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Thread: Cylinder Power in the 180-degree meridian

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    Cylinder Power in the 180-degree meridian

    Apparently, the formula "D=Dcyl sin^2 Ax" isn't the most accurate for finding the power of the cylinder in the 180-degree meridian. Is there a more accurate formula that can be used, without getting too technical?

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    Master OptiBoarder Darryl Meister's Avatar
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    Is there a more accurate formula that can be used, without getting too technical?


    It depends on what your application is. If you are just calculating thickness, your formula will suffice. If you are wanting to calculate prism at off-center points, you will have to get more technical.
    Darryl J. Meister, ABOM

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    ATO Member HarryChiling's Avatar
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    Will the equation gain more accuracy if measured in radians?
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    Quote Originally Posted by Darryl Meister

    It depends on what your application is. If you are just calculating thickness, your formula will suffice. If you are wanting to calculate prism at off-center points, you will have to get more technical.
    My goal is to be able to figure out everything on a workticket without the use of our computer software. Right now I'm using a graphing calculator and hope to upgrade to an IPac. Hopefully this will allow me to have access to an Excel spreadsheet (that I can carry around) where I can combine all of the formula's into one organized spreadsheet. Just enter the Rx and all of the information that is needed to process a lens is calculated. This is the easiest way for me to learn and I have found it to be quite fun, but challenging. It seems that all of the optical calculators I have found come up with slightly different values. This is frustrating, but is it because of my question above, that there are different formulas to get the final result? On my question, it appears that by using the formula I mentioned for calculating decentration, then it loses accuracy and another formula needs to be used.

    I don't know if you remember me calling you abut 5 years ago, when you helped me with the equation for figuring the tool needed. Somehow I erased the equation and now have renewed interest, since figuring out how to get it back in the calculator. Thanks for your help.
    Last edited by Bob Price; 01-11-2006 at 10:49 AM.

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    Master OptiBoarder Darryl Meister's Avatar
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    Quote Originally Posted by Harry
    Will the equation gain more accuracy if measured in radians?
    If you mean using radians for Ax, you will actually get the same answer either way...?

    Quote Originally Posted by Bob
    My goal is to be able to figure out everything on a workticket without the use of our computer software.
    Then you will definitely need a more "technical" approach. Calculating prism accurately requires a couple of simultaneous equations that use quite a bit of trigonometry. Fortunately, you can just copy them from a textbook or something, but you would still need to verify the sign convention of your input and output and that sort of thing.

    Quote Originally Posted by Bob
    This is frustrating, but is it because of my question above, that there are different formulas to get the final result?
    Yes, but if they're correct, you'll get the same answer either way.
    Darryl J. Meister, ABOM

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    ATO Member HarryChiling's Avatar
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    I am not sure how else you could go about finding the power in that meridian without using the the

    D=Dsph+Dcyl*sin^2(180-axis)

    What other method could you use to more accurately get this power? And where does this equation lose it's accuracy?
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    Master OptiBoarder Darryl Meister's Avatar
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    Quote Originally Posted by Harry
    What other method could you use to more accurately get this power? And where does this equation lose it's accuracy?
    It's not so much the power he's worried about, but rather calculating the prism at points away from the optical center (or, conversely, calculating how to induce prism in order decenter the optical center).

    Also, the sine-squared formula just approximates the curvature of the surface through a particular meridian. It doesn't actually represent the "power," since a sphero-cylindrical lens technically has no power in meridians other than the principal meridians (the rays of light refracted in these other meridians are skew rays, and do not actually intersect at a focus). Also, recall from our earlier discussions of Keating's Dioptric Power matrix that the power of a cylinder has both a curvital component (i.e., the sine-squared component) and a torsional component.
    Darryl J. Meister, ABOM

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    [QUOTE=Darryl Meister]Then you will definitely need a more "technical" approach. Calculating prism accurately requires a couple of simultaneous equations that use quite a bit of trigonometry. Fortunately, you can just copy them from a textbook or something, but you would still need to verify the sign convention of your input and output and that sort of thing.


    Would you please lead me in the right direction for the equations. I might already have them under my nose and will look for them tonight. The books I am working with are "Understanding Lens Surfacing" and "Introduction to Ophthalmic Optics". Thanks again for your input.

