NEW GRAPHIC METlfOD FOR DETERMINING THE DEPTH AND THICK~ 

NESS OF STRATA AND THE PROJECTION OF DIP. 

By HAROLD s. pALMER. 

INTRODUCTION. (NoTE.·--Diagram I may be used not only 

Geologists, b9th in the field and in the office, for the problem above stated but also to 

frequently encounter trigonometric problems obtain the difference in elevation between two 

the solution of which, though si1nple enough, points by omitting E and letting D equal the 

is somewhat laborious by the use of trigono- difference in elevation, H the horizontal dismetric and l~garithmic tables. Charts, tables, tance, and A the vertical angle.) · 

and diagrams of various types for facilitating II. To find the thickness of a stratum, given 

the computations have been published, and the dip of the stratum, the horizontal distance 

a new method may seem to be a superfluous across its outcrop at right angles to the strike, 

addition to the literature. llowever, it is felt and the difference in elevation of the upper 

that the simplicity and accuracy of diagrams and lower boundaries of the stratum. There 

FIGURE 15.-Diagram showing three cases in computing depth. 

suph as are presented in this paper warrant the 

addition. The alinement diagram. is used to a 

very considerable extent by engineers, especially mechanical engineers. The principle of 

the alinement diagram is explained herein, 

but rigorous demonstration is not attempted. 

Alinement diagrams are given for the following problems, which are numbered to agree 

with the numbering of the diagrams. The 

charts published herewith are suitable for use. 

I. To find the depth to a stratum, given the 

dip of the stratum, the horizontal distance 

from the outcrop, and the difference in elevation between the outcrop and the point at 

which the depth is desired. (See fig. 15.) 

The formula is 

D=HtanA ± E ________ ____ (1 

in which D =depth to stratum, H =horizontal 

distance frorn the outcrop, A= dip, E = difference in elevation. 

are three cases, ~pending Otn the relation of the 

dip of the beds to the slope of the ground. 

(See fig. 16, a, b, and c.) 

Case 1, where the ground is level. By inspection of figure 16, a, it is seen that the 

fornmla is 

T= H sin A ____________ (2) 

in which T = thickness, H = horizontal distance 

from outcrop, and A= dip. 

Case 2, where the ground slopes opposite to 

the dip. In figure 16, b, 

but 

Now 

or, ·as 

T=m-+-l 

m=H sin A 

l=E cos A 

cos X ·= sin (90°- X) 

Z=E sin (90°-A) 

123 
124 SHORTER CONTRIBUTIONS TO GENERAL GEOLOGY, 1918. 

Therefo.re 

T = H sin A+ E sin (90° -A) _____ (3) 

in which T, H, and A have the same ~alues as 

before, and E is the difference in elevation. 

Case 3, where the ground slopes with the 

dip. In figure 16, c, T=m- l_; making .the same 

substitutions as in case 2, we find 

T=H sin A- E sin (90° -A) _____ (4) 

in which T, H, A; and E have the same values 

as before. 

H 

~ / T 

a 

c 

plane of a supposed structure section, is desired. C is the desired projection of A. B is 

the angle between the plane of the section and 

the strike of the beds~ and angle PFS is a 

tight angle. The plane PFS is horizontal 

and the lines PQ, FG, and SR are normal to it 

and of equal length, so that the triangle QGR 

is also horizontal and is equal to PFS. 

Now, 

FS=PS sin B 

FS=FG cot A 

PS=PQ cot C 

Therefore 

FG cot A=PQ cot C sin B 

but · 

FG=PQ 

so 

cot A=cot C sin B · 

and, as the tangent is the reciprocal of the cotangent, 

1 1 . B 

tan A= tan C sin 

tan C = tan A sin B ___ ( 5) 

FIGURE 16.-Diagram showing three cases in computing thickness. 

in which C =dip of the projected 

angle, A = dip of the bed, B = 

angle between strike and pl~ne 

of section. 

