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Thread: Induced Prism in a direction I would not expect?

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    Induced Prism in a direction I would not expect?

    This has been puzzling me for a number of years, I'm sure someone smarter than me knows the answer. If I have a lens with the RX +5.00-5.00 x 45 and I move it 10mm, I understand and expect to induce 2.5 diopters of prism in the horizontal direction. What I don't understand is why the OptiCampus Induced Prism calculator tells me that I also induce the same prism in the vertical direction without moving the lens vertically?

    Am I missing something very simple?

    Thanks!!
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    I would guess you are calculating the power at 180 and then using Prentices rule for prism. That does not yield the correct answer except if the axis is 180 or 90.Opti campus answer is the correct answer. You have to calculate the prism created in each principle axis and then cross the prisms for the resultant prism. Using your method with this Rx +2.00 - 4.00 x 45 will tell you the lens is Plano at 180 and you create no prism. That is incorrect you will create vertices prism. This fact is completely foreign to the majority of opticians.

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    Hi Lensman, thank you for your response. To be sure, I assumed the OptiCampus calculation was correct. In years past, I would have picked up the phone and called Darryl. :(

    I've been using this calculator for a number of years for an account that prefers his progressives to have no prism (not even thinning prism) at the fitting cross instead of the PRP. Using the calculator with a value of 4mm in the decentration up box would give the prism needed to give this account what he requested. Typically I would see a vertical prism adjustment close to what I'd expect to see, however, as you indicated, if the axis was not 180 or 90, then I'd see a small amount of horizontal prism. The examples prescriptions that you and I gave above are unusual..and maybe why the majority of opticians never realized this? Maybe I should have posted the question in the general forum. Through the years, I have seen this effect in the lensometer. Lenses with high cylinders are sometimes difficult to spot in a location that eliminates both horizontal and vertical prism...especially if the surfacing process has left a small amount of unwanted prism on the lens.

    I'm still trying to wrap my head around how to calculate the prism created in each principle axis...my head is stuck on Prentices rule for prism that tells me if there the decentration is zero in a given direction, then the prism in that direction is also zero.

    How exactly is the calculator coming up with the answers it does?

    Thanks!!
    Last edited by EdSheridan; 01-16-2021 at 11:30 AM.

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    Take the two parts separately. 5.00 sphere decentered in 10 mm gives you 5D BI prism. Then do the cylinder. Prism created by decentering a cylinder requires a little trig or graph paper. A -5.00 cylinder at 45 in the right eye, decentered in 10 mm will yield 2.50D base out and 2.50D BI. Add all that up and you get 2.5 BI and 2.5 BD.

    You don't really need trig for your example since the axis is at 45. But you would for anything other than 0-30-45-60-90 degrees.

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    Kwill212
    please give a more detailed explanation of your solution. The cylinder on a lens has no power it only represents the difference between the principal meridians in this case +5.00 and 0.00

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    Quote Originally Posted by Lensman11 View Post
    Kwill212
    The cylinder on a lens has no power it only represents the difference between the principal meridians in this case +5.00 and 0.00
    You've lost me. How does the power go from +5.00 to 0.00 is the cylinder doesn't have any power?

    A more detailed response can be found here, in post #22, strangely enough in response to your own comment in that thread.

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    The cylinder represents the difference in power between the two principal meridians. When neutralizing the lens in a vertometer you will not find a reading of -5.00 only 0.00 and +5.00 the powers in the principal meridians. The -5.00 is the difference between +5.00 and 0.00

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    Quote Originally Posted by Lensman11 View Post
    The cylinder represents the difference in power between the two principal meridians. When neutralizing the lens in a vertometer you will not find a reading of -5.00 only 0.00 and +5.00 the powers in the principal meridians. The -5.00 is the difference between +5.00 and 0.00

    You seem to have answered you own question? I'm not sure what else I can add. The cylinder power is the difference between the principal meridians, 5 diopters. In this case it's -5.00 because we all use minus cylinder lenses. If you want to transpose the Rx to plus cyl, it is still 5 diopters and you'll still get the same answer for the induced prism. I'm still baffled by your "the cylinder doesn't have power" comment. Does a plano -3.00 x 90 have cylinder power?

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    Not to be repetitive the cylinder is the value of the difference of the two Primary powers if you want to say it has power that is fine. If it has a power than it has to have a value plus or minus. We all know you can transpose an Rx from plus cyl to minus cyl and the value of the cyl changes to make my point you can’t change the power of the lens or the powers in the principal meridians the sign of the cyl changes depending on the starting point of your reading if it had power you could not change the power from plus to minus.
    The resultant prism in this lens is derived entirely from the primary +5 meridian as the other primary is Plano which induces no prism. Refer back to any text book with ophthalmic lens formulas for a precise lesson on how this works. Luckily today you can use opticampus without understanding what is behind the numbers. When I went to school 50 years ago we did this with a slide rule and quite a few minutes of calculations and a good understanding to solve the problem.

