Moving Picture World (Jan-Feb 1927)

158 MOVING PICTURE WORLD January 8, 1927 Bluebook School Answers 545 and 546 Note: — This “School” is designed to arouse interest in the study of those many problems which constantly arise in motion picture projection, AND to cause men to really study the Bluebook and assimilate the vast amount of knowledge contained within its covers. Question No. 545 was a “practical” question submitted by Brother Allan Gengenbeck. New Orleans, La. It is as follows: What size wire should be used on a 250 volt circuit to transmit 200 amperes for a distance of 350 feet with a 3 per cent, loss under full load? All the “Doe” family of Chicago, viz., G. L., John, “Bill” and Jack, W. C. Budge, Springfield Gardens, N. Y. ; Richard Keuster, Brooklyn, N. Y. ; C. H. Hanover, Burlington, Iowa; George Lawrence, Jr., Sackville, New Brunswick (Are all you Nova Scotia chaps dead?); W. R. Gwynn, Longmont, Colorado; F. D. Orenbacher, Truesdale, Mo.; Charles Curie, Chattanooga, Tenn. ; E. Fergodo, Livermore, California; Frank Dudiak, Fairmont, W. Va. ; L. L. Ball, Empress Theatre, Arma, Kansas; D. G. Henderson, Quincy, 111.; T. R. Bankerton, Wentzville, Mo.; G. R. Hahn, Memphis, Tenn., and Charles Colby, Santa Fe, New Mexico, made more or less good on this one. I think Brother Gwynn is most nearly exactly correct and he is the only one who caught the point that induction and power factor must be taken into consideration where A. C. is to be transmitted — a calculation beyond any but the expert electrician I think. Let it be clearly understood that I often name men as answering correctly when, as a matter of fact, they have not answered exactly so. I consider whether the man really seems to understand the principle involved, rather than whether he may have made some error in calculation. Gwynn says : At the outset it must be stated that this calculation is for direct current and not for A. C., as induction and power factor are factors to be taken into consideration in calculating' the size of wire to be used in alternating current circuits. In as much as the frequency, character of the load, and certain physical conditions of the circuit must be known in order to calculate the size of the wire in A. C. circuits, it is evident that the author of the question did not intend that this problem should apply to A. C. circuits. One of the formulas for this calculation is reduced from five other successive formulas which appertain only indirectly to the solution of this problem, and would consume considerable space at no particular advantage. They are therefore omil ted. The final formula of the series may be stated as follows ; Circular mils = amperes x feet x 21.6 voltage drop. In which “circular mils” is the cross section area or size of the wire in circular mils; “feet” is the one-way length of the circuit; 21.6 is a constant representing twice the resistance of one mil foot of copper wire or the resistance of a copper wire 1 mil in diameter and a foot long at 75 degrees Fahrenheit. (This resistance is 10.79 ohms, which is doubled to allow for both wires of the circuit.) Voltage “drop” is the figure obtained by multiplying the impressed voltage by the percentage of allowable drop. With the above terms explained, the calculation of the wire size required is as follows; The voltage drop or pressure lost is 3 per cent, of 250 volts, or 7.5 volts. Substituting the known values for the unknowns of the formula. Circular mils = 200 amperes x350 feet x 21.6 7.5 voltage drop. Diameter of wire in mils equals square root of circular mils, or, Diameter = v 201,600 or 449 mils, or .449 inch. The size of the wire has now been determined in both mils and decimal parts of an inch. These must be converted into a standard commercial wire size. There are no less than seven standard wire gauges, among which the Brown & Sharpe Gauge is the standard, most commonly employed in the United States and Canada. Conversion tables from which wire sizes in circular mils and decimal parts of an inch may be converted into standard Brown and Sharpe sizes are to be found in Richardson’s Handbook of Projection and in all electrical engineering textbooks. Consulting these tables it will be found that B. and S. .0000 (211,600 circular mils, 460 mils in diameter, and .46 inch in diameter) is the standard size of wire which has the required area of cross section, though with some surplus. However, the job is not yet finished. The Fire Underwriters’ table of carrying capacity of wire sizes must be consulted, as this board, for reasons better known to itself, may demand an addition margin of safety. In this case this table shows that B. & S. .0000 may have a