# Optical Forums > Ophthalmic Optics >  Oakley's secret

## drk

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.

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## mrba

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.

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## drk

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!" :Cool:   :Cool:   :Cool:   :Cool:

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## mrba

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...

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## Darryl Meister

> 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

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## JRS

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

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## Darryl Meister

[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

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## JRS

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.

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## Darryl Meister

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

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## JRS

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

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## Darryl Meister

> 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

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## JRS

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.

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## HarryChiling

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

*D*total*=D*sph*+D*cyl*(sinX)²*


Power at 90th meridian is -4.59


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

*S*com*=S*90th*[1+(sinT)²/2n]*
*C*com*=S*com*(tanT)²*

giving you -0.23DS -0.35DC x090

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

*C²=C*1*²+C*2*²+2C*1*C*2*(cos2y)*
*S=S*1*+S*2*+[(C*1+*C*2*-C)/2]*
*tan2Ø=(C*2*sin2y)/(C*1*+C*2*cos2y)*

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.

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## rolandclaur

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!!

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## Darryl Meister

> 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.

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## HarryChiling

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.

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## Darryl Meister

> 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.

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## HarryChiling

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.

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## Darryl Meister

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.

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## HarryChiling

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.

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## Darryl Meister

> 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.

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## Alvaro Cordova

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

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## HarryChiling

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.

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## Darryl Meister

> 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._




> 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.




> 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.

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## Alvaro Cordova

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)

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## HarryChiling

I Have ordered Keating book, but would still like to know what his equation was I am sue I would still have to determine how to fit that equation into the whole scheme of things.  When I figure it out I will post it to the board so hopefully everyone has an answer.

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## Darryl Meister

> I Have ordered Keating book, but would still like to know what his equation was I am sue I would still have to determine how to fit that equation into the whole scheme of things.


Remember that his approach is completely different from your own (he decomposes the prescription into a kind of "vergence matrix"). His approach allows him to 1) Solve the equations for the correct prescription, which is slightly more accurate than subtracting out a cross cylinder, and 2) Compute the torsional component of the power error, which is considerably more accurate than using the power of an oblique meridian. What you're looking for appears in sections 20.17 and 20.18 of his latest text.

By the way, I just noticed you are using the power in the 90 as the _tangential_ power, when you should probably be using the power in the 180, since is the meridian that is actually tilted for face-form wrap. Consequently, your original error (using sin^2 to calculate power in the 90) would actually get you closer to the correct answer, since you were inadvertently calculating the power through the 180. (Who says two wrongs can't make a right?! ;) ) You also need to apply the sagittal calculations to the power in the 90 meridian using your approach, so ultimately both are needed.




> I totaly understand that you would consider the equation intelectual property, but the knowledge should be free in my opinion.


I missed this point of yours earlier. I don't consider the _equation_ (or, in this case, the system of equations) to be intellectual property (at least of mine), but I do consider my _code_ to be intellectual property, which is why I was careful to distinguish between the two. As for "knowledge," it often costs both time and money. ;) And keep in mind that we only have these equations in the first place because someone put forth a considerable amount of time and effort to derive them. Besides, knowledge comes from _reading_ the book, not just pulling a formula from one of its pages.

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## HarryChiling

I have ordered the book, and do enjoy the knowledge.  My library is extensive and most of it is on physics not just optics I want to be well rounded however you can probably tell by my paper that I have been trying to work on this problem for a while and am quite frustrated at the fact that all I have to go to is books and no one from optometrists to ophthalmologists have any clue aas to what I am asking them.  So you will have to excuse me if I come across as blunt because I share knowledge and find it frustrating when people do not do the same.  I noticed you quoted out of his new text I have ordered the older version and hope that the information is still provided within.  I have also run into problems with aspherics wich interest me greatly and see that you have quite some knowledge on them as well.  I am curious what equation I could use if I were to take a lens convert the diotric power to radius and plot it in cartesian coordinates, how the formula would be set up.  I understand the fact that it would be different depending on the asphericity of a lens(eg hyperbola, ellipse) I am wondering if it would have a specific format to the equation.  I have looked into some calculus books and am working on figuring out how this is put together I know that I am looking for eccentricity in the books, do you know where I can find a clear and concise text relating to the subject.

