International projectionist (Oct 1931-Sept 1933)

September 1933 INTERNATIONAL PROJECTIONIST 17 divide 7,345.729 by 63.23, both of these numbers would be multiplied by 100, making the actual numbers 734,572.9 divided by 6,323. This operation is done in accordance with the algebraic rule (which will be taken up in detail in the third lesson), which states that both divisor and dividend (or multiplier and multiplicand) may be multiplied by the same number without altering the fundamental relation between them. In calculation in connection with sound systems, decimals to many places will oftentimes be encountered, especially in considerations of resistance, capacity and inductance. The commonest decimal units follow: .1 equals 1/iO (one-tenth). .01 equals 1/100 (one one-hundredth). .001 equals 1/1000 (one one-thousandth). .0001 equals 1/10,000 (one ten-thousandth). .00001 equals 1/100,000 (one one-hundred-thousandth) . .000001 equals 1/1,000,000 (one-niil lionth). .0000001 equals 1/10,000,000 (one tenmillionth). The usual method of speaking of decimal figures is to regard .1 as "point one," or "one tenth" ; while .007 would be regarded as "point 0-0 seven," and .0000023 as "point 5-0's two three." In order to apply the foregoing considerations let us discuss for a moment a typical problem, such as might be encountered during an ordinary working day of a sound engineer in the recording room or the projection room: A milliameter is on hand, having a resistance of 40 ohms and registering a maximum of .25 amperes. It is desired to use the milliameter as a voltmeter, to measure a potential slightly under 300 volts, which can of course be done, provided a suitable resistance is placed in series with the meter. By Ohm's Law (and a set of entirely arithmetical calculations), the necessary resistance may be determined. (Ohm's Law: — E equals I x R, or R E equals — ) I. First, let us calculate the necessary total resistance which will be needed to limit the current to .25 amperes at a potential of 300 volts: R equals 300 , or R equals 1,200 ohms. As we .25 know (given above) that the resistance of , the meter itself is 40 ohms, it will be necessary to add a total external resistance of 1,160 ohms to the meter in order to safely measure a potential of 300 volts. In the next paper, the subject of logarithms, and the many ways in which they may be used to shorten what would otherwise be tedious mathematical operations, and the application of arithmetic and logarithms to the operation of the slide rule will be discussed, to be followed in succeeding issues by papers dealing with algebra, geometry, trigonometry and the simple calculus. [NOTE: Meanwhile there is appended hereto a series of problems the solving of which will be excellent practice and will serve to show how well you have absorbed the foregoing information. Numbers will be applied to all questions, the answers to which will be given in the following issue. As was stated in a foreword to this installment, readers are invited to submit their answers in advance of publication of the following issue, in which will also appear a list of the names of those who have correctly answered the questions. These question and answers will be a feature of this series. — Editor.} Problems : 1. In a group of resistors, there are units having the following resistances: 184.75 ohms, 25,000 ohms, 22.39 ohms, 6.5 ohms, 4.450 ohms, 50 ohms, 100.33 ohms, and 75.75 ohms. What is the total resistance of the group? 2. The formula for power in any circuit is Power in watts equals E (voltage in volts) X I (current in amperes). What is the power in a circuit carrying 25 A amperes at 32.33 volts? 3. It is desired to divide a total resistance of 20,000 ohms into 3 smaller resistances, one of which will be .23 of the total, one .57, and the third .20. What are th& three resistances? 4. Ohm's Law states that the current equals the quotient of the voltage divided by the resistance (with current in amperes, voltage in volts, and resistance in ohms). With a voltage of 110 volts, and a resistance of 5.5 ohms, what current will flow in the circuit? "SEEING" SOUND BY MEANS OF NOVEL PROJECTION PROCESS R. F. Mallina MEMBER, TECHNICAL STAFF, BELL TELEIPHONE LABORATORIES TN ACQUAINTING the visitors to ■'■ the Century of Progress Exposition with some of the essentials of modern electrical communication, it seemed very desirable to allow them both to hear and see the forms of the signal waves sent over certain types of telephone circuits, and to compare them with actual speech waves. By using an available type of loud speaker, the signals could easily be heard by a large audience. No apparatus was available, however, which would make the signals ea:sily visible to the larger groups that gather at the exhibit, although an ordinary oscilloscope would serve to make the signals visible to one or two people at a time. But by adapting telephone apparatus to serve as the actuating element of the oscilloscope, there was developed in Bell Laboratories a new oscilloscope, shown in Fig. 1, which throws the image of the signals on a large screen where they may be seen by a large number of people. The function of an oscilloscope is to represent an electric current, that varies in strength with time, by a point of light that varies in vertical distance above some base line with distance along that base.. The transformation is accomplished by two separately acting mechanisms, and by employing a powerful source of light, reflected by a mirror, as the visible image. The current to be depicted is passed through a telephone receiver such as is used in the ordinary hand set. A small spherical mirror is attached to the diaphragm of the receiver so that variations in the current tilt the mirror up and down. The mirror reflects a ray of light from a small electric lamp, like those used in automobile headlights, onto a small motion picture screen, where it may be seen by the spectators. The up and down motion of the reflected ray of light accurately corresponds to the variations of the current, and thus to the variation in pressure of the original sound. The Projection Process The motion of the light across the screen is provided by a rotating mirror as shown in Fig. 2. Here the mirror Figure 1