Oakley's secret

**drk** · 04-05-2004, 02:14 PM

Unlike Victoria's Secret, this is more technical.

Oakley labs seem to have a formula to alter a spectacle Rx so that when they make it in a "wrap" frame, there is less prism induced and less distortion of optics. I would like to know what formulae they use, so I can create the same effect when I do wrap lenses (heaven help my optical lab).

If I recall my optics correctly, there is a formula that derives induction of power by tilting on a horizontal or vertical axis (like with too much pantoscopic tilt) that is something like: change in sphere = Fsin(squared)theta/2, and cylinder is: blah, blah, blah. (I know these are not the actual equations, and I can look them up. My question is theoretical, more than needing to have the formulas just yet.)

I would assume that since the formulae can compute change of power in the horizontal meridian with lens tilting on a vertical axis (frame wrapping, I contend), the general concept would be to combine the spherocylindrical change with the orginal spherocylindrical Rx power to give a new Rx. (I would have to make an assumption as to how much angular lens tilting there is with an 8 base frame, as well, but it could be derived empirically with a protractor, but I'm guessing about 15 degrees.)

I think that there would be some prism induction as well, and I've tried tilting trial lenses of various power in the same direction as a wrap frame, and I seem to get results that are independent of type or degree of power. It seems as though any lens tilted temporally, like in a wrap frame, displaces objects temporally, like a base in prism would. I would assume that there is some formula for this based on lens thickness and degree of rotation, and not power, which makes sense, because even plano wrap lenses from sunglass companies can give this effect, and that "prism correction" is what Oakley touts even in their plano eyewear. I would assume it's on the magnitude of 1-2 pd total.

Do you brains have anything on this? I'd appreciate your general thoughts, and any formulas that may be echoing in your heads.

**mrba** · 04-05-2004, 02:46 PM

I only have peripheral comments.

This is probably not Oakley's tech. The Sola Spazio Lens comes to mind with your explanation.

Your Equation for tilting a lens does exist. We use that equation and then cut the resultant Cyl, and have had success. Generally you are inducing Cyl. I know Addidas reccomends decentering the lens differently to control for prism but this also increases distortion.

I think you should call Sola Lab or maybe a sola person can get on here and talk about their craaazy lens desighns.

**drk** · 04-05-2004, 05:04 PM

Ok, here are the formulae (I can't type a decent equation):

delta sphere in lens approx = F sin squared theta, total quantity divided by 3.

F= power in horizontal meridian
theta=angle lens is tilted from planar to eye

delta cylinder approx = F tan squared theta. (Answer will be extra power in the horizontal meridian, or axis 90)

Thus, the overall spherocylinder component is added to the original Rx for a resultant to see what the new power will be. I would attempt to use one of my contact lens crossed cylinder calculators, but there are formulas and graphical vector methods which are a pain in the bad *** optical.

Not only that, but if the original Rx is other than a sphere or strict axis 90 or 180, we'd have to calculate the power in the horizontal meridian using the formula for power in an oblique meridian:

F at theta = F of cylinder power times sin squared theta

theta being defined as the angle from the specified axis of the lens to the oblique meridian where power is being determined.

then we'd use that power for the horizontal meridian power in the equations above.

But really, we want to work backwards and reduce the power of the lens we surface, so that the final product will be what the Dr. writes. So I have yet to figure that out. It's just algebra, though, and work backwards, but the cross cylinder formula backwards might be a "biatch".

Then we'd still need a decent prism formula. I really think its dependent on the index, thickness, and degree of tilt, so it's roughly:
induced prism = index of refraction x thickness x tilt
but I don't know what the units are. Prism diopters are an impure measurement.

It's possible the equation is related to:
deviation by a prism in degrees = apical angle times (index-1)
and then we'd convert to prism diopters:
prism diopters = 100 tangent deviation in degrees

I just don't yet know how to get the apical angle of a prism created by tilting a lens!

"Hello, dude, this is Oakley World Headquarters. Radical!"

