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  1. #101
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    Quote Originally Posted by ml43 View Post
    Sorry for the confusion, when you said franklin bofocal, I assumed you meant executive bifocal(old habits die hard).


    you don't need to surface anything to prove the math.

    Easiest way is to take two FSV lenses with opposing powers(i.e. +1.00 ds and -1.00 ds)

    Put them back to back, total power of plano.
    If you moved them together in a lensmeter there should be no prism induced.
    But if you move one separately from another, it will behave like a +/- 1.00 DS in terms of prism
    Ok that makes sense, I will have to play with that on Monday, so for arguments sake I will grab a -2.00, plano, +2.00, and for kicks and giggles a +4.00 and use these for the distance Rx. I will also grab a +2.50 for my super imposed add and in order to get 2 diopters of base in prism I will need to move the +2.50 lens 8mm for each example?

  2. #102
    Master OptiBoarder optical24/7's Avatar
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    Ok Wes, I had an ah ha moment. I think this explains my -2/+2 question better to understand the mechanics of what's going on:

    To decenter the D OC out from the seg, you'd have to grind B in prism. The resulting B in would show up in the seg. Do I have it now?

  3. #103
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    Quote Originally Posted by optical24/7 View Post
    Ok Wes, I had an ah ha moment. I think this explains my -2/+2 question better to understand the mechanics of what's going on:

    To decenter the D OC out from the seg, you'd have to grind B in prism. The resulting B in would show up in the seg. Do I have it now?
    Sorry, no. The distance OC would have zero prism in this case, yet prism is ground into the carrier to move THE OC away from the bifocal. But from our perspective, the prism is created by sliding/decentering the seg over, (while keeping the distance OC in the same location) so that rather than looking through the seg's OC as the eye dropped down, you would be looking through the seg 5 mm away from the seg OC, thus creating prism at near only.
    Remember, the eye isn't looking through the middle of the seg where the OC is, in this case. It's looking through the seg 5mm out from the OC of the decentered seg.
    Last edited by Wes; 11-05-2015 at 09:56 PM.
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  4. #104
    Master OptiBoarder optical24/7's Avatar
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    Yes, no prism at the D OC, but when grinding the above Rx for a say 66/54 pd wouldn't you grind B in to move the D OC out so you could decenter the seg in to 54?

  5. #105
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    Quote Originally Posted by optical24/7 View Post
    Yes, no prism at the D OC, but when grinding the above Rx for a say 66/54 pd wouldn't you grind B in to move the D OC out so you could decenter the seg in to 54?
    I suppose you may have to, depending on the blank, from a manufacturing perspective. But considering that most ft blanks have the seg inset a few mm (5 is common) and dropped a few mm (again, 5 is common), perhaps not. It depends on your layout methodology. Do you block on center? Do you block on seg? I think this train of thought is as likely to add confusion to the issue as it is to add clarification. Do we want to discuss the question at hand, or lens surfacing in general?
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  6. #106
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    Quote Originally Posted by optical24/7 View Post
    Yes, no prism at the D OC, but when grinding the above Rx for a say 66/54 pd wouldn't you grind B in to move the D OC out so you could decenter the seg in to 54?
    you're decentering the seg to avoid prism at the distance while achieving decentered prism at the near/seg.

    easiest way to model this is like I stated about with the two lenses with apposing powers.

    if you want to get really graphic.

    take a -2.00 DS FSV lens,
    then edge a +2.00 DS to say 35mm.

    place the +2.00 lens on the -2.00 then read the power, now move the +2.00 lens around while keeping the -2.00 stationary.

    you'll notice a prism effect equal to +2.00, even though your lensometer reading is technically Plano, and you aren't grinding any prism

  7. #107
    Master OptiBoarder optical24/7's Avatar
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    Quote Originally Posted by Wes View Post
    I suppose you may have to, depending on the blank, from a manufacturing perspective. But considering that most ft blanks have the seg inset a few mm (5 is common) and dropped a few mm (again, 5 is common), perhaps not. It depends on your layout methodology. Do you block on center? Do you block on seg? I think this train of thought is as likely to add confusion to the issue as it is to add clarification. Do we want to discuss the question at hand, or lens surfacing in general?
    Yes, in my lab days, you surfaced on geo center, you used prism rings to move the D OC to a position you wanted in relation to the seg OC. On minus lenses, usually base out to move the OC in from geo center. With a D OC going out instead of in from geo center, I'd grind base in to move the OC out. Am I making sense?

  8. #108
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    Quote Originally Posted by optical24/7 View Post
    Yes, in my lab days, you surfaced on geo center, you used prism rings to move the D OC to a position you wanted in relation to the seg OC. On minus lenses, usually base out to move the OC in from geo center. With a D OC going out instead of in from geo center, I'd grind base in to move the OC out. Am I making sense?
    makes perfect sense,

    now translate that "prism" into decentration and the problem should make sense

  9. #109
    Master OptiBoarder MakeOptics's Avatar
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    Tests deal with theory. All the equations used in this thread so far have been first order approximations without taking any additional measurements into account. The question even specifies near only. Ifvany tester puts this much effort into any single question they would fail by running out of time.

