Using the following assumptions below I used excel to calculate the percentage of hearing one has when compared to normal. I dont know if this is considered valid and I get two different answers. This is only to get the magnitude of the hearing loss in percentage of says nothing about the different effects of differently shaped curves with the same area. In both cases is assumed normal hearing is at 0 db and total loss is at 100 db (I dont know what number to use here, maybe have worse than 100db) and the hearing range for both is 0hz to 8000 hz. If you plot the RSHL (reverse curve hearing loss) on a or graph, you can calculate the area under a normal graph and the RSHL graph and compare to get a percentage.
Method 1 - Take for example 0-1000 hz. Normal hearing would be from 0 to 100 db here so to speak so that area would be 1000 hz * (100-0) = 100,000 db-hz. Then you take an example RSHL value, say 60 db so that area would be 1000 hz * (100-60) = 40,000 db-hz or 40% of normal by this method. Repeat this every 1000 hz up to 8000 hz.
For the example RHSL shown below I show a percentage loss of 61%.
Method 2 - Instead of using db on the graph, go to Wikipedia lookup decibel and use the sound pressure ratio p1/p0 as calculated by db=10log(p1/p0). For the example RSHL below I show a percentage loss of 87%.
Hopefully I didn’t make any math errors in excel here as is very possible
I really dont want to review how to do log math from 20 years ago so if anyone is still familiar, should these results be different or the same, and is the result valid? Given that power of 10 is involved, I would guess that should be different at first glance.
Using the following assumptions below I used excel to calculate the percentage of hearing one has when compared to normal. <snip>
No need to reinvent the wheel. There are accepted formulas for calculating percent hearing loss, here is a link304. As you noted, this is not an overly useful metric.
Your methods seem fatally flawed in that you are essentially taking ‘the area under the curve’ of the . Since the frequency axis is logrithmic, you really need be using log(freq). In other words, your method weights hearing ability from 4K-8K equally to all hearing from 0-4K, and this is obviously not what you want.
First of all, I understand the freq is on a log scale. If I were to color in the area under the curve with a pencil and measure it directly off the paper, y es you would not be taking into acount the logarithmic scale. However, I am not doing that, I am instead using the actual numbers so the scale the freq is plotted on is irrelevant.
The Db scale is logarithmic, but so is our hearing. Notice the graph does not plot it on a log scale. Again, direct numbers can be used if the ares is calculated using the numbers, not the actual area on the paper graph.
On the , it shows a o db at the top throughout the frequency range 0 to 8Khz. It is implied to me that a normal hearing person would have thiere scores plotted at the zero line. If that is the case, then a comparison can be made at each frequency, the normal person at 0 db and a hearing impaired person at whatever they have.
I dont know if the pure tone audiogragh takes into account equal loudness curves, I have the feeling that the two are related, but I dont know how yet.
Thank you for sending the link. To me, this is a LEGAL and not really a medical mesurement of hearing loss for the purposes of compensation, not for the purposes of measuring hearing loss.
I found the way they measured hearing handicap (HH) to be quite disturbing.
They first measure the thresholds of hearing at several frequencies, take those numbers and then average them. Thats fine that makes sense to me.
If your good ear measured 30 db average and your bad ear measured 40 db average then the overal weighted HH was calculated giving a 5 times weighted score to the good ear. Therefore the weighted HH was (5(30)+1(40))/6 = 31.67. Thats makes sense to me too. The 5 multiplycation is an emperical result probably determined thru research that takes into account the good ear compensating for the bad.
The disturbing parts - Initially they used only the frequencies of 0.5, 1 & 2 khz, finally later they added 3 khz. This makes absolutely no sense to me at all. In fact it actually seems to purposfuly ignore all the frequencies above 3 K, the very frequencies that get damaged by age and exposure to noise! The very same frequencies that carry most of the speech information! This formula does not even look at that at all! I wonder if the purpose of doing this is because they figure as people get older they lose hearing above th 3 khz so it would make it impossible to determine what hearing was lost due to exposure to noise on the job for example. Determing what hearing loss was done due to exposure to noise and what was lost due to the aging process seems impossible though. Maybe the low frequencies are more stable and are better able to measure hearing loss due to noise.
In any even though, this criteria does not measure hearing handicap or percentage hearing loss at all. It is a legal criterial used to determine hearing loss only in the very l low frequencies and is used as a tool for compenstion for hearing loss, not to give a picture of the hearing impaired persons hearing as compared to a normal persons hearing.
I did notice in thier formula HH capping out at 92 db. This makes sense in that 92 db minus the 25 db threshold they used is 67 db. If you have 1.5 % more loss for every db, then with a cap at 67 db you get 67 db *1.5% = “100”% loss. Many people can hear and function somewhat past 92 db, but you have to draw the profound line in somewhere. I think at 90 db for me, I can actually feel the speaker vibrating against my dead ear so thats the only reason I would get a score there.
I summary. The HH appears to be a legal criteria used for award compensation, not a complete medical measurement to get at least the magnitude of hearing loss accross the entire accepted frequency range of 0 to 8 K. I also still dont see where my calc’s might have had an error, although I am sure they are there.
PS - I wish they would test past 8 K. I know I hear as high as 16 K.
Agreed: I always thought it was silly to only measure 500, 1000, and 2000 Hz.
