I'm losing it. What's up with PDs and high wrap frames?

[Lensman11]

Barry is correct the issue is how much do you compensate for the wrap. This was an issue before wrap frames became popular when we had to fabricate high base lenses I.e. cataract lenses 12 base and above. The solution we used was to have metal pd rulers curved to different base curves and measure the frame pd with the corresponding ruler to the base cover of the lenses It was even necessary with 8 base curve lenses though the impact was not as great.

[Prentice Pro 9000]

First you have to distinguish between the optics of a wrap glass and the mechanics of the same.
The optics pretty well covered above.
For the mechanics, try only thinking about a lens with a non-symmetrical front surface, such as a progressive or fully optimized digital single vision.
When you wrap a frame, you effectively narrow the chord of it’s “PD.”
This must be addressed in blocking by either:
1. Widening the blocking PD to compensate, or
2. Narrowing the DBL, which is functionally the same.

Both of these are mechanical adjustments in blocking purposes only. They would also hold true for bifocals, in theory.

B

If all is edged correctly with a PAL, then the PD should measure correctly? for instance, take a look at the pic below. It is a PAL and it's supposed to have a PD of 63 put end to end it is measuring about 57

[Robert Martellaro]

If all is edged correctly with a PAL, then the PD should measure correctly? for instance, take a look at the pic below. It is a PAL and it's supposed to have a PD of 63 put end to end it is measuring about 57

[ATTACH=CONFIG]15568[/ATTACH]

The fitting and prism reference point should measure 63 with a tolerance of 0.33 Δ or 1.0 mm, depending on the power. Use a cutout chart to measure.

https://www.essilorpro.com/content/dam/essilor-pro/varilux/documents/124991_PRO_VAR.pdf

Best regards,

Robert Martellaro

[drk]

True/false?

On PP9000's job, if the lab simply hasn't compensated for the ~15 degree wrap, the 63 ordered would come back about 61 due to lack of compensation.


True/false?
That huge a deviation from what was ordered exceeds the effect of merely having a wrapped frame. In other words, what could possibly be the excuse for that?

[Robert Martellaro]

I'm not a wrap expert, but to the best of my knowledge the FP and PRP should match the IPD regardless of the degree of tilt.

[drk]

Absolutely right.

The question in the above case is: why didn't that happen? Lab failed to compensate? Lab thought it was already compensated? But under no circumstances should it be off by 6 mm.

[Kwill212]

So I took a +3.50 and -3.50 and tilted them wrap style against a flat plane. I have to say... the mystery only gets thicker; the plus lens cleary produces a BO result, while the minus appears to keep everything still and just adds a pincushion effect to the image.


[ATTACH=CONFIG]15566[/ATTACH][ATTACH=CONFIG]15567[/ATTACH]

In Introduction to Ophthalmic Optics (Meister and Sheedy) this is referred to as Prism by Obliquity, Systems for Ophthalmic Dispensing has a section titled "Induced Prism with Wrap-Around Eyewear". This prismatic effect is due to the wrap angle, the front base curve, index of refraction, and the thickness of the lens. The induced prism can be approximated by the equation given by Meister and Sheedy:



Δ = the prism induced
θ = the angle of tilt
t = the thickness of the lens at the reference point
in meters
n = lens refractive index
F1 = the front curve of the lens




As you can see power of the lens for this type of induced prism is irrelevant. The reason there is such a difference between your +3.50 and -3.50 experiment is due solely to the base curve and thickness. The +3.50 is obviously going to have a thicker center and be on a higher base curve. You can see the effect of this type of prism by just picking up any plano lens(high base is better) and tilting it left to right about the 90 degree axis.

As for your other question about the Costa frames. I would not accept that. In my expertise sunglass companies are awful at making and measuring lenses. The way I measure high wrap jobs is to place the 34mm mark of the PD stick on the right lens nasal engraving then measure the the left nasal marking. The influence of wrap between these two points in going to be negligible. You can also just measure the distance between nasal engravings and add 34mm.

*edit fixed equation picture

[Prentice Pro 9000]

Absolutely right.

The question in the above case is: why didn't that happen? Lab failed to compensate? Lab thought it was already compensated? But under no circumstances should it be off by 6 mm.

This is all I wanna know!!!!

[Robert Martellaro]

Human error, resulting in deferred success. Let them go if it happens again.

[Prentice Pro 9000]

As for your other question about the Costa frames. I would not accept that. In my expertise sunglass companies are awful at making and measuring lenses. The way I measure high wrap jobs is to place the 34mm mark of the PD stick on the right lens nasal engraving then measure the the left nasal marking. The influence of wrap between these two points in going to be negligible. You can also just measure the distance between nasal engravings and add 34mm.

Thank you! This company is gaslighting me. It’s some company in Italy that does these glass lenses for Costa. The biggest kicker is that he has a pair of poly costas that came out fine. I even sent them pictures of both dotted up with a PD ruler in front of them and they’re still insisting that these are correct.

[Kwill212]

First you have to distinguish between the optics of a wrap glass and the mechanics of the same.
The optics pretty well covered above.
For the mechanics, try only thinking about a lens with a non-symmetrical front surface, such as a progressive or fully optimized digital single vision.
When you wrap a frame, you effectively narrow the chord of it’s “PD.”
This must be addressed in blocking by either:
1. Widening the blocking PD to compensate, or
2. Narrowing the DBL, which is functionally the same.

Both of these are mechanical adjustments in blocking purposes only. They would also hold true for bifocals, in theory.

B

I'd love to get some takes on the highlighted comment. Obviously increasing the wrap angle will reduce the frame PD as measured in reference to the 0° wrap line. In other words the temporal edges of the lenses get closer together as you described above. However, depending on the base curve, and seemingly the back curve of the lens, the MRP may be displaced inward OR outward when inducing wrap. I have attached a few diagrams to show what I mean. There are 2 lenses, both 8 base, one plus one minus, both with the same A measurement. Both lenses in the diagrams are right lenses from a top down view. The first diagram shows what they look with zero degrees of wrap. The second set shows what they look like with 15 degrees of wrap. The last two are just close up of the MRP displacement.

As you can see both have shortened the chord length of the frame pd by ~2.1mm. But, the minus lens MRP has been displaced toward the temporal side by about 0.6mm and the plus lens has been displaced toward the nasal side by about 2.0mm. The MRP of the minus lens travels in an arc first displacing it temporally then as the wrap increases it eventually travels back nasally until it reaches it original decentration. In this case about 27 degrees of wrap would be needed to return to its original decentration. Increasing the wrap further with then start to decenter the MRP nasal of its original position.

For the plus lens with the plano back curve, the MRP only displaces nasally, because the back curve of the lens has already started at the zero degree wrap line. I've used plus and minus here but it seems that any lens with a minus back curve would initially move the MRP temporally. The steeper the back curve the more magnitude of displacement temporally, and the more wrap required to displace the MRP nasally. Someone tell me I'm nuts and please show me why.

[Barry Santini]

Methinks the reference for these calcs should be the anterior/front surface and not the back surface

[drk]

My first reaction is:

Yes, you have displaced inward the MRP/optic axis on the plus lens by wrapping it. So outset it.

Well, the effect on the minus lens is less so. Maybe outsetting is not as important on a minus lens.

I would assume that it's calculated mathematically. Maybe a rule of thumb is not a great approach.

[Robert Martellaro]

Cosine of tilt angle seems to work.

I don't think ± matters as long as the lens shape is meniscus.