From the data given, and by assuming some of the dimensions not given, we should conclude that the voltage of the dynamo would be increased to about 0 volts by the change in the winding. * ** (320) Kindly inform me how to determine the amount of heat conducted away, in a given time, from a liquid metallic bath whose temperature is 800° F., by a copper rod inch diameter by 30 inches long, the rod having one end in the bath, and the other connected to copper plates of such size as to remain at the temperature of the room. C. L. J., Jersey City, N. J. ANS.-The proper way to make the determination is by actual experiment as follows: First determine the rate at which the molten metal will cool with the copper rod absent. Then insert the rod, and note the difference in the rate of cooling. The difference between the two results will show the cooling effect of the rod; so that, knowing the specific heat of the molten metal and its weight, the heat carried away by the rod can be calculated. Several determinations should be made and the average taken. * * * (321) Will you please inform me how much power it will require to propel a flat, scow-shaped boat, 10 feet wide by 40 feet long, by means of two side wheels, at a rate of 6 or 8 miles per hour? The carrying capacity is to be 20 tons. Kindly explain the method of estimating the above. A. R., Pueblo, Col. ANS-The exact amount of power required to propel a vessel at a given speed cannot be deduced very readily from the elementary principles of mechanics. Instead, we must rely upon empirical rules, based upon the actual performances of vessels. An empirical rule often used is ANS.-(a) The negative for the half-tone process is made by the insertion of a grating between the lens of the camera and the sensitized plate. The grating consists of two plates of glass that are ruled, by means of a diamond, in parallel lines at an angle of 45° with the edges of the plate. These two plates are placed so that the ruling on one plate is at right angles to the ruling on the other. The position of the grating is immediately in front of the sensitized plate, usually about inch away. The copperplate is coated with a sensitive enamel composed of glue, and sensitized with a solution of ammonium bichromate with excess of ammonia. After the plate receives a photographic impression from the negative by means of the ordinary photographic printing process, it is developed, washed, and the glue is "burned" hard by holding over a gas or oil stove. The impression is then etched on the copper with a solution of perchloride of iron to a depth slightly less than the thickness of ordinary writing paper. (b) Galvanoplasty is another term for electrotyping. The term also applies to the production of bronze statuary by electrodeposition of the metal on a plaster model; the model being afterwards broken up and removed. (c) Clock ornaments are usually made of spelter. The process called slushing consists of pouring the hot spelter into a brass mold, twirling the mold about until the metal nearest the mold is set, and pouring out the surplus metal. The result is a hollow casting. * ** (323) Kindly give me all the necessary rules and formulas for designing "solenoids." F. T. S., Allegheny, Pa. ANS.-The rules and formulas for designing a solenoid are the same as those which apply to the electromagnet. We have not the space to devote to this subject in the inquiry columns. Such formulas and information as are required for solenoid design may be found in "The Electromagnet" by Sylvanus P. Thompson, price $1.00. It may be obtained from The Technical Supply Co., Scranton, Pa. * * * (324) Is the following formula correct for ascertaining the maximum load that a derrick engine can lift maximum load (or effective pull at the hook) = steam pressure × piston area twice the stroke X gear ratio. I notice in "Mines and Minerals" for May, 1898, page 453, that Mr. Jacobs makes use of a formula by which he obtains a much larger pull than I do. He does not take into account the fact that the load has to be raised, while mine does, I think-the result being in foot-pounds. W. H. W., Toledo, O. ANS.