Only adding the 3000 Hz to the formula is better, but still not the best.
I wonder (does anyone know?): can you keep the formula equation the same but include test frequencies 500, 1000, 2000, 3000, 4000, 6000, 8000?
Instead of finding just the puretone average at 500, 1000, 2000 (and maybe 3000), can you find the average of all the mentioned thresholds and use THAT number in the calculation for percent hearing loss? I would imagine “yes”. If that’s the case, it would make it more accurate.
On the flip side… Yes I know… calculating percentage hearing loss is like “finding out how much I weigh in inches.”
First of all, I understand the freq is on a log scale. If I were to color in the area under the curve with a pencil and measure it directly off the paper, y es you would not be taking into acount the logarithmic scale. However, I am not doing that, I am instead using the actual numbers so the scale the freq is plotted on is irrelevant.
You have this backwards! Since the graph already uses a log scale, if you measure the area under this curve, the graph has already taken the log(freq) for you. If you are using the actual values to calculate, try using log(freq) wherever you use freq.
It is implied to me that a normal hearing person would have thiere scores plotted at the zero line. If that is the case, then a comparison can be made at each frequency, the normal person at 0 db and a hearing impaired person at whatever they have.
Well, I would say if your hearing is plotted across the at the zero line, you do not have normal hearing, you have perfect hearing!
The disturbing parts - Initially they used only the frequencies of 0.5, 1 & 2 khz, finally later they added 3 khz. This makes absolutely no sense to me at all. In fact it actually seems to purposfuly ignore all the frequencies above 3 K, the very frequencies that get damaged by age and exposure to noise! The very same frequencies that carry most of the speech information!
Yes, these are the frequencies that are frequently lost, but I cannot agree with you that this is where most of the speech information is. The telephone is a good example of this, probably only good to 2-4KHz, but in general speech comprehension is good over phone lines.
Using the following assumptions below I used excel to calculate the percentage of hearing one has when compared to normal. I dont know if this is considered valid and I get two different answers. This is only to get the magnitude of the hearing loss in percentage of says nothing about the different effects of differently shaped curves with the same area. In both cases is assumed normal hearing is at 0 db and total loss is at 100 db (I dont know what number to use here, maybe have worse than 100db) and the hearing range for both is 0hz to 8000 hz. If you plot the RSHL (reverse curve hearing loss) on a audiogram or graph, you can calculate the area under a normal graph and the RSHL graph and compare to get a percentage.
Method 1 - Take for example 0-1000 hz. Normal hearing would be from 0 to 100 db here so to speak so that area would be 1000 hz * (100-0) = 100,000 db-hz. Then you take an example RSHL value, say 60 db so that area would be 1000 hz * (100-60) = 40,000 db-hz or 40% of normal by this method. Repeat this every 1000 hz up to 8000 hz.
For the example RHSL shown below I show a percentage loss of 61%.
Method 2 - Instead of using db on the graph, go to Wikipedia lookup decibel and use the sound pressure ratio p1/p0 as calculated by db=10log(p1/p0). For the example RSHL below I show a percentage loss of 87%.
Hopefully I didn’t make any math errors in excel here as is very possible
I really dont want to review how to do log math from 20 years ago so if anyone is still familiar, should these results be different or the same, and is the result valid? Given that power of 10 is involved, I would guess that should be different at first glance.
nal has a program that calculates % loss of hearing based
Umm no. Log scale is only used for convience to get the graph to fit on paper as well as to make it easier to read. Taking the area off the paper graph would be misleading and incorrect. Unfortunately I cant agree with a single point you have made. Oh well agree with disagree.
You have this backwards! Since the graph already uses a log scale, if you measure the area under this curve, the graph has already taken the log(freq) for you. If you are using the actual values to calculate, try using log(freq) wherever you use freq.
Well, I would say if your hearing is plotted across the audiogram at the zero line, you do not have normal hearing, you have perfect hearing!
Yes, these are the frequencies that are frequently lost, but I cannot agree with you that this is where most of the speech information is. The telephone is a good example of this, probably only good to 2-4KHz, but in general speech comprehension is good over phone lines.
What does the area under a threshold hearing curve have to do with developing a measure of the most important parameter…speech comprehension. ?
With moderate to profound losses, keep in mind how flawed the entire scheme of pure tone thresholds as base data, is, as related to recruitment, internal distortion, dead or degenerated frequencies…etc. All of which materially determine the comprehension metric.
I have trouble with the idea that any single measure of loss is useful for comparative purposes… Ed
Well I am not trying to connect the two, I didnt bring that up I dont think.
I was just curious if % hearing loss could be calculated, and if so how that could be done.
Umm no. Log scale is only used for convience to get the graph to fit on paper as well as to make it easier to read. Taking the area off the paper graph would be misleading and incorrect. Unfortunately I cant agree with a single point you have made. Oh well agree with disagree.
Plot your hearing on a linear X axis from 128 to 20KHz and compare it to the log graph. Which do you think has more useful information?
Unfortunately I could not run the program, I think it said because I had a 64 bit operating system.
If I understand right they measured the speech inteligibility of hundreds of people and correlated that with thier . So if you input yours, it uses that data to say what your speech inteligibility likely is? Good idea, I wonder how accurate it is.
