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Thread: vertical imbalance (when dist o.c. and eyes at diff level)

  1. #1
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    vertical imbalance (when dist o.c. and eyes at diff level)

    I have a question about prism and vertical imbalance using a patients O.C. ht vs where their eyes sit vs what the seg ht is.


    For example: rx o.d. -5.00 -2.00 x 162
    o.s -3.00 -3.00 x 097 add +3.00 o.u.

    seg ht is 18mm, eyes sit at 25mm and the reading level is 14mm.

    I know you have to solve for the lens power at 90degr meridian.

    o.d.=-6.81 o.s.=-3.04

    Where should I go from here?

    Normally I would simply take the reading level and multiply by power at 90degrees and divide by 10.

    prism o.d.=(6.81 x 14)/10=9.5 prism bd

    prism o.s.=(3.04 x 14)/10=4.3 prism bd

    VI=5.2prism

    but what about the prism for the eyes sitting higher than the distance o.c., which I figure to be 4mm above seg ht. (18+4=22mm)

    So eyes at 25mm and o.c. at 22mm. we have a 3mm difference still.

    I hope Darryl is around. Thank you :hammer:

  2. #2
    Master OptiBoarder Darryl Meister's Avatar
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    If the eyes are sitting higher than the OCs, just ignore the pupil heights and concentrate on the distance from the actual optical centers to the reading point (which is generally around 4 to 5 mm below the ledge of a flat-top bifocal). Remember that vertical imbalance at near is caused by induced prism, which is in turn caused by looking away from the optical center of the lens.

    For instance, consider an extreme case of your Rx in which the optical centers had been ground at the same level as the reading point... In this situation, there would be no vertical imbalance at near (since the eyes are looking through the optical centers while reading), but a considerable amount of vertical imbalance at distance (since the wearer is now looking well above the optical centers during distance vision).

    So, consider the following job:

    B (Depth) = 40 mm
    Seg Height = 15 mm
    Pupil Height = 23 mm

    The optical centers would normally be surfaced to coincide with the midline ("180" or datum line) of the frame. This would make the optical center heights:

    1/2 B = 1/2 (40) = 20 mm

    Now, assuming that the wearer will look at least 5 mm below the ledge of the flat-top segment, the reading level distance becomes:

    OC Height - Seg Height + 5 = 20 - 15 + 5 = 10 mm

    You would then use this value of 10 mm for your vertical imbalance prism calculations, regardless of what the pupil height is.

    Best regards,
    Darryl

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