; f ee t N45OE &om point A and 500 f ee t lower and dips 30°W NW (N67.5'W). Find the azimuth, slope, and length of the shortes t l i ne connecting these two incl ined l ines and the distance from point A and point C t o the in tersec t ion of t h i s con- necting Zine.

The plan view showing the true azimuth of the two lines is drawn in view 1, Figure 10. Points B1 and Dl are arbitrary points plotted to provide two points on a line so the line may he projected ko other views. View 3 is drawn to show the true length and slope of line CD. View 2 is drawn to show the true length and slope of AB, and CD is also projected to this view. View 4 (FL 214) is projected perpendicular to line A2-B2. In view 4,sthis line is shown as point ~ 4 - ~ 4 .

The shortest distance from a line to a point is perpendicular to the line. In view 4, ~ 4 - ~ 4 is perpendicular to ~ b D 4 and is the shortest distance between line ~ 4 - ~ 4 and point ~ 4 ~ 4 . It is also shown in its true length (540 feet). Point ~4 is projected back to view 2 and the shortest distance ( ~ 2 - ~ 2 ) from a point ( ~ 2 ) to a line ( ~ 2 - ~ 2 ) is again perpendicular to the line. Line X2-Y2 is not, however, shown in its true length in this view. Points X2 and Y2 are projected to view 1 and line X1-Y1 is the true bearing (~46'~) of the intersecting line con- necting A1-B1 and C1-Dl. The true slope of this intersecting line (57 1/2O) is shown by projecting to view 5 (FL 115 is parallel to ~ l - ~ l ) and the true length is again shown and checks (540 feet) with view 4. As a further check on the accuracy of the drawing, view 6 perpendicular to line C3-D3 could be made and same procedure of projections repeated.

0 200 ,100 600 1000 -- 4 . - : - + : .I---

Scale 1 1 feet

Distance From a Poin t t o a Plane Points A, B, and C are three point s on a wlane. Point A h as an eleva- t i on of 300 feet . Point B ha> an elevati'on of 500 feet and i s 1500 feet due east of A. Point C has an elevation of 800 feet and i s 1750 feet from point A and 1500 feet from point B. Point E i s 300 feet N4S0W of point A a t an elevation of 300 feet . Find the distance from point E t o the plane i n the direction S4S0E with a plunge of 20'. Locate the piercing point of the l ine and plane and the angle between the l ine and plane.

The l o c a t i o n of t h e p i e rc ing po in t of t h e l i n e and p lane is found i n an edge view of t h e plane. The d i s t a n c e from po in t E t o t h e p i e rc ing po in t i s found i n t h e p r o j e c t i o n t h a t shows a t r u e view of t h e l i n e . The angle between t h e l i n e and p lane i s seen i n t h e view t h a t shows a t r u e view of t h e l i n e and an edge view of t h e plane.

F igure 11 i s t h e s o l u t i o n of t h i s problem. The p lan view (view 1) and view 2 ( ~ 4 5 ' ~ ) a r e drawn. The 20' angle i s l a i d o f f i n view 2 and a r b i t r a r y po in t F2 picked and p ro j ec t ed t o view 1. View 3 i s any v e r t i c a l s e c t i o n and drawn by p r o j e c t i n g p o i n t s A , B , and C and p l o t t i n g them a t t h e i r proper e l eva t ion . Po in t s E and F a r e p ro j ec t ed wi th measurements obta ined from view 2. A l e v e l ( s t r i k e ) l i n e B3-D3 p a r a l l e l t o F1 113 i s drawn i n view 3 and po in t D p ro j ec t ed t o t h e p lan view. Line B1-Dl i s t h e s t r i k e of t h e p lane . To f i n d t h e p i e rc ing po in t of t h e l i n e i n t h e p l ane , an edge view of t h e p lane i s needed. This i s done i n view 4 with FL 1 / 4 perpendicular t o t h e s t r i k e ( ~ 1 - ~ 1 ) determined i n view 1. Note t h a t i n view 4 t h e p lane a s def ined by p o i n t s A , B , and C must be extended t o l o c a t e t h e p i e r c i n g po in t ( ~ 4 ) . The p i e r c i n g po in t (~4) can be p ro j ec t ed back t o view 2 ( through view 1) t o show t h e l eng th of EP (150 f e e t ) i n i t s t r u e view.

The angle t h e l i n e EP makes wi th t h e plane can only be seen i n i t s t r u e p o s i t i o n i n a t r u e view of t h e l i n e and an edge view of t h e plane. This r equ i r e s two a d d i t i o n a l p ro j ec t ions . F i r s t view 5 i s drawn p a r a l l e l t o ~ 4 - ~ 4 - ~ 4 , t o show t h e plane i n a t r u e view and then view 6 i s drawn p a r a l l e l t o E5-P5-F5 t o show t h e l i n e i n a t r u e view and t h e plane a s an edge view. The d i s t a n c e ~ 6 - ~ 6 , t h e t r u e l e n g t h , checks wi th d i s t ance E2-P2 (150 f e e t ) and angle ~ 6 - ~ 6 - ~ 6 can be measured (36') which i s t h e t r u e angle between t h e l i n e and plane.

The nomenclature for fault displacements as used in this technical release is illustrated in Figure 12. This is the same nomenclature as used by Billings (1954).

CA net s h p DA d ~ p s l ~ p B A st r ike s l ~ p E A throw

ED heave Angle ACD rake Angle E C 4 plunge

FIGURE 12,-Nomenclature o f faul t displacement.

Determine t h e Line of I n t e r s e c t i o n of Two Oblique Planes I n the following example determine the bearing and plunge of the l i ne of in tersec t ion between two planes and the rake ( p i t c h ) of t ha t l i ne i n each plane. The given information i s : a bed dips 30°N100W and a faul t dips 20 OS45 OW.

