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Thread: Theres an easier way I just know it.

  1. #1
    Independent Owner OptiBoard Silver Supporter kcount's Avatar
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    Theres an easier way I just know it.

    Hello all!

    I regularly have to 'Adjust' an Rx for a specific focal length for patients and have a quick question. Historically I would do a focal length computation (P=1/f) and then match their Rx as close as possible but I'd love a more exact approach. Is there a more exact way to take a persons Rx (distance or near) and compute it to a specific focal length? I am hoping for something I could build into an excel spread sheet and simply plug in the Rx and hit go.

    I can't believe I'm the only one with patients needing very task specific products or doing therapeutic products.

    Any ideas?

    KC
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  2. #2
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    The dioptric demand of a given working distance is equal to the formula you are using (P=1/f). For an absolute presbyope, with no accommodation, this would be equal to the amount of addition power required to see clearly at this distance. However, for most patients, some amplitude of accommodation remains. The refractionist usually factors this in when arriving at the specified add power for a given patient. Of course, the add power determined by the refractionist is also based upon a specific working distance, often at 16 inches (40 cm).

    Frequently, the refractionist will select an add power that requires no more than 50% of the patient's amplitude of accommodation for the intended reading distance, leaving the other 50% in reserve. So, for instance, if the refractionist prescribes a +2.00 D add power for a 40 cm reading distance, which has a dioptric demand of 2.50 D, the patient is expected to make up the +0.50 D difference, and probably has an amplitude of accommodation of roughly 1.00 D (that is, twice the reserve of 0.50 D).

    If you would like to produce a lens for a working distance other than the prescribed reading distance, the simplest approach would be to adjust the add power for the difference in dioptric demand. A working distance of 66.67 cm, for instance, would have a dioptric demand of 1.50 D. Given our previous example with a +2.00 D add power determined for a reading distance of 40 cm, a demand of 2.50 D, the new working distance has 2.50 - 1.50 = 1.00 D less dioptric demand. Consequently, the add power would be reduced by 1.00 D, to 2.00 - 1.00 = +1.00 D.

    In symbolic terms, this would be:


    Another alternative is to simply "scale" the original add power by an amount equal to the ratio of the old working distance to the new working distance. So, for instance, if the working distance doubles compared to the original intended reading distance, you could reduce the original add power by 1/2, one-half, or 50%. This approach will reduce the amount of accommodation that the patient is expected to exert for a given working distance by the same factor though (by 50%, in this example).

    Of course, before making any modifications to the original prescription, a discussion with the refractionist is probably in order.

    Best regards,
    Darryl
    Last edited by Darryl Meister; 03-17-2012 at 09:08 AM.
    Darryl J. Meister, ABOM

  3. #3
    Enjoying the education drk's Avatar
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    I hate the term "refractionist"

    What is easier than all that is to measure the deficit in accommodation at a given distance..say with the binocular cross cylinder. The answer is an add! Trained monkey refractionist stuff!

    But think about it: like Darryl said, if dude is 60, just add the inverse of the working distance. If non-absolute presbyope, use a functional measure such as I said.

    Good rule of thumb: +0.75 is always a good amount! (Borysko would approve!)
    Last edited by drk; 03-14-2012 at 10:55 AM.

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    "Refractionist" is must shorter than writing out "optometrist, ophthalmologist, or refracting optician."
    Darryl J. Meister, ABOM

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    opti-tipster harry a saake's Avatar
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    ???

    Quote Originally Posted by Darryl Meister View Post
    "Refractionist" is must shorter than writing out "optometrist, ophthalmologist, or refracting optician."
    Darryl, you going to be at VEE, ill be there 4 days

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    Harry, I won't be attending VisionExpo East this year. I'm just too busy right now. Maybe next year though.

    Best regards,
    Darryl
    Darryl J. Meister, ABOM

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    OD or OMD is shorter than "refractionist".
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    Not every refractionist is an "OD" or "OMD" though. And these designations generally do not apply to optometrists and similar optical professionals in other countries.

    I should add that you guys are welcome to use whatever terms you like in your own posts.