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    Master OptiBoarder Darryl Meister's Avatar
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    Would you please lead me in the right direction for the equations. I might already have them under my nose and will look for them tonight.
    Just about any book on ophthalmic or geometrical optics should have them. Clinical Optics, Principles of Ophthalmic Optics, Optics of Ophthalmic Lenses, Geometrical, Physical, & Visual Optics, and so on.
    Darryl J. Meister, ABOM

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    Master OptiBoarder Darryl Meister's Avatar
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    For instance, you can find these equations to calculate the horizontal (Ph) and vertical (Pv) prism effects at a horizontal distance x and a vertical distance y from the optical center (compliments of Clinical Optics):

    Ph = y * Fc * sin A * cos A + x (Fs + Fc * sin^2 A)
    Pv = y (Fs + Fc * cos^2 A) + x * Fc * sin A * cos A

    With some consideration, and proper adherence to a sign convention, these equations can also be used to solve the for the horizontal prism and vertical prism necessary to move the optical center to a desired horizontal or vertical location.
    Darryl J. Meister, ABOM

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    ATO Member HarryChiling's Avatar
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    So let me see if this is correct if I would only want to move a lens horizontally then the equation would be

    Ph = y * Fc * sin A * cos A + x(Fs + Fc * sin^2 A)
    Ph = "0" * Fc * sin A * cos A + x(Fs + Fc * sin^2 A)
    Ph = x(Fs + Fc * sin^2 A)

    And similar would be used for the vertical prism.

    Pv = y(Fs + Fc * cos^2 A)

    So if this is true then what we were missing before was the decentration in the formula. Is that correct, and what measure does the x,y variable use mm or m?

    -or-

    Is the x,y component using a cartesian co-ordinate, where the Optical Center of the lens is equivalent to (0,0) or the center of the grid?
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    Master OptiBoarder Darryl Meister's Avatar
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    The variable x would represent the horizontal decentration (in cm, if I recall correctly) and y would represent the vertical decentration (again, in cm).

    The basic sine-squared formula is a just an approximation of curvature, not prism. It will give you a good ballpark estimate of the prism along a particular meridian, but it fails to compute the prism that occurs perpendicularly to that meridian. For instance, consider the prescription +1.00 -2.00 x 045. Decentering this lens horizontally will actually introduce a significant vertical prism component.
    Darryl J. Meister, ABOM

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    Quote Originally Posted by Darryl Meister
    It will give you a good ballpark estimate of the prism along a particular meridian, but it fails to compute the prism that occurs perpendicularly to that meridian. For instance, consider the prescription +1.00 -2.00 x 045. Decentering this lens horizontally will actually introduce a significant vertical prism component.

    Then going back to my opening question, is there a workable formula that will calculate this induced prism caused by an oblique axis? I would think that it isn't very criticle when you are blocking SV lenses using on center blocking, unless you have a large eye size and cut out is a problem. However,for multifocals using on center blocking, this must be more criticle when you need to move the OC horizontally and vertically to a desired location.

    Going back to another question I have, I still don't know how to figure the prism axis needed for grinding slab off prism in the surface room (on center blocking). The whole goal is to get the prism in the exact location to achieve a strait line to your segment. This might be an easy calculation, but I must admit, I am still in the learning stages and have a lot of questions. Thanks.

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    Master OptiBoarder Darryl Meister's Avatar
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    Then going back to my opening question, is there a workable formula that will calculate this induced prism caused by an oblique axis?
    Yes, I posted them earlier in this thread.

    Regarding your slab-off question, you don't position the slab line using prism. It is controlled by lens thickness.
    Darryl J. Meister, ABOM

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    Sorry to be so persistent, but I might not be asking the right questions. I think I need to be asking how to compensate for the prism angle, when trying to counter the OC moving vertically.

    Here is a right lens example.

    Poly Sv
    -4.75 -3.75 Axis60
    4.5 mm Decentration

    x(Fs+Fc*Sin(Axis)^2)
    .45(-4.75+-3.75*Sin(60)^2)=3.40 (Is this correct?)