I n Figure 1 3 t h e p l an view, view 1, i s drawn from t h e given information. Draw t h e s t r i k e of t h e bed and f a u l t and i n d i c a t e t h e d i r e c t i o n of dip. Next, views 2 and 3 a r e drawn perpendicular t o t h e s t r i k e and t h e amount of d i p i s p l o t t e d f o r each. A t an a r b i t r a r y d i s t a n c e X below and p a r a l l e l t o fo ld ing l i n e s 211 and 113 an a u x i l l a r y p lane i s drawn. This a u x i l l a r y p lane de f ines a common d i s t ance below t h e p lan view and remains cons tan t throughout t h e s o l u t i o n of t h e problem. Po in t s A and B a r e t h e p ro j ec t ions t o t h e p lan view of t h e i n t e r s e c t i o n of t h e f a u l t and t h e bed wi th t h e a u x i l l a r y p lane . These p o i n t s ( A and B) d e f i n e , i n t h e p lan view, a po in t on t h e f a u l t and t h e bed t h a t i s X d i s t ance beneath t h e su r f ace . A l i n e drawn from A t o C p a r a l l e l t o t h e s t r i k e of t h e f a u l t i s a s t r u c t u r e contour on t h e f a u l t plane. The l i n e from B t o C i s a l s o a s t r u c t u r e contour on t h e bed a t t h e same e l eva t ion a s t h e s t r u c t u r e contour 'on t h e f a u l t . These two s t r u c t u r e contours i n t e r s e c t a t po in t C. Po in t 0 i s t h e i n t e r s e c t i o n of t h e f a u l t and bed i n t h e p l an view and po in t C i s t h e p r o j e c t i o n i n t o t h e p l an view of t h e i n t e r s e c t i o n of t h e a u x i l l a r y p lane o r s t r u c t u r e contours . Two p o i n t s on t h e i n t e r s e c t i o n determine t h e bear ing of t h e i n t e r s e c t i o n , t h e r e f o r e , l i n e OC connecting t h e s e p o i n t s i s t h e bear ing ( ~ 7 9 ~ ~ ) of t h e i n t e r s e c t i o n of t h e two planes. The plunge of t h e i n t e r s e c t i o n i s determined i n a v e r t i c a l s ec t ion . FL 411 i s drawn p a r a l l e l t o N7g0W, po in t s 0 and C p ro j ec t ed perpendicular t o FL 411 and d i s t ance X l a i d o f f on t h e p r o j e c t i o n of po in t C and t h e angle measured.

To determine t h e rake ( p i t c h ) of t h e i n t e r s e c t i o n i n t h e p lane of t h e f a u l t , it i s necessary t o r o t a t e t h e f a u l t i n t o t h e p lan view. Use GH as a r ad ius and G a s t h e c e n t e r , draw an a r c t o i n t e r s e c t FL 113. This po in t on FL 113 i s t h e l o c a t i o n of po in t H when t h e f a u l t i s r o t a t e d i n t o a h o r i z o n t a l p o s i t i o n . A l i n e i s drawn from t h i s po in t on FL 113 p a r a l l e l t o t h e s t r i k e of t h e f a u l t ( t h i s l i n e i s a l s o a s t r u c t u r e contour l i n e X d i s t ance beneath t h e su r f ace r o t a t e d t o t h e su r f ace ) . A perpendicular from t h i s l i n e t o po in t C de f ines po in t D. This i s t h e same r e l a t i o n s h i p a s po in t A has on t h e GH a r c on FL 113. Poin t D i s t h e p r o j e c t i o n of po in t C when t h e f a u l t i s r o t a t e d i n t o t h e h o r i z o n t a l ( p l a n ) view. The rake of t h e i n t e r s e c t i o n of t h e s e two p lanes i n t h e p lane of t h e f a u l t i s measured between t h e s t r i k e of t h e f a u l t and l i n e OD, which i s on t h e p lane of t h e f a u l t r o t a t e d i n t o t h e p l a n view.

The rake of t h e i n t e r s e c t i o n i n t h e plane of t h e bed i s determined i n t h e same manner s t a r t i n g from view 2 and r o t a t i n g t h e a u x i l l a r y plane i n t o t h e ho r i zon ta l .

Displacement of a Vertical Fault

The displacement of a vertical fault can be determined if the attitude and location of two displaced horizons on each side of the fault are known. The location of additional horizons on one side of the fault can be found if their location on the other side is known.

Figure 14 is the graphical solution of the following problem: A ve r t i ca l faul t s t r i k e s east-west and i s exposed a t point A; a ve in with a dip of 30°S300E outcrops a t point B on the north s ide of the faul t and point C on the south side of the faul t . Another v e in with a dip of 45OS45OW outcrops a t point D on the north side of the fau l t and point E on the south s ide of the faul t . A th i rd vein with a dip of 20°S700W outcrops a t point F.

Point B i s 300 fee t north of A Point C i s 300 fee t south of A Point D i s 2500 fee t eas t of B P ~ i n t E i s 2500 fee t eas t of C Point F i s 2500 fee t eas t of E

Find the true displacement of the faul t and find continuation of th ird vein on north side of faul t .

The vertical fault and the six points are drawn in the plan view as shown in Figure 14. Through points B y C, D, E, and F strike lines are drawn for the veins and extended to intersect the vertical fault. The direction of dip is indicated on each strike line.

Next draw views 2, 3, and 4 with the folding lines perpendicular to the strike. The angle of dip is laid off in the proper direction in each view and an auxiliary plane "h" distance below the folding line is drawn. This h distance is the same wherever used in the solution of this problem. It represents the elevation of a structure contour line on the vein at h distance below the surface. Points M and L are the location projected into the plan view of the intersection of these structure contours and the vertical fault.

