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Thread: Vertical imbalance and PD problems with zero power???

  1. #1
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    Confused Vertical imbalance and PD problems with zero power???

    Academic question where you can show off your knowledge. Please do!!!
    Academic Senario: :shiner: I know it's a long one, but I am desperate!!!
    the RX OD +3.00 -4.00 x120 / OS +1.50 -3.00 x045. PD 64
    Frame measurements 56/14 Aviator Shape.

    I make the glasses and they check out perfectly. The patient trys to get use to the new rx but felt uncomfortable. The patient returns to the dr instead of me. The dr. assistant checks the glasses and says the PD is at 74, off by 10mm and there is 1^BU prism. The patient returns to me with a note from the Dr. that says "Dear Optician...fix it."

    The questions that follow the assignment are: Can you fix it? SHould you fix it? Stuff like that. (trying to get us to look deeper than the surface.)

    When I did the calculations (power at 180 and 90) I found;
    OD 0.00@180 and +2.00@90
    OS 0.00 @ 180 & 90.
    {D = DSph + DCyl (sin^)2} formula

    (THE INSTRUCTOR STRICKLY WANTS US TO DEAL WITH THE FACTS GIVEN IN THE SENARIO, :finger:NO ASSUMPTIONS ARE TO BE MADE.) This is where my problem comes in.

    1. Can you tell me the only way you can verify the PD of this kind of RX? I am completely stumped here, no ideas!!!

    2. And why is there vertical imbalance?
    [I have already submitted answers of the following... 1. Dr. assistant checked them incorrectly in the lensometer according to ANZI Standars. 2. The OC could be sitting 5mm above the patients line of sight. 3. The patient could be experiencing Anisometropic problems (difference in power between the eyes that may cause differing amounts of prism when the wearer looks off center, as in reading.), 4. The prism dial could have been bumped between lenses or frame table moved accidently. ]

    The instructor says there are more reasons????:angry: WHAT ARE THEY??????
    He says "There is also something else that can cause vertical imbalance especially in this RX." <direct quote. Well, I can't think of any more!!!
    Can anybody else???

    3. And finally, what do you percieve to be the problem? WHy? (this question is especially difficult one for me to remember the stipulation that you must to ONLY go off the facts given in the senario.:finger: )

    Please let me assure you that I am not looking for an easy way out of this assignment. I take my education very seriously. But I am truly stumped. I have communicated back and forth with the instructor and I am getting nowhere with him. I've talked to all the Optician in my office, even some of the doctors and all I got was blank stares.
    My brain is frying as I type this.
    PLLLLEEEEAAASSEE give it a shot. Let me know if you can give ANy input into ANY of these questions. I would greatly appreciate it!!!
    (can you tell I'm begging???)

    Desperate for some Optician wisdom,

    TracyElise:o

  2. #2
    Master OptiBoarder Joann Raytar's Avatar
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    Aviator Box measurements

    Quote Originally Posted by TracyElise
    The instructor says there are more reasons????:angry: WHAT ARE THEY??????
    He says "There is also something else that can cause vertical imbalance especially in this RX." <direct quote. Well, I can't think of any more!!!
    Can anybody else???
    The layout could have been wrong - bad box measurements. The person laying the lenses out could have measured straight down from the pupil instead of measuring to the bottom of the frame; in an aviator that can be especially noticable. Also, the person inspecting the lenses could have forgotten to start with the highest power at 90 when looking for vertical prism. A bad box measurement could also create the horizontal prism problems. I've seen folks measure aviators using the PD from the datum instead of the widest measurement across the A and go straight across instead of using the narrowest across the bridge.

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    Lensometer markings............................

    Quote Originally Posted by Jo
    The layout could have been wrong - bad box measurements..
    I agree with Jo. Just would like to add that often mistakes are made with spheres and cyl powers that have a near zero power at 90 degrees while marking the centers off center with the lensometer.
    Chris Ryser
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    http://optochemicals.com............................. http://arcoatings.com

  4. #4
    Pomposity! Spexvet's Avatar
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    It's possible to have prism in a plano lens, vertically and/or horizontally.

    High, oblique cyl like this can make the target in the lensometer do funky things. Notice that when you move the lens horizontally, the target moves vertically, even though it may not move horizontally at all, due to plano @ 180. If the vertical imbalance is not checked at the OC horizontal point, it may look like the center is high or low (vertical prism).

    IMHO, the REAL issue will have to do with cyl power or axis. More than Likely, one or the other is not exactly like his old Rx, and that messes up people like this (I'm married to one).

    A less likely reason could be that they lenses are on different base curves.

