Optical Theory question

[Dustin.B]

Afternoon folks,

I am hoping you can enlighten me on something, I am working my through the N.A.O's OCPP and have come across questions about a particular optical rule but I cannot find the rule in any of the chapters I have nor online. What I am searching for is the following...

"Gerstman's Three Quarter Rule"

Q= "Using Gerstman's three-quarter rule, what is the near PD for a patient who has a distance PD of 64 with an Rx of OU -1.00sph, +2.50 add, and is using a working distance of 30cm?"

A= " 1/.3 = 3.33 .75 x 3.33 = 2.49 or 2.5mm in per eye NPD = 59"

Normally I could review the answer as this is just a practice question and figure it out from there but I just am not seeing it here. Help a fella out?

~Dustin

[Robert Martellaro]

Dustin,

The three quarter rule says that for every diopter of dioptric demand, the segment/OC should be inset .75mm. The dioptric demand is the inverse of the working distance, in this case 3.33 D. It is an approximation for distance PDs between 62mm and 68mm and distance RXs with low power on the horizontal meridian.

Hope this helps,

Robert Martellaro

[drk]

So for a lower add it will spit out this:

+1.00 add
Inset 0.75mm per eye?
So DPD in this case is 64, and NPD would be 62.5? That's an "under inset".

That rule doesn't work UNDER +2.50 add.


At the +2.50 add, it would say inset ~2mm per eye, ergo 64/60, which is closer to reality. But on a narrow DPD of, say 57, it would over-inset (53) which could be a problem with higher-power lenses.

That rule doesn't work outside "average" DPDs.

So it's a rough rule of thumb indeed.

There is a secret formula...but it will take a PM to get it. (No consumers or onliners allowed.)

[Robert Martellaro]

drk,

The dioptric demand is the inverse of the working distance, not the add power.

If you need an exact value, PAL design for instance, you'll probably want something like Ellerbrock's formula.

Best regards,

Robert Martellaro

[drk]

Oh, I get it...working distance!

Makes more sense, now.