Next it is necessary to find the line of intersection of the veins on the fault. To do this the fault is rotated into the horizontal or plan view about its trace at the surface. Since this is a vertical fault the structure contour at h elevation on the fault when rotated into the horizontal will be h distance from the trace and is drawn as RR on Figure 14. Points M and L which are the location of the intersection of the structure contours of the veins and the fault must also be rotated into the horizontal. This is done by drawing perpendiculars from M and L to RR. Lines are drawn from J through the intersection of M on RR and K through the intersection of L on RR to their inter- section at S. These lines are the trace rotated into the horizontal of the veins on the south wall of the fault and S is their point of intersection. Lines parallel to JS and KS are drawn from the inter- section of the veins on the north side of the fault and the fault to their intersection at N. These lines are the trace of the veins on the

0 500 - Scale in feet

FiGJRE 14.- Displacement o f a v e r t i c a l faul t .

nor th w a l l of t h e f a u l t and N i s t h e i r i n t e r s e c t i o n . Since S and N were toge the r before f a u l t i n g l i n e SN i s t h e n e t s l i p . The d i p and s t r i k e components can be determined by cons t ruc t ing t h e h o r i z o n t a l and v e r t i c a l components a s shown.

Poin t N i s down and t o t h e e a s t of po in t S so t h e r e l a t i v e movement of t h e f a u l t i s t h e nor th block moved down and t o t h e e a s t i n r e l a t i o n t o t h e south block. The ne t s l i p i s 330 f e e t , t h e s t r i k e s l i p i s 90 f e e t , and the d i p s l i p i s 310 f e e t .

To f i n d t h e ex tens ion of t h e t h i r d ve in on t h e nor th s i d e of t h e f a u l t , view 4 i s drawn with angle of d i p and h d i s t ance l a i d o f f and po in t P found by .pro jec t ion . A perpendicular from P t o RR i s made and t h e l i n e from Q through t h e p r o j e c t i o n drawn. SN i n i t s proper o r i e n t a t i o n and l eng th i s t ransposed t o some convenient l o c a t i o n such a s S'N'. A l i n e p a r a l l e l t o S'Q i s drawn from N' t o t h e v e r t i c a l f a u l t . This i s t h e po in t where t h e ve in on t h e nor th s i d e of t h e f a u l t i n t e r s e c t s t h e f a u l t . The s t r i k e of t h e ve in i s drawn and po in t G on t h e ve in i s found t o be 2,280 f e e t e a s t of po in t D .

0 500 1- -

Scale In feet

FIGURE 14. - (repeated)

Displacement of an Inc l ined Fau l t

The g raph ica l s o l u t i o n of an i n c l i n e d f a u l t problem i s much t h e same a s wi th the v e r t i c a l f a u l t . Figure 15 i s the solut ion of the same problem as given for Figure 1 4 except i n t h i s problem the fau l t dips 45' south.

The p l an view i s l a i d out and views 2 , 3, and 4 drawn a s before . The d i s t ance h , an a r b i t r a r y d i s t a n c e below t h e fo ld ing l i n e s , de f ines an auxiliary plane o r s t r u c t u r e contour and h remains cons tan t wherever used throughout t h e problem.

View 5 of t h e f a u l t i s drawn, t h e angle of t h e f a u l t (45') and t h e h d i s t ance p l o t t e d . Line RR i s t h e t r a c e on t h e f a u l t of t h e s t r u c t u r e contour h d i s t a n c e below t h e su r f ace p r ~ j ~ e c t e d i n t o t h e p l an view. Poin ts M , L , and P a r e t h e p r o j e c t i o n i n t o t h e p lan view of t h e i n t e r - s e c t i o n of t h e s t r u c t u r e contours on t h e ve ins with t h e s t r u c t u r e contour on t h e f a u l t .

To f i n d t h e n e t s l i p of t h e f a u l t , t h e f a u l t must be r o t a t e d about W i n t o t h e h o r i z o n t a l o r p lan view. This i s done by swinging an a r c i n view 5 us ing t h e i n t e r s e c t i o n of VV and FL 115 a s t h e cen te r and t h e i n t e r s e c t i o n of h and t h e dipping f a u l t a s t h e r ad ius . Line TT i s drawn p a r a l l e l t o W through t h e po in t where t h e a r c i n t e r s e c t s FL 115.

Perpendiculars a r e dropped from p o i n t s M and P t o TT. The po in t S' i s def ined by t h e i n t e r s e c t i o n of l i n e s from J through t h e p r o j e c t i o n of M on TT and K through t h e p r o j e c t i o n of L on TT. N ' i s def ined by drawing l i n e s from t h e i n t e r s e c t i o n of t h e ve ins on t h e no r th s i d e of t h e f a u l t with W p a r a l l e l t o JS' and LS'. S'N' i s t h e n e t s l i p of t h e f a u l t (1050 f e e t ) .

The p r o j e c t i o n i n t o t h e p l an view of t h e n e t s l i p of t h e f a u l t i s SN. This i s found by drawing l i n e s J M and KL t o S and l i n e s p a r a l l e l t o JS and KS from t h e ve ins on t h e nor th s i d e of t h e f a u l t t o N. The r e l a t i v e movement along t h e f a u l t i s t h e no r th s i d e moved down and t o t h e e a s t i n r e l a t i o n t o t h e south s i d e .

To f i n d t h e plunge of t h e n e t s l i p , a view p a r a l l e l t o NS t o show NS i n i t s t r u e p o s i t i o n can be cons t ruc ted . This view can be moved t o an unc lu t t e r ed p a r t of t h e paper and cons t ruc ted i n t h e fol lowing s t e p s . Find t h e d i f f e r ence i n e l e v a t i o n between p o i n t s N and S. This i s accomplished by p r o j e c t i n g p o i n t s N and S p a r a l l e l t o VV t o t h e i r i n t e r - s e c t i o n wi th t h e f a u l t i n view 5 . Since N and S were t o g e t h e r before f a u l t i n g , p r o j e c t i n g t h e i r i n t e r s e c t i o n on t h e f a u l t t o l i n e W gives t h e i n t e r v a l 1, 2, which i s t h e i r d i f f e r e n c e i n e l eva t ion . On a sepa- r a t e p a r t of t h e paper l a y o f f t h e d i s t a n c e (850 f e e t ) SN. From N drop a perpendicular equal t o t h e d i f f e r e n c e i n e l e v a t i o n 1-2, t hen draw S-2. The plunge of t h e n e t s l i p i s 4b0, t h e n e t s l i p (S-2) i s 1050 f e e t and i s equal t o S'N'.

