Which Is Louder, 2x10 or 1x15?

[alexclaber]

Accurate measurement of bass enclosures (that is representative of how you actually perceive the output from multiple drivers, etc.) has been one of the biggest challenges, no doubt.

I share your pain!

Alex

 

[Chris K]

In physics class I woke up just in time to be able to say here and now that something is definitely missing in the homework above: distance, room properties, audience. I just use my big ears. They tell me: 10'' gives more power when you are close to the cabinet, in a small room, with hard walls/floor, low ceiling, small audience. 15'' (or larger) the opposite, because they favour lows. The reason is: lower sounds have wider wavelenghts; the wider the waves, the more they have the ablity to curve around objects. QED

 

[greenboy]

In physics class I woke up just in time to be able to say here and now that something is definitely missing in the homework above: distance, room properties {...}

Evidently physics class didn't prepare you with the concept of eliminating as many variables as possible when doing comparisons. Why should this be done? Because in this case, all you have to do is change rooms and positioning and what you are comparing can seem entirely different, even in direct opposition from earlier results.

Thus, absolutely nothing is learned FOR CERTAIN about the things you are supposedly comparing, and you might find that more than anything you are comparing the rooms as much as the speakers.

The reason is: lower sounds have wider wavelenghts; the wider the waves, the more they have the ablity to curve around objects.

English class and not physics would give one the first clue: waveLENGTHS are not WIDE, they are LONG. Or SHORT ; }

And physics:acoustics would say they don't "curve around" things so much as they reflect.

 

[Implosion]

Remember your high school - pi r2

15X15 = 225
10X10 = 100 (X2 = 200)

r = radius. The radius of a 15" speaker is 7.5" and for the ten inch speaker it's 5".

One 15" has the same area as two and a half 10"s. Area is proportional to the diameter squared! (And the working diameter is always less than the nominal diameter because the frame and surround are not part of the radiating area).

Alex

Proprtional to the radius squared. If you are using diameter you also have divide your answer by four. Therefore a 15" speaker only has 12.5% more surface area than two 10" speakers combined.

This should be a lesson to you kids. Don't sleep thru physics.

Yeah.

 

[alexclaber]

lots of incorrect statements... QED

What did Tower of Power sing? A little knowledge is a dangerous thing!

Alex

 

[nodestiny]

Remember your high school - pi r2

15X15 = 225
10X10 = 100 (X2 = 200)

Many speaker manufacturers list "Sd" which is going to be your effective cone area and is a much more accurate way to compare two speakers in terms of total cone area. Of course, cone area is only a part of the equation on "being loud" and has little to nothing to do with how it will actually sound.

 

[markfretless]

Go to [Invalid or Expired Link Removed] and read "The Theory Behind Using 5" Drivers For bass" and "How We hear Bass" -very interesting.
Best Regards,
Mark

 

[nodestiny]

Go to [Invalid or Expired Link Removed] and read "The Theory Behind Using 5" Drivers For bass" and "How We hear Bass" -very interesting.
Best Regards,
Mark

Interesting concept... has anybody actually tested these for their claimed 25-15K linear response?

 

[alexclaber]

r = radius. The radius of a 15" speaker is 7.5" and for the ten inch speaker it's 5".

The working radius of a 15" speaker is ~ 6.5" whilst for a 10" speaker it's ~ 4".

Proprtional to the radius squared. If you are using diameter you also have divide your answer by four.

It's proportional to radius or diameter squared - the /4 is a constant and thus comes out in the wash.

Alex

 

[greenboy]

Many speaker manufacturers list "Sd" which is going to be your effective cone area and is a much more accurate way to compare two speakers in terms of total cone area.

Yep. From early in the thread:

Look at the figure called Sd on a driver datasheet to see the actual surface area of a cone. Not that it matters as much as the figure that describes how much air the driver can displace: Vd, which is Sd * Xmax.

I hope this knowledge will help us escape from all the bad math, and faulty assumptions about how much of that nominal diameter of 15 or 10 is actually not frame edge, but actual movable mass ; }

 

[billfitzmaurice]

Accurate measurement of bass enclosures (that is representative of how you actually perceive the output from multiple drivers, etc.) has been one of the biggest challenges, no doubt.

Tom.

Measuring speakers of any configuration is neither difficult, nor unreasonably expensive, nor does it even require a special environment. What is required is knowledge, an abundance of which may be found here:

Invalid Link Removed

Interesting concept... has anybody actually tested these for their claimed 25-15K linear response?

It's not beyond the realm of possibility, and lying about your test results is sheer stupidity, as anyone with a computer, a microphone and a clue can out you. OTOH linear response over that bandwidth off-axis is exceedingly difficult to achieve, and might explain why they only show on-axis measurements. Caveat Emptor.

As to the original question: 2x10. A dual driver cab has higher radiating efficiency than a single driver cab.

 

[Implosion]

The working radius of a 15" speaker is ~ 6.5" whilst for a 10" speaker it's ~ 4".

It's proportional to radius or diameter squared - the /4 is a constant and thus comes out in the wash.

Alex

I was pointing out the difference between a radius and a diameter. Some people seem to get lost with these and yes, you can leave constants out but that leaves a bit room for misunderstanding and people may jump the gun (like I did apparently). With the working radiuses the approximation of One 15" has the same area as two and a half 10"s is correct.

 

[MikeyB]

Remember your high school - pi r2

15X15 = 225
10X10 = 100 (X2 = 200)

Whoops, better re-read that high-school text!

The "R" is radius, so:
Area of a 15" circle = pi * 7.5 * 7.5 = ~177 sq in.
Area of a 10" circle = pi * 5 * 5 = ~79 sq in.

This is slightly different than the surface area of the typical driver, since the driver is a conic section with a dust cap and not a circle... but close enough.

So, in simple terms, a 115 = 177 sq in, and a 210 = 158 sq in.

-M

 

[billfitzmaurice]

Whoops, better re-read that high-school text!

The "R" is radius, so:
Area of a 15" circle = pi * 7.5 * 7.5 = ~177 sq in.
Area of a 10" circle = pi * 5 * 5 = ~79 sq in.

This is slightly different than the surface area of the typical driver, since the driver is a conic section with a dust cap and not a circle... but close enough.

So, in simple terms, a 115 = 177 sq in, and a 210 = 158 sq in.

-M

All moot, as piston area (Sd) alone means nothing and is only significant as a contributing factor to cone displacement (Vd), which is the limiting factor to low frequency output. And since driver data sheets list Vd you don't need to be concerned about the math involved in arriving at it.

 

[Lync]

Still, it has to be said:

HOLY MATH FAIL!

 

[tallboybass]

As to the original question: 2x10. A dual driver cab has higher radiating efficiency than a single driver cab.

So this is the "correct" answer, all things being equal?

 

[KJung]

So this is the "correct" answer, all things being equal?

All things are never equal. So, no. Each cab must be looked at independantly, and thinking about driver 'size' impacting general volume between 'relatively' similarly sized cabs is a losing game IMO and IME.

 

[billfitzmaurice]

So this is the "correct" answer, all things being equal?

Yes, though you have no way of knowing if all things are equal, as no manufacturer provides the data necessary to make that determination.

 

[tallboybass]

OK so can we think of them as very similar in output generally?

 

[KJung]

OK so can we think of them as very similar in output generally?

See posts 7 and 8