PD on wrap frames

[Bob Price]

I hear a lot of talk about Rx calculator's for wrap frames, but not much about verifying PD. The common practice of verifying a PD is to take a ruler and measure the distance from the front of the frame. After tracing these frames we have found that the frame PD comes out narrow, so we will take a frame PD prior to blocking and inset the lens more depending on the difference. The PD then verifies correct.

I am questioning whether this is correct and how does it affect the patient? If you take a PD from the minus side you will get a narrower reading. By doing what we are doing is this creating too narrow of a PD for the patient? So, what is correct? Am I hallucinating here or is this a valid point?

Bob

[Barry Santini]

I hear a lot of talk about Rx calculator's for wrap frames, but not much about verifying PD. The common practice of verifying a PD is to take a ruler and measure the distance from the front of the frame. After tracing these frames we have found that the frame PD comes out narrow, so we will take a frame PD prior to blocking and inset the lens more depending on the difference. The PD then verifies correct.

I am questioning whether this is correct and how does it affect the patient? If you take a PD from the minus side you will get a narrower reading. By doing what we are doing is this creating too narrow of a PD for the patient? So, what is correct? Am I hallucinating here or is this a valid point?

Bob

Depending on the degree of wrap angle, the *effective* frame PD will narrow. You can use simple trigonometry to calculate it exactly. Just visualize that the wrapping of a frame would effectively tighten it against a client's temples, and you'll undertand conceptually how the finished frame PD becomes more narrow

Barry

[bt5050]

wrap frames - pd

I have found that when i get jobs in from zeiss - and they use their spazio software - the pds - does seem to come out exact - with the base in prism they induce - as well as the adjustements made to the rx-

do have a few ?'s -

1- when you are finaling a rx - that uses a adjusted pd - and rx -
do you apply the adjusted PD + rx to apply the tolerances - or do you the actual perscribed rx- and pd ??

2- is it correct to measure these pd's from the back side - do u get a more actuarate measurment - since you are not compensating or reading the ruler at such an angle - ?

3- when laying out this work - to be edged - should you also - take the frame pd - from the backside - and use that for your decentration- or should you - take the a + the dbl - and then calulate the decentration on these wrap frames ?

look forward to your answers -

b

[lensgrinder]

The shortest distance between two points is a straight line.
When taking a Frame PD the key is not to bend your PD stick, if this requires you to take the measurement from the back (this method seems to be problematic due to the temples, sometimes), take DBL + A, use Frames Book measurements, then do so.
The patient's PD is the same, do not bend your PD stick, take it from the back if needed.

If you are using a compensated Rx, then you would apply ANSI to the compensated Rx, because this is what they are made to, not the original Rx.

[Samuel Jong]

PD on wrap frames

Depending on the degree of wrap angle, the *effective* frame PD will narrow. You can use simple trigonometry to calculate it exactly. Just visualize that the wrapping of a frame would effectively tighten it against a client's temples, and you'll undertand conceptually how the finished frame PD becomes more narrow

Barry

Is there any calculator to determine the PD on wrap frames? Thanks.

Samuel

[Leo Hadley Jr]

This is a great subject,
not only for the compensated PD, but the RX as well. I cant tell you how many times I have seen older licensed opticians rejecting these jobs because they do not believe compensations have to be made for the wrap. I think it is very important this subject be covered extensively in continuing education, trade magazines, and any other resource like Optiboard for the non licensed states.
I believe Harry Chiling has a tilt and wrap calculator posted in the file section of OptiBoard.
That may help out a great deal.

[AWTECH]

Wrap frames and PD

I hear a lot of talk about Rx calculator's for wrap frames, but not much about verifying PD. The common practice of verifying a PD is to take a ruler and measure the distance from the front of the frame. After tracing these frames we have found that the frame PD comes out narrow, so we will take a frame PD prior to blocking and inset the lens more depending on the difference. The PD then verifies correct.

I am questioning whether this is correct and how does it affect the patient? If you take a PD from the minus side you will get a narrower reading. By doing what we are doing is this creating too narrow of a PD for the patient? So, what is correct? Am I hallucinating here or is this a valid point?

Bob

Consider the a spherical lens with a mono PD of 30mm. With a high wrap frame you may get a needed PD for lens mounting of about 31.5mm. This will place the optical center in front of the pupil. The object of this compensation would be to place the lens in front of the pupil and to do this you need to increase the PD during lens fitting. If you measure a wraped frame link this you would find the distance, when measured on the bench, from the center of the frame to the center of the lens would be 31.5mm. When placed on the patient and then measured keeping the PD stick on the same plane as traditional flat dress eyewear you would measure the PD as 30.0mm.

There are a lot more calculations that go into mounting lenses in wrap frames, but this should help you understand the basics of the PD.

Base in prism accomplishes the same thing as increasing the PD.

[bt5050]

i think these explanations ( the photos where helpful for me ) - that r found on the ziess spazio site -
it shows how not compensating the rx- or pd - alters the actual rx - and the pupil- oc relationship changes -

http://www.solatechnologies.com/products/sunoptics/spazio/

at what power - should compensation start being used ? - it appears they may use - 3d- but i am unclear if this is just their example - or that is what they suggest

at 1st - i was a bit concerned using -these altered rx and induced prism - jobs - and it came from my exp. that a percribing MD and od - todl the pts -the glasses where made way wrong - and was not happy about it -

although i got a little satisfaction - by calling him - and showing him how wraps are done - and that we where actually correct - - it is not always conveyed to the the pts very well - and all they hear is - they made them wrong -

so now i make it a point to mention - at the time of sale - that the rx- will be compensated for the wrap - so they are actually made to the correct rx- just in case in the small % - they go back for a recheck - without seeing me 1st - and get misinformation

have a great weekend -
b

[HarryChiling]

I have suggested in te past that when doing a wrap the lenses be engraved or laser marked with something to give other a way of reversing the Rx, for example if the lens had engraving T15W20 an optician can determine that the Rx was compensated for 15^o tilt and 20^o wrap compensation. Then if the formulas used to determine the compensations were reversed (simple as the compensations would be multiplying instead of dividing into the power matrix the compensations) then the patient upon dispense is given a small business card size CD with a compensation program on it (example: Darryl's or mine) wich also has a reverse function so that anyone down the line can reverse the compensations and see what the original Rx was. This would add great value to a lens package and cost very little to the manufacturer.

