The Moving picture world (November 1926-December 1926)

178 MOVING PICTURE WORLD November 20, 1926 Bluebook School Answers 530 and 531 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. 530 — A practical question submitted l^iy John Griffith, Ansonia, Conn. Suppose you have two 60watt, 110volt incandescent lamps burning in series on 220 volts and one of them bums out. You find you have not another, but instead have a 40watt, 110volt lamp, which would supply sufficient illumination. Could you use it in scries with the 60-watt lamp temporarily, while a lamp of the proper wattage is secured? l^vidently this one was a goatgetter i)ar excellence. John Griffith, Ansonia, Conn.: C. H. Hanover, Burlington, la.; C. E. Curie. Chattanooga, Tenn. and George Einzinger, New York City, are the only ones whose replies I could consider as really correct and Curie and Griffith disagree in their answer as a whole, because Griffith says it would not work because of the overload of the smaller lamp, whereas Curie says it would work, but would greatly decrease the life of the smaller lamp. The latter is correct. Griffith proposed this problem to me verljally, and I made the mistake of not selecting lamps further apart in capacity. Curie is correct. Griffith is correct in what he really had in mind when he proposed the question. Hanover puts the matter in this wav. He says : "This question may be answered with a 'yes,' but such an answer would not, I think, convey the real meat of the 'nut.' It would probably not be correct if we view the matter in accordance with what I believe Brother Grlfflth really had in mind. "The wattage consumed by an incandescent lamp is equal to the E. M. F. multiplied by the amperage used (E x C), hence 60 110 = 0.5454.'5 amperes, which is the normal current consumption of a 60 watt, 110 volt lamp. If we then divide the voltag-6 of the lamp. (110) by its normal amperage, we find it to have 201.6666 plus ohms resistance. "Proceeding similarly we find that the resistance of a 40 watt, 110 volt lamp is 302.22.'30 plus ohms and its normal amperage 0.36.36 plus amperes. "The combined resistance of one 40 watt and one 60 watt, 110 volt lamp would there be 201.6666 + 302.2250, or .503. 8916 plus ohms, which would allow f220 -= .')03.8016 plus) 0.4366 plus amperes to flow under a pressure of 220 volts. "We therefore would have a current flow of that amount with the 60 and 40 watt lamp in series, and that, while not sufficient to bring the 60 watt lamp up to candle power, would overload the smaller lamp, •and while probably not sufficiently so to jiecessarily cause it to burn out immediately, it would shorten its life greatly, while at the same time the 60 watter would be working very inefficiently. "What I believe Brother Griffith had in view was a difference so great that the smaller lamp would burn out at once." In that latter presumption you are, as I have already said, correct. The combined resistance of the filament of the low wattage lamp and of the higher wattage lamp does just as Hanover says, as Griffith says and as Curie says. If the difference were greater, the smaller lamp would probably burn out at once; 110 volt lamps of different wattage must not be used in series on 220, and they cannot be so used if the difference is very great. Apology to W. R. Gwynn, Longmont, Col. : I just discovered that a page containing correct reply to this question had adhered to the back of a letter from Brother Gwynn. He shows by diagram that the 60 watt lamp would consume 88 volts and the 40 watter 132 volts of the total pressure. I shall print his answer, which is in excellent form. He says : "Contrary to the wording of the question, the 40 watt lamp would supply sufficient illumination, assuming the 60 watt lamp to be in good condition, until it burned out. Its life would, however, be very .short, due to excess voltage it would receive, as shown below. J3Z iblLS 40 w fiSLlt 110 V 60 w 605 'Amperes Amperes, Current consumed at 110 V. pressure. fi = — Amperes, Current consumed at 110 V 11 110 V. pressure. Resistance of 40w lamp E llOv = — = 302% C 4 ohms or 005 ohms. Resistance of 60w lamp ohms or 0O."> olims. 11 llOv 6 11 = 201% Resistances in series = sum of Individual resistances. Total Resistance in circuit 605 605 or 2^3 3025 ohms. 6 E 220v 264 f = — or = Amperes. R 3025 605 6 E — CR, or E at 40w lamp ; 605 ohms = 132 volts. 2 i: ^ OR, or E at OOw lamp fi05 ohms = 88 volts. 264a 605 264a 605 Overload at 40w lamp = 132v. — llOv. = 22 volts. Underload 60\v lamp = llOv. — 88 v. = 22 volts. Underload, therefore equals overload. Total volt'ige accounted for: 132v. + 88v. = 220v. Friend Einzinger replies correctly as to voltage, amperage and results to the smaller lamp. He then adds : "Considering the varying resistances of the lamp filaments, after a more complicated calculation, we get the following exact valuation in illumination powers etc.: For the 00 watt lamp, 72.5 volts; 31.1 w.; 0.430 amperes and 10.5 candle power. For the 40 watt lamp we have 147.5 volts; 63.5 W. ; 0.430 amperes and 92 candle power, so that our first calculation is proved not to be far from the facts, except in the matter of Illumination delivered. The 60 watt lamp burns with one-fifth of Its normal brightness, while the other is nearly three times as bright as normal. This latter should burn out in about two hours, the tungsten being destroyed by its high temperature. "By the way, a few words about candle (also called candlepower) and lumens of a light source. "We can define one lumen as the amount of light which falls upon a spherical surface of one square foot, held at a distance of one foot from the center of the light source, which sends out one international candle uniformly in all directions. As there is space in the total surface of a sphere of one foot radius a number of 4 pi =4x3, 14159 = 12,57 square foot, a light of the amount of one international candle signifies 12,57 lumens. "The international candle Is used In U. S. A., Great Britain and France; Germany uses the one-tenth smaller hefner candle. "Incandescent Mazda lamps give, at normal wattage, 10.4 lumens, or 0.83 candles per watt, that means a 60 watt lamp gives normal 60 x 0.83 — 50 candles or 628 lumens. A 40 watt lamp gives normal 40 x 0,83 = 33 candles or 416 lumens." Question No. 531. — What is meant by "normal temperature'' as applied to electric conductors? What is normal temperature? C. H. Hanover, Burlington, la. ; Frank Dudiak, Fairmont Theatre, Fairmont, W. Va. ; C. E. Curie, Chattanooga Tenn.; W. C. B'udge Springfield Gardens, N. Y. ; A. L. Hutchinson, Paducah, Ky. ; "Bill" Doe, John Doe and G. L. Doe, Chicago, III, ; T. R. Guimond, Mobile, Ala. ; Allan Gengenbeck, New Orleans, La.; D. G. Henderson, Quincy, 111.; .Mbert Hancock, Dallas, Texas ; G. L. .'Mbertson, St. John, New Brunswick; Andrew Pauldon, St. Louis, Mo. ; and T. R. Dickenson, San Francisco, Cal., all answered this one correctly. Many, somewhat to my surprise, answered that normal temperature was a temperature equal to the surrounding air, without further remark. That is both correct and incorrect. Temperature of the air is "normal" temperature all right, but a standard has been set up, known as "normal temperature," because air temperature varies widely. I think Brother Curie covers the matter fully, thus : "Strictly speaking, normal temperature would be the temperature of the conductor when heated only by the surrounding air under ordinary conditions — the "open'' air. It would, however, be impossible to use this in practice, due to wide variations, seasonable and climatic, hence authorities have adopted 75 degrees Fahr. or 24 degrees Cent., as the standard normal temperature." Which is eminently correct in its entirety, hence Brother Curie may, having first removed his headgear, wiped his nose and dusted off his ears, waddle up to the head of the class, remaining there until some enterprising genius bumps him back again.