Understanding Neutron Radiography Reading V-Kodak Part 2 of 3

The antilog of 0.63 is 4.3, which means that 4.3 times more exposure is
required to produce a density of 3.5 than of 1.0. It is therefore desired that the
thinnest portion of the object to be radiographed transmit exactly 4.3 times
more radiation than the thickest part, so that with the proper adjustment of
radiographic exposure, all parts of the object will be rendered within the
density range 1.0 to 3.5. The ratios of x-ray intensities transmitted by different
portions of the object will depend on kilovoltage; examination of the exposure
chart of the x-ray machine reveals the proper choice of kilovoltage. For
example, in the chart shown in the figure below, the 180 Kv line shows that a
thickness range of about 7/8 to about 11/4 inches of steel corresponds to an
exposure ratio of 35 mA-min to 8 mA-min, or 4.3, which is the ratio required.
The next problem is to determine the radiographic exposure needed. The
chart shown below gives the exposure to produce a density of 1.0 on Film X.
Since it is desired to produce a density 1.0 under the thick section (11/4
inches), the exposure time would be 35 mA-min.
Charlie Chong/ Fion Zhang
Radiography in Modern Industry. Rochester, NY: Eastman Kodak Co. 1980