    Good luck.
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  10. #110
    Master OptiBoarder optical24/7's Avatar
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    When I thought of it from a manufacturing angle (surfacing it). It made much more sense.

  11. #111
    Master OptiBoarder MakeOptics's Avatar
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    There are a lot of shortcuts in manufacturing. Software in this day and age takes the complexity out of the process and adds additional layers of accuracy. As a former lab person, a lot of what is learned is rule of thumb, some in the lab will start looking into the why's that make the answers what they are. This quest for knowledge is a good thing, with schooling and theory it works the other way around you learn the fundamentals as building blocks and build on them until the picture becomes more clear. Sure we can add accuracy to this equation and even get it to the point of near flawless, but the question still remains simple and the answer is only graded one way. The discussion is nice, but the question and answer remain brief with only 4 choices. I liked the test Darryl put together and enjoyed taking it. I am glad to see others have as well, I think this thread will live on and constantly have additional questions and explanations as to how the questions can be made better, stronger. I enjoy reading.
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  12. #112
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    I am enjoying it as well, as I do not want to be "right" but to know "right"?! I understand completely the idea of two separate lenses and the image being bent/displaced by the add of the lens in the front and then passing the second lens through the optical center so no deviation. But looking at it from a FT-35 standpoint that is not really the case because I am not pulling the bifocal off the lens and replacing it. Would it be safe to say the optical center of the distance RX is being pushed further out in order to be able to decenter the lens said amount in order to gain said near prism? And if that be the case then I'm not sure how one wouldn't need to take into consideration the total power? Anyway I hope it's safe to say "C" will be the answer on the test even if I do not agree with it nor have been able to recreate it in an actual pair of glasses. Though I have a pair being made from an outside lab to take myself out of the equation. I am as hard headed as they come and sometimes need a sludge hammer to get it to sink in so thank you all in advance for your patience with me. :)

  13. #113
    Master OptiBoarder MakeOptics's Avatar
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    Quote Originally Posted by King.Matthew View Post
    I am enjoying it as well, as I do not want to be "right" but to know "right"?! I understand completely the idea of two separate lenses and the image being bent/displaced by the add of the lens in the front and then passing the second lens through the optical center so no deviation. But looking at it from a FT-35 standpoint that is not really the case because I am not pulling the bifocal off the lens and replacing it. Would it be safe to say the optical center of the distance RX is being pushed further out in order to be able to decenter the lens said amount in order to gain said near prism? And if that be the case then I'm not sure how one wouldn't need to take into consideration the total power? Anyway I hope it's safe to say "C" will be the answer on the test even if I do not agree with it nor have been able to recreate it in an actual pair of glasses. Though I have a pair being made from an outside lab to take myself out of the equation. I am as hard headed as they come and sometimes need a sludge hammer to get it to sink in so thank you all in advance for your patience with me. :)
    You're making the assumption that the near PD specified in the question doesn't take the power into account. Since the question supplies a near PD the assumption should be the other way around, we should assume that this near PD is accurate for the same scenario given 0 prism, so now the question is a bit simplified. With your assumption we are throwing out the near PD and working the whole scenario from scratch. Again theory vs reality are two separate beasts. In reality most opticians I have seen even the ones that say they don't just subtract 3mm from the binocular distance PD and supply that as the near PD.
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  14. #114
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    Quote Originally Posted by MakeOptics View Post
    You're making the assumption that the near PD specified in the question doesn't take the power into account. Since the question supplies a near PD the assumption should be the other way around, we should assume that this near PD is accurate for the same scenario given 0 prism, so now the question is a bit simplified. With your assumption we are throwing out the near PD and working the whole scenario from scratch. Again theory vs reality are two separate beasts. In reality most opticians I have seen even the ones that say they don't just subtract 3mm from the binocular distance PD and supply that as the near PD.
    Now were talking about a whole other ball game, taking the question as it is none of the answer's seem correct as it can be done so D is not right but 54 will induce too much prism to be right as well. But like I said I will post what comes tomorrow.

  15. #115
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    Quote Originally Posted by King.Matthew View Post
    Would it be safe to say the optical center of the distance RX is being pushed further out in order to be able to decenter the lens said amount in order to gain said near prism? And if that be the case then I'm not sure how one wouldn't need to take into consideration the total power?
    You have a degree in engineering, right?

    Let's do this with mostly somewhat basic mathematics.

    Given:
    - Base curve of the distance of the bifocal is spherical
    - Base curve of the seg is spherical
    - Rear curve of the the lens is spherical

    Let's define what a bifocal is:
    A smaller lens fused to a larger lens where the (distance between the two lenses) = n, where n is the limit as x approaches zero(0).
    if and only if the front curve of the base lens(carrier) matches the rear curve of the top lens(seg)


    Since we are only dealing with horizontal prism, we can take half of a horizontal cross section by dividing the entire lens into little slices(n), as n approaches infinity.

