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Thread: calculation for optotypes

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    calculation for optotypes

    Does anyone now how to calculate different size from some optotype, conversions (Snellen to logMAR…) …?

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    Thanks for links, but I need formulas...

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    First it's easier to use the decimal acuity when converting:

    20/10 = 2.00
    20/12 = 1.67
    20/16 = 1.25
    20/20 = 1.00
    20/25 = 0.80
    20/30 = 0.67
    20/40 = 0.50
    20/50 = 0.40
    20/60 = 0.33
    20/80 = 0.25
    20/100 = 0.20
    20/120 = 0.17
    20/160 = 0.13
    20/200 = 0.10

    Now since 6m/6m (20ft/20ft) is a letter that subtends 5 minutes of arc over a 6m range and 60 minutes of arc are contained in one degree, our formula:

    x = size of the letter in mm

    tan(5/60) = x / 6000mm
    0.001454 = x / 6000mm
    x = 0.001454 * 6000mm
    x = 8.73mm

    So now that we know 20/20 is the equivalent of 8.73mm @ 6m and we know the various decimal acuities, our next step would be to find the other letter sizes.

    20/10 = 8.73 / 2 = 4.37mm
    20/12 = 8.73 / 1.67 = 5.23mm
    20/16 = 8.73 / 1.25 = 5.82mm
    20/20 = 8.73 / 1.00 = 8.73mm
    etc...

    Another simpler version:

    8.73mm / (20/10) = 8.73 *10 / 20 = 87.3 / 20 = 4.37mm Basically divide the 20/20 letter size by the snellen fraction.
    Last edited by YrahG; 01-31-2010 at 03:54 AM.

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    I think it's like this:

    Decimal-ize the Snellen fraction:
    20/400 = 0.05

    Then take the log (base 10) of that, and add a minus sign to denote "reduced":

    -1.3

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    Master OptiBoarder Darryl Meister's Avatar
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    Flip the numerator with the denominator of your Snellen fraction, so that 20/400 becomes 400/20.

    Solve the fraction in decimal form, so that 400/20 = 20. This is your minimum angle of resolution (MAR) in arcminutes.

    Finally, determine the base-10 logarithm of the answer using the "LOG" function, so that log 20 = 1.301.

    This means that a Snellen acuity of 20/400 is equivalent to a logMAR acuity of 1.301.

    To compute the Snellen equivalent from logMAR notation, first raise a base of 10 to the power of the logMAR value, so that 10^1.301 = 20.

    Then multiply this result by your Snellen testing distance used for the numerator (e.g., 20 feet) to determine the new Snellen denominator, so that 20 * 20 = 400 or 20/400.

    Also note that smaller logMAR values represent better visual acuity, so a logMAR value of -0.3 (20/10) is better than a logMAR value of 1.0 (20/200).
    Darryl J. Meister, ABOM

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    What is relationship between Jaeger’s optotype and N- optotype for near?
    Size, VA?

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    What's up? drk's Avatar
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    Ok, now you're just being ungracious.

    What's in it for us, Seva?:D

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    I am sorry. I loose my control:hammer:.
    Ok,;) thanks for all!

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    Quote Originally Posted by YrahG View Post
    Now since 6m/6m (20ft/20ft) is a letter that subtends 5 minutes of arc over a 6m range and 60 minutes of arc are contained in one degree, our formula:

    x = size of the letter in mm

    tan(5/60) = x / 6000mm
    0.001454 = x / 6000mm
    x = 0.001454 * 6000mm
    x = 8.73mm

    So now that we know 20/20 is the equivalent of 8.73mm @ 6m and we know the various decimal acuities, our next step would be to find the other letter sizes.

    .
    Those this applies to calculate the letter size for 20/20 on a near chart to be used at 14 inches (35cm or 350mm)?
    0.001454*350= 0.5089mm. I print the capital E from the snellen fonts and the letter size is so small that i can not read it even at a close distance.

    I must be wrong. Does anyone knows which is the correct size of 20/20 for near chart at 14 inches?

  12. #12
    Master OptiBoarder Darryl Meister's Avatar
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    For reasons that I have yet to understand, visual acuity at near testing distances is based upon the height of lowercase letters, called the "x-height." Note that this represents a discrepancy from distance vision testing, which is based upon uppercase letters. And the x-height of the 20/20 Snellen letter at 40 cm (16") is 0.58 mm, so at 14" it will be slightly smaller. You can use the mathematics, above, to calculate the final value at 14".

    Best regards,
    Darryl
    Last edited by Darryl Meister; 05-28-2011 at 08:57 AM.
    Darryl J. Meister, ABOM

  13. #13
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    Thanks for your explanation. Now i undestand why the letters i printed where so small.

    Thanks again

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