-You cannot determine the hoisting power of an engine by the formula given. Mr. Jacobs, in the answer to which you refer, is correct in the method he uses to arrive at the dimensions of the required engine. A hoisting engine should always be double cylindered, so that if one is on the dead center, the other will be in a position favorable to lifting the load. This fact, Mr. Jacobs was careful to explain. In order to lift the load the force exerted by the steam, acting-through the piston-rod and connecting-rod-at the crank pin, must be sufficient to overcome the frictional resistances of the gearing, rope, and sheave, and to transmit a pull at the hook slightly in Sheave excess of the load. The power of an engine bears no direct relation to the dead load it can lift. It is a question simply of leverage, and for that reason hoisting engines are not fitted with heavy fly wheels, as are engines used for driving machinery. In the latter case the engine is always started under a very light load-resistance of line shafting, countershafting, belts, etc. When it attains its regular working speed, the governor either throttles the steam or varies the cut-off, so keeping the velocity practically constant. When a machine is "put on "the momentum of the flywheel prevents the speed of the engine from being materially or suddenly reduced. With a hoisting engine it is different-the start having to be made under full load; for which reason two cylinders are used and the cranks are set at right angles. Of course, the horsepower of the engine is all used; for example: If the indicated horsepower of the engine is 100 and its efficiency is 85 per cent., then 85 horsepower are delivered to the hoisting mechanism. If the hoisting mechanism has an efficiency of 75 per cent, then the horsepower left for raising the load is 85 X .75 63.75. That is The power used in overcoming resistances in the engine = The power used in overcoming resistances in the gearing, etc. = The power used in raising the load 15 H. P. 21.25 H. P. 63.75 H. P. 100.00 H. P. (325) I should be pleased to learn your explanation of the loop in the cards taken from air pumps of steam engines. The loop is to be found in the cards from many vertical air pumps, which pumps are single acting and have three sets of valves-foot, bucket, and head valves. A. F. H., Boston, Mass. ANS.-If point c in the diagram is the beginning of the outward stroke, it is seen that along part d the Loop HOME STUDY C pressure is gradually diminishing until the end of the stroke is reached; before the piston begins its return movement the air contained in the cylinder is very likely, for a short time, exposed to the cooling influence of the cylinder walls, in which case the pressure would fall, and the return stroke begin with a reduced pressure that would gradually rise along part a up to point b, when it would be sufficient to open the valves, with a resulting decrease in pressure up to the point c. * ** (326) (a) What is the percentage of gas and air used in gas engines? (b) What is the pressure in the cylinder at the moment explosion takes place? (c) I enclose sketch of gas-engine cylinder in which the dimensions are given in millimeters. What would be the power of such an engine receiving an impulse every other stroke? (d) What weight of balance wheel would you advise and what diameter? (e) The cylinder is to be made of seamless brass tubing 19 mm. inside diameter, 2.3 mm. thick, and 80 mm. long; what would be its bursting strength? (f) How many degrees should the crank be below center when the gas is ignited? (g) You will notice that in the sketch I show the air compressed from about 65 mm. to 12 mm; will that be too much? (h) Can you tell me how to make gold-colored soft solder? (1) What can I put in potato juice to prevent it from smelling bad and souring. W. C. B., Keokuk, Iowa. ANS. (a) For ordinary coal gas, 1 volume of gas to 8 volumes of air. (b) This pressure varies from 140 to 200 pounds per square inch; your engine will give about 150 pounds just after explosion. (c) The engine should run at from 750 to 1,000 revolutions per minute. At 750 revolutions the power will be approximately horsepower and at 1,000 revolutions horsepower. (d) Use two balance wheels, one on each side of the crank, of the following dimensions: outside diameter 7 inches, width of face 4 inch, depth of rim inch. These flywheels, if made of brass, will give the weight required. (e) It is impossible to determine this quantity with any degree of accuracy, because of the peculiar nature of brass tubing. It would probably stand 2,000 pounds per square inch without bursting. (f) 20°. (g) Yes; compression is too great: make length of compression space one-half that of stroke for flame ignition and one-third for electric or tube ignition. (h) Soft solder should be colored after application. Either plate with brass, or if gold is desired, deposit copper first before placing in gold plating solution. We know of nothing that would color the solder in bulk, without