    Best regards,
    Darryl
    Darryl J. Meister, ABOM

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    Master OptiBoarder OptiBoard Silver Supporter
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    I have found that formulas are not always effective with everyone. Two different +2.50 patients may need a different add power at 4 feet. I suggest trial lenses or refracting at a specific distance for the most consistent results.
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    Master OptiBoarder MakeOptics's Avatar
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    Quote Originally Posted by sharpstick777 View Post
    I have found that formulas are not always effective with everyone. Two different +2.50 patients may need a different add power at 4 feet. I suggest trial lenses or refracting at a specific distance for the most consistent results.
    I agree with the trial lens, especially since the working distance is hardly ever on the script. As an aside different doctors use different ratios for the amount of accommodative reserve, Darryl mentioned 1/2, but I have seen 2/3 and even 1/3. Depending on the amount left in reserve by the time they get to you the script may need tweaking unless it's fresh. (I don't mean tweaking as changing the script just adjusting the working distance for the add and the associated power)

  11. #11
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    I agree that refracting for the correct working distance is always the best solution when adjusting the lens power for a new working distance, although it may not necessarily be feasible in every case, particularly for the optician.

    Best regards,
    Darryl
    Darryl J. Meister, ABOM

  12. #12
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    To address a separate request, I am also including a version of the formula for compensating the add power for a new near working distance that also takes into account the position of the spectacle lens relative to the principal plane of the eye, which will result in more accurate results in higher Rx powers (from System for Ophthalmic Dispensing):



    where Vertex is the vertex distance of the lens in meters (allowing for an extra 1.5 mm to the principal plane of the eye), Power is the specified distance Rx power of the lens, AddNEW is the new addition power for the new working distance, DemandNEW is the dioptric demand associated with the new working distance, AddOLD is the original near addition power, and DemandOLD is the dioptric demand associated with the original near refraction distance.

    The dioptric demand is equal to 100 / D, where D is the working distance in centimeters. For a typical near refraction distance of 40 cm (16 in), the dioptric demand is 100 / 40 = 2.50 D.

    Best regards,
    Darryl
    Darryl J. Meister, ABOM

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    Master OptiBoarder MakeOptics's Avatar
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    Quote Originally Posted by Darryl Meister View Post
    To address a separate request, I am also including a version of the formula for compensating the add power for a new near working distance that also takes into account the position of the spectacle lens relative to the principal plane of the eye, which will result in more accurate results in higher Rx powers (from System for Ophthalmic Dispensing):



    where Vertex is the vertex distance of the lens in meters (allowing for an extra 1.5 mm to the principal plane of the eye), Power is the specified distance Rx power of the lens, AddNEW is the new addition power for the new working distance, DemandNEW is the dioptric demand associated with the new working distance, AddOLD is the original near addition power, and DemandOLD is the dioptric demand associated with the original near refraction distance.

    The dioptric demand is equal to 100 / D, where D is the working distance in centimeters. For a typical near refraction distance of 40 cm (16 in), the dioptric demand is 100 / 40 = 2.50 D.

    Best regards,
    Darryl
    You could make that formula longer if you replace all the dioptric powers with their equivalent focal powers 1/f. Vertex compensated powers including the actual distance power, nice pull.

  14. #14
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    Yeah, Brooks's original version has several nested fractions that can become difficult to work with or implement in a calculator or spreadsheet, if you're not paying close attention, so I tried to clean it up a bit. It also makes it a little easier to understand what is actually going on in the formula.
    Darryl J. Meister, ABOM

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    Master OptiBoarder MakeOptics's Avatar
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    Quote Originally Posted by Darryl Meister View Post
    Yeah, Brooks's original version has several nested fractions that can become difficult to work with or implement in a calculator or spreadsheet, if you're not paying close attention, so I tried to clean it up a bit. It also makes it a little easier to understand what is actually going on in the formula.
    I prefer my math like my code more modular so that expressions can be reused and more formulas and theories memorized. I find it simpler to understand a concept instead of a long drawn out formula that is essentially 3 or 4 simple formulas nested together. Either way thanks for sharing it, I have the book but I am sure many others don't. You may have encouraged someone to go out and buy a book.

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