To f i n d t h e l o c a t i o n of t h e t h i r d ve in on t h e nor th s i d e of t h e f a u l t , a l i n e i s drawn from Q through t h e p r o j e c t i o n of P on TT. This l i n e i n t e r s e c t s S'N' a t S ' . A l i n e p a r a l l e l t o QS' i s drawn from N ' t o W. The s t r i k e of t h e t h i r d ve in i s drawn from t h i s i n t e r s e c t i o n on W. Poin t G on t h e t h i r d ve in i s found 3,620 f e e t e a s t of po in t D.

FIGURE 15 -Displacement o f an ~ n c l l n e i i f au l t

Stereographic P r o j e c t i o n

Stereographic p r o j e c t i o n i s a r a p i d method of so lv ing some geologic problems i f angles and s p a t i a l r e l a t i o n s between l i n e s and p lanes a r e needed. The fol lowing examples i l l u s t r a t e some of t h e uses of s te reo- graphic p ro j ec t ions .

The s te reographic o r Wulff meridional s t e r e o n e t i s shown i n Figure 16. Ex t r a copies a r e provided at t h e back of t h i s t e c h n i c a l r e l e a s e . I f a sphere wi th meridional o r g r e a t c i r c l e s and pole o r smal l c i r c l e s drawn two degrees a p a r t on i t s su r f ace w a s c u t i n h a l f through t h e p o l e s , Figure 16 i s a p r o j e c t i o n of t h e s e a r c s on t h e e q u a t o r i a l p lane . The bear ing of l i n e s o r planes i s measured from t h e nor th and south poles along t h e smal l c i r c l e s . The d i p of l i n e s and planes i s measured along t h e g r e a t c i r c l e s , t h e amount (deg rees ) of d i p be ing counted i n from t h e per iphery of t h e ne t a long t he east-west a x i s .

F I G U R E 16. - W~lff net

True Dip from Two Apparent Dips

I f , from a common point, two apparent dips are measured, 30°N400E and 15°N150E, detemine the bearing and amount of true dip.

Figure 17 i s t h e s o l u t i o n of t h i s problem. The s t e r e o n e t , Figure 16, i s t aped t o a desk o r drawing board and o v e r l a i d by t r a c i n g paper . The t r a c i n g paper is f a s t ened a t t h e c e n t e r of t h e n e t e i t h e r by a p i n o r a reversed thumb t a c k p laced beneath t h e n e t s o t h a t it may be r o t a t e d . The no r th , sou th , e a s t , and west p o i n t s on t h e perimeter of t h e n e t a r e marked on t h e t r a c i n g paper .

With t h e fou r c a r d i n a l compass po in t s marked on t h e t r a c i n g paper and i n t h e i r t r u e p o s i t i o n s wi th r e s p e c t t o t h e s t e r e o n e t , l i n e s i n d i c a t i n g t h e bear ing of t h e two apparent d ips ( ~ 1 5 ' ~ and ~ 4 0 ' ~ ) a r e drawn from t h e c e n t e r of t h e n e t t o t h e edge. Next t h e paper i s r o t a t e d so t h e ~ 4 0 ' ~ l i n e co inc ides wi th t h e e a s t l i n e of t h e n e t . The amount of d i p (30') i s counted i n from t h e perimeter and marked. The ~ 1 5 ' E l i n e i s t hen r o t a t e d t o t h e e a s t diameter and 15' counted i n from t h e per imeter and marked. The paper i s t hen r o t a t e d u n t i l t h e two apparent d ips (15' and 30') l i e on t h e same g r e a t c i r c l e . The g r e a t c i r c l e i s t r a c e d and t h e north-south ( s t r i k e ) and e a s t ( d i p ) diameter drawn. The amount of t r u e d i p i s 40' counted i n from per imeter of g r e a t c i r c l e on t h e e a s t diameter . The paper i s t hen r o t a t e d t o i t s o r i g i n a l p o s i t i o n wi th t h e two apparent d ips i n t h e N15OE and ~ 4 0 ' ~ d i r e c t i o n and t h e bear ing of t r u e d i p i s read a s ~ 8 6 ' ~ .

FIGURE 17.-True d ip from two apparent d ips.

Amarent Dix , from True D i x ,

When drawing cross s e c t i o n s and o t h e r type i l l u s t r a t i o n s , it i s not always poss ib l e t o draw them perpendicular t o t h e s t r i k e . I n t h e s e cases t h e apparent d i p should be p l o t t e d , not t r u e d ip . The s t e r eo - ne t i s a f a s t method of ob ta in ing apparent d ips when t h e t r u e d i p i s known.

The fol lowing problem i s an example. A bed dips 30°!i400W. What i s the apparent dip i n the S70°W direct ion?

Figure 18 i s t h e s o l u t i o n of t h i s problem. The bear ing of t r u e d i p (~40'~) and d i r e c t i o n of apparent d i p (~70'~) a r e p l o t t e d ; t h e bear ing of t r u e d ip i s r o t a t e d t o t h e west diameter ; 30' counted i n from the per imeter ; and t h e g r e a t c i r c l e i s drawn. The paper i s then r o t a t e d s o t h e l i n e of t h e bear ing of t h e apparent d i p d e s i r e d (~70'~) i s on t h e west diameter and t h e amount of apparent d i p (11') counted i n from t h e perimeter .