making it too hard or unfit for use. (i) Add a little of Condy's red disinfecting fluid (a solution of potassium permanganate in water). We cannot, very well, give advice of this kind without knowing the use you intend to make of the potato juice. (327) Will you kindly inform me as to the proper manner to proceed in retracing section lines as shown on enclosed sketch? The land south of the river was first sectionized and corners of fractional sections set on south bank of stream where the lines intersected it. The south bank of the river was also meandered from one of these corners to another. Some time after, by another man, the lands north of the river were surveyed. His notes indicate that he began on the south bank at these fractional corners and extended the lines thence. As the stones on the north side are both north and west of where they would be were this the case, it is apparent that it was not done this way. Now where is the line crossing the river? Is it a straight line between the nearest corners on opposite banks of the river as shown between sections 14 and 15, or is it projected straight on either side to the river bank and a straight line between these points, as shown between sections 14 and 23, or is it projected across from the north side till it meets the meander line, as shown between sections 4 and 9? F. T. L., La Junta, Colo. ANS.-This is one of the problems encountered in the practice of surveying that is very difficult to solve satisfactorily in all cases. The surveyor must be guided largely by the conditions as he finds them and by the decisions of the courts in similar cases. In retracing the lines of the government surveys, he should adhere strictly to the infallible and inflexible rule that the lines must be relocated in the positions where originally run when it is possible to do so, regardless of any inaccuracies in the original lines, or of any considerations as to where they should be. It should also be understood that the properties bounded by a non-navigable stream extend to the center of the stream, or filum aquæ as it is termed. The filum aquæ, or thread of the stream," is midway between the lines of ordinary low-water mark, without regard to the depth of water or position of the channel. It is the boundary line for all properties limited by the stream. The lines between adjacent properties bordering on the stream on the same side terminate at the filum aquæ. Such lines are not properly considered as extending across the stream. If any such line is not at right angles to the general course of the stream, it is deflected at the meander line so as to extend from the meander line to the center of the stream at right angles to the stream's general course. The legal rule governing the case has been stated as follows by the Chief Justice of the Supreme Court of Michigan: Extend the line of division between the two parcels from the meander line to the center line of the river, as nearly as possible at right angles to the general course of the river at that point. It will thus be seen that none of the three methods suggested in the question would be correct, except that the method indicated for the line between sections 14 and 23 might be correct in some cases. The lines of the government survey must be relocated in their true positions as originally run. These lines properly terminate in the meander lines on the respective banks of the river, from which intersections, the boundary lines should be extended at right angles to the stream, terminating at the center of the stream. If there are no meander corners, or starting corners for the lines of the more recent government survey on the north side of the river, it is our opinion that it would be proper to extend the lines of that survey as far as ordinary low-water mark. * ** (328) (a) Will you please give instructions for charging magnets? (b) Are magnets that are built up in layers better than those made of a single piece of steel? (c) What kind of steel is best for magnets? (d) How long does a good magnet retain its magnetism? (e) In what numbers of HOME STUDY MAGAZINE are magnets or magneto-machines written up? H. A. H., Bellevue, Ky. ANS.