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## Darryl Meister

Hi Harry,




> I noticed you quoted out of his new text I have ordered the older version and hope that the information is still provided within.


I can't say whether these formulas appear in his older edition or not; it's been awhile since I've actually seen it (and it's been out-of-print for some time). The sections I cited are from his current edition, and I would certainly recommend ordering his current edition.




> So you will have to excuse me if I come across as blunt because I share knowledge and find it frustrating when people do not do the same.


Rest assured, I have shared quite a bit of knowledge with the optical community over the past 15 years. But I also understand the importance of "doing your own homework," as they say. This is more of an issue for me with wrapped lenses, since my employer has a great deal of intellectual property in this area (as I mentioned earlier).




> I am curious what equation I could use if I were to take a lens convert the diotric power to radius and plot it in cartesian coordinates, how the formula would be set up.


You would simply use the sagitta formula, using X as the semi-diameter. (Y is generally not necessary, since these lenses are rotationally symmetrical.) General aspheric surfaces use polynomials that also provide the sagitta as a function of X. Conicoid aspherics can be computed this way, as well. You have to know how to describe your surface in at least _some_ manner first, but then converting it to a cartesian system is pretty straightforward.

As for the best text on this particular topic, I would recommend Jalie's _Principles of Ophthalmic Lenses_, which has an extensive section on ray-tracing aspheric lenses.

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## Alvaro Cordova

I only have the first edtion of the Keating book.  I looked on Amazon and it's like 95 bucks.

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## HarryChiling

EDITED


-2.00 -3.00 x 025 tilt 20 in a hard resin

I come up with

-1.59 -2.66 x 028

your calculator comes up with

-1.74 DS -2.98 DC × 22
They are very similar, I am waiting for the book you suggested to arrive, but I would have to say that if your equations are not open to public scrutiny then I cannot for sure say that yours is incorrect, but since mine is I am left vurnerable, But I am also left with a situatiuon where if my equations are wrong I openly invite anyone to change and shape the equation closer to accuracy. When I recieve the book you suggested I will of course post it's relevance and update any changes I make.

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## Darryl Meister

> They are very similar,


Well, they're both very similar to the original prescription, at least. ;)




> I am waiting for the book you suggested to arrive


Please keep in mind that I suggested you order his _current_ edition. I don't know that the necessary equations appear in his first edition, which has also been out-of-print for some time.




> but I would have to say that if your equations are not open to public scrutiny then I cannot for sure say that yours is incorrect,


The equations I used have been open to public scrutiny for ten years now (since Keating originally presented them). Besides, unless you understand the principles or mathematics involved, how could you really ever say that one approach is more accurate than the other? In any event, I have compared the results of Keating's equations against accurate optical ray-tracing for thin lens cases, and they are surprisingly close.

I should also add that, when calculating facial wrap compensation, it's important to consider the tilt induced by decentering a meniscus lens (which is often of the same magnitude as the frame tilt, itself). These equations assume that you are providing the exact _lens_ tilt, though this is relatively difficult to predict by simply looking at _frame_ tilt.




> When I recieve the book you suggested I will of course post it's relevance and update any changes I make


If you receive his book, and the necessary equations _aren't_ present (for the _single_ tilt case; he only treats combined tilts in a separate journal article, as I described earlier), I will post them at that point when I have some more time. But I would still encourage you to read the necessary material in order to understand what is happening.

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## HarryChiling

Daryl


why does your rxcompesator program give a different answer than your online form for the very same problem you presented me with?

I have received the book you suggested and am looking into it.

Please explain the difference between your program and web site.  If you can please explain more on the matrix equation used.

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## Darryl Meister

> why does your rxcompesator program give a different answer than your online form for the very same problem you presented me with? Please explain the difference between your program and web site


The online program, like all of the equations you'll run across for calculating astigmatism of oblique incidence, is based on _lens_ tilt (that is, the tilt of the optical axis). Unfortunately, eyecare professionals generally only measure _frame_ tilt (in fact, you can't measure lens tilt beforehand). However, decentering a meniscus lens actually introduces tilt as well, which is independent of the frame tilt. The program takes _frame_ tilt and then calculates _lens_ tilt from it.

