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Thread: Optician in India just asked a great question

  1. #1
    ATO Member HarryChiling's Avatar
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    Post Optician in India just asked a great question

    Darryl,

    I know you enjoy the math and I recently got a question about lenticulars from an optician and thought I'd share it.

    The type of lenticular you are asking about cannot be done using
    traditional processing techniques, if you needed to make a minus
    lenticular or myodisc that can be done and is fairly easy. First you
    would need to determine your minimum blank size of the lens

    mbs = minimum blank size
    dec = decentration
    DPD = distance PD
    ED = effective diameter

    mbs = ED + 2*dec + 1mm

    Now that we have our minimum bank size we can determine the saggital
    height of the lens using the front curve and minimum blank size.

    F1 = front curve in diopters
    n = material index

    sag(F1)@mbs = [F1*(mbs/2)^2]/[2000*(n-1)]

    Now using the back curve and the optical zone we need to find out
    where the optical zone intersects the peripheral curve.

    F2 = back curve in diopters
    oz = optic zone

    sag(F2)@oz = [F2*(oz/2)^2]/[2000*(n-1)]

    Now we need to know the diference between the plate height of the
    front curve and the sag of the back curve.

    diff = difference
    ct = center thickness

    diff = sag(F1)@mbs + ct - sag(F2)@oz

    Now this difference will help us detremine the peripheral curve, for
    the peripheral curve to be correct the differnce between the
    peripheral curve sag at the minimum blank size minus the peripheral
    curve at the optical zone size will be equal to the difference.

    F3 = peripheral curve
    sag(F3)@mbs = [F3*(mbs/2)^2]/[2000*(n-1)]
    sag(F3)@oz = [F3*(oz/2)^2]/[2000*(n-1)]

    diff = sag(F3)@mbs - sag(F3)@oz

    If we substitute out the forumulas above and simplify we get:

    diff = sag(F1)@mbs + ct - sag(F2)@oz
    diff = [F1*(mbs/2)^2]/[2000*(n-1)] + ct + [F2*(oz/2)^2]/[2000*(n-1)]

    diff = sag(F3)@mbs - sag(F3)@oz
    diff = [F3*(mbs/2)^2]/[2000*(n-1)] - [F3*(oz/2)^2]/[2000*(n-1)]

    Now we set them equal to each other:

    [F3*(mbs/2)^2]/[2000*(n-1)] - [F3*(oz/2)^2]/[2000*(n-1)] =
    [F1*(mbs/2)^2]/[2000*(n-1)] + ct + [F2*(oz/2)^2]/[2000*(n-1)]

    Now we simplify, start by multiplying both sides by 2000*(n-1) which
    woudl give us:

    [F3*(mbs/2)^2] - [F3*(oz/2)^2] = [F1*(mbs/2)^2] + [ct*2000*(n-1)] + [F2*(oz/2)^2]
    F3*[(mbs/2)^2 - (oz/2)^2] = [F1*(mbs/2)^2] + [ct*2000*(n-1)] + [F2*(oz/2)^2]
    F3 = { [F1*(mbs/2)^2] + [ct*2000*(n-1)] + [F2*(oz/2)^2] } / [(mbs/2)^2 - (oz/2)^2]

    So now we know what our peripheral curve is so we need to know what
    thickness to cut the first cut our peripherl curve onto.

    pt = peripheral curve thickness

    pt = sag(F1)@mbs + ct - sag(F3)@mbs

    So then your steps would be to:
    1.) generate the F3 curve at a thickness of pt onto your blank and run
    it through the finer and polisher
    2.) tape the back surface with surface saver tape
    3.) cut the F2 curve at a thickness of ct onto your blank and run it
    through the finer and polisher

    At no point should the lenses be deblocked they should remain on the
    blocks through the entire processing steps then removed at the end.

  2. #2
    Master OptiBoarder Darryl Meister's Avatar
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    Good stuff, Harry.

    Alternatively, since any cylinder power must be surfaced onto the front side of minus-lenticular and myodisc lenses anyway, you could begin with a semi-finished lens blank that has a base curve equal to the desired peripheral or carrier curve, such as a Plano Base for a myodisc or an 8.00 or 10.00 Base for a minus-lenticular. Just flip the lens around, in order to use the front curve of the lens blank as the back carrier curve of the final lens, and you've reduced the number of surfacing operations necessary to make these.
    Darryl J. Meister, ABOM

  3. #3
    ATO Member HarryChiling's Avatar
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    Quote Originally Posted by Darryl Meister View Post
    Good stuff, Harry.