-(a) It is supposed that in the present instance the magnet to be charged is a horseshoe magnet. The magnetization may then be accomplished either by means of a permanent horseshoe magnet or an electromagnet; they are both used in the same manner. The horseshoe is fastened to a table, and an armature laid across its ends; the magnet is then placed at right angles to it with one pole on each leg near the end, and is stroked over it towards the curved part, then returned through the air and the stroke repeated, both sides being treated in the same manner. Ten strokes will complete the magnetization, and the polarity of either leg will be the same as the one with which it was in contact. An electromagnet charged with not less than 3 cells may be used in place of the steel magnet. (b) Yes. (c) Tool steel drawn to a straw color or a little lower. All shaping and filing must be done before magnetization. (d) Practically forever; though there is a gradual decrease in strength. (e) In the July, 1896, HOME STUDY MAGAZINE, there is an article entitled "Does the Magnetic Needle Point North?"; in the August, 1896, number, “Magnets"; in the November, 1897, number, "The Design of Hoisting Magnets." * ** (329) Will you please let me know (a) What the discharge per minute is, in gallons, from the nozzle of a hose 700 feet long, 24 inches in diameter, with 60 pounds pressure per square inch at hydrant, to which the hose is attached, the opening of the nozzle at discharge point being 14 inches in diameter? (b) What is the longest board 12 inches wide that can be placed in a room 18 ft. X 24 ft., the ends of the board being cut square, and such board to be laid flat on the floor? Please show how to work out both the above problems. P. W., Lake Linden, Mich. ANS.-(a) A pressure of 60 pounds per square inch corresponds to a head of 60.434 138.25 feet. part of this head is absorbed in forcing the water from the hydrant through the hose to the nozzle, while the remainder appears at the nozzle as pressure, and acts to force the water through the nozzle and give the issuing jet its velocity. The part of the head absorbed in forcing the water from the hydrant through the hose, which we will call z, depends on the method of connection between the hydrant and hose, and the condition of the hose, whether rough or smooth, etc. We will assume that the connection A (2) V2 = .98 √ 2g hq = .98164.32 (138.25 -- x). The quantity of water in cubic feet per second Q flowing through both the hose and the nozzle is the same, and is equal in each case to the area in square feet multiplied by the velocity in feet per second. The area of the hose is (24) X .7854 ÷ 144 = .03408 square feet, and of the nozzle (14) X.7854 ÷ 144 =.01227 square feet; therefore, Q = 1 × .03408 from which v1 = these values of and 2 in equations (1) and (2), and squaring, we have Q2 (૨ .03408 and = ૨ .01227' 5.36 XxX 2.5 X .00116 .03 X 700 1 X .01227, Substituting (3) (4) and Q2 = .9604 × 64.32 ×(138.25 — a) × .00015. Solving for Q, we have Q = .308 cubic feet per second, from which the discharge in gallons per minute is found to be .308 X 7.48 X 60 = 138.23 gallons. (b) See HOME STUDY MAGAZINE, May, 1897. Answers to Inquiries, No. 140, for a similar question and solution. (330) (a) What is the process of panning out gold in placer mining? (b) What is meant by "drop forging"? REX, Scofield, Utah. ANS.-(a) See July number, Answers to Inquiries, No. 259. (b) Drop forging is a process of forging in which the metal is formed by being compressed between two dies. One of the dies forms a part of the anvil, while the other is attached to the hammer. The hammer used is generally a "drop hammer," that is, a hammer consisting of a heavy weight, which is lifted and then allowed to drop on the anvil. ** * (331) Can you give me any information regarding the causes that tend to make an are lamp flash? I. J., Brenham, Tex. ANS.-Arc lamps flash because of poor regulation of the carbon-feeding mechanism, or because of a bad spot in the carbon. * ** (332) Can you give me or tell where I may find explicit directions for making a 12-inch electric fan which can be driven from one or two cells of a battery? G. H. B., Cleveland, Ohio. ANS.-Write to the Leavitt Motor Co., Providence, R. I. They will sell you castings and all supplies for a battery fan, and furnish complete instructions. (333) (a) Fig. 1 is a drawing of an elbow for a water pipe. The layout was made by our foreman. Every rivet hole was put in on the flat plate, and after the plates were rolled and put together the holes matched perfectly. The elbow is 10 feet in diameter on the inside; it is made of 3-inch plate, and the rivets are inch in diameter, about 2-inch pitch. The ten sections of which the elbow is composed are alike, and are tapered so that the water in flowing through in the direction of the arrow will not strike and also one onto c e. This will give the points d', f', and c', e', as shown in Fig. B. Draw a circular arc through d', d, c, c', and extend it a little beyond c', and do the same through f', f, e, e'. With center i and radius id, describe a semicircle, and do the same with center j and radius jf. Divide each semicircle into a convenient number of equal parts, not less than 6, and draw perpendiculars to ab through all CIRCUMFERENTIAL SEAMS. 