Consequently, the website assumes you know the actual lens tilt, while the program calculates it for you. If you look at a print-out from the program, you will see a "True Wrap" value; plug this value into the website and the results should be the same. In the future, I may update the website with this added functionality.




> I have received the book you suggested and am looking into it... If you can please explain more on the matrix equation used.


It's actually explained in great detail in the book (in the chapter on Diopter Power and Off-Axis Meridians). If you still have any questions after having a look at it, let me know.

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## HarryChiling

I have read over the chapter you are talking about and see the equations however they look like the same formulas I used just in different formats.  I did also read over the chapter on power matrixs and the theory is pretty much similar to what I understand on the topic.  All I have managed to learn is another way of doing exactly the same thing, and have 2 new variables that I don't exactly understand (how they would fit into the eqaution if they would at all).  

What is the importance of the Trace?  How does that fit into anything?

What is the relevance of the determinant? How does this fit into anything?




> It's actually explained in great detail in the book (in the chapter on Diopter Power and Off-Axis Meridians). If you still have any questions after having a look at it, let me know.


This chapter has almost no new information that I have not covered in my paper, and does not touch on tilts, which is wher you said my equation is flawed.

In the abberations chapter they do touch on RA and tilts.  It is getting late and I am starting to fudge numbers and lose track on the page so I will continue tommorow.  

I will say one thing the book is very in depth and does make for a great read.  I received it yesterday and have 3-4 chapters left.  That is probably about 400 pages read and understood in one day.  I would definately rate it as one of the best books  will be adding to the library.

I will post later on what I find on tilts.

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## Darryl Meister

> What is the importance of the Trace? How does that fit into anything?... What is the relevance of the determinant? How does this fit into anything


The determinant associates a real number with the terms of a matrix, and can be used to help solve the system of linear equations that make up the matrix. The trace relates to a plane, parallel to a coordinate axis (X,Y,Z), that is described by the solutions to a series of equations -- and can also be determined from the terms of a matrix. Their importance in this context is primarily that they provide convenient quantities from the dioptric power matrix. For instance, the trace happens to coincide with Euler's constant, which is the sum of the principal curvatures of a toric surface, and is used to convert a power matrix back to normal sphere and cylinder values.




> All I have managed to learn is another way of doing exactly the same thing, and have 2 new variables that I don't exactly understand (how they would fit into the eqaution if they would at all


Their use is covered later in the chapter. And you certainly aren't learning to do things exactly the same way.

I dug my old copy of Keating's book out. However, again, keep in mind that the specific equations for compensating for the astigmatism produced by the tilt of a spherocylindrical lens are covered in his _current_ edition, as I stated before.




> This chapter has almost no new information that I have not covered in my paper, and does not touch on tilts, which is wher you said my equation is flawed.


Harry, I said that this particular chapter would explain the _dioptrix power matrix_, which was what you asked about in this specific case, not lens tilt.

Further, the chapter _does indeed_ provide information not convered in your paper -- and information directly related to the inaccuracies of your equations. On page 337, for instance, you might recognize that the dioptric power matrix comprises four terms:

Px = S + C sin^2 A
Py = S + C cos^2 A
Pt = -C sin A cos A
Pt = -C sin A cos A

Note that your equations for tangential and sagittal power, when done properly, use the first two terms. These are referred to as the _curvital_ components by Keating. However, your equations neglect the second two (identical) terms of the power matrix. These are referred to as the _torsional_ components.




> I would definately rate it as one of the best books will be adding to the library.


Yes, it is very comprehensive, particularly from an ophthalmic point of view. I'm glad that you're enjoying it.

Now, that said, I did agree to provide the equations from his new text if you didn't get what you need from the one you ordered.

First, we calculate the astigmatism change factors for the tangential and sagittal powers, as well as a change factor for the power halfway between:

T = (2n + sin^2 W) / (2n cos^2 W)
S = 1 + (sin^2 W) / (2n)
H = (T + S) / 2

where W is the wrap tilt angle.