carrying capacity of 210 amperes for rubber covered wire and 312 amperes for wires with other insulation. This size, therefore, meets this board’s requirements, regardless of the type of its insulation, hence the answer is B. & S. .0000 copper wire. A very complete and a most excellent answer. Question No. 546 — What objections are there to overloading and thus overheating rheostat coils or grids ? It is hard to say just who has made the best answer, but I think I shall publish the reply of Brother “Bill” Doe, who says : Briefly, the objections to overloading rheostat coils or grids are: (a) the life of the resistance element will be shortened. (b) The characteristics of the metal will be slightly altered, and in such manner that the resistance offered thereafter will be somewhat increased, (c) The overload may, and if sufficient, will fuse the metal at some point, thus stopping all current flow until a repair is made, or if it be a multiple coil rheostat, then the current flow will be reduced by the capacity of the coil or coils burned out. (d) High temperature is objectionable from every viewpoint, and may, under some conditions, be highly dangerous, (e) Overloading any electrical apparatus is bad practice. Please understand, men, that if space permitted, I would very much like to publish several answers to nearly every question. It would be well to do so, because it supplies the element of different viewpoint's. Space will not permit, however, so it can’t be done. Note : Brother Curie questions Hanover’s answer to Question 535 in which he says the mil foot standard has been found to be 10.79 ohms, but ordinarily is quoted at 10.5 ohms. Says tables in various reliable text books give it at 10.505 ohms. Wants to know who is right, Hawkins or the other text books. The other text books are correct, but unless we are dealing with very large or very delicate problems the additional .005 of an ohm is too small to be of any importance, and it complicates calculations considerably. In any projection problem I can think of it is entirely negligible, therefore, I have never criticized its omission. As to Vitaphone (Continued from preceding page) both the description and any published instructions would be misleading almost at once, because of constant changes being made in the apparatus, which is still in what he termed a “fluid” state of development. That, gentlemen, is why I have said nothing. When the time comes that the apparatus has been developed to a stage which gives promise of being at least fairly permanent, this department will give you full information concerning it. It now is delivering very wonderful results, which may or may not be further improved, but the apparatus as now used will have slight resemblance to the apparatus used within a comparatively short time. That much I have on the highest authority. Worden Submits A Real Puzzler AE. WORDEN, local representative Courtland and Temple Theatres, # Courtland, N. Y., puts up one I am ■ free to say has me very well puzzled. He says they use Powers projectors with Mazda equipment. Left side of picture is out of focus, so much that it is very noticeable on titles. Have tried moving screens and projectors. The trouble developed recently. Prior to its development the picture was sharp and clear all over. My immediate assumption was, of course, that the trouble was in the projection lens, so I wrote them to try rotating the lenses. They write that this was tried and without helping anything. If one side cleared up the other was thrown out of focus. Fine! It’s the lens alright! Then I read on and got this heaved at me: “In fact the trouble is the same when we tried one of the lenses from the other house!” So lens trouble seems to be “out.” Ho hum! Dad bing these chaps anyhow! They’re always springing something which gets the editorial brain all frothed up. The fact that rotating the lens clears up the bad side and fogs the good one automatically places the trouble in the projection lens, BLTT the lens from the other theatre gives, presumably, a perfect picture, both before and after, so there you are! I called up Herbert Griffin, International Projector Corporation, and read the letters to him. His comment, was: “Well, Rich, that’s one of the hot ones. The thing MUST be in the lens. It is one of those things one must be on the ground to answer.” There was another possibility, viz: that the aperture plates were a bit further from the lens on one side than on the other, but that possibility was automatically discarded when rotating the lens cleared one side and fogged the other. So I dunno! If any one can offer any suggestions, for the love of Mike stand up and broadcast it.”