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Thread: Tilting Polarized Lenses

  1. #1
    ATO Member OptiBoard Bronze Supporter HarryChiling's Avatar
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    Tilting Polarized Lenses

    I had some one contact me about the effects of tilting a polarized lens. The patient said that the tilted lenses were not as effective as his regular polarized lenses. So I got to thinking (I know I'll try not to hurt myself). If we are compensateing the Rx for the effects of tilt, wouldn't we need to further compensate the axis of the polarized filter in the wrap. Here's an example:

    Malus' Law
    I = intensity of light through the lens
    Io = initial intensity
    a = angle between the light entering and the polarized ray leaving the lens

    I = Iocos2(a)

    Since we are blocking glare from above and below or alighned on the 90th meridian the initial intensity would be canceled out due to the cosine of 90 being "0" and the rays aligned with the 180th meridian would be 100% as the cosine of 180 is "-1" then squared it's "1". Assuming we live in a perfect world and their would be no absorbtion through the fiter and it's 100% effective. If we were to tilt the polarized film would we need to apply compensations to Malus' Law similar to power?

    Would using stokes parameters and applying some sort of compensation to the vector a good way to approach it? Could it be compensated for? Does the tilt effect the film?
    Last edited by HarryChiling; 09-28-2007 at 01:21 AM.

  2. #2
    Rising Star lensgrinder's Avatar
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    Well Harry I could be way off base here, but I will try anyway.
    My theory has been that when we put polarized lenses into a frame with enough tilt and wrap the polarization is not aligned along the 0 - 180, it is also tilted by a certain amount. Which would cause the polarizing filter to not do its job.
    So in my theory if you took Keatings equation for the effective tilt angle to find out how much you need to adjust the lens. For example if you have
    15oof faceform and 10oof pantoscopic tilt then your effective tilt angle is about 18o. So before we surface the lens we need to rotate the polarization 18o for each lens.
    Let me know your thoughts.

  3. #3
    ATO Member OptiBoard Bronze Supporter HarryChiling's Avatar
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    Quote Originally Posted by lensgrinder View Post
    Well Harry I could be way off base here, but I will try anyway.
    My theory has been that when we put polarized lenses into a frame with enough tilt and wrap the polarization is not aligned along the 0 - 180, it is also tilted by a certain amount. Which would cause the polarizing filter to not do its job.
    So in my theory if you took Keatings equation for the effective tilt angle to find out how much you need to adjust the lens. For example if you have
    15oof faceform and 10oof pantoscopic tilt then your effective tilt angle is about 18o. So before we surface the lens we need to rotate the polarization 18o for each lens.
    Let me know your thoughts.
    In the keating equations that same tilt is then further used to compensate the dioptric power matrix for the tangential, saggital, and torsional components, which are then further used to compensate the cylinder axis. I would assume that the polarized filter should be treated more as a cylinder axis than as a effective tilt angle.

    I think I may have a solution:

    From the article
    1. Use equation (17) to find the effective tilt angle.
    2. Use equation (19) to find the tangential meridian expressed in the lens' local co-ordinate system.
    Since the effective coordinate system is A degrees off from the local coordinate system wouldn't we jst block up the lenses -A degrees off? And if this thinking is correct then wouldn't it mean that technically progressives and FT's with their films aligned along the 180 with no way of realigning them are not a good option in high wrap frames?

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    Allen Weatherby OptiBoard Gold Supporter
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    Quote Originally Posted by HarryChiling View Post
    In the keating equations that same tilt is then further used to compensate the dioptric power matrix for the tangential, saggital, and torsional components, which are then further used to compensate the cylinder axis. I would assume that the polarized filter should be treated more as a cylinder axis than as a effective tilt angle.

    I think I may have a solution:

    From the article
    1. Use equation (17) to find the effective tilt angle.
    2. Use equation (19) to find the tangential meridian expressed in the lens' local co-ordinate system.
    Since the effective coordinate system is A degrees off from the local coordinate system wouldn't we jst block up the lenses -A degrees off? And if this thinking is correct then wouldn't it mean that technically progressives and FT's with their films aligned along the 180 with no way of realigning them are not a good option in high wrap frames?
    As you know dealing with the vision and specticale lenses you are dealing with a static item, (the lens) and a dymanic item the eye, (the eye). At some point the precision you are calculating to will serve no useful purpose due to the difference in placement by the patient each time they wear their glasses.

