pal pt w/ 15Δ BO

[optical24/7]

oh daaaaang.... two questions:

what about opposing prisms in each eye (one up and one down) and single vision lenses?

Applies to all prism. Prism not only displaces viewed object, it also displaces the pupil.

[drk]

OK, I admit that this stuff confuses me, but the prism doesn't displace the pupil except for maybe the wearer's face in the mirror. Only incident image displacement is what we're talking about.

You're confusing me. Stop it.

When measuring for fitting height, you're using demo lenses without prism, so it doesn't matter.

[Prentice Pro 9000]

Applies to all prism. Prism not only displaces viewed object, it also displaces the pupil.

So to be clear if I had a patient with a 5 BU and a 5 BD pal rx I would move the right lens down about 1.5mm and the left lens up about 1.5mm. And this is something that is done every time, correct?

[drk]

This stuff makes no sense to my brain. Can we start at the beginning?

Let's say a person has a "higher right eye/hyperorbit". You put the fitting height at the correct individual height, and we're done.

Let's say a person has no hyperorbit and their eyes are level. But the right eye turns upward due to a muscle problem. So, say 6^ BD prism is prescribed.

You put the patient in a frame, you probably should occlude one eye then the other (like a pupillometer does) and dot the pupil center with a non-dried-out marker. (Normally you wouldn't need to occlude an eye for a vertical MRP, but maybe this patient's right eye is shooting up because his left eye is dominant and looking at your eyes so you force the right eye to look at you by covering the left.)

In this scenario, the dots would be the same fitting height: 24 high, let's say, because you occluded and the guy has normal orbital symmetry.

So, for sake of simplicity let's load all the prism in the right eye, and not split it.

When verifying, you get out a ruler and mark the specified vertical MRP, and put the dot on the lensometer. You should read 6^ BD OD, none OS.

You put it on his face, and the right image shoots up. The right eye then either "relaxes" (because it's a phoria and he was straining to maintain binocular fusion) or "stops seeing double" (because it's a strabismus and he couldn't fuse to get a single image).

Where is there a need to compensate anything?

[Prentice Pro 9000]

So IIRC 1 prism diopter will move an object 1cm at 1meter. that works on both sides of the lenses, but I thought it was only a cosmetic thing. wtf. Somebody needs to draw this out

[AndyOptom]

Old but gold, a thread about this particular type of compensation, with input from the late (and great) Daryl Meister:

Fitting PALs with prescribed prism

TLDR: compensate the centration by 0.3 mm towards the apex of the prism, per prism dioptre. 'Pretty inconsequential' below 3 prism dioptres (per the linked thread), which makes sense since that's less than 1 mm of compensation.

[Kwill212]

So IIRC 1 prism diopter will move an object 1cm at 1meter.

Yes. Because the object image is displaced towards the apex of the prism, the eye has to deviate .27mm towards the apex of the prism, at the lens surface for an average stop distance of 27mm. For simplicity, assume 30mm, or round, and you get .3mm per diopter.

Forget prism for a moment. Look at target on a wall at a distance of 1000mm, now physically move that target up 10mm. Your eye will have to physically rotate upward to look at the new target location. The amount your line of sight travels across the lens to the new target can be calculated with simple trig.

Same thing applies whether it is a real object moving or the image of the object being displaced by prism.

Pick up a copy of System for Ophthalmic Dispensing. It has very detailed explanations and a lot of pictures.

[drk]

Still makes no sense to me.

If we're talking about PALs, we're talking about binocularity through skinny-winny corridors. Maybe skinny near zones.

I don't know who but Andy can run with me on this, but hear me out: If there is prism, one eye will, in the binocular state, deviate outwards (let's just use the monocular exo case). So the first-order thinking is "well, the eye is pointing out, now, so it can't use the corridor...".

But here's my obstacle: Sure, the eye is turned out because the prism moved the image out there, but it also moved the corridor out there as well, optically.

The way to test it would be this: take a frame off the board and use scotch tape to make a sliver of clear lens simulating a corridor. Then look through 15^ BI at a near object. (Your eyes will not be converged anymore, but dead straight like you're looking at distance.) Can you see through the openings in either eye (have to judge this in a binocular condition, not close one eye then another)? Or do you see clear in one eye and fuzz in the other?

I"m gonna do it.

But here's what I don't get, and never did (and it goes with the whole short corridor for myopia argument):

[drk]

Dang it, i ruined my ar-coating with the scotch tape!

[Prentice Pro 9000]

Yes. Because the object image is displaced towards the apex of the prism, the eye has to deviate .27mm towards the apex of the prism, at the lens surface for an average stop distance of 27mm. For simplicity, assume 30mm, or round, and you get .3mm per diopter.

Forget prism for a moment. Look at target on a wall at a distance of 1000mm, now physically move that target up 10mm. Your eye will have to physically rotate upward to look at the new target location. The amount your line of sight travels across the lens to the new target can be calculated with simple trig.

Same thing applies whether it is a real object moving or the image of the object being displaced by prism.

Pick up a copy of System for Ophthalmic Dispensing. It has very detailed explanations and a lot of pictures.

I'm going to crack that open as soon as I get home!

[Robert_S]

Still makes no sense to me.

If we're talking about PALs, we're talking about binocularity through skinny-winny corridors. Maybe skinny near zones.

I don't know who but Andy can run with me on this, but hear me out: If there is prism, one eye will, in the binocular state, deviate outwards (let's just use the monocular exo case). So the first-order thinking is "well, the eye is pointing out, now, so it can't use the corridor...".

But here's my obstacle: Sure, the eye is turned out because the prism moved the image out there, but it also moved the corridor out there as well, optically.

The way to test it would be this: take a frame off the board and use scotch tape to make a sliver of clear lens simulating a corridor. Then look through 15^ BI at a near object. (Your eyes will not be converged anymore, but dead straight like you're looking at distance.) Can you see through the openings in either eye (have to judge this in a binocular condition, not close one eye then another)? Or do you see clear in one eye and fuzz in the other?

I"m gonna do it.

But here's what I don't get, and never did (and it goes with the whole short corridor for myopia argument):

It's not to do with measuring. It's to do with glazing. When blocking the lenses, the prism causes the perceived placement of the uncut lens to shift, and so you must compensate for that.

[drk]

Hmm. So you're saying that you're trying to (forgive my ignorance!) block up an uncut? And you are using that lighted grid thingy?

[Robert Martellaro]

Scroll down to "Puzzle" and the differences between virtual and real images.

https://www.eophtha.com/posts/fun-with-prisms-in-ophthalmology

[drk]

First year optometry school.

[AndyOptom]

It's not to do with measuring. It's to do with glazing. When blocking the lenses, the prism causes the perceived placement of the uncut lens to shift, and so you must compensate for that.

This was my understanding of it as well, yes.

As Robert (S, not M) has mentioned, during blocking, the view of the blocker mires through the lenses are displaced by the prismatic Rx, and so we need to compensate the centration accordingly.