If any manufacturer applies any of these techniques to their lenses and comes up with some ultra premium product as a thank you please give me a hefty discount on it.

[HarryChiling]

To find the Rx that should be surfaced into the lens use this calculator:

http://www.optiboard.com/forums/showpost.php?p=155243&postcount=1

Once the Rx is figured out Use this calculator to come up with you surfaceing ticket:

http://www.optiboard.com/forums/showpost.php?p=159616&postcount=1

Now with the back curves figured out use this excell sheet to find out what second cuts (lenticular) to make to thin out the edges:

http://www.optiboard.com/forums/showpost.php?p=183504&postcount=3

I will eventually incorporate all these calculators into one huge wrap and tilt calculator that will blockout the sun and bring about another ice age, but time is a prescious commodity right now. ;)

[lensgrinder]

Base in prism accomplishes the same thing as increasing the PD.

Please correct me if I am wrong, but the prism is due to the lens tilt, thickness of the lens, and base curve. If you increase the PD, the angle still remains the same. The person is never going to look over the OC, because the light is coming in at an angle to the line of sight, not straight. This is why you need the prism, to correct for the displacement and based on a 14mm vertex and 15 degrees of wrap you would need to increase the PD about 7mm per eye. Let me know your thoughts.

[bt5050]

cd

then the patient upon dispense is given a small business card size CD with a compensation program on it

i think your idea with a engraving and something for the pt at disp ( ie card) - would be great - so it is apparent when the rx has been changed from the orginal - provided by the persriber -

however - i dothink going as far as giving a pt a cd- is kinda over board - ( at least here where i am ) - since the od.md's expire there rx's - the pt would not have the rx for years and years - and need the rx converted back -

furthermore - if we wanted to make another pr - it is most likely it will be in a differant frame - with differant wrap- and therefore we could easily work off the real rx -


BUT I DO LOVE THE ENGRAVINGS - and dipense card - idea -!!!

b

[HarryChiling]

Please correct me if I am wrong, but the prism is due to the lens tilt, thickness of the lens, and base curve. If you increase the PD, the angle still remains the same. The person is never going to look over the OC, because the light is coming in at an angle to the line of sight, not straight. This is why you need the prism, to correct for the displacement and based on a 14mm vertex and 15 degrees of wrap you would need to increase the PD about 7mm per eye. Let me know your thoughts.

The prism is for alighning the ray when leaving the lens to be parralel with the optical axis, the compensation to the PD would be for the ray from lens front through the lens IMO. So if we were to consider this the compensation would be:

tan(wrap)=dec/thickness

for example a lens 2mm thick wraped 15^o, would be:

tan(15)=dec/2mm
tan(15)*2mm=dec
dec=tan(15)*2mm
dec=0.2679*2
dec=0.5358

So add 0.5mm to your lens PD.

however - i dothink going as far as giving a pt a cd- is kinda over board - ( at least here where i am ) - since the od.md's expire there rx's - the pt would not have the rx for years and years - and need the rx converted back -

The idea is so that the prescribing doctor can reverse the Rx themselves and would have the software for further use. It would be more to add value than anything else. You could hand out a card with the Prescribed Rx and the Compensated Rx, but if the card is lost the glasse are wrong again. However if the engravings catch on then when a preescriber sees it they will know to use the software to get the original Rx. The patient doesn't have to worry about anything and the prescriber now has a tool at their disposal whihc would further prommote their product. Just an idea, I can dream can't I. :D

[Darryl Meister]

The prism is for alighning the ray when leaving the lens to be parralel with the optical axis, the compensation to the PD would be for the ray from lens front through the lens IMO. So if we were to consider this the compensation would be

This only applies to prism induced by decentration. The tilt of the actual lens, with or without any decentration, also induces a prismatic effect that is independent of decentration and proportional to the front curve of the lens. Moreover, the tilt of a lens is further exacerbated by decentration, even when the PD is correct, since the decentration of a meniscus lens occurs about an arc. While I provide a fairly accurate first order approximation of these effects in my Rx Compensator program, my impression is that very few "wrap calculators" take these variables into account.

And this all assumes that the PD was correct in the first place. Since wrap tilt effectively reduces the optical center distance, by an amount roughly proportional to the cosine of the angle of tilt, this needs to be factored in before edging the lenses -- either in the compensation software or by the person laying out the lens for edging.

[Barry Santini]

And this all assumes that the PD was correct in the first place.

Yes, we all need to review whether our traditional methods for "taking a PD" are, in fact...*representative* of the actual visual axis...

Since wrap tilt effectively reduces the optical center distance, by an amount roughly proportional to the cosine of the angle of tilt, this needs to be factored in before edging the lenses -- either in the compensation software or by the person laying out the lens for edging.

This type of PD compensation is already on my wish list for Darryl's next version of his Rx compensator...

How 'bout it, Darryl????!


barry

me