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Thread: ANSI Z80.1 tolerance question

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    Smilie ANSI Z80.1 tolerance question

    Hey folks!

    I'm not sure how I have never, ever noticed it but now as I'm looking at the ANSI Z80.1 2015 tolerance I noticed that the Sphere tolerance is based on Sphere Meridian Power (Minus Cylinder Convention). Does this mean that I don't have to find and use the power at Meridian of Highest Absolute Power to see which tolerance I need to use? If I'm reading it correctly, as long as the Rx is minus cylinder form I can just use the Sph power (whether its more or less than +/- 6.50) to determine which tolerance to use. Is that right?

    Much appreciated!

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    One eye sees, the other feels. OptiBoard Gold Supporter
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    Quote Originally Posted by azhanli22 View Post
    Hey folks!

    I'm not sure how I have never, ever noticed it but now as I'm looking at the ANSI Z80.1 2015 tolerance I noticed that the Sphere tolerance is based on Sphere Meridian Power (Minus Cylinder Convention). Does this mean that I don't have to find and use the power at Meridian of Highest Absolute Power to see which tolerance I need to use? If I'm reading it correctly, as long as the Rx is minus cylinder form I can just use the Sph power (whether its more or less than +/- 6.50) to determine which tolerance to use. Is that right?

    Much appreciated!
    Welcome to Optiboard.

    I've always used the sphere power in the strongest meridian.

    I just found this...

    http://cnpg.comparenetworks.com/1783...015-Explained/

    I haven't had a chance to to listen yet, but should be worth a look.

    Best regards,

    Robert Martellaro
    Roberts Optical Ltd.
    Wauwatosa Wi.
    www.roberts-optical.com
    ~~~~~~~~~~~~~~~~~~
    Science is a way of trying not to fool yourself. - Richard P. Feynman

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    OptiBoard Novice teej's Avatar
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    It's a long video but I thought it had a lot of information that I was wondering about! Thanks Robert Martellaro!

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    One eye sees, the other feels. OptiBoard Gold Supporter
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    Quote Originally Posted by teej View Post
    It's a long video but I thought it had a lot of information that I was wondering about! Thanks Robert Martellaro!
    You're welcome.

    I noticed that the explanation given for relaxing the axis recommendations for low cylinder powers was due, at least in part, to a "Focimeter Error for Cylinder Axis" study by G. Fry. However, my understanding was that power error, affecting the wearer's quality of vision, as described by D. Meister HERE, was the primary reason for the change, assuming that mechanical tolerances were also realistic.

    I was wrong to say that we should use the meridian of highest power tolerance, that was changed in one of the earlier revisions.

    Best regards,

    Robert Martellaro
    Roberts Optical Ltd.
    Wauwatosa Wi.
    www.roberts-optical.com
    ~~~~~~~~~~~~~~~~~~
    Science is a way of trying not to fool yourself. - Richard P. Feynman

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    Master OptiBoarder OptiBoard Silver Supporter lensmanmd's Avatar
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    Robert,

    The meridian of highest power, referenced @ the 90 is apropos. This power would be used for vertical imbalance reference.
    However, the video does state that power tolerances are defined by errors in the sphere and cylinder separately, meaning that the sph could be .13 off AND the cyl could be .13 off, with a net of .26 total power. I believe this is what you were originally referring to

    Regards.
    I bend light. That is what I do.

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    One eye sees, the other feels. OptiBoard Gold Supporter
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    Quote Originally Posted by lensmanmd View Post
    Robert,

    The meridian of highest power, referenced @ the 90 is apropos. This power would be used for vertical imbalance reference.
    However, the video does state that power tolerances are defined by errors in the sphere and cylinder separately, meaning that the sph could be .13 off AND the cyl could be .13 off, with a net of .26 total power. I believe this is what you were originally referring to

    Regards.
    I believe the OP was looking for the sphere power tolerance. For example, -3.00 -2.00 x 180. The older recommendations asked for the power in the strongest meridian, -5.00, the newer standards (2010?) use the -3.00 value, and as you stated the cylinder power is calculated independent of the sphere power, where -5.26 +2.13 x 90 would pass if I'm interpeting this correctly.

    Best regards,

    Robert Martellaro
    Roberts Optical Ltd.
    Wauwatosa Wi.
    www.roberts-optical.com
    ~~~~~~~~~~~~~~~~~~
    Science is a way of trying not to fool yourself. - Richard P. Feynman

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