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Thread: Oakley's secret

  1. #26
    ATO Member HarryChiling's Avatar
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    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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  2. #27
    Master OptiBoarder Darryl Meister's Avatar
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    Quote Originally Posted by Harry
    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.

    Quote Originally Posted by Harry
    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.
    Last edited by Darryl Meister; 09-03-2005 at 06:00 PM. Reason: Added point about sagittal power
    Darryl J. Meister, ABOM

  3. #28
    ATO Member HarryChiling's Avatar
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    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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  4. #29
    Master OptiBoarder Darryl Meister's Avatar
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    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.
    Darryl J. Meister, ABOM

  5. #30
    OptiBoardaholic OptiBoard Silver Supporter Alvaro Cordova's Avatar
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    I only have the first edtion of the Keating book. I looked on Amazon and it's like 95 bucks.

  6. #31
    ATO Member HarryChiling's Avatar
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    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.
    Last edited by HarryChiling; 09-05-2005 at 07:50 PM.
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  7. #32
    Master OptiBoarder Darryl Meister's Avatar
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    Quote Originally Posted by Harry
    They are very similar,
    Well, they're both very similar to the original prescription, at least. ;)

    Quote Originally Posted by Harryl
    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.

    Quote Originally Posted by Harry
    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.

    Quote Originally Posted by Harry
    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.
    Darryl J. Meister, ABOM

  8. #33
    ATO Member HarryChiling's Avatar
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    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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  9. #34
    Master OptiBoarder Darryl Meister's Avatar
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    Quote Originally Posted by Harry
    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.

    Quote Originally Posted by Harry
    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.
    Darryl J. Meister, ABOM

  10. #35
    ATO Member HarryChiling's Avatar
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    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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  11. #36
    Master OptiBoarder Darryl Meister's Avatar
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    Quote Originally Posted by Harry
    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.

    Quote Originally Posted by Harry
    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.

    Quote Originally Posted by Harry
    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.

    Quote Originally Posted by Harry
    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).
    Last edited by Darryl Meister; 09-11-2005 at 11:39 AM.
    Darryl J. Meister, ABOM

  12. #37
    ATO Member HarryChiling's Avatar
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    I see so then the trace and determinat are necesary to convert the power matrix back to the spherocylindrical paramaters.
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  13. #38
    Master OptiBoarder Darryl Meister's Avatar
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    Quote Originally Posted by Harry
    I see so then the trace and determinat are necesary to convert the power matrix back to the spherocylindrical paramaters.
    Yes.
    Darryl J. Meister, ABOM

  14. #39
    ATO Member HarryChiling's Avatar
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    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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  15. #40
    Master OptiBoarder Darryl Meister's Avatar
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    T = 1.18
    S = 1.04
    Darryl J. Meister, ABOM

  16. #41
    Master OptiBoarder rbaker's Avatar
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    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.

  17. #42
    Master OptiBoarder Darryl Meister's Avatar
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    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.
    Darryl J. Meister, ABOM

  18. #43
    ATO Member HarryChiling's Avatar
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    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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  19. #44
    OptiBoardaholic OptiBoard Silver Supporter Alvaro Cordova's Avatar
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    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

  20. #45
    ATO Member HarryChiling's Avatar
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    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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  21. #46
    Master OptiBoarder Darryl Meister's Avatar
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    Quote Originally Posted by Al
    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.

    Quote Originally Posted by Harry
    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.

    Quote Originally Posted by Harry
    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).
    Darryl J. Meister, ABOM

  22. #47
    ATO Member HarryChiling's Avatar
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    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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  23. #48
    OptiBoardaholic OptiBoard Silver Supporter Alvaro Cordova's Avatar
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    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

  24. #49
    Master OptiBoarder Darryl Meister's Avatar
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    Quote Originally Posted by Al
    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.
    Darryl J. Meister, ABOM

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    ATO Member HarryChiling's Avatar
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    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.
    Last edited by HarryChiling; 09-12-2005 at 07:41 PM.
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