% or dB?

Discussion in 'Our World, Our Culture' started by Sunshine, Sep 11, 2011.

  1. Sunshine

    Sunshine New Member

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    I always think hearing loss only measure in dB...only I see many post on AD which say "I am 50% deaf..." etc. 50% =\= 50dB right? Are both correct to use (%/dB)?
     
  2. Beach girl

    Beach girl Active Member

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    dB is more scientific.
     
  3. purplewowies

    purplewowies Member

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    My mom uses percent, and I could never quite determine how many dB that was. I once asked her and she said 70-75 dB, which is also the amount of percent loss she says she has. I like dB better.
     
  4. Beach girl

    Beach girl Active Member

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    I think it depends on context to some extent, and to whom you're speaking. People who have never had occasion to study hearing might not have any idea what a 20 dB loss is, or an 80 dB loss. So I can understand in casual conversation someone might say "I have a moderate loss" or "I have a 50%" loss, or whatever, just to get the general point across.

    But for people who know the subject, it's more accurate to use dB. Of course complicating things is that a lot of people have more of a high-frequency loss and a lesser degree of loss in the lower frequencies, so even saying a certain dB loss doesn't mean that's necessarily true across all frequencies.
     
  5. Sunshine

    Sunshine New Member

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    Lol...confusing :)
     
  6. Bottesini

    Bottesini Old Deaf Ranter Premium Member

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    DB and percent don't really correlate either. Someone put up a chart one time, but I don't remember what thread it was.
     
  7. lovezebras

    lovezebras New Member

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    I've read numerous times online and in my medical books and from my audi that hearing loss if said as a percent is not accurate. Hearing loss is measured in dB not percent
     
  8. posts from hell

    posts from hell New Member

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    youre correct... its =/=... dB is the proper way to use it.
     
  9. AJWSmith

    AJWSmith New Member

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    I don't think 50dB = 50%. Firstly, the audi charts I've seen go to 120dB. Secondly, and this is the most important point, dB is a logarithmic scale. This means that 2dB hearing loss (HL) isn't twice as bad as a 1dB loss, but in fact 10x worse, and 10dB HL is 100x bigger than 1dB. Not sure of my maths but if 120dB is total deafness, I think 50% of total deafness is 119dB not 60dB. As a rough guess, I'd say 60dB is about 1% deafness compared to 120dB. Therefore somebody with 90dB HL is many times worse than somebody with 60dB HL, even though on the audi chart it is only a 50% increase from 60dB to 90dB.
    Here's an example of a logarithmic curve:
    [​IMG]
     
  10. Speedy Hawk

    Speedy Hawk New Member

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    Techically you cannot measure dB in percentage cos it goes over 100dB and more. Really percentage is meanless. Stick with dB.

    I noticed that old people with HOH and parents with new deaf baby say in percentage cos audi seems to think that dB is too hard to explain and dumb it down to percentage as if 100dB is max. Some audi are good and explain fully properly.
     
  11. Speedy Hawk

    Speedy Hawk New Member

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    Techically you cannot measure dB in percentage cos it goes over 100dB and more. Really percentage is meanless. Stick with dB.

    I noticed that old people with HOH and parents with new deaf baby say in percentage cos audi seems to think that dB is too hard to explain and dumb it down to percentage as if 100dB is max. Some audi are good and explain fully properly.
     
  12. posts from hell

    posts from hell New Member

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    10 db = 10x. so 20db is 100x of 1db.

    100db is 100x of 80db....

    50% loss? from 100 db - it would be 95 db, from my total guess.

    My explanation would be: 50x of 80db is 95db. 100x is 100db. cut that back by 50x, and you get 95db.

    regardless - at 100 and 95db, they're still loud so we can't accurately say 95-100db is actually 50% of eachother. It's just how much energy that is put out. I am no acoustics engineer so I can't tell you for a fact.

    Then again, I can see it being 85dB being 50% less.... And logically that would make sense.
     
