Sunday, February 6, 2011

Inverse Square vs. Direct Square

Learning about the inverse square law and the direct square law can be quite confusing at first. Once you learn the formulas and dust the cobwebs off of your algebra skills to be able to solve for the variable using the formulas, knowing WHEN to use WHICH formula is sometimes the biggest challenge you will face. In order to determine this, we need to know what kind of information the question is asking you for. Let's take a close look at each formula:

Inverse Square Law states: "The intensity is inversely proportional to the square of the distance."

Notice that the value for original intensity (I1) is in the numerator, and the value for the original distance (D1) is in the denominator, thus it is "inversely proportional to the square of the distance."

Use this formula when the problem asks you to solve for a unit of radiation intensity, dose, or exposure. Also, remember that radiation "intensity" is not measured in units of mAs, so if the question is asking you for a mAs value, this is not the formula for you. Units of radiation exposure or radiation dose are required for this formula (R - Roentgen, mR - milliRoentgen, rad, rem, Gy - gray, or Sv - Seivert).

Still confused? This simple tip could save you... you should now already have the fundamental knowledge that radiation intensity will decrease as the distance from its source increases. So look at your distance values: If the distance increases, then I2 should be a smaller number than I1. The opposite is true as well; if the distance decreases, the intensity will be stronger, and I2 will be a larger number than I1. This is important to remember when we discuss the direct square law:

Direct Square Law / Density Maintenance Formula:

Two main differences with this formula are: Instead of radiation intensity, we are using mAs values. Also, the original mAs and the original distance are both in the numerator - "direct" vs. "inverse." We need to be using this formula when the question asks for a mAs value.

*** side note: You know that radiation exposure is directly proportional to mAs. In other words, if I double my mAs value, the radiation exposure value will double. Remember, these two units are distinct and separate, but related to one another.

After performing a few practice problems, you may notice that as the distance increases, the mAs value will increase. This is due to the "direct" relationship. As the distance from the radiation source to the image receptor increases, the mAs required to maintain density (density maintenance formula) will increase. So, if your D1 value is smaller than your D2 value, then your mAs1 value should be smaller than your mAs2 value. Conversely, if your D1 value is larger than your D1 value, then your mAs1 value needs to be larger than your mAs2 value.

What if the values presented in the question provide units of radiation exposure/intensity/dose AND mAs values? Don't panic... just find out what the question is asking for, and apply what we have discussed.

Example: A radiographic exposure of the chest was taken at a distance of 72" using 10 mAs and had an exposure of 50 mR. What would the exposure be at a distance of 80"?

The question is asking "What would the exposure be ...?" Key word: exposure. This is your key term that determines we are looking for a unit of radiation intensity. First, fill in your variables:

I1 = 50mR
I2 = "x" or unknown
D1 = 72"
D2 = 80"

I trust your ability to solve once the equation is set up properly ;-) Remember that your distance is increasing, so your value for I2 should be smaller with this formula (this is how you can tell if you forgot to invert the distances - it will be a larger value if you forget).

Example: A radiograph of the knee produced 100mR of exposure when 70 kVp and 10 mAs was used at 40". What new mAs would be required at a distance of 60 to maintain density"?

Key term: "What new mAs...to maintain density?" This one screams, "Density maintenance formula!!!"

Fill in your variables and solve:

mAs1 = 10
mAs2 = x
D1 = 40"
D2 = 60"

Your mAs2 value should be greater than mAs1 because your distance is increasing.

On a final note - if you have time, double-check your work by plugging in your answers into the original equation. Don't forget to square your distances, and reduce fractions into the lowest terms before squaring to save you from having large numbers to deal with.

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23 comments:

  1. Hi, Can you please comment on an examination that need CR to be angulated? will the angulated CR affect the true FFD needed for particular examination?

    ReplyDelete
  2. Hi there, Quick question
    if I were to view a fine fracture in bone which one of the 2 would help and why?
    1. increasing kvp
    2. decreasing kvp

    great blog btw!

    ReplyDelete
  3. Hi there, Quick question
    if I were to view a fine fracture in bone which one of the 2 would help and why?
    1. increasing kvp
    2. decreasing kvp

    great blog btw!

    ReplyDelete
  4. Hi Jeremy,

    Can you email me please?
    I work in affiliation with a health care school and wanted to chat.

    Thanks,
    Vanessa

    ReplyDelete
  5. Hi Jeremy,

    Can you email me please?
    I work in affiliation with a health care school and wanted to chat.

