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MultiWay Conventional loudspeakers with crossovers 

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21st May 2008, 11:08 PM  #1 
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Join Date: Jun 2007

Equivalent Speaker Sizes
This is sort of a simple question, almost too simple for this group, but i suspect if anyone knows the answer, people here do.
I'm trying to get a general rule of thumb for how many small multiple speakers equal one large speaker. So for example, do two 6.5" woofers equal ...what?... a 10" speaker? Do two 8" woofers equal a 12" woofer? Again, I'm not saying this is true, these are illustrations. I'm asking you if you know roughly what the equivalent small speakers are relative to the cone area of a large speaker? So, for 4", 5.25", 6.5", 8", and 10" speakers in pairs, what are the equivalent in 8", 10", 12", and 15" speakers? Does this question make sense? Steve/bluewizard 
21st May 2008, 11:50 PM  #2 
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Join Date: Feb 2004
Location: Queensland Australia

No its a fair Q. It is really just maths after an informed guess. You want the actual area that radiates sound. A good approximation to start with is the distance between half way across the roll surround. This is the material that joins the cone to the frame. it will often be a "half round" section. Now say you have a 6" driver. The actual cone will be about 4" and the surround takes up the other 1" on the circumferance (1" x2=2") so the distance we use is 5". (That is an informed guess for the effective cone area.) All area formulas involve 'squaring' so 5 squared is 25. The pi component is common to all the calculations so it can be ingnored. If we use the same system for a 8" unit we get 7 squared i.e. approx 49. And so on. Now that is a rough system but it is quite sensible and reveals the fact that although an 8" driver seems at first sight to be 33% larger in linear terms than a 6" it has nearly twice as effective area.
It is fun to play around with scaling in other areas of life. If you double a dimension the area squares but the volume (of a solid) cubes. That is why airplanes can't get too big. If you double the dimensions the area of the wings goes up 4 times but the volume and hence weight goes up 8! Hope that makes sense. Early here. I'm a bit brain dead.
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21st May 2008, 11:58 PM  #3 
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Join Date: Jan 2005
Location: San Diego, CA

In purely mathmatical terms, you just calculate the area of a circle  the formula is pi x radius (r) squared. (Somebody please check my math here.)
If we round off pi to 3.142, and we want to calculate the area of a 7" woofer, we take the radius (half the diameter), which is 3.5, square it, which is 12.25, then multiply by 3.142, we get (rounded off) about 38.5 square inches. Using the same formula for say a 15" woofer we have 7.5 x 7.5 x 3.142 = 176.7 sq. in. How many 7" woofers does it take to equal a 15" woofer? 176.7 x 38.5 = about 4.6. So to compare the area of any two drivers (circles) just use these formulae. Now, this is just pure mathematics  comparing drivers in only this way will not tell you much about performance  knowing what you want your speakers to do and what individual drivers are capable of is far more useful. EDIT: posted at the same time as Jonathan's
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22nd May 2008, 12:08 AM  #4 
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Join Date: Feb 2004
Location: Queensland Australia

sdclc 126. Your post might have gone up during one of my "edits" so you may have missed my early contibution.
You are perfectly correct in the maths but I still think the area that actually works in moving the air is the important figure and for comparison purposes pi can be ignored. Its just makes it simpler. Rough back of envelope calculations show that an 8" driver will be about twice as effective as a 6" unit and a 12" driver double that of an 8" But I agree we shouldn't get too hung up on measurements as the ear is the final arbiter. I remember a German firm 35 yrs ago having a bank of multiple small drives to get a faster rise time than one larger equivalent......... There is quantity and then there's quality.
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"It was the Springtime of the year when aunt is calling to aunt like mastodons bellowing across primeval swamps." P.G. Wodehouse. 
22nd May 2008, 12:10 AM  #5 
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Location: Queensland Australia

Sorry sdclc126....you beat me that time!
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22nd May 2008, 03:20 AM  #6 
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Join Date: Jun 2003
Location: Chamblee, Ga.

FWIW I just use the effective diameter, so comparing two drivers such as a 15" (~13" eff. dia.) to an 18" (~16" eff. dia.):
dt = 16"^2/13"^2 = ~1.5x larger For calculating equivalent sizes such as two 15" (~13" eff. dia.): d2 = 2^0.5*13" = ~18.38", so a bit larger than a single 18" GM
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22nd May 2008, 03:35 AM  #7  
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Quote:
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22nd May 2008, 04:14 AM  #8 
Speakerholic
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Well, I had nothing to do except go say hi to the speakers so I took my trusty tape measure with me and this is what I came up with. From 4.5" to 15" From TB, CSS, MCM and OEM to Pioneer and Selenium, from PA to lofi so there's going to be some differences but this'll work in a pinch.
I reduced it to the reference point of the 4.5" It is 1 and the rest are that many times larger. 4.5"  1 5.0  1.3 6.5  2.6 8.0  4.3 10  6.8 12  9.9 15  16.0 
22nd May 2008, 07:56 AM  #9 
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Join Date: Jan 2002
Location: Wellington, New Zealand

Sd = cone area  phase plug area (if your drivers have phase plugs).
One thing I don't know (yes it is just one thing, I know everything else does Sd take into account cone depth  therefore you really need to calculate Sd as a tapered cylinder to be precise? David. 
22nd May 2008, 12:53 PM  #10 
diyAudio Member

I'm not really sure what the actual point of this is, but if its to see how many smaller woofers it would take to equal the "output" of the larger woofer, then cone displacement is what matters, not cone area. Displacement will take into account the excursion and cone area.
This could be incorrect, but I was once told that the cone profile had little impact on displacement and thus output. The cone profile, I was told, affected directionality of upper frequencies (if being used that high), fr response, and otherwise was of little importance. Now having said that, you do have to massively increase stroke to equal the increase of even a small cone area increase. I have a TC Sounds high excursion woofer with an 11" effective cone area and 32mm of linear xmax one way. This will output more than many 15" woofers, however, as soon as we get into a 15" woofer with roughly 15mm of xmax, it starts to go to the 15 (This is based on simulations and measurements). A good point to add here is that in order for the 12" woofer to stay linear for 32mm of xmax is very difficult. The woofer was very expensive, is very very deep, and very very heavy. It's also very inefficient. Just remember the old engine adage that muscle cars like to tought, there is no replacement for displacement. We are talking about air pumps here basically. Unlike engines which can break that rule with increases in pressure (turbo or compression), speakers are pretty limited in what they can do to increase output. 
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