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CONSULTING

DEPARTMENT

In order to receive attention, all inquiries must be addressed to the Consulting Department, and must be accompanied
with the full name and address of the sender. In view of the fact that the inquiries already received have
far exceeded the available space, the Editors reserve the right to select for publication

only those inquiries likely to be of greatest interest to the readers of the mag-
azine in general. Other inquiries will be answered by letter.

C. G. S. System-Units of Force and

Work

Question: Explain the "C. G. S." system used in Mechanics. Also the units of force and work, and give their English values.J. C. V.

Answer: The units of mass, length, and time are said to be "fundamental" units, since all other units, such as those of area, velocity, acceleration, power, etc., are referred to them. These latter units are therefore called "derived" units. The magnitudes of the fundamental units are, however, arbitrary. A large class of writers use the centimeter, gramme, and second, and this system is usually called the "C. G. S." system; others use the foot, pound, and second. It thus becomes important to have a systematic method of reducing measurements from one system of units to another.

For strictly scientific purposes, a unit of work is taken to be the work done by a unit of force when its point of application moves through one foot in the direction of its action; but, as a convenient and sufficiently accurate standard for practical purposes, the quantity of work which is done in lifting I pound through the height of 1 foot is commonly adopted as the unit, and this quantity of work. is spoken of as one foot-pound. It is, however, important to observe that the

foot-pound is not perfectly invariable, since the weight of a pound, and therefore the work done in lifting it through a given height, differs at different places, being a little greater near the Poles than near the Equator.

On the metrical system, the kilogrammeter is the unit; it is the work done when the weight of one kilogramme is raised through a height of one meter. This is equal to 7.24 foot-pounds, and one foot-pound.1381 of a kilogrammeter.

The unit of force is the poundal. The poundal may be defined as the force necessary to produce an acceleration or gain of velocity of one foot per second, when it acts on a mass of one pound for one second. In the "C. G. S." system, it is called the dyne, and represents a gain of velocity of 1 centimeter per second when it acts on a mass of I gramme for one second. In one system the units are given as so many foot-pounds, and in the other so many gramme-seconds.

How to Find a Positive or Negative
Terminal

Question: Please explain if there is any way of finding the positive or negative terminal of a generator if we know the direction of motion, the machine being at rest.-H. G. A.

CONSULTING DEPARTMENT-(Continued)

Answer: This can be determined by Fleming's rule, which states that if we hold the thumb, first finger, and second finger of the right hand, all three at right angles, and the thumb represents the direction of motion, the first finger the direction of lines of force, then the second finger will represent the direction of current flow. By testing with a compass you can determine which is the north pole of your machine, and knowing that the lines of force pass from the north pole to the south pole through the air, this will give you their direction. Then apply the rule given you, and you can obtain the direction of current flow. The terminal toward which the current flows will be the positive.

same speed as the gear A. But when passing a curve, it is quite possible for either of these gears to turn more slowly than the other and thus permit the desired compensation in the movement of the wheels.

How to Clean a Boiler

Question: What is the best method of cleaning a badly scaled boiler, and what compound should be used?-A. C. P.

Answer: First open boiler up, and note where the loose scale, if any, has lodged. Wash out thoroughly, and put in the required amount of compound. While the boiler is in service, open the blow-off valve for a few seconds, two

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COMPENSATING GEAR FOR AUTOMOBILE AXLES.

Differential Gear

Question: What is the construction and method of working of the differential gear such as is used on the rear axle of an automobile?-N. P. W.

Answer: The accompanying figure shows a form of compensating gear which, with various modifications of detail, has received a very extensive application. The axle is divided at the center, and, riding on the two abutting ends, is the hub E of the gear or sprocket A. This hub rides loosely upon the axle ; and in the body of the gear are the two freely moving bevel pinions C and D. To the two ends of the shafts, the bevel gears B and M are respectively keyed. It is evident that under ordinary conditions on a straight line, the two bevel gears B and M will be rotated with their shafts in the same direction and at the

or three times a day, to be assured that it does not become stopped up with scale.

After running the boiler for a week, shut it down; and, when the pressure is down and the boiler cooled off, run the water out and take off the hand-hole plates. Note what effect the compound has had on the scale, and where the disengaged scale has lodged. Wash out thoroughly, and use judgment as to whether it is advisable to use a less or greater quantity of compound, or to add a small quantity daily.

Continue the washing out at short intervals, as many boilers have been burned by large quantities of scale dropping on the crown sheets and not being removed.

It is necessary to have a chemical analysis of the water used in order fully to

CONSULTING DEPARTMENT-(Continued)

determine the kind and quantity of compound to be used. An alkaline compound that has given satisfaction, consists of water, 58 per cent soda-ash, and 70 per cent caustic soda. Proportions, 3,750 gallons of water, 1,600 pounds of soda-ash, and 2,600 pounds of caustic soda. This has been used where the principal incrusting solids have been calcium carbonate and sulphate and magnesium sulphate.

Reversing Rotation and Current Connections of Compensator

Question 1: How do you reverse the direction of rotation of a revolving three-phase synchronous motor?

Question 2: If you reverse the direction of the current in the fields, will that have any effect on the motor?

Question 3: Beginning at the right and lettering the three armature leads A, B, C, if you

Answer 5: The accompanying sketch shows connections required for a compensator with a three-phase motor. The compensator consists of coils a, b, and c, wound on a laminated iron core, each coil being provided with a number of taps, 1, 2, 3, 4, etc. When starting the switch is in the lower position, as shown; and when the motor has reached its rated speed, the switch is thrown to the upper or running position, so that the primary terminals are connected directly to the line.

Storage Batteries

Question 1: Why are there more negative plates than positive plates in a storage battery? Is it absolutely necessary?

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CONNECTIONS OF COMPENSATOR WITH 3-PHASE MOtor. leave A connected as it was but transpose the wires running to B and C, what will be the effect?

Question 4: Is a compensator a transformer or a reactance coil?

Question 5: Please describe the necessary connections and switches for connecting up a compensator with a three-phase motor.-F. J. M.

Answer 1: The direction of rotation can be reversed by interchanging any two of the leads.

Answer 2: This will have no effect on the motor.

Answer 3: Transposing wires B and C will reverse the direction of rotation.

Answer 4: A compensator is an autotransformer, or a transformer with only one coil; or again, it may be considered a reactance coil.

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HANDBOOK ON ENGINEERING

T

By

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CONSULTING DEPARTMENT-(Continued)

Answer: We think that you must refer to different positions of the boom rather than to different heights of the load, for it is a fact that the height of the load in no way affects the tendency of the derrick to overturn. Referring to the figure, you will see that the load is shown in two places in full lines, and in two places in dotted lines. For any one position of the boom, the turning moment is constant for a given load. In

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the first case, where the boom is shown nearly vertical, the tendency to turn about the center line of the mast would be Wd; and, supposing the derrick car C to be supported on the line passing through the center of the mast, the tendency to overturn about this support would be expressed by the moment Wd. The tendency of the car to resist this overturning, is expressed by its weight multiplied by the distance L, which represents the distance of the center of gravity from the center line of support under consideration. If the weight W is such that when the boom is nearly vertical the car just balances the weight, then, when the boom is swung to the horizontal position shown, the car will overturn. Since the moment now is WxB, in which B represents the distance from the end of the boom to the center line of the mast, the tendency to turn in the horizontal position is as much greater than that in the nearly vertical position, as the distance B is to D.

For formulas relating to derricks, see THE TECHNICAL WORLD, Vol. 1, No. 4 (June, 1904), page 474.

Mention The Technical World.

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