The Moving picture world (November 1926-December 1926)

December 11, 1926 MOVING PICTURE WORLD 457 Bluehook School Answers 538 and 539 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. Note — -Through error Friend W. C. Budge was not credited with correct reply to ■question No. 530. Apologies, Brother Budge. Will see to it that this doesn't happen again — until next time, anyhow. I try to avoid ■error, but am only 'uman — or partly so, anyihow. Question No. 538 — What would be the voltage drop of a copper circuit of No. 6 wire working at capacity or less, if the one-way length of the circuit be eighty-five feet? W. C. Budge, Springfield Gardens, N. Y. ; C. H. Hanover, Burlington, la.; G. L. "Bill," John and William Doe, Chicago, 111.; A. L. Lehman, Glenside, Pa.; Charles E. Curie, Chattanooga, Tenn. ; E. Fergedo, Livermore, Calif.; Allan Gengenbeck, New Orleans, La.; G. R. Hahn, Memphis, Tenn. ; F. D. Orenbacher, Truesdale, Mo.; Gilbert Atkinson, ■Cleveland, Ohio; D. G. Henderson, Quincy, 111.; Albert C. Hancock, Dallas, Texas, and Gilbert OfFenbecker, Springfield, Mass., all either came through with a correct answer ■or evidenced the fact that while they might "be bum "figgerers," anyhow they understand how the thing is done. In addition, George Einzinger, New York ■City, makes a reply which I shall quote, though I am not so sure as to whether or not it is correct in principle. I would appreciate a statement of the source of his .authority for calculating voltage drop based •on one ampere and multiplying by the total amperage. It has always seemed to me that is how it really ought to be, but I have missed the rule which reads that way, if there is one. Brother Curie says : Having a circuit of No. 6 copper wire with a one-way leng-th of 85 feet, wishing to ascertain tlie voltage drop at capacity or less, we proceed as follows; Using our Bluebook, we find from Table No. 1. page 70, that the area of cross section of a No. C wire is 62,250 circular mils. Applying formula No. 3, Page 75 of the Bluebook, which 21xIL. is as follows: e = . In this formula A 21 is twice the mil foot standard of resistance, which we use instead of doubling the wire length because it is more convenient. "Li" is the one-way length of the circuit, "I" the current in amperes and "A" the area of -cross section of the conductors in circular mils. Substituting our own figures we have 21x50x83 voltage drop equals = 3.4, which 26,250 is the voltage drop of the circuit at capacity. Friend Einzinerer savs : Copper wire No. G has a cross section of 262.50 circ. mils. By length 2x8.5=170 feet and mil foot resistance being 10.80 ohms, we get Resistance = R = (170 x 10.80) -^ 2G250 = 0,27 ohms voltage drop in general accord. Ohms law: V = 1 x R. Therefore voltage drop for 1 amphere: V = lxR=lx 0,07=: 0,07 volts, to be multiplied with the number of the amperes flowing through the considered circuit. You will note that at capacity of a No. 6 wire (60 amperes for other than R. C. insulation) this would give 4.2 volts drop whereas at 20 it would only be 1.4. That is a new one on me and if it is right I want to know it. I could easily communicate with the authorities and find out, but would rather settle it among ourselves. Question No. 539 — Suppose you are asked what size wires you wish to have installed for the projection room circuit in a new theatre in which you will be Chief Projectionist.. The length of the circuit will be 80 feet. (Printer left out some words in the question). There will .be three M. P. projectors, each using 70 amperes, one spot using 60 amperes and one Brinkert effect projector using a total of 160 amperes. (They certainly did ball this question up RIGHT, but anyhow they could not well alter the principle, so it's all right and we will proceed.) Current is 8 cents per K. W. hour. In a general way tell us what size wires you would recommend, and on what you would base your recommendation. Budge remarks that the question is not complete in that it does not designate the kind of current or circuit — 2-wire or 3-wire. which is true. That was my own fault. I had in mind merely the bringing out of the underlying principles — the various things to be considered in such a matter, and did not word the question as carefully as I should. Then the printer helped some and it really is a pretty well mixed up mess. Taken as a whole, brother Hanover has handled the matter remarkably well, everything considered. He grasped the real intent of the question and handled the whole matter very well indeed. He answers thus : I do not believe it is the intention that we actually work out the problem in figures, but merely that we set forth our views as to what various things should be considered in such a matter, and I shall therefore answer along those lines. The question sets up a condition in which there is a possible total maximum amperage of 70 plus 70 plus 70 plus 60 plus 80 plus 80 (I assume the Brinkert is a dissolver, which I think they all are) equals 430 amperes. That is the total however it is possible to use. except for the fact that the variou.s lamps might, due to any one of several causes, use say ten percent higher amperage, also there is a small, and for the purpose