The above formula is strictly true only where 

the dip of the beds is greater than the slope of 

the ground. (See fig. 16, d.) Where the 

dip is less than the slope the formula becomes 

T=E sin (90° -A) ~H sin A _____ (4a) 

The difference is algebraic and not arithmetic, 

so that the formula is practically workable for 

either subcase. 

III. Projection of dips-that is, to find the 

slope of the trace of an inclined plane upon a 

vertical plane oblique to the strike. Consider 

figure 17, a, which is a plan or map; at A is. an 

inclined plane the strike and dip of which are 

shown in the conventional way, and along 

S-P is a vertical plane up~n which this dip is to 

be projected. B is the angle between the 

strike and the section plane. In figure 17, b, 

A is a dip whose projection on SPQR, the 

USE OF THE ALINEMENT DIAGRAMS. 

The alinement diagram is worked in an 

extremely simple way. The points on the 

outer scales representing the given quantities 

are joined by a line, and the unknown q uantity, or result, is read at the intersection with 

the middle scale. Interpolation between the 

values marked on the scales is effected just as 

on a slide rule. · It is suggested that the edge 

of a draftsman's transparent triangle or a piece 

of slender black thread be used for projecting 

the cross line. · 

USE OF THE DEPTH DIAG~AM. 

In using the depth diagram (Pl. XIV and 

problem I), find the point on the left-hand scale 

that represents the distance from the outcrop 

to the point at which the depth is desired, and 

join it to the point on the right-hand scale that 
U. S. GEOLOGICAL SURVEY 

Distance 

1,000 

900 

800 

700 

500 

400 

300 

200 

PROFESSIONAL PAPER 120 PLATE XIY 

' ' 

Depth 

to bed 

10,000 

2,000 

1,000 

700 

500 

400 

300 

...... 

zoo 

100 

70 

50 

40 

30 

20 

10 

I 

1 J 

ALINEMENT DIAGRAM I, FOR COMPUTING DEPTH. 

Dip 

DETERMINING DEPTH AND THJ:CKNESS OF STRATA AND PROJECTION OF DIP. 125 

represents the dip of the beds. At the intersection of the line determined by these points 

with the middle scale the depth may be read. 

The dotted line on Plate XIV shows the solution when the distance is 600 feet and the dip 

20°. The depth to the stratum is found to be 

2.20 feet. 

The above rule holds only if the surface is 

level. If the point at which the depth is to be 

determined is higher than the outcrop-, the 

difference in their elevation is added to the 

depth as read by the above-described method. 

If the outcrop is the higher of the two points, 

the difference of elevation is subtracted from the reading. 

The diagram may be used for 

the converse problem, namely, to 

find the dip of the bed when the 

depth and distance from the outcrop are given. In this case the 

given quantities are set on the 

middle and left scale, and the line 

joining them is produced till it 

cuts the right scale, at which the 

dip is read. Another converse 

problem arises when the distance 

from the outcrop is unknown but 

the dip and depth to the stratum 

the point on the left scale that represents the 

difference in elevation of the higher and lower 

boundaries of the outcrop, and at the intersection of the connecting line with the middle 

scale read the correction. The correction is 

added or subtracted as directed above. 

In Plate XV the broken line shows the ~t 

step in solving the problem when the horizontal width of the outcrop is 600 feet and the 

dip is 20°. The thickness is read on the middle 

scale as 206 feet. The dashed line shows the 

correction to be introduced if the difference in 

elevation between the two edges of the outare known. In this case known 

points on the right and middle FIGURE 17.-Diagram for derivation of formula for projected dip. 

scale are joined by a line the extension of 

which gives the result at its· intersection with 

the left scale. 

USE OF THE THICKNESS DIAGRAM. 

In using the thickness diagram (Pl. XV and 

problem II), find the point on the left scale 

that represents the width of. the outcrop and 

join it to the point on the right scale that 

represents the dip of the beds. - At the intersection of the line determined by these points 

with the middle scale the thickness may be 

read. 