    Now, I wouldn't just grind 3.4 diopters @ 180, because the OC will move vertically. So, to keep the OC from moving perpendicular to the 180, I would have to alter the prism axis (and possibly the prism, I'm not sure). Is this the right way of thinking, and if so, how would I calulate this?

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    Master OptiBoarder lensgrinder's Avatar
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    You would grind 3.4 D of prism in the horiziontal meridian to move the OC 4.5mm, because that is where you want to move the OC. In this case you want the base of the prism @ 180, because it is a minus lens. Of course it is SV so it really does not matter. If you ground the lens at an oblique axis it would move the OC in the horizontal and vertical meridian. The question does arise, do you really want to grind prism with only 4.5mm of decentration? That would depend on what blank size you need.

    Hope this helps.

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    Quote Originally Posted by lensgrinder
    The question does arise, do you really want to grind prism with only 4.5mm of decentration? That would depend on what blank size you need.

    Hope this helps.
    I realize that for SV this isn't extremely important, but would it be for Bi's and Tri's? When I try to duplicate our workticket with these formulas - SQRT(V^2+H^2)=Prism and Tan(V/H)=Axis - I don't get the same results. When your trying to move the OC to a desired location, there should be only one right answer. This is why I'm so persistent on trying to figure out what affect an oblique axis has when doing the calculations. I'm probably way off base and just didn't do the figuring correctly, using the formulas mentioned above.

    Also, does the index of refraction have any affect on figuring for prism? Or, is it always Power*Decentration=Prism?

    Thanks for all of the input.

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    Master OptiBoarder lensgrinder's Avatar
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    Yes refractive index does have an effect on prism in lower powers. I would recommend that you read Don Whitney's paper on prism found in the file download area. The name is Prentice's Rule - It's Applications and Limitations by Don Whitney

    Could you give an example of what the software computes versus what you get, using a bifocal?


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    Master OptiBoarder Darryl Meister's Avatar
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    Grinding base in or out prism, only, when trying to decenter the optical center of an oblique axis cylinder will move the optical center up or down, as well.
    Darryl J. Meister, ABOM

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    Quote Originally Posted by lensgrinder
    Yes refractive index does have an effect on prism in lower powers. I would recommend that you read Don Whitney's paper on prism found in the file download area. The name is Prentice's Rule - It's Applications and Limitations by Don Whitney.
    Thanks for leading me to this article. I'm not having much luck yet, but will continue to work on it.

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    RETIRED JRS's Avatar
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    Yes, the prism has a tendency to follow (or track) along the cylinder axis. So to get 2 diopters in on a lens with a cylinder axis of 45, would probably need a prism axis of about -174 (354) degrees. Obviously eye dependent, etc. I got the math around here somewhere Bob. I'll try to dig it up for you. And I tried to leave you a bit of info on EXCEL functions in your other post.
    J. R. Smith


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    ATO Member HarryChiling's Avatar
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    Their is an article that would be relevant to your question


    Analysis of Clinical Approximation in Applying Prentice's Rule to Decentration of Spherocylinder Lenses - Ophthalmic Physiological Optics 1984;4(3) pages 265-273

    It should have relevant information, if I can get the article I will let you know and send you off a copy.
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    Rising Star Bezza's Avatar
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    Jalie's Principals of Ophthalmic Lenses covers this topic very well, theres trigonometrical solutions and some neat graphical constructions that you can use also. I had the pleasure of spending a week with Mo Jalie on a revision course for my SMC(tech), what that guy doesnt know about ophthalmic lenses isnt worth knowing

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    Master OptiBoarder Darryl Meister's Avatar
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    I agree.
    Darryl J. Meister, ABOM

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    ATO Member HarryChiling's Avatar
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    It can be done by using prentice's rule in the form of a power matrix. This will account for the torsional component of the spherocylindrical lens.
    for the example -4.75 -3.75 x 060 with a decentration of 4.5mm

    The horizontal prism is 3.4 and the vertical prism is 0.73

    It is a little complicated to explain but if you pm me I will send you a scan of the equations that I used or if you look in Geometric, Physical, and Visual Optics by Keating you will find it.
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