N

I

W - - E

I S

FIGURE 18 - Apparent d ~ p froni true d r ~ p

Line of I n t e r s e c t i o n of Two Oblique Planes

A bed dips 30°N1 O O W and a faul t dips 20 OS45 O W . What i s the bearing and plunge the l i ne of in tersec t ion betueen the bed and the fau l t and what i s the rake of t h i s l i ne i n the plane of the bed and i n the plane of the fau l t?

Figure 19 i s t h e s o l u t i o n . The bea r ing of t h e d i p of t h e bed and t h e f a u l t a r e drawn. The N I O O W l i n e i s r o t a t e d t o t h e west diameter and 30° counted i n and t h e g r e a t c i r c l e drawn. This i s r epea t ed f o r t h e ~ 4 5 ~ ~ l i n e . A l i n e from t h e c e n t e r of t h e n e t through t h e po in t of i n t e r s e c t i o n of t h e two g r e a t c i r c l e s i s t h e l i n e of i n t e r s e c t i o n of t h e two p lanes . Rota t ing t h e t r a c i n g paper t o i t s o r i g i n a l p o s i t i o n t h e bear ing of t h e i n t e r s e c t i o n i s ~ 8 1 ~ ~ . Rota te t h e paper so t h e N81°w l i n e i s on t h e west diameter and t h e plunge of t h e i n t e r s e c t i o n counting i n from t h e per iphery i s 12'.

To determine t h e rake of t h e l i n e of i n t e r s e c t i o n i n t h e p lane of t h e f a u l t r o t a t e t h e t r a c i n g paper s o t h e s t r i k e of t h e f a u l t i s a long t h e north-south a x i s . The rake (36') i s found by counting t h e smal l c i r c l e s from t h e no r th pole along t h e g r e a t c i r c l e of t h e f a u l t t o t h e po in t of i n t e r s e c t i o n determined above. To determine t h e rake of t h e l i n e f o r t h e bed, t h e s t r i k e of t h e bed i s p laced on t h e north-south a x i s and t h e angle found by aga in counting t h e smal l c i r c l e t o t h e po in t of i n t e r s e c t i o n .

FIGURE 19 - In tersect ion of t w o obl ique planes

Rota t ion of a Bed

Occasional ly it i s d e s i r a b l e t o r e s t o r e t o t h e i r o r i g i n a l p o s i t i o c beds , f a u l t s , and j o i n t s t h a t have been r o t a t e d . This i s e a s i l y and quickly done by s te reographic p ro j ec t ion . The fol lowing problem i s an example.

!lbo beds are separated by an unconformity. The top bed dips 15°S200E and the lower bed dips 40°N200E. Find the dip and s t r i k e o f the lower bed when the top bed was horizontal (being deposited).

I n Figure 20 t h e g r e a t c i r c l e s f o r t h e d i p and s t r i k e of t h e two beds a r e drawn as i n t h e previous examples. To f i n d t h e d i p and s t r i k e of t h e lower bed when t h e t o p bed was h o r i z o n t a l it i s necessary t o r o t a t e t h e t o p bed i n t o t h e h o r i z o n t a l and t h e bottom bed through t h e same amount of r o t a t i o n . This i s accomplished by moving t h e paper s o t h e s t r i k e l i n e of t h e t o p bed i s on t h e north-south diameter . A l l po in t s on t h e g r e a t c i r c l e of t h e t o p bed when it i s r o t a t e d 15O i n t o t h e h o r i z o n t a l w i l l f a l l on t h e perimeter of t h e ne t . This inc ludes po in t A which i s t h e i n t e r s e c t i o n of t h e two g r e a t c i r c l e s . Likewise a l l o t h e r po in t s on t h e g r e a t c i r c l e of t h e lower bed w i l l a l s o r o t a t e 15' a long t h e small c i r c l e s . A few of t h e s e p o i n t s a r e i nd ica t ed by t h e dashed l i n e s . Poin t A i s a p o s i t i o n of zero d i p , so it i s r o t a t e d t o t h e no r th pole . A g r e a t c i r c l e i s drawn from po in t A and connecting t h e ends of t h e dashed a r c s . This g r e a t c i r c l e (dashed) r e p r e s e n t s t h e p o s i t i o n of t h e lower bed ( 5 2 O ~ 1 2 ' ~ ) when t h e t o p bed was ho r i zon ta l .

Rota t ion of a Fau l t

I f t h e same bed has a d i f f e r e n t d i p and s t r i k e on oppos i te s i d e s of a f a u l t , r o t a t i o n along t h e f a u l t has occurred. I n t h e previous example r o t a t i o n of t h e bed was about a h o r i z o n t a l a x i s . To so lve problems of r o t a t i o n about an i n c l i n e d a x i s , an a d d i t i o n a l s t e p of r o t a t i n g t h e a x i s i n t o a h o r i z o n t a l o r v e r t i c a l p o s i t i o n be fo re r o t a t i n g t h e beds i s requi red .

The fol lowing problem i s an example. A faul t dips 30°N200E. A bed i n the south block dips 1 0 " ~ 2 0 ~ ~ , and a bed i n the north block dips 28°N300W. What has been the rotat ion of the north block with respect t o the south . block (angle and clockwise or counter-clockwise) and i s the faul t moue- ment simple rotation?

The problem can be solved i n t h r e e s t e p s . F i r s t t h e f a u l t p lane i s r o t a t e d i n t o t h e h o r i z o n t a l , t h e beds a r e moved through t h e same angle of r o t a t i o n , t h e angle of r o t a t i o n between t h e beds can then be measured. Second t h e f a u l t p lane i s r o t a t e d t o t h e v e r t i c a l and t h e beds aga in move t h e same amount. Third t h e no r th bed i s r o t a t e d through t h e angle of r o t a t i o n determined i n t h e f i rs t s t e p about an a x i s perpendicular t o t h e f a u l t plane t o s e e i f it coinc ides with t h e south bed. If it does, t h e f a u l t movement i s simple r o t a t i o n .