The above rule holds only if the surface is 

level. A correction must be added if the 

ground surface slopes in the opposite direction 

to the dip, and the same correction must be 

subtracted if the surface slope and the dip· are 

in the same direction. This is the correction 

E sin (90° -A) in formulas (3) and (4). Subtract the dip from 90 ° and find the corresponding point on the right scale. Join it to 

28288°-19-9 

crop is 50 feet. As there is no 50-foot graduation on the left scale, the decimal point is 

shifted to the right and the 500-foot graduation is joined by a line to the 70 ° mark on the 

left scale (90°-20°=70°). The reading is 470 

feet, but the decimal point must be shifted 

back, so the true correction is 4.7 feet. Then 

if the bed dips "into the hill" the thickness is 

206 + 47 = 253 feet; if it dips "with the hill" 

the thickness is 206-4 7 = 159 feet. 

Like the depth diagram, the thickness diagram has converse uses, but they can be applied 

only when the ground surface is level. Thus 

if the width of the outcrop and thickness of the 

bed are known the dip can be computed, or if 

the dip and thickness are known the width of 

outcrop can be computed. 

USE OF THE PROJECTION DIAGRAM. 

In using the projection diagram (Pl. XVI 

and problem III) find the point on the left 

scale that represents the dip and the point on 
126 SHORTER CONTRIBUTIONS TO GENERAL GEOLOGY, 1918. 

the right scale that represents the angle between the strike and the plane upon which the 

dip is to be projected. Join these points by a 

line, and at its intersection with the middle 

scale read the projected dip. As this problem 

involves only one case there are no corrections 

to be added or subtracted. 

In Plate XVI the broken line shows the solution of the problem of projecting a dip of 43° 

upon a plane at an angle of 35 ° to the strike. 

The result, 28°, is read on the middle scale. 

As with the other diagrams, the converse 

problems may be solved. Thus, if the trace of 

a bed along a vertical plane and the angle between the strike and the vertical plane are set 

on the middle and right scales, respectively, 

10 

-- -- --

0 

good for only one formula, whereas the slide 

rule is applicable to many formulas. 

Consider figure 18, a, in which ab, cd, and ef 

are parallel lines cut by two secants, ace and 

bdf From e and f drop two line~, eg and hj, 

perpendicular to · ab. By inspection of the 

diagram it is seen that 

cd = Y2 ( ab + ef) 

This equation is equivalent to saying that 

the space cut off on the middle line is equal to 

half the sum of the spaces cut on the outer 

lines. 

Now, if, as in figure 18, b, the outer lines are 

graduated evenly with equal divisions and the 

middle line is graduated with divisions half as 

20 10 10 100 10 

50 

- 15 -- -- 5 

----

10 5 10 

5 

s 

0 

b c 

FIGURE 18.-Diagram showing three stages in derivation of principle of the ~linement diagram. 

the dip of the plane may be read on the left 

scale. · Again, if the dip of a bed and the slope 

of its trace on a vertical plane are known, and 

these points are set on the left and middle 

scales, respectively, the angle between the 

strike and the vertical plane may be read on 

the tight scale. 

PRINCIPLE OF THE ALINEMENT DIAGRAM. 

The alinement diagram is a device by means 

of which certain arithmetic processes (addition, subtraction, multiplication, division, involutiqp., and evolution). may be mechanically 

or graphically performed. It is similar in principle to the slide rule in that by means of it 

distances proportional to numbers. or to log a-

. rithms . are mechanically added; subtracted, 

multiplied, or divided, but this work is done by 

a different mechanical process. It differs further from the slide rule in. that each diagram is 

great, addition can be graphically performed. 

The bases of the three scales are put on the 

same level for convenience only. The secant 

line is taken so as to cut off six divisions on 

the left scale and eight on the right scale. It 

must, perforce, cut 14 divisions on the middle 

scale, and the ·addition of 6 and 8 is graphically 

accomplished. 

It will be remembered that logarithms are 

essentially exponents (to the base 10 in the 

common system), and that multiplication is 

accomplished by adding the logarithms. This 

is by the algebraic law 

xmxn = x<m+n)---- - ~- ----- - ( 6) 

wh~ch may be verbally stated as follows: The 

multiplication of powers of the same quantity 

is effected by using the sum ·of the exponents 

as the exponent for the product. 