    The missing information, that would be helpful:
    - old Rx (always critical)
    - seeing how the Dr assistant checked glasses
    - does the patient see well through glasses? (feeling "uncomfortable and seeing poorly are very different)
    - base curves

    Good luck. Hope this helps. Don't bother asking for help when you're in the simple camera or mirror units - it's been tooooooo long!
    ...Just ask me...

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    Master OptiBoarder Darryl Meister's Avatar
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    I may not entirely follow the problem, since your first assertion is that they "check out perfectly." After all, if they check out perfectly, you should not have any vertical imbalance. And, since they are single vision lenses, vertical imbalance at near shouldn't be an issue, either. However, it is entirely possible that the doctor's assistant measured the lenses incorrectly, especially given the amount of oblique cylinder involved (which, as the other posters have already pointed out, will induce a significant amount of vertical prism if the PD is off).
    Darryl J. Meister, ABOM

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    Bad address email on file QDO1's Avatar
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    Quote Originally Posted by Spexvet
    It's possible to have prism in a plano lens, vertically and/or horizontally.
    Yes, you can order a plano prism - they would, for example, be plano prisims in a testing trial set.

    Perhaps a look at the design philosophy of XYZ optics in Oakley sunspecs might point you in the right direction for your original question.

    In this senario you will get both horizontal and vertical prism if the centres are set too high or low. The centres might be aligned to each other, when checked on the focimeter, but if they are not alligned for position of wear, the patient will be presented with prism. You will also need to adjust the height to compensate for the frontal angle of the frame etc.

    An inexperienced glazer might induce some prisim in the job, when blocking up the pair of lenses, as stock lenses might have a little prism in them. an experienced glazer would make sure that any prism in the right lens was offset by any prisim in the left lens

    In the example it sounds like the job was ordered on datum, and the DR checked for position of wear

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    Master OptiBoarder lensgrinder's Avatar
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    Greetings,
    We know that with a +3.00 -4.00 @ 120 our full cylinder power is @ 30. So we have a -1.00 @ 30. We have 1^ BU. This is like a resultant prism question. We can resolve this in to vertical and horizontal components, except we do not need the horizontal component. If we move this lens horizontally the lens will move vertically in the 30th meridian. Think of a right triangle. What we are going to figure for is the hypotenuse. Which will be the total amout we will move this lens to induce 1^ of BU prism.
    We know the angle is 30 and since we know the lens has a diopter of prism in the 90th meridian with +2.00 D in the same meridian, how much do we need to move this lens to create that?
    1^=2 X f (Prentice's Rule)
    f=.5cm or 5mm

    Now we know our vertical component is 5mm which is opposite our angle. So to find the hypotenuse we will use sine=Opposite/Hypotenuse

    sin30=5/Hpotenuse
    Hypotenuse=5/0.5
    Hypotenuse=10
    Or 10mm. So it is possible for a lens with no power @ the horizontal meridian to have prism vertically.

    This was figured using the right eye since no information was given as to whether the lens had 0.5^BU in one eye and 0.5^ BD in the other. So I used 1^ BU just in the right.
    Hope this helps.
    Last edited by lensgrinder; 10-09-2005 at 10:23 PM.

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    Quote Originally Posted by Chris Ryser
    I agree with Jo. Just would like to add that often mistakes are made with spheres and cyl powers that have a near zero power at 90 degrees while marking the centers off center with the lensometer.

    Is this marking before or after edging the lens? If there is not power in the 90 or 180, how can it be marked wrong? The oc will be centered in the lensometer regardless. Right?

  9. #9
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    The instructor specifically wants us not to make assumptions at all. There is no prescribed prism in the rx, so the 1^BU has to be a mistake in the lensmeter. I understand that if the OC were to be sitting 5mm above the line of sight then the patient could experience the 1^BU, but according to the instructor this is not what he is looking for. When he resubmitted my answers back to me about the vertical imbalance, he wrote: "First the weakest lens could have been centered and dotted first. Second, the prism dial on the lensmeter could have been in the wrong position or accidently turned while working with the glasses. The lensmeter table could have been moved between lenses. And a forth one (that's the extra credit one)."
    I suppose, from looking at the reasons he gave that he is looking for a "surface" reason. Sorry to be stressing this so much but he is serious about NO assumptions. "Work only with the facts!" he is really getting under my skin with this one.

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    Quote Originally Posted by lensgrinder
    If we move this lens horizontally the lens will move vertically in the 30th meridian.
    I like where you are going with this one. Concerning the question, How can you verify the PD on this rx? Remember there is no power in either lenses 180 meridian. Do you think he could be talking about neutralizing the reflection on the lens to find the OC, by eyeballing it(no lensometer)? Where two reflection s become one? Does an optical center exist on a lens with 0 power in the 180 and 90 meridia, even if there is power elsewhere in the lens????