**mrba** · 04-05-2004, 05:29 PM

Robert Shanbaum is better at this... In the meantime I will consult Jalie and get back to you inbetween the 7 zillion more things I have to do today...

**Darryl Meister** · 04-06-2004, 07:32 AM

In the meantime I will consult Jalie and get back to you

Unfortunately, Jalie's books won't really show you any more more than DRK's already posted.

And, really, these are all just approximations to begin with. The process of compensating "wrapped" lenses for the as-worn position is more complex than you might first imagine, and to do it accurately requires sophisticated computer ray tracing.

For instance, the formulas for lens tilt specifically apply to thin spherical lenses, and you must calculate the formulas using both of the actual principal powers of the lens if it has cylinder, which is generally the case, and your calculations need to consider the deviation of the cylinder axis from the axes of the lens tilt, as DRK suggested. Some more approximations will help you simplify the work a bit, but then you must take those results and subtract them from the initial prescription using cross cylinder formulas... And those are a considerably bigger pain than the tilt formulas.

And, yes, prism is another consideration too, since tilting a lens introduces some, even if the person is looking through the optical center. Pantoscopic tilt generally won't introduce prism imbalance, but face form tilt will. This prism is also affected by things like front the curve, in addition to the thickness and refractive index of the lens. And, keep in mind that all of these factors are intrisically related, which means that if you compensate for one (e.g., prism) you should adjust the other (e.g., power) slightly if you want to be as precise as possible. For example, a computer program would "iterate" until it arrives at the correct combination of values. Consequently, undertaking these as-worn compensations, yourself, is certainly no small task.

The Sola Spazio Lens comes to mind with your explanation.

Spazio is a special type of lens available from SOLA Technologies, not just a compensated "wrapped" lens. The prescription is certainly adjusted precisely using computer ray tracing software for Spazio, but the lens, itself, also has several proprietary features. For instance, the front surface is atoric to improve off-center performance, since 8-base lenses generally represent a departure from "best form" optics for most prescriptions (unless you're a +2.00 hyperope). Additionally, in high minus powers, the back surface uses a special lenticular/carrier curve in order to reduce edge thickness, which would become quite excessive for an 8-base lens. You can see a picture of the concept at the Spazio website.

Best regards,
Darryl

**JRS** · 04-06-2004, 11:33 AM

Darryl (long time no see),

I messed around with some of this stuff a couple of years ago, but never got a chance to compare notes with others. Below is an example, and results. If you ran this through the 'Enigma' calcs, or if you know another way, can you desk check this?

Original Rx:
OD -2.00 -2.00 x 90, DPD = 32.0
OS -3.00 -4.00 x 90, DPD = 32.0

Refracted @ 13.0, Normal Fit @ 11.0 (vertex)
Normal BC = 4.00, Final BC (for wrap) = 7.00
Horizontal Tilt (face form) = 12 degrees

Results:
Final Fit = 12.9 (vertex), using 7.00 BC

OD -2.06 -2.18 x 87, DPD = 32.4, 0.17D base out (prismatic)
OS -3.10 -4.32 x 88, DPD = 32.4, 0.31D base out (prismatic)

Just trying to see if my thought process on this was in the right direction. Or I'm all wet and need to start over (maybe).

Thanks for any help you can provide.

John

**Darryl Meister** · 04-07-2004, 07:56 AM

[qutoe]Results:
Final Fit = 12.9 (vertex), using 7.00 BC

OD -2.06 -2.18 x 87, DPD = 32.4, 0.17D base out (prismatic)
OS -3.10 -4.32 x 88, DPD = 32.4, 0.31D base out (prismatic)[/quote]

Is this the resultant prescription you calculated in the as-worn position, or the compensated prescription you would actually surface?

Best regards,
Darryl

**JRS** · 04-07-2004, 08:21 AM

Hopefully, it is the Rx in 'as worn'.

