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Thread: Help with the math!

  1. #1
    ATO Member HarryChiling's Avatar
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    Redhot Jumper Help with the math!

    I am working on something and I have an equation that I am having a hard time solving. I think I might be makign a stupid mistake in the math so I am hopeing someone might catch it as I am having a tough time catching it.

    The variables:

    D = Back Vertex Power
    F = Front Curve
    n = Index of Lens
    q = Back Vertex to Center of Rotation

    F2(n + 2) + F[2/q(n2 - 1) + D(n + 2)] + n[D + (n - 1)/q]2 = 0

    It looks daunting but heres how far I've gotten:

    F2(n + 2) + F[2/q(n2 - 1) + D(n + 2)] + n[D + (n - 1)/q]2 = 0

    A = (n+2)
    B = [2/q(n2 - 1) + D(n + 2)]
    C = n[D + (n - 1)/q]2

    So now I can rewrite the equation to read:

    AF2 + BF + C = 0

    now it's in a simple quadratic form, no I would assume to find the + and - root I would use the quadratci equation:

    F- = (-B - [B2 - 4AC]1/2)/2A
    F+ = (-B + [B2 - 4AC]1/2)/2A

    So now I will get the positive and negative root, which BTW will be the Ostwalt and Wollastan branches of the Tscherning Ellipse. Now here's my delema I have been plugging in numbers left and right and I can't get them to pan out. I will try with an example:

    D = 1.00
    F = ?
    n = 1.5
    q = 25mm = 0.025m

    A = (1.5 + 2) = 3.5
    B = 2/0.025(2.25 - 1) + 1(1.5 + 2)
    B = 80(1.25) + 3.5
    B = 103.5
    C = 1.5(1 + 2(1(1.5 - 1)/0.025) + ((1.5 - 1)2/0.0252)
    C = 1.5(1 + 40 + (0.25/0.000625)
    C = 1.5(1 + 40 + 400) = 1.5(441) = 661.5

    A = 3.5
    B = 103.5
    C = 661.5

    F- = (-103.5 - [103.52 - 4*3.5*661.5]1/2)/2*3.5
    F- = (-103.5 - [10712.25 - 9261]1/2)/7
    F- = (-103.5 - [1451.25]1/2)/7
    F- = (-103.5 - 38.1)/7 = -141.6/7 = -20.2
    F+ = (-103.5 + 38.1)/7 = -65.4/7 = -9.3

    For some reason those numbers don't look right to me, any suggestions?
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    Harry:
    You know that there aren't but about 4 people on Optiboards anywhere near as good a you with numbers.

    Chip

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    Quote Originally Posted by HarryChiling View Post
    I am working on something and I have an equation that I am having a hard time solving. I think I might be makign a stupid mistake in the math so I am hopeing someone might catch it as I am having a tough time catching it.

    The variables:

    D = Back Vertex Power
    F = Front Curve
    n = Index of Lens
    q = Back Vertex to Center of Rotation

    F2(n + 2) + F[2/q(n2 - 1) + D(n + 2)] + n[D + (n - 1)/q]2 = 0

    It looks daunting but heres how far I've gotten:

    F2(n + 2) + F[2/q(n2 - 1) + D(n + 2)] + n[D + (n - 1)/q]2 = 0

    A = (n+2)
    B = [2/q(n2 - 1) + D(n + 2)]
    C = n[D + (n - 1)/q]2

    So now I can rewrite the equation to read:

    AF2 + BF + C = 0

    now it's in a simple quadratic form, no I would assume to find the + and - root I would use the quadratci equation:

    F- = (-B - [B2 - 4AC]1/2)/2A
    F+ = (-B + [B2 - 4AC]1/2)/2A

    So now I will get the positive and negative root, which BTW will be the Ostwalt and Wollastan branches of the Tscherning Ellipse. Now here's my delema I have been plugging in numbers left and right and I can't get them to pan out. I will try with an example:

    D = 1.00
    F = ?
    n = 1.5
    q = 25mm = 0.025m

    A = (1.5 + 2) = 3.5
    B = 2/0.025(2.25 - 1) + 1(1.5 + 2)
    B = 80(1.25) + 3.5
    B = 103.5
    C = 1.5(1 + 2(1(1.5 - 1)/0.025) + ((1.5 - 1)2/0.0252)
    C = 1.5(1 + 40 + (0.25/0.000625)
    C = 1.5(1 + 40 + 400) = 1.5(441) = 661.5

    A = 3.5
    B = 103.5
    C = 661.5

    F- = (-103.5 - [103.52 - 4*3.5*661.5]1/2)/2*3.5
    F- = (-103.5 - [10712.25 - 9261]1/2)/7
    F- = (-103.5 - [1451.25]1/2)/7
    F- = (-103.5 - 38.1)/7 = -141.6/7 = -20.2
    F+ = (-103.5 + 38.1)/7 = -65.4/7 = -9.3

    For some reason those numbers don't look right to me, any suggestions?
    Hi Harry,

    At least your derivation for F seems to be totally OK, I just plugged your result in the top equation and get very small numbers for the result (< 0.03), for both -141.6/7 and -65.4/7.

