File:The American annual of photography (1922) (14758622656).jpg

Identifier: americanannualof36newy (find matches)
Title: The American annual of photography
Year: 1922 (1920s)
Authors:
Subjects: Photography
Publisher: New York : Tennant and Ward
Contributing Library: Harold B. Lee Library
Digitizing Sponsor: Brigham Young University

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given in terms ofinches of spark gap (back-up), milli-amperes of current,seconds of time and inches of distance. If the parts to beradiographed are of uniform size and thickness for a greatlength of time, the distance and spark-gap being the same,the average exposure may be expressed in terms of milli-ampere-seconds and time: E = Exposure; M. A. = Milli-Amperes ; S ^ Seconds Time. Or E = M. A. S. in seconds.It has further been found the reaction produced on aphotographic plate by the X-Ray varied directly as thecurrent and the time, i. e., an increase in either one of thesefactors will cause a proportionate increase in the photographicreaction. Let us for example assume that for a given nega-tive a correct result is obtained by having the tube at a certaindistance from the body, the back-up and the milli-amperes allcorrectly computed. The same result may be obtained bygiving an exposure of half the time provided we double themilli-amperes, and so on any desired time may be obtained 90
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STONE IN THE KIDNEY. Illustrating article Photography as Applied to Radiography, by S. A. Schwarz. 91 provided we make a proportionate increase in the milli-amperage. Taking a concrete example let us assume the following:Conditions:— Distance = 20 inches M. A. = 20 Spark = 3 inchesCorrect exposure time = 12 sec. Required: The expos-ure time if the distance is changed to 15 inches, no otherfactors being changed. 12 : X = 2o2 : 152The product of the means being equal to the product of theextremes we have:— 400X = 12 X 225Solving for X we have:— X = 6% seconds.Problem No. 2. What would be the correct exposure if the current weredoubled ? Taking the original example as given above we have:Since doubling the current doubles the intensity of the raythe time requirement for the exposure will only be one half,in other words the correct exposure will be 6 seconds. In conclusion it may be said that although these variousformulae and other methods will to a certain degree helpthe Roentgeno

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