But at this point I got diverted to other things and left it - about 2 years ago. The thread made me think about it again, so I posted what I did. I don't get as much time (recently almost zero time) anymore to dink around making experimental lenses, so I never finalized this.

Many, many years ago I had a program that AO had written for the military. It was for the inserts (wraps) in pilot helmets. BLEP's is what I think they called it (ballistic laser eyewear protection). But I've long since lost that data.

If I'm way out in left field, thats ok... just say so.

**Darryl Meister** · 04-07-2004, 09:28 AM

When you tilt a lens, be it with pantoscopic tilt or face-form tilt, remember that you effectively increase the sphere power and introduce a cylinder component (of the same sign as the sphere power) at the axis of the tilt into the prescription. DRK wasn't too far off. Consequently, when compensating for this effect, your prescription would have to back down on the sphere power and add some cylinder of opposite power.

This sort of thing was more common back in the days of cataract lenses. Not many people think about it anymore.

Best regards,
Darryl

**JRS** · 04-07-2004, 12:48 PM

Should I expect both PLUS & MINUS sphere/cylinder powers to decrease? Or does the change to back vertex still play a part in the compensation - or completely ignore that aspect?

I suppose it's possible that my major components may be alright, but my +/- math could be reversed. So instead of a sphere power increase of -.06 (R eye) it would be a decrease of .06D. Same with the cylinder too.

I guess, fundamentally, I was just looking to see if I was pointed in the right direction using the old cross-cylinder calc.

Thanks for responding Darryl

**Darryl Meister** · 04-07-2004, 03:35 PM

Should I expect both PLUS & MINUS sphere/cylinder powers to decrease? Or does the change to back vertex still play a part in the compensation - or completely ignore that aspect?

All of these particular formulas, which are more or less approximations, pertain to the back vertex power. You would have to back down on the sphere power of both plus and minus lenses (e.g., less plus or less minus). A tilted minus lens introduces minus cylinder at axis 090, while a tilted plus lens introduces plus cylinder at axis 090.

Keep in mind that to do this stuff accurately, you really need to ray trace the lens in its position of wear. The simple formulas you find in textbooks are a start, but it takes extremely complex algorithms to do it "just right." And once you have to deal with cylinders at an oblique axis, or pantoscopic and face-form tilt, it can get really hairy fast. Also remember that these effects are all interrelated. For instance, if you grind prism into the lens to compensate for as-worn prism, you effectively change the tilt of the lens, which means that your power compensation would need to be adjusted accordingly.

And, really, one of the most complex aspects of this process is accurately determining how much tilt the lens will have in the first place, which will vary with a number of factors. Like I said, this is definitely no small task. The guy who wrote the software we use for Spazio is an expert in differential geometry, which his algorithms rely heavily on. I, personally, know just enough differential geometry to be really, really frightened of it.

I could probably provide you with an algorithm for combining crossed cylinders though. I wrote a simple one up in Visual Basic not too long ago, which you could probably easily rewrite into C or whatever.

Best regards,
Darryl

**JRS** · 04-07-2004, 07:01 PM

Thanks for help Darryl (as always you are a wealth of knowledge). I originally started this 'task' just to show how doing certain things... produces some unexpected results, sometimes. Never planned on writing a absolute calculation. I think I'll just make a few minor adjustments, and let it die a natural death.

**HarryChiling** · 04-19-2005, 07:29 AM

I have found a aproximate formula to compensate for the power.

Example -4.00DS -1.00DC x 140 n=1.498(CR-39) tilt=15°

Steps

1.) find the power in the 90th meridian using Oblique Meridian Formula

Dtotal=Dsph+Dcyl(sinX)²

Power at 90th meridian is -4.59

2.) Use power at 90th meridian in Martins Lens Formula to account for tilt

Scom=S90th[1+(sinT)²/2n]
Ccom=Scom(tanT)²

giving you -0.23DS -0.35DC x090

3.) Then use Thompson's formula for Obliquely Crossed Cylinders to add them together