I n Figure 21A t h e g r e a t c i r c l e s r ep re sen t ing t h e a t t i t u d e and d i p of t h e two beds and t h e f a u l t a r e p l o t t e d as before . Rota t ion has occurred along t h e f a u l t , t h e r e f o r e , t h e a x i s of r o t a t i o n of t h e f a u l t must be perpendicular t o t h e f a u l t . To measure t h e angular d i f f e r e n c e ( ang le of r o t a t i o n ) between t h e two beds t h e a x i s o r r o t a t i o n of t h e f a u l t i s r o t a t e d t o t h e v e r t i c a l ( f a u l t r o t a t e d t o h o r i z o n t a l ) . The two beds a r e r o t a t e d through t h e same angle . To do t h i s t h e s t r i k e of t h e f a u l t i s placed on t h e north-south diameter and t h e d i p on t h e e a s t diameter. When t h e f a u l t i s r o t a t e d 30' i n t o t h e h o r i z o n t a l i t s g r e a t c i r c l e co inc ides wi th t h e per iphery of t h e n e t . With t h e t r a c i n g paper h e l d i n t h e same p o s i t i o n t h e two g r e a t c i r c l e s r ep re sen t ing t h e two beds a r e r o t a t e d 30' i n t h e same d i r e c t i o n a s i nd ica t ed by t h e dashed l i n e s . Great c i r c l e s , i nd i ca t ed by t h e hachure l i n e s , a r e found by r o t a t i n g t h e t r a c i n g paper u n t i l t h e po in t s p ro j ec t ed by do t t ed l i n e s l i e on t h e same g r e a t c i r c l e . The angle between t h e two beds can be measured, a s i n d i c a t e d , on t h e per iphery of t h e n e t .

FIGURE 21 A.-Rotation about an incl ined ax is .

Figure 21B shows t h e second and t h i r d s t e p s . I n t h e second s t e p t h e f a u l t i s r o t a t e d t o a v e r t i c a l p o s i t i o n , t h e a x i s of r o t a t i o n of t h e f a u l t w i l l then be h o r i z o n t a l and t h e beds can then be r o t a t e d about t h e h o r i z o n t a l a x i s . Rotat ion of t h e f a u l t t o t h e v e r t i c a l i s oppo- s i t e of t h e h o r i z o n t a l r o t a t i o n done i n t h e f i r s t s t e p . The s t r i k e of t h e f a u l t ( o r i g i n a l p l o t t e d p o s i t i o n aga in ) is p laced on t h e north- south a x i s and d i p on t h e e a s t diameter . To r o t a t e t h e f a u l t t o t h e v e r t i c a l it i s r o t a t e d 60' down ( e a s t t o w e s t ) , t h e g r e a t c i r c l e w i l l t hen co inc ide wi th t h e north-south a x i s of t h e n e t . With t h e t r a c i n g paper he ld i n t h e same p o s i t i o n t h e two beds a r e r o t a t e d 60° i n t h e same d i r e c t i o n . Thei r new l o c a t i o n s a r e i n d i c a t e d by t h e hachured g r e a t c i r c l e s .

The t h i r d s t e p involves r o t a t i n g t h e nor th bed through t h e angle determined i n t h e f i r s t s t e p about t h e axis of r o t a t i o n normal t o t h e f a u l t . To do t h i s t h e t r a c i n g paper i s r o t a t e d so t h e s t r i k e of t h e f a u l t i s placed on t h e east-west diameter . The nor th bed i s r o t a t e d 4 7 O a s i nd ica t ed by t h e do t t ed l i n e s .

When r o t a t e d through t h e angle of 47' t h e n o r t h bed co inc ides wi th t h e south bed, t h e r e f o r e , t h e movement of t h e f a u l t has been simple r o t a t i o n of 47' of t h e no r th block counter-clockwise wi th r e spec t t o t h e south block.

Poles

Rotat ion of beds on t h e s t e r eone t can be made more expedi t ious ly by us ing po in t s r ep re sen t ing t h e poles of planes i n s t e a d of t h e p lanes themselvt :~. The pole of a plane i s a l i n e perpendicular t o t h e plane and passing through t h e cen te r of t h e s t e r e o n e t . Every p lane , repre- sen ted by a g r e a t c i r c l e on t h e s t e r e o n e t , has a unique po in t a l s o on t h e ne t t h a t r ep re sen t s t h e pole of t h e plane.

Figure 22 i l l u s t r a t e s t h e r e l a t i o n s h i p between po le s and p lanes . Figure 22A i s a t h r e e dimensional drawing of t h e lower r e f e rence hemisphere. A p lane (rock s t r a tum) def ined by p o i n t s N B $ 0 d i p s 45O due e a s t . Line A0 perpendicular t o t h e p lane and pass ing through t h e cen te r of t h e n e t i s t h e pole . P ro j ec t ing t h e pole and p lane t o t h e s t e r eone t i n Figure 21B t h e plane i s def ined by t h e g r e a t c i r c l e N C S and t h e pole by po in t P. I f t h e plane had dipped 3 0 ~ ~ 4 5 ' ~ t h e l o c a t i o n of t h e po in t de f in ing t h e pole would be 60' i n from t h e

- periphery o r 30' out from t h e c e n t e r of t h e n e t and have a ~45'~ bear ing .

B

0 FIGURE 22.-Relationship between planes and poles.

Rota t ion of a Bed

The problem i l l u s t r a t e d i n Figure 20 can be so lved us ing poles i n s t e a d of planes. The problem r e i t e r a t e d is : Tuo beds are separated by an unconfomity. The top bed dips 1S0S20 O E and the lower bed dips 40 ON20 O E .

Find the dip and s t r i k e of the Zower bed when the top bed was horizontal .

