interpupillary distance in cm =
= 0.17 + (0.59 * (inner canthal distance in cm)) + (0.41 * (outer canthal distance in cm))
Derivation:
(1) The lines in the figure are parallel, with outer canthal distance the X axis, interpupillary distance the Y axis and the slope of 0.41.
(2) The Y-axis intercept varies for each line and is a function of inner distance. When these are solved for, the line with slope 0.59 and Y axis intercept 0.17 is derived.
NOTE: Feingold and Bossert's paper has a typographical error, with 0.7 printed instead of 0.17
Lateral displacement of the inner canthus can be identified if:
(1) The ratio of the (inner canthal distance in cm) / (interpupillary distance in cm) > 0.6 (from page 146 Hall).
(2) The examiner observes that a vertical line passing through the inferior lacrimal point crosses the iris.
NOTE: I noted that some values from the graph of normal relations gave a ratio > 0.6, so it may be prudent to use both criteria before making the diagnosis of lateral displacement of the inner canthus.
Limitations:
• The patient must hold the eyes perfectly still and looking directly forward. This may be difficult to achieve in an infant or small child.
• I am not sure if the equation is valid for all races.