Next, we compute the components (Px, Py, Pt, and Pt) of the dioptric power matrix as described by the equations above.

Px = S + C sin^2 A
Py = S + C cos^2 A
Pt = -C sin A cos A
Pt = -C sin A cos A

Now, we can solve directly for a new dioptric power matrix that represents the compensated prescription.

Cx = Px / T
Cy = Py / S
Ct = Pt / H
Ct = Pt / H

Note that these equations actually solve _for_ the compensated prescription, which is more accurate than computing the effective prescription change and subtracting it from the original prescription.

Lastly, you need to convert these new (compensated) power matrix components back into traditional sphere, cylinder, and axis values. This is described in that same chapter I referred you to earlier on dioptric power (there are several equations involved, but they're pretty straightforward).

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## HarryChiling

I see so then the trace and determinat are necesary to convert the power matrix back to the spherocylindrical paramaters.

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## Darryl Meister

> I see so then the trace and determinat are necesary to convert the power matrix back to the spherocylindrical paramaters.


Yes.

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## HarryChiling

So I tool the problem -2.00 -3.00 x 025 with a 20 tilt in hard resin (n=1.498)

put it into a power matrix


p = (-2.54  +1.15)
     ( 1.15   -4.46)

then put it through the equations mentioned

T=(2*1.498+sin^2(20))/(2*1.498+cos^2(20) = .8025
S=1+sin^2(20)/2*1.498 = .8366
H=(T+S)/2 = .8196

then I put it through the compensation

Cx = Px/T = -2.54/.8025 = -3.17
Cy = Py/S = -4.46/.8366 = -5.33
Ct = Pt/H = 1.15/.8196 = 1.40
Ct = Pt/H = 1.15/.8196 = 1.40

then put it back into matrix

Pc = (-3.17  1.40)
       (1.40  -5.33)

Then I went to find the trace and determinant

t= Px+Py = (-3.17)+(-5.33) = -8.50
d=PxPy-Pt^2 = (-3.17)(-5.33)-(1.40)^2 = 14.9361

C= - SquareRoot(t^2-4d) = -SquareRoot((-8.50)^2-4(14.9361) = -3.54
S= (t-C)/2 = [(-8.50)-(-3.54)]/2 = -2.48
tan@=(S-Px)/Pt = [(-2.48)-(-3.17)]/1.40 = .4928
@=26

So the Rx I got is -2.48 -3.54 x 026

again your calculator spit out something different and my earlier equation although slightly off was more accurate than the equation described.  I am still going to continue the reading on the tilts and see if I am doing something wrong here, but as it stands I am getting farther from the correct answer then when I started.

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## Darryl Meister

T = 1.18
S = 1.04

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## rbaker

Going back many years through my emerging veil of senile dementia I recall doing something like this in geometric optics as relating to camera lenses, particularly calculations with very large format view cameras where the plane of the lens and the film could be positioned at many different angles. This was done to alter the perspective of architectural and engineering photos. 

Ill bet you could find some interesting mathematics at the Eastman Library in Rochester or a like source. You might also find some interesting stuff in the papers of Doc Tillyer or Louise Rowe of American Optical.

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## Darryl Meister

> You might also find some interesting stuff in the papers of Doc Tillyer or Louise Rowe of American Optical.


Indeed, there are stacks of papers on ophthalmic optics sitting over at American Optical. Dick Whitney has been kind enough to share some of these in the past.

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## HarryChiling

I was reading over the text and found my mistakes.  I was using 2n+cos^2W in the denominator for T with this adjusted

T=1.1767 = 1.18

The format I had the equation for the S was wrong adjusted

S = 1.0390 = 1.04

These mistakes would also change my H

H = 1.1079 = 1.11

Then the compensated power matrix would be

Pc = (-2.16  1.04)
       (1.04  -4.29)

So the tilt and determinant would be

t=-6.45
d=8.18

so the new Rx would be

C=-2.98
S=-1.74
angle=22

-1.74 -2.98 x 022

EUREKA, the number of mistakes I kept making was horrible.  Chalk it up to late nights I guess.  Darryl thank you for the help, I would never had gotten my mind off of that until I got it right.  I appreciate you walking me through it.  Also the text is a god send, thank you for suggesting it.  If you know of any other good texts you can recommend I am all ears.

