Do 2 4" = 1 8" - AVS Forum
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Old 08-18-2014, 04:45 AM - Thread Starter
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Do 2 4" = 1 8"

I have an 8" speaker amplifier I use for my guitar. It is a nice warm sound but compare to say a 10" amp it sounds very compressed.

I am trying to figure a way to open the sound without buying a new amp.

If I change the 8" to some type of multiple speaker baffle would that make it sound more open without losing the warmth?

I know Ampeg for instance has a bass amp that has 8 x 10" speakers that a lot of professional bands use but there are other bass players who prefer 15" speakers.

So is it possible to use multiple speakers to achieve the sound of a single speaker adding a more open sound without sacrificing warmth and all the low end?

Any help greatly appreciated.
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Old 08-18-2014, 07:16 AM
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No it doesn't. Not even close. Also guitar speakers normally don't enough excursion to handle bass notes.
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Old 08-18-2014, 07:17 AM
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Quote:
Originally Posted by Diggar View Post
I have an 8" speaker amplifier I use for my guitar. It is a nice warm sound but compare to say a 10" amp it sounds very compressed.

I am trying to figure a way to open the sound without buying a new amp.

If I change the 8" to some type of multiple speaker baffle would that make it sound more open without losing the warmth?

I know Ampeg for instance has a bass amp that has 8 x 10" speakers that a lot of professional bands use but there are other bass players who prefer 15" speakers.

So is it possible to use multiple speakers to achieve the sound of a single speaker adding a more open sound without sacrificing warmth and all the low end?

Any help greatly appreciated.
What's the amp? Changing the baffle to a multiple speaker configuration probably isn't going to give you what you're looking for. With smaller speakers, you're not going to have as much possible movement of the speakers as you will with the 8" you've got now. And I'm guessing it'd sound quite a bit thinner.

An extension cabinet will give you a bigger sound, but won't change your tone much. I occasionally used to use an extension cabinet with my JC-120 depending on the venue back in the day. But that was primarily due to stage configuration, and not any specific tone I was looking for.

I'm guessing the 8" amp is a practice amp? Are you looking for better tone while practicing, but yet not wanting to disturb the rest of the house?
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Old 08-18-2014, 11:31 AM
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Even if all parameters were exactly the same, the area of two 4" circles is about 25"sq and the area of a single 8" circle is about 50"sq.

As others have said there are so many variables that speaker size is a poor determination of most anything except in a very general way. Displacement, various speaker parameters, enclosure size and type, etc. play into the sound.

"After silence, that which best expresses the inexpressible, is music" - Aldous Huxley
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Old 08-18-2014, 03:11 PM
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Find the Piston area of the speakers using the area of a circle. If you do this, you will get a chart that looks something like this -

4.5" - 1
5.0 - 1.3
6.5 - 2.6
8.0 - 4.3
10 - 6.8
12 - 9.9
15 - 16.0


This was actually given to me when I asked a similar question in another forum.

Area of a Circle equals PI times the Radius Squared.

A = (PI) R²

So, let's work this out.

A = (pi) R² = (3.14159) x 2² = 3.14159 x 4 = 12.566 square inches

A = (pi) R² = (3.14159) x 4² = 3.14159 x 16 = 50.265 square inches


So, if we divide the two 50.265/12.566 = 4.00

So, one 8" speakers equal FOUR 4" speakers.

Here is another chart that I calculated myself -

4.00" = 12.57 sq.in. = 1.00
5.00" = 19.64 sq.in. = 1.56
5.25" = 21.65 sq.in. = 1.72
6.00" = 28.27 sq.in. = 2.49
6.50" = 33.18 sq.in. = 2.64
7.00" = 38.49 sq.in. = 3.06
8.00" = 50.27 sq.in. = 3.999 = 4.0
10.0" = 78.54 sq.in. = 6.25
12.0" = 113.1 sq.in. = 8.997 = 9.0
15.0" = 176.7 sq.in. = 14.06

Using this chart, we divide 4 (for an 8" speaker) by 1 (for a 4" speaker) and we get the same result; FOUR 4" speakers equals ONE 8" speaker.

