What is the projected image width of a part 6 inches wide at an SID of 44 inches and 9 inches from the IR?

To determine the projected image width of a part that is 6 inches wide at a Source to Image Distance (SID) of 44 inches and an object distance of 9 inches from the image receptor (IR), you can apply the concept of magnification.

Magnification in radiography is calculated using the formula:

[ \text{Magnification} = \frac{\text{SID}}{\text{SOD}} ]

where SOD (Source to Object Distance) is the distance from the X-ray source to the object. The SOD can be calculated as:

[ \text{SOD} = \text{SID} - \text{Object Distance} ]

In this case:

• SID = 44 inches

• Object Distance = 9 inches

• Therefore, SOD = 44 inches - 9 inches = 35 inches.

Now calculating the magnification:

[ \text{Magnification} = \frac{44 \text{ inches}}{35 \text{ inches}} \approx 1.257 ]

Now to find the projected image width, you multiply the original object width by the magnification factor:

[ \text{Projected Image Width} = \text{Object Width} \times \