Geophysics - Temperature Compensation of Old Type Askania Magnetometers

The theory of the Askania mag-netometer, as well as a complete discussion of all factors influencing magnetometer readings, is very ably described by J. Wallace Joyce.1 We will assume that the reader is thoroughly familiar with this theory and we will use whenever possible the same notations for the physical values involved in the theory. On the other hand, we believe that the principal formulas of the magnetometer can be deduced much more easily by the use of differential calculus than by the methods employed by Joyce. Consequently, we are presenting herewith a resumé of the theory of the vertical magnetometer in order to determine the correct formula of the temperature coefficient which we propose to compensate. Short Theory of the Vertical Magnetometers We shall accept the following notations: m = total mass of the magnet system, in grams. g = acceleration of gravity, in dynes. a = projection on the axis of the magnet system of the distance between the axis of support and the center of gravity of the magnet system, in centimeters. h = projection on the perpendicular to the axis of the magnet system of the distance between the axis of support and the center of gravity of the magnet system, in centimeters. M = 2pL = M, magnetic moment of the magnet system. Z = vertical component of the earth's magnetic field. ? = deflection angle of the magnet system. = focal length of objective lens in the optical system S = scale reading. In a vertical magnetometer the axis of the magnet system is kept horizontal within 2". Therefore, referring to Fig 1, we see that the magnetic torque of the magnet system influenced by Z must be counterbalanced by the gravity torque. MZ cos ? = mga cos ? + mgh sin ? or dividing by cos ? MZ = mga + mgh tan ? If we now move to another magnetic station and the temperature conditions change, the values of all the variables will change also: M, a, and h under the influence of temperature. 2, both with time and space. ? and S, as a result of all the above changes. Taking increments in Eq 2 we obtain: (Mo + ?M)(Z + ?Z) = mgao + mg?a + mg(h + ?h) tan (? +??) but tan (? +??), because ? is small, can be replaced by tan ? + tan ??; in fact tan (?+ ??) = sin ?.cos ?? + cos ?.sin ??/ cos ?.cos ?? — sin ?sin ?? but sin ? sin ???0 tan (8 + ??) ? tan ? + tan ??. Fur-thermore, neglecting all terms of the equation containing two differentials becomes: or a differential and a tan ??, Eq 3