headphone comparison model
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Earregular

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I think I may have come up with a solution to the long debated question of comparing headphones to speakers of similiar performance. People often state that in order to get the same performance from speakers as a decent pair of headphones you would need to spend three or four times as much. I believe that the relationship is not that simple. A pair of Grado SR125s are obviously going to sound a lot better than a $450 speaker. Stereophile rated the SR125s class B in the last recommended components and a pair of class B speakers are going to be probraly around two grand, so such a linear comparison will not work. I believe the relationship would look more like this

P(h)= he^3 e being eulers # not the letter
and h being the price of the
headphones
If you throw in a number such as 69 ( the price of a pair of SR60s ) you come up with what I remember was around 1300. This is about the price of a pair of new class C loudspeakers. I don't know, it makes sense to me, what do you think ?
 
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markl

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A Class B headphone means it's Class B *in the category of headphones* relative to other cans they've tested. These ratings are not comparable to speaker ratings.

Quote:

I believe that the relationship is not that simple. A pair of Grado SR125s are obviously going to sound a lot better than a $450 speaker.


I beg to differ. A low-end set of PSBs will sound better to some, just an one example.

I don't think there's any way to come up with a formula that says, "for every X dollar you spend on phones, you get Y dollars worth of speaker value". It's going to depend on every set of ears, and how much you can give up the advantages of speakers for the sonic benefits of headphones.

Apples and oranges.

Mark
 
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Eagle_Driver

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It's way too complex to figure out a scale. Some headphones (Sony's "consumer" MDR-VCRAPDJ models come to mind) are so mediocre in performance that the "two to three times as much" calculation is more accurate. But others, such as the Grados that you pointed out, your calculation is closer to the truth.
 
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Earregular

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Yes, I realize that there are far too many variables to make direct comparisons, but how does it work as a generalization ?
 
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Aboogwa

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Quote:

Originally posted by Earregular
I believe the relationship would look more like this

P(h)= he^3 e being eulers # not the letter
and h being the price of the
headphones
If you throw in a number such as 69 ( the price of a pair of SR60s ) you come up with what I remember was around 1300. This is about the price of a pair of new class C loudspeakers. I don't know, it makes sense to me, what do you think ?



Uhh.. first you say that a linear relationship will not work... but then you suggest another linear relationship as a correction??


P(h)= he^3

this is still linear... it's just that instead of speakers being 2-3 times more expensive they are now e^3 (this is still a constant) times more expensive..

Unless you meant something different

And what's the point of coming up with a "mathematical" formula for this anyway?
 
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zzz

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How many dollars worth of broccoli you get by spending ten bucks on bananas?

Quote:

Originally posted by Earregular
I think I may have come up with a solution to the long debated question of comparing headphones to speakers of similiar performance. People often state that in order to get the same performance from speakers as a decent pair of headphones you would need to spend three or four times as much. I believe that the relationship is not that simple.


 
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HappymaN

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Quote:

Originally posted by Aboogwa
Uhh.. first you say that a linear relationship will not work... but then you suggest another linear relationship as a correction??


P(h)= he^3

this is still linear... it's just that instead of speakers being 2-3 times more expensive they are now e^3 (this is still a constant) times more expensive..


Somebody just got schooled in Maths...
 
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Aboogwa

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Quote:

Originally posted by Earregular
Is that a linear function or an exponential one ?


It's linear, the variable is never raised to a power anywhere.

It would be exponential if it were something like
P(h) = 3e^h for example
 
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DanG

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You guys are having so much trouble finding a relationship! I would think it should be a sinh.
 
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MacDEF

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Quote:

Originally posted by Aboogwa

P(h)= he^3

this is still linear


Unless he meant

P(h) = (he)^3

 
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purk

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Aboogwa,
I approve of your finding that e^3 is not a variable. It is a constant of 20.086.

Thus,

P(h)= he^3

P(h)=20.086*h, since we don't have a square or cubic...anywhere, it is indeed a linear relationship. Expect a positive slope on this fuction.

However, if
P(h) = 3e^h
dp/dh=3(1)e^h<---- that's an exponential relationship.

Nice work dude, sorry for post this...I'm just having fun.

Purk
 
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HappymaN

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Quote:

Originally posted by purk
Aboogwa,
I approve of your finding that e^3 is not a variable. It is a constant of 20.086.

Thus,

P(h)= he^3

P(h)=20.086*h, since we don't have a square or cubic...anywhere, it is indeed a linear relationship. Expect a positive slope on this fuction.

However, if
P(h) = 3e^h
dp/dh=3(1)e^h<---- that's an exponential relationship.

Nice work dude, sorry for post this...I'm just having fun.

Purk


Just let me get my Texas Instuments TI-83+ out of the drawer...
 
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Dusty Chalk

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Quote:

Originally posted by DanG
You guys are having so much trouble finding a relationship! I would think it should be a sinh.


Groan!
 
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Joe Bloggs

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How is an exponential like that ever going to work? Good cheap headphones beat speakers many times its price but towards the top (e.g. Orpheus) you can buy a good speaker system for the same price. If anything I think it should be something like

y = C log x or
y = C sqrt x
where x is the cost of the headphones, y is the cost of the comparable speaker, and C is some large constant > 1.

And this only applies for x > ~50USD
 
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