C²=C1²+C2²+2C1C2(cos2y)
S=S1+S2+[(C1+C2-C)/2]
tan2Ø=(C2sin2y)/(C1+C2cos2y)

this gives me the power of -4.35DS -1.00DC x130

I have put it into a lensmeter to verify the formula on many occasions and it is fairly accurate. It does not require very much accuracy because lets face it you only have but so many Rx's that can go into the high wrap frames anyway. As for the Vertex Distance compensation that should be a no brainer. The prism compensation, I think it would be simple algebra since the lens is in this case 2.2mm thick in the center and the tilt can also be used as the angle of change you could use

tan15=x/2.2

I came up with .59D of prism once again I verifyed it through the lensmeter and my calculations seemed accurate. I have been using this formula to make wrap sunglasses for quite some time and it seems to work well for me. I know it is slightly crude but its simplicity makes it elegant for everyday use. I also have created a TI-83 program for my calculator to compute these powers.

Now you have Harry's Secret. My lab is one of few in the area that will put Rx's into high wrap frames. Other labs in the area are afraid of touching them so patients are told that they cannot be done. Also I work in Maryland an optician does not need to be certified so many optical locations pay poorly and hire inexperienced staff. They see themselves saving money on labor however the amount of work we get from opticians who cannot do certain jobs astounds me. It makes it worth my time to always hunt for more knowledge. I have recently began sand blasting designs onto lenses. The old heads may know this as polaris. Again no one in my area offers it or even knows that it can be done. I am still getting the hang of it so I have not actually tried to offer it yet however that will be coming soon.

**rolandclaur** · 06-17-2005, 02:33 AM

Just one quick question guys... Where the hell do all these formulaes exist? Is there a body of literature out there that discusses intensely opthalmic optics and the formulae that accompany it?? If so I would defintely like to know!!

**Darryl Meister** · 06-17-2005, 03:17 AM

The prism compensation, I think it would be simple algebra since the lens is in this case 2.2mm thick in the center and the tilt can also be used as the angle of change you could use

Keep in mind that you are only adjusting for prism induced by decentration with this approach. Tilting a lens introduces prism, even without decentration. This prism is roughly proportional to the base curve, center thickness, and tilt of the lens. Also, using the meridional or "curvital" powers at 90 and 180 to compensate for a tilted lens does not consider the "torsional" component of cylinder power (as Keating would call it), though this approach is better than not compensating at all.

**HarryChiling** · 08-28-2005, 10:18 PM

This equation is similar to the equation that you use on your site opticampus.com for tilt and wrap compensation. The step between 2 and 3 I have listed is modified neutralizing the unwanted induced power and astigmatism then put through step three to come up with the correct power before surfaceing. I don't claim that the equation compensates for everything and the prism calculation is definately wrong I fudged on that but it is a great starting point and I have used this equation for some time now with great success. I have programed it into a TI-83 calculator so that I can surface these lenses on the fly. But I do still write out the calculation on the ticket so my customers get a glimpse of the magic behind the lenses that I make. I have written a paper on this subject and actually hope that it will get me the masters in opthalmic optic. I think since you use the very same equation on your site this did not warrant any critique however if you would like to critique my writing please feel free to. I will send you my paper and hope to hear your opinions on that.

**Darryl Meister** · 08-28-2005, 10:32 PM

Originally Posted by Harry

I think since you use the very same equation on your site this did not warrant any critique

Hi Harry,

The code at OptiCampus uses an entirely different method of solving this problem. It considers both the curvital and torsional powers of the tilted lens, and also computes the power compensation required for simultaneous pantoscopic and face-form tilt. Your equations will lose accuracy for lenses with cylinder power at an oblique axis and will not allow you to compensate for both pantoscopic tilt and face-form tilt. But, as I said, they are certainly better than using nothing at all, and I applaud your effort.

