inches. The total strain on the guy is about 18,000 pounds, or 9 tons; and allowing a factor of 5, it would require a galvanized iron wire rope 51⁄2 inches in circumference to sustain this strain. These sizes can be obtained from any manufacturer's catalogue. The above figures are approximate only, and do not take into consideration the weights of the various members. These weights, of course, should always be considered in the case of heavy derricks; and, when the sizes have been approximately determined, the calculation should be repeated, allowing for the extra weight. Location of Cut-Out Block Question: Please inform me why the Board of Underwriters require cut-out block between mains and service switch instead of the reverse. Switch first and then cut-out is safer for all concerned, in my opinion.-J. P. D. Answer: The Underwriters require all wiring and apparatus in the building, including the service switch, to be protected by the fuses or cut-out. Therefore the cut-out should come before the switch; otherwise the switch would not be protected, and hence the requirements would not be fulfilled. Torque and Lagging Current Question 1: Of which of the following is the torque of a D. C. motor considered to be the result:-Magnetic poles of opposite name being brought in proximity; or the tendency of a current-bearing wire to rotate about a magnetic pole? most directly as the field current, leaving aside the change of permeability of the magnetic paths. For series motors, the torque varies approximately as the square of the current, because the same current flows in both the armature and the field. If the brushes were shifted around 90° from the normal point-supposing that to be possible-then the two fields would be in the same plane, and the torque would be zero. This condition, however, could not be realized, because of the sparking and racing of the motor which would take place. Answer 2: Under non-inductive load conditions, the primary current and E. M. F. are almost in phase, and are displaced nearly 180° from the secondary current and E. M. F., the two latter being in phase with each other. Under very light load conditions, provided the core. losses are small and the magnetizing component of the current comparatively large, there may be much less than 180° difference between the primary and secondary current; but this difference can never become as small as 90°. Methods of Graduating Instruments Question: Please tell me how such instruments as are graduated on the face are manufactured-for instance, protractors or scales. Are they stamped or cut? Also where can I get detailed information on the subject?— W. H. K. ment. Does the current in a secondary Question 2: of an alternating-current transformer alternate a quarter of a cycle behind that in the primary?—E. O. M. Answer 1: The torque of a motor is due to the attempt of two magnetic fields to coincide. One of these magnetic fields is due to the field coils, and the other is due to the armature current. These fields are kept at right angles by the location of the brushes which govern the direction of the field of the armature. As stated, the tendency of the two fields to come into coincidence produces the torque; hence the torque varies almost directly as the armature current and al Answer: There are several ways of making graduated instruments. The specific manner adopted depends a great deal upon the intended use of the instruThe best mathematical instruments are graduated by dividing engines. Frequently, however, protractors and scales are simply made by stamping, or by comparison with some standard. Instruments thus made are to be avoided for accurate work. Graduations are made for so many different purposessuch as thermometers, scales, steam gauges, etc.-that it is impossible to give one general description of the method of the manufacturing. For a detailed description of the dividing engine, which is really essential for accurate graduations, you may refer to some good cyclopedia, such as "Appleton's Cyclopedia of Ap During his recent visit East our manager was fortunate in securing Unusual Bargains in Remainder Book Sale Those who order early will be sure of getting the books. They are not soiled or shop-worn but SEND TO-DAY FOR COMPLETE PRICE-LIST. Send to us for any book you see advertised, we will supply promptly and at LOWEST PRICE Please send me your Remainder Sale price list and free duplicating postal card order book. Date. plied Mechanics." There is an article in the American Machinist of May 7, 1896, describing several dividing tools. There are also a number of other articles on this same line in the June and July issues of 1898. turning the fly wheel until the mark E is exactly opposite the end of the tram, the engine will be on crank-end dead center. Placing an Engine on Dead Center Question: Kindly explain by means of a sketch, the method of placing an engine on dead center.-J. D. Answer: When setting the valves of a steam engine, it is necessary to place the engine on dead center. The operation of setting a valve is rather delicate; and, therefore, one must be able to place the engine exactly on dead center. M To do this, place the crank in a position similar to position A in the sketch. Scratch a mark on the crosshead, and another exactly opposite it on the guide, as shown at M. Make a center punch mark on the floor or any other convenient place, as P. With one end of a tram placed on the point P, scratch a mark C on the rim of the fly wheel with the other. Now turn the engine over so that the crank occupies a position similar to its original position, but on the opposite side of the center line, as shown in the dotted position B. When the crank is in the correct position, the mark on the crosshead will again be exactly opposite the mark M on the guide. Again place one end of the tram on point P, and scratch a mark on the rim of the fly wheel D. Now bisect the arcs between the points C and D, obtaining the points E and F. To bring the engine on head-end dead center, it will be necessary only to place one end of the tram on the point P, and bring the fly wheel to rest when the mark F is exactly opposite the other end of the tram. In the same way, by holding the tram in this position, and Heating of Water Question: I have a boiler containing 150 two-inch seamless copper tubes 4 feet long. I have a sufficient quantity of exhaust steam at 2 pounds' pressure to supply this, and should like to know how many gallons of water per hour I can raise to a temperature of 130°, the water being taken in at a temperature of 65° F. Will you please give all the calculations for this, so that I can follow them and thus be able to work out the quantity for other temperatures?