Now, if the markings of the scales are not 

equidistant as in figurel8, b,"but are laid off as 
U. S. GEOLOGICAL SURVl~Y PROFESSIONAL PAPER 120 PLATE XV 

Width of outcrop Thickness Dip 

-900 

-80° 

---- '700 ---- 60° 

50° 

300 

----:::::-~--------------zoo 

100 

90 

80 

70 10 00 

400 60 

50 

40 

30 

300 50 

20 40 

30 

10 

9 

200 8_: 

7 - 20 

6 

4 

3 

I - !0 

,J 

2 

ALINEMENT DIAGRAM II, FOR COMPUTING THICKNESS" 

I 

DETERMI~ING DEPTH AND THICKNESS OF STRAT AND PROJECTION OF DIP. 12 7 

in figure 18, c, so that the distances from the 

base are proportiJnal to the logarithms of 

numbers, we can in effect add logarithms. 

But this is equiyalent to multiplying the 

numbers. -.In the figure the multiplication of 

4 and 5 is shown, the result being 20. In 

figure 18, c, the dli.stance on the outer scales 

proportional to a logarithmic difference equal 

to 1 is. 2 inches; olfl the middle scale it is half 

as great, or 1 inch.l 

The scales may be graduated in proportion 

to logarithms of tf·igonometric or other functions as well as in proportion to logarithms of 

numbers. It is therefore possible to manipulate a variety of factors. 

Sometimes it is desirable to compress a scale, 

as for example a log. sin. scale in which the divisions between 1° and 20° come relatively far 

apart. In the charrs here presented (Pls. XIVXVI) the scales for logarithms of numbers were 

drawn with a modulus 1 of 10 inches. A log. 

sin. scale with a modulus of 10 inches would 

have to be 17.5 inbhes long in order to range 

from 1° to 90°. 1log. tan. scale would have 

to be 35.2 inches lqng to range from 1° to 89°. 

When the moduli of the outer scales are not 

equal, the modulus of the middle scale is determined by the f<;>llo'fing formula, in which M3 = 

modulus of middle lscale, M1 =modulus of right 

scale, and M2 = moci:lu lus of left scale: 

At the same time, the middle scale must be 

shifted toward thJ1t outer scale which has the 

smaller modulus. The position of the middle 

scale is determined by the following equation, 

in which M1 and *2 have the same values as 

before, D1 =distance from middle scale to right 

scale, and D2 =distance from middle scale to 

left scale. 

modulus t~ t any two lines which intersect on 

one of th outer scales and cut off a distance 

on the mi dJe s.cale equal to its modulus wiJl 

also cut o a distance on the third scale equal 

to its mo ulus. 

Let M11 2 , and M3 be the moduli and D1 and 

D2 the di t ances of the outer scales from the 

middle se les. Assume the lettering as indicated. 

B 

E 

A, G, ,.F' r ----:D2--- --~ ---~n,-- --?>j 

FIGURE 19.-Dr·agram for derivatjon of the interrelation of M1, M2, Ma, 

D1, and D2. 

Then i1 the triangles ACG and AEF 

:M3 J\1:1 

D2 =n1 +Dz 

and 

M3 (D1 +D2 ) =M1D2 

Si1nilar y in the triangles FCG and FBA 

Therefore 

and 

Ma Mz 

D1 D1 +D2 

M3 (D1 +D2) =M2D1 

M2D1 =MlD2 

D1 M1 

D2=M2 (q. e. d.) 

Taking this proportion by composition and 

transposi g, we have ' 

I g 1

= 1 - - - -- -- -- -- - - - - ( 8) M1 D 1 

2 2 

M1 +M~=D1 +D2 

The principles underlying this and . the last 

equation may be deduced from figure 19. The 

principles of the diagram require that the middle scale must be so placed and have such a 

1 The term "modulus" is used to mean the length on a scale proportional to a unit difference of logarithms. For example, the outer scales 

in figure 18 are said to have~ modulus of 2 inches because the distance 

from the point representing 1 to the point representing 10 is 2 inches. 