    Alternatively, since any cylinder power must be surfaced onto the front side of minus-lenticular and myodisc lenses anyway, you could begin with a semi-finished lens blank that has a base curve equal to the desired peripheral or carrier curve, such as a Plano Base for a myodisc or an 8.00 or 10.00 Base for a minus-lenticular. Just flip the lens around, in order to use the front curve of the lens blank as the back carrier curve of the final lens, and you've reduced the number of surfacing operations necessary to make these.
    Nice tip DragonLensManWV does that, I have found for cylindrical lenses if you use the power matrix of the lens in the formula and then use the negative torsion of the lens alogn with the the new lenticular matrix the cylinder curves cancel each other out to create an oval oriented along the 180 or 090, much more cosmetically appealing. You could also just use the opposite axis (mirrored alogn the 180) in spherocylindrical form. ex:

    Original Rx: -6.00 -2.00 x 045
    Front Curve: 2 Base
    Back Base: 8.00
    Back Cross: 10.00
    Axis of Base: 045

    Lenticular Base: 4.00
    Lenticular Cross: 6.00
    Axis of Base: (315) 135

    The numbers are of course hypothetical. I have run a few jobs in the wraps and they have consistently come out with an Oval OZ oriented along the 180.

  4. #4
    Master OptiBoarder Darryl Meister's Avatar
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    Yes, you could certainly place the horizontal/vertical component of the cylinder on the back, if you're willing to work out the additional computations. Just keep in mind that if the cylindrical components are not equal in both lenses, you will get ellipses with two different sizes, which may not be cosmetically acceptable. You also increase the likelihood of a reject due to the cylinder axis being off.

    If you do go this route though, rather than using the dioptric power matrix, you'll save yourself a little time by using Humphrey's system, since it decomposes the cylinder into an oblique component (C45) at axis 45, which would be surfaced on the front side, and a horizontal component (C0) at axis 0, which would be surfaced on the back:





    This circumvents the need to combine the vertical and horizontal cylinder powers into a single cylinder power (C).
    Darryl J. Meister, ABOM

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    ATO Member HarryChiling's Avatar
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    Nice then I could use the MRE in his equations to compute the lenticular in my above formula's. Thanks D, that's the kind of path I like to go down.

  6. #6
    Underemployed Genius Jacqui's Avatar
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    Sounds like you guys are trying to do it the hard way.

  7. #7
    Master OptiBoarder Darryl Meister's Avatar
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    Sounds like you guys are trying to do it the hard way.
    Hey, I'm all for just slapping all of the cylinder on the front side!

    On a side-note, if you are willing to put a little extra effort into these, you can actually get better cosmetic results with minus-lenticular lenses by decentering the bowl or optical zone by an amount equal to at least the decentration required for the frame. Otherwise, the nasal edge of the lens will actually be thicker than the temporal edge, which not only looks awkward but also makes it more difficult to adjust the nose pad arms.
    Darryl J. Meister, ABOM

  8. #8
    ATO Member HarryChiling's Avatar
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    Its my approach for putting it on a wrap so that you could use an 8 base lens.

    Darryl,
    I hav actualy though of taking the thickness of the nasal and temporal sides and using thes figures to incorporate prism into the peripheral curve to eliminate the thickness of both edges, but I would think that this would have an effect on the OZ size and shape. Still working on it.

  9. #9
    Master OptiBoarder Darryl Meister's Avatar
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    I would think that this would have an effect on the OZ size and shape. Still working on it
    I haven't given the shape of the optical zone a great deal of thought, but I suspect that since we're talking about the intersection of sections of two spheres, the shape of the bowl should remain nominally circular, regardless of the amount of prism surfaced into the bowl. Perhaps Robert can shed some light, if his experience has indicated otherwise though...
    Darryl J. Meister, ABOM

  10. #10
    ATO Member HarryChiling's Avatar
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    Quote Originally Posted by Darryl Meister View Post
    I haven't given the shape of the optical zone a great deal of thought, but I suspect that since we're talking about the intersection of sections of two spheres, the shape of the bowl should remain nominally circular, regardless of the amount of prism surfaced into the bowl. Perhaps Robert can shed some light, if his experience has indicated otherwise though...
    I shoud clarify, if I used a toric OZ, a toric peripherall curve to match the OZ will be a circle (tried it and it works and it's crazy expensive if you order your RGP contacts liek this which is where I got the idea), but my thought is that the prism may throw off the match between the peripheral curve and the OZ's, I thought I'd give it a try sometime soon so when I do I'll definately keep you in the know.

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