Radiography in Modern Industry - Kodak

Nomogram MethodsIn Figure 54, the scales at the far left and far right are relative exposure values. They do notrepresent milliampere-minutes, curie-hours, or any other exposure unit; they are to be consideredmerely as multiplying (or dividing) factors, the use of which is explained below. Note, also, thatthese scales are identical, so that a ruler placed across them at the same value will intersect thevertical lines, in the center of the diagram, at right angles.On the central group of lines, each labeled with the designation of a film whose curve is shown inthe Figure 47, the numbers represent densities.The use of Figure 54 will be demonstrated by a re-solution of the same problems used asillustrations in both of the preceding sections. Note that in the use of the nomogram, thestraightedge must be placed so that it is at right angles to all the lines--that is, so that it cuts theoutermost scales on the left and the right at the same value.Example 1: Suppose a radiograph made on Film Z (See Figure 47) with an exposure of 12 mAminhas a density of 0.8 in the region of maximum interest. It is desired to increase the density to2.0 for the sake of the increased contrast there available.Place the straightedge across Figure 54 so that it cuts the Film Z scale at 0.8. The reading on theoutside scales is then 9.8. Now move the straightedge upward so that it cuts the Film Z scale at2.0; the reading on the outside scales is now 41. The original exposure (12 mA-min) must bemultiplied by the ratio of these two numbers--that is, by 41/9.8 = 4.2. Therefore, the new exposureis 12 x 4.2 mA-min or 50 mA-min.Example 2: Film X has a higher contrast than Film Z at D = 2.0 (See Figure 47) and also lowergraininess. Suppose that, for these reasons, it is desired to make the aforementioned radiographon Film X with a density of 2.0 in the same region of maximum interest.Place the straightedge on Figure 54 so that it cuts the scale for Film Z at 2.0. The reading on theoutside scales is then 41, as in Example 1. When the straightedge is placed across the Film Xscale at 2.0, the reading on the outside scale is 81. In the previous example, the exposure for adensity of 2.0 on Film Z was found to be 50 mA-min. In order to give a density of 2.0 on Film X,this exposure must be multiplied by the ratio of the two scale readings just found--81/41 = 1.97.The new exposure is therefore 50 x 1.97 or 98 mA-min.Example 3: The types of problems given in Examples 1 and 2 are often combined in actualpractice. Suppose, for example, that a radiograph was made on Film X (See Figure 47) with anexposure of 20 mA-min and that a density of 1.0 was obtained. A radiograph at the samekilovoltage on Film Y at a density of 2.5 is desired for the sake of the higher contrast and thelower graininess obtainable. The problem can be solved graphically in a single step.The reading on the outside scale for D = 1.0 on Film X is 38. The corresponding reading for D =2.5 on Film Y is 420. The ratio of these is 420/38 = 11, the factor by which the original exposuremust be multiplied. The new exposure to produce D = 2.5 on Film Y is then 20 x 11 or 220 mAmin.Sliding Scales For Exposure ChartsAn exposure chart is an exceedingly useful radiographic tool. However, as pointed out in"Preparing An Exposure Chart", it has the limitations of applying only to a specific set ofradiographic conditions. These are:1. The x-ray machine usedRadiography in Modern Industry 77