I used the 1.25 because if the Deviation Value is a Positive 1 then by the way the deviation index is calculated the mAs used is 125% of the mAs required (mAs that would result in a Deviation Index of 0) . When the Deviation value is a negative 1 the mAs used is 80% of the mAs required (mAs that would result in a Deviation Index of 0). Hope that explains it. XRayBob
can we adjust the kVp to reach a DI of 0?
Turns out you can use New mAs = (mAs)(0.8)^DI for all values of DI! Just one exposure indicator and just one formula.
Thank you.
Ooops another error was pointed out. For problem 3 at time 7 minutes the slide erroneously says the technique was overexposed. Sorry
I just don't understand where ypu are getting the 1.25 and the 0.8 from
I used the 1.25 because if the Deviation Value is a Positive 1 then by the way the deviation index is calculated the mAs used is 125% of the mAs required (mAs that would result in a Deviation Index of 0) . When the Deviation value is a negative 1 the mAs used is 80% of the mAs required (mAs that would result in a Deviation Index of 0).
Hope that explains it.
XRayBob
@@bobgrossman1 How do you figure 125% and 80%?
@@ashleyjones8665 5/4 = 125% while 4/5 = 80%. A DI of -1 means you used 80% of the optimal mAs a DI of +1 means you used 125% of the desired mAs.