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    My perspective on this discussion. Is Opticampus applying the 10 mm decentration to the power in total listed as sphere (+5.00) and not +2.50 @ the 180 as we assume on a optical cross. Perhaps since axis is 45 the 10mm is split and applied 5mm x +5.00 along the 90 and 180 axis. Both answers yield +2.50 Diopters Prism this would conclude with Base IN and Base Up.
    Attached Thumbnails Attached Thumbnails IMG_5138.jpg  
    Last edited by PRECISIONLAB; 01-18-2021 at 07:45 AM.

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    Quote Originally Posted by EdSheridan View Post
    This has been puzzling me for a number of years, I'm sure someone smarter than me knows the answer. If I have a lens with the RX +5.00-5.00 x 45 and I move it 10mm, I understand and expect to induce 2.5 diopters of prism in the horizontal direction. What I don't understand is why the OptiCampus Induced Prism calculator tells me that I also induce the same prism in the vertical direction without moving the lens vertically?

    Am I missing something very simple?

    Thanks!!
    Click image for larger version. 

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    Hi Ed,

    Lenses with cylinder power also produce a separate prismatic effect through the meridian of the lens containing the cylinder power, 90° away from the cylinder axis. If the cylinder axis is at an oblique angle, moving the lens to the right or left will actually induce a vertical prismatic effect in addition to the horizontal prismatic effect. Furthermore, if one meridian of the lens is plano in power, the lens will not have a true optical center, because the plano‐powered meridian of the lens will have zero prismatic effect. These optical characteristics can sometimes make verifying the prism in lenses with cylinder power challenging.

    http://experiencevelocity.com/static...7b3ecc2612.pdf

    Hope this helps,

    Robert Martellaro
    Roberts Optical Ltd.
    Wauwatosa Wi.
    www.roberts-optical.com
    ~~~~~~~~~~~~~~~~~~
    Science is a way of trying not to fool yourself. - Richard P. Feynman

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    Thank you to everyone who has helped with this. At least I was able to use the math provided by Kwill212 to replicate for myself what I get from the OptiCampus Calculator.

    From a different thread, Kwill212 wrote...

    Vertical prism is induced when horizontally decentering a cylinder. You can calculate it using Prentice's Rule and some Trig. The sphere power is irrelevant for this situation. Find the cylinder power at 90°.
    sin²(135-90)*-4.00D=-2.00D. Then calculate the vertical decentration of the cylinder solving a right triangle. Assume you decenter horizontally 5mm. Tan(135)=X/5) which means X=-5 where X = vertical decentration. Then simply use Prentice's Rule to determine the induced prism. .5cm*2.00D=1D vertical prism. Obviously you would need to correctly input the signs to get the correct direction up or down, or draw it out. Your example doesn't really require trig if you know the basics, but this same method works for cylinder at any axis and power.

    For example.

    -5.00 -3.25 x 062 decentered 6mm horizontally. Cylinder power at 90° is
    sin²(62-90)*-3.25D=-0.72D. Vertical decentration is Tan(62)=X/6mm. X=11.28mm. Prentice's Rule is 1.128cm*.72D=.81D vertical prism.

    What I'm really having a hard time wrapping my head around is where the vertical decentration of 5mm and 11.28mm in the examples above come from? In the second example, we only moved the lens horizontally 6mm, yet we use 11.28mm of vertical decentration in prentice's rule for to calculate vertical prism. I know that is correct, but it's very hard for me to visualize.

    Sometimes I start to see it when I think of a +5.00-5.00 x 45 lens as 2 different lenses placed on top of one another. One +5.00 sph and one plano - 5.00 x 45. As the -5.00 cylinder is moved horizontally at 45 degrees, it pushes the center down and out at equal rates.

    I'm going to ponder some more. Thanks for all the help!!

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    Quote Originally Posted by Lensman11 View Post
    Not to be repetitive the cylinder is the value of the difference of the two Primary powers if you want to say it has power that is fine. If it has a power than it has to have a value plus or minus. We all know you can transpose an Rx from plus cyl to minus cyl and the value of the cyl changes to make my point you can’t change the power of the lens or the powers in the principal meridians the sign of the cyl changes depending on the starting point of your reading if it had power you could not change the power from plus to minus.
    The resultant prism in this lens is derived entirely from the primary +5 meridian as the other primary is Plano which induces no prism. Refer back to any text book with ophthalmic lens formulas for a precise lesson on how this works. Luckily today you can use opticampus without understanding what is behind the numbers. When I went to school 50 years ago we did this with a slide rule and quite a few minutes of calculations and a good understanding to solve the problem.
    Lets just agree to leave the "I had to learn with a slide rule after I walked uphill through 3 feet of snow backwards to school" BS at the door OK?

    I think the point you are trying to get across is that the cylinder doesn't have an inherent + or - attached to it? It depends on your reference point. Whether or not the Rx is written in plus or minus cylinder. A person doesn't have plus or minus cylinder, a prescription doesn't have plus or minus cylinder. That's all fine and dandy and correct. But, after you make the physical lens, it's going to be minus cylinder power. Because as far as I am aware no one has made plus cylinder lenses for decades. If you want to call it "amount of cylinder" instead of "cylinder power", go ahead, it's the same thing. BTW the phrase "cylinder power(s)" appears 159 times in System for Ophthalmic Dispensing 3rd edition.

