I was trying to find a mathematical way of determining the amount of power that different frequencies use, or..... where a crossover aught to be purly from a power distribution point of view. Since audio response is logarithmic, I thought about finding the halfway point useing:
1/2 Log(20000) [for a 20-20KHz signal].
However the answer is only 141Hz. Even if you add in the 20Hz at the bottom, 161Hz doesn't seem right. Any educated thoughts?
1/2 Log(20000) [for a 20-20KHz signal].
However the answer is only 141Hz. Even if you add in the 20Hz at the bottom, 161Hz doesn't seem right. Any educated thoughts?
Pbassred said:
I've yet to find a proper answer to this.
I decided instead to pay attention to the idea of avoiding crossing in the range 300 to 3000Hz.
My system will be 40-200, 200-20k. I'd quite like to know whether one of those bands deserves a bigger amplifier than the other. 50W is more than enough for the 200-20k driver but I could up the power to the woofer if it was required to match the higher range.
If you listen at very high levels (have to shout to be heard) then you may benefit from higher power amplifiers for low frequencies, especially if you have inefficient speakers.
OTOH, some of the world's most highly-regarded (tube) amplifiers have fewer than 10 Watts and most folks probably use less than a single watt more than they realize.
OTOH, some of the world's most highly-regarded (tube) amplifiers have fewer than 10 Watts and most folks probably use less than a single watt more than they realize.
Thanks for the varied replies:
Rich - yes there is an article about bi-amping. which has a table showing 350Hz. Rod tantalisingly says he worked it out but doesn't show the formula. The rest of the table was worked out by Fane so I'm going to take 350 as the righ answer, but I don't know how they got there.
soongsc - Your formula is incorrect unless the answer is 9.8Hz.
Wiz - close, but I don't follow the logic. The root(?) off1 times(?)f2
Rich - yes there is an article about bi-amping. which has a table showing 350Hz. Rod tantalisingly says he worked it out but doesn't show the formula. The rest of the table was worked out by Fane so I'm going to take 350 as the righ answer, but I don't know how they got there.
soongsc - Your formula is incorrect unless the answer is 9.8Hz.
Wiz - close, but I don't follow the logic. The root(?) off1 times(?)f2
Pbassred said:
N is the number of octaves. 20K~20Hz is 10 Octaves (well 9.96 if calculated with the formula) in a two way system you want to have about 5 octaves for each driver, this would mean you should have an XO point of around 600Hz. I think you can calculate the rest from there.
Reference: Popular Electronics April 1980, page 48
Pbassred,
I couldn't help but notice noone is directly answering your question. Unfortunately there is no direct answer. It all depends on the music you are playing.
I googled, and of course there was some data HERE of all places.
This was posted years ago by FORR:
The following data helped me a lot, they were published in a french audio magazine by Pierre Etienne Sirder about twenty five years ago. They are based on Harwood's works at the BBC. For a maximum power of 55 W, the power distribution in bands is:
Band (Hz) -> Power (W)
32 - 63 -> 4.4 W
63 - 125 -> 8.8 W
crossover at 125 Hz, total bass = 13.2 W
125 - 250 -> 10 W
250 - 500 -> 10 W
500 - 750 -> 7 W
750 - 1000 -> 5 W
1000 - 1500 -> 3.5 W
1500 - 2000 -> 2.5 W
crossover at 1500 Hz, total medium = 35.5 W
2000 - 3000 -> 1.75 W
3000 - 4000 -> 0.875 W
4000 - 6000 -> 0.44 W
6000 - 8000 -> 0.22 W
8000 - 12000 -> 0.11 W
cross over at 1500 Hz, total tweeter = 5.9 W
If your crossover is at 125 Hz and 1500 Hz, then the electric power reaching each unit is the sum of the powers in the band 32 - 125 Hz (bass), 125 - 1500 Hz (medium), 1500 - 12000 Hz (tweeter).
By scaling to your maximum power and your crossvoer frequencies, you can estimate the power requirements of your amplifier and speakers.
~~~~~~~ Forr
It is generally agreed that 300-350 Hz splits the power distributuion in half. I would be a bit suspicious of the power figures for the real low end... these figures were probably gather before they put REAL bass in recorded music.
I couldn't help but notice noone is directly answering your question. Unfortunately there is no direct answer. It all depends on the music you are playing.
I googled, and of course there was some data HERE of all places.