The pole f o r t h e t o p bed i s l o c a t e d 15' out from t h e cen te r of t h e n e t (75' i n from per iphery) wi th a bear ing of N20°w. The pole f o r t h e lower bed i s 40' ou t from t h e cen te r of t h e n e t (50' i n from per iphery) with a bear ing of ~ 2 0 ' ~ . Bearings a r e l a i d o f f on t h e per imeter of t h e n e t and d ip counted o f f on t h e g r e a t c i r c l e s on t h e east-west diameter and t h e l o c a t i o n of t h e two poles a r e p l o t t e d . Poin t T i s t h e pole of t h e t o p bed and po in t L t h e lower bed. To determine t h e a t t i t u d e of t h e lower bed when t h e t o p bed was h o r i z o n t a l t h e t o p bed must be r o t a t e d i n t o t h e ho r i zon ta l . To accomplish t h i s t h e t r a c i n g paper i s r o t a t e d u n t i l po in t T i s on t h e east-west diameter (west s i d e of c e n t e r ) . When t h e bed i s r o t a t e d i n t o t h e h o r i z o n t a l t h e pole w i l l be v e r t i c a l ; t h e r e f o r e , p o i n t T moves 15' t o 0 and, without moving t h e t r a c i n g paper , po in t L i s r o t a t e d 15' a long t h e smal l c i r c l e i n t h e same d i r e c t i o n a t T t o po in t L1. The po in t L1 i s t h e pole of t h e lower bed when t h e t o p bed was h o r i z o n t a l . The bea r ing of p o i n t L1 i s S 1 2 ' ~ and t h e d i p counted i n from t h e per iphery along t h e east-west diameter i s 38'. The a t t i t u d e of t h e lower bed, t h e r e f o r e , i s 52' (90'-38') ~12'~.

FIGURE 23 - R o t a t ~ o n using poles

Rota t ion of a Fau l t

The problem solved i n Figure 21 ( r o t a t i o n about a f a u l t ) can a l s o be solved us ing p o i n t s r ep re sen t ing t h e poles of p lanes . Figure 24 i s t h e s o l u t i o n us ing t h i s method.

As given be fo re , a faul t dips 30 O h 7 2 0 O E , a bed i n the south block dips 10°N200W, and a bed i n the north block dips 2 8 O N 3 O O E . What has been the ro ta t ion of the north block wi th respect t o the south block and i s the faul t movement s i p Z e rotat ion.

The pole of t h e f a u l t (PF) i s ~ O ~ S ~ O O W ( a l l angles of d i p counted i n from per iphery of n e t ) , t h e pole of t h e south bed (PSB) is 80°S200E, and t h e pole of t h e nor th bed (PNB) i s 620s30°~ .

The po le s a r e p l o t t e d on t h e t r a c i n g paper a s shown i n Figure 24. To r o t a t e t h e f a u l t i n t o t h e h o r i z o n t a l ( a x i s of r o t a t i o n v e r t i c a l ) , t h e t r a c i n g paper i s r o t a t e d u n t i l t h e pole of t h e f a u l t (PF) i s on t h e east-west diameter (west s i d e ) . When t h e f a u l t i s r o t a t e d i n t o t h e h o r i z o n t a l t h e pole (PF) w i l l move 30' t o t h e c e n t e r of t h e n e t . Po in t s PSB and PNB w i l l a l s o move 30' along t h e i r r e s p e c t i v e smal l c i r c l e s a s shown by dashed l i n e s t o p o i n t s PSBH and PNBH. The angle of r o t a t i o n of t h e f a u l t i s measured between l i n e s from t h e c e n t e r through p o i n t s PSBH and PNBH. This angle can be convenient ly counted along t h e smal l c i r c l e s on t h e per iphery as ind ica t ed .

To r o t a t e t h e f a u l t i n t o t h e v e r t i c a l po in t PF i s aga in loca t ed on t h e west r ad ius of t h e n e t and r o t a t e d out 60' o r u n t i l po in t PF i s on t h e per iphery of t h e n e t . The p o i n t s PSB and PNB a r e r o t a t e d along t h e i r r e s p e c t i v e smal l c i r c l e s through t h e same 60° of r o t a t i o n t o p o i n t s PSBV and PNBV.

To determine i f simple r o t a t i o n has occurred t h e nor th bed must be r o t a t e d 47O about t h e a x i s of r o t a t i o n which i s normal t o t h e f a u l t . This i s accomplished by r o t a t i n g t h e t r a c i n g paper u n t i l po in t PF i s on t h e no r th r ad ius of t h e n e t . Poin t PNBV i s r o t a t e d 47' a long i t s smal l c i r c l e where it coinc ides wi th po in t PSBV confirming t h e movement was simple r o t a t i o n and t h a t t h e nor th block was r o t a t e d 47' counter- clockwise wi th r e spec t t o t h e south block.

FIGURE 24 -Rotat ion about an i n c l ~ n e d ax is

Vertical Drill Holes

The stereographic technique can be used to solve dip and strike problems involving unoriented cores from drill holes. The following is an example.

Points A and B are the locations of two ve r t i ca l core t e s t holes. The top of a key marker bed i s encountered a t e levat ion 157 .5 i n hole A and the core obtained shows t ha t the bed dips a t an angle of 45O. Since the core has been rotated i n the core barrel , the d irect ion of dip i s unknown. Hole B is located 100 fee t N60°W of hole A and the key marker bed m s encountered a t e levat ion 100.0. What i s the a t t i t ude of the key marker bed?

From the information given (100 feet horizontally and 57.5 feet verti- cally) the apparent dip ( 3 0 ~ ~ 6 0 ~ ~ ) from A to B of the key marker bed can be determined either trigonometrically or graphically. With an apparent dip and bearing and the true dip known, two possibilities of the bearing of true dip can be found. More information, such as a third test core hole, is necessary to provide the unique solution of the bearing of true dip.