    However, regarding what it really means, I just checked another publication

    http://www.drdrbill.com/downloads/op...ens_Design.pdf

    On page 6 there is a diagram, where I can read the solutions of about -19 and -8.5 for D = 1, but if I understand correctly, this is for a working distance of 33cm, not infinity(!?) as in your case.

    Hope this helps

    Added: There seems to be no definite standard for the sign of the solution, Iīve found another article by Darryl, where you can quite
    correctly read off the respective solutions +9 and +20 for D=1 in the diagram, in another one they are again negative, but slightly off because calculated for q=27mm.Itīs all here in this thread

    http://www.optiboard.com/forums/showthread.php?t=19865
    Last edited by xiaowei; 03-27-2008 at 12:14 PM.

  4. #4
    ATO Member HarryChiling's Avatar
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    Quote Originally Posted by xiaowei View Post
    Hi Harry,

    At least your derivation for F seems to be totally OK, I just plugged your result in the top equation and get very small numbers for the result (< 0.03), for both -141.6/7 and -65.4/7.

    However, regarding what it really means, I just checked another publication

    http://www.drdrbill.com/downloads/op...ens_Design.pdf

    On page 6 there is a diagram, where I can read the solutions of about -19 and -8.5 for D = 1, but if I understand correctly, this is for a working distance of 33cm, not infinity(!?) as in your case.

    Hope this helps

    Added: There seems to be no definite standard for the sign of the solution, Iīve found another article by Darryl, where you can quite
    correctly read off the respective solutions +9 and +20 for D=1 in the diagram, in another one they are again negative, but slightly off because calculated for q=27mm.Itīs all here in this thread

    http://www.optiboard.com/forums/showthread.php?t=19865
    So what your saying is I'm not crazy. So the sign doesn't matter in the root? Is there any case where the best curve would be a negative front curve and if so then how would you be able to tell if the computations really needed a negative front or not. I am more asking from a programming standpoint since it would be hard to implement a program that dispalys these results without a concrete formula although I could set a limit and then assume that all above the limit would be negative values but somehow I think that this would create problems.
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    Quote Originally Posted by HarryChiling View Post
    So what your saying is I'm not crazy. So the sign doesn't matter in the root? Is there any case where the best curve would be a negative front curve and if so then how would you be able to tell if the computations really needed a negative front or not. I am more asking from a programming standpoint since it would be hard to implement a program that dispalys these results without a concrete formula although I could set a limit and then assume that all above the limit would be negative values but somehow I think that this would create problems.
    NONONO, that&#180;s not what I mean to say. Just checked with my literature and there only seems to be a confusion of terms (as with also some of the papers I cited). Where did you get the initial formula from?

    Compared to my sources,

    Your F is actually B (or whatever), the BACK surface power, hence negative (there are diagrams for Tschernings ellipse that show this and not the front curve, hence the confusion)

    D is clearly not Back Vertex power but simply lens power

    Hence it seems you need to calculate F = D-B, which would give +10.3 and +21.2 in your test case!

    Wacha think?

  6. #6
    ATO Member HarryChiling's Avatar
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    Quote Originally Posted by xiaowei View Post
    NONONO, thatīs not what I mean to say. Just checked with my literature and there only seems to be a confusion of terms (as with also some of the papers I cited). Where did you get the initial formula from?

    Compared to my sources,

    Your F is actually B (or whatever), the BACK surface power, hence negative (there are diagrams for Tschernings ellipse that show this and not the front curve, hence the confusion)

    D is clearly not Back Vertex power but simply lens power

    Hence it seems you need to calculate F = D-B, which would give +10.3 and +21.2 in your test case!