    So we are left with two very thin prisms that for all intensive purposes are the exact height of a ray of light and symmetrical at both ends in regards to the vertical.
    Further more, we can model(test) these prisms using any vertical height we want, because we are only concerned with horizontal prism and one ray of light.

    Now, as the apex of the rear prism(OC of the distance) moves towards or away from the base of the front prism(geometric center of the seg),
    it displaces(causes a prismatic effect) a ray of light at a rate that is equal to the curve difference between the front curve of the Seg, and the base curve of the carrier.
    Which is equal to the power of the Seg.

    In other words, moving the OC of the rear curve, or moving the inset of the Seg has the same effect.
    Further more, this has the exact same effect as moving the Seg itself on the carrier lens.


    The only reason you would use total power(front curve of the seg, minus the rear curve of the carrier),
    is if you are moving both the seg and the distance oc at the same rate in opposite directions, and you are measuring from the origin to one spot.
    Not the actual difference between the two.

    i.e. if you take a +2.00 DS, mounted on a +2.00DS. If you only move one lens, you only need to calculated prism for +2.00D, not +4.00D.
    You would only use +4.00D if you are moving both lenses together


    edit:

    would it make more sense to say you are decentering 2D BI of prism in the near, then "grinding" 1D of base out prism?
    Last edited by ml43; 07-27-2015 at 11:27 PM.

  16. #116
    Master OptiBoarder optical24/7's Avatar
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    Quote Originally Posted by King.Matthew View Post
    Now were talking about a whole other ball game, taking the question as it is none of the answer's seem correct as it can be done so D is not right but 54 will induce too much prism to be right as well. But like I said I will post what comes tomorrow.
    Matt, I think your making the same mistake I did...Not accounting for the canceling prismatic effect in the near due to the prismatic effect of decentering the distance. Lets surface the test question using prism rings instead of blocking to move the D OC.

    +2.00/+2.50 add; PD 66/54 intended final OC placements. If you block the lens with the usual 2mm inset, you'd need to move the D OC out another 4mm. To move the OC out 4mm from that point, you'd use a .80 prism ring B out. When you inset the seg 4mm more than the N PD you have 1.8 B in (+4.50 x .4 cm), BUT you HAVE to account for the .80 out you ground to move the D oc out those 4mm... 1.8 in + .80 out = 1 B in. at near in the seg

    When I thought about it this way, it makes perfect sense. The same way it works with my -2.00/+2.00add scenario. If I wanted the same 1 D base in near only I would be grinding 1 B in at the D oc to move it out, the residual prism I used in the distance would show up in the seg (plano power) as 1 base in.

    So once again, the prism you created to move the D oc shows up in the near and cancels the prism induced by moving the seg in extra mm's.

  17. #117
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    exactly,

    when you finish decenter, you are moving both the front and rear curves of the lens. Thus, you calculate using the total power.

    when you surface decenter you are only changing the rear curve relative to the front(or vise versa). in a spherical SV, this just moves the OC(creates prism relative to the GC).

    in a bifocal, this is the same as moving just the seg, given you are placing the new OC in front of the pupil. therefore you calculate induced prism using only the seg power because that's technically all that moved relative to the rear curve.

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    Eating Crow I am!!!!! First please let me say thank you for your patience!! I had my light bulb moment after receiving said job and noticed I was wrong. So I was able to continue down the train of thought past where I was stopping and realize that ultimately we are just moving one of the two lenses, I could not get past that as I did not see them as two lenses, I do now, we ultimately are moving the back lens if you will by pushing the OC out and thus theoretically are decentering the seg by itself by doing so and thus only need to take the seg power into consideration.
    Again thank you all for your input and patience!!

  19. #119
    Master OptiBoarder MakeOptics's Avatar
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    Knowledge is best approached like a dog on a bone.
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  20. #120
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    Pretty tasty bone I might add?!!

  21. #121
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    I'm glad that's all worked out. Welcome to Optiboard, Matthew King.
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  22. #122
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    Quote Originally Posted by Wes View Post
    I'm glad that's all worked out. Welcome to Optiboard, Matthew King.
    Thank you, glad to be be here Wes.
    Last edited by King.Matthew; 07-30-2015 at 06:13 PM.

  23. #123
    Master OptiBoarder MakeOptics's Avatar
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    Quote Originally Posted by King.Matthew View Post
    Pretty tasty bone I might add?!!
    Yum ; ) Welcome and great questions to cut your teeth on. I know many don't like older posts rekindled but your questions are good ones that sparked a lot of thought.
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    I love this forum, but often get frustrated with the lack of substantive posts. It is always filled with folks wondering why their ABO or NCLE grade has not come in yet, of I xyz frames come in blue. Both important topics, but this is the kind of dialogue I enjoy. Best thread we have experienced here in some time, and I was pleased to read it. Welcome, Matthew. There are some very bright folks here, as you have seen already.

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    Ans for Ques 17 is a) 71mm. Reason: MBS = (frame PD/geometric center distance - person PD) + ED. mbs = ((54 +16) - 64) + 64. mbs = (70-64) + 64. mbs = 6 + 64 = 70. 71mm is within 1mm tolerance

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