The logarithm of 1 is 0.000 and the logarithm of 10 is 1.000, and the difference between these logarithms is unity. 

Now, in t e triangles FOG and FBA 

M3 Dt 

Mz Dl +D2 

Substitut · g from the prevwus equation, we 

have 
128 SHORTER CONTRIBUTIONS TO GENERAL GEOLOGY, 1~18. 

and solving for M3, we have 

M = MtMz 

a Mt +M2 (q. e. d.) 

The length of any scale may be determined 

by the following equation: 

L=M (log F1 -log F2

) ________ (9) 

in which L =the length of the scale, M =its 

modulus, and F1 and F2 =the maximum and 

minimum values of the variable to be represented on the scale. 

This may be explained by an example. 

Suppose it is desired to fit a log. tan. scale 

into a space about 12 inches long. Let F2 = 

tan 1°, and F1 =tan 89°. Then 

log F2 =log tan 1°=8.2419 

log F1 =log tan 89° = 11.7581 

then 

log F1 -log F2 = 11.7581-8.2419 =3.5162 

and 

or 

L=M 3.5162 

L 

1\{=3.5162 

Substitute 12 for L: 

1\{=12--;.:-3.5162=3.41 + 

As an engineer's scale is used in drafting the 

scales and can be conveniently used only for 

:.noduli of 1, 1%, 2, 2.5, 373, or 5 inches or 

these lengths multi:Wied by 10, we choose a / 

modulus of 3 73 inches, which is nearest to 

the inconvenient irrational decimal 3.41 +. 

Substitute 373 for M and we find L=373 X 

3.5162 = 11.72 inches. 

ACCURACY AND ADVANTAGES. 

The point may be raised that the calculations made by these charts. are not accurate, 

and in the strictest sense this is undoubtedly 

true. But there are two facts that should be 

borne in mind: (1) Measurements made by 

pacing or by a hand compass or clinometer are 

not accurate; (2) the surfaces with which the 

geologist deals are not true plan_es. Measurements of angles are seldom accurate within 1°, 

which is approximately equivalent to an error 

of 1 per cent, and pacing under the ~ost favorable conditions can not be trusted closer 

than to 1 per cent. The undulations of fault 

or bedding planes \vill usually introduce an 

error of the same or greater magnitude. Therefore it is unsound logic to demand that the 

error in the calculations based upon such 

measurements be less than 1 per cent. If the 

chart is used with reasonable care, the error 

is somewhat less than 1 per cent and is therefore negligible. Moreover, the use of the chart 

gives less opportunity for the introduction of 

gross errors, such as come from a mistake in 

multiplication or in the addition of logarithms. 

The charts, therefore, have the following 

advantages over the other methods for solving 

these equations: 

1. They are more speedy. As a test of the 

speed of this method the writer used the first 

draft of the three charts for a series of test 

problems. The time required to solve the 

problems by means of the charts and by 

logarithms in the ordinary way was taken 

with a stop watch. It was found that the 

time with the charts ranged from one-sixth to 

one-fourth of that required by the logarithmic 

method. The test was fair to the logarithms, 

as the writer is reasonably proficient in their 

use. 

2. The chances of gross errors are eliminated. 

In the test mentioned in the preceding paragraph the correctness of the answers was 

checked, and a mean error of about 0.2 per 

cent was found, with a maximum of 0.7 per 

cent. One gross error of about 40 per cent 

was made in one of the logarithmic solutions 

and was first dete'cted by the check given by 

the chart solution. 

3. It is unnecessary to keep the formula in 

mind, as it is incorporated in the construction 

of the chart. 
U. S. GEOLOGICAL SURVEY PROFESSIONAL PAPER 120 PLATE XVI 

Dip 

Projected 

dip 

ALINEMENT DIAGRAM III, FOR COMPUTING PROJECTED DIP. 

An~le '?f 

proJeCtiOn 

I 

l ..