-M. A. M. Answer: In heaters of this kind, it is customary to assume that each square foot will give 200 B. T. U. to the water per hour for each degree difference in temperature between the surrounding water and the steam. In the case at hand, the average temperature of the water is 2x (130+65), or 98 degrees. The temperature of steam at 2 pounds' pressure is 219° F. This can be obtained from any table of the properties of steam. Consequently the total difference in temperature between the steam and water is 219 98 (= 121) degrees. We may allow approximately 1.6 feet of pipe for each square foot of heating surface. To determine this accurately, the pipe should be measured and the circumference calculated. Since there are 150 tubes, each 4 feet long, we have 600 feet of tubing: and the total surface, allowing 1.6 feet per square foot, is 600 1.6, or 375. The total quantity of heat given off per hour, then, is equal to the number of square feet, times the efficiency, times the difference in temperature, or 375 × 200 × 121 (= 9,079,000). The water is to be raised from 65° to 130°, or 65° in all; and each pound which is heated to this extent will require 65 heat units. Since there are 8.3 pounds to the gallon, every gallon raised 65° will require 8.3 x 65 (= 540, approximately) heat units. Since the total number of heat units is 9,079,000, the number of gallons which can be heated in one hour is 9,079,000 divided by the heat necessary for one gallon, or by 540. This gives 16,810 for the number of gallons which can be heated in one hour. shown in the accompanying sketch. Two transformers are used, one wound as shown for transformation of 100 to 1,000, and the other for transformation of 86.7 to 1,000. In other words, the transformation ratio must, in the first case, be I to 10, and in the second case .867 to 10. Answer 2: We may treat the threephase system as three separate singlephase systems, connecting a single-phase transformer across two of the mains, thereby stepping the voltage down to 110, and treating the circuit as a single-phase 110-volt parallel system. When this method is used, care, of course, should be taken to keep the system as nearly balanced as possible. This is the simplest method. Another method consists in using three transformers, the primaries being connected to the three lines, delta or star, as the case may be; the secondaries being likewise connected up, and carried to the cut-out boxes; and the load being then balanced up on the three phases. There are several other methods of steppingdown and connecting to the lamps, but the above are perhaps the most common. Cause of Pump Failure Question: Why does a pump fail to work sometimes when pumping against heavy pressure?-B. R. T. Answer: There may be a number of reasons; but it often happens on account of full pressure resting on the discharge valves. Under these conditions the water in the pump cylinder is simply compressed and prevents water entering the cylinder. By suitable connections in a pipe, and a waste delivery, the water can be expelled without any pressure, if the pump is started and run for a while. Center of Gravity-Water-Gauge Pressure-Size of Air Fan Question 1: How do you find the center of gravity of a wrought-iron lever 30 inches long. 3/4 inch thick, 24 inches deep at one end, and 17 inches at the other? Question 2: A water gauge shows 2.4 inches; what is the equivalent per square inch in gauge pressure? Give a rule for changing the same. Question 3: How do you find the size of a fan required for lifting a given quantity of air from a shaft at a given pressure?-D. B. Answer 1: The center of gravity of solid, is the same as its geometrical cenany regular figure, whether plane or ter. Since the face of the lever is a trapezoid, we know that its center of gravity must lie on a straight line joining from one end can be determined by the the middle of the two bases. Its distance method of moments. Consider the trapezoid to be made up of a parallelogram each of these figures is easily obtained, and a triangle, the center of gravity of since it is the geometrical center. That of the parallelogram will be 15 inches will be 20 inches from one end, and 10 from either end; and that of the triangle inches from the other. The sum of these two moments about an axis through the center of the parallelogram, is 235. Dividing this sum by the sum of the weights of the parallelogram and triangle, we obtain 235 495 (=1) inches for the distance of the center of gravity from the axis of reference. In other words, the center of gravity is 15 5-11 inches from the smaller end of the bar. Since the lever is of the same thickness, 34 inch throughout, the exact center will be 3% of an inch from the surface of the bar. Practically the quickest method would be by balancing the lever on a kni fe edge. Answer 2: The equivalent of 2.4 inches of water is .0864 pounds per square inch. One inch of water at 62° F. is equivalent to a pressure of .03608 pounds per square inch. |