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## Alvaro Cordova

Can any of you do a problem involving both panto and faceform?  I am still not quite there yet.  Like -2.00 -1.00 x 55  with 20 of panto and 10 of facefrom.  How would this be figured out using the equations above.

Al

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## HarryChiling

Darryl I see where you got the torsional and saggital meidians formula from.  It is in the abberations chapter pg 440 although it was slightly modified by you and put into 2 steps.

Pt=P[(2n+sin^2W)/(2n*cos^2W)]

Ps=P[1+(sin^2W/2n)]

RA=Pt-Ps

Of course he talks about what the Rx would be after you tilt the prescription.  I like the way you reworked it to use the Rx backwards.  I have not seen the new edition but am very tempted to get it.  I have a suspicious feeling that it was you that reworked the equation.

the equivalent in your formula is

Px=Cx[(2n+sin^2W)/(2n*cos^2W)]

Py=Cy[1+(sin^2W/2n)]

I am not sure how the risidual astigmatism was reworked but again I will post when I figure it out.

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## Darryl Meister

> Can any of you do a problem involving both panto and faceform?.. How would this be figured out using the equations above


Unfortunately, the process is a great deal more complicated.




> I appreciate you walking me through it. Also the text is a god send, thank you for suggesting it. If you know of any other good texts you can recommend I am all ears


I'm glad you like it.

The big difference between Keating's book and most books on ophthalmic-related optics is the extensive use of power matrices, which I once found to be a bit of a turn-off. (Though I actually run into similar _curvature_ matrices quite regularly when dealing with our lens designers.)

However, if you play around with these power matrices enough, you'll find that they provide a rather convenient way of describing power that can be easily manipulated in several useful ways. For instance, they're additive, so you can compute _crossed cylinders_ by simply adding two power matrices together. You can also compute the prism at any point on a spherocylinder lens by multiplying a power matrix by a "decentration" column matrix.




> I have not seen the new edition but am very tempted to get it. I have a suspicious feeling that it was you that reworked the equation.


I reworked one or two bits of his other expressions for calculating the combined (panto and wrap) tilt, but the formulas I have posted here I have taken directly from the 2nd Ed of his book. I think you understand what is happening here, but I'll summarize it for you.

Consider the X component of a dioptric power matrix, which represents the "curvital" power through the 180. We'll call this component Px, and it represents our "desired" prescription. Assume that we have already calculated our astigmatism change factor T for the tangential meridian.

The _effective_ Px component, which we'll call Ex, represents the curvital power through the 180 that the wearer would experience once the lens is tilted, and is given by:

Ex = T * Px

Note that this represents the _altered_ power produced by tilting the lens (generally, tilting a lens produces an increase in sphere power and induced cylinder of the same sign). Now, if we know that we want to achieve our desired prescription component, Px,  once the lens is tilted, we can simply rearrange the above expression as follows by dividing through by T. This would represent the _compensated_ Px component, which we'll call Cx:

Cx = Px / T

This represents the curvital power through the 180 that we would need to specify in order to provide the correct effective power once the lens is tilted.

By the way, for _pantoscopic_ tilt, you should interchange T and S (i.e., Cx = Px / S and Cy = Py / T).

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## HarryChiling

I thought that panto would be switching T and S.  because all that changes is the planes tha are being tilted.  I have tried reworking the compensated Rx matrix back through I thought just maybe it might be that simple to calculate both tilts together, however it does not work.  I will attepmt to read a little more and see if I can maybe rework or apply some of the formulas. 




> However, if you play around with these power matrices enough, you'll find that they provide a rather convenient way of describing power that can be easily manipulated in several useful ways. For instance, they're additive, so you can compute _crossed cylinders_ by simply adding two power matrices together. You can also compute the prism at any point on a spherocylinder lens by multiplying a power matrix by a "decentration" column matrix.


I agree it does seem a great way of handling an Rx I feel like a whole new side to optics has just opened up and with the understanding of Power Matrix as well as the other forms of matrix described in the book I feel I have a better grasp of things.

