Reprinted from BMR Journal of Australian Geology & Geophysics, 10, 143-144 © Commonwealth of Australia 1987

Direction of magnetisation in core samples from drill holes Terry Lee1

A solution is presented for the practical determination of the direction of magnetism in core samples from drill holes.

Introduction This note treats the problem of finding the common direction of magnetisation in a number of cores obtained from drill holes. The problem arises because, while the direction of magnetisation may be measured in the sample, the core sample is rotated during sampling.

Some related problems are to be found in structural geology. In those cases, graphical methods of solution, employing a stereographic net, have been known for some time. Such methods of solution are in widespread use amongst geologists. A discussion of some of these problems and their solutions can be found in works by Tex (1954), Whitten (1966, pp 10-31) and Phillips (1960). In these, the reader may find a discussion of the use of the stereographic net and a statement of some of its mathematical properties. Some of these properties will be made use of here.

Solution It is clear, when looking down the drill hole, that all the possible angles for the direction of magnetisation will trace out a cone. A cross section of one such cone, of angle l/Ji, is shown in Figure I . Here, l/Ji is the angle that the direction of magnetisation makes with the drill hole axis . At the position where the sample is taken, the drill hole has a hade of Iii ' The cross section of the cone can be projected, stereographically, onto the equatorial plane OABC by means of the construction shown in the Figure 1.

Figure 1.

0+ rJ;i

0- rJ;i Oi

1/2 (Oi - rJ;i) 1 /2 Oi

1/2 (Oi + rJ;i)

z

24-2/41 p

I Division of Geophysics, Bureau of Mineral Resources, GPO Box 378, Canberra, ACT 2601

In general, the direction of the ith drill hole, will be fixed by specifying not only the hade but also the angle "Yi' Figure 2 shows the projection of the direction of such a drill hole. Also shown is the projection of the base of the cone, which specifies the various possible directions of magnetisation. The fact that the base of the cone projects as a circle, of radius ri, follows from a theorem concerning the stereographic projection. This theorem states that any circle on the sphere, a cross section of which is shown in Figure 1, will project as a circle (Hilton, 1963, p.3).

Suppose that the centre of such a circle is at a distance Ri from the centre of the projection (Fig . 2). The problem of finding the direction of the magnetisation vector can now be reduced to finding the point on the equatorial plane for the direction of the magnetisation vector. A moment's thought will show us that, in general, the results of at least three samples will be required.

y

Ri

x 24-2/40

Figure 2.

From Figure 1, it can be readily seen that, for the ith drill hole, the various distances OB, OC and BC are given by the equations 1-3 below.

OB = tan (1i/2) = Bi . . . . . . . . . . . . . . . . . . . . . . . . .. (1) OC = tan {(Iii + l/Ji)l2} = Ci ................... (2) BC = {tan (l/J/2) + tan2 (1i/2) tan (l/J/2)} /

{I-tan (1i/2) tan (l/J/2)J = Ai .... . .......... (3) Also AB = {tan (l/J/2) sec (1i/2)} / {I + tan (1i/2)

tan (l/J/2)} ................ .............. .. (4) and rj = {tan (l/J/2) sec2 (1i/2)} / {I -tan2 (1i/2) tan2

(l/J/2) } ................................ , ... (5)

144 T. LEE

From Figure 2, we can see that, if (xi> Yi) are the coordinates of the centre of the circle, then:

Xi. = Ri cos ('Yi) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. (6) Yi = Ri sin hi) . . ..... . ........ . . . ... .......... (7) Ri = ri - tan {( ¥ti - Oi)l2} ... . .... ... .. . . . . ...... (8)

Now it is easy to write down the equations for the various circles. Suppose that the equations are numbered for i 1, 2 ... N and that the equation for the Nth circle is subtracted from all the other equations.

When this is done:

X(XN - Xi) + Y(YN - Yi) x~ + y~) = {3i

Y2 (r? - r~ - X? - Y? +

1, 2 .. N-I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. (9)

When N = 3, the point on the yox plane, for the direction of magnetisation (x, y), is described by:

x = {31 (Y3 - Y2) - {32 (Y3 - Yl) (x3 - xl) (Y3 - Y2) - (X3 - X2) (Y3 - Yl) . .. (10)

Y = {32 (X3 - Xl) - {31 (X3 - X2)

(X3 - Xl) (Y3 - Y2) - (X3 - X2) (Y3 - Yl) ' " (11)

While equations 9,10, and 11 might be suitable for the situation where the errors in the measurements are negligible, this may not be so in the practical situation. In this case, the alternative method, based on the method of least squares (given below) should be used. In situations where the mutual inclination of drill holes is small, say the order of 10° , then the method of least squares should be used in preference to the method defined by equations 9, 10; and 11.

Where there are more than three equations, they may be solved by the method of least squares .

Writing (XN - Xi) l'/i

YN - yi

N - 1

SI =L l'/f

i = 1

N - 1

S3 = L{3i l'/i

i = 1

N - 1

S5 = LI?

i = 1

Ii

N - 1

S2 =L Ii l'/i

i = 1

N - 1

S4 =L{3i Ii

i = 1

one finds that

X = S3 S5 - S4 S2 ... . . . ..... . ... . .. . .... ... . . (12) SI S5 - S22

Y = SI S4 - S2 S3 ..... . ..... . .. .. . . . .. . ... . .. (13) SI S5 - S22

Any reader not familiar with the method of least squares can find all the details he needs in Froberg (1965).

Once the coordinates for the projection of the direction of magnetisation are known; the hade, 0, and the azimuth, 'Y, of the direction of magnetisation can be found from equations (14) and (15) below:

'Y o

arctran (y/ x) ....... .. . .. .... .............. (14) 2 arctran Jx2 + y2 . . .. .. . ........ . . . . . . ... (15)

Example As an example, a synthetic data set (Table 1) was generated on a stereographic net. The direction as specified on the net was determined by 'Y = 30° and 0 = 59° . Using the method described above, the values of 30.6° and 59.6° were found for the hade and azimuth of the direction of magnetisation, respectively .

Table 1. Synthetic data set for worked example.

0, 'Y, v" 20° 66° 11 ° 40° 74° 16 \1, ° 70° 54° 33 \1, °

340° 48° 41 y, 0 0° 60° 26 \1, °

Acknowledgements The writer would like to thank Phil McFadden and Mart Idnurm for their comments. Chris Torlowski computed the example .

References Froberg, C.-E., 1965-lntroduction to numerical analysis. Addison•

Wesley . Hilton, H ., 1963-Mathematical crystallography and the theory of

groups of movements. Dover. Phillips, F .e., 1960-The use of stereographic projection in

structural geology (2nd edition) . Edward Arnold. Tex, E. den, 1954-Stereographic distinction of linear and planar

structures from apparent lineations in random exposure planes. Journal oj the Geological Society oj Australia, 1, 55-66.

Whitten, E.M., 1966-Structural geology of folded rocks. Rand McNally.