  13. AJWSmith

    AJWSmith New Member

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    I have read some more stuff on measuring sound (see links below). It's clear that 50dB doesn't equal 50% hearing loss. However they suggest that the human brain perceives sound in a logarithmic way, so 100dB supposedly sounds twice as loud as 50dB even though measuring sound pressure itself, it's 100x greater. This is probably due to the brain manipulating & leveling out sounds, amplifying the quiet noises and dampening down the loud ones. But I think from a hearing loss point of view it's more linear than logarithmic as the brain needs to receive the sound before it can boost/dampen it. So the brain of a person with a moderate HL is receiving considerably more sound information than the brain of a person with severe HL, even though the audi scale only shows a small drop.
    dB: What is a decibel?
    Decibel - Wikipedia, the free encyclopedia

    posts from hell - like you I've tried to manually work out 50% of 100dB but my maths skills are too rusty. My best attempt is that 90dB is 50% of 100dB, but I can't explain my reasoning mathematically - it's just an educated guess. My more logical attempt is to say 99dB is 50% of 100dB but that seems too extreme...
     
  14. AJWSmith

    AJWSmith New Member

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    We need daredevel7's mathematical skills here...
     
  15. naisho

    naisho Forum Disorders M.D.,Ph.D

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    I recall an audiologist saying that the reason why the charts go up to ~120 db is because from there onward, it is 'medically' considered the threshold for experiencing pain and/or getting the nerves/drums damaged. Sensitivity may possibly vary in real life for different individuals, but I think generally 120 is the accepted maximum.

    Beyond that point is like standing next to a jet engine accelerating, being next to a firing gun, 'loud cases' that if you can't hear it, you can definitely feel it.

    And yeah, the formula is logarithmic, which is why it's an exponential curve rather than progressive.
     
  16. naisho

    naisho Forum Disorders M.D.,Ph.D

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  17. deafskeptic

    deafskeptic Active Member Premium Member

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    Yeah. I don't like using percentages so I generally tell people that my loss is around 115 across most frequencies except for a couple of lower frequencies; it is around 90 db at the 250 and 500 frequenies.
     
  18. naisho

    naisho Forum Disorders M.D.,Ph.D

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    AJW, the formula for calculating db loss was in your links, so
    db = 10log10 (power value 1/ power value0) where the numbers in (x / y) is your power ratio from the source.

    there was an example given that 30 db is calculated by 10log10(1000w/1w) = 30db.

    50% of that power source would have been 500w, 10log10(500w/1w) = 26.989db.
     
  19. Daredevel7

    Daredevel7 Adrenaline Junky Premium Member

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    Someone say my name?

    This is how I see it (I could be wrong, btw):

    To me,
    1 dB represents 10^0 = 1x
    10 db represents 10^1 = 10x
    20 db represents 10^2 = 100x
    30 db represents 10^3 = 1000x
    40 db represents 10^4 = 10000x
    ....
    80 db represents 10^8 = 100000000x
    100db represents 10^10 = 10000000000x

    So.. PFH was correct when he said:
    Because:
    10^10 / 10^8 = 10^2 = 100x
    (100dB) / (80dB)

    Meaning 100 dB is definitely 100x bigger than 80dB

    Now, about the 50% loss. That assumes that there is an "optimum" or a "maximum". Because you can't put a percentage of loss without a reference. So let's assume that 100dB is the "maximum".

    100 dB is equal to 10^10.
    So 50% of 10^10 is actually just 5^10.

    Using my calculator to find the log of 5^10 using a base of 10... that equals to 7.6 ish. **

    This means 50% loss from 100 dB is about 76 dB, which supports PFH's theory:

    And I'd have to agree. 50% loss from 120 dB is actually 117 dB. So looks like 100dB is a better point of reference.

    ** If you want to know what the hell I am talking about with the log stuff, here is an explanation. For those who don't care, end of post! :)

    If you use your calculator, you can use log to solve for X if you know Y for 10^X=Y. (This is assuming log is based on 10, but most are.)

    For example, 10^2 = 100, right? So if I type in 100, then press log, my answer will be 2. Obviously, you don't need a calculator if you have something like 100, 10000, 100000000000000, you only have to count the number of zeros, but when you have a more complicated number, you will need your calculator for that. So, that's what I did for 5^10.
     
  20. naisho

    naisho Forum Disorders M.D.,Ph.D

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    I did the calcs based on the x kW to y W given on the wiki:

    To calculate 100db loss that is 10 zero's, 10log10 (10000000000w/1w) = 100db
    50% of that would be 10log10 (5000000000w/1w) = 96.989db

    Somehow, I don't think this is necessarily gauged as percentage of hearing loss, it's rather change in percentage in the amount of noise.
     

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