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## Alvaro Cordova

Hey Darryl, would you be able to give me the name of the paper by Keating you mentioned that dealt with both the computing of faceform and panto?  I could probably get it from interlibrary loan

Thanks

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## Darryl Meister

> Hey Darryl, would you be able to give me the name of the paper by Keating you mentioned that dealt with both the computing of faceform and panto


"Oblique Central Refraction in Spherocylindrical Corrections with Both Faceform and Pantoscopic Tilt." Optometry and Visual Science. Vol. 72, No. 4, pp. 258-265.

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## HarryChiling

I have the article and wouldn't mind sending it to anyone who PM's me their e-mail address.

NEVER MIND I JUST BOUGHT AN ARTICLE THAT CITED KEATINGS ARTICLE.

Still kinda interesting, but already discussed on this board.  It does give an alternate to a keating equation however it does explain that you would need to be correcting a 20D lens for any significant difference.

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## HarryChiling

How could I get a copy of that article?

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## Alvaro Cordova

The  most direct way is to find a local Optometry school and ask to use the library.  The easiest way is to go to your local library and fill out a form for an inter-library loan.

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## Darryl Meister

> How could I get a copy of that article?


Harry, I am sending you an e-mail regarding this.

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## HarryChiling

I have tried working a problem out and can get the right answer when I do the left eye, but for some reason I keep making a mistake somewhere when I try to figure it out for the right.


problem being faceform and panto using Keatings article.  Any suggestions?

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## Darryl Meister

> Any suggestions?


I'd just have to see your math. As you've probably noticed by now, the procedure for calculating combined tilt is considerably more involved than for calculating simple panto or wrap. Your mistake could be an order-of-operations issue, a sign convention issue, etcetera.

I'll give you a big hint for handling Keating's equations for right and left eyes, though: Ignore his conversions for the left eye, and simply change the wrap angle from positive to negative for the left lens. You'll get the same answer without having to use all of those additional equations to compensate for the left-eye coordinate system.

Lastly, all of these equations rely heavily on the dioptric power matrix. You should ensure that you've read that particular chapter in Keating's book thoroughly enough to understand what is happening (after all, you'll have to explain it, yourself, in your paper).

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## Alvaro Cordova

I am going to review the article and tell you my first impressions on it  (May take me a few days as I work all day tomorrow and have tons of other stuff to do to, but I'm really looking forward to it).  I am pretty good at linear algebra etc.. so I don't believe I'll have a problem with the math just the initial setting up of the problem.

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## HarryChiling

I got it.  Whenever I do math that involves a few steps I always make errors.  I will show you my answers following allong with the article

example -2.00 -3.00 x 25 panto=10 face=20 n=1.586
(same exact problem Darryl presented me with for just faceform)
Solving for the right eye

theta=22.27

Tc=1.22
Sc=1.05
Hc=1.13

A=27.27

theta sub A=-2.2732  -or-  177.73

new Rx on the tangential and sagital meridian of the eye
-2.00 -3.00 x 178

P sub x=-2.00
P sub y=-5.00
P sub t=-0.12

P" sub x=-1.64
P" sub y=-4.78
P" sub t=-0.11

t=-6.42
d=7.83

C"=-3.15
S"=-1.64
theta sub A"= 0

A"=25.51
theta"=25.51

compensated Rx is -1.64 -3.15 x 26

Thank you again Darryl the article and book are very informative.  I totaly understand the part in the article where he has you switch the tangential and sagittal meridians by subtracting 90 degrees.  I had a felling that their might be a better way that is why I was dead set in working the equation out using the right eye even though the article walks you step by step through the left eye with an example.  I usually end up messing with numbers to check for patterns or easier ways of doing things.  I guess this is what you did as well.  I have a few more areas in optis I have not explored yet and am curious how many other equations can be applied to the power matrix?  I will do more reading and post back.

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## Darryl Meister

> I usually end up messing with numbers to check for patterns or easier ways of doing things. I guess this is what you did as well.