**HarryChiling** · 08-29-2005, 06:47 PM

This is the paper I have written and plan on submitting to the ABO for the masters designation and so far every Rx I could think of that I plug into the equations comes to the same final out put as your calculator. Since I don't know your equation I can't take what you say that your formula compensates for torsional astigmatism as fact. Please show your equations. Not only am I really interested but I have been using this formula for quite a few years now and if it not correct as you say and you have one that is better, I would like to now it.

**Darryl Meister** · 08-30-2005, 08:56 AM

Hi Harry,

The equations I used were derived from an article written by Dr. Michael Keating (a physicist who teaches ophthalmic optics at one of the colleges of optometry). They are based on a Matrix algebra representation of lens power, and also involve the computation of oblique tilt angles (that is, combined pantoscopic and face form tilt, resulting in a new tilt around a new axis), which requires rotating the coordinate system of the calculations.

Just out of curiosity, what do your equations output as the compensated Rx for the following prescription?

-2.00 DS -3.00 DC x 025 with 20 Deg of wrap tilt

By the way, I noticed in your formulas that you are calculating the power through the 90 meridian using sin^2, which actually calculates the power through the 180; you should change this to cos^2.

**HarryChiling** · 08-30-2005, 08:29 PM

My equation calculates

-1.12 -3.75 x 019
after the correction you mentioned cos^2

your calculator shows

-1.74 -2.98 × 022

But how do I know that your calculator is correct if you will not share the equations you used to come up with it. We must rely on blind faith instead of knowledge, not me.

**Darryl Meister** · 08-30-2005, 08:48 PM

But how do I know that your calculator is correct

The point of this exercise was simply to illustrate that our calculations will in fact not come out the same, as you asserted earlier.

if you will not share the equations you used to come up with it. We must rely on blind faith instead of knowledge

Not really. I have already explained to you exactly why your equations will not be accurate. It's not a matter of faith; it's a consequence of the fact that your equations ignore a certain optical component of lens surfaces (the "torsional" component I mentioned earlier). I have also provided you with the reference I used. You can check out Keating's Geometrical, Physical, and Visual Optics for a thorough discussion of these topics.

I may post some equations when I have more time, but I consider all of the actual algorithms from OptiCampus -- which took me sometime to develop, test, and implement -- to be an important form of intellectual property. Of course, I make their use freely available to the public.

**Alvaro Cordova** · 08-30-2005, 10:54 PM

Hey Darryl, when you get a chance. Do you know what chapter discusses the combined usage of tilt and faceform? I have the Keating book. I tried to find it, but it just gives examples of one or the other not both as far as I can tell (I did see the stuff about the torsional component in chapter sixteen

Thanks

**HarryChiling** · 08-31-2005, 05:18 AM

Well at least lead me in the right direction as to where you picked this information up from. I totaly understand that you would consider the equation intelectual property, but the knowledge should be free in my opinion. After all I am not asking you to give me the source code for your web pages. All the other formulas used on your site were developed by others and coded by you.

Give me a book or clue or something.

**Darryl Meister** · 08-31-2005, 07:45 AM

Originally Posted by Harry

Well at least lead me in the right direction as to where you picked this information up from... Give me a book or clue or something.

But I did:

You can check out Keating's Geometrical, Physical, and Visual Optics for a thorough discussion of these topics.

Originally Posted by Harry

All the other formulas used on your site were developed by others and coded by you.

If only it were so simple.

However, I'll consider posting his basic equations when I return.

Originally Posted by Al

Do you know what chapter discusses the combined usage of tilt and faceform?

He actually wrote an article about combined tilts that appeared in a journal. I'm traveling right now, but I'll post the actual reference once I'm back home.
You might need a subscription to the AAO's Optometry and Vision Science magazine to get a reprint. That said, the process of compensating for combined tilts is considerably more complex than the process used for single tilt compensation, and you gain very little for most jobs.

**Alvaro Cordova** · 08-31-2005, 09:48 AM

Yeah, but it's for posterity's sake :D (I've been using that line for everything from going to the restroom to ordering food. So if it doesn't make sense don't worry, I'm just being silly)