    Thanks,
    Vanessa

    ReplyDelete
  6. Liyana: a good rule of thumb to keep FFD constant during tube angulation (like for an AP sacrum) would be for every five degrees of tube angle, reduce the vertical tube distance by one inch. For a tube angle of 15 degrees, bring the FFD to 37 inches. At the center of the image receptor, this should keep the 40 inch FFD constant.

    ReplyDelete
  7. D: I have always said that the kVp should be appropriate for the part. You can reduce exposure by increasing kVp, but visualization of a fine fracture should be obtainable by having an adequate exposure technique, doing anything you can to control the contrast (good collimation for example), and selection of the appropriate processing alogorithm. Did you have a specific body part in mind?

    ReplyDelete
  8. Goody: I don't think your email address is visible... if you would like to shoot me an email, my current address is jeremy.enfinger@gmail.com

    ReplyDelete
  9. Your inverse square law formula has a typo. "d1" is not squared.

    Patrick

    ReplyDelete
  10. Your inverse square law formula has a typo. "d1" is not squared.

    Patrick

    ReplyDelete
  11. In regards to patient dose, which is better:
    1)an increased SID, with an increase in mAs to maintain OD
    2)a decrease in SID, with a decrease in mAs to maintain OD
    3)#1 and #2 are the same in patient dose

    I would really appreciate your understanding of this. :)

    ReplyDelete
  12. In regards to patient dose, which is better:
    1)an increased SID, with an increase in mAs to maintain OD
    2)a decrease in SID, with a decrease in mAs to maintain OD
    3)#1 and #2 are the same in patient dose

    I would really appreciate your understanding of this. :)

    ReplyDelete
  13. @Anonymous: Thanks for your comment... it really depends what type of exam you're doing. If the body part you are imaging has a lot of OID, then you would want to use the higher SID to improve spatial resolution. If it doesn't, you could probably use a shorter SID and minimize wear and tear on the x-ray tube.

    I could be wrong, but I thought Mayo Clinic was trying to change the U.S. standard SID to 48" a few years ago relating it to patient dose, but I'm not certain what happened with it. If I can find any info, I'll post.

    ReplyDelete
  14. @Anonymous: Thanks for your comment... it really depends what type of exam you're doing. If the body part you are imaging has a lot of OID, then you would want to use the higher SID to improve spatial resolution. If it doesn't, you could probably use a shorter SID and minimize wear and tear on the x-ray tube.

    I could be wrong, but I thought Mayo Clinic was trying to change the U.S. standard SID to 48" a few years ago relating it to patient dose, but I'm not certain what happened with it. If I can find any info, I'll post.

    ReplyDelete
  15. Thanks for your quick response! My fellow student & I are are experimenting with 72"SID and 40"SIDs on a phantom for the posterior obliques of C-spine (so there is a lot of OID). The DR unit returns images with minimal difference in magnification and spatial resolution. The differences really seem to be whether the patient would receive less dose at either SID because of the change in SOD and the adjusted mAs.

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  16. @Anonymous: No problem... other things to consider with C-spine obliques: which projection would allow for best visualization of intervertebral joint spaces? Also, the thyroid gland is very sensitive to radiation - which position will reduce exposure to the thyroid gland? Trying not to give away too much, but good food for thought.

    ReplyDelete
  17. No problem... thanks for commenting!

    ReplyDelete
  18. In 'Direct Square', is "Conversely, if your D1 value is larger than your D1 value, "; OOPS!

    ReplyDelete
  19. Along the same question as Anonymous...
    Is ESE increased, decreased, or the same if mAs is increased to compensate for an increase in SID.

    Review texts offer increased in one edition as answer, another edition says decreased (stating because of beam divergence makes distance best protection). Another person says ESE remains the same. Can you elaborate? Thanks!

    ReplyDelete
  20. Along the same question as Anonymous...
    Is ESE increased, decreased, or the same if mAs is increased to compensate for an increase in SID.

    Review texts offer increased in one edition as answer, another edition says decreased (stating because of beam divergence makes distance best protection). Another person says ESE remains the same. Can you elaborate? Thanks!

    ReplyDelete
  21. I would have to review my textbooks for a precise answer, but I'll fill you in on my thoughts. As SID increases, the beam divergence (with no change in collimation) spreads over a larger surface area. When we increase the mAs to compensate, it is because there are fewer photons at higher SID reaching the patient in the original collimated field size (14x17 for example). If the SID is doubled, the intensity is 1/4 the original value. Because intensity is directly proportional to quantity of photons, we multiply the mAs by 4 to compensate, which means that the same number of photons have been produced at the shorter SID. We know the kVp has not changed, so I would tend to believe that if the technologists maintains the field size between old and new SID's, it would produce the same ESE. Let's open it up for discussion if anyone else has a differing opinion. Great question!

    ReplyDelete

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