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Thread: compensating prism using dissimilar segs

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    compensating prism using dissimilar segs

    Help. We can't seem to remember how to do this. Can anyone please give us the steps to calculate for this?
    Thank you,
    mtillman

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    No math but you get base up prism from ST type bifocals, base down from round type bifocals. The amount will vary with the add, I think. Usually measured at the optical center of seg.
    Boils down to use the ST on the lens with the highest minus (or lowest plus power) and the round one on the lowest minus (or highest plus).
    It's seldome done today but most plus lens patients would be better served with a round top seg.

    Chip

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    ATO Member HarryChiling's Avatar
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    Give an example and I'll show you.
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    ATO Member HarryChiling's Avatar
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    mtillman check your e-mail.
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    Awe, I was hoping to see some of the examples. I love seeing harry whip out some sweet maths :D

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    ATO Member HarryChiling's Avatar
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    For the following RX, determine the dissimilar segment solution:
    OD +3.50 -2.50 x 045
    OS +2.00 -2.00 x 150
    Add +2.75
    RL 9mm
    Step 1) Find out the prism difference between both eyes:

    A: Get the Rx along the 90th meridian for both eyes

    D@90 = Sph + Cyl * sin^2(difference to axis)
    ODD@90 = +3.50 + (-2.50) * sin^2(90-45)
    ODD@90 = +3.50 + (-2.50) * 0.50
    ODD@90 = +3.50 + (-1.25)
    ODD@90 = +2.25

    OSD@90 = +2.00 + (-2.00) * sin^2(150-90)
    OSD@90 = +2.00 + (-2.00) * 0.75
    OSD@90 = +2.00 + (-1.50)
    OSD@90 = +0.50

    B: Use Prentices rule to figure out the amount of prism created at the Reading Level

    Prism = Power * Decentration(mm) / 10

    OD:Prism = +2.25 * 9mm / 10
    OD:Prism = 2.03 and since it is a plus lens it's BU

    OS:Prism = +0.50 * 9mm / 10
    OS:Prism = +0.45 again since it is a plus lens it's BU

    The prism difference is:

    +2.03 - +0.45 = +1.60 BU OD or -1.60 BD OS

    Step 2: Find a combination of segs that will eliminate the difference in prism

    A: This can be done by deconstructing the prism differnce into decentration or seg drop using again the prentices rule

    Decentration(mm) = Prism * 10 / Power(Add)
    Decentration(mm) = +1.60 * 10 / +2.25
    Decentration(mm) = +16 / +2.25
    Decentration(mm) = 7mm

    B: Now we pick a combination of segs that have a difference of 7mm in their seg tops to their optical centers

    Difference(FT or CT) = Segment Depth - Segment Width/2
    Difference(Round or Ultex) = Segment Width/2

    FT22 = 8mm
    FT25 = 6.5mm
    FT28 = 5mm
    FT35 = 4.5
    FT40 = On the line
    FT45 = On the line
    Rnd22 = 11mm
    Rnd25 = 12.5 mm
    Ultex = 14 mm
    Exec = On the line

    You could go with a FT25 and a FT40 or FT45 which would leave about a 0.12 diopter of prism unchecked or you could leave up to about 0.50 diopters of prism unchecked you could go with a FT28 and a FT40 and leave 0.45 unchecked, you could also go with a FT28 and a RND22 and leave 0.23 unchecked, you could go with a FT28 and a FT40 or FT45 leaving 0.45 enchecked at this point I think you get the drift.

    C: Place the correct lens with the correct eye, since the difference is 1.6 BU OD or 1.6 BD OS, if we were to choose the option of the FT28 and RND22:

    RND22 = 11mm * +2.25 / 10 = 2.48 BD
    FT28 = 5mm * +2.25 / 10 = 1.13 BD

    More BD in the RND22 and the imbalance calls for more BD in the OS so the RND22 would be for the left eye.

    Quote Originally Posted by AdmiralKnight
    Awe, I was hoping to see some of the examples. I love seeing harry whip out some sweet maths :D
    I got the example my e-mail so resonded that way, but here you go some sweet math. I love doing math it occupies my mind.
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    ATO Member HarryChiling's Avatar
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    Now you can find data for various segments here:

    http://onlineopticianry.com/wordpress/?p=157
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    [quote=HarryChiling;232887]
    B: Now we pick a combination of segs that have a difference of 7mm in their seg tops to their optical centers

    Difference(FT or CT) = Segment Depth - Segment Width/2
    Difference(Round or Ultex) = Segment Width/2


    FT25 = 6.5mm
    Ultex = 14 mm

    /quote]

    Your FT25 example being 25mm wide and 17mm deep/2 = 4mm, although as you later listed, many manufacturers place the OC of FT25 5mm below line. (And Ultex A is 38mm wide/2 = 19mm).
    So a required seg OC difference of 7mm would favour FT25 and Round Top 22 for dissimilar segs. This dissimilar pair choice also has the advantage of looking the best cosmetically.:cheers:

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    ATO Member HarryChiling's Avatar
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    [quote=tmorse;237703]
    Quote Originally Posted by HarryChiling View Post
    B: Now we pick a combination of segs that have a difference of 7mm in their seg tops to their optical centers

    Difference(FT or CT) = Segment Depth - Segment Width/2
    Difference(Round or Ultex) = Segment Width/2


    FT25 = 6.5mm
    Ultex = 14 mm

    /quote]

    Your FT25 example being 25mm wide and 17mm deep/2 = 4mm, although as you later listed, many manufacturers place the OC of FT25 5mm below line. (And Ultex A is 38mm wide/2 = 19mm).
    So a required seg OC difference of 7mm would favour FT25 and Round Top 22 for dissimilar segs. This dissimilar pair choice also has the advantage of looking the best cosmetically.:cheers:
    Thanks tmorse, the problem wiht working in a lab is I don't have to try and remember the seg to OC I could just grab a blank and measure it which is how I came up with that number. It was more an estimate but it would work all the same. I did however take the time to compile and list on the ATO site which is the link in the previous post a table that contains the data so now no one has to try and figure it out again. All is well again.
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