On the tracing paper overlying the stereonet plot the vector representing the direction and amount of dip from point A to B (30~~60'~). This is 0-AB in Figure 25. Next rotate the tracing paper until the end of the vector (point AB) lies on a great circle representing 45O of dip. There are only two great circles of 45' dip that point AB will fall on as shown in Figure 25. Only one of these gives the bearing of true dip, but until more information is provided we cannot determine which one. The two possibilities of true dip are 45O~68'~ and 45'~7OW.

S

FIGURE 25 -True d ~ p from ve r t~ca l core holes.

I n c l i n e d D r i l l Hole

As another example, consider the same pmb lem except only the e levat ion of the top of the bed a t 257.5 i n hole A i s known and hole B i s incl ined from the ve r t i ca l 40' i n a S60°W direct ion (dips 50 'S60°W) and the beds make an angle of 45' with the core ax i s (d ip 45'1.

This i s e s s e n t i a l l y t h e same problem a s t h e previous one except it w i l l a l s o involve r o t a t i o n . F i r s t p l o t t h e vec to r s r ep re sen t ing d i r e c t i o n and d i p of t h e bed from A t o B and t h e d i r e c t i o n and d i p of t h e d r i l l ho le a t B. These a r e p o i n t s AB and DH on Figure 26. Next, t h e i n c l i n e d d r i l l ho le i s r o t a t e d t o t h e v e r t i c a l . The t r a c i n g paper i s r o t a t e d s o t h a t po in t DH i s on t h e west r ad ius . When DH i s r o t a t e d t o t h e v e r t i c a l it moves t o 0 and po in t AB moves through 40' t o ABV. Next, as i n t h e previous problem, r o t a t e t h e t r a c i n g paper and draw t h e two 45' g r e a t c i r c l e s through ABV. The l a s t s t e p i s r o t a t e t h e p r o j e c t i o n back t o i t s ' o r i g i n a l p o s i t i o n . P lace t h e o r i g i n a l bear ing of DH on t h e west r ad ius and r o t a t e DH from 0 t o 40'. The two g r e a t c i r c l e s through ABV w i l l a l s o r o t a t e 40' t o t h e p o s i t i o n s i n d i c a t e d by t h e hachured g r e a t c i r c l e . These two g r e a t c i r c l e s a r e t h e two p o s s i b i l i t i e s of d i p and s t r i k e of t h e key bed. When t h e t r a c i n g paper i s r o t a t e d t o i t s o r i g i n a l p o s i t i o n over t h e n e t , t hey a r e 4 2 ' ~ l l ' ~ and 70°N18'E.

S

FIGURE 25.-True d i p fro111 an ~ r l c l ~ r l e d core l i o le

Combination Orthographic and Stereographic Technique

When solving problems involving displacement of non-rotational faults a combination of orthographic and stereographic procedures are often- times simpler to use than straight orthographic projections. The inclined fault problem in Figure 15 can be solved in the following manner by using this combination method.

Figure 27 is the plan view. The third vein at point F on the south side of the fault-has not been shown.

The great circles representing the fault and on the stereonet in Figure 28. Great circle BXB' the vein Located at points B and C, and points D and E. Line XO on the stereonet is of the trace of the intersection of the vein

the two veins are plotted EXYW represents the fault,

L

DYD' the vein.located at the horizontal projection at B and C with fault.

Likewise, line OY is the trace of the intersection of the vein at D and E with the fault.

To find the horizontal projection of the net slip, plot the bearings (from the stereonet) of line OX and OY on the plan view. Lines XS and X'N are the bearing of OX (90' - 59' = 31°), and YS and Y ~ N are the bearing of OY (90' - 23' = 67'). Their intersection is at S and N. Since points S and N were together before faulting, SN is the horizontal projection of the net slip (850 feet). The bearing of the net slip (SN) is SlgOE. This bearing is plotted on the stereonet as 0-SN. By rotating 0-SN until SN is on the South pole and counting in on the small circles to the intersection of 0-SN with the great circle of the fault the plunge of the net slip is 44'.

Next, rotate the tracing paper and place the strike of the fault on the north-south diameter. Counting the angle in on the small circles, EX (38O) is the rake of vein B in the plane of the fault and WY (74') is the rake of vein D in the plane of the fault. The rake of vein B is southeast and vein D southwest.

To determine the total net slip, return to the plan view (Figure 27) and rotate the fault into the horizontal. Since points X' , X, Y, and Y' are at the surface, they do not move when the fault is rotated. The angle of rake of a fault is measured in the plane of the fault (see Figure 12). If the angle of rake, as net, of the veins on the fault ar.e plotted will be rotated into the horizontal. From rake angle (38' ) in a southeast direction, angle (74') in a southwest direction. The at S' and N~ determines line SIN' which is the fault .

determined from the stereo- in the plan view, the fault points X' and X lay off the and at Y and Y' the rake intersection of these lines the net slip (1150 feet) of

0 200 400 . . J Scale In feet

FIGURE 2 7 , - I n c h e d fau l t problem

References

Billings, M. P., 1954, Structural Geology: Prentice-Hall, Inc., 514 pp., illus.

Bucher, W. H., 1944, The Stereographic Projection, A Handy Tool for the Practical Geologist: Journal of Geology, Vol. 52, pp. 191-212.

Donn, W. L., and J. A. Shirner, 1958, Graphic Methods in Structural Geology: Appleton-Century-Crafts, Inc., 180 pp., illus.

Fisher, D. J., 1938, Problem of Two Tilts and the Stereographic Projection: Bull. Amer. Assoc. Pet. Geol., Vol. XXII, pp. 1261-71.

LeRoy, L. W., and J. W. Low, 1954, Graphic Problems in Petroleum ~ e b l o ~ ~ : Harper & Brothers, 238 pp. , illus.

Nevin, C. M. , 1949, Principles of Structural Geology: John Wiley & Sons, Inc. , 410 pp. , illus.

ir U. S. GOVERNMENT PRINTING O F F I C E : 1980-626-988/ Lo