This was posted years ago by FORR:
The following data helped me a lot, they were published in a french audio magazine by Pierre Etienne Sirder about twenty five years ago. They are based on Harwood's works at the BBC. For a maximum power of 55 W, the power distribution in bands is:
Band (Hz) -> Power (W)
32 - 63 -> 4.4 W
63 - 125 -> 8.8 W
crossover at 125 Hz, total bass = 13.2 W
125 - 250 -> 10 W
250 - 500 -> 10 W
500 - 750 -> 7 W
750 - 1000 -> 5 W
1000 - 1500 -> 3.5 W
1500 - 2000 -> 2.5 W
crossover at 1500 Hz, total medium = 35.5 W
2000 - 3000 -> 1.75 W
3000 - 4000 -> 0.875 W
4000 - 6000 -> 0.44 W
6000 - 8000 -> 0.22 W
8000 - 12000 -> 0.11 W
cross over at 1500 Hz, total tweeter = 5.9 W
If your crossover is at 125 Hz and 1500 Hz, then the electric power reaching each unit is the sum of the powers in the band 32 - 125 Hz (bass), 125 - 1500 Hz (medium), 1500 - 12000 Hz (tweeter).
By scaling to your maximum power and your crossvoer frequencies, you can estimate the power requirements of your amplifier and speakers.
~~~~~~~ Forr
It is generally agreed that 300-350 Hz splits the power distributuion in half. I would be a bit suspicious of the power figures for the real low end... these figures were probably gather before they put REAL bass in recorded music.
Noah, Yes I agree that all music will have make different demands and that drivers will be different. However I was thinking in terms of knowledge based on science rather than guessing and assumption. Youre absolutly right. The mathematics will always be wrong to a degree. You have to start somewhere and make allowances.
the answers so far:
350, 600, 632
We're all smart people, and yet we don't know
the answers so far:
350, 600, 632
We're all smart people, and yet we don't know
Pbassred said:Noah, Yes I agree that all music will have make different demands and that drivers will be different. However I was thinking in terms of knowledge based on science rather than guessing and assumption. Youre absolutly right. The mathematics will always be wrong to a degree. You have to start somewhere and make allowances.
the answers so far:
350, 600, 632
We're all smart people, and yet we don't know
Actually 632 is the more acurate number based on the equation.
Assuming driver effeciency is the same, the same amount of power per octave will results in equal power distribution throughout the spectrum.
WE know!
Kelticwizard wrote:
Pbassred wrote:
Pbassred:
That's the square root of the product of the two frequencies.
square root (20 times 20,000) = 632.456
I got that formula from David Weems' excellent book, Designing, Building and Testing Loudspeakers, the original edition back in 1980 or so.
Bellieve it or not, 1980 was before handheld calculators became cheap to buy-they were like $100. People still had to do their mathematics by hand, (frightening, isn't it?
}
So Weems filled his book with formulas that would not require calculators, let alone computers, to use. That formula was one of them.
Pbassred wrote:
Pbassred:
That's the square root of the product of the two frequencies.
square root (20 times 20,000) = 632.456
I got that formula from David Weems' excellent book, Designing, Building and Testing Loudspeakers, the original edition back in 1980 or so.
Bellieve it or not, 1980 was before handheld calculators became cheap to buy-they were like $100. People still had to do their mathematics by hand, (frightening, isn't it?
} So Weems filled his book with formulas that would not require calculators, let alone computers, to use. That formula was one of them.
Yes wiz - back in the day I think I bought the last of my college's stock of slide rules! With those things you needed a good idea of the answer before you started. Otherwise you couldn't get the right number of zeros on the end.
The reason I wrote "root(?)" was that I couldn't see what place a root had in a logritmic function ( I was assuming that power distribution across frequencies would be logrithmic.).
Noah - OK then lets suppose we take white noise as our benchmark. If we knew what the mid point ( or 3 or 4 splits) was for that, its not a great leap to talor that to a particular music source. Let make no mistake; we would still need to taylor it, but science beats guessing.
The reason I wrote "root(?)" was that I couldn't see what place a root had in a logritmic function ( I was assuming that power distribution across frequencies would be logrithmic.).
Noah - OK then lets suppose we take white noise as our benchmark. If we knew what the mid point ( or 3 or 4 splits) was for that, its not a great leap to talor that to a particular music source. Let make no mistake; we would still need to taylor it, but science beats guessing.
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