120°/4 10' DIA DIA HOME STUDY. 24 -30′′ the edges of the plates. How can the sections be laid out without the use of triangulation? (b) Fig. 2 represents a smokestack. I want to make a layout of the two parts, without the use of triangulation; in large jobs like this one, the method of triangulation already described in HOME STUDY MAGAZINE has to be done with great accuracy, otherwise serious errors creep in. (c) Fig. 3 represents a piece of a stack to be made of 2 pieces, the seam being at A. Please explain the shortest and best method of doing this without triangulation. F. A. B., Charleston, S. C. ANS. (a) Each section is a frustum of a cone with equally inclined bases. Draw an elevation, 4, of one section, and draw a central section line ab. This central section is to be circular. The bases will be slightly elliptic. Draw ed and ef parallel to ab. Now lay out the development of the right frustum de ef, and take off the parts c d g and ehf in the following manner: Add another figure like d c e fonto d HOME STUDY. the points of division, intersecting the diameter of each circle, as shown in 1', 2', etc. Draw 1'-1', '-, etc., intersecting g e in 1", 2", etc. Set off d' c" so that the length of the arc d' dcc" will be the same as twice the length of the semicircle d-3-c. Set off fe" so that the length of the arc ffee" will be the same as twice the length of the semicircle f-s-e. Divide the arcs d'e" and fe" each into as many equal parts as the semicircles d-3-c and f-3-e were divided-6 in this example-in 1", 2", etc., and draw the lines 1"-1", "-", etc. Make 1"-1"" = 1'-1", 2-2 2'-2', etc. Also make d'g' and c'g" each = d g, etc. Draw a curve through g', 1"", आ etc. to g", and another one through h', 1"", 2011 etc. to h", as shown. The figure g'g' h" h' is the required layout. (b) See HOME STUDY MAGAZINE, August, 1897, Answers to Inquires, No. 279. (c) See Answers to Inquiries, No. 334 (a), in this number. = 3/4" RIVETS. 8 STEEL (334) (a) I would like to know how to lay out Fig. 1 in four sections, and (b) how to find the radius Cin Fig. 2, also the chord E, and the dimension D. W. B., Pittsburg, Pa. ANS. (a) The sides of this transition piece consist of 4 equal parts. Draw a top view or plan A, of one of the four equal parts, and an elevation B, of this part. The center line for both views is a b; and the conjugate center line of the plan is b c. Divide the quadrant e-6 into a convenient number of equal parts; and divide the smaller quadrant c d into the same number of equal parts. Project the point d from the plan into the elevation B to 6'. Draw 6-6', and ec, B, and continue them to meet in f. Draw projectors through all the points of division of both quadrants: 5-5, 4-4, etc., A, intersecting bc in 5", 4", etc., and intersecting 6-e of B, in 5, 4, etc.; also intersecting ac, as shown. Draw a center line ec, C, for the lay-out. Make ec, C, ec, B, and cf, C, cf, B. 335 Draw lines from all found in a similar manner. these points to f. Make 1'-1'""', A, f-1', B; and make f-1', C, equal to the distance from 1""" to the first point of division on the quadrant cd. The points 2',3', etc., C, are found in a similar manner. Make ag, B,a-6, A; and with 6-g, B, as radius, and 6, C, With 6'-a, B, as radius, as center, describe the arc a. and 6', C, as center, describe an arc cutting area in a. Draw the outline e-6-a-6'-c; this is one-quarter of the required layout. (b) See HOME STUDY MAGAZINE, Your August, 1897, Answers to Inquiries No. 279. Fig. 2 is not reproduced here. * ** (335) (a) What is the name of the electrical Make 1"-1"", A, f-1, B; and with 1-1, A, as radius and with f, C, as center describe the arc 1. Make e-1, C, =e-1, A. The points 2, 3, etc., C, are 65432 HOME STUDY. HOME STUDY. machine of which the enclosed is H. J. W., London, B. C. ANS.-(a) In the accompanying figure, we have shown the apparIt is atus mentioned by you. merely an experimental transto illustrate former, intended some of the peculiar properties of alternating currents, and the effect of induction. The ends of the coil of fine wire care soldered to the terminals a, b, and an alternating current is passed through this coil. The humming sound produced when a pail is placed upon the iron core is due to the vibration |