negligible, incandescent light current consumption. I think, however, -we may assume 430 to cover it, but we must, due to incompletness of the question, also assume the current to be 110 v. D. C controlled entirely by rheostats, since if it be A. C. or 220 D. C. fken through a motor generator set or a rectifier, or A. C. through transformers, that would alter the necessary pro.ieotion room feeder size very materially. And again the question is incomplete in that it does not specify 2-wire or 3-wire circuit, therefore I shall assume a 2wire. First, we must use a bit of common sense. It is not within the range of reasonableness to presume that all arcs will ever be used at one time. If that were to he we would merely apply formula No. 3, page 75 of the Bluebook, or formula No. 5. same page, in either case selecting the voltage drop we believe will represent the highest efficiency, everything considered. And right here I might remark that 8 cents is very high where so much power is to be used therefore we will be fully justified in selecting a low percentage of voltage drop — say not to exceed two or at most three volts. It will be high in first cost (installation), but that will all come back in time through the constant saving by minimi^iing the loss, which like the interest on the mortgage, works constantly, or at least every moment the circuit is in use. And now let us consider the probabilities. Would three M. P. projectors ever be used at the same time? Most emphatically no! Would two of them be used at once. Again no, but qualified by the fact that their arcs might be burned for a considerable time simultaneously, while, with care, a minute or two would suffice, the fact remains that where a man has his hands pretty full in handling the equipment, as is the case in many modern installations these hectic days, he cannot and will not give close attention to the lighting the arc of th projector next to be used, at just the right time, or may not immediately cut off the arc of the projector that has just finished. We therefore must for the sake of safety supply full wire capacity for two M. P. projector arcs, which is 70 X 70 = 140. Next we must consider the spot and the Brinkert, and here I shall take a position which some may hold unreasonable, but I do not think it is. This is a new theatre, and after all a few dollars more or less in construction cost is not a serious matter, BUT a possible permanent injury to wires, after installation by overloading IS a very serious matter. I have known of a motion picture projector projecting a picture or cartoon, with an effect by a Brinkert and a spot on a musician at the same time. Unusual, true, but it has been done, and may be done again, hence I think we must add to our 140 the spot amperage (GO), and at least one Brinkert lamp, which also is 80 making a total of 60 plus 80 plus 140 equals 280 amperes. And now here is why I said what I shall recommend may seem unreasonable. I shall add the other Brinkert lamp, making a total o£ 280 plus 80 equals 360 amperes. My reason for this is because it is not altogether impossible, however improbable, that all these lamps may not be sometime used at one and the same time, and anyhow it is quite entirely possible that the amperage at the motion picture projector arcs may be increased in the future, and if there is no surplus projection room circuit capacity where would we be "at?" Under the conditions named I do not believe it would be really good practice to install feeders of less than 360 amperes capacity, which certainly would take care of future amperage incieases without overloading, provided the two Brinkert's lamps and the spot be not used while a M. P. projector is in operation and that stunt may always be side stepped I think without serious loss. As to the percentage drop, I would not consider a drop in excess of three volts with current at 8 cents, and two would. I believe be the better practice, though that would depend upon the added cost of the larger feeders. I could not determine that until I knew just what it would be, therefore I would consult with the electrical contractor as to the cost of wires to supply two and three volt drop. The Bluebook advises us that the determining factor in such a matter is whether or no the loss in power by reason of voltage drop would be sufllcient to pay interest on additional installation charges of wires large enough to reduce the drop. That is mighty good and sound advice, but it must be applied with common sense, and it is not. I think, possible to take up and discuss all phases oi sucli a matter in a book, unless space there in is unlimited. The possibility of future increase in amperage is one important factor, especially when the present M. P. projector amperage is only 70, because that is rather low nowadays. Gentlemen, I think you will all agree with me that this is a very competent answer to the question. In the main, the question was well answered by Curie, John and "Bill" Doe, Fergodo and some others, but I believe even they will agree that Hanover has handled the whole proposition best. As to the method Brother Enzinger uses in calculating resistance by amperage, what do YOU think about it or do you know authority for it.