Also notice that a 6.5" speaker using this last chart, is 70% larger than a 5" speaker. (2.64/1.56 = 1.70)

A 12" speaker is 125% larger than a 8" speaker. (9/4 =2.25)

And so on...

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Last edited by bluewizard; 08-20-2014 at 03:37 PM.
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Old 08-18-2014, 03:27 PM
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Quote:
Originally Posted by bluewizard View Post

A 12" speaker is 125% larger than a 8" speaker. (9/4 =2.25)

And so on...

Steve/bluewizard
I think you meant to say "225% larger".

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Old 08-18-2014, 03:27 PM
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If you want the difference between two specific speakers rather than a generalization, look up the full specs on the speakers, and look at the Sd specification, which is the surface area of the actual cone.

Here is a chart I makes using the Sd of Dayton speakers. Keeping in mind that the cones could be very different. On could be shallow and flat, another very deep.

Using Dayton woofers as an example, and using their Sd spec, we determine the relative effective size of various speakers.

4.00" = 37.4 cm² = 1.00
5.00" = 52.8 cm² = 1.41
5.25" = 93.3 cm² = 2.49
6.00" = 84.9 cm² = 2.27
6.50" = 134.8cm² = 3.60
7.00" = 132.7cm² = 3.55 (more typical 125cm²)
8.00" = 211.2cm² = 5.65
10.0" = 343.1cm² = 9.17
12.0" = 498.8cm² = 13.3
15.0" = 843.7cm² = 22.6

Further because the surround is not part of the Pistonic action of the driver, we have to subtract the surrounds. Now a typical surround is about 0.75". But part, about half of the surround, does have some pistonic action; about half.

To allow for this, we subtract 0.75" from the rated dimension of the speakers.

So, an 8" speaker minus 0.75" is now 7.25". To find the area we still use the same formula (7.25/2 = 3.626) -

A = (pi) R² = 3.14159 x 3.625² = 41.28 square inches.

Now a 4" driver, which would now be 3.25" (3.25/2 = 1.625) -

A = (pi) R² = 3.14159 x 1.625² =8.296 square inches

So, we divide the value for the 8" by the value for the 4" and we have -

41.28 / 8.296 = 4.96

Using this method, it takes 4.96 FOUR inch drivers to equal ONE 8" driver.

This is probably the most accurate as a general estimation for all drivers, as it uses the actual cone diameter plus half the Surround, and assumes Pistonic action.

Sorry haven't worked this chart out yet.

The corrected formula, though more confusing, would be - where "D" is diameter -

A = (pi) ((D - 0.75) / 2)²

A = (pi) ((8" - 0.75) / 2)²

A = (pi) ((7.25) / 2)²

A = (pi) (3.625)² = 3.14159 (3.625)² = 41.28

Which is exactly what we had before

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Last edited by bluewizard; 08-18-2014 at 03:53 PM.
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Old 08-18-2014, 03:32 PM
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I think you meant to say "225% larger".

Yeah, that is confusing. If you divide one by the other, the first 100% they are the same, it is only the addition 125% that it is larger.

For example, if the answer were 1.25, that would only be 25% larger.

Another example, we can clearly see that 625 is only 25% larger than 500.

But if you divide 625 / 500 you get 1.25. Clearly it is not 125% larger, just 25% LARGER.

Look at it this way, I think it is less confusing -

Also notice that a 6.5" speaker using this last chart, is 70% larger than a 5" speaker. (2.64/1.56 = 1.70)

A 12" speaker is 125% larger than a 8" speaker. (9/4 =2.25)


Restated, and the way we would commonly look at it -

It takes 1.7 FIVE inch divers to equal ONE 6.5" driver.

It takes 2.25 EIGHT inch drivers to equal ONE 12" driver.


What we functionally want to know is NOT how much bigger is this than that, but rather how many of these does it take to equal one of those.

Steve/bluewizard

Last edited by bluewizard; 08-18-2014 at 03:48 PM.
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Old 08-18-2014, 04:33 PM
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Quote:
Originally Posted by bluewizard View Post
I think you meant to say "225% larger".