    Wacha think?
    I had thought about that and it doesn't work the ellipse gets outta whack. I am attaching the file I am createing and I got the formula originally from:

    https://green.eyes.arizona.edu/~jsch...es/class10.pdf

    Those documents you cited in the previous posts left the equation ambiguous so I searched for another. The percival I pieced together from the equation in the:

    http://www.drdrbill.com/downloads/op...ens_Design.pdf

    Document, the form of the formula was a little jacked up so it gave me issues but I finally thought I got it right I am just getting wierd numbers from it.
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    Last edited by HarryChiling; 05-17-2008 at 04:43 PM.
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    I honestly have no idea what you're tying to figure out, so I'm not sure what a "correct" a, b and c value would look like, but I had a quick question about your formula B equation. Is it 2/[q(n^2-1) + D(n+2)], or is it (2/q)[(n^2-1) + D(n+2)]? Because, I read it the first way and got a b value much much lower... which makes no sense when putting it in the quadratic formula. Damn. *scribbles over page* ...ok, so I tried it your way, but your division sign is still plaguing me.


    b = 2/0.025(1.25) + 3.5

    You worked that as (2/0.025)(1.25) + 3.5 = 103.5
    I saw it as 2/[(0.025)(1.25)] + 3.5 = 67.5

    Working that further in:

    aF^2 + bF + c
    (3.5)F^2 + (67.5)F + (661.5) = 0

    Err... I really need to do the math before typing. This way gives another negative square. Bleh. Ok.

    Nope! Nothing wrong with your math Harry, once I stopped being stupid, I got what you got. Nuts to this computer math though. Nothing beats scribbling it out on paper! :D
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    ATO Member HarryChiling's Avatar
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    If you look at the Excel sheet the columns for the base curve of the percival lenses are inversed, as an example (+4.00 lens is on a 4base instead of an 8 base, and a -4.00 lens is on an 8 base instead of a 4 base), the plano lens is correct, but on either side the bases flip flop, which leads me to believe that I made a mistake with a sign where the power is concerned, the two documents I cited above in a post have one discrepency bewteen their quadratic equations:

    Doc 1 = AF2 + BF + C = 0
    Doc 2 = AF2 - BF + C = 0

    So I made the Doc 2 into:

    AF2 + (-B)F + C = 0

    I tried the formula the other way around and it doesn't correct the problem. Also when I think about it in the descriminant the B variable is squared so the sign wouldn't mean squat there and in the rest of the quadratic equation the B is added anyway so a difference in sign here just makes the (-) root and the (+) root flip flop. I thought also that maybe the C variable in the percival lens could be where the probelm lies since the power of the lens is in here as well. I can't tell where, but I will be checking that out and post back tommorow.

    Admiral Knight,

    LOL, I had to go to paper as well when I originally ran into problems. I don't think a computer will ever replace pencil and paper. I am glad you finally came to the same conclusion I must say that i did all the same things you did, but you find that if you change the way the formuals written you get an imaginary number in the descriminant i = -11/2.
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    I'll be the first too admit I'm not the best at math. But other that the teeeeeeeny (math skill wise) optics portion when I was in school, I haven't done any hardcore calc since highschool. Quite honestly, I wouldn't know what to do with an imaginary number if it bit me. :)

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    Master OptiBoarder Darryl Meister's Avatar
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    I haven't double-checked your actual calculations, but keep in mind that the form of the equation that you're using is for the back surfaces of the lens not for the front surfaces. You would need to add the power of the lens to arrive at the front surfaces...
    Darryl J. Meister, ABOM

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    ATO Member HarryChiling's Avatar
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    Quote Originally Posted by Darryl Meister View Post
    I haven't double-checked your actual calculations, but keep in mind that the form of the equation that you're using is for the back surfaces of the lens not for the front surfaces. You would need to add the power of the lens to arrive at the front surfaces...
    Then my math must be off, cause that doesn't make sense with the numbers I'm getting.
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    Master OptiBoarder Darryl Meister's Avatar
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    You're either calculating the back curves or flipping the sign of the results; you certainly won't get front curves that are equal to -20.2 D with Tscherning's solutions...
    Darryl J. Meister, ABOM

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    Old Optician to New OD Aarlan's Avatar
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    Quote Originally Posted by AdmiralKnight View Post
    . Quite honestly, I wouldn't know what to do with an imaginary number if it bit me. :)
    bite it back


    AA

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    Master OptiBoarder Darryl Meister's Avatar
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    Or multiply it by another imaginary number to get -1. ;)
    Darryl J. Meister, ABOM

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    Pomposity! Spexvet's Avatar
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    Quote Originally Posted by Darryl Meister View Post
    Or multiply it by another imaginary number to get -1. ;)
    i, i, captain! ;)
    ...Just ask me...

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    Is it bad that I laughed at that? :D

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    Master OptiBoarder Darryl Meister's Avatar
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    Is it bad that I laughed at that?
    There are about ten people in the optical industry who think this kind of stuff is funny, and half of them are posting in this thread...
    :cheers:
    Darryl J. Meister, ABOM

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