If you’ve made it as far as optometry school, there’s no doubt that you have been able to conquer some pretty tough math classes: trig, algebra, calculus, stats, the works. And yet, when it comes to applying the near-far AC/A equation in clinic, I know I can’t be the only one whose brain seems to shut down. The equation I learned during 2nd-year is absolutely riddled with mathematical land mines – especially when it comes time to do a calculation in your head:
AC/A = [(V2 +Ph2) – (V1 + Ph1)] / [(A2 – A1)]
Calculating vergence demand with an interpupillary distance (PD) other than 6.0 cm or accommodative demand for a near working distance other than 40 cm can be tough to do without a pen and paper – especially with the added need to remember which units to use and which sign convention is appropriate. That’s why I love the other AC/A method outlined below – the less mental math, the better!
One of my main logic checks with calculating AC/A is to do a quick “PD check” right off the bat. This will help me know right away whether the AC/A is going to be larger or smaller than the interpupillary distance (PD in centimeters).
- If the patient is more exophoric (or less esophoric) at near than at far, the AC/A is going to be low (i.e., a smaller value than their PD).
- If the patient is more esophoric (or less exophoric) at near than at far, the AC/A is going to be high.
The below method of AC/A calculation incorporates this kind of thinking to help you understand why you might add or subtract from the distance PD depending on the phorias.
AC/A = PD (cm) +/- [(Diff between far and near phoria) * Working distance (m)]
- If the phoria at near is more eso than at far, add it to the distance PD.
|Ex. Distance phoria is 2 exo, near phoria (at 40 cm) is 5 eso, distance PD is 60mmCalculate: AC/A = 6.0 + (7*0.40) = 8.8PD check: The patient is more eso at near than at far, so the AC/A should be larger in magnitude than the interpupillary distance in centimeters – and it is!|
- If phoria at near is more exo than at far, subtract is from the distance PD.
|Ex. Distance phoria is 2 exo, near phoria is 6 exo (at 30 cm), distance PD is 55mmCalculate: AC/A = 5.5 – (4*0.3) = 4.3PD check: The patient is more exo at near than at far, so the AC/A should be smaller in magnitude than the interpupillary distance in centimeters – and it is!|