Yeah, that is confusing. If you divide one by the other, the first 100% they are the same, it is only the addition 125% that it is larger.

For example, if the answer were 1.25, that would only be 25% larger.

Another example, we can clearly see that 625 is only 25% larger than 500.

But if you divide 625 / 500 you get 1.25. Clearly it is not 125% larger, just 25% LARGER.

Look at it this way, I think it is less confusing -

Also notice that a 6.5" speaker using this last chart, is 70% larger than a 5" speaker. (2.64/1.56 = 1.70)

A 12" speaker is 125% larger than a 8" speaker. (9/4 =2.25)

Restated, and the way we would commonly look at it -

It takes 1.7 FIVE inch divers to equal ONE 6.5" driver.

It takes 2.25 EIGHT inch drivers to equal ONE 12" driver.

What we functionally want to know is NOT how much bigger is this than that, but rather how many of these does it take to equal one of those.

Steve/bluewizard
Steve,


I hope you didn't think I was criticizing.; Actually I appreciate that you took the time to show the calculations. There are a lot of people here who don't understand formulas for calculating areas, and, that goes along way when trying to determine relative driver size equivalents.

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Old 08-18-2014, 05:20 PM
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Quote:
Originally Posted by bluewizard View Post
Here is a chart I makes using the Sd of Dayton speakers...
Further because the surround is not part of the Pistonic action of the driver, we have to subtract the surrounds.
Sd accounts for the surround, there's no subtraction to be made.

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Old 08-18-2014, 11:38 PM
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Quote:
Originally Posted by Bill Fitzmaurice View Post
Sd accounts for the surround, there's no subtraction to be made.
Because Sd is so variable, even among the Dayton brand I didn't find consistency.

The subtractions only occurs relative to the Rated Size of a speaker. Even that is an estimate because the surround on my 5" desktop speakers are smaller than the surrounds on my 12" speakers. So, I make a best guess on what is typical.

From the outer edge of my 12" driver to the center of the surround is actually about 7/8". I guess I should actually use twice that amount because there is a Surround on both edges of the cone. So, subtract 1.75 for both edges of the surround.

On my 5" bookshelf speaker, it is about 0.5" from the edge to the center of the surround, so now that you have me thinking about it, I should subtract TWICE that or 1".

You can subtract 0.75" or 1.0" or 1.5" as you please, but you subtract this from the Rated Size of the speaker. Not the Sd. You are converting the Rated size (5", 5.25", 6.5", 8", etc...) to the functional Piston size.

The Sd is what the Sd is. No need for any conversions or adjustments. And in my calculations based on Sd, I didn't make any adjustments, I just used the Sd numbers.

For the purposes of this discussion, we are interested in the general size of a 4" relative to the general size of a 8".

If you have specific speakers you can measure them, and use the numbers you find in your own calculations.

Hopefully I haven't confused the issue more.

Steve/bluewizard

Last edited by bluewizard; 08-19-2014 at 02:07 PM.
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Old 08-19-2014, 12:27 AM
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Quote:
Originally Posted by jdlynch View Post
Steve,


I hope you didn't think I was criticizing.....
We're cool, even as I wrote it, I had a feeling it wasn't that clear. But I think people see it now.

Steve/bluewizard

Last edited by bluewizard; 08-20-2014 at 03:10 PM.
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Old 08-19-2014, 10:21 AM
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I assume the OP was talking more about SPL...which is more of a loaded question.

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Old 08-19-2014, 02:33 PM
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I don't think it is so much about SPL as it is about the general tone of the speakers.

There are a lot of details missing from the original post. The 8" sound more compressed, but is it in a small closed cabinet? Is the 10" in a larger or possibly open cabinet?

The smaller and tighter a given cabinet is, the less output you get from it.

Generally you will get more output and more impact for more and larger speakers.

For bass, 12" and 15" are the most common. For Lead, 10" and 12" are the most common. Typically 8" are used in practice amps.

One has to now ask, how much money do you have to spend?