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    Quote Originally Posted by Darryl Meister
    I may not entirely follow the problem, since your first assertion is that they "check out perfectly." After all, if they check out perfectly, you should not have any vertical imbalance. And, since they are single vision lenses, vertical imbalance at near shouldn't be an issue, either. However, it is entirely possible that the doctor's assistant measured the lenses incorrectly, especially given the amount of oblique cylinder involved (which, as the other posters have already pointed out, will induce a significant amount of vertical prism if the PD is off).

    But the PD is not off and there is no vertical imbalance, the Dr. Assistant checked the glasses incorrectly(0 power at 180 and 0 power in one eye at 90). And when I submitted the idea that the glasses could be sitting 5mm above the line of sight, the instructor blows me down with the fact that I am making assumptions. I can only go off the information given, unfortunately. I have submitted answers concerning adjustments, base curves, vertex distances, pantoscopic tilt, face form, everything that I thought could POSSIBLY be wrong. To this he replies that I am making assumptions.
    He is going to send me to the crazy house if I am not carefull.:angry:

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    Quote Originally Posted by Jo
    The layout could have been wrong - bad box measurements. The person laying the lenses out could have measured straight down from the pupil instead of measuring to the bottom of the frame; in an aviator that can be especially noticable. Also, the person inspecting the lenses could have forgotten to start with the highest power at 90 when looking for vertical prism. A bad box measurement could also create the horizontal prism problems. I've seen folks measure aviators using the PD from the datum instead of the widest measurement across the A and go straight across instead of using the narrowest across the bridge.
    I tried this answer but the instructor was looking for something else. Can you think of anything else?

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    Master OptiBoarder lensgrinder's Avatar
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    Quote Originally Posted by TracyElise
    I like where you are going with this one. Concerning the question, How can you verify the PD on this rx? Remember there is no power in either lenses 180 meridian. Do you think he could be talking about neutralizing the reflection on the lens to find the OC, by eyeballing it(no lensometer)? Where two reflection s become one? Does an optical center exist on a lens with 0 power in the 180 and 90 meridia, even if there is power elsewhere in the lens????
    To verify the PD take the lenses out of the frame and put them on a Box-O-Graph, find geometric center and then move back or to left with the right eye and to the right with the left eye 3mm for each eye (This is the decentration) place a dot at that location and then put the lenses back in the frame and place that dot over the lensometer and spot to make sure you have it in the right location. Then look in the lensometer to see what kind of prism you have. Also sorry for the long winded answer before, but if you take the power in the 30th meridian (-1.00) and use Prentice's rule :
    1^=1.00D X f
    f=1cm or 10mm.
    So again moving this lens 10mm will create 1^ of prism.

    Hope this helps.

  14. #14
    Master OptiBoarder Darryl Meister's Avatar
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    But the PD is not off and there is no vertical imbalance, the Dr. Assistant checked the glasses incorrectly (0 power at 180 and 0 power in one eye at 90).
    Then, like I said, that should be the answer to the problem. ;) Really, anything else would require additional assumptions, which your instructor claims are not needed. For that matter, the fact that the patient is experiencing a problem -- given that the glasses were made correctly -- requires you to make a few assumptions, since this problem provides very little background information.

    I think what the instructor is trying to get you to understand is the nature of prism in an oblique cylinder. While these two lenses have no "power" through the 180 (which is a discussion I'll save for another day), there is still prism and even an optical center.

    The problem is that, because there is virtually no horizontal prism induced through the 180, it is easy to measure the lenses at the wrong location. That is to say, you can slide the lenses to the right or left, which will induce vertical prism, but the optical center will still be centered horizontally in the lensomter / vertometer.

    For instance, if you measure these two lenses at 5 mm out (equivalent to reading the lenses at 74 mm instead of 64), the right lens will show 0.87 D base down and the left lens will show 0.75 D base up, but neither will show any horizontal prism or OC movement. Consequently, you have produced 1.62 D of vertical imbalance with virtually no horizontal prism (that is, the optical centers haven't moved to the left or right in the lensometer / vertometer at all).

    Unfortunately, none of this explains your patient's problem, but I don't know that you necessarily need to in this exercise. I think you just need to say, "They're right, and the assistant is measuring them incorrectly and unwittingly inducing vertical prism. Etcetera."
    Darryl J. Meister, ABOM

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    Idea Thank you all

    Just wanted to say thanks to everyones input. The instructor was looking for the effect the horizontal movement has on the vertical. I didn't even know that existed. It's good to know.
    I love this place!!!:cheers: Thanks for all your help!

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