    Now I will say that I don't have any formal training, just learning the practical aspect from old timers. All the technical and math aspects I got from text books, the internet, and optiboard. I'm always trying to learn more. So if you would like to give a detailed answer as to how to come to the correct answer to this type of calculation, a "precise lesson" as it were, I'd be more than happy to listen. The way it stands I don't find this to even be an optical question, it's just fairly straight forward math. I find the way I do the calculation is fairly simple, works for any power of spherocylinder lens at any axis, and gives the correct answer. If there is an easier way I'm all ears.

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    Your request was posted directly above your last post

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    Quote Originally Posted by Lensman11 View Post
    Your request was posted directly above your last post
    Am I taking crazy pills? Are you talking about comment #12?

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    Quote Originally Posted by EdSheridan View Post
    Thank you to everyone who has helped with this. At least I was able to use the math provided by Kwill212 to replicate for myself what I get from the OptiCampus Calculator.

    From a different thread, Kwill212 wrote...

    Vertical prism is induced when horizontally decentering a cylinder. You can calculate it using Prentice's Rule and some Trig. The sphere power is irrelevant for this situation. Find the cylinder power at 90°.
    sin²(135-90)*-4.00D=-2.00D. Then calculate the vertical decentration of the cylinder solving a right triangle. Assume you decenter horizontally 5mm. Tan(135)=X/5) which means X=-5 where X = vertical decentration. Then simply use Prentice's Rule to determine the induced prism. .5cm*2.00D=1D vertical prism. Obviously you would need to correctly input the signs to get the correct direction up or down, or draw it out. Your example doesn't really require trig if you know the basics, but this same method works for cylinder at any axis and power.

    For example.

    -5.00 -3.25 x 062 decentered 6mm horizontally. Cylinder power at 90° is
    sin²(62-90)*-3.25D=-0.72D. Vertical decentration is Tan(62)=X/6mm. X=11.28mm. Prentice's Rule is 1.128cm*.72D=.81D vertical prism.

    What I'm really having a hard time wrapping my head around is where the vertical decentration of 5mm and 11.28mm in the examples above come from? In the second example, we only moved the lens horizontally 6mm, yet we use 11.28mm of vertical decentration in prentice's rule for to calculate vertical prism. I know that is correct, but it's very hard for me to visualize.

    Sometimes I start to see it when I think of a +5.00-5.00 x 45 lens as 2 different lenses placed on top of one another. One +5.00 sph and one plano - 5.00 x 45. As the -5.00 cylinder is moved horizontally at 45 degrees, it pushes the center down and out at equal rates.

    I'm going to ponder some more. Thanks for all the help!!
    Here are some pictures that might help you visualize it better. 2 are 3d renderings to help with the visualization of what is happening, and 2 are just 2d to look at the math. I made these in plano plus cylinder form to simplify it for me. I threw one in at the end in minus cyl too if that helps. This is just the cylinder portion of the equation. You still need to add these prism values to the prism created by decentering the sphere power horizontally to get the correct total horizontal prism for the whole lens. Obviously the + or - signs are important when adding the prism together to get the correct amount and direction. I prefer to just visualize the lens for that part though.

    As I said I am just flying by the seat of my pants here from text books and math. According to this thread(be warned its a rabbit hole of higher level math) the sin² equation is an approximation but it suffices for this particular issue.


    Click image for larger version. 

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    Wow...Thank you for taking the time to do that for me Kwill!

    I am only able to see the first picture (the Optiboard says the rest are an invalid format), but I already think I finally already get it from the first picture. I think the plus cylinder example makes it easier for me to visualize :) This is exactly what I was looking for.

    I ground a plano -5 yesterday to help myself see it. I put it in the lensometer at axis 45 and moved it horizontally. It did what I expected...prism out and down. For a moment I thought I understood.....and I lost it!!

    I've been doing this for 40 years, and it became so counter-intuitive to me that my brain was stuck. It's like my own personal Monty Hall paradox. ;)

    Thanks again!

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    No problem. Hopefully I can get the pictures to work on this try.
    Attached Thumbnails Attached Thumbnails iso cylinder.jpg   top cylinder.jpg   cylinder@45.jpg   cylinder@62.jpg   minus cylinder.jpg  


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    They worked! You were correct in your first response when you said to break it into 2 different problems. It's so much easier to see when you look at it as a cylinder and a sphere. The plus cylinder helped even more. The torus shape is what was blowing my mind.

    I looked through my database of all the 5 diopter cylinder jobs we did in 2020 and it's interesting that the axis is usually at or very close to 180 or 90. I bet it's a physiological thing with the way an eye can be deformed. It's probably why focus is primarily given to the horizontal imbalance created by an improper PD, but no one ever talks about the vertical imbalance it creates. With an axis close to 180 or 90, the vertical prism created is minimal.

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