For example, you could build a Marshall half-stack using 10" speakers, but just the raw drivers alone in a common brand is going to cost roughly $65 each in brands like Celestion or Eminence. That means $260 on just drivers, then you have the additional cost of the cabinet.

Typically, the bigger the driver, and the more drivers you use, the bigger the cabinet. No way around that.

Then you have to consider Impedance; both the total for the number of speakers you have, and the capabilities of your amp.

There are two configurations that I know of where you can combine several speakers, and in the end have the same impedance as one single speaker. These are 4 speakers, and 9 speakers.

With 4 drivers, you combine 2 in parallel, the wire those two gangs in series. Or the alternative that will produce the same result. Place 2 drivers in series, the wire those 3 gangs in parallel. If you have 8 ohms speakers, the when the four are combine the total will still be 8 ohms.

The other with 9 drivers, you wire 3 drivers wired in series, that will give you 3 gangs, wire those gangs in parallel and the final impedance will be the same as that of one speaker.

Three 8 ohm in series is 24 ohms.

Three 24 ohm gangs in parallel is 8 ohms.

So, yes it is possible to do what you want to do, the question is, how much money are you willing spend and how much work are you willing to do, to make it happen?

It would also be helpful if you told us specifically what equipment you have. Also, it this a self-contained amp; speaker and amp in one, or is it a separate head and bottom?


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Old 08-19-2014, 04:13 PM
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It really would help if we know the specific equipment you were talking about. We could look up specs and photos and get some sense of the possible differences.

A closed cabinet is going to be less efficient, meaning less output, than an fully or partly open back cabinet.

The common self-contained pick it up and carry it amps like the Fender Twin Reverb, were most often Open back cabinets. Where as larger multi-driver cabinets like a typical Marshal stack were - large and closed.

It could be the power of the respective amps.

It could be the quality of the respective amps and speakers.

A number of factor could come into play.

Steve/bluewizard

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Old 08-20-2014, 03:06 PM
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Just following up on the Relative Size aspect.

OK, I recalculated using this formula -

A = (pi) ((D - 1.5) / 2)²

I used 1.5 as the adjustment factor, based on the fact that my 5" speakers would be 1 and my 12" speaker would be 2, and this was half-way between. The only way to get more accurate, would be to measure actual specific drivers between the outer edge of the frame and the center of the surround, then double that number and use it in the calculations. Alternately simply measure from the center of the surround to the center of the surround to get the true piston diameter of the driver.

I also performed the calculations in an EXCEL spreadsheet, and verified the first calculations using a calculator to make sure I had entered the formula correctly in the spreadsheet.

Here are the results. This is a closer approximation of the true piston size of a range of drivers -

4.00" = 4.91 sq.in. = 1.00
5.00" = 9.62 sq.in. = 1.96
5.25" = 11.05 sq.in. = 2.25
6.00" = 15.90 sq.in. = 3.24
6.50" = 19.64 sq.in. = 4.00
7.00" = 23.76 sq.in. = 4.48
8.00" = 33.18 sq.in. = 6.76
10.0" = 56.75 sq.in. = 11.56
12.0" = 86.59 sq.in. = 17.64
15.0" = 143.14 sq.in. = 29.16

If we compare a 6.5" and a 5" using this new chart, we have 4/1.96 = 2.04. So, a 6.5" is 2.04 times bigger than a 5" driver.

If we use the previous chart that simply used the Rated Dimension and made no adjustment, then compare the two same two drivers, we have 2.64/1.56 = 1.69 or 1.7. Implying that a 6.5" is is 1.7 times bigger than a 5".

If we compare 3x6.5" against 2x8" we now have -

three 6.5" -
3 x 4 = 12

two 8" -
2 x 6.76 = 13.53

13.53 / 12 = 1.1275 = 1.13

So, using these new numbers, TWO 8" drivers are about 13% larger than THREE 6.5" drivers.

Using the original Rated Size unadjusted chart we have 8 / 7.92 = 1.01, or the 2x8" is 1% larger.

I think either way, you can get a rough idea of the relative difference between two different driver configurations.

Which you use is up to you.