Basically, it occurred to me that if Keating's equations were indeed general enough (that is, if they would work with both positive and negative tilts without producing an ambiguous reference angle or something), you could replicate the performance of the left eye by simply changing the sign of the wrap angle. Since cylinder axis is prescribed using TABO notation, which means that the axis is measured counter-clockwise from 0 in both eyes, there is really no difference in power specification. Pantoscopic tilt also has the same coordinate system for both eyes. Further, since the left eye coordinate system for lens tilt basically represents a horizontally flipped right eye system, you could replicate a positive face-form wrap angle in the left eye by imaging a right eye with face-form wrap in the opposite direction (i.e., negative tilt). I double-checked the results and, indeed, I was correct, and this much simpler approach works just as well.

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## HarryChiling

That makes sense and simplifys the equation.

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## VVizard

Darryl, does your algorithm give the same result as exact ray-tracing would, even for tilt angles of 20-30deg?

I am asking this because I do agree with your approach to the subject and I wonder what the point to code all this when it seems like it is not less work then just to adjust some existing ray-tracing algorithm, isn't it?

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## Darryl Meister

> Darryl, does your algorithm give the same result as exact ray-tracing would, even for tilt angles of 20-30deg?


For thin-ish lenses, it's actually quite close for the range of tilts that you are likely to see with a spectacle frame.




> I am asking this because I do agree with your approach to the subject and I wonder what the point to code all this when it seems like it is not less work then just to adjust some existing ray-tracing algorithm, isn't it?


If you have access to ray-tracing algorithms, and the capability to make such adjustments, No, you wouldn't need a program like this.

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## HarryChiling

In KeaTings article he does mention that it is an approximation and that it losses accuracy at tilts of more than 30 degrees, but the likely hood of running into a frame with that much tilt is pretty slim and at that point it does give you a good approximation.

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## Lonewolf_21187

:Nerd:  "Of course" leave it too us great  the "Italians" to develop a monomer with another chemical maker of whom will be not mentioned at this time, but it is also under experiment right now that this particular monomer will be available for Rx soon. Several sunglass companies now use it. NXT technology. From sources.

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## Jedi

> "Of course" leave it too us great  the "Italians" to develop a monomer with another chemical maker of whom will be not mentioned at this time, but it is also under experiment right now that this particular monomer will be available for Rx soon. Several sunglass companies now use it. NXT technology. From sources.


Isn't NXT just a fancy name for Trivex?

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## AWTECH

Sort of the same, since both Trivex and NXT can be made with different variations each, Trivex or NXT is not exactly the same. They can each make the material more or less flexible for example.

These are both products developed originally by Simula Technologies and they are both marketed under license from Simula.

NXT I think is restricted to plano and Trivex is restricted to prescription.

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## lensgrinder

Where can I find the copy of Keating's article that is referred?  I have the latest copy of his book and would like to read this article.  I looked on the Optometry and Vision Sciecne web site, but the archives do not go back that far.

Thanks in advance

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## HarryChiling

Lensgrinder send me an e-mail and I WIll forward it to you.

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## billy_ntu

> I got it. Whenever I do math that involves a few steps I always make errors. I will show you my answers following allong with the article
> 
> example -2.00 -3.00 x 25 panto=10 face=20 n=1.586
> (same exact problem Darryl presented me with for just faceform)
> Solving for the right eye
> 
> theta=22.27
> 
> Tc=1.22
> ...


I tried to understand the tilt formula, like Darryl presented before, without any panto tilt, and it is ok.

Now with panto,wrap tilt i couldn't follow it, can someone explain, how to find (formula) the "theta"=22.27, A=27.27, and "theta sub A".

Thank you.

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## HarryChiling

> I tried to understand the tilt formula, like Darryl presented before, without any panto tilt, and it is ok.
> 
> Now with panto,wrap tilt i couldn't follow it, can someone explain, how to find (formula) the "theta"=22.27, A=27.27, and "theta sub A".
> 
> Thank you.


Billy,

Do you have a copy of the paper the formulas come from?

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## billy_ntu

Hi Harry,

No i don't have it, could you please send it to me?

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## HarryChiling

PM me your e-mail address.

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## loratz

Add Heat Bend Lens,done.

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