    As far as polarized film alignment, although we compensate for this in our processing of ICE-TECH lenses, it is not what I would say is a potential significant problem. On our PALs it is a bit more exacting since we can properly align the polarized film for any frame angle correctly. Our PALs designs are cut individually according to the known frame information. That is the polarized blank is laid out according to the frame and patient needs and the intermediate is properly oriented.

  5. #5
    ATO Member OptiBoard Bronze Supporter HarryChiling's Avatar
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    Quote Originally Posted by AWTECH View Post
    As you know dealing with the vision and specticale lenses you are dealing with a static item, (the lens) and a dymanic item the eye, (the eye). At some point the precision you are calculating to will serve no useful purpose due to the difference in placement by the patient each time they wear their glasses.

    As far as polarized film alignment, although we compensate for this in our processing of ICE-TECH lenses, it is not what I would say is a potential significant problem. On our PALs it is a bit more exacting since we can properly align the polarized film for any frame angle correctly. Our PALs designs are cut individually according to the known frame information. That is the polarized blank is laid out according to the frame and patient needs and the intermediate is properly oriented.
    I understand that their comes a point where any more precise is like beating a dead horse, but the polarized filter even tilted slightly can make a significant difference.

    Thanks Allen for chiming in, becuase I didn't even consider internal free form progressives. It seems that if my calculations are correct that the internal designs would be the only ones that would give you the best effective polarization.

    When you say you compensate the polarized filter then you do rotate to offset the effects of tilt and panto? You don't have to go into detail, but this would differentiate you from the flock and you should probably let the world know as any other designs other than free form in a polarized would be flawed compared to yours.

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    Allen Weatherby OptiBoard Gold Supporter
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    Quote Originally Posted by HarryChiling View Post
    I understand that their comes a point where any more precise is like beating a dead horse, but the polarized filter even tilted slightly can make a significant difference.

    Thanks Allen for chiming in, becuase I didn't even consider internal free form progressives. It seems that if my calculations are correct that the internal designs would be the only ones that would give you the best effective polarization.

    When you say you compensate the polarized filter then you do rotate to offset the effects of tilt and panto? You don't have to go into detail, but this would differentiate you from the flock and you should probably let the world know as any other designs other than free form in a polarized would be flawed compared to yours.
    Yes Harry we do compensate the film axis as a part of our calculations.

    We just did a warp job to replace a poorly done compensation for a ECP and found they did a partial compensation, but were off by 5 mm on total PD and about 10 degrees in each eye on axis. There are not that many doing wrap arounds that understand the math involved in even getting close. Fortunately you do.

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    Rising Star lensgrinder's Avatar
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    Quote Originally Posted by HarryChiling View Post
    In the keating equations that same tilt is then further used to compensate the dioptric power matrix for the tangential, saggital, and torsional components, which are then further used to compensate the cylinder axis. I would assume that the polarized filter should be treated more as a cylinder axis than as a effective tilt angle.

    I think I may have a solution:

    From the article
    1. Use equation (17) to find the effective tilt angle.
    2. Use equation (19) to find the tangential meridian expressed in the lens' local co-ordinate system.
    Since the effective coordinate system is A degrees off from the local coordinate system wouldn't we jst block up the lenses -A degrees off? And if this thinking is correct then wouldn't it mean that technically progressives and FT's with their films aligned along the 180 with no way of realigning them are not a good option in high wrap frames?
    I would think that you would just find theta, because the other equation deals with the tangential meridian which would deal with the axis of the Rx and you need to find out how much to rotate the polarization of the lens so it will line up on the 180.

  8. #8
    ATO Member OptiBoard Bronze Supporter HarryChiling's Avatar
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    Quote Originally Posted by lensgrinder View Post
    I would think that you would just find theta, because the other equation deals with the tangential meridian which would deal with the axis of the Rx and you need to find out how much to rotate the polarization of the lens so it will line up on the 180.
    Here is my thinking and correct me if I'm wrong since the A angle from the equation is the rotation of the {x', y', z'} coordinate (tilted lens) from the {x, y, z} coordinate then I was thinking this would also be the correct amount of negative rotation to apply to the film before surfaceing. Really can't tell you as I am still working on it myself, but it was an interesting problem that obviously needs to be addressed with wrap eyewear. I just mentioned the tangential meridian since the formula references it and in this case it would be the match to the 180 which is where we want our polarized film to lie on.

    To get the e = A + A the A is the roation I believe?

  9. #9
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    This will only work on single vision lenses. Progressives and FTs will have to be special made to order.

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