Steve/bluewizard
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Old 08-20-2014, 03:39 PM
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Again, with no additional information, the best I can say is that a better 8" driver is going to be better than two 4" drivers.

Without being able to see the amps and speakers, there is not much anyone can really say.

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Old 08-23-2014, 01:07 PM
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If you're going to all that trouble to calculate relative area (much more than I would have done for this thread, but nice job!) you might as well find the angle of the cone as it slopes to the middle and add that bit of area in, plus the extra area of the dust dome if present, since none of that is usually a simple flat (2D) circle...

Still think my original answer was close enough (post 4) since there are so many variables involved in figuring out actual output, let alone the tone of that output...

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Old 08-24-2014, 02:48 PM
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Using Dayton Drivers and the Sd specification, I came up with this, however, give the many variables, this falls short of a generalization. Meaning, it is not a generalization that can be applied to all driver, because its specific, it can only be applied to these drivers.

If you have two specific drivers and if you have their Sd specs, then you can compare them.

But what if you don't?

Using Dayton woofers as an example, and using their SD spec, we determine the relative effective size of various speakers.

4.00" = 37.4 cm² = 1.00
5.00" = 52.8 cm² = 1.41
5.25" = 93.3 cm² = 2.49
6.00" = 84.9 cm² = 2.27
6.50" = 134.8cm² = 3.60
7.00" = 132.7cm² = 3.55 (more typical 125cm²)
8.00" = 211.2cm² = 5.65
10.0" = 343.1cm² = 9.17
12.0" = 498.8cm² = 13.3
15.0" = 843.7cm² = 22.6

Using these numbers, if we compare 3x6.5" to 2x8", what do we come up with?

3x6.5" = 3 x 3.60 = 10.8

2x8" = 2 x 5.65 = 11.3

11.3 / 10.8 = 1.0463

Or 2x8" is about 5% larger. But that only applies to these specific drivers.

But regardless of the actual numbers, the result is the same, 3x6.5" are roughly equal to 2x8".

Steve/bluewizard

Last edited by bluewizard; 08-24-2014 at 02:52 PM.
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Old 08-25-2014, 06:08 AM - Thread Starter
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Thank you for all your help. One of the things I am taking away from all of this is the cabinet influence.

In any event the cabinet volume is .43cu/ft with 2 1" x 2" ports. So I am not sure of I should consider a bigger cab or maybe play with the ports.

I do appreciate all the help
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Old 08-25-2014, 12:24 PM
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PM Bill Fitzmaurice. He posted to you earlier and design speakers. He can suggest a program or two that would help you out.

"After silence, that which best expresses the inexpressible, is music" - Aldous Huxley
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Old 08-25-2014, 01:30 PM
 
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Quote:
Originally Posted by jdlynch View Post
Steve,


I hope you didn't think I was criticizing.; Actually I appreciate that you took the time to show the calculations. There are a lot of people here who don't understand formulas for calculating areas, and, that goes along way when trying to determine relative driver size equivalents.
Cone area doesn't describe the bass dynamic range of a driver, you have to include Xmax.

Maximum possible Xmax tends to go up with the diameter of the cone.

So a 12" driver has about 250% the cone area of an 8 inch, and can also have 150% as much Xmax, further increasing its potential bass dynamic range. These are of course rough approximations, but they make the point that just like "There is no substitute for cubic inches (in car engines)", "There is no substitute for cone diameter (in subwoofers)."
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Old 08-25-2014, 04:59 PM
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As a n00b to this place who has hung out on guitar forums for years, It's kind of funny to run into a thread like this on an AV forum.

I can't think of a guitar amp/cab with 4" speakers that sounds good. My old Fender Champ has an 8" speaker and it sounds wonderful, but some people mod them with 10 inch speakers to get more volume or low end. I think that would be a much better way to go than modding it for smaller speakers. Also, a larger speaker should be able to handle more signal without overloading or compressing the sound (ceteris paribus). As for "warmth", that is going to depend a lot on your choice of speaker, but it shouldn't be hard to retain warmth with the right larger speaker.
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